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		<title>Elementary Approach to Modular Equations: Ramanujan&#8217;s Theory 4</title>
		<link>http://paramanands.wordpress.com/2012/01/15/elementary-approach-to-modular-equations-ramanujans-theory-4/</link>
		<comments>http://paramanands.wordpress.com/2012/01/15/elementary-approach-to-modular-equations-ramanujans-theory-4/#comments</comments>
		<pubDate>Sun, 15 Jan 2012 10:30:26 +0000</pubDate>
		<dc:creator>paramanands</dc:creator>
				<category><![CDATA[Elliptic Functions]]></category>
		<category><![CDATA[Mathematical Analysis]]></category>

		<guid isPermaLink="false">http://paramanands.wordpress.com/?p=2677</guid>
		<description><![CDATA[Lambert Series In this post we will focus our attention on series of the form: which are more popularly known as Lambert Series. We will not deal with the general theorems concerning such series but will restrict ourselves to the Lambert series for the theta functions and study some identities involving these series. To understand [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=paramanands.wordpress.com&amp;blog=8565079&amp;post=2677&amp;subd=paramanands&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p><strong>Lambert Series</strong></p>
<p>In this post we will focus our attention on series of the form:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7Da_%7Bn%7D%5Ccdot%5Cfrac%7Bq%5E%7Bb_n%7D%7D%7B1+%5Cpm+q%5E%7Bc_n%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;sum_{n = 0}^{&#92;infty}a_{n}&#92;cdot&#92;frac{q^{b_n}}{1 &#92;pm q^{c_n}}' title='&#92;displaystyle &#92;sum_{n = 0}^{&#92;infty}a_{n}&#92;cdot&#92;frac{q^{b_n}}{1 &#92;pm q^{c_n}}' class='latex' /></p>
<p>which are more popularly known as Lambert Series. We will not deal with the general theorems concerning such series but will restrict ourselves to the Lambert series for the theta functions and study some identities involving these series.</p>
<p>To understand their origin let&#8217;s start with the following infinite product:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+-+q%5E%7Bn%7D%29+%26%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+-+q%5E%7B2n%7D%29%281+-+q%5E%7B2n+-+1%7D%29%5C%5C++++%26%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+-+q%5E%7Bn%7D%29%281+%2B+q%5E%7Bn%7D%29%281+-+q%5E%7B2n+-+1%7D%29%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned}&#92;prod_{n = 1}^{&#92;infty}(1 - q^{n}) &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 - q^{2n})(1 - q^{2n - 1})&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 - q^{n})(1 + q^{n})(1 - q^{2n - 1})&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned}&#92;prod_{n = 1}^{&#92;infty}(1 - q^{n}) &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 - q^{2n})(1 - q^{2n - 1})&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 - q^{n})(1 + q^{n})(1 - q^{2n - 1})&#92;end{aligned}' class='latex' /></p>
<p>so that we arrive at</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+%2B+q%5E%7Bn%7D%29%281+-+q%5E%7B2n+-+1%7D%29+%3D+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;prod_{n = 1}^{&#92;infty}(1 + q^{n})(1 - q^{2n - 1}) = 1' title='&#92;displaystyle &#92;prod_{n = 1}^{&#92;infty}(1 + q^{n})(1 - q^{2n - 1}) = 1' class='latex' /></p>
<p>Taking logarithms we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Clog%281+%2B+q%5E%7Bn%7D%29+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Clog%281+-+q%5E%7B2n+-+1%7D%29+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;sum_{n = 1}^{&#92;infty}&#92;log(1 + q^{n}) + &#92;sum_{n = 1}^{&#92;infty}&#92;log(1 - q^{2n - 1}) = 0' title='&#92;displaystyle &#92;sum_{n = 1}^{&#92;infty}&#92;log(1 + q^{n}) + &#92;sum_{n = 1}^{&#92;infty}&#92;log(1 - q^{2n - 1}) = 0' class='latex' /></p>
<p>and a differentiation with respect to <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=fff&amp;fg=222&amp;s=0' alt='q' title='q' class='latex' /> yields</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bnq%5E%7Bn+-+1%7D%7D%7B1+%2B+q%5E%7Bn%7D%7D+%3D+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%282n+-+1%29q%5E%7B2n+-+2%7D%7D%7B1+-+q%5E%7B2n+-+1%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;sum_{n = 1}^{&#92;infty}&#92;frac{nq^{n - 1}}{1 + q^{n}} = &#92;sum_{n = 1}^{&#92;infty}&#92;frac{(2n - 1)q^{2n - 2}}{1 - q^{2n - 1}}' title='&#92;displaystyle &#92;sum_{n = 1}^{&#92;infty}&#92;frac{nq^{n - 1}}{1 + q^{n}} = &#92;sum_{n = 1}^{&#92;infty}&#92;frac{(2n - 1)q^{2n - 2}}{1 - q^{2n - 1}}' class='latex' /></p>
<p>Multiplying by <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=fff&amp;fg=222&amp;s=0' alt='q' title='q' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bnq%5E%7Bn%7D%7D%7B1+%2B+q%5E%7Bn%7D%7D+%3D+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%282n+-+1%29q%5E%7B2n+-+1%7D%7D%7B1+-+q%5E%7B2n+-+1%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;sum_{n = 1}^{&#92;infty}&#92;frac{nq^{n}}{1 + q^{n}} = &#92;sum_{n = 1}^{&#92;infty}&#92;frac{(2n - 1)q^{2n - 1}}{1 - q^{2n - 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(1)' title='&#92;displaystyle &#92;sum_{n = 1}^{&#92;infty}&#92;frac{nq^{n}}{1 + q^{n}} = &#92;sum_{n = 1}^{&#92;infty}&#92;frac{(2n - 1)q^{2n - 1}}{1 - q^{2n - 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(1)' class='latex' /></p>
<p>This represents a non-obvious identity between two Lambert series.</p>
<p><strong>Theta Functions and Their Lambert series</strong></p>
<p>Let&#8217;s recall the <a title="Elliptic Functions: Fourier Series" href="http://paramanands.wordpress.com/2011/02/16/elliptic-functions-fourier-series/" target="_blank">Fourier series for the elliptic functions</a>:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%28u%2C+k%29+%3D+%5Cfrac%7B2%5Cpi%7D%7BKk%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn+%2B+1+%2F+2%7D%5Csin%282n+%2B+1%29z%7D%7B1+-+q%5E%7B2n+%2B+1%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}(u, k) = &#92;frac{2&#92;pi}{Kk}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n + 1 / 2}&#92;sin(2n + 1)z}{1 - q^{2n + 1}}' title='&#92;displaystyle &#92;text{sn}(u, k) = &#92;frac{2&#92;pi}{Kk}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n + 1 / 2}&#92;sin(2n + 1)z}{1 - q^{2n + 1}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bcn%7D%28u%2C+k%29+%3D+%5Cfrac%7B2%5Cpi%7D%7BKk%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn+%2B+1+%2F+2%7D%5Ccos%282n+%2B+1%29z%7D%7B1+%2B+q%5E%7B2n+%2B+1%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{cn}(u, k) = &#92;frac{2&#92;pi}{Kk}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n + 1 / 2}&#92;cos(2n + 1)z}{1 + q^{2n + 1}}' title='&#92;displaystyle &#92;text{cn}(u, k) = &#92;frac{2&#92;pi}{Kk}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n + 1 / 2}&#92;cos(2n + 1)z}{1 + q^{2n + 1}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bdn%7D%28u%2C+k%29+%3D+%5Cfrac%7B%5Cpi%7D%7B2K%7D+%2B+%5Cfrac%7B2%5Cpi%7D%7BK%7D%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%5Ccos%282nz%29%7D%7B1+%2B+q%5E%7B2n%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{dn}(u, k) = &#92;frac{&#92;pi}{2K} + &#92;frac{2&#92;pi}{K}&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(2nz)}{1 + q^{2n}}' title='&#92;displaystyle &#92;text{dn}(u, k) = &#92;frac{&#92;pi}{2K} + &#92;frac{2&#92;pi}{K}&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(2nz)}{1 + q^{2n}}' class='latex' /></p>
<p>where <img src='http://s0.wp.com/latex.php?latex=z+%3D+%5Cpi+u+%2F+2K&amp;bg=fff&amp;fg=222&amp;s=0' alt='z = &#92;pi u / 2K' title='z = &#92;pi u / 2K' class='latex' />. As we shall see in this post these Fourier series become the basis of many identities involving Lambert series for theta functions.</p>
<p>Putting <img src='http://s0.wp.com/latex.php?latex=u+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = 0' title='u = 0' class='latex' /> in the last equation we see that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B2K%7D%7B%5Cpi%7D+%3D+1+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B1+%2B+q%5E%7B2n%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{2K}{&#92;pi} = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n}}' title='&#92;displaystyle &#92;frac{2K}{&#92;pi} = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n}}' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctheta_%7B3%7D%5E%7B2%7D%28q%29+%3D+1+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B1+%2B+q%5E%7B2n%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;theta_{3}^{2}(q) = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n}}' title='&#92;displaystyle &#92;theta_{3}^{2}(q) = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n}}' class='latex' /></p>
<p>Putting <img src='http://s0.wp.com/latex.php?latex=u+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = 0' title='u = 0' class='latex' /> in the series for <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bcn%7D%5C%2C+u&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{cn}&#92;, u' title='&#92;text{cn}&#92;, u' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B2K%7D%7B%5Cpi%7D+%3D+%5Cfrac%7B4%7D%7Bk%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn+%2B+1%2F2%7D%7D%7B1+%2B+q%5E%7B2n+%2B+1%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{2K}{&#92;pi} = &#92;frac{4}{k}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n + 1/2}}{1 + q^{2n + 1}}' title='&#92;displaystyle &#92;frac{2K}{&#92;pi} = &#92;frac{4}{k}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n + 1/2}}{1 + q^{2n + 1}}' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctheta_%7B3%7D%5E%7B2%7D%28q%29+%3D+%5Cfrac%7B%5Ctheta_%7B3%7D%5E%7B2%7D%28q%29%7D%7B%5Ctheta_%7B2%7D%5E%7B2%7D%28q%29%7D%5C%2C4%5Csqrt%7Bq%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B1+%2B+q%5E%7B2n+%2B+1%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;theta_{3}^{2}(q) = &#92;frac{&#92;theta_{3}^{2}(q)}{&#92;theta_{2}^{2}(q)}&#92;,4&#92;sqrt{q}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n + 1}}' title='&#92;displaystyle &#92;theta_{3}^{2}(q) = &#92;frac{&#92;theta_{3}^{2}(q)}{&#92;theta_{2}^{2}(q)}&#92;,4&#92;sqrt{q}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n + 1}}' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctheta_%7B2%7D%5E%7B2%7D%28q%29+%3D+4%5Csqrt%7Bq%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B1+%2B+q%5E%7B2n+%2B+1%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;theta_{2}^{2}(q) = 4&#92;sqrt{q}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n + 1}}' title='&#92;displaystyle &#92;theta_{2}^{2}(q) = 4&#92;sqrt{q}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n + 1}}' class='latex' /></p>
<p>The series for <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsn%7D%5C%2Cu&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{sn}&#92;,u' title='&#92;text{sn}&#92;,u' class='latex' /> is bit complicated. We divide both sides by <img src='http://s0.wp.com/latex.php?latex=u+%3D+2Kz+%2F+%5Cpi&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = 2Kz / &#92;pi' title='u = 2Kz / &#92;pi' class='latex' /> and take limits as <img src='http://s0.wp.com/latex.php?latex=u+%5Cto+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='u &#92;to 0' title='u &#92;to 0' class='latex' /> to get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+1+%3D+%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2K%7D%5Cright%29%5E%7B2%7D%5Cfrac%7B4%7D%7Bk%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%282n+%2B+1%29q%5E%7Bn+%2B+1%2F2%7D%7D%7B1+-+q%5E%7B2n+%2B+1%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 1 = &#92;left(&#92;frac{&#92;pi}{2K}&#92;right)^{2}&#92;frac{4}{k}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n + 1/2}}{1 - q^{2n + 1}}' title='&#92;displaystyle 1 = &#92;left(&#92;frac{&#92;pi}{2K}&#92;right)^{2}&#92;frac{4}{k}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n + 1/2}}{1 - q^{2n + 1}}' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctheta_%7B3%7D%5E%7B4%7D%28q%29+%3D+%5Cfrac%7B%5Ctheta_%7B3%7D%5E%7B2%7D%28q%29%7D%7B%5Ctheta_%7B2%7D%5E%7B2%7D%28q%29%7D%5C%2C4%5Csqrt%7Bq%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%282n+%2B+1%29q%5E%7Bn%7D%7D%7B1+-+q%5E%7B2n+%2B+1%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;theta_{3}^{4}(q) = &#92;frac{&#92;theta_{3}^{2}(q)}{&#92;theta_{2}^{2}(q)}&#92;,4&#92;sqrt{q}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}' title='&#92;displaystyle &#92;theta_{3}^{4}(q) = &#92;frac{&#92;theta_{3}^{2}(q)}{&#92;theta_{2}^{2}(q)}&#92;,4&#92;sqrt{q}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctheta_%7B2%7D%5E%7B2%7D%28q%29%5Ctheta_%7B3%7D%5E%7B2%7D%28q%29+%3D+4%5Csqrt%7Bq%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%282n+%2B+1%29q%5E%7Bn%7D%7D%7B1+-+q%5E%7B2n+%2B+1%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;theta_{2}^{2}(q)&#92;theta_{3}^{2}(q) = 4&#92;sqrt{q}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}' title='&#92;displaystyle &#92;theta_{2}^{2}(q)&#92;theta_{3}^{2}(q) = 4&#92;sqrt{q}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}' class='latex' /></p>
<p>To summarize</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctheta_%7B3%7D%5E%7B2%7D%28q%29+%3D+%5Cphi%5E%7B2%7D%28q%29+%3D+1+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B1+%2B+q%5E%7B2n%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%282%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;theta_{3}^{2}(q) = &#92;phi^{2}(q) = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(2)' title='&#92;displaystyle &#92;theta_{3}^{2}(q) = &#92;phi^{2}(q) = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(2)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctheta_%7B2%7D%5E%7B2%7D%28q%29+%3D+4%5Csqrt%7Bq%7D%5C%2C%5Cpsi%5E%7B2%7D%28q%5E%7B2%7D%29+%3D+4%5Csqrt%7Bq%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B1+%2B+q%5E%7B2n+%2B+1%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%283%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;theta_{2}^{2}(q) = 4&#92;sqrt{q}&#92;,&#92;psi^{2}(q^{2}) = 4&#92;sqrt{q}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(3)' title='&#92;displaystyle &#92;theta_{2}^{2}(q) = 4&#92;sqrt{q}&#92;,&#92;psi^{2}(q^{2}) = 4&#92;sqrt{q}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(3)' class='latex' /></p>
<p>i.e. <img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cpsi%5E%7B2%7D%28q%5E%7B2%7D%29+%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B1+%2B+q%5E%7B2n+%2B+1%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%284%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;psi^{2}(q^{2}) = &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(4)' title='&#92;displaystyle &#92;psi^{2}(q^{2}) = &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(4)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctheta_%7B2%7D%5E%7B2%7D%28q%29%5Ctheta_%7B3%7D%5E%7B2%7D%28q%29+%3D+4%5Csqrt%7Bq%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%282n+%2B+1%29q%5E%7Bn%7D%7D%7B1+-+q%5E%7B2n+%2B+1%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%285%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;theta_{2}^{2}(q)&#92;theta_{3}^{2}(q) = 4&#92;sqrt{q}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(5)' title='&#92;displaystyle &#92;theta_{2}^{2}(q)&#92;theta_{3}^{2}(q) = 4&#92;sqrt{q}&#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(5)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cleft%281+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B1+%2B+q%5E%7B2n%7D%7D%5Cright%29%5Cleft%28%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B1+%2B+q%5E%7B2n+%2B+1%7D%7D%5Cright%29+%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%282n+%2B+1%29q%5E%7Bn%7D%7D%7B1+-+q%5E%7B2n+%2B+1%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%286%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;left(1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n}}&#92;right)&#92;left(&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n + 1}}&#92;right) = &#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(6)' title='&#92;displaystyle &#92;left(1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n}}&#92;right)&#92;left(&#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n + 1}}&#92;right) = &#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(6)' class='latex' /></p>
<p>From the last equation we obtain</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28q%29%5Cpsi%5E%7B2%7D%28q%5E%7B2%7D%29+%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%282n+%2B+1%29q%5E%7Bn%7D%7D%7B1+-+q%5E%7B2n+%2B+1%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(q)&#92;psi^{2}(q^{2}) = &#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}' title='&#92;displaystyle &#92;phi^{2}(q)&#92;psi^{2}(q^{2}) = &#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cpsi%5E%7B4%7D%28q%29+%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%282n+%2B+1%29q%5E%7Bn%7D%7D%7B1+-+q%5E%7B2n+%2B+1%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%287%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;psi^{4}(q) = &#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(7)' title='&#92;displaystyle &#92;psi^{4}(q) = &#92;sum_{n = 0}^{&#92;infty}&#92;frac{(2n + 1)q^{n}}{1 - q^{2n + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(7)' class='latex' /></p>
<p>Again from equation <img src='http://s0.wp.com/latex.php?latex=%284%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(4)' title='(4)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D%5Cpsi%5E%7B2%7D%28q%29+%26%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%2F2%7D%7D%7B1+%2B+q%5E%7Bn+%2B+1%2F2%7D%7D%5C%5C++++%26%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%2F2%7D%5Csum_%7Bk+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bk%7Dq%5E%7Bk%28n+%2B+1%2F2%29%7D%5C%5C++++%26%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Csum_%7Bk+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bk%7Dq%5E%7Bnk+%2B+k%2F2+%2B+n%2F2%7D%5C%5C++++%26%3D+%5Csum_%7Bk+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bk%7Dq%5E%7Bk%2F2%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7Dq%5E%7Bnk+%2B+n%2F2%7D%5C%5C++++%26%3D+%5Csum_%7Bk+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bk%7Dq%5E%7Bk%2F2%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7Dq%5E%7B%28k+%2B+1%2F2%29n%7D%5C%5C++++%26%3D+%5Csum_%7Bk+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%28-1%29%5E%7Bk%7Dq%5E%7Bk%2F2%7D%7D%7B1+-+q%5E%7Bk+%2B+1%2F2%7D%7D%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned}&#92;psi^{2}(q) &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n/2}}{1 + q^{n + 1/2}}&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}q^{n/2}&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}q^{k(n + 1/2)}&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}q^{nk + k/2 + n/2}&#92;&#92;    &amp;= &#92;sum_{k = 0}^{&#92;infty}(-1)^{k}q^{k/2}&#92;sum_{n = 0}^{&#92;infty}q^{nk + n/2}&#92;&#92;    &amp;= &#92;sum_{k = 0}^{&#92;infty}(-1)^{k}q^{k/2}&#92;sum_{n = 0}^{&#92;infty}q^{(k + 1/2)n}&#92;&#92;    &amp;= &#92;sum_{k = 0}^{&#92;infty}&#92;frac{(-1)^{k}q^{k/2}}{1 - q^{k + 1/2}}&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned}&#92;psi^{2}(q) &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n/2}}{1 + q^{n + 1/2}}&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}q^{n/2}&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}q^{k(n + 1/2)}&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}q^{nk + k/2 + n/2}&#92;&#92;    &amp;= &#92;sum_{k = 0}^{&#92;infty}(-1)^{k}q^{k/2}&#92;sum_{n = 0}^{&#92;infty}q^{nk + n/2}&#92;&#92;    &amp;= &#92;sum_{k = 0}^{&#92;infty}(-1)^{k}q^{k/2}&#92;sum_{n = 0}^{&#92;infty}q^{(k + 1/2)n}&#92;&#92;    &amp;= &#92;sum_{k = 0}^{&#92;infty}&#92;frac{(-1)^{k}q^{k/2}}{1 - q^{k + 1/2}}&#92;end{aligned}' class='latex' /></p>
<p>Hence we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D2%5Cpsi%5E%7B2%7D%28q%29+%26%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%2F2%7D%7D%7B1+%2B+q%5E%7Bn+%2B+1%2F2%7D%7D+%2B+%5Cfrac%7B%28-1%29%5E%7Bn%7Dq%5E%7Bn%2F2%7D%7D%7B1+-+q%5E%7Bn+%2B+1%2F2%7D%7D%5C%5C++++%26%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B2q%5E%7Bn%7D%7D%7B1+-+q%5E%7B4n+%2B+1%7D%7D+-+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B2q%5E%7B3n+%2B+2%7D%7D%7B1+-+q%5E%7B4n+%2B+3%7D%7D%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned}2&#92;psi^{2}(q) &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n/2}}{1 + q^{n + 1/2}} + &#92;frac{(-1)^{n}q^{n/2}}{1 - q^{n + 1/2}}&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;frac{2q^{n}}{1 - q^{4n + 1}} - &#92;sum_{n = 0}^{&#92;infty}&#92;frac{2q^{3n + 2}}{1 - q^{4n + 3}}&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned}2&#92;psi^{2}(q) &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n/2}}{1 + q^{n + 1/2}} + &#92;frac{(-1)^{n}q^{n/2}}{1 - q^{n + 1/2}}&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;frac{2q^{n}}{1 - q^{4n + 1}} - &#92;sum_{n = 0}^{&#92;infty}&#92;frac{2q^{3n + 2}}{1 - q^{4n + 3}}&#92;end{aligned}' class='latex' /></p>
<p>so that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cpsi%5E%7B2%7D%28q%29+%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B1+-+q%5E%7B4n+%2B+1%7D%7D+-+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7B3n+%2B+2%7D%7D%7B1+-+q%5E%7B4n+%2B+3%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%288%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;psi^{2}(q) = &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 - q^{4n + 1}} - &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{3n + 2}}{1 - q^{4n + 3}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(8)' title='&#92;displaystyle &#92;psi^{2}(q) = &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 - q^{4n + 1}} - &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{3n + 2}}{1 - q^{4n + 3}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(8)' class='latex' /></p>
<p>Again starting with the Lambert series for <img src='http://s0.wp.com/latex.php?latex=%5Cphi%5E%7B2%7D%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;phi^{2}(q)' title='&#92;phi^{2}(q)' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D%5Cphi%5E%7B2%7D%28q%29+%26%3D+1+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B1+%2B+q%5E%7B2n%7D%7D%5C%5C++++%26%3D+1+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%7D%5Csum_%7Bk+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bk%7Dq%5E%7B2nk%7D%5C%5C++++%26%3D+1+%2B+4%5Csum_%7Bk+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bk%7D%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Dq%5E%7B%282k+%2B+1%29n%7D%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned}&#92;phi^{2}(q) &amp;= 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n}}&#92;&#92;    &amp;= 1 + 4&#92;sum_{n = 1}^{&#92;infty}q^{n}&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}q^{2nk}&#92;&#92;    &amp;= 1 + 4&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}&#92;sum_{n = 1}^{&#92;infty}q^{(2k + 1)n}&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned}&#92;phi^{2}(q) &amp;= 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{1 + q^{2n}}&#92;&#92;    &amp;= 1 + 4&#92;sum_{n = 1}^{&#92;infty}q^{n}&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}q^{2nk}&#92;&#92;    &amp;= 1 + 4&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}&#92;sum_{n = 1}^{&#92;infty}q^{(2k + 1)n}&#92;end{aligned}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+%5Cphi%5E%7B2%7D%28q%29+%3D+1+%2B+4%5Csum_%7Bk+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%28-1%29%5E%7Bk%7Dq%5E%7B2k+%2B+1%7D%7D%7B1+-+q%5E%7B2k+%2B+1%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%289%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;Rightarrow &#92;phi^{2}(q) = 1 + 4&#92;sum_{k = 0}^{&#92;infty}&#92;frac{(-1)^{k}q^{2k + 1}}{1 - q^{2k + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(9)' title='&#92;displaystyle &#92;Rightarrow &#92;phi^{2}(q) = 1 + 4&#92;sum_{k = 0}^{&#92;infty}&#92;frac{(-1)^{k}q^{2k + 1}}{1 - q^{2k + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(9)' class='latex' /></p>
<p>In the above manipulations of the Lambert series we were expressing the series as a double series and interchanging the order of summation. There are other general procedures to rearrange a double series one of which is called the Clausen&#8217;s procedure. This is given by the following formula:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum_%7Bm+%3D+0%7D%5E%7B%5Cinfty%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7Da_%7Bmn%7D+%3D+%5Csum_%7Bm+%3D+0%7D%5E%7B%5Cinfty%7D%5Cleft%28a_%7Bmm%7D+%2B+%5Csum_%7Bn+%3E+m%7D%5E%7B%5Cinfty%7D%28a_%7Bmn%7D+%2B+a_%7Bnm%7D%29%5Cright%29%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%2810%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;sum_{m = 0}^{&#92;infty}&#92;sum_{n = 0}^{&#92;infty}a_{mn} = &#92;sum_{m = 0}^{&#92;infty}&#92;left(a_{mm} + &#92;sum_{n &gt; m}^{&#92;infty}(a_{mn} + a_{nm})&#92;right)&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(10)' title='&#92;displaystyle &#92;sum_{m = 0}^{&#92;infty}&#92;sum_{n = 0}^{&#92;infty}a_{mn} = &#92;sum_{m = 0}^{&#92;infty}&#92;left(a_{mm} + &#92;sum_{n &gt; m}^{&#92;infty}(a_{mn} + a_{nm})&#92;right)&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(10)' class='latex' /></p>
<p>Applying this technique to the series for <img src='http://s0.wp.com/latex.php?latex=%5Cpsi%5E%7B2%7D%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;psi^{2}(q)' title='&#92;psi^{2}(q)' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D%5Cpsi%5E%7B2%7D%28q%29+%26%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B1+-+q%5E%7B4n+%2B+1%7D%7D+-+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7B3n+%2B+2%7D%7D%7B1+-+q%5E%7B4n+%2B+3%7D%7D%5C%5C++++%26%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%7D%5Csum_%7Bm+%3D+0%7D%5E%7B%5Cinfty%7Dq%5E%7Bm%284n+%2B+1%29%7D+-+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7Dq%5E%7B3n+%2B+2%7D%5Csum_%7Bm+%3D+0%7D%5E%7B%5Cinfty%7Dq%5E%7Bm%284n+%2B+3%29%7D%5C%5C++++%26%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Csum_%7Bm+%3D+0%7D%5E%7B%5Cinfty%7Dq%5E%7Bn+%2B+m%284n+%2B+1%29%7D+-+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Csum_%7Bm+%3D+0%7D%5E%7B%5Cinfty%7Dq%5E%7B3n+%2B+2+%2B+m%284n+%2B+3%29%7D%5C%5C++++%26%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cleft%28q%5E%7B2n%282n+%2B+1%29%7D+%2B+2q%5E%7Bn%7D%5Csum_%7Bm+%3E+n%7D%5E%7B%5Cinfty%7Dq%5E%7Bm%284n+%2B+1%29%7D%5Cright%29+-+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cleft%28q%5E%7B%282n+%2B+1%29%282n+%2B+2%29%7D+%2B+2q%5E%7B3n+%2B+2%7D%5Csum_%7Bm+%3E+n%7D%5E%7B%5Cinfty%7Dq%5E%7Bm%284n+%2B+3%29%7D%5Cright%29%5C%5C++++%26%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cleft%28q%5E%7B2n%282n+%2B+1%29%7D+%2B+2q%5E%7Bn%7D%5Ccdot%5Cfrac%7Bq%5E%7B%28n+%2B+1%29%284n+%2B+1%29%7D%7D%7B1+-+q%5E%7B4n+%2B+1%7D%7D%5Cright%29+-+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cleft%28q%5E%7B%282n+%2B+1%29%282n+%2B+2%29%7D+%2B+2q%5E%7B3n+%2B+2%7D%5Ccdot%5Cfrac%7Bq%5E%7B%28n+%2B+1%29%284n+%2B+3%29%7D%7D%7B1+-+q%5E%7B4n+%2B+3%7D%7D%5Cright%29%5C%5C++++%26%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7Dq%5E%7B2n%282n+%2B+1%29%7D%5Ccdot%5Cfrac%7B1+%2B+q%5E%7B4n+%2B+1%7D%7D%7B1+-+q%5E%7B4n+%2B+1%7D%7D+-+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7Dq%5E%7B%282n+%2B+1%29%282n+%2B+2%29%7D%5Ccdot%5Cfrac%7B1+%2B+q%5E%7B4n+%2B+3%7D%7D%7B1+-+q%5E%7B4n+%2B+3%7D%7D%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned}&#92;psi^{2}(q) &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 - q^{4n + 1}} - &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{3n + 2}}{1 - q^{4n + 3}}&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}q^{n}&#92;sum_{m = 0}^{&#92;infty}q^{m(4n + 1)} - &#92;sum_{n = 0}^{&#92;infty}q^{3n + 2}&#92;sum_{m = 0}^{&#92;infty}q^{m(4n + 3)}&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;sum_{m = 0}^{&#92;infty}q^{n + m(4n + 1)} - &#92;sum_{n = 0}^{&#92;infty}&#92;sum_{m = 0}^{&#92;infty}q^{3n + 2 + m(4n + 3)}&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;left(q^{2n(2n + 1)} + 2q^{n}&#92;sum_{m &gt; n}^{&#92;infty}q^{m(4n + 1)}&#92;right) - &#92;sum_{n = 0}^{&#92;infty}&#92;left(q^{(2n + 1)(2n + 2)} + 2q^{3n + 2}&#92;sum_{m &gt; n}^{&#92;infty}q^{m(4n + 3)}&#92;right)&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;left(q^{2n(2n + 1)} + 2q^{n}&#92;cdot&#92;frac{q^{(n + 1)(4n + 1)}}{1 - q^{4n + 1}}&#92;right) - &#92;sum_{n = 0}^{&#92;infty}&#92;left(q^{(2n + 1)(2n + 2)} + 2q^{3n + 2}&#92;cdot&#92;frac{q^{(n + 1)(4n + 3)}}{1 - q^{4n + 3}}&#92;right)&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}q^{2n(2n + 1)}&#92;cdot&#92;frac{1 + q^{4n + 1}}{1 - q^{4n + 1}} - &#92;sum_{n = 0}^{&#92;infty}q^{(2n + 1)(2n + 2)}&#92;cdot&#92;frac{1 + q^{4n + 3}}{1 - q^{4n + 3}}&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned}&#92;psi^{2}(q) &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{n}}{1 - q^{4n + 1}} - &#92;sum_{n = 0}^{&#92;infty}&#92;frac{q^{3n + 2}}{1 - q^{4n + 3}}&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}q^{n}&#92;sum_{m = 0}^{&#92;infty}q^{m(4n + 1)} - &#92;sum_{n = 0}^{&#92;infty}q^{3n + 2}&#92;sum_{m = 0}^{&#92;infty}q^{m(4n + 3)}&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;sum_{m = 0}^{&#92;infty}q^{n + m(4n + 1)} - &#92;sum_{n = 0}^{&#92;infty}&#92;sum_{m = 0}^{&#92;infty}q^{3n + 2 + m(4n + 3)}&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;left(q^{2n(2n + 1)} + 2q^{n}&#92;sum_{m &gt; n}^{&#92;infty}q^{m(4n + 1)}&#92;right) - &#92;sum_{n = 0}^{&#92;infty}&#92;left(q^{(2n + 1)(2n + 2)} + 2q^{3n + 2}&#92;sum_{m &gt; n}^{&#92;infty}q^{m(4n + 3)}&#92;right)&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}&#92;left(q^{2n(2n + 1)} + 2q^{n}&#92;cdot&#92;frac{q^{(n + 1)(4n + 1)}}{1 - q^{4n + 1}}&#92;right) - &#92;sum_{n = 0}^{&#92;infty}&#92;left(q^{(2n + 1)(2n + 2)} + 2q^{3n + 2}&#92;cdot&#92;frac{q^{(n + 1)(4n + 3)}}{1 - q^{4n + 3}}&#92;right)&#92;&#92;    &amp;= &#92;sum_{n = 0}^{&#92;infty}q^{2n(2n + 1)}&#92;cdot&#92;frac{1 + q^{4n + 1}}{1 - q^{4n + 1}} - &#92;sum_{n = 0}^{&#92;infty}q^{(2n + 1)(2n + 2)}&#92;cdot&#92;frac{1 + q^{4n + 3}}{1 - q^{4n + 3}}&#92;end{aligned}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+%5Cpsi%5E%7B2%7D%28q%29+%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%7Dq%5E%7Bn%28n+%2B+1%29%7D%5Ccdot%5Cfrac%7B1+%2B+q%5E%7B2n+%2B+1%7D%7D%7B1+-+q%5E%7B2n+%2B+1%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%2811%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;Rightarrow &#92;psi^{2}(q) = &#92;sum_{n = 0}^{&#92;infty}(-1)^{n}q^{n(n + 1)}&#92;cdot&#92;frac{1 + q^{2n + 1}}{1 - q^{2n + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(11)' title='&#92;displaystyle &#92;Rightarrow &#92;psi^{2}(q) = &#92;sum_{n = 0}^{&#92;infty}(-1)^{n}q^{n(n + 1)}&#92;cdot&#92;frac{1 + q^{2n + 1}}{1 - q^{2n + 1}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(11)' class='latex' /></p>
<p>We can look back at the Fourier series of <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bdn%7D%28u%2C+k%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{dn}(u, k)' title='&#92;text{dn}(u, k)' class='latex' /> to derive further identities:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B2K%7D%7B%5Cpi%7D%5C%2C%5Ctext%7Bdn%7D%28u%2C+k%29+%3D+1+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%5Ccos%282nz%29%7D%7B1+%2B+q%5E%7B2n%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{2K}{&#92;pi}&#92;,&#92;text{dn}(u, k) = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(2nz)}{1 + q^{2n}}' title='&#92;displaystyle &#92;frac{2K}{&#92;pi}&#92;,&#92;text{dn}(u, k) = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(2nz)}{1 + q^{2n}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+%5Ctheta_%7B3%7D%5E%7B2%7D%28q%29%5C%2C%5Cfrac%7B%5Ctheta_%7B4%7D%28q%29%7D%7B%5Ctheta_%7B3%7D%28q%29%7D%5C%2C%5Cfrac%7B%5Ctheta_%7B3%7D%28z%2C+q%29%7D%7B%5Ctheta_%7B4%7D%28z%2C+q%29%7D+%3D+1+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%5Ccos%282nz%29%7D%7B1+%2B+q%5E%7B2n%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;Rightarrow &#92;theta_{3}^{2}(q)&#92;,&#92;frac{&#92;theta_{4}(q)}{&#92;theta_{3}(q)}&#92;,&#92;frac{&#92;theta_{3}(z, q)}{&#92;theta_{4}(z, q)} = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(2nz)}{1 + q^{2n}}' title='&#92;displaystyle &#92;Rightarrow &#92;theta_{3}^{2}(q)&#92;,&#92;frac{&#92;theta_{4}(q)}{&#92;theta_{3}(q)}&#92;,&#92;frac{&#92;theta_{3}(z, q)}{&#92;theta_{4}(z, q)} = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(2nz)}{1 + q^{2n}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+%5Ctheta_%7B3%7D%28q%29%5Ctheta_%7B4%7D%28q%29%5C%2C%5Cdfrac%7B%7B%5Cdisplaystyle+1+%2B+2%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%5E%7B2%7D%7D%5Ccos%282nz%29%7D%7D%7B%7B%5Cdisplaystyle+1+%2B+2%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%7Dq%5E%7Bn%5E%7B2%7D%7D%5Ccos%282nz%29%7D%7D+%3D+1+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%5Ccos%282nz%29%7D%7B1+%2B+q%5E%7B2n%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;Rightarrow &#92;theta_{3}(q)&#92;theta_{4}(q)&#92;,&#92;dfrac{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}q^{n^{2}}&#92;cos(2nz)}}{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}(-1)^{n}q^{n^{2}}&#92;cos(2nz)}} = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(2nz)}{1 + q^{2n}}' title='&#92;displaystyle &#92;Rightarrow &#92;theta_{3}(q)&#92;theta_{4}(q)&#92;,&#92;dfrac{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}q^{n^{2}}&#92;cos(2nz)}}{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}(-1)^{n}q^{n^{2}}&#92;cos(2nz)}} = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(2nz)}{1 + q^{2n}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+%5Ctheta_%7B4%7D%5E%7B2%7D%28q%5E%7B2%7D%29%5C%2C%5Cdfrac%7B%7B%5Cdisplaystyle+1+%2B+2%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%5E%7B2%7D%7D%5Ccos%282nz%29%7D%7D%7B%7B%5Cdisplaystyle+1+%2B+2%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%7Dq%5E%7Bn%5E%7B2%7D%7D%5Ccos%282nz%29%7D%7D+%3D+1+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%5Ccos%282nz%29%7D%7B1+%2B+q%5E%7B2n%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;Rightarrow &#92;theta_{4}^{2}(q^{2})&#92;,&#92;dfrac{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}q^{n^{2}}&#92;cos(2nz)}}{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}(-1)^{n}q^{n^{2}}&#92;cos(2nz)}} = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(2nz)}{1 + q^{2n}}' title='&#92;displaystyle &#92;Rightarrow &#92;theta_{4}^{2}(q^{2})&#92;,&#92;dfrac{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}q^{n^{2}}&#92;cos(2nz)}}{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}(-1)^{n}q^{n^{2}}&#92;cos(2nz)}} = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(2nz)}{1 + q^{2n}}' class='latex' /></p>
<p>Replacing <img src='http://s0.wp.com/latex.php?latex=z&amp;bg=fff&amp;fg=222&amp;s=0' alt='z' title='z' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=z%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='z/2' title='z/2' class='latex' /> and switching to Ramanujan&#8217;s <img src='http://s0.wp.com/latex.php?latex=%5Cphi&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;phi' title='&#92;phi' class='latex' /> function we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28-q%5E%7B2%7D%29%5Cdfrac%7B%7B%5Cdisplaystyle+1+%2B+2%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%5E%7B2%7D%7D%5Ccos%28nz%29%7D%7D%7B%7B%5Cdisplaystyle+1+%2B+2%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%7Dq%5E%7Bn%5E%7B2%7D%7D%5Ccos%28nz%29%7D%7D+%3D+1+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%5Ccos%28nz%29%7D%7B1+%2B+q%5E%7B2n%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%2812%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(-q^{2})&#92;dfrac{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}q^{n^{2}}&#92;cos(nz)}}{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}(-1)^{n}q^{n^{2}}&#92;cos(nz)}} = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(nz)}{1 + q^{2n}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(12)' title='&#92;displaystyle &#92;phi^{2}(-q^{2})&#92;dfrac{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}q^{n^{2}}&#92;cos(nz)}}{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}(-1)^{n}q^{n^{2}}&#92;cos(nz)}} = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(nz)}{1 + q^{2n}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(12)' class='latex' /></p>
<p>Replacing <img src='http://s0.wp.com/latex.php?latex=z&amp;bg=fff&amp;fg=222&amp;s=0' alt='z' title='z' class='latex' /> with <img src='http://s0.wp.com/latex.php?latex=%5Cpi+-+z&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;pi - z' title='&#92;pi - z' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28-q%5E%7B2%7D%29%5Cdfrac%7B%7B%5Cdisplaystyle+1+%2B+2%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%7Dq%5E%7Bn%5E%7B2%7D%7D%5Ccos%28nz%29%7D%7D%7B%7B%5Cdisplaystyle+1+%2B+2%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%5E%7B2%7D%7D%5Ccos%28nz%29%7D%7D+%3D+1+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%28-q%29%5E%7Bn%7D%5Ccos%28nz%29%7D%7B1+%2B+q%5E%7B2n%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(-q^{2})&#92;dfrac{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}(-1)^{n}q^{n^{2}}&#92;cos(nz)}}{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}q^{n^{2}}&#92;cos(nz)}} = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{(-q)^{n}&#92;cos(nz)}{1 + q^{2n}}' title='&#92;displaystyle &#92;phi^{2}(-q^{2})&#92;dfrac{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}(-1)^{n}q^{n^{2}}&#92;cos(nz)}}{{&#92;displaystyle 1 + 2&#92;sum_{n = 1}^{&#92;infty}q^{n^{2}}&#92;cos(nz)}} = 1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{(-q)^{n}&#92;cos(nz)}{1 + q^{2n}}' class='latex' /></p>
<p>Multiplying the last two equations we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B4%7D%28-q%5E%7B2%7D%29+%3D+%5Cleft%281+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%5Ccos%28nz%29%7D%7B1+%2B+q%5E%7B2n%7D%7D%5Cright%29%5Cleft%281+%2B+4%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%28-q%29%5E%7Bn%7D%5Ccos%28nz%29%7D%7B1+%2B+q%5E%7B2n%7D%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{4}(-q^{2}) = &#92;left(1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(nz)}{1 + q^{2n}}&#92;right)&#92;left(1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{(-q)^{n}&#92;cos(nz)}{1 + q^{2n}}&#92;right)' title='&#92;displaystyle &#92;phi^{4}(-q^{2}) = &#92;left(1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}&#92;cos(nz)}{1 + q^{2n}}&#92;right)&#92;left(1 + 4&#92;sum_{n = 1}^{&#92;infty}&#92;frac{(-q)^{n}&#92;cos(nz)}{1 + q^{2n}}&#92;right)' class='latex' /></p>
<p>It is a surprise now that the RHS is indeed independent of <img src='http://s0.wp.com/latex.php?latex=z&amp;bg=fff&amp;fg=222&amp;s=0' alt='z' title='z' class='latex' />. We note that the functions <img src='http://s0.wp.com/latex.php?latex=%5Ccos%28nz%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;cos(nz)' title='&#92;cos(nz)' class='latex' /> for <img src='http://s0.wp.com/latex.php?latex=n+%3D+1%2C+2%2C+3%2C+%5Cldots&amp;bg=fff&amp;fg=222&amp;s=0' alt='n = 1, 2, 3, &#92;ldots' title='n = 1, 2, 3, &#92;ldots' class='latex' /> are orthogonal on the interval of <img src='http://s0.wp.com/latex.php?latex=%5B-%5Cpi%2C+%5Cpi%5D&amp;bg=fff&amp;fg=222&amp;s=0' alt='[-&#92;pi, &#92;pi]' title='[-&#92;pi, &#92;pi]' class='latex' /> and hence we can multiply the series on the right and integrate the whole equation term by term with respect to <img src='http://s0.wp.com/latex.php?latex=z&amp;bg=fff&amp;fg=222&amp;s=0' alt='z' title='z' class='latex' /> on the interval <img src='http://s0.wp.com/latex.php?latex=%5B-%5Cpi%2C+%5Cpi%5D&amp;bg=fff&amp;fg=222&amp;s=0' alt='[-&#92;pi, &#92;pi]' title='[-&#92;pi, &#92;pi]' class='latex' /> to get the following:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+2%5Cpi%5Ccdot%5Cphi%5E%7B4%7D%28-q%5E%7B2%7D%29+%3D+2%5Cpi+%2B+16%5Cpi%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%28-q%5E%7B2%7D%29%5E%7Bn%7D%7D%7B%281+%2B+q%5E%7B2n%7D%29%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 2&#92;pi&#92;cdot&#92;phi^{4}(-q^{2}) = 2&#92;pi + 16&#92;pi&#92;sum_{n = 1}^{&#92;infty}&#92;frac{(-q^{2})^{n}}{(1 + q^{2n})^{2}}' title='&#92;displaystyle 2&#92;pi&#92;cdot&#92;phi^{4}(-q^{2}) = 2&#92;pi + 16&#92;pi&#92;sum_{n = 1}^{&#92;infty}&#92;frac{(-q^{2})^{n}}{(1 + q^{2n})^{2}}' class='latex' /></p>
<p>On replacing <img src='http://s0.wp.com/latex.php?latex=-q%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='-q^{2}' title='-q^{2}' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=fff&amp;fg=222&amp;s=0' alt='q' title='q' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B4%7D%28q%29+%3D+1+%2B+8%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B%281+%2B+%28-q%29%5E%7Bn%7D%29%5E%7B2%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%2813%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{4}(q) = 1 + 8&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{(1 + (-q)^{n})^{2}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(13)' title='&#92;displaystyle &#92;phi^{4}(q) = 1 + 8&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{(1 + (-q)^{n})^{2}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(13)' class='latex' /></p>
<p>Again putting <img src='http://s0.wp.com/latex.php?latex=qe%5E%7Biz%7D+%3D+a%2C+qe%5E%7B-iz%7D+%3D+b&amp;bg=fff&amp;fg=222&amp;s=0' alt='qe^{iz} = a, qe^{-iz} = b' title='qe^{iz} = a, qe^{-iz} = b' class='latex' /> in equation <img src='http://s0.wp.com/latex.php?latex=%2812%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(12)' title='(12)' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28-ab%29%5C%2C%5Cfrac%7Bf%28a%2C+b%29%7D%7Bf%28-a%2C+-b%29%7D+%3D+1+%2B+2%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Ba%5E%7Bn%7D+%2B+b%5E%7Bn%7D%7D%7B1+%2B+a%5E%7Bn%7Db%5E%7Bn%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%2814%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(-ab)&#92;,&#92;frac{f(a, b)}{f(-a, -b)} = 1 + 2&#92;sum_{n = 1}^{&#92;infty}&#92;frac{a^{n} + b^{n}}{1 + a^{n}b^{n}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(14)' title='&#92;displaystyle &#92;phi^{2}(-ab)&#92;,&#92;frac{f(a, b)}{f(-a, -b)} = 1 + 2&#92;sum_{n = 1}^{&#92;infty}&#92;frac{a^{n} + b^{n}}{1 + a^{n}b^{n}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(14)' class='latex' /></p>
<p>From equation <img src='http://s0.wp.com/latex.php?latex=%2813%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(13)' title='(13)' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D%5Cphi%5E%7B4%7D%28q%29+%26%3D+1+%2B+8%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bq%5E%7Bn%7D%7D%7B%281+%2B+%28-q%29%5E%7Bn%7D%29%5E%7B2%7D%7D%5C%5C++++%26%3D+1+%2B+8%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%7D%5Csum_%7Bk+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bk%7D%28k+%2B+1%29%28-q%29%5E%7Bnk%7D%5C%5C++++%26%3D+1+%2B+8%5Csum_%7Bk+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bk%7D%28k+%2B+1%29%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%28%28-1%29%5E%7Bk%7Dq%5E%7Bk+%2B+1%7D%29%5E%7Bn%7D%5C%5C++++%26%3D+1+%2B+8%5Csum_%7Bk+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bk%7D%28k+%2B+1%29%5Cfrac%7B%28-1%29%5E%7Bk%7Dq%5E%7Bk+%2B+1%7D%7D%7B1+-+%28-1%29%5E%7Bk%7Dq%5E%7Bk+%2B+1%7D%7D%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned}&#92;phi^{4}(q) &amp;= 1 + 8&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{(1 + (-q)^{n})^{2}}&#92;&#92;    &amp;= 1 + 8&#92;sum_{n = 1}^{&#92;infty}q^{n}&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}(k + 1)(-q)^{nk}&#92;&#92;    &amp;= 1 + 8&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}(k + 1)&#92;sum_{n = 1}^{&#92;infty}((-1)^{k}q^{k + 1})^{n}&#92;&#92;    &amp;= 1 + 8&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}(k + 1)&#92;frac{(-1)^{k}q^{k + 1}}{1 - (-1)^{k}q^{k + 1}}&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned}&#92;phi^{4}(q) &amp;= 1 + 8&#92;sum_{n = 1}^{&#92;infty}&#92;frac{q^{n}}{(1 + (-q)^{n})^{2}}&#92;&#92;    &amp;= 1 + 8&#92;sum_{n = 1}^{&#92;infty}q^{n}&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}(k + 1)(-q)^{nk}&#92;&#92;    &amp;= 1 + 8&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}(k + 1)&#92;sum_{n = 1}^{&#92;infty}((-1)^{k}q^{k + 1})^{n}&#92;&#92;    &amp;= 1 + 8&#92;sum_{k = 0}^{&#92;infty}(-1)^{k}(k + 1)&#92;frac{(-1)^{k}q^{k + 1}}{1 - (-1)^{k}q^{k + 1}}&#92;end{aligned}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+%5Cphi%5E%7B4%7D%28q%29+%3D+1+%2B+8%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bnq%5E%7Bn%7D%7D%7B1+%2B+%28-q%29%5E%7Bn%7D%7D%5C%2C%5C%2C%5C%2C%5C%2C%5Ccdots%5C%2C%2815%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;Rightarrow &#92;phi^{4}(q) = 1 + 8&#92;sum_{n = 1}^{&#92;infty}&#92;frac{nq^{n}}{1 + (-q)^{n}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(15)' title='&#92;displaystyle &#92;Rightarrow &#92;phi^{4}(q) = 1 + 8&#92;sum_{n = 1}^{&#92;infty}&#92;frac{nq^{n}}{1 + (-q)^{n}}&#92;,&#92;,&#92;,&#92;,&#92;cdots&#92;,(15)' class='latex' /></p>
<p>Since this post has already grown quite long, it is time to conclude. By now the reader would have got a flavor of the identities relating theta functions to their Lambert series. In the next post we will continue our journey by establishing more identities of the similar form. Some of them would later be used to derive modular equations in the manner of Ramanujan.</p>
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		<title>Elementary Approach to Modular Equations: Ramanujan&#8217;s Theory 3</title>
		<link>http://paramanands.wordpress.com/2011/12/26/elementary-approach-to-modular-equations-ramanujans-theory-3/</link>
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		<pubDate>Mon, 26 Dec 2011 12:19:11 +0000</pubDate>
		<dc:creator>paramanands</dc:creator>
				<category><![CDATA[Elliptic Functions]]></category>
		<category><![CDATA[Elliptic Integrals]]></category>
		<category><![CDATA[Mathematical Analysis]]></category>

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		<description><![CDATA[Connection between Theta Functions and Hypergeometric Functions Let&#8217;s recall the Gauss Transformation formula from an earlier post: where is the hypergeometric function . Putting we get or If we set where is one of the theta functions defined by Ramanujan (see this post) so that and Therefore we finally obtain: The beauty of this formula [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=paramanands.wordpress.com&amp;blog=8565079&amp;post=2602&amp;subd=paramanands&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p><strong>Connection between Theta Functions and Hypergeometric Functions</strong></p>
<p>Let&#8217;s recall the Gauss Transformation formula from <a title="Elementary Approach to Modular Equations: Hypergeometric Series 1" href="http://paramanands.wordpress.com/2011/10/22/elementary-approach-to-modular-equations-hypergeometric-series-1/" target="_blank">an earlier post</a>:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+F%5Cleft%28a%2C+b%3B+2b%3B+%5Cfrac%7B4x%7D%7B%281+%2B+x%29%5E%7B2%7D%7D%5Cright%29+%3D+%281+%2B+x%29%5E%7B2a%7DF%5Cleft%28a%2C+a+-+b+%2B+%5Cfrac%7B1%7D%7B2%7D%3B+b+%2B+%5Cfrac%7B1%7D%7B2%7D%3B+x%5E%7B2%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle F&#92;left(a, b; 2b; &#92;frac{4x}{(1 + x)^{2}}&#92;right) = (1 + x)^{2a}F&#92;left(a, a - b + &#92;frac{1}{2}; b + &#92;frac{1}{2}; x^{2}&#92;right)' title='&#92;displaystyle F&#92;left(a, b; 2b; &#92;frac{4x}{(1 + x)^{2}}&#92;right) = (1 + x)^{2a}F&#92;left(a, a - b + &#92;frac{1}{2}; b + &#92;frac{1}{2}; x^{2}&#92;right)' class='latex' /></p>
<p>where <img src='http://s0.wp.com/latex.php?latex=F&amp;bg=fff&amp;fg=222&amp;s=0' alt='F' title='F' class='latex' /> is the hypergeometric function <img src='http://s0.wp.com/latex.php?latex=_%7B2%7DF_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='_{2}F_{1}' title='_{2}F_{1}' class='latex' />. Putting <img src='http://s0.wp.com/latex.php?latex=a+%3D+b+%3D+1%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='a = b = 1/2' title='a = b = 1/2' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+%5Cfrac%7B4x%7D%7B%281+%2B+x%29%5E%7B2%7D%7D%5Cright%29+%3D+%281+%2B+x%29%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+x%5E%7B2%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{4x}{(1 + x)^{2}}&#92;right) = (1 + x)&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; x^{2}&#92;right)' title='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{4x}{(1 + x)^{2}}&#92;right) = (1 + x)&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; x^{2}&#92;right)' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+1+-+%5Cleft%28%5Cfrac%7B1+-+x%7D%7B1+%2B+x%7D%5Cright%29%5E%7B2%7D%5Cright%29+%3D+%281+%2B+x%29%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+x%5E%7B2%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;left(&#92;frac{1 - x}{1 + x}&#92;right)^{2}&#92;right) = (1 + x)&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; x^{2}&#92;right)' title='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;left(&#92;frac{1 - x}{1 + x}&#92;right)^{2}&#92;right) = (1 + x)&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; x^{2}&#92;right)' class='latex' /></p>
<p>If we set</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1+-+x%7D%7B1+%2B+x%7D+%3D+%5Cfrac%7B%5Cphi%5E%7B2%7D%28-q%29%7D%7B%5Cphi%5E%7B2%7D%28q%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{1 - x}{1 + x} = &#92;frac{&#92;phi^{2}(-q)}{&#92;phi^{2}(q)}' title='&#92;displaystyle &#92;frac{1 - x}{1 + x} = &#92;frac{&#92;phi^{2}(-q)}{&#92;phi^{2}(q)}' class='latex' /></p>
<p>where <img src='http://s0.wp.com/latex.php?latex=%5Cphi%28q%29+%3D+%5Csum_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;phi(q) = &#92;sum_{-&#92;infty}^{&#92;infty}q^{n^{2}}' title='&#92;phi(q) = &#92;sum_{-&#92;infty}^{&#92;infty}q^{n^{2}}' class='latex' /> is one of the theta functions defined by Ramanujan (see <a title="Elementary Approach to Modular Equations: Ramanujan’s Theory 1" href="http://paramanands.wordpress.com/2011/11/06/elementary-approach-to-modular-equations-ramanujans-theory-1/" target="_blank">this post</a>) so that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+x+%3D+%5Cfrac%7B%5Cphi%5E%7B2%7D%28q%29+-+%5Cphi%5E%7B2%7D%28-q%29%7D%7B%5Cphi%5E%7B2%7D%28q%29+%2B+%5Cphi%5E%7B2%7D%28-q%29%7D+%3D+%5Cfrac%7B4q%5Cpsi%5E%7B2%7D%28q%5E%7B4%7D%29%7D%7B%5Cphi%5E%7B2%7D%28q%5E%7B2%7D%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle x = &#92;frac{&#92;phi^{2}(q) - &#92;phi^{2}(-q)}{&#92;phi^{2}(q) + &#92;phi^{2}(-q)} = &#92;frac{4q&#92;psi^{2}(q^{4})}{&#92;phi^{2}(q^{2})}' title='&#92;displaystyle x = &#92;frac{&#92;phi^{2}(q) - &#92;phi^{2}(-q)}{&#92;phi^{2}(q) + &#92;phi^{2}(-q)} = &#92;frac{4q&#92;psi^{2}(q^{4})}{&#92;phi^{2}(q^{2})}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+1+%2B+x+%3D+%5Cfrac%7B2%5Cphi%5E%7B2%7D%28q%29%7D%7B%5Cphi%5E%7B2%7D%28q%29+%2B+%5Cphi%5E%7B2%7D%28-q%29%7D+%3D+%5Cfrac%7B%5Cphi%5E%7B2%7D%28q%29%7D%7B%5Cphi%5E%7B2%7D%28q%5E%7B2%7D%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;Rightarrow 1 + x = &#92;frac{2&#92;phi^{2}(q)}{&#92;phi^{2}(q) + &#92;phi^{2}(-q)} = &#92;frac{&#92;phi^{2}(q)}{&#92;phi^{2}(q^{2})}' title='&#92;displaystyle &#92;Rightarrow 1 + x = &#92;frac{2&#92;phi^{2}(q)}{&#92;phi^{2}(q) + &#92;phi^{2}(-q)} = &#92;frac{&#92;phi^{2}(q)}{&#92;phi^{2}(q^{2})}' class='latex' /></p>
<p>and</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+x%5E%7B2%7D+%3D+%5Cfrac%7B16q%5E%7B2%7D%5Cpsi%5E%7B4%7D%28q%5E%7B4%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7B2%7D%29%7D+%3D+%5Cfrac%7B%5Cphi%5E%7B4%7D%28q%5E%7B2%7D%29+-+%5Cphi%5E%7B4%7D%28-q%5E%7B2%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7B2%7D%29%7D+%3D+1+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7B2%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7B2%7D%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle x^{2} = &#92;frac{16q^{2}&#92;psi^{4}(q^{4})}{&#92;phi^{4}(q^{2})} = &#92;frac{&#92;phi^{4}(q^{2}) - &#92;phi^{4}(-q^{2})}{&#92;phi^{4}(q^{2})} = 1 - &#92;frac{&#92;phi^{4}(-q^{2})}{&#92;phi^{4}(q^{2})}' title='&#92;displaystyle x^{2} = &#92;frac{16q^{2}&#92;psi^{4}(q^{4})}{&#92;phi^{4}(q^{2})} = &#92;frac{&#92;phi^{4}(q^{2}) - &#92;phi^{4}(-q^{2})}{&#92;phi^{4}(q^{2})} = 1 - &#92;frac{&#92;phi^{4}(-q^{2})}{&#92;phi^{4}(q^{2})}' class='latex' /></p>
<p>Therefore we finally obtain:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+1+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29+%3D+%5Cfrac%7B%5Cphi%5E%7B2%7D%28q%29%7D%7B%5Cphi%5E%7B2%7D%28q%5E%7B2%7D%29%7D%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+1+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7B2%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7B2%7D%29%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) = &#92;frac{&#92;phi^{2}(q)}{&#92;phi^{2}(q^{2})}&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;frac{&#92;phi^{4}(-q^{2})}{&#92;phi^{4}(q^{2})}&#92;right)' title='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) = &#92;frac{&#92;phi^{2}(q)}{&#92;phi^{2}(q^{2})}&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;frac{&#92;phi^{4}(-q^{2})}{&#92;phi^{4}(q^{2})}&#92;right)' class='latex' /></p>
<p>The beauty of this formula is that it can be applied recursively. The hypergeometric function on the right is the same as that on the left except that the parameter <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=fff&amp;fg=222&amp;s=0' alt='q' title='q' class='latex' /> is replaced by <img src='http://s0.wp.com/latex.php?latex=q%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='q^{2}' title='q^{2}' class='latex' />. Also the first factor on the right consists of parameter <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=fff&amp;fg=222&amp;s=0' alt='q' title='q' class='latex' /> in numerator and <img src='http://s0.wp.com/latex.php?latex=q%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='q^{2}' title='q^{2}' class='latex' /> in denominator so that the recursion can be utilized quite nicely to lead us to the following result:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+1+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29+%3D+%5Cfrac%7B%5Cphi%5E%7B2%7D%28q%29%7D%7B%5Cphi%5E%7B2%7D%28q%5E%7B2%5E%7Bn%7D%7D%29%7D%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+1+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7B2%5E%7Bn%7D%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7B2%5E%7Bn%7D%7D%29%7D%5Cright%29%5C%2C%5C%2C%5C%2C%5Ccdots+%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) = &#92;frac{&#92;phi^{2}(q)}{&#92;phi^{2}(q^{2^{n}})}&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;frac{&#92;phi^{4}(-q^{2^{n}})}{&#92;phi^{4}(q^{2^{n}})}&#92;right)&#92;,&#92;,&#92;,&#92;cdots (1)' title='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) = &#92;frac{&#92;phi^{2}(q)}{&#92;phi^{2}(q^{2^{n}})}&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;frac{&#92;phi^{4}(-q^{2^{n}})}{&#92;phi^{4}(q^{2^{n}})}&#92;right)&#92;,&#92;,&#92;,&#92;cdots (1)' class='latex' /></p>
<p>When <img src='http://s0.wp.com/latex.php?latex=n+%5Cto+%5Cinfty%2C+q%5E%7B2%5E%7Bn%7D%7D+%5Cto+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='n &#92;to &#92;infty, q^{2^{n}} &#92;to 0' title='n &#92;to &#92;infty, q^{2^{n}} &#92;to 0' class='latex' /> and therefore <img src='http://s0.wp.com/latex.php?latex=%5Cphi%28q%5E%7B2%5E%7Bn%7D%7D%29+%5Cto+1%2C%5C%2C%5Cphi%28-q%5E%7B2%5E%7Bn%7D%7D%29+%5Cto+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;phi(q^{2^{n}}) &#92;to 1,&#92;,&#92;phi(-q^{2^{n}}) &#92;to 1' title='&#92;phi(q^{2^{n}}) &#92;to 1,&#92;,&#92;phi(-q^{2^{n}}) &#92;to 1' class='latex' /> so that we obtain</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28q%29+%3D%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+1+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29%5C%2C%5C%2C%5C%2C%5Ccdots+%282%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(q) =&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)&#92;,&#92;,&#92;,&#92;cdots (2)' title='&#92;displaystyle &#92;phi^{2}(q) =&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)&#92;,&#92;,&#92;,&#92;cdots (2)' class='latex' /></p>
<p>We can use the Gauss transformation formula again by setting</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+x+%3D+%5Cfrac%7B%5Cphi%5E%7B2%7D%28-q%29%7D%7B%5Cphi%5E%7B2%7D%28q%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle x = &#92;frac{&#92;phi^{2}(-q)}{&#92;phi^{2}(q)}' title='&#92;displaystyle x = &#92;frac{&#92;phi^{2}(-q)}{&#92;phi^{2}(q)}' class='latex' /></p>
<p>and noting that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B4x%7D%7B%281+%2B+x%29%5E%7B2%7D%7D+%3D+4%5C%2C%5Cfrac%7B%5Cphi%5E%7B2%7D%28-q%29%7D%7B%5Cphi%5E%7B2%7D%28q%29%7D%5Cfrac%7B%5Cphi%5E%7B4%7D%28q%29%7D%7B%28%5Cphi%5E%7B2%7D%28q%29+%2B+%5Cphi%5E%7B2%7D%28-q%29%29%5E%7B2%7D%7D+%3D+%5Cfrac%7B%5Cphi%5E%7B2%7D%28q%29%5Cphi%5E%7B2%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7B2%7D%29%7D+%3D+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7B2%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7B2%7D%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{4x}{(1 + x)^{2}} = 4&#92;,&#92;frac{&#92;phi^{2}(-q)}{&#92;phi^{2}(q)}&#92;frac{&#92;phi^{4}(q)}{(&#92;phi^{2}(q) + &#92;phi^{2}(-q))^{2}} = &#92;frac{&#92;phi^{2}(q)&#92;phi^{2}(-q)}{&#92;phi^{4}(q^{2})} = &#92;frac{&#92;phi^{4}(-q^{2})}{&#92;phi^{4}(q^{2})}' title='&#92;displaystyle &#92;frac{4x}{(1 + x)^{2}} = 4&#92;,&#92;frac{&#92;phi^{2}(-q)}{&#92;phi^{2}(q)}&#92;frac{&#92;phi^{4}(q)}{(&#92;phi^{2}(q) + &#92;phi^{2}(-q))^{2}} = &#92;frac{&#92;phi^{2}(q)&#92;phi^{2}(-q)}{&#92;phi^{4}(q^{2})} = &#92;frac{&#92;phi^{4}(-q^{2})}{&#92;phi^{4}(q^{2})}' class='latex' /></p>
<p>and</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+1+%2B+x+%3D+%5Cfrac%7B2%5Cphi%5E%7B2%7D%28q%5E%7B2%7D%29%7D%7B%5Cphi%5E%7B2%7D%28q%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 1 + x = &#92;frac{2&#92;phi^{2}(q^{2})}{&#92;phi^{2}(q)}' title='&#92;displaystyle 1 + x = &#92;frac{2&#92;phi^{2}(q^{2})}{&#92;phi^{2}(q)}' class='latex' /></p>
<p>Thus we arrive at</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7B2%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7B2%7D%29%7D%5Cright%29+%3D+%5Cfrac%7B2%5Cphi%5E%7B2%7D%28q%5E%7B2%7D%29%7D%7B%5Cphi%5E%7B2%7D%28q%29%7D%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{&#92;phi^{4}(-q^{2})}{&#92;phi^{4}(q^{2})}&#92;right) = &#92;frac{2&#92;phi^{2}(q^{2})}{&#92;phi^{2}(q)}&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)' title='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{&#92;phi^{4}(-q^{2})}{&#92;phi^{4}(q^{2})}&#92;right) = &#92;frac{2&#92;phi^{2}(q^{2})}{&#92;phi^{2}(q)}&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29+%3D+%5Cfrac%7B%5Cphi%5E%7B2%7D%28q%29%7D%7B2%5Cphi%5E%7B2%7D%28q%5E%7B2%7D%29%7D%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7B2%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7B2%7D%29%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) = &#92;frac{&#92;phi^{2}(q)}{2&#92;phi^{2}(q^{2})}&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{&#92;phi^{4}(-q^{2})}{&#92;phi^{4}(q^{2})}&#92;right)' title='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) = &#92;frac{&#92;phi^{2}(q)}{2&#92;phi^{2}(q^{2})}&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{&#92;phi^{4}(-q^{2})}{&#92;phi^{4}(q^{2})}&#92;right)' class='latex' /></p>
<p>By applying recursion we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29+%3D+%5Cfrac%7B%5Cphi%5E%7B2%7D%28q%29%7D%7B2%5E%7Bn%7D%5Cphi%5E%7B2%7D%28q%5E%7B2%5E%7Bn%7D%7D%29%7D%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7B2%5E%7Bn%7D%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7B2%5E%7Bn%7D%7D%29%7D%5Cright%29%5C%2C%5C%2C%5C%2C%5Ccdots+%283%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) = &#92;frac{&#92;phi^{2}(q)}{2^{n}&#92;phi^{2}(q^{2^{n}})}&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{&#92;phi^{4}(-q^{2^{n}})}{&#92;phi^{4}(q^{2^{n}})}&#92;right)&#92;,&#92;,&#92;,&#92;cdots (3)' title='&#92;displaystyle _{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) = &#92;frac{&#92;phi^{2}(q)}{2^{n}&#92;phi^{2}(q^{2^{n}})}&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; &#92;frac{&#92;phi^{4}(-q^{2^{n}})}{&#92;phi^{4}(q^{2^{n}})}&#92;right)&#92;,&#92;,&#92;,&#92;cdots (3)' class='latex' /></p>
<p>Dividing <img src='http://s0.wp.com/latex.php?latex=%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(1)' title='(1)' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=%283%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(3)' title='(3)' class='latex' /> and replacing <img src='http://s0.wp.com/latex.php?latex=2%5E%7Bn%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='2^{n}' title='2^{n}' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=m&amp;bg=fff&amp;fg=222&amp;s=0' alt='m' title='m' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cdfrac%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C+%5Cdfrac%7B1%7D%7B2%7D%3B+1%3B+1+-+%5Cdfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29%7D%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C+%5Cdfrac%7B1%7D%7B2%7D%3B+1%3B+%5Cdfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29%7D+%3D+m%5C%2C%5Cdfrac%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C+%5Cdfrac%7B1%7D%7B2%7D%3B+1%3B+1+-+%5Cdfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7Bm%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7Bm%7D%29%7D%5Cright%29%7D%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C+%5Cdfrac%7B1%7D%7B2%7D%3B+1%3B+%5Cdfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7Bm%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7Bm%7D%29%7D%5Cright%29%7D%5C%2C%5C%2C%5C%2C%5Ccdots+%284%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2}, &#92;dfrac{1}{2}; 1; 1 - &#92;dfrac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2}, &#92;dfrac{1}{2}; 1; &#92;dfrac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)} = m&#92;,&#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2}, &#92;dfrac{1}{2}; 1; 1 - &#92;dfrac{&#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2}, &#92;dfrac{1}{2}; 1; &#92;dfrac{&#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}&#92;right)}&#92;,&#92;,&#92;,&#92;cdots (4)' title='&#92;displaystyle &#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2}, &#92;dfrac{1}{2}; 1; 1 - &#92;dfrac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2}, &#92;dfrac{1}{2}; 1; &#92;dfrac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)} = m&#92;,&#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2}, &#92;dfrac{1}{2}; 1; 1 - &#92;dfrac{&#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2}, &#92;dfrac{1}{2}; 1; &#92;dfrac{&#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}&#92;right)}&#92;,&#92;,&#92;,&#92;cdots (4)' class='latex' /></p>
<p>Mutiplying by <img src='http://s0.wp.com/latex.php?latex=-%5Cpi&amp;bg=fff&amp;fg=222&amp;s=0' alt='-&#92;pi' title='-&#92;pi' class='latex' /> and taking exponentials on both sides we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+F%5Cleft%28%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29+%3D+%5Cleft%5C%7BF%5Cleft%28%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7Bm%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7Bm%7D%29%7D%5Cright%29%5Cright%5C%7D%5E%7Bm%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle F&#92;left(&#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) = &#92;left&#92;{F&#92;left(&#92;frac{&#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}&#92;right)&#92;right&#92;}^{m}' title='&#92;displaystyle F&#92;left(&#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) = &#92;left&#92;{F&#92;left(&#92;frac{&#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}&#92;right)&#92;right&#92;}^{m}' class='latex' /></p>
<p>where <img src='http://s0.wp.com/latex.php?latex=m&amp;bg=fff&amp;fg=222&amp;s=0' alt='m' title='m' class='latex' /> is a positive integral power of <img src='http://s0.wp.com/latex.php?latex=2&amp;bg=fff&amp;fg=222&amp;s=0' alt='2' title='2' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=F%28x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='F(x)' title='F(x)' class='latex' /> represents the Ramanujan&#8217;s function defined in <a title="Elementary Approach to Modular Equations: Ramanujan’s Theory 2" href="http://paramanands.wordpress.com/2011/12/18/elementary-approach-to-modular-equations-ramanujans-theory-2/" target="_blank">previous post</a>.</p>
<p>The same equation <img src='http://s0.wp.com/latex.php?latex=%284%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(4)' title='(4)' class='latex' /> on taking reciprocals leads us to</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cleft%5C%7BF%5Cleft%281+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29%5Cright%5C%7D%5E%7Bm%7D+%3D+F%5Cleft%281+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7Bm%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7Bm%7D%29%7D%5Cright%29%5C%2C%5C%2C%5C%2C%5Ccdots+%285%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;left&#92;{F&#92;left(1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)&#92;right&#92;}^{m} = F&#92;left(1 - &#92;frac{&#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}&#92;right)&#92;,&#92;,&#92;,&#92;cdots (5)' title='&#92;displaystyle &#92;left&#92;{F&#92;left(1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)&#92;right&#92;}^{m} = F&#92;left(1 - &#92;frac{&#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}&#92;right)&#92;,&#92;,&#92;,&#92;cdots (5)' class='latex' /></p>
<p><strong>The Inversion Fomula</strong></p>
<p>Using the above result we can prove the fundamental inversion formula</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+F%5Cleft%281+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29+%3D+q%5C%2C%5C%2C%5C%2C%5Ccdots+%286%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle F&#92;left(1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) = q&#92;,&#92;,&#92;,&#92;cdots (6)' title='&#92;displaystyle F&#92;left(1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) = q&#92;,&#92;,&#92;,&#92;cdots (6)' class='latex' /></p>
<p>Clearly if we set</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+x_%7Bm%7D+%3D+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7Bm%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7Bm%7D%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle x_{m} = &#92;frac{&#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}' title='&#92;displaystyle x_{m} = &#92;frac{&#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}' class='latex' /></p>
<p>then <img src='http://s0.wp.com/latex.php?latex=x_%7Bm%7D+%5Cto+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='x_{m} &#92;to 1' title='x_{m} &#92;to 1' class='latex' /> as <img src='http://s0.wp.com/latex.php?latex=m+%5Cto+%5Cinfty&amp;bg=fff&amp;fg=222&amp;s=0' alt='m &#92;to &#92;infty' title='m &#92;to &#92;infty' class='latex' /> and hence <img src='http://s0.wp.com/latex.php?latex=1+-+x_%7Bm%7D+%5Cto+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='1 - x_{m} &#92;to 0' title='1 - x_{m} &#92;to 0' class='latex' />. Clearly then we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=F%281+-+x_%7Bm%7D%29+%5Csim+%281+-+x_%7Bm%7D%29%2F16&amp;bg=fff&amp;fg=222&amp;s=0' alt='F(1 - x_{m}) &#92;sim (1 - x_{m})/16' title='F(1 - x_{m}) &#92;sim (1 - x_{m})/16' class='latex' /> and hence</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7DF%5Cleft%281+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29+%26%3D+%5Csqrt%5Bm%5D%7BF%281+-+x_%7Bm%7D%29%7D%5C%5C++++%26%3D+%5Clim_%7Bm+%5Cto+%5Cinfty%7D%5Csqrt%5Bm%5D%7BF%281+-+x_%7Bm%7D%29%7D%5C%5C++++%26%3D+%5Clim_%7Bm+%5Cto+%5Cinfty%7D%5Csqrt%5Bm%5D%7B%5Cfrac%7B1+-+x_%7Bm%7D%7D%7B16%7D%7D%5C%5C++++%26%3D+%5Clim_%7Bm+%5Cto+%5Cinfty%7D%5Csqrt%5Bm%5D%7B1+-+x_%7Bm%7D%7D+%3D+A%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned}F&#92;left(1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) &amp;= &#92;sqrt[m]{F(1 - x_{m})}&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;sqrt[m]{F(1 - x_{m})}&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;sqrt[m]{&#92;frac{1 - x_{m}}{16}}&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;sqrt[m]{1 - x_{m}} = A&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned}F&#92;left(1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right) &amp;= &#92;sqrt[m]{F(1 - x_{m})}&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;sqrt[m]{F(1 - x_{m})}&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;sqrt[m]{&#92;frac{1 - x_{m}}{16}}&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;sqrt[m]{1 - x_{m}} = A&#92;end{aligned}' class='latex' /></p>
<p>Then we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D+%5Clog+A+%26%3D+%5Clim_%7Bm+%5Cto+%5Cinfty%7D%5Cfrac%7B1%7D%7Bm%7D%5Clog%281+-+x_%7Bm%7D%29%5C%5C++++%26%3D+%5Clim_%7Bm+%5Cto+%5Cinfty%7D%5Cfrac%7B1%7D%7Bm%7D%5Clog%5Cleft%281+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%5E%7Bm%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7Bm%7D%29%7D%5Cright%29%5C%5C++++%26%3D+%5Clim_%7Bm+%5Cto+%5Cinfty%7D%5Cfrac%7B1%7D%7Bm%7D%5Clog%5Cleft%28%5Cfrac%7B%5Cphi%5E%7B4%7D%28q%5E%7Bm%7D%29+-+%5Cphi%5E%7B4%7D%28-q%5E%7Bm%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7Bm%7D%29%7D%5Cright%29%5C%5C++++%26%3D+%5Clim_%7Bm+%5Cto+%5Cinfty%7D%5Cfrac%7B1%7D%7Bm%7D%5Clog%5Cleft%28%5Cfrac%7B16q%5E%7Bm%7D%5Cpsi%5E%7B4%7D%28q%5E%7B2m%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7Bm%7D%29%7D%5Cright%29%5C%5C++++%26%3D+%5Clim_%7Bm+%5Cto+%5Cinfty%7D%5Clog+q+%2B+%5Cfrac%7B%5Clog+16%7D%7Bm%7D+%2B+%5Clog%5Cleft%28%5Cfrac%7B%5Cpsi%5E%7B4%7D%28q%5E%7B2m%7D%29%7D%7B%5Cphi%5E%7B4%7D%28q%5E%7Bm%7D%29%7D%5Cright%29%5C%5C++++%26%3D+%5Clog+q+%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned} &#92;log A &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;frac{1}{m}&#92;log(1 - x_{m})&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;frac{1}{m}&#92;log&#92;left(1 - &#92;frac{&#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}&#92;right)&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;frac{1}{m}&#92;log&#92;left(&#92;frac{&#92;phi^{4}(q^{m}) - &#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}&#92;right)&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;frac{1}{m}&#92;log&#92;left(&#92;frac{16q^{m}&#92;psi^{4}(q^{2m})}{&#92;phi^{4}(q^{m})}&#92;right)&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;log q + &#92;frac{&#92;log 16}{m} + &#92;log&#92;left(&#92;frac{&#92;psi^{4}(q^{2m})}{&#92;phi^{4}(q^{m})}&#92;right)&#92;&#92;    &amp;= &#92;log q &#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned} &#92;log A &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;frac{1}{m}&#92;log(1 - x_{m})&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;frac{1}{m}&#92;log&#92;left(1 - &#92;frac{&#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}&#92;right)&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;frac{1}{m}&#92;log&#92;left(&#92;frac{&#92;phi^{4}(q^{m}) - &#92;phi^{4}(-q^{m})}{&#92;phi^{4}(q^{m})}&#92;right)&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;frac{1}{m}&#92;log&#92;left(&#92;frac{16q^{m}&#92;psi^{4}(q^{2m})}{&#92;phi^{4}(q^{m})}&#92;right)&#92;&#92;    &amp;= &#92;lim_{m &#92;to &#92;infty}&#92;log q + &#92;frac{&#92;log 16}{m} + &#92;log&#92;left(&#92;frac{&#92;psi^{4}(q^{2m})}{&#92;phi^{4}(q^{m})}&#92;right)&#92;&#92;    &amp;= &#92;log q &#92;end{aligned}' class='latex' /></p>
<p>And thus we have <img src='http://s0.wp.com/latex.php?latex=A+%3D+q&amp;bg=fff&amp;fg=222&amp;s=0' alt='A = q' title='A = q' class='latex' /> and the inversion formula is established. In other words if</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+x+%3D+1+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle x = 1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}' title='&#92;displaystyle x = 1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}' class='latex' /></p>
<p>then</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+q+%3D+F%28x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle q = F(x)' title='&#92;displaystyle q = F(x)' class='latex' /></p>
<p>Using <img src='http://s0.wp.com/latex.php?latex=%286%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(6)' title='(6)' class='latex' /> in <img src='http://s0.wp.com/latex.php?latex=%282%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(2)' title='(2)' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%5Cleft%5C%7BF%5Cleft%281+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29%5Cright%5C%7D+%3D%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+1+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}&#92;left&#92;{F&#92;left(1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)&#92;right&#92;} =&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)' title='&#92;displaystyle &#92;phi^{2}&#92;left&#92;{F&#92;left(1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)&#92;right&#92;} =&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; 1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}&#92;right)' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28F%28x%29%29+%3D%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C+%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+x%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(F(x)) =&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; x&#92;right)' title='&#92;displaystyle &#92;phi^{2}(F(x)) =&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2}, &#92;frac{1}{2}; 1; x&#92;right)' class='latex' /></p>
<p>where</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+x+%3D+1+-+%5Cfrac%7B%5Cphi%5E%7B4%7D%28-q%29%7D%7B%5Cphi%5E%7B4%7D%28q%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle x = 1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}' title='&#92;displaystyle x = 1 - &#92;frac{&#92;phi^{4}(-q)}{&#92;phi^{4}(q)}' class='latex' /></p>
<p>Using the Ramnujan&#8217;s notation <img src='http://s0.wp.com/latex.php?latex=x%2C+y%2C+z&amp;bg=fff&amp;fg=222&amp;s=0' alt='x, y, z' title='x, y, z' class='latex' /> we now have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%28F%28x%29%29+%3D+%5Cphi%28e%5E%7B-y%7D%29+%3D+%5Csqrt%7Bz%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi(F(x)) = &#92;phi(e^{-y}) = &#92;sqrt{z}' title='&#92;displaystyle &#92;phi(F(x)) = &#92;phi(e^{-y}) = &#92;sqrt{z}' class='latex' /></p>
<p><strong>Transformation Formula for <img src='http://s0.wp.com/latex.php?latex=%5Cphi%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;phi(q)' title='&#92;phi(q)' class='latex' /></strong></p>
<p>If we have a look at the definitions of <img src='http://s0.wp.com/latex.php?latex=x%2C+y&amp;bg=fff&amp;fg=222&amp;s=0' alt='x, y' title='x, y' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=F%28x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='F(x)' title='F(x)' class='latex' /> then we can see that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Clog+F%28x%29+%3D+-y&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;log F(x) = -y' title='&#92;log F(x) = -y' class='latex' /></p>
<p>and</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Clog+F%28x%29+%5Ccdot+%5Clog+F%281+-+x%29+%3D+%5Cpi%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;log F(x) &#92;cdot &#92;log F(1 - x) = &#92;pi^{2}' title='&#92;log F(x) &#92;cdot &#92;log F(1 - x) = &#92;pi^{2}' class='latex' /></p>
<p>so that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Clog+F%281+-+x%29+%3D+-%5Cpi%5E%7B2%7D+%2F+y&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;log F(1 - x) = -&#92;pi^{2} / y' title='&#92;log F(1 - x) = -&#92;pi^{2} / y' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=F%281+-+x%29+%3D+e%5E%7B-%5Cpi%5E%7B2%7D+%2F+y%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='F(1 - x) = e^{-&#92;pi^{2} / y}' title='F(1 - x) = e^{-&#92;pi^{2} / y}' class='latex' /></p>
<p>Let <img src='http://s0.wp.com/latex.php?latex=%5Calpha%2C%5Cbeta&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;alpha,&#92;beta' title='&#92;alpha,&#92;beta' class='latex' /> be positive numbers such that <img src='http://s0.wp.com/latex.php?latex=%5Calpha%5Cbeta+%3D+%5Cpi&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;alpha&#92;beta = &#92;pi' title='&#92;alpha&#92;beta = &#92;pi' class='latex' /> and let <img src='http://s0.wp.com/latex.php?latex=y+%3D+%5Calpha%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = &#92;alpha^{2}' title='y = &#92;alpha^{2}' class='latex' /> so that <img src='http://s0.wp.com/latex.php?latex=%5Cpi%5E%7B2%7D%2Fy+%3D+%5Cbeta%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;pi^{2}/y = &#92;beta^{2}' title='&#92;pi^{2}/y = &#92;beta^{2}' class='latex' />. Then we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B%5Cphi%5E%7B2%7D%28F%281+-+x%29%29%7D%7B%5Cphi%5E%7B2%7D%28F%28x%29%29%7D+%3D+%5Cdfrac%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%3B+1%3B+1+-+x%5Cright%29%7D%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%3B+1%3B+x%5Cright%29%7D+%3D+%5Cfrac%7By%7D%7B%5Cpi%7D+%3D+%5Cfrac%7B%5Calpha%5E%7B2%7D%7D%7B%5Calpha%5Cbeta%7D+%3D+%5Cfrac%7B%5Calpha%7D%7B%5Cbeta%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{&#92;phi^{2}(F(1 - x))}{&#92;phi^{2}(F(x))} = &#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2}; 1; 1 - x&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2}; 1; x&#92;right)} = &#92;frac{y}{&#92;pi} = &#92;frac{&#92;alpha^{2}}{&#92;alpha&#92;beta} = &#92;frac{&#92;alpha}{&#92;beta}' title='&#92;displaystyle &#92;frac{&#92;phi^{2}(F(1 - x))}{&#92;phi^{2}(F(x))} = &#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2}; 1; 1 - x&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2}; 1; x&#92;right)} = &#92;frac{y}{&#92;pi} = &#92;frac{&#92;alpha^{2}}{&#92;alpha&#92;beta} = &#92;frac{&#92;alpha}{&#92;beta}' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B%5Cphi%5E%7B2%7D%28e%5E%7B-%5Cbeta%5E%7B2%7D%7D%29%7D%7B%5Cphi%5E%7B2%7D%28e%5E%7B-%5Calpha%5E%7B2%7D%7D%29%7D+%3D+%5Cfrac%7B%5Calpha%7D%7B%5Cbeta%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{&#92;phi^{2}(e^{-&#92;beta^{2}})}{&#92;phi^{2}(e^{-&#92;alpha^{2}})} = &#92;frac{&#92;alpha}{&#92;beta}' title='&#92;displaystyle &#92;frac{&#92;phi^{2}(e^{-&#92;beta^{2}})}{&#92;phi^{2}(e^{-&#92;alpha^{2}})} = &#92;frac{&#92;alpha}{&#92;beta}' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Csqrt%7B%5Calpha%7D%5C%2C%5Cphi%28e%5E%7B-%5Calpha%5E%7B2%7D%7D%29+%3D+%5Csqrt%7B%5Cbeta%7D%5C%2C%5Cphi%28e%5E%7B-%5Cbeta%5E%7B2%7D%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;sqrt{&#92;alpha}&#92;,&#92;phi(e^{-&#92;alpha^{2}}) = &#92;sqrt{&#92;beta}&#92;,&#92;phi(e^{-&#92;beta^{2}})' title='&#92;sqrt{&#92;alpha}&#92;,&#92;phi(e^{-&#92;alpha^{2}}) = &#92;sqrt{&#92;beta}&#92;,&#92;phi(e^{-&#92;beta^{2}})' class='latex' /></p>
<p>In the classical notation this is</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Csqrt%7Bs%7D%5C%2C%5Ctheta_%7B3%7D%28e%5E%7B-%5Cpi+s%7D%29+%3D+%5Ctheta_%7B3%7D%28e%5E%7B-%5Cpi+%2F+s%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;sqrt{s}&#92;,&#92;theta_{3}(e^{-&#92;pi s}) = &#92;theta_{3}(e^{-&#92;pi / s})' title='&#92;sqrt{s}&#92;,&#92;theta_{3}(e^{-&#92;pi s}) = &#92;theta_{3}(e^{-&#92;pi / s})' class='latex' /> (see <a title="The Magic of Theta Functions: Contd." href="http://paramanands.wordpress.com/2010/10/20/the-magic-of-theta-functions-contd/" target="_blank">this post</a>)</p>
<p>and therefore the above constitutes a proof of the transformation formula for theta functions without the use of <em>Poisson Summation formula</em>.</p>
<p>In the next post we will have a look various identities relating the theta functions of <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=fff&amp;fg=222&amp;s=0' alt='q' title='q' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=q%5E%7Bn%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='q^{n}' title='q^{n}' class='latex' /> and these will be finally transcribed in the form of modular equations. Most of the identities will be derived using <em>Lambert series</em> for the theta functions.</p>
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		<title>Elementary Approach to Modular Equations: Ramanujan&#8217;s Theory 2</title>
		<link>http://paramanands.wordpress.com/2011/12/18/elementary-approach-to-modular-equations-ramanujans-theory-2/</link>
		<comments>http://paramanands.wordpress.com/2011/12/18/elementary-approach-to-modular-equations-ramanujans-theory-2/#comments</comments>
		<pubDate>Sun, 18 Dec 2011 07:38:21 +0000</pubDate>
		<dc:creator>paramanands</dc:creator>
				<category><![CDATA[Elliptic Functions]]></category>
		<category><![CDATA[Elliptic Integrals]]></category>
		<category><![CDATA[Mathematical Analysis]]></category>

		<guid isPermaLink="false">http://paramanands.wordpress.com/?p=2558</guid>
		<description><![CDATA[Ramanujan&#8217;s Theory of Elliptic Functions Ramanujan used the letter in place of and studied the function in great detail and developed his theory of elliptic integrals and functions. Following Ramanujan let&#8217;s define For Ramanujan elliptic function theory consisted of finding relations among . As can be easily seen from the above definitions corresponds to and [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=paramanands.wordpress.com&amp;blog=8565079&amp;post=2558&amp;subd=paramanands&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p><strong>Ramanujan&#8217;s Theory of Elliptic Functions</strong></p>
<p>Ramanujan used the letter <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> in place of <img src='http://s0.wp.com/latex.php?latex=k%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='k^{2}' title='k^{2}' class='latex' /> and studied the function <img src='http://s0.wp.com/latex.php?latex=_%7B2%7DF_%7B1%7D%281%2F2%2C+1%2F2%3B+1%3B+x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='_{2}F_{1}(1/2, 1/2; 1; x)' title='_{2}F_{1}(1/2, 1/2; 1; x)' class='latex' /> in great detail and developed his theory of elliptic integrals and functions. Following Ramanujan let&#8217;s define</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+x+%3D+k%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle x = k^{2}' title='&#92;displaystyle x = k^{2}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+z+%3D%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C%5Cfrac%7B1%7D%7B2%7D%3B+1%3B+x%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle z =&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2},&#92;frac{1}{2}; 1; x&#92;right)' title='&#92;displaystyle z =&#92;,_{2}F_{1}&#92;left(&#92;frac{1}{2},&#92;frac{1}{2}; 1; x&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+y+%3D+%5Cpi%5Ccdot%5Cdfrac%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%3B+1%3B+1+-+x%5Cright%29%7D%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%3B+1%3B+x%5Cright%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle y = &#92;pi&#92;cdot&#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2}; 1; 1 - x&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2}; 1; x&#92;right)}' title='&#92;displaystyle y = &#92;pi&#92;cdot&#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2}; 1; 1 - x&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2}; 1; x&#92;right)}' class='latex' /></p>
<p>For Ramanujan elliptic function theory consisted of finding relations among <img src='http://s0.wp.com/latex.php?latex=x%2C+y%2C+z&amp;bg=fff&amp;fg=222&amp;s=0' alt='x, y, z' title='x, y, z' class='latex' />. As can be easily seen from the above definitions <img src='http://s0.wp.com/latex.php?latex=z&amp;bg=fff&amp;fg=222&amp;s=0' alt='z' title='z' class='latex' /> corresponds to <img src='http://s0.wp.com/latex.php?latex=K&amp;bg=fff&amp;fg=222&amp;s=0' alt='K' title='K' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> corresponds to <img src='http://s0.wp.com/latex.php?latex=%5Cpi+K%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;pi K&#039;/K' title='&#92;pi K&#039;/K' class='latex' /> so that the nome <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=fff&amp;fg=222&amp;s=0' alt='q' title='q' class='latex' /> is given by <img src='http://s0.wp.com/latex.php?latex=q+%3D+e%5E%7B-y%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='q = e^{-y}' title='q = e^{-y}' class='latex' />.</p>
<p>Ramanujan in fact studied the function <img src='http://s0.wp.com/latex.php?latex=F%28x%29+%3D+e%5E%7B-y%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='F(x) = e^{-y}' title='F(x) = e^{-y}' class='latex' /> and obtained the fundamental formulas connecting <img src='http://s0.wp.com/latex.php?latex=x%2C+y%2C+z&amp;bg=fff&amp;fg=222&amp;s=0' alt='x, y, z' title='x, y, z' class='latex' />. In doing so Ramanujan went far ahead of his predecessors and developed the <em>theory of theta functions to alternative bases</em>. Using various identities in the theory of hypergeometric functions he studied the properties of <img src='http://s0.wp.com/latex.php?latex=u_%7Bx%7D+%3D%5C%2C_%7B2%7DF_%7B1%7D%281+-+n%2C+n%3B+1%3B+x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='u_{x} =&#92;,_{2}F_{1}(1 - n, n; 1; x)' title='u_{x} =&#92;,_{2}F_{1}(1 - n, n; 1; x)' class='latex' /> for non-integral <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=fff&amp;fg=222&amp;s=0' alt='n' title='n' class='latex' /> and the associated function (which is nome)</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cexp%5Cleft%28-%5Cfrac%7B%5Cpi%7D%7B%5Csin%28%5Cpi+n%29%7D%5Cfrac%7Bu_%7B1+-+x%7D%7D%7Bu_%7Bx%7D%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;exp&#92;left(-&#92;frac{&#92;pi}{&#92;sin(&#92;pi n)}&#92;frac{u_{1 - x}}{u_{x}}&#92;right)' title='&#92;displaystyle &#92;exp&#92;left(-&#92;frac{&#92;pi}{&#92;sin(&#92;pi n)}&#92;frac{u_{1 - x}}{u_{x}}&#92;right)' class='latex' /></p>
<p>For <img src='http://s0.wp.com/latex.php?latex=n+%3D+1%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='n = 1/2' title='n = 1/2' class='latex' /> this reduces to the classical theory. Also the factor <img src='http://s0.wp.com/latex.php?latex=%5Cpi%2F%5Csin%28%5Cpi+n%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;pi/&#92;sin(&#92;pi n)' title='&#92;pi/&#92;sin(&#92;pi n)' class='latex' /> is actually <img src='http://s0.wp.com/latex.php?latex=%5CGamma%28n%29%5CGamma%281+-+n%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;Gamma(n)&#92;Gamma(1 - n)' title='&#92;Gamma(n)&#92;Gamma(1 - n)' class='latex' />. This is the source of the constant <img src='http://s0.wp.com/latex.php?latex=%5Cpi&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;pi' title='&#92;pi' class='latex' /> in the formula for nome. Apart from the classical case <img src='http://s0.wp.com/latex.php?latex=n+%3D+1%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='n = 1/2' title='n = 1/2' class='latex' /> Ramanujan developed the theories for <img src='http://s0.wp.com/latex.php?latex=n+%3D+1%2F3%2C+1%2F4%2C+1%2F6&amp;bg=fff&amp;fg=222&amp;s=0' alt='n = 1/3, 1/4, 1/6' title='n = 1/3, 1/4, 1/6' class='latex' /></p>
<p>We will restrict ourselves however to the classical case when <img src='http://s0.wp.com/latex.php?latex=n+%3D+1%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='n = 1/2' title='n = 1/2' class='latex' />.</p>
<p>Let us then define along the lines of Ramanujan the function <img src='http://s0.wp.com/latex.php?latex=F%28x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='F(x)' title='F(x)' class='latex' /> by</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+F%28x%29+%3D+e%5E%7B-y%7D+%3D+%5Cexp%5Cleft%28-%5Cpi%5C%2C%5Cdfrac%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%3B1%3B+1+-+x%5Cright%29%7D%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%3B1%3B+x%5Cright%29%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle F(x) = e^{-y} = &#92;exp&#92;left(-&#92;pi&#92;,&#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; 1 - x&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; x&#92;right)}&#92;right)' title='&#92;displaystyle F(x) = e^{-y} = &#92;exp&#92;left(-&#92;pi&#92;,&#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; 1 - x&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; x&#92;right)}&#92;right)' class='latex' /></p>
<p><strong>Behavior of <img src='http://s0.wp.com/latex.php?latex=F%28x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='F(x)' title='F(x)' class='latex' /> Near <img src='http://s0.wp.com/latex.php?latex=x+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = 0' title='x = 0' class='latex' /></strong></p>
<p>We first establish the important result that <img src='http://s0.wp.com/latex.php?latex=F%28x%29%2Fx&amp;bg=fff&amp;fg=222&amp;s=0' alt='F(x)/x' title='F(x)/x' class='latex' /> tends to positive constant value as <img src='http://s0.wp.com/latex.php?latex=x+%5Cto+0%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='x &#92;to 0+' title='x &#92;to 0+' class='latex' />. In other words we wish to prove that <img src='http://s0.wp.com/latex.php?latex=F%28x%29+%3D+x%28A+%2B+o%281%29%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='F(x) = x(A + o(1))' title='F(x) = x(A + o(1))' class='latex' /> where <img src='http://s0.wp.com/latex.php?latex=A&amp;bg=fff&amp;fg=222&amp;s=0' alt='A' title='A' class='latex' /> is some positive constant. This is equivalent to proving</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+-%5Cpi%5C%2C%5Cdfrac%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%3B1%3B+1+-+x%5Cright%29%7D%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%3B1%3B+x%5Cright%29%7D+%3D+%5Clog+x+%2B+%5Clog+%28A+%2B+o%281%29%29+%3D+%5Clog+A+%2B+%5Clog+x+%2B+o%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle -&#92;pi&#92;,&#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; 1 - x&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; x&#92;right)} = &#92;log x + &#92;log (A + o(1)) = &#92;log A + &#92;log x + o(1)' title='&#92;displaystyle -&#92;pi&#92;,&#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; 1 - x&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; x&#92;right)} = &#92;log x + &#92;log (A + o(1)) = &#92;log A + &#92;log x + o(1)' class='latex' /></p>
<p>Since the denominator on the left tends to <img src='http://s0.wp.com/latex.php?latex=1&amp;bg=fff&amp;fg=222&amp;s=0' alt='1' title='1' class='latex' /> as <img src='http://s0.wp.com/latex.php?latex=x+%5Cto+0%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='x &#92;to 0+' title='x &#92;to 0+' class='latex' /> it is sufficient to prove that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+-%5Cpi%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%3B1%3B+1+-+x%5Cright%29+%3D+%5Clog+x+%2B+C+%2B+o%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle -&#92;pi&#92;,_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; 1 - x&#92;right) = &#92;log x + C + o(1)' title='&#92;displaystyle -&#92;pi&#92;,_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; 1 - x&#92;right) = &#92;log x + C + o(1)' class='latex' /></p>
<p>where <img src='http://s0.wp.com/latex.php?latex=C+%3D+%5Clog+A&amp;bg=fff&amp;fg=222&amp;s=0' alt='C = &#92;log A' title='C = &#92;log A' class='latex' /> is some constant. Replacing <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=%281+-+x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(1 - x)' title='(1 - x)' class='latex' /> we see that we need to prove</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cpi%5C%2C_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%3B1%3B+x%5Cright%29+%3D+-%5Clog+%281+-+x%29+-+C+%2B+o%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;pi&#92;,_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; x&#92;right) = -&#92;log (1 - x) - C + o(1)' title='&#92;displaystyle &#92;pi&#92;,_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; x&#92;right) = -&#92;log (1 - x) - C + o(1)' class='latex' /></p>
<p>when <img src='http://s0.wp.com/latex.php?latex=x+%5Cto+1-&amp;bg=fff&amp;fg=222&amp;s=0' alt='x &#92;to 1-' title='x &#92;to 1-' class='latex' />. In the language of elliptic integrals this means that</p>
<p><img src='http://s0.wp.com/latex.php?latex=2K%28%5Csqrt%7Bx%7D%29+%3D+-%5Clog%281+-+x%29+-+C+%2B+o%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='2K(&#92;sqrt{x}) = -&#92;log(1 - x) - C + o(1)' title='2K(&#92;sqrt{x}) = -&#92;log(1 - x) - C + o(1)' class='latex' /></p>
<p>Replacing <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=1+-+k%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='1 - k^{2}' title='1 - k^{2}' class='latex' /> we see that</p>
<p><img src='http://s0.wp.com/latex.php?latex=2K%28k%27%29+%3D+-2%5Clog+k+-+C+%2B+o%281%29+&amp;bg=fff&amp;fg=222&amp;s=0' alt='2K(k&#039;) = -2&#92;log k - C + o(1) ' title='2K(k&#039;) = -2&#92;log k - C + o(1) ' class='latex' /> where <img src='http://s0.wp.com/latex.php?latex=k+%5Cto+0%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='k &#92;to 0+' title='k &#92;to 0+' class='latex' /> or in other words</p>
<p><img src='http://s0.wp.com/latex.php?latex=K%28k%27%29+%3D+K%27+%3D+-%5Clog+k+-+%28C%2F2%29+%2B+o%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='K(k&#039;) = K&#039; = -&#92;log k - (C/2) + o(1)' title='K(k&#039;) = K&#039; = -&#92;log k - (C/2) + o(1)' class='latex' />.</p>
<p>We will now establish that <img src='http://s0.wp.com/latex.php?latex=K%27+%3D+%5Clog+%284%2Fk%29+%2B+o%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039; = &#92;log (4/k) + o(1)' title='K&#039; = &#92;log (4/k) + o(1)' class='latex' /> so that <img src='http://s0.wp.com/latex.php?latex=C+%3D+-%5Clog+16&amp;bg=fff&amp;fg=222&amp;s=0' alt='C = -&#92;log 16' title='C = -&#92;log 16' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=A+%3D+e%5E%7BC%7D+%3D+1%2F16&amp;bg=fff&amp;fg=222&amp;s=0' alt='A = e^{C} = 1/16' title='A = e^{C} = 1/16' class='latex' /></p>
<p>We proceed by starting with the definition of <img src='http://s0.wp.com/latex.php?latex=K%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;' title='K&#039;' class='latex' />:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D+K%27+%26%3D+K%28k%27%29+%3D+%5Cint_%7B0%7D%5E%7B%5Cpi+%2F+2%7D+%5Cfrac%7Bd%5Ctheta%7D%7B%5Csqrt%7B1+-+%281+-+k%5E%7B2%7D%29%5Csin%5E%7B2%7D%5Ctheta%7D%7D%5C%5C++++%26%3D+%5Cint_%7B0%7D%5E%7B%5Cpi+%2F+2%7D+%5Cfrac%7Bk%27%5Csin%5Ctheta%5C%2Cd%5Ctheta%7D%7B%5Csqrt%7B1+-+%281+-+k%5E%7B2%7D%29%5Csin%5E%7B2%7D%5Ctheta%7D%7D+%2B+%5Cint_%7B0%7D%5E%7B%5Cpi+%2F+2%7D+%5Cfrac%7B1+-+k%27%5Csin%5Ctheta%5C%2Cd%5Ctheta%7D%7B%5Csqrt%7B1+-+%281+-+k%5E%7B2%7D%29%5Csin%5E%7B2%7D%5Ctheta%7D%7D%5C%5C++++%26%3D+%5Cint_%7B0%7D%5E%7B%5Cpi+%2F+2%7D+%5Cfrac%7Bk%27%5Csin%5Ctheta%5C%2Cd%5Ctheta%7D%7B%5Csqrt%7Bk%5E%7B2%7D+%2B+k%27%5E%7B2%7D%5Ccos%5E%7B2%7D%5Ctheta%7D%7D+%2B+%5Cint_%7B0%7D%5E%7B%5Cpi+%2F+2%7D+%5Cfrac%7B1+-+k%27%5Csin%5Ctheta%5C%2Cd%5Ctheta%7D%7B%5Csqrt%7B1+-+k%27%5E%7B2%7D%5Csin%5E%7B2%7D%5Ctheta%7D%7D%5C%5C++++%26%3D+%5Cint_%7B0%7D%5E%7Bk%27%7D%5Cfrac%7Bdt%7D%7B%5Csqrt%7Bt%5E%7B2%7D+%2B+k%5E%7B2%7D%7D%7D+%2B+%5Cint_%7B0%7D%5E%7B%5Cpi+%2F+2%7D%5Csqrt%7B%5Cfrac%7B1+-+k%27%5Csin%5Ctheta%7D%7B1+%2B+k%27%5Csin%5Ctheta%7D%7D%5C%2Cd%5Ctheta%5C%5C++++%26%3D+%5Clog%5Cleft%28%5Cfrac%7B1+%2B+k%27%7D%7Bk%7D%5Cright%29+%2B+%5Cint_%7B0%7D%5E%7B%5Cpi+%2F+2%7D%5Csqrt%7B%5Cfrac%7B1+-+k%27%5Csin%5Ctheta%7D%7B1+%2B+k%27%5Csin%5Ctheta%7D%7D%5C%2Cd%5Ctheta%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned} K&#039; &amp;= K(k&#039;) = &#92;int_{0}^{&#92;pi / 2} &#92;frac{d&#92;theta}{&#92;sqrt{1 - (1 - k^{2})&#92;sin^{2}&#92;theta}}&#92;&#92;    &amp;= &#92;int_{0}^{&#92;pi / 2} &#92;frac{k&#039;&#92;sin&#92;theta&#92;,d&#92;theta}{&#92;sqrt{1 - (1 - k^{2})&#92;sin^{2}&#92;theta}} + &#92;int_{0}^{&#92;pi / 2} &#92;frac{1 - k&#039;&#92;sin&#92;theta&#92;,d&#92;theta}{&#92;sqrt{1 - (1 - k^{2})&#92;sin^{2}&#92;theta}}&#92;&#92;    &amp;= &#92;int_{0}^{&#92;pi / 2} &#92;frac{k&#039;&#92;sin&#92;theta&#92;,d&#92;theta}{&#92;sqrt{k^{2} + k&#039;^{2}&#92;cos^{2}&#92;theta}} + &#92;int_{0}^{&#92;pi / 2} &#92;frac{1 - k&#039;&#92;sin&#92;theta&#92;,d&#92;theta}{&#92;sqrt{1 - k&#039;^{2}&#92;sin^{2}&#92;theta}}&#92;&#92;    &amp;= &#92;int_{0}^{k&#039;}&#92;frac{dt}{&#92;sqrt{t^{2} + k^{2}}} + &#92;int_{0}^{&#92;pi / 2}&#92;sqrt{&#92;frac{1 - k&#039;&#92;sin&#92;theta}{1 + k&#039;&#92;sin&#92;theta}}&#92;,d&#92;theta&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{1 + k&#039;}{k}&#92;right) + &#92;int_{0}^{&#92;pi / 2}&#92;sqrt{&#92;frac{1 - k&#039;&#92;sin&#92;theta}{1 + k&#039;&#92;sin&#92;theta}}&#92;,d&#92;theta&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned} K&#039; &amp;= K(k&#039;) = &#92;int_{0}^{&#92;pi / 2} &#92;frac{d&#92;theta}{&#92;sqrt{1 - (1 - k^{2})&#92;sin^{2}&#92;theta}}&#92;&#92;    &amp;= &#92;int_{0}^{&#92;pi / 2} &#92;frac{k&#039;&#92;sin&#92;theta&#92;,d&#92;theta}{&#92;sqrt{1 - (1 - k^{2})&#92;sin^{2}&#92;theta}} + &#92;int_{0}^{&#92;pi / 2} &#92;frac{1 - k&#039;&#92;sin&#92;theta&#92;,d&#92;theta}{&#92;sqrt{1 - (1 - k^{2})&#92;sin^{2}&#92;theta}}&#92;&#92;    &amp;= &#92;int_{0}^{&#92;pi / 2} &#92;frac{k&#039;&#92;sin&#92;theta&#92;,d&#92;theta}{&#92;sqrt{k^{2} + k&#039;^{2}&#92;cos^{2}&#92;theta}} + &#92;int_{0}^{&#92;pi / 2} &#92;frac{1 - k&#039;&#92;sin&#92;theta&#92;,d&#92;theta}{&#92;sqrt{1 - k&#039;^{2}&#92;sin^{2}&#92;theta}}&#92;&#92;    &amp;= &#92;int_{0}^{k&#039;}&#92;frac{dt}{&#92;sqrt{t^{2} + k^{2}}} + &#92;int_{0}^{&#92;pi / 2}&#92;sqrt{&#92;frac{1 - k&#039;&#92;sin&#92;theta}{1 + k&#039;&#92;sin&#92;theta}}&#92;,d&#92;theta&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{1 + k&#039;}{k}&#92;right) + &#92;int_{0}^{&#92;pi / 2}&#92;sqrt{&#92;frac{1 - k&#039;&#92;sin&#92;theta}{1 + k&#039;&#92;sin&#92;theta}}&#92;,d&#92;theta&#92;end{aligned}' class='latex' /></p>
<p>If we denote the integral on the right as <img src='http://s0.wp.com/latex.php?latex=a%28k%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='a(k)' title='a(k)' class='latex' /> we can see that by uniformity arguments <img src='http://s0.wp.com/latex.php?latex=a%28k%29+%5Cto+a%280%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='a(k) &#92;to a(0)' title='a(k) &#92;to a(0)' class='latex' /> as <img src='http://s0.wp.com/latex.php?latex=k+%5Cto+0%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='k &#92;to 0+' title='k &#92;to 0+' class='latex' />. And value of <img src='http://s0.wp.com/latex.php?latex=a%280%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='a(0)' title='a(0)' class='latex' /> is easily calculated as:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+a%280%29+%3D+%5Cint_%7B0%7D%5E%7B%5Cpi+%2F+2%7D%5Csqrt%7B%5Cfrac%7B1+-+%5Csin%5Ctheta%7D%7B1+%2B+%5Csin%5Ctheta%7D%7D%5C%2Cd%5Ctheta+%3D+%5Cint_%7B0%7D%5E%7B%5Cpi+%2F+2%7D%5Cfrac%7B%5Ccos%5Ctheta%7D%7B1+%2B+%5Csin%5Ctheta%7D%5C%2Cd%5Ctheta+%3D+%5Clog+2&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle a(0) = &#92;int_{0}^{&#92;pi / 2}&#92;sqrt{&#92;frac{1 - &#92;sin&#92;theta}{1 + &#92;sin&#92;theta}}&#92;,d&#92;theta = &#92;int_{0}^{&#92;pi / 2}&#92;frac{&#92;cos&#92;theta}{1 + &#92;sin&#92;theta}&#92;,d&#92;theta = &#92;log 2' title='&#92;displaystyle a(0) = &#92;int_{0}^{&#92;pi / 2}&#92;sqrt{&#92;frac{1 - &#92;sin&#92;theta}{1 + &#92;sin&#92;theta}}&#92;,d&#92;theta = &#92;int_{0}^{&#92;pi / 2}&#92;frac{&#92;cos&#92;theta}{1 + &#92;sin&#92;theta}&#92;,d&#92;theta = &#92;log 2' class='latex' /></p>
<p>Therefore we have the following relation</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7DK%27+%26%3D+%5Clog%5Cleft%28%5Cfrac%7B2%281+%2B+k%27%29%7D%7Bk%7D%5Cright%29+%2B+o%281%29%5C%5C++++%26%3D+%5Clog%5Cleft%28%5Cfrac%7B4%7D%7Bk%7D%5Cright%29+%2B+%5Clog%5Cleft%28%5Cfrac%7B1+%2B+k%27%7D%7B2%7D%5Cright%29+%2B+o%281%29%5C%5C++++%26%3D+%5Clog%5Cleft%28%5Cfrac%7B4%7D%7Bk%7D%5Cright%29+%2B+%5Clog%5Cleft%28%5Cfrac%7B1+%2B+%5Csqrt%7B1+-+k%5E%7B2%7D%7D%7D%7B2%7D%5Cright%29+%2B+o%281%29%5C%5C++++%26%3D+%5Clog%5Cleft%28%5Cfrac%7B4%7D%7Bk%7D%5Cright%29+%2B+%5Clog%5Cleft%28%5Cdfrac%7B1+%2B+1+-+%5Cdfrac%7Bk%5E%7B2%7D%7D%7B2%7D%7D%7B2%7D%5Cright%29+%2B+o%281%29%5C%5C++++%26%3D+%5Clog%5Cleft%28%5Cfrac%7B4%7D%7Bk%7D%5Cright%29+%2B+%5Clog%5Cleft%281+-+%5Cfrac%7Bk%5E%7B2%7D%7D%7B4%7D%5Cright%29+%2B+o%281%29%5C%5C++++%26%3D+%5Clog%5Cleft%28%5Cfrac%7B4%7D%7Bk%7D%5Cright%29+-+%5Cfrac%7Bk%5E%7B2%7D%7D%7B4%7D+%2B+o%281%29%5C%5C++++%26%3D+%5Clog%5Cleft%28%5Cfrac%7B4%7D%7Bk%7D%5Cright%29+%2B+o%281%29+%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned}K&#039; &amp;= &#92;log&#92;left(&#92;frac{2(1 + k&#039;)}{k}&#92;right) + o(1)&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{4}{k}&#92;right) + &#92;log&#92;left(&#92;frac{1 + k&#039;}{2}&#92;right) + o(1)&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{4}{k}&#92;right) + &#92;log&#92;left(&#92;frac{1 + &#92;sqrt{1 - k^{2}}}{2}&#92;right) + o(1)&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{4}{k}&#92;right) + &#92;log&#92;left(&#92;dfrac{1 + 1 - &#92;dfrac{k^{2}}{2}}{2}&#92;right) + o(1)&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{4}{k}&#92;right) + &#92;log&#92;left(1 - &#92;frac{k^{2}}{4}&#92;right) + o(1)&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{4}{k}&#92;right) - &#92;frac{k^{2}}{4} + o(1)&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{4}{k}&#92;right) + o(1) &#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned}K&#039; &amp;= &#92;log&#92;left(&#92;frac{2(1 + k&#039;)}{k}&#92;right) + o(1)&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{4}{k}&#92;right) + &#92;log&#92;left(&#92;frac{1 + k&#039;}{2}&#92;right) + o(1)&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{4}{k}&#92;right) + &#92;log&#92;left(&#92;frac{1 + &#92;sqrt{1 - k^{2}}}{2}&#92;right) + o(1)&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{4}{k}&#92;right) + &#92;log&#92;left(&#92;dfrac{1 + 1 - &#92;dfrac{k^{2}}{2}}{2}&#92;right) + o(1)&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{4}{k}&#92;right) + &#92;log&#92;left(1 - &#92;frac{k^{2}}{4}&#92;right) + o(1)&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{4}{k}&#92;right) - &#92;frac{k^{2}}{4} + o(1)&#92;&#92;    &amp;= &#92;log&#92;left(&#92;frac{4}{k}&#92;right) + o(1) &#92;end{aligned}' class='latex' /></p>
<p>It now follows that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Clim_%7Bx+%5Cto+0%2B%7D%5Cfrac%7BF%28x%29%7D%7Bx%7D+%3D+%5Cfrac%7B1%7D%7B16%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;lim_{x &#92;to 0+}&#92;frac{F(x)}{x} = &#92;frac{1}{16}' title='&#92;displaystyle &#92;lim_{x &#92;to 0+}&#92;frac{F(x)}{x} = &#92;frac{1}{16}' class='latex' /> or in a grander form</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Clim_%7Bx+%5Cto+0%2B%7D%5Cfrac%7B1%7D%7Bx%7D%5C%2C%5Cexp%5Cleft%28-%5Cpi%5C%2C%5Cdfrac%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%3B1%3B+1+-+x%5Cright%29%7D%7B_%7B2%7DF_%7B1%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%3B1%3B+x%5Cright%29%7D%5Cright%29+%3D+%5Cfrac%7B1%7D%7B16%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;lim_{x &#92;to 0+}&#92;frac{1}{x}&#92;,&#92;exp&#92;left(-&#92;pi&#92;,&#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; 1 - x&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; x&#92;right)}&#92;right) = &#92;frac{1}{16}' title='&#92;displaystyle &#92;lim_{x &#92;to 0+}&#92;frac{1}{x}&#92;,&#92;exp&#92;left(-&#92;pi&#92;,&#92;dfrac{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; 1 - x&#92;right)}{_{2}F_{1}&#92;left(&#92;dfrac{1}{2},&#92;dfrac{1}{2};1; x&#92;right)}&#92;right) = &#92;frac{1}{16}' class='latex' /></p>
<p>This is the key result which is used in Ramanujan&#8217;s elliptic function theory. In the next post we use this result and the transformation fomulas of hypergeometric functions to develop Ramanujan&#8217;s theory.</p>
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		<title>Elementary Approach to Modular Equations: Ramanujan&#8217;s Theory 1</title>
		<link>http://paramanands.wordpress.com/2011/11/06/elementary-approach-to-modular-equations-ramanujans-theory-1/</link>
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		<pubDate>Sun, 06 Nov 2011 07:49:23 +0000</pubDate>
		<dc:creator>paramanands</dc:creator>
				<category><![CDATA[Elliptic Functions]]></category>
		<category><![CDATA[Elliptic Integrals]]></category>
		<category><![CDATA[Mathematical Analysis]]></category>

		<guid isPermaLink="false">http://paramanands.wordpress.com/?p=2416</guid>
		<description><![CDATA[Ramanujan developed his theory of modular equations using the theory of theta functions independently of Jacobi. A complete understanding of his approach is unfortunately not possible till now because he did not publish something like Fundamenta Nova containing detailed explanations of his approach. What we have today is his Notebooks edited by Bruce C. Berndt [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=paramanands.wordpress.com&amp;blog=8565079&amp;post=2416&amp;subd=paramanands&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>Ramanujan developed his theory of modular equations using the theory of theta functions independently of Jacobi. A complete understanding of his approach is unfortunately not possible till now because he did not publish something like <em>Fundamenta Nova </em>containing detailed explanations of his approach. What we have today is his Notebooks edited by Bruce C. Berndt and his Collected Papers. His Notebooks are just statements of various mathematical formulas without any proof. A large part of these notebooks is concerned with modular equations and modern authors have not been able to discern his methods fully. Hence I will not be able to present a true picture of his approach. Rather I will try to present whatever I understand from his Collected Papers and his Notebooks and only focus on the elementary aspects.</p>
<p><strong>Ramanujan&#8217;s Theta Functions<br />
</strong></p>
<p>Ramanujan started his theory of modular equations from the relation <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+nK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = nK&#039;/K' title='L&#039;/L = nK&#039;/K' class='latex' /> and worked with the parameter <img src='http://s0.wp.com/latex.php?latex=q+%3D+e%5E%7B-%5Cpi+K%27%2FK%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='q = e^{-&#92;pi K&#039;/K}' title='q = e^{-&#92;pi K&#039;/K}' class='latex' />. The modulus <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> can be expressed in form of theta functions of parameter <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=fff&amp;fg=222&amp;s=0' alt='q' title='q' class='latex' />. Thus we have <img src='http://s0.wp.com/latex.php?latex=k+%3D+k%28q%29+%3D+%5Ctheta_%7B2%7D%5E%7B2%7D%28q%29%2F%5Ctheta_%7B3%7D%5E%7B2%7D%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='k = k(q) = &#92;theta_{2}^{2}(q)/&#92;theta_{3}^{2}(q)' title='k = k(q) = &#92;theta_{2}^{2}(q)/&#92;theta_{3}^{2}(q)' class='latex' />. From the equation <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+nK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = nK&#039;/K' title='L&#039;/L = nK&#039;/K' class='latex' /> we immediately see that <img src='http://s0.wp.com/latex.php?latex=l+%3D+%5Ctheta_%7B2%7D%5E%7B2%7D%28q%5E%7Bn%7D%29%2F%5Ctheta_%7B3%7D%5E%7B2%7D%28q%5E%7Bn%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='l = &#92;theta_{2}^{2}(q^{n})/&#92;theta_{3}^{2}(q^{n})' title='l = &#92;theta_{2}^{2}(q^{n})/&#92;theta_{3}^{2}(q^{n})' class='latex' />. Thus an algebraic relation between <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> is actually a relation between theta functions of parameter <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=fff&amp;fg=222&amp;s=0' alt='q' title='q' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=q%5E%7Bn%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='q^{n}' title='q^{n}' class='latex' />. Ramanujan was such an expert in formal manipulations of series, products and continued fractions that he readily obtained many relations between theta functions of <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=fff&amp;fg=222&amp;s=0' alt='q' title='q' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=q%5E%7Bn%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='q^{n}' title='q^{n}' class='latex' /> and then transformed them into the traditional relation between <img src='http://s0.wp.com/latex.php?latex=k%2C+l&amp;bg=fff&amp;fg=222&amp;s=0' alt='k, l' title='k, l' class='latex' />. Thus his modular equations are quite diverse in nature and sometimes they also contain more than 2 moduli.</p>
<p>Following Ramanujan we define the the function <img src='http://s0.wp.com/latex.php?latex=f%28a%2C+b%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(a, b)' title='f(a, b)' class='latex' /> by</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+f%28a%2C+b%29+%3D+%5Csum_%7Bn+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7Da%5E%7Bn%28n+%2B+1%29%2F2%7Db%5E%7Bn%28n+-+1%29%2F2%7D+%3D+1+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%28ab%29%5E%7Bn%28n+-+1%29%2F2%7D%28a%5E%7Bn%7D+%2B+b%5E%7Bn%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle f(a, b) = &#92;sum_{n = -&#92;infty}^{&#92;infty}a^{n(n + 1)/2}b^{n(n - 1)/2} = 1 + &#92;sum_{n = 1}^{&#92;infty}(ab)^{n(n - 1)/2}(a^{n} + b^{n})' title='&#92;displaystyle f(a, b) = &#92;sum_{n = -&#92;infty}^{&#92;infty}a^{n(n + 1)/2}b^{n(n - 1)/2} = 1 + &#92;sum_{n = 1}^{&#92;infty}(ab)^{n(n - 1)/2}(a^{n} + b^{n})' class='latex' /></p>
<p>for all complex numbers <img src='http://s0.wp.com/latex.php?latex=a%2C+b&amp;bg=fff&amp;fg=222&amp;s=0' alt='a, b' title='a, b' class='latex' /> with <img src='http://s0.wp.com/latex.php?latex=%7Cab%7C+%3C+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='|ab| &lt; 1' title='|ab| &lt; 1' class='latex' />. This function has the following elementary properties</p>
<ul>
<li><img src='http://s0.wp.com/latex.php?latex=f%28a%2C+b%29+%3D+f%28b%2C+a%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(a, b) = f(b, a)' title='f(a, b) = f(b, a)' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=f%28-1%2C+a%29+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(-1, a) = 0' title='f(-1, a) = 0' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=f%281%2C+a%29+%3D+2f%28a%2C+a%5E%7B3%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(1, a) = 2f(a, a^{3})' title='f(1, a) = 2f(a, a^{3})' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=f%28a%2C+b%29+%3D+a%5E%7Bn%28n+%2B+1%29%2F2%7Db%5E%7Bn%28n+-+1%29%2F2%7Df%28a%28ab%29%5E%7Bn%7D%2C+b%28ab%29%5E%7B-n%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(a, b) = a^{n(n + 1)/2}b^{n(n - 1)/2}f(a(ab)^{n}, b(ab)^{-n})' title='f(a, b) = a^{n(n + 1)/2}b^{n(n - 1)/2}f(a(ab)^{n}, b(ab)^{-n})' class='latex' /> where <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=fff&amp;fg=222&amp;s=0' alt='n' title='n' class='latex' /> is an integer</li>
</ul>
<p>The first property is quite obvious and follows from the observation that changing the index of summation in definition of <img src='http://s0.wp.com/latex.php?latex=f%28a%2C+b%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(a, b)' title='f(a, b)' class='latex' /> from <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=fff&amp;fg=222&amp;s=0' alt='n' title='n' class='latex' /> to <img src='http://s0.wp.com/latex.php?latex=-n&amp;bg=fff&amp;fg=222&amp;s=0' alt='-n' title='-n' class='latex' /> leads to exchange of <img src='http://s0.wp.com/latex.php?latex=a&amp;bg=fff&amp;fg=222&amp;s=0' alt='a' title='a' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=b&amp;bg=fff&amp;fg=222&amp;s=0' alt='b' title='b' class='latex' />.</p>
<p>For the second property we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D+f%28-1%2C+a%29+%26+%3D+f%28a%2C+-1%29+%3D+%5Csum_%7Bn+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D%28-1%29%5E%7Bn%28n+-+1%29%2F2%7D%5C%5C++++%26+%3D+%5Csum_%7Bn+%3D+-%5Cinfty%7D%5E%7Bn+%3D+0%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D%28-1%29%5E%7Bn%28n+-+1%29%2F2%7D+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D%28-1%29%5E%7Bn%28n+-+1%29%2F2%7D%5C%5C++++%26+%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%28n+%2B+1%29%2F2%7Da%5E%7Bn%28n+-+1%29%2F2%7D+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%28n+-+1%29%2F2%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D%5C%5C++++%26+%3D+1+-+1+%2B+%5Csum_%7Bn+%3D+2%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%28n+%2B+1%29%2F2%7Da%5E%7Bn%28n+-+1%29%2F2%7D+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%28n+-+1%29%2F2%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D%5C%5C++++%26+%3D+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7B%28n+%2B+1%29%28n+%2B+2%29%2F2%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%28n+-+1%29%2F2%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned} f(-1, a) &amp; = f(a, -1) = &#92;sum_{n = -&#92;infty}^{&#92;infty}a^{n(n + 1)/2}(-1)^{n(n - 1)/2}&#92;&#92;    &amp; = &#92;sum_{n = -&#92;infty}^{n = 0}a^{n(n + 1)/2}(-1)^{n(n - 1)/2} + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2}(-1)^{n(n - 1)/2}&#92;&#92;    &amp; = &#92;sum_{n = 0}^{&#92;infty}(-1)^{n(n + 1)/2}a^{n(n - 1)/2} + &#92;sum_{n = 1}^{&#92;infty}(-1)^{n(n - 1)/2}a^{n(n + 1)/2}&#92;&#92;    &amp; = 1 - 1 + &#92;sum_{n = 2}^{&#92;infty}(-1)^{n(n + 1)/2}a^{n(n - 1)/2} + &#92;sum_{n = 1}^{&#92;infty}(-1)^{n(n - 1)/2}a^{n(n + 1)/2}&#92;&#92;    &amp; = &#92;sum_{n = 1}^{&#92;infty}(-1)^{(n + 1)(n + 2)/2}a^{n(n + 1)/2} + &#92;sum_{n = 1}^{&#92;infty}(-1)^{n(n - 1)/2}a^{n(n + 1)/2}&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned} f(-1, a) &amp; = f(a, -1) = &#92;sum_{n = -&#92;infty}^{&#92;infty}a^{n(n + 1)/2}(-1)^{n(n - 1)/2}&#92;&#92;    &amp; = &#92;sum_{n = -&#92;infty}^{n = 0}a^{n(n + 1)/2}(-1)^{n(n - 1)/2} + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2}(-1)^{n(n - 1)/2}&#92;&#92;    &amp; = &#92;sum_{n = 0}^{&#92;infty}(-1)^{n(n + 1)/2}a^{n(n - 1)/2} + &#92;sum_{n = 1}^{&#92;infty}(-1)^{n(n - 1)/2}a^{n(n + 1)/2}&#92;&#92;    &amp; = 1 - 1 + &#92;sum_{n = 2}^{&#92;infty}(-1)^{n(n + 1)/2}a^{n(n - 1)/2} + &#92;sum_{n = 1}^{&#92;infty}(-1)^{n(n - 1)/2}a^{n(n + 1)/2}&#92;&#92;    &amp; = &#92;sum_{n = 1}^{&#92;infty}(-1)^{(n + 1)(n + 2)/2}a^{n(n + 1)/2} + &#92;sum_{n = 1}^{&#92;infty}(-1)^{n(n - 1)/2}a^{n(n + 1)/2}&#92;end{aligned}' class='latex' /></p>
<p>Now <img src='http://s0.wp.com/latex.php?latex=%28n+%2B+1%29%28n+%2B+2%29%2F2+-+n%28n+-+1%29%2F2+%3D+%28n%5E%7B2%7D+%2B+3n+%2B+2+-+n%5E%7B2%7D+%2B+n%29%2F2+%3D+2n+%2B+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='(n + 1)(n + 2)/2 - n(n - 1)/2 = (n^{2} + 3n + 2 - n^{2} + n)/2 = 2n + 1' title='(n + 1)(n + 2)/2 - n(n - 1)/2 = (n^{2} + 3n + 2 - n^{2} + n)/2 = 2n + 1' class='latex' /> so that <img src='http://s0.wp.com/latex.php?latex=%28-1%29%5E%7B%28n+%2B+1%29%28n+%2B+2%29%2F2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='(-1)^{(n + 1)(n + 2)/2}' title='(-1)^{(n + 1)(n + 2)/2}' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%28-1%29%5E%7Bn%28n+-+1%29%2F2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='(-1)^{n(n - 1)/2}' title='(-1)^{n(n - 1)/2}' class='latex' /> are of opposite signs and hence the sum cancels to zero.</p>
<p>To handle <img src='http://s0.wp.com/latex.php?latex=f%281%2C+a%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(1, a)' title='f(1, a)' class='latex' /> we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D+f%281%2C+a%29+%26+%3D+%5Csum_%7Bn+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D%5C%5C++++%26+%3D+1+%2B+1+%2B+%5Csum_%7Bn+%3D+-%5Cinfty%7D%5E%7B-2%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D%5C%5C++++%26+%3D+2+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D+%2B+%5Csum_%7Bn+%3D+2%7D%5E%7B%5Cinfty%7Da%5E%7Bn%28n+-+1%29%2F2%7D%5C%5C++++%26+%3D+2+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D%5C%5C++++%26+%3D+2%5Cleft%281+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D%5Cright%29%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned} f(1, a) &amp; = &#92;sum_{n = -&#92;infty}^{&#92;infty}a^{n(n + 1)/2}&#92;&#92;    &amp; = 1 + 1 + &#92;sum_{n = -&#92;infty}^{-2}a^{n(n + 1)/2} + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2}&#92;&#92;    &amp; = 2 + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2} + &#92;sum_{n = 2}^{&#92;infty}a^{n(n - 1)/2}&#92;&#92;    &amp; = 2 + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2} + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2}&#92;&#92;    &amp; = 2&#92;left(1 + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2}&#92;right)&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned} f(1, a) &amp; = &#92;sum_{n = -&#92;infty}^{&#92;infty}a^{n(n + 1)/2}&#92;&#92;    &amp; = 1 + 1 + &#92;sum_{n = -&#92;infty}^{-2}a^{n(n + 1)/2} + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2}&#92;&#92;    &amp; = 2 + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2} + &#92;sum_{n = 2}^{&#92;infty}a^{n(n - 1)/2}&#92;&#92;    &amp; = 2 + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2} + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2}&#92;&#92;    &amp; = 2&#92;left(1 + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2}&#92;right)&#92;end{aligned}' class='latex' /></p>
<p>Splitting the sum on right into even and odd indices we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D+f%281%2C+a%29+%26+%3D+2%5Cleft%281+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Da%5E%7Bn%282n+%2B+1%29%7D+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Da%5E%7Bn%282n+-+1%29%7D%5Cright%29%5C%5C++++%26+%3D+2%5Cleft%281+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Da%5E%7Bn%28n+-+1%29%2F2%7D%28a%5E%7B3%7D%29%5E%7Bn%28n+%2B+1%29%2F2%7D+%2B+%5Csum_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7Da%5E%7Bn%28n+%2B+1%29%2F2%7D%28a%5E%7B3%7D%29%5E%7Bn%28n+-+1%29%2F2%7D%5Cright%29+%3D+2f%28a%2C+a%5E%7B3%7D%29%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned} f(1, a) &amp; = 2&#92;left(1 + &#92;sum_{n = 1}^{&#92;infty}a^{n(2n + 1)} + &#92;sum_{n = 1}^{&#92;infty}a^{n(2n - 1)}&#92;right)&#92;&#92;    &amp; = 2&#92;left(1 + &#92;sum_{n = 1}^{&#92;infty}a^{n(n - 1)/2}(a^{3})^{n(n + 1)/2} + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2}(a^{3})^{n(n - 1)/2}&#92;right) = 2f(a, a^{3})&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned} f(1, a) &amp; = 2&#92;left(1 + &#92;sum_{n = 1}^{&#92;infty}a^{n(2n + 1)} + &#92;sum_{n = 1}^{&#92;infty}a^{n(2n - 1)}&#92;right)&#92;&#92;    &amp; = 2&#92;left(1 + &#92;sum_{n = 1}^{&#92;infty}a^{n(n - 1)/2}(a^{3})^{n(n + 1)/2} + &#92;sum_{n = 1}^{&#92;infty}a^{n(n + 1)/2}(a^{3})^{n(n - 1)/2}&#92;right) = 2f(a, a^{3})&#92;end{aligned}' class='latex' /></p>
<p>The last equation can be handled as follows:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D+f%28a%2C+b%29+%26%3D+%5Csum_%7Bk+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7Da%5E%7Bk%28k+%2B+1%29%2F2%7Db%5E%7Bk%28k+-+1%29%2F2%7D%5C%5C++++%26+%3D+%5Csum_%7Bk+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7Da%5E%7B%28k+%2B+n%29%28k+%2B+n+%2B+1%29%2F2%7Db%5E%7B%28k+%2B+n%29%28k+%2B+n+-+1%29%2F2%7D%5C%5C++++%26+%3D+a%5E%7Bn%28n+%2B+1%29%2F2%7Db%5E%7Bn%28n+-+1%29%2F2%7D%5Csum_%7Bk+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7Da%5E%7Bk%28k+%2B+2n+%2B+1%29%2F2%7Db%5E%7Bk%28k+%2B+2n+-+1%29%2F2%7D%5C%5C++++%26+%3D+a%5E%7Bn%28n+%2B+1%29%2F2%7Db%5E%7Bn%28n+-+1%29%2F2%7D%5Csum_%7Bk+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7D%5C%7Ba%28ab%29%5E%7Bn%7D%5C%7D%5E%7Bk%28k+%2B+1%29%2F2%7D%5C%7Bb%28ab%29%5E%7B-n%7D%5C%7D%5E%7Bk%28k+-+1%29%2F2%7D%5C%5C++++%26+%3D+a%5E%7Bn%28n+%2B+1%29%2F2%7Db%5E%7Bn%28n+-+1%29%2F2%7Df%28a%28ab%29%5E%7Bn%7D%2C+b%28ab%29%5E%7B-n%7D%29%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned} f(a, b) &amp;= &#92;sum_{k = -&#92;infty}^{&#92;infty}a^{k(k + 1)/2}b^{k(k - 1)/2}&#92;&#92;    &amp; = &#92;sum_{k = -&#92;infty}^{&#92;infty}a^{(k + n)(k + n + 1)/2}b^{(k + n)(k + n - 1)/2}&#92;&#92;    &amp; = a^{n(n + 1)/2}b^{n(n - 1)/2}&#92;sum_{k = -&#92;infty}^{&#92;infty}a^{k(k + 2n + 1)/2}b^{k(k + 2n - 1)/2}&#92;&#92;    &amp; = a^{n(n + 1)/2}b^{n(n - 1)/2}&#92;sum_{k = -&#92;infty}^{&#92;infty}&#92;{a(ab)^{n}&#92;}^{k(k + 1)/2}&#92;{b(ab)^{-n}&#92;}^{k(k - 1)/2}&#92;&#92;    &amp; = a^{n(n + 1)/2}b^{n(n - 1)/2}f(a(ab)^{n}, b(ab)^{-n})&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned} f(a, b) &amp;= &#92;sum_{k = -&#92;infty}^{&#92;infty}a^{k(k + 1)/2}b^{k(k - 1)/2}&#92;&#92;    &amp; = &#92;sum_{k = -&#92;infty}^{&#92;infty}a^{(k + n)(k + n + 1)/2}b^{(k + n)(k + n - 1)/2}&#92;&#92;    &amp; = a^{n(n + 1)/2}b^{n(n - 1)/2}&#92;sum_{k = -&#92;infty}^{&#92;infty}a^{k(k + 2n + 1)/2}b^{k(k + 2n - 1)/2}&#92;&#92;    &amp; = a^{n(n + 1)/2}b^{n(n - 1)/2}&#92;sum_{k = -&#92;infty}^{&#92;infty}&#92;{a(ab)^{n}&#92;}^{k(k + 1)/2}&#92;{b(ab)^{-n}&#92;}^{k(k - 1)/2}&#92;&#92;    &amp; = a^{n(n + 1)/2}b^{n(n - 1)/2}f(a(ab)^{n}, b(ab)^{-n})&#92;end{aligned}' class='latex' /></p>
<p>If we put <img src='http://s0.wp.com/latex.php?latex=a+%3D+qe%5E%7B2iz%7D%2C+b+%3D+qe%5E%7B-2iz%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='a = qe^{2iz}, b = qe^{-2iz}' title='a = qe^{2iz}, b = qe^{-2iz}' class='latex' /> we obtain <img src='http://s0.wp.com/latex.php?latex=f%28a%2C+b%29+%3D+%5Csum_%7Bn+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%5E%7B2%7D%7De%5E%7B2inz%7D+%3D+%5Ctheta_%7B3%7D%28z%2C+q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(a, b) = &#92;sum_{n = -&#92;infty}^{&#92;infty}q^{n^{2}}e^{2inz} = &#92;theta_{3}(z, q)' title='f(a, b) = &#92;sum_{n = -&#92;infty}^{&#92;infty}q^{n^{2}}e^{2inz} = &#92;theta_{3}(z, q)' class='latex' /> which gives the link between Ramanujan&#8217;s theta function and the classical Jacobi&#8217;s theta function.</p>
<p>The <a title="Elliptic Functions: Theta Functions Contd." href="http://paramanands.wordpress.com/2011/02/01/elliptic-functions-theta-functions-contd/" target="_blank">Jacobi&#8217;s Triple Product</a> identity is given by</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctheta_%7B3%7D%28z%2C+q%29+%3D+%5Csum_%7Bn+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%5E%7B2%7D%7De%5E%7B2inz%7D+%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+-+q%5E%7B2n%7D%29%281+%2B+q%5E%7B2n+-+1%7De%5E%7B2iz%7D%29%281+%2B+q%5E%7B2n+-+1%7De%5E%7B-2iz%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;theta_{3}(z, q) = &#92;sum_{n = -&#92;infty}^{&#92;infty}q^{n^{2}}e^{2inz} = &#92;prod_{n = 1}^{&#92;infty}(1 - q^{2n})(1 + q^{2n - 1}e^{2iz})(1 + q^{2n - 1}e^{-2iz})' title='&#92;displaystyle &#92;theta_{3}(z, q) = &#92;sum_{n = -&#92;infty}^{&#92;infty}q^{n^{2}}e^{2inz} = &#92;prod_{n = 1}^{&#92;infty}(1 - q^{2n})(1 + q^{2n - 1}e^{2iz})(1 + q^{2n - 1}e^{-2iz})' class='latex' /></p>
<p>and this transforms to the  following when we put <img src='http://s0.wp.com/latex.php?latex=a+%3D+qe%5E%7B2iz%7D%2C+b+%3D+qe%5E%7B-2iz%7D%2C+q%5E%7B2%7D+%3D+ab&amp;bg=fff&amp;fg=222&amp;s=0' alt='a = qe^{2iz}, b = qe^{-2iz}, q^{2} = ab' title='a = qe^{2iz}, b = qe^{-2iz}, q^{2} = ab' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7Df%28a%2C+b%29+%26%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+-+%28ab%29%5E%7Bn%7D%29%281+%2B+q%5E%7B2n+-+2%7Da%29%281+%2B+q%5E%7B2n+-+2%7Db%29%5C%5C++++%26%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+-+%28ab%29%5E%7Bn%7D%29%281+%2B+a%28ab%29%5E%7Bn+-+1%7D%29%281+%2B+b%28ab%29%5E%7Bn+-+1%7D%29%5C%5C++++%26%3D+%28-a%3Bab%29_%7B%5Cinfty%7D%28-b%3Bab%29_%7B%5Cinfty%7D%28ab%3Bab%29_%7B%5Cinfty%7D%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned}f(a, b) &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 - (ab)^{n})(1 + q^{2n - 2}a)(1 + q^{2n - 2}b)&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 - (ab)^{n})(1 + a(ab)^{n - 1})(1 + b(ab)^{n - 1})&#92;&#92;    &amp;= (-a;ab)_{&#92;infty}(-b;ab)_{&#92;infty}(ab;ab)_{&#92;infty}&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned}f(a, b) &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 - (ab)^{n})(1 + q^{2n - 2}a)(1 + q^{2n - 2}b)&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 - (ab)^{n})(1 + a(ab)^{n - 1})(1 + b(ab)^{n - 1})&#92;&#92;    &amp;= (-a;ab)_{&#92;infty}(-b;ab)_{&#92;infty}(ab;ab)_{&#92;infty}&#92;end{aligned}' class='latex' /></p>
<p>where we have the notation</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%28a%3B+q%29_%7Bn%7D+%3D+%5Cprod_%7Bk+%3D+1%7D%5E%7Bn%7D%281+-+aq%5E%7Bk+-+1%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle (a; q)_{n} = &#92;prod_{k = 1}^{n}(1 - aq^{k - 1})' title='&#92;displaystyle (a; q)_{n} = &#92;prod_{k = 1}^{n}(1 - aq^{k - 1})' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%28a%3B+q%29_%7B%5Cinfty%7D+%3D+%5Cprod_%7Bk+%3D+1%7D%5E%7B%5Cinfty%7D%281+-+aq%5E%7Bk+-+1%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle (a; q)_{&#92;infty} = &#92;prod_{k = 1}^{&#92;infty}(1 - aq^{k - 1})' title='&#92;displaystyle (a; q)_{&#92;infty} = &#92;prod_{k = 1}^{&#92;infty}(1 - aq^{k - 1})' class='latex' /></p>
<p>Next Ramanujan defines his other theta functions in terms of <img src='http://s0.wp.com/latex.php?latex=f%28a%2C+b%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(a, b)' title='f(a, b)' class='latex' />:</p>
<ul>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%28q%29+%3D+f%28q%2C+q%29+%3D+%5Csum_%7Bn+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%5E%7B2%7D%7D+%3D+%5Ctheta_%7B3%7D%28q%29+%3D+%5Cfrac%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi(q) = f(q, q) = &#92;sum_{n = -&#92;infty}^{&#92;infty}q^{n^{2}} = &#92;theta_{3}(q) = &#92;frac{(-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}}' title='&#92;displaystyle &#92;phi(q) = f(q, q) = &#92;sum_{n = -&#92;infty}^{&#92;infty}q^{n^{2}} = &#92;theta_{3}(q) = &#92;frac{(-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}}' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cpsi%28q%29+%3D+f%28q%2C+q%5E%7B3%7D%29+%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7Dq%5E%7Bn%28n+%2B+1%29%2F2%7D+%3D+%5Cfrac%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;psi(q) = f(q, q^{3}) = &#92;sum_{n = 0}^{&#92;infty}q^{n(n + 1)/2} = &#92;frac{(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}}' title='&#92;displaystyle &#92;psi(q) = f(q, q^{3}) = &#92;sum_{n = 0}^{&#92;infty}q^{n(n + 1)/2} = &#92;frac{(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}}' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+f%28-q%29+%3D+f%28-q%2C+-q%5E%7B2%7D%29+%3D+%5Csum_%7Bn+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%7Dq%5E%7Bn%283n+%2B+1%29%2F2%7D+%3D+%28q%3Bq%29_%7B%5Cinfty%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle f(-q) = f(-q, -q^{2}) = &#92;sum_{n = -&#92;infty}^{&#92;infty}(-1)^{n}q^{n(3n + 1)/2} = (q;q)_{&#92;infty}' title='&#92;displaystyle f(-q) = f(-q, -q^{2}) = &#92;sum_{n = -&#92;infty}^{&#92;infty}(-1)^{n}q^{n(3n + 1)/2} = (q;q)_{&#92;infty}' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cchi%28q%29+%3D+%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D+%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+%2B+q%5E%7B2n+-+1%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;chi(q) = (-q;q^{2})_{&#92;infty} = &#92;prod_{n = 1}^{&#92;infty}(1 + q^{2n - 1})' title='&#92;displaystyle &#92;chi(q) = (-q;q^{2})_{&#92;infty} = &#92;prod_{n = 1}^{&#92;infty}(1 + q^{2n - 1})' class='latex' /></li>
</ul>
<p>Clearly from the definitions of <img src='http://s0.wp.com/latex.php?latex=f%28a%2C+b%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(a, b)' title='f(a, b)' class='latex' /> and Jacobi&#8217;s Triple Product identity we can see that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%28q%29+%3D+%28%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%29%5E%7B2%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi(q) = ((-q;q^{2})_{&#92;infty})^{2}(q^{2};q^{2})_{&#92;infty}' title='&#92;displaystyle &#92;phi(q) = ((-q;q^{2})_{&#92;infty})^{2}(q^{2};q^{2})_{&#92;infty}' class='latex' /></p>
<p>Ramanujan changes one of the factors <img src='http://s0.wp.com/latex.php?latex=%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='(-q;q^{2})_{&#92;infty}' title='(-q;q^{2})_{&#92;infty}' class='latex' /> in a form so that the product formula for <img src='http://s0.wp.com/latex.php?latex=%5Cphi%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;phi(q)' title='&#92;phi(q)' class='latex' /> looks more symmetrical. Thus</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D+%26%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+%2B+q%5E%7B2n+-+1%7D%29+%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%281+%2B+q%5E%7Bn%7D%29%7D%7B%281+%2B+q%5E%7B2n%7D%29%7D+%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%281+-+q%5E%7B2n%7D%29%7D%7B%281+-+q%5E%7Bn%7D%29%281+%2B+q%5E%7B2n%7D%29%7D%5C%5C++++%26%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7B%281+-+q%5E%7B2n+-+1%7D%29%281+%2B+q%5E%7B2n%7D%29%7D+%3D+%5Cfrac%7B1%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned}(-q;q^{2})_{&#92;infty} &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 + q^{2n - 1}) = &#92;prod_{n = 1}^{&#92;infty}&#92;frac{(1 + q^{n})}{(1 + q^{2n})} = &#92;prod_{n = 1}^{&#92;infty}&#92;frac{(1 - q^{2n})}{(1 - q^{n})(1 + q^{2n})}&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}&#92;frac{1}{(1 - q^{2n - 1})(1 + q^{2n})} = &#92;frac{1}{(q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}}&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned}(-q;q^{2})_{&#92;infty} &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 + q^{2n - 1}) = &#92;prod_{n = 1}^{&#92;infty}&#92;frac{(1 + q^{n})}{(1 + q^{2n})} = &#92;prod_{n = 1}^{&#92;infty}&#92;frac{(1 - q^{2n})}{(1 - q^{n})(1 + q^{2n})}&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}&#92;frac{1}{(1 - q^{2n - 1})(1 + q^{2n})} = &#92;frac{1}{(q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}}&#92;end{aligned}' class='latex' /></p>
<p>Again we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%28-q%29+%3D+%5Ctheta_%7B3%7D%28-q%29+%3D+%5Ctheta_%7B4%7D%28q%29+%3D+%5Cfrac%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D+%3D+%5Cfrac%7B%28q%3Bq%29_%7B%5Cinfty%7D%7D%7B%28-q%3Bq%29_%7B%5Cinfty%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi(-q) = &#92;theta_{3}(-q) = &#92;theta_{4}(q) = &#92;frac{(q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(-q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}} = &#92;frac{(q;q)_{&#92;infty}}{(-q;q)_{&#92;infty}}' title='&#92;displaystyle &#92;phi(-q) = &#92;theta_{3}(-q) = &#92;theta_{4}(q) = &#92;frac{(q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(-q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}} = &#92;frac{(q;q)_{&#92;infty}}{(-q;q)_{&#92;infty}}' class='latex' /></p>
<p>Again the series expansion for <img src='http://s0.wp.com/latex.php?latex=%5Cpsi%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;psi(q)' title='&#92;psi(q)' class='latex' /> is obvious and by the triple product identity</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D%5Cpsi%28q%29+%26%3D+f%28q%2C+q%5E%7B3%7D%29+%3D+%28-q%3Bq%5E%7B4%7D%29_%7B%5Cinfty%7D%28-q%5E%7B3%7D%3Bq%5E%7B4%7D%29_%7B%5Cinfty%7D%28q%5E%7B4%7D%3Bq%5E%7B4%7D%29_%7B%5Cinfty%7D%5C%5C++++%26%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+%2B+q%5E%7B4n+-+3%7D%29%281+%2B+q%5E%7B4n+-+1%7D%29%281+-+q%5E%7B4n%7D%29%5C%5C++++%26%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+%2B+q%5E%7B4n+-+3%7D%29%281+%2B+q%5E%7B4n+-+1%7D%29%281+-+q%5E%7B2n%7D%29%281+%2B+q%5E%7B2n%7D%29%5C%5C++++%26%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+%2B+q%5E%7B2%282n+-1%29+-+1%7D%29%281+%2B+q%5E%7B2%282n%29+-+1%7D%29%281+-+q%5E%7B2n%7D%29%281+%2B+q%5E%7B2n%7D%29%5C%5C++++%26%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+%2B+q%5E%7B2n+-+1%7D%29%281+-+q%5E%7B2n%7D%29%281+%2B+q%5E%7B2n%7D%29%5C%5C++++%26%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+-+q%5E%7B2n%7D%29%281+%2B+q%5E%7Bn%7D%29+%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%281+-+q%5E%7B2n%7D%29%281+-+q%5E%7B2n%7D%29%7D%7B1+-+q%5E%7Bn%7D%7D%5C%5C++++%26%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1+-+q%5E%7B2n%7D%7D%7B1+-+q%5E%7B2n+-+1%7D%7D+%3D+%5Cfrac%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned}&#92;psi(q) &amp;= f(q, q^{3}) = (-q;q^{4})_{&#92;infty}(-q^{3};q^{4})_{&#92;infty}(q^{4};q^{4})_{&#92;infty}&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 + q^{4n - 3})(1 + q^{4n - 1})(1 - q^{4n})&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 + q^{4n - 3})(1 + q^{4n - 1})(1 - q^{2n})(1 + q^{2n})&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 + q^{2(2n -1) - 1})(1 + q^{2(2n) - 1})(1 - q^{2n})(1 + q^{2n})&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 + q^{2n - 1})(1 - q^{2n})(1 + q^{2n})&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 - q^{2n})(1 + q^{n}) = &#92;prod_{n = 1}^{&#92;infty}&#92;frac{(1 - q^{2n})(1 - q^{2n})}{1 - q^{n}}&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}&#92;frac{1 - q^{2n}}{1 - q^{2n - 1}} = &#92;frac{(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}}&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned}&#92;psi(q) &amp;= f(q, q^{3}) = (-q;q^{4})_{&#92;infty}(-q^{3};q^{4})_{&#92;infty}(q^{4};q^{4})_{&#92;infty}&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 + q^{4n - 3})(1 + q^{4n - 1})(1 - q^{4n})&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 + q^{4n - 3})(1 + q^{4n - 1})(1 - q^{2n})(1 + q^{2n})&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 + q^{2(2n -1) - 1})(1 + q^{2(2n) - 1})(1 - q^{2n})(1 + q^{2n})&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 + q^{2n - 1})(1 - q^{2n})(1 + q^{2n})&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}(1 - q^{2n})(1 + q^{n}) = &#92;prod_{n = 1}^{&#92;infty}&#92;frac{(1 - q^{2n})(1 - q^{2n})}{1 - q^{n}}&#92;&#92;    &amp;= &#92;prod_{n = 1}^{&#92;infty}&#92;frac{1 - q^{2n}}{1 - q^{2n - 1}} = &#92;frac{(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}}&#92;end{aligned}' class='latex' /></p>
<p>Finally <img src='http://s0.wp.com/latex.php?latex=f%28-q%29+%3D+f%28-q%2C+-q%5E%7B2%7D%29+%3D+%28q%3Bq%5E%7B3%7D%29_%7B%5Cinfty%7D%28q%5E%7B2%7D%3Bq%5E%7B3%7D%29_%7B%5Cinfty%7D%28q%5E%7B3%7D%3Bq%5E%7B3%7D%29_%7B%5Cinfty%7D+%3D+%28q%3Bq%29_%7B%5Cinfty%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(-q) = f(-q, -q^{2}) = (q;q^{3})_{&#92;infty}(q^{2};q^{3})_{&#92;infty}(q^{3};q^{3})_{&#92;infty} = (q;q)_{&#92;infty}' title='f(-q) = f(-q, -q^{2}) = (q;q^{3})_{&#92;infty}(q^{2};q^{3})_{&#92;infty}(q^{3};q^{3})_{&#92;infty} = (q;q)_{&#92;infty}' class='latex' /></p>
<p>Also it should be noted that the function <img src='http://s0.wp.com/latex.php?latex=%5Cpsi%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;psi(q)' title='&#92;psi(q)' class='latex' /> is the equivalent of classical <img src='http://s0.wp.com/latex.php?latex=%5Ctheta_%7B2%7D%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;theta_{2}(q)' title='&#92;theta_{2}(q)' class='latex' /> and we have <img src='http://s0.wp.com/latex.php?latex=%5Ctheta_%7B2%7D%28q%29+%3D+2q%5E%7B1%2F4%7D%5Cpsi%28q%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;theta_{2}(q) = 2q^{1/4}&#92;psi(q^{2})' title='&#92;theta_{2}(q) = 2q^{1/4}&#92;psi(q^{2})' class='latex' />. The functions <img src='http://s0.wp.com/latex.php?latex=%5Cchi%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;chi(q)' title='&#92;chi(q)' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%5Cchi%28-q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;chi(-q)' title='&#92;chi(-q)' class='latex' /> will be used later in the definitions Ramanujan&#8217;s invariants.</p>
<p><strong>Theta Function Identities</strong></p>
<p>The above formulas presented above provide the series as well as product expansions of these functions and it turns out that their product expansions are quite helpful in deriving an amazing number of results. Ramanujan uses this idea to the fullest extent as is shown by the following results:</p>
<ul>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bf%28q%29%7D%7Bf%28-q%29%7D+%3D+%5Cfrac%7B%5Cpsi%28q%29%7D%7B%5Cpsi%28-q%29%7D+%3D+%5Cfrac%7B%5Cchi%28q%29%7D%7B%5Cchi%28-q%29%7D+%3D+%5Csqrt%7B%5Cfrac%7B%5Cphi%28q%29%7D%7B%5Cphi%28-q%29%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{f(q)}{f(-q)} = &#92;frac{&#92;psi(q)}{&#92;psi(-q)} = &#92;frac{&#92;chi(q)}{&#92;chi(-q)} = &#92;sqrt{&#92;frac{&#92;phi(q)}{&#92;phi(-q)}}' title='&#92;displaystyle &#92;frac{f(q)}{f(-q)} = &#92;frac{&#92;psi(q)}{&#92;psi(-q)} = &#92;frac{&#92;chi(q)}{&#92;chi(-q)} = &#92;sqrt{&#92;frac{&#92;phi(q)}{&#92;phi(-q)}}' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+f%5E%7B3%7D%28-q%29+%3D+%5Cphi%5E%7B2%7D%28-q%29%5Cpsi%28q%29+%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%7D%282n+%2B+1%29q%5E%7Bn%28n+%2B+1%29%2F2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle f^{3}(-q) = &#92;phi^{2}(-q)&#92;psi(q) = &#92;sum_{n = 0}^{&#92;infty}(-1)^{n}(2n + 1)q^{n(n + 1)/2}' title='&#92;displaystyle f^{3}(-q) = &#92;phi^{2}(-q)&#92;psi(q) = &#92;sum_{n = 0}^{&#92;infty}(-1)^{n}(2n + 1)q^{n(n + 1)/2}' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cchi%28q%29+%3D+%5Cfrac%7Bf%28q%29%7D%7Bf%28-q%5E%7B2%7D%29%7D+%3D+%5Csqrt%5B3%5D%7B%5Cfrac%7B%5Cphi%28q%29%7D%7B%5Cpsi%28-q%29%7D%7D+%3D+%5Cfrac%7B%5Cphi%28q%29%7D%7Bf%28q%29%7D+%3D+%5Cfrac%7Bf%28-q%5E%7B2%7D%29%7D%7B%5Cpsi%28-q%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;chi(q) = &#92;frac{f(q)}{f(-q^{2})} = &#92;sqrt[3]{&#92;frac{&#92;phi(q)}{&#92;psi(-q)}} = &#92;frac{&#92;phi(q)}{f(q)} = &#92;frac{f(-q^{2})}{&#92;psi(-q)}' title='&#92;displaystyle &#92;chi(q) = &#92;frac{f(q)}{f(-q^{2})} = &#92;sqrt[3]{&#92;frac{&#92;phi(q)}{&#92;psi(-q)}} = &#92;frac{&#92;phi(q)}{f(q)} = &#92;frac{f(-q^{2})}{&#92;psi(-q)}' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+f%5E%7B3%7D%28-q%5E%7B2%7D%29+%3D+%5Cphi%28-q%29%5Cpsi%5E%7B2%7D%28q%29%2C%5C%2C%5C%2C%5C%2C+%5Cchi%28q%29%5Cchi%28-q%29+%3D+%5Cchi%28-q%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle f^{3}(-q^{2}) = &#92;phi(-q)&#92;psi^{2}(q),&#92;,&#92;,&#92;, &#92;chi(q)&#92;chi(-q) = &#92;chi(-q^{2})' title='&#92;displaystyle f^{3}(-q^{2}) = &#92;phi(-q)&#92;psi^{2}(q),&#92;,&#92;,&#92;, &#92;chi(q)&#92;chi(-q) = &#92;chi(-q^{2})' class='latex' /></li>
</ul>
<p>To establish the first of these we can start with <img src='http://s0.wp.com/latex.php?latex=f%28q%29+%3D+%28-q%3B-q%29_%7B%5Cinfty%7D+%3D+%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(q) = (-q;-q)_{&#92;infty} = (-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}' title='f(q) = (-q;-q)_{&#92;infty} = (-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}' class='latex' /> and therefore</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bf%28q%29%7D%7Bf%28-q%29%7D+%3D+%5Cfrac%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%3Bq%29_%7B%5Cinfty%7D%7D+%3D+%5Cfrac%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D+%3D+%5Cfrac%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{f(q)}{f(-q)} = &#92;frac{(-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(q;q)_{&#92;infty}} = &#92;frac{(-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}} = &#92;frac{(-q;q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}}' title='&#92;displaystyle &#92;frac{f(q)}{f(-q)} = &#92;frac{(-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(q;q)_{&#92;infty}} = &#92;frac{(-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}} = &#92;frac{(-q;q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}}' class='latex' /></p>
<p>and</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B%5Cpsi%28q%29%7D%7B%5Cpsi%28-q%29%7D+%3D+%5Cfrac%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%5Cfrac%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D+%3D+%5Cfrac%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{&#92;psi(q)}{&#92;psi(-q)} = &#92;frac{(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}}&#92;frac{(-q;q^{2})_{&#92;infty}}{(q^{2};q^{2})_{&#92;infty}} = &#92;frac{(-q;q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}}' title='&#92;displaystyle &#92;frac{&#92;psi(q)}{&#92;psi(-q)} = &#92;frac{(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}}&#92;frac{(-q;q^{2})_{&#92;infty}}{(q^{2};q^{2})_{&#92;infty}} = &#92;frac{(-q;q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}}' class='latex' /></p>
<p>Similarly we can establish that <img src='http://s0.wp.com/latex.php?latex=%5Csqrt%7B%5Cphi%28q%29%2F%5Cphi%28-q%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;sqrt{&#92;phi(q)/&#92;phi(-q)}' title='&#92;sqrt{&#92;phi(q)/&#92;phi(-q)}' class='latex' /> is also equal to <img src='http://s0.wp.com/latex.php?latex=%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%2F%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='(-q;q^{2})_{&#92;infty}/(q;q^{2})_{&#92;infty}' title='(-q;q^{2})_{&#92;infty}/(q;q^{2})_{&#92;infty}' class='latex' /> and thereby the first result is established. The second set of identities is established in the following fashion</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28-q%29%5Cpsi%28q%29+%3D+%5Cfrac%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B2%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B2%7D%7D%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B2%7D%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B2%7D%7D%5Cfrac%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D+%3D+%5Cfrac%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B3%7D%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28-q%3Bq%29_%7B%5Cinfty%7D%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(-q)&#92;psi(q) = &#92;frac{(q;q^{2})_{&#92;infty}^{2}(q^{2};q^{2})_{&#92;infty}^{2}}{(-q;q^{2})_{&#92;infty}^{2}(-q^{2};q^{2})_{&#92;infty}^{2}}&#92;frac{(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}} = &#92;frac{(q^{2};q^{2})_{&#92;infty}^{3}(q;q^{2})_{&#92;infty}}{(-q;q)_{&#92;infty}^{2}}' title='&#92;displaystyle &#92;phi^{2}(-q)&#92;psi(q) = &#92;frac{(q;q^{2})_{&#92;infty}^{2}(q^{2};q^{2})_{&#92;infty}^{2}}{(-q;q^{2})_{&#92;infty}^{2}(-q^{2};q^{2})_{&#92;infty}^{2}}&#92;frac{(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}} = &#92;frac{(q^{2};q^{2})_{&#92;infty}^{3}(q;q^{2})_{&#92;infty}}{(-q;q)_{&#92;infty}^{2}}' class='latex' /></p>
<p>Since</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%28-q%3Bq%29_%7B%5Cinfty%7D+%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+%2B+q%5E%7Bn%7D%29+%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1+-+q%5E%7B2n%7D%7D%7B1+-+q%5E%7Bn%7D%7D+%3D+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7B1+-+q%5E%7B2n+-+1%7D%7D+%3D+%5Cfrac%7B1%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle (-q;q)_{&#92;infty} = &#92;prod_{n = 1}^{&#92;infty}(1 + q^{n}) = &#92;prod_{n = 1}^{&#92;infty}&#92;frac{1 - q^{2n}}{1 - q^{n}} = &#92;prod_{n = 1}^{&#92;infty}&#92;frac{1}{1 - q^{2n - 1}} = &#92;frac{1}{(q;q^{2})_{&#92;infty}}' title='&#92;displaystyle (-q;q)_{&#92;infty} = &#92;prod_{n = 1}^{&#92;infty}(1 + q^{n}) = &#92;prod_{n = 1}^{&#92;infty}&#92;frac{1 - q^{2n}}{1 - q^{n}} = &#92;prod_{n = 1}^{&#92;infty}&#92;frac{1}{1 - q^{2n - 1}} = &#92;frac{1}{(q;q^{2})_{&#92;infty}}' class='latex' /></p>
<p>it follows that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28-q%29%5Cpsi%28q%29+%3D+%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B3%7D%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B3%7D+%3D+%28q%3Bq%29_%7B%5Cinfty%7D%5E%7B3%7D+%3D+f%5E%7B3%7D%28-q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(-q)&#92;psi(q) = (q^{2};q^{2})_{&#92;infty}^{3}(q;q^{2})_{&#92;infty}^{3} = (q;q)_{&#92;infty}^{3} = f^{3}(-q)' title='&#92;displaystyle &#92;phi^{2}(-q)&#92;psi(q) = (q^{2};q^{2})_{&#92;infty}^{3}(q;q^{2})_{&#92;infty}^{3} = (q;q)_{&#92;infty}^{3} = f^{3}(-q)' class='latex' /></p>
<p>To get the series expansion requires some effort. We note the series and product expansions of <img src='http://s0.wp.com/latex.php?latex=%5Ctheta_%7B1%7D%28z%2C+q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;theta_{1}(z, q)' title='&#92;theta_{1}(z, q)' class='latex' /> given by</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D+%5Ctheta_%7B1%7D%28z%2C+q%29+%26%3D+2q%5E%7B1%2F4%7D%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%7Dq%5E%7Bn%28n+%2B+1%29%7D%5Csin%282n+%2B+1%29z%5C%5C++++%26%3D+2q%5E%7B1%2F4%7D%5Csin+z%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+-+q%5E%7B2n%7D%29%281+-+2q%5E%7B2n%7D%5Ccos+2z+%2B+q%5E%7B4n%7D%29%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned} &#92;theta_{1}(z, q) &amp;= 2q^{1/4}&#92;sum_{n = 0}^{&#92;infty}(-1)^{n}q^{n(n + 1)}&#92;sin(2n + 1)z&#92;&#92;    &amp;= 2q^{1/4}&#92;sin z&#92;prod_{n = 1}^{&#92;infty}(1 - q^{2n})(1 - 2q^{2n}&#92;cos 2z + q^{4n})&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned} &#92;theta_{1}(z, q) &amp;= 2q^{1/4}&#92;sum_{n = 0}^{&#92;infty}(-1)^{n}q^{n(n + 1)}&#92;sin(2n + 1)z&#92;&#92;    &amp;= 2q^{1/4}&#92;sin z&#92;prod_{n = 1}^{&#92;infty}(1 - q^{2n})(1 - 2q^{2n}&#92;cos 2z + q^{4n})&#92;end{aligned}' class='latex' /></p>
<p>Dividing by <img src='http://s0.wp.com/latex.php?latex=2q%5E%7B1%2F4%7D%5Csin+z&amp;bg=fff&amp;fg=222&amp;s=0' alt='2q^{1/4}&#92;sin z' title='2q^{1/4}&#92;sin z' class='latex' /> and taking limits when <img src='http://s0.wp.com/latex.php?latex=z+%5Cto+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='z &#92;to 0' title='z &#92;to 0' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cprod_%7Bn+%3D+1%7D%5E%7B%5Cinfty%7D%281+-+q%5E%7B2n%7D%29%5E%7B3%7D+%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%7D%282n+%2B+1%29q%5E%7Bn%28n+%2B+1%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;prod_{n = 1}^{&#92;infty}(1 - q^{2n})^{3} = &#92;sum_{n = 0}^{&#92;infty}(-1)^{n}(2n + 1)q^{n(n + 1)}' title='&#92;displaystyle &#92;prod_{n = 1}^{&#92;infty}(1 - q^{2n})^{3} = &#92;sum_{n = 0}^{&#92;infty}(-1)^{n}(2n + 1)q^{n(n + 1)}' class='latex' /></p>
<p>Replacing <img src='http://s0.wp.com/latex.php?latex=q&amp;bg=fff&amp;fg=222&amp;s=0' alt='q' title='q' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=%5Csqrt%7Bq%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;sqrt{q}' title='&#92;sqrt{q}' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%28q%3Bq%29_%7B%5Cinfty%7D%5E%7B3%7D+%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn%7D%282n+%2B+1%29q%5E%7Bn%28n+%2B+1%29%2F2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle (q;q)_{&#92;infty}^{3} = &#92;sum_{n = 0}^{&#92;infty}(-1)^{n}(2n + 1)q^{n(n + 1)/2}' title='&#92;displaystyle (q;q)_{&#92;infty}^{3} = &#92;sum_{n = 0}^{&#92;infty}(-1)^{n}(2n + 1)q^{n(n + 1)/2}' class='latex' /></p>
<p>For the 3rd result we can see that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bf%28q%29%7D%7Bf%28-q%5E%7B2%7D%29%7D+%3D+%5Cfrac%7B%28-q%3B-q%29_%7B%5Cinfty%7D%7D%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D+%3D+%5Cfrac%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D+%3D+%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D+%3D+%5Cchi%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{f(q)}{f(-q^{2})} = &#92;frac{(-q;-q)_{&#92;infty}}{(q^{2};q^{2})_{&#92;infty}} = &#92;frac{(-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(q^{2};q^{2})_{&#92;infty}} = (-q;q^{2})_{&#92;infty} = &#92;chi(q)' title='&#92;displaystyle &#92;frac{f(q)}{f(-q^{2})} = &#92;frac{(-q;-q)_{&#92;infty}}{(q^{2};q^{2})_{&#92;infty}} = &#92;frac{(-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(q^{2};q^{2})_{&#92;infty}} = (-q;q^{2})_{&#92;infty} = &#92;chi(q)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B%5Cphi%28q%29%7D%7Bf%28q%29%7D+%3D+%5Cfrac%7B1%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D+%3D+%5Cfrac%7B%28-q%3Bq%29_%7B%5Cinfty%7D%7D%7B%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D+%3D+%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D+%3D+%5Cchi%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{&#92;phi(q)}{f(q)} = &#92;frac{1}{(q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}} = &#92;frac{(-q;q)_{&#92;infty}}{(-q^{2};q^{2})_{&#92;infty}} = (-q;q^{2})_{&#92;infty} = &#92;chi(q)' title='&#92;displaystyle &#92;frac{&#92;phi(q)}{f(q)} = &#92;frac{1}{(q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}} = &#92;frac{(-q;q)_{&#92;infty}}{(-q^{2};q^{2})_{&#92;infty}} = (-q;q^{2})_{&#92;infty} = &#92;chi(q)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bf%28-q%5E%7B2%7D%29%7D%7B%5Cpsi%28-q%29%7D+%3D+%5Cfrac%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D+%3D+%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D+%3D+%5Cchi%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{f(-q^{2})}{&#92;psi(-q)} = &#92;frac{(q^{2};q^{2})_{&#92;infty}(-q;q^{2})_{&#92;infty}}{(q^{2};q^{2})_{&#92;infty}} = (-q;q^{2})_{&#92;infty} = &#92;chi(q)' title='&#92;displaystyle &#92;frac{f(-q^{2})}{&#92;psi(-q)} = &#92;frac{(q^{2};q^{2})_{&#92;infty}(-q;q^{2})_{&#92;infty}}{(q^{2};q^{2})_{&#92;infty}} = (-q;q^{2})_{&#92;infty} = &#92;chi(q)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B%5Cphi%28q%29%7D%7B%5Cpsi%28-q%29%7D+%3D+%5Cfrac%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%5Cfrac%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D+%3D+%5Cfrac%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B2%7D%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{&#92;phi(q)}{&#92;psi(-q)} = &#92;frac{(-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}}&#92;frac{(-q;q^{2})_{&#92;infty}}{(q^{2};q^{2})_{&#92;infty}} = &#92;frac{(-q;q^{2})_{&#92;infty}^{2}}{(q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}}' title='&#92;displaystyle &#92;frac{&#92;phi(q)}{&#92;psi(-q)} = &#92;frac{(-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}}&#92;frac{(-q;q^{2})_{&#92;infty}}{(q^{2};q^{2})_{&#92;infty}} = &#92;frac{(-q;q^{2})_{&#92;infty}^{2}}{(q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}}' class='latex' /></p>
<p>Since we have <img src='http://s0.wp.com/latex.php?latex=%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%3Bq%29_%7B%5Cinfty%7D+%3D+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='(q;q^{2})_{&#92;infty}(-q;q)_{&#92;infty} = 1' title='(q;q^{2})_{&#92;infty}(-q;q)_{&#92;infty} = 1' class='latex' /> it follows that <img src='http://s0.wp.com/latex.php?latex=%28q%5E%7B2%7D%3Bq%5E%7B4%7D%29_%7B%5Cinfty%7D%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D+%3D+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='(q^{2};q^{4})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty} = 1' title='(q^{2};q^{4})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty} = 1' class='latex' />  and therefore</p>
<p><img src='http://s0.wp.com/latex.php?latex=%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D+%3D+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='(q;q^{2})_{&#92;infty}(-q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty} = 1' title='(q;q^{2})_{&#92;infty}(-q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty} = 1' class='latex' /></p>
<p>Putting pieces together we get <img src='http://s0.wp.com/latex.php?latex=%5Cphi%28q%29%2F%5Cpsi%28-q%29+%3D+%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B3%7D+%3D+%5Cchi%5E%7B3%7D%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;phi(q)/&#92;psi(-q) = (-q;q^{2})_{&#92;infty}^{3} = &#92;chi^{3}(q)' title='&#92;phi(q)/&#92;psi(-q) = (-q;q^{2})_{&#92;infty}^{3} = &#92;chi^{3}(q)' class='latex' /></p>
<p>Also note that the relation <img src='http://s0.wp.com/latex.php?latex=%28q%5E%7B2%7D%3Bq%5E%7B4%7D%29_%7B%5Cinfty%7D+%3D+%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='(q^{2};q^{4})_{&#92;infty} = (q;q^{2})_{&#92;infty}(-q;q^{2})_{&#92;infty}' title='(q^{2};q^{4})_{&#92;infty} = (q;q^{2})_{&#92;infty}(-q;q^{2})_{&#92;infty}' class='latex' /> used above can be expressed as <img src='http://s0.wp.com/latex.php?latex=%5Cchi%28q%29%5Cchi%28-q%29+%3D+%5Cchi%28-q%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;chi(q)&#92;chi(-q) = &#92;chi(-q^{2})' title='&#92;chi(q)&#92;chi(-q) = &#92;chi(-q^{2})' class='latex' />.</p>
<p>Finally</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D+%5Cphi%28-q%29%5Cpsi%5E%7B2%7D%28q%29+%26%3D+%5Cfrac%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%5Cfrac%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B2%7D%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B2%7D%7D%5C%5C++++%26%3D+%5Cfrac%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B3%7D%7D%7B%28-q%3Bq%29_%7B%5Cinfty%7D%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D+%3D+%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%5E%7B3%7D+%3D+f%5E%7B3%7D%28-q%5E%7B2%7D%29%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned} &#92;phi(-q)&#92;psi^{2}(q) &amp;= &#92;frac{(q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(-q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}}&#92;frac{(q^{2};q^{2})_{&#92;infty}^{2}}{(q;q^{2})_{&#92;infty}^{2}}&#92;&#92;    &amp;= &#92;frac{(q^{2};q^{2})_{&#92;infty}^{3}}{(-q;q)_{&#92;infty}(q;q^{2})_{&#92;infty}} = (q^{2};q^{2})_{&#92;infty}^{3} = f^{3}(-q^{2})&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned} &#92;phi(-q)&#92;psi^{2}(q) &amp;= &#92;frac{(q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(-q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}}&#92;frac{(q^{2};q^{2})_{&#92;infty}^{2}}{(q;q^{2})_{&#92;infty}^{2}}&#92;&#92;    &amp;= &#92;frac{(q^{2};q^{2})_{&#92;infty}^{3}}{(-q;q)_{&#92;infty}(q;q^{2})_{&#92;infty}} = (q^{2};q^{2})_{&#92;infty}^{3} = f^{3}(-q^{2})&#92;end{aligned}' class='latex' /></p>
<p>What we see in the proofs of the above results is that they are really elementary and depend upon formal manipulation of series and products. The only non-trivial identity we have used till now is the Jacobi&#8217;s Triple product which forms the basis of all the above product expansions and the relations between these Ramanujan theta functions.</p>
<p>Next we establish some identities connecting theta functions of <img src='http://s0.wp.com/latex.php?latex=q%2C+q%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='q, q^{2}' title='q, q^{2}' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=q%5E%7B4%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='q^{4}' title='q^{4}' class='latex' />. These can be seen as equivalents of the classical theta function identities established <a title="The Magic of Theta Functions" href="http://paramanands.wordpress.com/2010/10/19/the-magic-of-theta-functions/" target="_blank">here</a>.</p>
<ul>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%28q%29+%2B+%5Cphi%28-q%29+%3D+2%5Cphi%28q%5E%7B4%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi(q) + &#92;phi(-q) = 2&#92;phi(q^{4})' title='&#92;displaystyle &#92;phi(q) + &#92;phi(-q) = 2&#92;phi(q^{4})' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%28q%29+-+%5Cphi%28-q%29+%3D+4q%5Cpsi%28q%5E%7B8%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi(q) - &#92;phi(-q) = 4q&#92;psi(q^{8})' title='&#92;displaystyle &#92;phi(q) - &#92;phi(-q) = 4q&#92;psi(q^{8})' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%28q%29%5Cphi%28-q%29+%3D+%5Cphi%5E%7B2%7D%28-q%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi(q)&#92;phi(-q) = &#92;phi^{2}(-q^{2})' title='&#92;displaystyle &#92;phi(q)&#92;phi(-q) = &#92;phi^{2}(-q^{2})' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cpsi%28q%29%5Cpsi%28-q%29+%3D+%5Cpsi%28q%5E%7B2%7D%29%5Cphi%28-q%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;psi(q)&#92;psi(-q) = &#92;psi(q^{2})&#92;phi(-q^{2})' title='&#92;displaystyle &#92;psi(q)&#92;psi(-q) = &#92;psi(q^{2})&#92;phi(-q^{2})' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%28q%29%5Cpsi%28q%5E%7B2%7D%29+%3D+%5Cpsi%5E%7B2%7D%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi(q)&#92;psi(q^{2}) = &#92;psi^{2}(q)' title='&#92;displaystyle &#92;phi(q)&#92;psi(q^{2}) = &#92;psi^{2}(q)' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28q%29+-+%5Cphi%5E%7B2%7D%28-q%29+%3D+8q%5Cpsi%5E%7B2%7D%28q%5E%7B4%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(q) - &#92;phi^{2}(-q) = 8q&#92;psi^{2}(q^{4})' title='&#92;displaystyle &#92;phi^{2}(q) - &#92;phi^{2}(-q) = 8q&#92;psi^{2}(q^{4})' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28q%29+%2B+%5Cphi%5E%7B2%7D%28-q%29+%3D+2%5Cphi%5E%7B2%7D%28q%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(q) + &#92;phi^{2}(-q) = 2&#92;phi^{2}(q^{2})' title='&#92;displaystyle &#92;phi^{2}(q) + &#92;phi^{2}(-q) = 2&#92;phi^{2}(q^{2})' class='latex' /></li>
<li><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B4%7D%28q%29+-+%5Cphi%5E%7B4%7D%28-q%29+%3D+16q%5Cpsi%5E%7B4%7D%28q%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{4}(q) - &#92;phi^{4}(-q) = 16q&#92;psi^{4}(q^{2})' title='&#92;displaystyle &#92;phi^{4}(q) - &#92;phi^{4}(-q) = 16q&#92;psi^{4}(q^{2})' class='latex' /></li>
</ul>
<p>The first two identities follow by using the series expansions of <img src='http://s0.wp.com/latex.php?latex=%5Cphi%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;phi(q)' title='&#92;phi(q)' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%5Cpsi%28q%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;psi(q)' title='&#92;psi(q)' class='latex' />. The next one is handled as follows:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%28q%29%5Cphi%28-q%29+%3D+%5Cfrac%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%5Cfrac%7B%28q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28-q%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D+%3D+%5Cleft%28%5Cfrac%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi(q)&#92;phi(-q) = &#92;frac{(-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}}&#92;frac{(q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(-q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}} = &#92;left(&#92;frac{(q^{2};q^{2})_{&#92;infty}}{(-q^{2};q^{2})_{&#92;infty}}&#92;right)^{2}' title='&#92;displaystyle &#92;phi(q)&#92;phi(-q) = &#92;frac{(-q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}}&#92;frac{(q;q^{2})_{&#92;infty}(q^{2};q^{2})_{&#92;infty}}{(-q;q^{2})_{&#92;infty}(-q^{2};q^{2})_{&#92;infty}} = &#92;left(&#92;frac{(q^{2};q^{2})_{&#92;infty}}{(-q^{2};q^{2})_{&#92;infty}}&#92;right)^{2}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%28-q%5E%7B2%7D%29+%3D+%5Cfrac%7B%28q%5E%7B2%7D%3Bq%5E%7B4%7D%29_%7B%5Cinfty%7D%28q%5E%7B4%7D%3Bq%5E%7B4%7D%29_%7B%5Cinfty%7D%7D%7B%28-q%5E%7B2%7D%3Bq%5E%7B4%7D%29_%7B%5Cinfty%7D%28-q%5E%7B4%7D%3Bq%5E%7B4%7D%29_%7B%5Cinfty%7D%7D+%3D+%5Cfrac%7B%28q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D%7B%28-q%5E%7B2%7D%3Bq%5E%7B2%7D%29_%7B%5Cinfty%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi(-q^{2}) = &#92;frac{(q^{2};q^{4})_{&#92;infty}(q^{4};q^{4})_{&#92;infty}}{(-q^{2};q^{4})_{&#92;infty}(-q^{4};q^{4})_{&#92;infty}} = &#92;frac{(q^{2};q^{2})_{&#92;infty}}{(-q^{2};q^{2})_{&#92;infty}}' title='&#92;displaystyle &#92;phi(-q^{2}) = &#92;frac{(q^{2};q^{4})_{&#92;infty}(q^{4};q^{4})_{&#92;infty}}{(-q^{2};q^{4})_{&#92;infty}(-q^{4};q^{4})_{&#92;infty}} = &#92;frac{(q^{2};q^{2})_{&#92;infty}}{(-q^{2};q^{2})_{&#92;infty}}' class='latex' /></p>
<p>Similarly we can prove the next two identities. Next one follows from the application of the first and second:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28q%29+-+%5Cphi%5E%7B2%7D%28-q%29+%3D+%28%5Cphi%28q%29+%2B+%5Cphi%28-q%29%29%28%5Cphi%28q%29+-+%5Cphi%28-q%29%29+%3D+8q%5Cphi%28q%5E%7B4%7D%29%5Cpsi%28q%5E%7B8%7D%29+%3D+8q%5Cpsi%5E%7B2%7D%28q%5E%7B4%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(q) - &#92;phi^{2}(-q) = (&#92;phi(q) + &#92;phi(-q))(&#92;phi(q) - &#92;phi(-q)) = 8q&#92;phi(q^{4})&#92;psi(q^{8}) = 8q&#92;psi^{2}(q^{4})' title='&#92;displaystyle &#92;phi^{2}(q) - &#92;phi^{2}(-q) = (&#92;phi(q) + &#92;phi(-q))(&#92;phi(q) - &#92;phi(-q)) = 8q&#92;phi(q^{4})&#92;psi(q^{8}) = 8q&#92;psi^{2}(q^{4})' class='latex' /></p>
<p>The second last identity requires us to analyze the series expansions:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28q%29+%2B+%5Cphi%5E%7B2%7D%28-q%29+%3D+%5Csum_%7Bm%2C+n+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7Dq%5E%7Bm%5E%7B2%7D+%2B+n%5E%7B2%7D%7D+%2B+%5Csum_%7Bm%2C+n+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7D%28-1%29%5E%7Bm+%2B+n%7Dq%5E%7Bm%5E%7B2%7D+%2B+n%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(q) + &#92;phi^{2}(-q) = &#92;sum_{m, n = -&#92;infty}^{&#92;infty}q^{m^{2} + n^{2}} + &#92;sum_{m, n = -&#92;infty}^{&#92;infty}(-1)^{m + n}q^{m^{2} + n^{2}}' title='&#92;displaystyle &#92;phi^{2}(q) + &#92;phi^{2}(-q) = &#92;sum_{m, n = -&#92;infty}^{&#92;infty}q^{m^{2} + n^{2}} + &#92;sum_{m, n = -&#92;infty}^{&#92;infty}(-1)^{m + n}q^{m^{2} + n^{2}}' class='latex' /></p>
<p>Clearly the terms in both sums cancel if <img src='http://s0.wp.com/latex.php?latex=%28m+%2B+n%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(m + n)' title='(m + n)' class='latex' /> is odd and add up when <img src='http://s0.wp.com/latex.php?latex=%28m+%2B+n%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(m + n)' title='(m + n)' class='latex' /> is even. Again if <img src='http://s0.wp.com/latex.php?latex=%28m+%2B+n%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(m + n)' title='(m + n)' class='latex' /> is even i.e <img src='http://s0.wp.com/latex.php?latex=m+%2B+n+%3D+2j&amp;bg=fff&amp;fg=222&amp;s=0' alt='m + n = 2j' title='m + n = 2j' class='latex' /> then <img src='http://s0.wp.com/latex.php?latex=m%2C+n&amp;bg=fff&amp;fg=222&amp;s=0' alt='m, n' title='m, n' class='latex' /> must be of same parity and hence <img src='http://s0.wp.com/latex.php?latex=m+-+n+%3D+2k&amp;bg=fff&amp;fg=222&amp;s=0' alt='m - n = 2k' title='m - n = 2k' class='latex' />. In that case <img src='http://s0.wp.com/latex.php?latex=m%5E%7B2%7D+%2B+n%5E%7B2%7D+%3D+2%28j%5E%7B2%7D+%2B+k%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='m^{2} + n^{2} = 2(j^{2} + k^{2})' title='m^{2} + n^{2} = 2(j^{2} + k^{2})' class='latex' />. Thus we have finally</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cphi%5E%7B2%7D%28q%29+%2B+%5Cphi%5E%7B2%7D%28-q%29+%3D+2%5Csum_%7Bj%2C+k+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7Dq%5E%7B2%28j%5E%7B2%7D+%2B+k%5E%7B2%7D%29%7D+%3D+2%5Csum_%7Bj%2C+k+%3D+-%5Cinfty%7D%5E%7B%5Cinfty%7D%28q%5E%7B2%7D%29%5E%7Bj%5E%7B2%7D+%2B+k%5E%7B2%7D%7D+%3D+2%5Cphi%5E%7B2%7D%28q%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;phi^{2}(q) + &#92;phi^{2}(-q) = 2&#92;sum_{j, k = -&#92;infty}^{&#92;infty}q^{2(j^{2} + k^{2})} = 2&#92;sum_{j, k = -&#92;infty}^{&#92;infty}(q^{2})^{j^{2} + k^{2}} = 2&#92;phi^{2}(q^{2})' title='&#92;displaystyle &#92;phi^{2}(q) + &#92;phi^{2}(-q) = 2&#92;sum_{j, k = -&#92;infty}^{&#92;infty}q^{2(j^{2} + k^{2})} = 2&#92;sum_{j, k = -&#92;infty}^{&#92;infty}(q^{2})^{j^{2} + k^{2}} = 2&#92;phi^{2}(q^{2})' class='latex' /></p>
<p>The last identity is obtained from the earlier ones as follows:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7D+%5Cphi%5E%7B4%7D%28q%29+-+%5Cphi%5E%7B4%7D%28-q%29+%26%3D+%28%5Cphi%5E%7B2%7D%28q%29+%2B+%5Cphi%5E%7B2%7D%28-q%29%29%28%5Cphi%5E%7B2%7D%28q%29+-+%5Cphi%5E%7B2%7D%28-q%29%29%5C%5C++++%26%3D+16q%5Cphi%5E%7B2%7D%28q%5E%7B2%7D%29%5Cpsi%5E%7B2%7D%28q%5E%7B4%7D%29+%3D+16q%5Cpsi%5E%7B4%7D%28q%5E%7B2%7D%29%5Cend%7Baligned%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;begin{aligned} &#92;phi^{4}(q) - &#92;phi^{4}(-q) &amp;= (&#92;phi^{2}(q) + &#92;phi^{2}(-q))(&#92;phi^{2}(q) - &#92;phi^{2}(-q))&#92;&#92;    &amp;= 16q&#92;phi^{2}(q^{2})&#92;psi^{2}(q^{4}) = 16q&#92;psi^{4}(q^{2})&#92;end{aligned}' title='&#92;displaystyle &#92;begin{aligned} &#92;phi^{4}(q) - &#92;phi^{4}(-q) &amp;= (&#92;phi^{2}(q) + &#92;phi^{2}(-q))(&#92;phi^{2}(q) - &#92;phi^{2}(-q))&#92;&#92;    &amp;= 16q&#92;phi^{2}(q^{2})&#92;psi^{2}(q^{4}) = 16q&#92;psi^{4}(q^{2})&#92;end{aligned}' class='latex' /></p>
<p>After these preliminary identities on theta functions it is time to relate them to the elliptic integral <img src='http://s0.wp.com/latex.php?latex=K&amp;bg=fff&amp;fg=222&amp;s=0' alt='K' title='K' class='latex' /> and the modulus <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' />. Ramanujan did this in a very ingenious way by studying the properties of hypergeometric function <img src='http://s0.wp.com/latex.php?latex=_%7B2%7DF_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='_{2}F_{1}' title='_{2}F_{1}' class='latex' />. This will be presented in the next post.</p>
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		<title>Elementary Approach to Modular Equations: Jacobi&#8217;s Transformation Theory 5</title>
		<link>http://paramanands.wordpress.com/2011/11/03/elementary-approach-to-modular-equations-jacobis-transformation-theory-5/</link>
		<comments>http://paramanands.wordpress.com/2011/11/03/elementary-approach-to-modular-equations-jacobis-transformation-theory-5/#comments</comments>
		<pubDate>Thu, 03 Nov 2011 05:47:30 +0000</pubDate>
		<dc:creator>paramanands</dc:creator>
				<category><![CDATA[Elliptic Functions]]></category>
		<category><![CDATA[Elliptic Integrals]]></category>
		<category><![CDATA[Mathematical Analysis]]></category>

		<guid isPermaLink="false">http://paramanands.wordpress.com/?p=2381</guid>
		<description><![CDATA[Jacobi&#8217;s Second Real Transformation The second transformation is obtained by taking so that . On the face of it the transformation thus involves imaginary quantities, but will be shown later to be a real transformation only. In the case of this transformation we will use in place of and in place of . Also we [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=paramanands.wordpress.com&amp;blog=8565079&amp;post=2381&amp;subd=paramanands&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p><strong>Jacobi&#8217;s Second Real Transformation</strong></p>
<p>The second transformation is obtained by taking <img src='http://s0.wp.com/latex.php?latex=m+%3D+0%2C+m%27+%3D+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='m = 0, m&#039; = 1' title='m = 0, m&#039; = 1' class='latex' /> so that <img src='http://s0.wp.com/latex.php?latex=%5Comega+%3D+iK%27%2Fp&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;omega = iK&#039;/p' title='&#92;omega = iK&#039;/p' class='latex' />. On the face of it the transformation thus involves imaginary quantities, but will be shown later to be a real transformation only. In the case of this transformation we will use <img src='http://s0.wp.com/latex.php?latex=l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='l_{1}' title='l_{1}' class='latex' /> in place of <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=M_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='M_{1}' title='M_{1}' class='latex' /> in place of <img src='http://s0.wp.com/latex.php?latex=M&amp;bg=fff&amp;fg=222&amp;s=0' alt='M' title='M' class='latex' />. Also we will keep the factor <img src='http://s0.wp.com/latex.php?latex=%28-1%29%5E%7B%28p+-+1%29%2F2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='(-1)^{(p - 1)/2}' title='(-1)^{(p - 1)/2}' class='latex' /> with the multiplier <img src='http://s0.wp.com/latex.php?latex=M_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='M_{1}' title='M_{1}' class='latex' />. We thus obtain the following by putting <img src='http://s0.wp.com/latex.php?latex=%5Comega+%3D+iK%27%2Fp&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;omega = iK&#039;/p' title='&#92;omega = iK&#039;/p' class='latex' /> in the <a title="Elementary Approach to Modular Equations: Jacobi’s Transformation Theory 4" href="http://paramanands.wordpress.com/2011/10/30/elementary-approach-to-modular-equations-jacobis-transformation-theory-4/" target="_blank">general formulas in <img src='http://s0.wp.com/latex.php?latex=2s%5Comega&amp;bg=fff&amp;fg=222&amp;s=0' alt='2s&#92;omega' title='2s&#92;omega' class='latex' /></a>:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM_%7B1%7D%7D%2C+l_%7B1%7D%5Cright%29+%3D+%5Cfrac%7B%5Ctext%7Bsn%7D%5C%2Cu%7D%7BM_%7B1%7D%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C%5Cdfrac%7B2siK%27%7D%7Bp%7D%7D%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C%5Cdfrac%7B2siK%27%7D%7Bp%7D%7D+%3D+%5Csqrt%7B%5Cfrac%7Bk%5E%7Bp%7D%7D%7Bl_%7B1%7D%7D%7D%5Cprod_%7Bs+%3D+-%28p+-+1%29%2F2%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5Cleft%28u+%2B+%5Cfrac%7B2siK%27%7D%7Bp%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M_{1}}, l_{1}&#92;right) = &#92;frac{&#92;text{sn}&#92;,u}{M_{1}}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;,&#92;dfrac{2siK&#039;}{p}}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,&#92;dfrac{2siK&#039;}{p}} = &#92;sqrt{&#92;frac{k^{p}}{l_{1}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{sn}&#92;left(u + &#92;frac{2siK&#039;}{p}&#92;right)' title='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M_{1}}, l_{1}&#92;right) = &#92;frac{&#92;text{sn}&#92;,u}{M_{1}}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;,&#92;dfrac{2siK&#039;}{p}}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,&#92;dfrac{2siK&#039;}{p}} = &#92;sqrt{&#92;frac{k^{p}}{l_{1}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{sn}&#92;left(u + &#92;frac{2siK&#039;}{p}&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bcn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM_%7B1%7D%7D%2C+l_%7B1%7D%5Cright%29+%3D+%5Ctext%7Bcn%7D%5C%2Cu%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28K+-+%5Cdfrac%7B2siK%27%7D%7Bp%7D%5Cright%29%7D%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C%5Cdfrac%7B2siK%27%7D%7Bp%7D%7D+%3D+%5Csqrt%7B%5Cfrac%7Bl%27_%7B1%7Dk%5E%7Bp%7D%7D%7Bl_%7B1%7Dk%27%5E%7Bp%7D%7D%7D%5Cprod_%7Bs+%3D+-%28p+-+1%29%2F2%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bcn%7D%5Cleft%28u+%2B+%5Cfrac%7B2siK%27%7D%7Bp%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{cn}&#92;left(&#92;frac{u}{M_{1}}, l_{1}&#92;right) = &#92;text{cn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;left(K - &#92;dfrac{2siK&#039;}{p}&#92;right)}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,&#92;dfrac{2siK&#039;}{p}} = &#92;sqrt{&#92;frac{l&#039;_{1}k^{p}}{l_{1}k&#039;^{p}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{cn}&#92;left(u + &#92;frac{2siK&#039;}{p}&#92;right)' title='&#92;displaystyle &#92;text{cn}&#92;left(&#92;frac{u}{M_{1}}, l_{1}&#92;right) = &#92;text{cn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;left(K - &#92;dfrac{2siK&#039;}{p}&#92;right)}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,&#92;dfrac{2siK&#039;}{p}} = &#92;sqrt{&#92;frac{l&#039;_{1}k^{p}}{l_{1}k&#039;^{p}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{cn}&#92;left(u + &#92;frac{2siK&#039;}{p}&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bdn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM_%7B1%7D%7D%2C+l_%7B1%7D%5Cright%29+%3D+%5Ctext%7Bdn%7D%5C%2Cu%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28K+-+%5Cdfrac%7B2siK%27%7D%7Bp%7D%5Cright%29%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C%5Cdfrac%7B2siK%27%7D%7Bp%7D%7D+%3D+%5Csqrt%7B%5Cfrac%7Bl%27_%7B1%7D%7D%7Bk%27%5E%7Bp%7D%7D%7D%5Cprod_%7Bs+%3D+-%28p+-+1%29%2F2%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bdn%7D%5Cleft%28u+%2B+%5Cfrac%7B2siK%27%7D%7Bp%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{dn}&#92;left(&#92;frac{u}{M_{1}}, l_{1}&#92;right) = &#92;text{dn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;left(K - &#92;dfrac{2siK&#039;}{p}&#92;right)}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,&#92;dfrac{2siK&#039;}{p}} = &#92;sqrt{&#92;frac{l&#039;_{1}}{k&#039;^{p}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{dn}&#92;left(u + &#92;frac{2siK&#039;}{p}&#92;right)' title='&#92;displaystyle &#92;text{dn}&#92;left(&#92;frac{u}{M_{1}}, l_{1}&#92;right) = &#92;text{dn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;left(K - &#92;dfrac{2siK&#039;}{p}&#92;right)}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,&#92;dfrac{2siK&#039;}{p}} = &#92;sqrt{&#92;frac{l&#039;_{1}}{k&#039;^{p}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{dn}&#92;left(u + &#92;frac{2siK&#039;}{p}&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+M_%7B1%7D+%3D+%28-1%29%5E%7B%28p+-+1%29%2F2%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cdfrac%7B%5Ctext%7Bsn%7D%5Cleft%28K+-+%5Cdfrac%7B2siK%27%7D%7Bp%7D%5Cright%29%7D%7B%5Ctext%7Bsn%7D%5C%2C%5Cdfrac%7B2siK%27%7D%7Bp%7D%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle M_{1} = (-1)^{(p - 1)/2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{&#92;text{sn}&#92;left(K - &#92;dfrac{2siK&#039;}{p}&#92;right)}{&#92;text{sn}&#92;,&#92;dfrac{2siK&#039;}{p}}&#92;right)^{2}' title='&#92;displaystyle M_{1} = (-1)^{(p - 1)/2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{&#92;text{sn}&#92;left(K - &#92;dfrac{2siK&#039;}{p}&#92;right)}{&#92;text{sn}&#92;,&#92;dfrac{2siK&#039;}{p}}&#92;right)^{2}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l_%7B1%7D+%3D+k%5E%7Bp%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5E%7B4%7D%5Cleft%28K+-+%5Cfrac%7B2siK%27%7D%7Bp%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l_{1} = k^{p}&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}^{4}&#92;left(K - &#92;frac{2siK&#039;}{p}&#92;right)' title='&#92;displaystyle l_{1} = k^{p}&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}^{4}&#92;left(K - &#92;frac{2siK&#039;}{p}&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l%27_%7B1%7D+%3D+%5Cdfrac%7Bk%27%5E%7Bp%7D%7D%7B%7B%5Cdisplaystyle%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bdn%7D%5E%7B4%7D%5C%2C%5Cfrac%7B2siK%27%7D%7Bp%7D%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l&#039;_{1} = &#92;dfrac{k&#039;^{p}}{{&#92;displaystyle&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{dn}^{4}&#92;,&#92;frac{2siK&#039;}{p}}}' title='&#92;displaystyle l&#039;_{1} = &#92;dfrac{k&#039;^{p}}{{&#92;displaystyle&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{dn}^{4}&#92;,&#92;frac{2siK&#039;}{p}}}' class='latex' /></p>
<p>Clearly the above formulas can be simplified by using the Jacobi&#8217; imaginary transformations to get rid of the imaginary unit <img src='http://s0.wp.com/latex.php?latex=i&amp;bg=fff&amp;fg=222&amp;s=0' alt='i' title='i' class='latex' /> and thus we obtain:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l_%7B1%7D+%3D+%5Cdfrac%7Bk%5E%7Bp%7D%7D%7B%7B%5Cdisplaystyle%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bdn%7D%5E%7B4%7D%5Cleft%28%5Cdfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l_{1} = &#92;dfrac{k^{p}}{{&#92;displaystyle&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{dn}^{4}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}}' title='&#92;displaystyle l_{1} = &#92;dfrac{k^{p}}{{&#92;displaystyle&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{dn}^{4}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l%27_%7B1%7D+%3D+k%27%5E%7Bp%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5E%7B4%7D%5Cleft%28K+-+%5Cfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l&#039;_{1} = k&#039;^{p}&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}^{4}&#92;left(K - &#92;frac{2sK&#039;}{p}, k&#039;&#92;right)' title='&#92;displaystyle l&#039;_{1} = k&#039;^{p}&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}^{4}&#92;left(K - &#92;frac{2sK&#039;}{p}, k&#039;&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+M_%7B1%7D+%3D+%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cdfrac%7B%5Ctext%7Bsn%7D%5Cleft%28K+-+%5Cdfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29%7D%7B%5Ctext%7Bsn%7D%5Cleft%28%5Cdfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle M_{1} = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{&#92;text{sn}&#92;left(K - &#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}{&#92;text{sn}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}&#92;right)^{2}' title='&#92;displaystyle M_{1} = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{&#92;text{sn}&#92;left(K - &#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}{&#92;text{sn}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}&#92;right)^{2}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM_%7B1%7D%7D%2C+l_%7B1%7D%5Cright%29+%3D+%5Cfrac%7B%5Ctext%7Bsn%7D%5C%2Cu%7D%7BM_%7B1%7D%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+%2B+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%7D%7B%5Ctext%7Bsc%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29%7D%7D%7B1+%2B+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsc%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M_{1}}, l_{1}&#92;right) = &#92;frac{&#92;text{sn}&#92;,u}{M_{1}}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 + &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}}{1 + k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}' title='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M_{1}}, l_{1}&#92;right) = &#92;frac{&#92;text{sn}&#92;,u}{M_{1}}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 + &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}}{1 + k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}' class='latex' /></p>
<p>and similar formulas for <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bcn%7D%2C%5C%2C%5Ctext%7Bdn%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{cn},&#92;,&#92;text{dn}' title='&#92;text{cn},&#92;,&#92;text{dn}' class='latex' />.</p>
<p>From the last equation we can see that the least positive value of <img src='http://s0.wp.com/latex.php?latex=u&amp;bg=fff&amp;fg=222&amp;s=0' alt='u' title='u' class='latex' /> which the right side vanishes is <img src='http://s0.wp.com/latex.php?latex=u+%3D+2K&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = 2K' title='u = 2K' class='latex' /> and the left side vanishes for <img src='http://s0.wp.com/latex.php?latex=u%2FM_%7B1%7D+%3D+2L_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='u/M_{1} = 2L_{1}' title='u/M_{1} = 2L_{1}' class='latex' /> and finally we get the relation <img src='http://s0.wp.com/latex.php?latex=K%2FM_%7B1%7D+%3D+L_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='K/M_{1} = L_{1}' title='K/M_{1} = L_{1}' class='latex' />.</p>
<p><strong>Jacobi&#8217;s Second Complementary Transformation</strong></p>
<p>From the first two equations above we can see that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsc%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM_%7B1%7D%7D%2C+l_%7B1%7D%5Cright%29+%3D+%5Csqrt%7B%5Cfrac%7Bk%27%5E%7Bp%7D%7D%7Bl%27_%7B1%7D%7D%7D%5Cprod_%7Bs+%3D+-%28p+-+1%29%2F2%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsc%7D%5Cleft%28u+%2B+%5Cfrac%7B2siK%27%7D%7Bp%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sc}&#92;left(&#92;frac{u}{M_{1}}, l_{1}&#92;right) = &#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;_{1}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{sc}&#92;left(u + &#92;frac{2siK&#039;}{p}&#92;right)' title='&#92;displaystyle &#92;text{sc}&#92;left(&#92;frac{u}{M_{1}}, l_{1}&#92;right) = &#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;_{1}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{sc}&#92;left(u + &#92;frac{2siK&#039;}{p}&#92;right)' class='latex' /></p>
<p>Replacing <img src='http://s0.wp.com/latex.php?latex=u&amp;bg=fff&amp;fg=222&amp;s=0' alt='u' title='u' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=iu&amp;bg=fff&amp;fg=222&amp;s=0' alt='iu' title='iu' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM_%7B1%7D%7D%2C+l%27_%7B1%7D%5Cright%29+%3D+%28-1%29%5E%7B%28p+-+1%29%2F2%7D%5Csqrt%7B%5Cfrac%7Bk%27%5E%7Bp%7D%7D%7Bl%27_%7B1%7D%7D%7D%5Cprod_%7Bs+%3D+-%28p+-+1%29%2F2%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5Cleft%28u+%2B+%5Cfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M_{1}}, l&#039;_{1}&#92;right) = (-1)^{(p - 1)/2}&#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;_{1}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{sn}&#92;left(u + &#92;frac{2sK&#039;}{p}, k&#039;&#92;right)' title='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M_{1}}, l&#039;_{1}&#92;right) = (-1)^{(p - 1)/2}&#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;_{1}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{sn}&#92;left(u + &#92;frac{2sK&#039;}{p}, k&#039;&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%28-1%29%5E%7B%28p+-+1%29%2F2%7D%5Csqrt%7B%5Cfrac%7Bk%27%5E%7Bp%7D%7D%7Bl%27_%7B1%7D%7D%7D%5C%2C%5Ctext%7Bsn%7D%28u%2C+k%27%29%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5Cleft%28u+%2B+%5Cfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29%5Ctext%7Bsn%7D%5Cleft%28u+-+%5Cfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = (-1)^{(p - 1)/2}&#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;_{1}}}&#92;,&#92;text{sn}(u, k&#039;)&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}&#92;left(u + &#92;frac{2sK&#039;}{p}, k&#039;&#92;right)&#92;text{sn}&#92;left(u - &#92;frac{2sK&#039;}{p}, k&#039;&#92;right)' title='&#92;displaystyle = (-1)^{(p - 1)/2}&#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;_{1}}}&#92;,&#92;text{sn}(u, k&#039;)&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}&#92;left(u + &#92;frac{2sK&#039;}{p}, k&#039;&#92;right)&#92;text{sn}&#92;left(u - &#92;frac{2sK&#039;}{p}, k&#039;&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%5Csqrt%7B%5Cfrac%7Bk%27%5E%7Bp%7D%7D%7Bl%27_%7B1%7D%7D%7D%5Cleft%28%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29%5E%7B2%7D%5Cright%29%5Ctext%7Bsn%7D%28u%2C+k%27%29%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%28u%2C+k%27%29%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29%7D%7D%7B1+-+k%27%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%28u%2C+k%27%29%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = &#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;_{1}}}&#92;left(&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}&#92;left(&#92;frac{2sK&#039;}{p}, k&#039;&#92;right)^{2}&#92;right)&#92;text{sn}(u, k&#039;)&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}(u, k&#039;)}{&#92;text{sn}^{2}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}}{1 - k&#039;^{2}&#92;,&#92;text{sn}^{2}(u, k&#039;)&#92;,&#92;text{sn}^{2}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}' title='&#92;displaystyle = &#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;_{1}}}&#92;left(&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}&#92;left(&#92;frac{2sK&#039;}{p}, k&#039;&#92;right)^{2}&#92;right)&#92;text{sn}(u, k&#039;)&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}(u, k&#039;)}{&#92;text{sn}^{2}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}}{1 - k&#039;^{2}&#92;,&#92;text{sn}^{2}(u, k&#039;)&#92;,&#92;text{sn}^{2}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}' class='latex' /></p>
<p>We can see that the factors on the right side of the above equation which are independent of <img src='http://s0.wp.com/latex.php?latex=u&amp;bg=fff&amp;fg=222&amp;s=0' alt='u' title='u' class='latex' /> must lead to the value <img src='http://s0.wp.com/latex.php?latex=1%2FM_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='1/M_{1}' title='1/M_{1}' class='latex' /> (as can be verified by taking limits as <img src='http://s0.wp.com/latex.php?latex=u+%5Cto+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='u &#92;to 0' title='u &#92;to 0' class='latex' />). Therefore we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM_%7B1%7D%7D%2C+l%27_%7B1%7D%5Cright%29+%3D+%5Cfrac%7B%5Ctext%7Bsn%7D%28u%2C+k%27%29%7D%7BM_%7B1%7D%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%28u%2C+k%27%29%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29%7D%7D%7B1+-+k%27%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%28u%2C+k%27%29%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2sK%27%7D%7Bp%7D%2C+k%27%5Cright%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M_{1}}, l&#039;_{1}&#92;right) = &#92;frac{&#92;text{sn}(u, k&#039;)}{M_{1}}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}(u, k&#039;)}{&#92;text{sn}^{2}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}}{1 - k&#039;^{2}&#92;,&#92;text{sn}^{2}(u, k&#039;)&#92;,&#92;text{sn}^{2}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}' title='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M_{1}}, l&#039;_{1}&#92;right) = &#92;frac{&#92;text{sn}(u, k&#039;)}{M_{1}}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}(u, k&#039;)}{&#92;text{sn}^{2}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}}{1 - k&#039;^{2}&#92;,&#92;text{sn}^{2}(u, k&#039;)&#92;,&#92;text{sn}^{2}&#92;left(&#92;dfrac{2sK&#039;}{p}, k&#039;&#92;right)}' class='latex' /></p>
<p>From the above we can see that the least positive value of <img src='http://s0.wp.com/latex.php?latex=u&amp;bg=fff&amp;fg=222&amp;s=0' alt='u' title='u' class='latex' /> for which the right side vanishes is <img src='http://s0.wp.com/latex.php?latex=u+%3D+2K%27%2Fp&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = 2K&#039;/p' title='u = 2K&#039;/p' class='latex' /> and the left side vanishes for <img src='http://s0.wp.com/latex.php?latex=u%2FM_%7B1%7D+%3D+2L%27_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='u/M_{1} = 2L&#039;_{1}' title='u/M_{1} = 2L&#039;_{1}' class='latex' /> so that we get <img src='http://s0.wp.com/latex.php?latex=L%27_%7B1%7D+%3D+K%27%2F%28pM_%7B1%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;_{1} = K&#039;/(pM_{1})' title='L&#039;_{1} = K&#039;/(pM_{1})' class='latex' />.</p>
<p>From the equations <img src='http://s0.wp.com/latex.php?latex=L_%7B1%7D+%3D+K%2FM_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='L_{1} = K/M_{1}' title='L_{1} = K/M_{1}' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=L%27_%7B1%7D+%3D+K%27%2F%28pM_%7B1%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;_{1} = K&#039;/(pM_{1})' title='L&#039;_{1} = K&#039;/(pM_{1})' class='latex' /> we get <img src='http://s0.wp.com/latex.php?latex=K%27%2FK+%3D+pL%27_%7B1%7D%2FL_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;/K = pL&#039;_{1}/L_{1}' title='K&#039;/K = pL&#039;_{1}/L_{1}' class='latex' />.</p>
<p>Finally from the Jacobi&#8217;s first and second real transformations (and their complementary forms) we see that corresponding to a <img src='http://s0.wp.com/latex.php?latex=k+%5Cin+%280%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='k &#92;in (0, 1)' title='k &#92;in (0, 1)' class='latex' /> there are two unique moduli <img src='http://s0.wp.com/latex.php?latex=l%2C+l_%7B1%7D+%5Cin+%280%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, l_{1} &#92;in (0, 1)' title='l, l_{1} &#92;in (0, 1)' class='latex' /> such that <img src='http://s0.wp.com/latex.php?latex=l+%3C+k+%3C+l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='l &lt; k &lt; l_{1}' title='l &lt; k &lt; l_{1}' class='latex' /> such that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1%7D%7Bp%7D%5Cfrac%7BL%27%7D%7BL%7D+%3D+%5Cfrac%7BK%27%7D%7BK%7D+%3D+p%5Cfrac%7BL%27_%7B1%7D%7D%7BL_%7B1%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{1}{p}&#92;frac{L&#039;}{L} = &#92;frac{K&#039;}{K} = p&#92;frac{L&#039;_{1}}{L_{1}}' title='&#92;displaystyle &#92;frac{1}{p}&#92;frac{L&#039;}{L} = &#92;frac{K&#039;}{K} = p&#92;frac{L&#039;_{1}}{L_{1}}' class='latex' /></p>
<p>Moreover the relation between <img src='http://s0.wp.com/latex.php?latex=k%2C+l&amp;bg=fff&amp;fg=222&amp;s=0' alt='k, l' title='k, l' class='latex' /> is algebraic and relation between <img src='http://s0.wp.com/latex.php?latex=k%2C+l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='k, l_{1}' title='k, l_{1}' class='latex' /> is also algebraic. If we analyze the forms of the Jacobi&#8217;s transformations, we will see that the first complementary tranformation is analogous to the second transformation and second complementary transformation is analogous to the first transformation. From these considerations it follows that the the relation between <img src='http://s0.wp.com/latex.php?latex=l%2C+k&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, k' title='l, k' class='latex' /> is same as that between <img src='http://s0.wp.com/latex.php?latex=k%2C+l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='k, l_{1}' title='k, l_{1}' class='latex' /> and the relation among <img src='http://s0.wp.com/latex.php?latex=l%2C+k%2C+l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, k, l_{1}' title='l, k, l_{1}' class='latex' /> is same as that among <img src='http://s0.wp.com/latex.php?latex=l%27_%7B1%7D%2C+k%27%2C+l%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='l&#039;_{1}, k&#039;, l&#039;' title='l&#039;_{1}, k&#039;, l&#039;' class='latex' />.</p>
<p>Combining both the first and second transformations we can obtain the multiplication formulas for elliptic functions. For example, since the relation between <img src='http://s0.wp.com/latex.php?latex=k%2C+l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='k, l_{1}' title='k, l_{1}' class='latex' /> is same as that between <img src='http://s0.wp.com/latex.php?latex=l%2C+k&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, k' title='l, k' class='latex' /> we can replace <img src='http://s0.wp.com/latex.php?latex=k%2C+l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='k, l_{1}' title='k, l_{1}' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=l%2C+k&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, k' title='l, k' class='latex' /> in the second transformation and let <img src='http://s0.wp.com/latex.php?latex=N_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='N_{1}' title='N_{1}' class='latex' /> denote the corresponding multiplier. Then we have <img src='http://s0.wp.com/latex.php?latex=N_%7B1%7D+%3D+L%2FK+%3D+1%2F%28pM%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='N_{1} = L/K = 1/(pM)' title='N_{1} = L/K = 1/(pM)' class='latex' /> so that <img src='http://s0.wp.com/latex.php?latex=MN_%7B1%7D+%3D+1%2Fp&amp;bg=fff&amp;fg=222&amp;s=0' alt='MN_{1} = 1/p' title='MN_{1} = 1/p' class='latex' />. The formula is now given by</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BN_%7B1%7D%7D%2C+k%5Cright%29+%3D+%5Cfrac%7B%5Ctext%7Bsn%7D%28u%2C+l%29%7D%7BN_%7B1%7D%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+%2B+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%28u%2C+l%29%7D%7B%5Ctext%7Bsc%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2sL%27%7D%7Bp%7D%2C+l%27%5Cright%29%7D%7D%7B1+%2B+l%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%28u%2C+l%29%5C%2C%5Ctext%7Bsc%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2sL%27%7D%7Bp%7D%2C+l%27%5Cright%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{N_{1}}, k&#92;right) = &#92;frac{&#92;text{sn}(u, l)}{N_{1}}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 + &#92;dfrac{&#92;text{sn}^{2}(u, l)}{&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sL&#039;}{p}, l&#039;&#92;right)}}{1 + l^{2}&#92;,&#92;text{sn}^{2}(u, l)&#92;,&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sL&#039;}{p}, l&#039;&#92;right)}' title='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{N_{1}}, k&#92;right) = &#92;frac{&#92;text{sn}(u, l)}{N_{1}}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 + &#92;dfrac{&#92;text{sn}^{2}(u, l)}{&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sL&#039;}{p}, l&#039;&#92;right)}}{1 + l^{2}&#92;,&#92;text{sn}^{2}(u, l)&#92;,&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sL&#039;}{p}, l&#039;&#92;right)}' class='latex' /></p>
<p>Replacing <img src='http://s0.wp.com/latex.php?latex=u&amp;bg=fff&amp;fg=222&amp;s=0' alt='u' title='u' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=u%2FM&amp;bg=fff&amp;fg=222&amp;s=0' alt='u/M' title='u/M' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%28pu%2C+k%29+%3D+pM%5C%2C%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+%2B+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29%7D%7B%5Ctext%7Bsc%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2sL%27%7D%7Bp%7D%2C+l%27%5Cright%29%7D%7D%7B1+%2B+l%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29%5Ctext%7Bsc%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2sL%27%7D%7Bp%7D%2C+l%27%5Cright%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}(pu, k) = pM&#92;,&#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#92;right)&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 + &#92;dfrac{&#92;text{sn}^{2}&#92;left(&#92;dfrac{u}{M}, l&#92;right)}{&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sL&#039;}{p}, l&#039;&#92;right)}}{1 + l^{2}&#92;,&#92;text{sn}^{2}&#92;left(&#92;dfrac{u}{M}, l&#92;right)&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sL&#039;}{p}, l&#039;&#92;right)}' title='&#92;displaystyle &#92;text{sn}(pu, k) = pM&#92;,&#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#92;right)&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 + &#92;dfrac{&#92;text{sn}^{2}&#92;left(&#92;dfrac{u}{M}, l&#92;right)}{&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sL&#039;}{p}, l&#039;&#92;right)}}{1 + l^{2}&#92;,&#92;text{sn}^{2}&#92;left(&#92;dfrac{u}{M}, l&#92;right)&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sL&#039;}{p}, l&#039;&#92;right)}' class='latex' /></p>
<p>Using the first Jacobi transformation we can replace the functions of modulus <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> on the right side with functions of modulus <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> and thereby we obtain the multiplication formula for the elliptic function <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsn%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{sn}' title='&#92;text{sn}' class='latex' />. We can obtain the multiplication formula in another way by using the first transformation to switch from modulus <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> to a larger modulus <img src='http://s0.wp.com/latex.php?latex=l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='l_{1}' title='l_{1}' class='latex' /> and then using the second transformation to switch from modulus <img src='http://s0.wp.com/latex.php?latex=l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='l_{1}' title='l_{1}' class='latex' /> to the smaller modulus <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' />.</p>
<p>To summarize Jacobi&#8217;s transformations:</p>
<p><em>Given a modulus <img src='http://s0.wp.com/latex.php?latex=k+%5Cin+%280%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='k &#92;in (0, 1)' title='k &#92;in (0, 1)' class='latex' /> and a positive prime <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> we have two unique moduli <img src='http://s0.wp.com/latex.php?latex=l%2C+l_%7B1%7D+%5Cin+%280%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, l_{1} &#92;in (0, 1)' title='l, l_{1} &#92;in (0, 1)' class='latex' /> such that <img src='http://s0.wp.com/latex.php?latex=l+%3C+k+%3C+l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='l &lt; k &lt; l_{1}' title='l &lt; k &lt; l_{1}' class='latex' /> and</em></p>
<p><em><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1%7D%7Bp%7D%5Cfrac%7BL%27%7D%7BL%7D+%3D+%5Cfrac%7BK%27%7D%7BK%7D+%3D+p%5Cfrac%7BL%27_%7B1%7D%7D%7BL_%7B1%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{1}{p}&#92;frac{L&#039;}{L} = &#92;frac{K&#039;}{K} = p&#92;frac{L&#039;_{1}}{L_{1}}' title='&#92;displaystyle &#92;frac{1}{p}&#92;frac{L&#039;}{L} = &#92;frac{K&#039;}{K} = p&#92;frac{L&#039;_{1}}{L_{1}}' class='latex' /></em></p>
<p><em>Such a series of moduli <img src='http://s0.wp.com/latex.php?latex=l%2C+k%2C+l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, k, l_{1}' title='l, k, l_{1}' class='latex' /> may be called an ascending series of order <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' />. It follows that <img src='http://s0.wp.com/latex.php?latex=l%27_%7B1%7D%2C+k%27%2C+l%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='l&#039;_{1}, k&#039;, l&#039;' title='l&#039;_{1}, k&#039;, l&#039;' class='latex' /> is also an ascending series of order <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' />. The relation between <img src='http://s0.wp.com/latex.php?latex=l%2C+k&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, k' title='l, k' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=k%2C+l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='k, l_{1}' title='k, l_{1}' class='latex' /> is algebraic in nature.</em></p>
<p><em>Jacobi&#8217;s first transformation allows us to express elliptic functions of a given modulus in the form of elliptic functions of a greater modulus. Its complementary transformation provides a similar relation but in form of complementary moduli and therefore helps us to express an elliptic function of a given modulus in terms of elliptic functions of a smaller modulus.</em></p>
<p><em>Jacobi&#8217;s second transformation allows us to express elliptic functions of a given modulus in the form of elliptic functions of a smaller modulus. Its complementary transformation provides a similar relation but in form of complementary moduli and therefore helps us to express an elliptic function of a given modulus in terms of elliptic functions of a greater modulus.</em></p>
<p><em>From these considerations it is seen that the Jacobi&#8217;s first complementary transformation is analogous to Jacobi&#8217;s second transformation and Jacobi&#8217;s second complementary transformation is analogous to Jacobi&#8217;s first transformation.</em></p>
<p><em>Together both the Jacobi&#8217;s transformations allow us express the multiplication formulas for elliptic functions either by first switching to a higher modulus and then back to the original modulus or by first switching to a lower modulus and back to original modulus.</em></p>
<p><em>In terms of the multiplier, we see that the first transformation gives <img src='http://s0.wp.com/latex.php?latex=L+%3D+K%2FpM&amp;bg=fff&amp;fg=222&amp;s=0' alt='L = K/pM' title='L = K/pM' class='latex' />, the first complementary transformation gives <img src='http://s0.wp.com/latex.php?latex=L%27+%3D+K%27%2FM&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039; = K&#039;/M' title='L&#039; = K&#039;/M' class='latex' />, the second transformation gives <img src='http://s0.wp.com/latex.php?latex=L_%7B1%7D+%3D+K%2FM_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='L_{1} = K/M_{1}' title='L_{1} = K/M_{1}' class='latex' />, the second complementary transformation gives <img src='http://s0.wp.com/latex.php?latex=L%27_%7B1%7D+%3D+K%27%2FpM_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;_{1} = K&#039;/pM_{1}' title='L&#039;_{1} = K&#039;/pM_{1}' class='latex' />.</em></p>
<p><em>The above relations also hold when <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> is not prime because then the transformations can be taken as composition of the transformations corresponding to prime factors of <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' />.<br />
</em></p>
<p><em></em>With these remarks we complete our presentation of the theory of transformation as developed by Jacobi in his masterpiece <em>Fundamenta Nova.</em> From the next post Ramanujan will take on the stage.</p>
<p>&nbsp;</p>
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		<title>Elementary Approach to Modular Equations: Jacobi&#8217;s Transformation Theory 4</title>
		<link>http://paramanands.wordpress.com/2011/10/30/elementary-approach-to-modular-equations-jacobis-transformation-theory-4/</link>
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		<pubDate>Sun, 30 Oct 2011 07:45:54 +0000</pubDate>
		<dc:creator>paramanands</dc:creator>
				<category><![CDATA[Elliptic Functions]]></category>
		<category><![CDATA[Elliptic Integrals]]></category>
		<category><![CDATA[Mathematical Analysis]]></category>

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		<description><![CDATA[Transformation of Elliptic Functions The relation and other variants of it as described in previous post lead to the differential equation when and Putting in the differential equation we see that . Hence the relation between and in effect expresses in terms of elliptic functions of a different but related modulus . Thus we obtain [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=paramanands.wordpress.com&amp;blog=8565079&amp;post=2324&amp;subd=paramanands&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p><strong>Transformation of Elliptic Functions</strong></p>
<p>The relation</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+y+%3D+%5Cfrac%7Bx%7D%7BM%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7Bx%5E%7B2%7D%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D%7D%7B1+-+k%5E%7B2%7Dx%5E%7B2%7D%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle y = &#92;frac{x}{M}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{x^{2}}{&#92;text{sn}^{2}&#92;,4s&#92;omega}}{1 - k^{2}x^{2}&#92;text{sn}^{2}&#92;,4s&#92;omega}' title='&#92;displaystyle y = &#92;frac{x}{M}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{x^{2}}{&#92;text{sn}^{2}&#92;,4s&#92;omega}}{1 - k^{2}x^{2}&#92;text{sn}^{2}&#92;,4s&#92;omega}' class='latex' /></p>
<p>and other variants of it</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+1+-+y+%3D+%281+-+x%29%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B%5Cleft%281+-+%5Cdfrac%7Bx%7D%7B%5Ctext%7Bsn%7D%28K+-+4s%5Comega%29%7D%5Cright%29%5E%7B2%7D%7D%7B1+-+k%5E%7B2%7Dx%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 1 - y = (1 - x)&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{&#92;left(1 - &#92;dfrac{x}{&#92;text{sn}(K - 4s&#92;omega)}&#92;right)^{2}}{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' title='&#92;displaystyle 1 - y = (1 - x)&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{&#92;left(1 - &#92;dfrac{x}{&#92;text{sn}(K - 4s&#92;omega)}&#92;right)^{2}}{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+1+%2B+y+%3D+%281+%2B+x%29%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B%5Cleft%281+%2B+%5Cdfrac%7Bx%7D%7B%5Ctext%7Bsn%7D%28K+-+4s%5Comega%29%7D%5Cright%29%5E%7B2%7D%7D%7B1+-+k%5E%7B2%7Dx%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 1 + y = (1 + x)&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{&#92;left(1 + &#92;dfrac{x}{&#92;text{sn}(K - 4s&#92;omega)}&#92;right)^{2}}{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' title='&#92;displaystyle 1 + y = (1 + x)&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{&#92;left(1 + &#92;dfrac{x}{&#92;text{sn}(K - 4s&#92;omega)}&#92;right)^{2}}{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+1+-+ly+%3D+%281+-+kx%29%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cfrac%7B%281+-+kx%5C%2C%5Ctext%7Bsn%7D%28K+-+4s%5Comega%29%29%5E%7B2%7D%7D%7B1+-+k%5E%7B2%7Dx%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 1 - ly = (1 - kx)&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{(1 - kx&#92;,&#92;text{sn}(K - 4s&#92;omega))^{2}}{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' title='&#92;displaystyle 1 - ly = (1 - kx)&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{(1 - kx&#92;,&#92;text{sn}(K - 4s&#92;omega))^{2}}{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+1+%2B+ly+%3D+%281+%2B+kx%29%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cfrac%7B%281+%2B+kx%5C%2C%5Ctext%7Bsn%7D%28K+-+4s%5Comega%29%29%5E%7B2%7D%7D%7B1+-+k%5E%7B2%7Dx%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 1 + ly = (1 + kx)&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{(1 + kx&#92;,&#92;text{sn}(K - 4s&#92;omega))^{2}}{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' title='&#92;displaystyle 1 + ly = (1 + kx)&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{(1 + kx&#92;,&#92;text{sn}(K - 4s&#92;omega))^{2}}{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' class='latex' /></p>
<p>as described in <a title="Elementary Approach to Modular Equations: Jacobi’s Transformation Theory 3" href="http://paramanands.wordpress.com/2011/10/29/elementary-approach-to-modular-equations-jacobi%e2%80%99s-transformation-theory-3/" target="_blank">previous post</a> lead to the differential equation</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7BMdy%7D%7B%5Csqrt%7B%281+-+y%5E%7B2%7D%29%281+-+l%5E%7B2%7Dy%5E%7B2%7D%29%7D%7D+%3D+%5Cfrac%7Bdx%7D%7B%5Csqrt%7B%281+-+x%5E%7B2%7D%29%281+-+k%5E%7B2%7Dx%5E%7B2%7D%29%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{Mdy}{&#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = &#92;frac{dx}{&#92;sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}' title='&#92;displaystyle &#92;frac{Mdy}{&#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = &#92;frac{dx}{&#92;sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}' class='latex' /></p>
<p>when</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+M+%3D+%28-1%29%5E%7B%28p+-+1%29%2F2%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cfrac%7B%5Ctext%7Bsn%7D%28K+-+4s%5Comega%29%7D%7B%5Ctext%7Bsn%7D%5C%2C4s%5Comega%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle M = (-1)^{(p - 1)/2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{&#92;text{sn}(K - 4s&#92;omega)}{&#92;text{sn}&#92;,4s&#92;omega}&#92;right)^{2}' title='&#92;displaystyle M = (-1)^{(p - 1)/2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{&#92;text{sn}(K - 4s&#92;omega)}{&#92;text{sn}&#92;,4s&#92;omega}&#92;right)^{2}' class='latex' /></p>
<p>and</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l+%3D+k%5E%7Bp%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5E%7B4%7D%28K+-+4s%5Comega%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l = k^{p}&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}^{4}(K - 4s&#92;omega)' title='&#92;displaystyle l = k^{p}&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}^{4}(K - 4s&#92;omega)' class='latex' /></p>
<p>Putting <img src='http://s0.wp.com/latex.php?latex=x+%3D+%5Ctext%7Bsn%7D%28u%2C+k%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = &#92;text{sn}(u, k)' title='x = &#92;text{sn}(u, k)' class='latex' /> in the differential equation we see that <img src='http://s0.wp.com/latex.php?latex=y+%3D+%5Ctext%7Bsn%7D%28u%2FM%2C+l%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = &#92;text{sn}(u/M, l)' title='y = &#92;text{sn}(u/M, l)' class='latex' />. Hence the relation between <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> in effect expresses <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsn%7D%28u%2FM%2C+l%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{sn}(u/M, l)' title='&#92;text{sn}(u/M, l)' class='latex' /> in terms of elliptic functions of a different but related modulus <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' />.</p>
<p>Thus we obtain</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%5Cfrac%7B%5Ctext%7Bsn%7D%5C%2Cu%7D%7BM%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;frac{&#92;text{sn}&#92;,u}{M}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;,4s&#92;omega}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' title='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;frac{&#92;text{sn}&#92;,u}{M}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;,4s&#92;omega}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' class='latex' /></p>
<p>Using the expressions of <img src='http://s0.wp.com/latex.php?latex=1+-+y%2C+1+%2B+y%2C+1+-+ly%2C+1+%2B+ly&amp;bg=fff&amp;fg=222&amp;s=0' alt='1 - y, 1 + y, 1 - ly, 1 + ly' title='1 - y, 1 + y, 1 - ly, 1 + ly' class='latex' /> it is easy to see that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csqrt%7B1+-+y%5E%7B2%7D%7D+%3D+%5Csqrt%7B1+-+x%5E%7B2%7D%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7Bx%5E%7B2%7D%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%28K+-+4s%5Comega%29%7D%7D%7B1+-+k%5E%7B2%7Dx%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;sqrt{1 - y^{2}} = &#92;sqrt{1 - x^{2}}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{x^{2}}{&#92;text{sn}^{2}(K - 4s&#92;omega)}}{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' title='&#92;displaystyle &#92;sqrt{1 - y^{2}} = &#92;sqrt{1 - x^{2}}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{x^{2}}{&#92;text{sn}^{2}(K - 4s&#92;omega)}}{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csqrt%7B1+-+l%5E%7B2%7Dy%5E%7B2%7D%7D+%3D+%5Csqrt%7B1+-+k%5E%7B2%7Dx%5E%7B2%7D%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cfrac%7B1+-+k%5E%7B2%7Dx%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%28K+-+4s%5Comega%29%7D%7B1+-+k%5E%7B2%7Dx%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;sqrt{1 - l^{2}y^{2}} = &#92;sqrt{1 - k^{2}x^{2}}&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}(K - 4s&#92;omega)}{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' title='&#92;displaystyle &#92;sqrt{1 - l^{2}y^{2}} = &#92;sqrt{1 - k^{2}x^{2}}&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}(K - 4s&#92;omega)}{1 - k^{2}x^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' class='latex' /></p>
<p>In the last relation if we put <img src='http://s0.wp.com/latex.php?latex=x+%3D+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = 1' title='x = 1' class='latex' /> so that <img src='http://s0.wp.com/latex.php?latex=y+%3D+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = 1' title='y = 1' class='latex' /> we get the expression for <img src='http://s0.wp.com/latex.php?latex=l%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='l&#039;' title='l&#039;' class='latex' /> as</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l%27+%3D+k%27%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cfrac%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%28K+-+4s%5Comega%29%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D+%3D+k%27%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cfrac%7B%5Ctext%7Bdn%7D%5E%7B2%7D%28K+-+4s%5Comega%29%7D%7B%5Ctext%7Bdn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D+%3D+k%27%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cfrac%7Bk%27%7D%7B%5Ctext%7Bdn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l&#039; = k&#039;&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{1 - k^{2}&#92;,&#92;text{sn}^{2}(K - 4s&#92;omega)}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega} = k&#039;&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{&#92;text{dn}^{2}(K - 4s&#92;omega)}{&#92;text{dn}^{2}&#92;,4s&#92;omega} = k&#039;&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{k&#039;}{&#92;text{dn}^{2}&#92;,4s&#92;omega}&#92;right)^{2}' title='&#92;displaystyle l&#039; = k&#039;&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{1 - k^{2}&#92;,&#92;text{sn}^{2}(K - 4s&#92;omega)}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega} = k&#039;&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{&#92;text{dn}^{2}(K - 4s&#92;omega)}{&#92;text{dn}^{2}&#92;,4s&#92;omega} = k&#039;&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{k&#039;}{&#92;text{dn}^{2}&#92;,4s&#92;omega}&#92;right)^{2}' class='latex' /></p>
<p>and so</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l%27+%3D+%5Cdfrac%7Bk%27%5E%7Bp%7D%7D%7B%7B%5Cdisplaystyle%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bdn%7D%5E%7B4%7D%5C%2C4s%5Comega%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l&#039; = &#92;dfrac{k&#039;^{p}}{{&#92;displaystyle&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{dn}^{4}&#92;,4s&#92;omega}}' title='&#92;displaystyle l&#039; = &#92;dfrac{k&#039;^{p}}{{&#92;displaystyle&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{dn}^{4}&#92;,4s&#92;omega}}' class='latex' /></p>
<p>Putting <img src='http://s0.wp.com/latex.php?latex=x+%3D+%5Ctext%7Bsn%7D%5C%2Cu&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = &#92;text{sn}&#92;,u' title='x = &#92;text{sn}&#92;,u' class='latex' /> in the expressions for <img src='http://s0.wp.com/latex.php?latex=%5Csqrt%7B1+-+y%5E%7B2%7D%7D%2C+%5Csqrt%7B1+-+l%5E%7B2%7Dy%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;sqrt{1 - y^{2}}, &#92;sqrt{1 - l^{2}y^{2}}' title='&#92;sqrt{1 - y^{2}}, &#92;sqrt{1 - l^{2}y^{2}}' class='latex' /> we get the following results:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bcn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%5Ctext%7Bcn%7D%5C%2Cu%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%28K+-+4s%5Comega%29%7D%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{cn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;text{cn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}(K - 4s&#92;omega)}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' title='&#92;displaystyle &#92;text{cn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;text{cn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}(K - 4s&#92;omega)}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bdn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%5Ctext%7Bdn%7D%5C%2Cu%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cfrac%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%28K+-+4s%5Comega%29%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C4s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{dn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;text{dn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}(K - 4s&#92;omega)}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' title='&#92;displaystyle &#92;text{dn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;text{dn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}(K - 4s&#92;omega)}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,4s&#92;omega}' class='latex' /></p>
<p>The above formulas remain unchanged if <img src='http://s0.wp.com/latex.php?latex=M&amp;bg=fff&amp;fg=222&amp;s=0' alt='M' title='M' class='latex' /> is replaced by <img src='http://s0.wp.com/latex.php?latex=-M&amp;bg=fff&amp;fg=222&amp;s=0' alt='-M' title='-M' class='latex' /> and hence in the above formulas we can keep</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+M+%3D+%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cfrac%7B%5Ctext%7Bsn%7D%28K+-+4s%5Comega%29%7D%7B%5Ctext%7Bsn%7D%5C%2C4s%5Comega%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle M = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{&#92;text{sn}(K - 4s&#92;omega)}{&#92;text{sn}&#92;,4s&#92;omega}&#92;right)^{2}' title='&#92;displaystyle M = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{&#92;text{sn}(K - 4s&#92;omega)}{&#92;text{sn}&#92;,4s&#92;omega}&#92;right)^{2}' class='latex' /></p>
<p>Now the values of <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsn%7D%28u+%2B+4%5Comega%29%2C+%5Ctext%7Bsn%7D%28u+%2B+8%5Comega%29%2C+%5Cldots%2C+%5Ctext%7Bsn%7D%28u+%2B+2%28p+-+1%29%5Comega%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{sn}(u + 4&#92;omega), &#92;text{sn}(u + 8&#92;omega), &#92;ldots, &#92;text{sn}(u + 2(p - 1)&#92;omega)' title='&#92;text{sn}(u + 4&#92;omega), &#92;text{sn}(u + 8&#92;omega), &#92;ldots, &#92;text{sn}(u + 2(p - 1)&#92;omega)' class='latex' /> are the same as the values <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsn%7D%28u+%2B+2%5Comega%29%2C+%5Ctext%7Bsn%7D%28u+%2B+4%5Comega%29%2C+%5Cldots%2C+%5Ctext%7Bsn%7D%28u+%2B+%28p+-+1%29%5Comega%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{sn}(u + 2&#92;omega), &#92;text{sn}(u + 4&#92;omega), &#92;ldots, &#92;text{sn}(u + (p - 1)&#92;omega)' title='&#92;text{sn}(u + 2&#92;omega), &#92;text{sn}(u + 4&#92;omega), &#92;ldots, &#92;text{sn}(u + (p - 1)&#92;omega)' class='latex' /> except for the <em>order</em> of terms and <em>sign</em> of each term. It is best to check this by putting some actual odd value of <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> (like 3, 5 etc). But upon squaring these values we see that the problem of sign no longer exists and therefore in each of the above relations we can actually replace <img src='http://s0.wp.com/latex.php?latex=4s&amp;bg=fff&amp;fg=222&amp;s=0' alt='4s' title='4s' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=2s&amp;bg=fff&amp;fg=222&amp;s=0' alt='2s' title='2s' class='latex' /> (by putting <img src='http://s0.wp.com/latex.php?latex=u+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = 0' title='u = 0' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=u+%3D+K&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = K' title='u = K' class='latex' /> and noting that <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsn%7D%28K+%2B+%5Calpha%29+%3D+%5Ctext%7Bsn%7D%28K+-+%5Calpha%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{sn}(K + &#92;alpha) = &#92;text{sn}(K - &#92;alpha)' title='&#92;text{sn}(K + &#92;alpha) = &#92;text{sn}(K - &#92;alpha)' class='latex' />. Thus we arrive at the following:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%5Cfrac%7B%5Ctext%7Bsn%7D%5C%2Cu%7D%7BM%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C2s%5Comega%7D%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C2s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;frac{&#92;text{sn}&#92;,u}{M}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;,2s&#92;omega}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,2s&#92;omega}' title='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;frac{&#92;text{sn}&#92;,u}{M}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;,2s&#92;omega}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,2s&#92;omega}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bcn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%5Ctext%7Bcn%7D%5C%2Cu%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%28K+-+2s%5Comega%29%7D%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C2s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{cn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;text{cn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}(K - 2s&#92;omega)}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,2s&#92;omega}' title='&#92;displaystyle &#92;text{cn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;text{cn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}(K - 2s&#92;omega)}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,2s&#92;omega}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bdn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%5Ctext%7Bdn%7D%5C%2Cu%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cfrac%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%28K+-+2s%5Comega%29%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C2s%5Comega%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{dn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;text{dn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}(K - 2s&#92;omega)}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,2s&#92;omega}' title='&#92;displaystyle &#92;text{dn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;text{dn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;frac{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}(K - 2s&#92;omega)}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,2s&#92;omega}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+M+%3D+%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cfrac%7B%5Ctext%7Bsn%7D%28K+-+2s%5Comega%29%7D%7B%5Ctext%7Bsn%7D%5C%2C2s%5Comega%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle M = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{&#92;text{sn}(K - 2s&#92;omega)}{&#92;text{sn}&#92;,2s&#92;omega}&#92;right)^{2}' title='&#92;displaystyle M = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{&#92;text{sn}(K - 2s&#92;omega)}{&#92;text{sn}&#92;,2s&#92;omega}&#92;right)^{2}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l+%3D+k%5E%7Bp%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5E%7B4%7D%28K+-+2s%5Comega%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l = k^{p}&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}^{4}(K - 2s&#92;omega)' title='&#92;displaystyle l = k^{p}&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}^{4}(K - 2s&#92;omega)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l%27+%3D+%5Cdfrac%7Bk%27%5E%7Bp%7D%7D%7B%7B%5Cdisplaystyle%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bdn%7D%5E%7B4%7D%5C%2C2s%5Comega%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l&#039; = &#92;dfrac{k&#039;^{p}}{{&#92;displaystyle&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{dn}^{4}&#92;,2s&#92;omega}}' title='&#92;displaystyle l&#039; = &#92;dfrac{k&#039;^{p}}{{&#92;displaystyle&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{dn}^{4}&#92;,2s&#92;omega}}' class='latex' /></p>
<p>In the development of these formulas so far we have assumed that <img src='http://s0.wp.com/latex.php?latex=%5Comega+%3D+%28mK+%2B+m%27iK%27%29%2Fp&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;omega = (mK + m&#039;iK&#039;)/p' title='&#92;omega = (mK + m&#039;iK&#039;)/p' class='latex' /> where <img src='http://s0.wp.com/latex.php?latex=m%2C+m%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='m, m&#039;' title='m, m&#039;' class='latex' /> are relatively prime to each other, but we have not given any specific values to <img src='http://s0.wp.com/latex.php?latex=m%2C+m%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='m, m&#039;' title='m, m&#039;' class='latex' />. It turns that by giving different values to <img src='http://s0.wp.com/latex.php?latex=m&amp;bg=fff&amp;fg=222&amp;s=0' alt='m' title='m' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=m%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='m&#039;' title='m&#039;' class='latex' /> we arrive at different transformations. Out of all these transformations two are real i.e. the relation between <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> is expressed using real numbers as coefficients. We are going to study these two transformations next.</p>
<p><strong>Jacobi&#8217;s First Real Transformation</strong></p>
<p>By choosing <img src='http://s0.wp.com/latex.php?latex=m+%3D+1%2C+m%27+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='m = 1, m&#039; = 0' title='m = 1, m&#039; = 0' class='latex' /> we get <img src='http://s0.wp.com/latex.php?latex=%5Comega+%3D+K%2Fp&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;omega = K/p' title='&#92;omega = K/p' class='latex' /> which is a real number and from this we instantly see that all the numbers involved in the formulas mentioned above are real numbers. The corresponding formulas are as follows:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%5Cfrac%7B%5Ctext%7Bsn%7D%5C%2Cu%7D%7BM%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C%5Cdfrac%7B2sK%7D%7Bp%7D%7D%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C%5Cdfrac%7B2sK%7D%7Bp%7D%7D%5C%2C%5C%2C%5C%2C%5Cldots+%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;frac{&#92;text{sn}&#92;,u}{M}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;,&#92;dfrac{2sK}{p}}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,&#92;dfrac{2sK}{p}}&#92;,&#92;,&#92;,&#92;ldots (1)' title='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;frac{&#92;text{sn}&#92;,u}{M}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;,&#92;dfrac{2sK}{p}}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,&#92;dfrac{2sK}{p}}&#92;,&#92;,&#92;,&#92;ldots (1)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bcn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%5Ctext%7Bcn%7D%5C%2Cu%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28K+-+%5Cdfrac%7B2sK%7D%7Bp%7D%5Cright%29%7D%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C%5Cdfrac%7B2sK%7D%7Bp%7D%7D%5C%2C%5C%2C%5C%2C%5Cldots+%282%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{cn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;text{cn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;left(K - &#92;dfrac{2sK}{p}&#92;right)}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,&#92;dfrac{2sK}{p}}&#92;,&#92;,&#92;,&#92;ldots (2)' title='&#92;displaystyle &#92;text{cn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;text{cn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;left(K - &#92;dfrac{2sK}{p}&#92;right)}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,&#92;dfrac{2sK}{p}}&#92;,&#92;,&#92;,&#92;ldots (2)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bdn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%5Ctext%7Bdn%7D%5C%2Cu%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28K+-+%5Cdfrac%7B2sK%7D%7Bp%7D%5Cright%29%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C%5Cdfrac%7B2sK%7D%7Bp%7D%7D%5C%2C%5C%2C%5C%2C%5Cldots+%283%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{dn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;text{dn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;left(K - &#92;dfrac{2sK}{p}&#92;right)}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,&#92;dfrac{2sK}{p}}&#92;,&#92;,&#92;,&#92;ldots (3)' title='&#92;displaystyle &#92;text{dn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;text{dn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;left(K - &#92;dfrac{2sK}{p}&#92;right)}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,&#92;dfrac{2sK}{p}}&#92;,&#92;,&#92;,&#92;ldots (3)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+M+%3D+%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cdfrac%7B%5Ctext%7Bsn%7D%5Cleft%28K+-+%5Cdfrac%7B2sK%7D%7Bp%7D%5Cright%29%7D%7B%5Ctext%7Bsn%7D%5C%2C%5Cdfrac%7B2sK%7D%7Bp%7D%7D%5Cright%29%5E%7B2%7D%5C%2C%5C%2C%5C%2C%5Cldots+%284%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle M = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{&#92;text{sn}&#92;left(K - &#92;dfrac{2sK}{p}&#92;right)}{&#92;text{sn}&#92;,&#92;dfrac{2sK}{p}}&#92;right)^{2}&#92;,&#92;,&#92;,&#92;ldots (4)' title='&#92;displaystyle M = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{&#92;text{sn}&#92;left(K - &#92;dfrac{2sK}{p}&#92;right)}{&#92;text{sn}&#92;,&#92;dfrac{2sK}{p}}&#92;right)^{2}&#92;,&#92;,&#92;,&#92;ldots (4)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l+%3D+k%5E%7Bp%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5E%7B4%7D%5Cleft%28K+-+%5Cfrac%7B2sK%7D%7Bp%7D%5Cright%29%5C%2C%5C%2C%5C%2C%5Cldots+%285%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l = k^{p}&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}^{4}&#92;left(K - &#92;frac{2sK}{p}&#92;right)&#92;,&#92;,&#92;,&#92;ldots (5)' title='&#92;displaystyle l = k^{p}&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}^{4}&#92;left(K - &#92;frac{2sK}{p}&#92;right)&#92;,&#92;,&#92;,&#92;ldots (5)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l%27+%3D+%5Cdfrac%7Bk%27%5E%7Bp%7D%7D%7B%7B%5Cdisplaystyle%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bdn%7D%5E%7B4%7D%5C%2C%5Cfrac%7B2sK%7D%7Bp%7D%7D%7D%5C%2C%5C%2C%5C%2C%5Cldots+%286%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l&#039; = &#92;dfrac{k&#039;^{p}}{{&#92;displaystyle&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{dn}^{4}&#92;,&#92;frac{2sK}{p}}}&#92;,&#92;,&#92;,&#92;ldots (6)' title='&#92;displaystyle l&#039; = &#92;dfrac{k&#039;^{p}}{{&#92;displaystyle&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{dn}^{4}&#92;,&#92;frac{2sK}{p}}}&#92;,&#92;,&#92;,&#92;ldots (6)' class='latex' /></p>
<p>It is easy to see that the least positive value of <img src='http://s0.wp.com/latex.php?latex=u&amp;bg=fff&amp;fg=222&amp;s=0' alt='u' title='u' class='latex' /> for which the right side of <img src='http://s0.wp.com/latex.php?latex=%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(1)' title='(1)' class='latex' /> vanishes is <img src='http://s0.wp.com/latex.php?latex=u+%3D+2K%2Fp&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = 2K/p' title='u = 2K/p' class='latex' /> and the left side of this equation vanishes when <img src='http://s0.wp.com/latex.php?latex=u%2FM+%3D+2L&amp;bg=fff&amp;fg=222&amp;s=0' alt='u/M = 2L' title='u/M = 2L' class='latex' />. Equating these two values of <img src='http://s0.wp.com/latex.php?latex=u&amp;bg=fff&amp;fg=222&amp;s=0' alt='u' title='u' class='latex' /> we get the relation <img src='http://s0.wp.com/latex.php?latex=K%2FpM+%3D+L&amp;bg=fff&amp;fg=222&amp;s=0' alt='K/pM = L' title='K/pM = L' class='latex' />.</p>
<p><strong>Jacobi&#8217;s First Complementary Transformation</strong></p>
<p>Next we need to recast the above formulas in another form by using the identities:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cv%7D%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cv%7D+%3D+-%5Cfrac%7B%5Ctext%7Bsn%7D%28u+%2B+v%29%5C%2C%5Ctext%7Bsn%7D%28u+-+v%29%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cv%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;,v}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,v} = -&#92;frac{&#92;text{sn}(u + v)&#92;,&#92;text{sn}(u - v)}{&#92;text{sn}^{2}&#92;,v}' title='&#92;displaystyle &#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}&#92;,v}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,v} = -&#92;frac{&#92;text{sn}(u + v)&#92;,&#92;text{sn}(u - v)}{&#92;text{sn}^{2}&#92;,v}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%28K+-+v%29%7D%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cv%7D+%3D+%5Cfrac%7B%5Ctext%7Bcn%7D%28u+%2B+v%29%5C%2C%5Ctext%7Bcn%7D%28u+-+v%29%7D%7B%5Ctext%7Bcn%7D%5E%7B2%7D%5C%2Cv%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}(K - v)}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,v} = &#92;frac{&#92;text{cn}(u + v)&#92;,&#92;text{cn}(u - v)}{&#92;text{cn}^{2}&#92;,v}' title='&#92;displaystyle &#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}&#92;,u}{&#92;text{sn}^{2}(K - v)}}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,v} = &#92;frac{&#92;text{cn}(u + v)&#92;,&#92;text{cn}(u - v)}{&#92;text{cn}^{2}&#92;,v}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1+-+k%5E%7B2%7D%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%28K+-+v%29%7D%7B1+-+k%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cu%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2Cv%7D+%3D+%5Cfrac%7B%5Ctext%7Bdn%7D%28u+%2B+v%29%5C%2C%5Ctext%7Bdn%7D%28u+-+v%29%7D%7B%5Ctext%7Bdn%7D%5E%7B2%7D%5C%2Cv%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{1 - k^{2}&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}(K - v)}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,v} = &#92;frac{&#92;text{dn}(u + v)&#92;,&#92;text{dn}(u - v)}{&#92;text{dn}^{2}&#92;,v}' title='&#92;displaystyle &#92;frac{1 - k^{2}&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}(K - v)}{1 - k^{2}&#92;,&#92;text{sn}^{2}&#92;,u&#92;,&#92;text{sn}^{2}&#92;,v} = &#92;frac{&#92;text{dn}(u + v)&#92;,&#92;text{dn}(u - v)}{&#92;text{dn}^{2}&#92;,v}' class='latex' /></p>
<p>On applying these identities to the formulas <img src='http://s0.wp.com/latex.php?latex=%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(1)' title='(1)' class='latex' /> and using <img src='http://s0.wp.com/latex.php?latex=%284%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(4)' title='(4)' class='latex' /> to replace <img src='http://s0.wp.com/latex.php?latex=M&amp;bg=fff&amp;fg=222&amp;s=0' alt='M' title='M' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%28-1%29%5E%7B%28p+-+1%29%2F2%7D%5C%2C%5Ctext%7Bsn%7D%5C%2Cu%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cdfrac%7B%5Ctext%7Bsn%7D%5C%2C%5Cdfrac%7B2sK%7D%7Bp%7D%7D%7B%5Ctext%7Bsn%7D%5Cleft%28K+-+%5Cdfrac%7B2sK%7D%7Bp%7D%5Cright%29%7D%5Cright%29%5E%7B2%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B%5Ctext%7Bsn%7D%5Cleft%28u+%2B+%5Cdfrac%7B2sK%7D%7Bp%7D%5Cright%29%5C%2C%5Ctext%7Bsn%7D%5Cleft%28u+-+%5Cdfrac%7B2sK%7D%7Bp%7D%5Cright%29%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5C%2C%5Cdfrac%7B2sK%7D%7Bp%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#92;right) = (-1)^{(p - 1)/2}&#92;,&#92;text{sn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{&#92;text{sn}&#92;,&#92;dfrac{2sK}{p}}{&#92;text{sn}&#92;left(K - &#92;dfrac{2sK}{p}&#92;right)}&#92;right)^{2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{&#92;text{sn}&#92;left(u + &#92;dfrac{2sK}{p}&#92;right)&#92;,&#92;text{sn}&#92;left(u - &#92;dfrac{2sK}{p}&#92;right)}{&#92;text{sn}^{2}&#92;,&#92;dfrac{2sK}{p}}' title='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#92;right) = (-1)^{(p - 1)/2}&#92;,&#92;text{sn}&#92;,u&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{&#92;text{sn}&#92;,&#92;dfrac{2sK}{p}}{&#92;text{sn}&#92;left(K - &#92;dfrac{2sK}{p}&#92;right)}&#92;right)^{2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{&#92;text{sn}&#92;left(u + &#92;dfrac{2sK}{p}&#92;right)&#92;,&#92;text{sn}&#92;left(u - &#92;dfrac{2sK}{p}&#92;right)}{&#92;text{sn}^{2}&#92;,&#92;dfrac{2sK}{p}}' class='latex' /></p>
<p>and using equation <img src='http://s0.wp.com/latex.php?latex=%285%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(5)' title='(5)' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%28-1%29%5E%7B%28p+-+1%29%2F2%7D%5Csqrt%7B%5Cfrac%7Bk%5E%7Bp%7D%7D%7Bl%7D%7D%5Cprod_%7Bs+%3D+-%28p+-+1%29%2F2%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5Cleft%28u+%2B+%5Cfrac%7B2sK%7D%7Bp%7D%5Cright%29%5C%2C%5C%2C%5C%2C%5Cldots+%287%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#92;right) = (-1)^{(p - 1)/2}&#92;sqrt{&#92;frac{k^{p}}{l}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{sn}&#92;left(u + &#92;frac{2sK}{p}&#92;right)&#92;,&#92;,&#92;,&#92;ldots (7)' title='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#92;right) = (-1)^{(p - 1)/2}&#92;sqrt{&#92;frac{k^{p}}{l}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{sn}&#92;left(u + &#92;frac{2sK}{p}&#92;right)&#92;,&#92;,&#92;,&#92;ldots (7)' class='latex' /></p>
<p>and similarly</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bcn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%5Csqrt%7B%5Cfrac%7Bl%27k%5E%7Bp%7D%7D%7Blk%27%5E%7Bp%7D%7D%7D%5Cprod_%7Bs+%3D+-%28p+-+1%29%2F2%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bcn%7D%5Cleft%28u+%2B+%5Cfrac%7B2sK%7D%7Bp%7D%5Cright%29%5C%2C%5C%2C%5C%2C%5Cldots+%288%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{cn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;sqrt{&#92;frac{l&#039;k^{p}}{lk&#039;^{p}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{cn}&#92;left(u + &#92;frac{2sK}{p}&#92;right)&#92;,&#92;,&#92;,&#92;ldots (8)' title='&#92;displaystyle &#92;text{cn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;sqrt{&#92;frac{l&#039;k^{p}}{lk&#039;^{p}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{cn}&#92;left(u + &#92;frac{2sK}{p}&#92;right)&#92;,&#92;,&#92;,&#92;ldots (8)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bdn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%5Csqrt%7B%5Cfrac%7Bl%27%7D%7Bk%27%5E%7Bp%7D%7D%7D%5Cprod_%7Bs+%3D+-%28p+-+1%29%2F2%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bdn%7D%5Cleft%28u+%2B+%5Cfrac%7B2sK%7D%7Bp%7D%5Cright%29%5C%2C%5C%2C%5C%2C%5Cldots+%289%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{dn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;sqrt{&#92;frac{l&#039;}{k&#039;^{p}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{dn}&#92;left(u + &#92;frac{2sK}{p}&#92;right)&#92;,&#92;,&#92;,&#92;ldots (9)' title='&#92;displaystyle &#92;text{dn}&#92;left(&#92;frac{u}{M}, l&#92;right) = &#92;sqrt{&#92;frac{l&#039;}{k&#039;^{p}}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{dn}&#92;left(u + &#92;frac{2sK}{p}&#92;right)&#92;,&#92;,&#92;,&#92;ldots (9)' class='latex' /></p>
<p>Dividing the equation <img src='http://s0.wp.com/latex.php?latex=%287%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(7)' title='(7)' class='latex' /> by equation <img src='http://s0.wp.com/latex.php?latex=%288%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(8)' title='(8)' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsc%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%5Cright%29+%3D+%28-1%29%5E%7B%28p+-+1%29%2F2%7D%5Csqrt%7B%5Cfrac%7Bk%27%5E%7Bp%7D%7D%7Bl%27%7D%7D%5Cprod_%7Bs+%3D+-%28p+-+1%29%2F2%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsc%7D%5Cleft%28u+%2B+%5Cfrac%7B2sK%7D%7Bp%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sc}&#92;left(&#92;frac{u}{M}, l&#92;right) = (-1)^{(p - 1)/2}&#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{sc}&#92;left(u + &#92;frac{2sK}{p}&#92;right)' title='&#92;displaystyle &#92;text{sc}&#92;left(&#92;frac{u}{M}, l&#92;right) = (-1)^{(p - 1)/2}&#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{sc}&#92;left(u + &#92;frac{2sK}{p}&#92;right)' class='latex' /></p>
<p>Replacing <img src='http://s0.wp.com/latex.php?latex=u&amp;bg=fff&amp;fg=222&amp;s=0' alt='u' title='u' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=iu&amp;bg=fff&amp;fg=222&amp;s=0' alt='iu' title='iu' class='latex' /> in the above equation and noting that <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsc%7D%28iu%2C+k%29+%3D+i%5C%2C%5Ctext%7Bsn%7D%28u%2C+k%27%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{sc}(iu, k) = i&#92;,&#92;text{sn}(u, k&#039;)' title='&#92;text{sc}(iu, k) = i&#92;,&#92;text{sn}(u, k&#039;)' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%27%5Cright%29+%3D+%5Csqrt%7B%5Cfrac%7Bk%27%5E%7Bp%7D%7D%7Bl%27%7D%7D%5Cprod_%7Bs+%3D+-%28p+-+1%29%2F2%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5Cleft%28u+-+%5Cfrac%7B2siK%7D%7Bp%7D%2C+k%27%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#039;&#92;right) = &#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{sn}&#92;left(u - &#92;frac{2siK}{p}, k&#039;&#92;right)' title='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#039;&#92;right) = &#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;}}&#92;prod_{s = -(p - 1)/2}^{(p - 1)/2}&#92;text{sn}&#92;left(u - &#92;frac{2siK}{p}, k&#039;&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%5Csqrt%7B%5Cfrac%7Bk%27%5E%7Bp%7D%7D%7Bl%27%7D%7D%5C%2C%5Ctext%7Bsn%7D%28u%2C+k%27%29%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5Cleft%28u+%2B+%5Cfrac%7B2siK%7D%7Bp%7D%2C+k%27%5Cright%29%5C%2C%5Ctext%7Bsn%7D%5Cleft%28u+-+%5Cfrac%7B2siK%7D%7Bp%7D%2C+k%27%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = &#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;}}&#92;,&#92;text{sn}(u, k&#039;)&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}&#92;left(u + &#92;frac{2siK}{p}, k&#039;&#92;right)&#92;,&#92;text{sn}&#92;left(u - &#92;frac{2siK}{p}, k&#039;&#92;right)' title='&#92;displaystyle = &#92;sqrt{&#92;frac{k&#039;^{p}}{l&#039;}}&#92;,&#92;text{sn}(u, k&#039;)&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}&#92;left(u + &#92;frac{2siK}{p}, k&#039;&#92;right)&#92;,&#92;text{sn}&#92;left(u - &#92;frac{2siK}{p}, k&#039;&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%28-1%29%5E%7B%28p+-+1%29%2F2%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28%5Cfrac%7B2siK%7D%7Bp%7D%2C+k%27%5Cright%29%5C%2C%5Ctext%7Bsn%7D%28u%2C+k%27%29%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%28u%2C+k%27%29%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2siK%7D%7Bp%7D%2C+k%27%5Cright%29%7D%7D%7B1+-+k%27%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%28u%2C+k%27%29%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2siK%7D%7Bp%7D%2C+k%27%5Cright%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = (-1)^{(p - 1)/2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}^{2}&#92;left(&#92;frac{2siK}{p}, k&#039;&#92;right)&#92;,&#92;text{sn}(u, k&#039;)&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}(u, k&#039;)}{&#92;text{sn}^{2}&#92;left(&#92;dfrac{2siK}{p}, k&#039;&#92;right)}}{1 - k&#039;^{2}&#92;,&#92;text{sn}^{2}(u, k&#039;)&#92;,&#92;text{sn}^{2}&#92;left(&#92;dfrac{2siK}{p}, k&#039;&#92;right)}' title='&#92;displaystyle = (-1)^{(p - 1)/2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;text{sn}^{2}&#92;left(&#92;frac{2siK}{p}, k&#039;&#92;right)&#92;,&#92;text{sn}(u, k&#039;)&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}^{2}(u, k&#039;)}{&#92;text{sn}^{2}&#92;left(&#92;dfrac{2siK}{p}, k&#039;&#92;right)}}{1 - k&#039;^{2}&#92;,&#92;text{sn}^{2}(u, k&#039;)&#92;,&#92;text{sn}^{2}&#92;left(&#92;dfrac{2siK}{p}, k&#039;&#92;right)}' class='latex' /></p>
<p>The factor on the right side which is independent of <img src='http://s0.wp.com/latex.php?latex=u&amp;bg=fff&amp;fg=222&amp;s=0' alt='u' title='u' class='latex' /> must be <img src='http://s0.wp.com/latex.php?latex=1%2FM&amp;bg=fff&amp;fg=222&amp;s=0' alt='1/M' title='1/M' class='latex' /> as can be easily seen when we divide both sides by <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsn%7D%28u%2C+k%27%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{sn}(u, k&#039;)' title='&#92;text{sn}(u, k&#039;)' class='latex' /> and take limits as <img src='http://s0.wp.com/latex.php?latex=u+%5Cto+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='u &#92;to 0' title='u &#92;to 0' class='latex' />.</p>
<p>Thus we finally obtain</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctext%7Bsn%7D%5Cleft%28%5Cfrac%7Bu%7D%7BM%7D%2C+l%27%5Cright%29+%3D+%5Cfrac%7B%5Ctext%7Bsn%7D%28u%2C+k%27%29%7D%7BM%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cdfrac%7B1+%2B+%5Cdfrac%7B%5Ctext%7Bsn%7D%5E%7B2%7D%28u%2C+k%27%29%7D%7B%5Ctext%7Bsc%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2sK%7D%7Bp%7D%2C+k%5Cright%29%7D%7D%7B1+%2B+k%27%5E%7B2%7D%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7D%28u%2C+k%27%29%5C%2C%5Ctext%7Bsc%7D%5E%7B2%7D%5Cleft%28%5Cdfrac%7B2sK%7D%7Bp%7D%2C+k%5Cright%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#039;&#92;right) = &#92;frac{&#92;text{sn}(u, k&#039;)}{M}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 + &#92;dfrac{&#92;text{sn}^{2}(u, k&#039;)}{&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sK}{p}, k&#92;right)}}{1 + k&#039;^{2}&#92;,&#92;text{sn}^{2}(u, k&#039;)&#92;,&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sK}{p}, k&#92;right)}' title='&#92;displaystyle &#92;text{sn}&#92;left(&#92;frac{u}{M}, l&#039;&#92;right) = &#92;frac{&#92;text{sn}(u, k&#039;)}{M}&#92;prod_{s = 1}^{(p - 1)/2}&#92;dfrac{1 + &#92;dfrac{&#92;text{sn}^{2}(u, k&#039;)}{&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sK}{p}, k&#92;right)}}{1 + k&#039;^{2}&#92;,&#92;text{sn}^{2}(u, k&#039;)&#92;,&#92;text{sc}^{2}&#92;left(&#92;dfrac{2sK}{p}, k&#92;right)}' class='latex' /></p>
<p>The only factor on the right side which can vanish is <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsn%7D%28u%2C+k%27%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{sn}(u, k&#039;)' title='&#92;text{sn}(u, k&#039;)' class='latex' /> and the smallest value for which this vanishes is <img src='http://s0.wp.com/latex.php?latex=u+%3D+2K%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = 2K&#039;' title='u = 2K&#039;' class='latex' />. The left side vanishes when <img src='http://s0.wp.com/latex.php?latex=u%2FM+%3D+2L%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='u/M = 2L&#039;' title='u/M = 2L&#039;' class='latex' /> and therefore upon equating these two values of <img src='http://s0.wp.com/latex.php?latex=u&amp;bg=fff&amp;fg=222&amp;s=0' alt='u' title='u' class='latex' /> we get <img src='http://s0.wp.com/latex.php?latex=K%27%2FM+%3D+L%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;/M = L&#039;' title='K&#039;/M = L&#039;' class='latex' />.</p>
<p>We have thus derived the two relations <img src='http://s0.wp.com/latex.php?latex=K%2FpM+%3D+L&amp;bg=fff&amp;fg=222&amp;s=0' alt='K/pM = L' title='K/pM = L' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=K%27%2FM+%3D+L%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;/M = L&#039;' title='K&#039;/M = L&#039;' class='latex' /> from which we obtain the fundamental relation <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+pK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = pK&#039;/K' title='L&#039;/L = pK&#039;/K' class='latex' />. From this it also follows that the value of <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> occurring in these equations is actually less than <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' />.</p>
<p>In the next post we will describe one more transformation which is also real and it leads to a value of <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> which is greater than <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> (which will be denoted by <img src='http://s0.wp.com/latex.php?latex=l_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='l_{1}' title='l_{1}' class='latex' />) and in this case <img src='http://s0.wp.com/latex.php?latex=K%27%2FK+%3D+pL%27_%7B1%7D%2FL_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;/K = pL&#039;_{1}/L_{1}' title='K&#039;/K = pL&#039;_{1}/L_{1}' class='latex' />.</p>
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		<title>Elementary Approach to Modular Equations: Jacobi’s Transformation Theory 3</title>
		<link>http://paramanands.wordpress.com/2011/10/29/elementary-approach-to-modular-equations-jacobi%e2%80%99s-transformation-theory-3/</link>
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		<pubDate>Sat, 29 Oct 2011 08:59:15 +0000</pubDate>
		<dc:creator>paramanands</dc:creator>
				<category><![CDATA[Elliptic Functions]]></category>
		<category><![CDATA[Elliptic Integrals]]></category>
		<category><![CDATA[Mathematical Analysis]]></category>

		<guid isPermaLink="false">http://paramanands.wordpress.com/?p=2257</guid>
		<description><![CDATA[Analytic Approach to Transformation Theory Jacobi understood that the algebraic approach for obtaining modular equations could not be applied easily in case of higher degrees. Hence he followed an analytic approach. The idea he used was to express the relation in a form where each of and appears as a product of various factors. Essentially [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=paramanands.wordpress.com&amp;blog=8565079&amp;post=2257&amp;subd=paramanands&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p><strong>Analytic Approach to Transformation Theory<br />
</strong></p>
<p>Jacobi understood that the algebraic approach for obtaining modular equations could not be applied easily in case of higher degrees. Hence he followed an analytic approach. The idea he used was to express the relation <img src='http://s0.wp.com/latex.php?latex=y+%3D+xN%281%2C+x%5E%7B2%7D%29%2FD%281%2C+x%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = xN(1, x^{2})/D(1, x^{2})' title='y = xN(1, x^{2})/D(1, x^{2})' class='latex' /> in a form where each of <img src='http://s0.wp.com/latex.php?latex=N&amp;bg=fff&amp;fg=222&amp;s=0' alt='N' title='N' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=D&amp;bg=fff&amp;fg=222&amp;s=0' alt='D' title='D' class='latex' /> appears as a product of various factors. Essentially he examined the roots of <img src='http://s0.wp.com/latex.php?latex=N%2C+D&amp;bg=fff&amp;fg=222&amp;s=0' alt='N, D' title='N, D' class='latex' /> and expressed them in form of a product where each factor corresponds to a given root. This approach was very useful for Jacobi as he used this relation to finally develop the theory of <em>Theta Functions</em> and their relation to elliptic functions. In fact the entire <em>Fundamenta Nova</em> is split into two sections, the first section dealing with transformation theory and the second section dealing with the expansion of elliptic functions into infinite series and products (which is basically the theory of theta functions).</p>
<p>The presentation here is based on the Calyley&#8217;s <em>An Elementary Treatise on Elliptic Functions</em> and our goal here will be to establish the existence of transformation of odd prime degree and to establish the fundamental relation <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+pK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = pK&#039;/K' title='L&#039;/L = pK&#039;/K' class='latex' /> which is induced by these transformations. We will not explore the development of theta functions based on this transformation theory as done in Jacobi&#8217;s or Cayley&#8217;s books.</p>
<p>To begin with lets fix the notation which is almost similar to that used by Jacobi/Cayley except that we use <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> in place of <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=fff&amp;fg=222&amp;s=0' alt='n' title='n' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> in place of <img src='http://s0.wp.com/latex.php?latex=%5Clambda&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;lambda' title='&#92;lambda' class='latex' />. Let&#8217;s assume that <img src='http://s0.wp.com/latex.php?latex=m&amp;bg=fff&amp;fg=222&amp;s=0' alt='m' title='m' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=m%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='m&#039;' title='m&#039;' class='latex' /> are some fixed integers relatively prime to each other not both zero and let</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Comega+%3D+%5Cfrac%7BmK+%2B+m%27iK%27%7D%7Bp%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;omega = &#92;frac{mK + m&#039;iK&#039;}{p}' title='&#92;displaystyle &#92;omega = &#92;frac{mK + m&#039;iK&#039;}{p}' class='latex' /></p>
<p>The transformation is given by <img src='http://s0.wp.com/latex.php?latex=y+%3D+U%2FV&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = U/V' title='y = U/V' class='latex' /> and the resulting equations</p>
<p><img src='http://s0.wp.com/latex.php?latex=V+%2B+U+%3D+%281+%2B+x%29A%5E%7B2%7D%2C+V+-+U+%3D+%281+-+x%29B%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='V + U = (1 + x)A^{2}, V - U = (1 - x)B^{2}' title='V + U = (1 + x)A^{2}, V - U = (1 - x)B^{2}' class='latex' /></p>
<p>i.e.</p>
<p><img src='http://s0.wp.com/latex.php?latex=1+%2B+y+%3D+%281+%2B+x%29A%5E%7B2%7D%2FV%2C+1+-+y+%3D+%281+-+x%29B%5E%7B2%7D%2FV&amp;bg=fff&amp;fg=222&amp;s=0' alt='1 + y = (1 + x)A^{2}/V, 1 - y = (1 - x)B^{2}/V' title='1 + y = (1 + x)A^{2}/V, 1 - y = (1 - x)B^{2}/V' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=V+%2B+lU+%3D+%281+%2B+kx%29C%5E%7B2%7D%2C+V+-+lU+%3D+%281+-+kx%29D%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='V + lU = (1 + kx)C^{2}, V - lU = (1 - kx)D^{2}' title='V + lU = (1 + kx)C^{2}, V - lU = (1 - kx)D^{2}' class='latex' /></p>
<p>i.e.</p>
<p><img src='http://s0.wp.com/latex.php?latex=1+%2B+ly+%3D+%281+%2B+kx%29C%5E%7B2%7D%2FV%2C+1+-+ly+%3D+%281+-+kx%29D%5E%7B2%7D%2FV&amp;bg=fff&amp;fg=222&amp;s=0' alt='1 + ly = (1 + kx)C^{2}/V, 1 - ly = (1 - kx)D^{2}/V' title='1 + ly = (1 + kx)C^{2}/V, 1 - ly = (1 - kx)D^{2}/V' class='latex' /></p>
<p>The polynomials <img src='http://s0.wp.com/latex.php?latex=U%2C+V&amp;bg=fff&amp;fg=222&amp;s=0' alt='U, V' title='U, V' class='latex' /> are given by</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+U+%3D+%5Cfrac%7Bx%7D%7BM%7D%5Cleft%281+-+%5Cfrac%7Bx%5E%7B2%7D%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%284%5Comega%2C+k%29%7D%5Cright%29%5Cleft%281+-+%5Cfrac%7Bx%5E%7B2%7D%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%288%5Comega%2C+k%29%7D%5Cright%29%5Ccdots%5Cleft%281+-+%5Cfrac%7Bx%5E%7B2%7D%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%282%28p+-+1%29%5Comega%2C+k%29%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle U = &#92;frac{x}{M}&#92;left(1 - &#92;frac{x^{2}}{&#92;text{sn}^{2}(4&#92;omega, k)}&#92;right)&#92;left(1 - &#92;frac{x^{2}}{&#92;text{sn}^{2}(8&#92;omega, k)}&#92;right)&#92;cdots&#92;left(1 - &#92;frac{x^{2}}{&#92;text{sn}^{2}(2(p - 1)&#92;omega, k)}&#92;right)' title='&#92;displaystyle U = &#92;frac{x}{M}&#92;left(1 - &#92;frac{x^{2}}{&#92;text{sn}^{2}(4&#92;omega, k)}&#92;right)&#92;left(1 - &#92;frac{x^{2}}{&#92;text{sn}^{2}(8&#92;omega, k)}&#92;right)&#92;cdots&#92;left(1 - &#92;frac{x^{2}}{&#92;text{sn}^{2}(2(p - 1)&#92;omega, k)}&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%5Cfrac%7Bx%7D%7BM%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%281+-+%5Cfrac%7Bx%5E%7B2%7D%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D%284s%5Comega%2C+k%29%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = &#92;frac{x}{M}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(1 - &#92;frac{x^{2}}{&#92;text{sn}^{2}(4s&#92;omega, k)}&#92;right)' title='&#92;displaystyle = &#92;frac{x}{M}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(1 - &#92;frac{x^{2}}{&#92;text{sn}^{2}(4s&#92;omega, k)}&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=V+%3D+%281+-+k%5E%7B2%7Dx%5E%7B2%7D%5Ctext%7Bsn%7D%5E%7B2%7D%284%5Comega%2C+k%29%29%281+-+k%5E%7B2%7Dx%5E%7B2%7D%5Ctext%7Bsn%7D%5E%7B2%7D%288%5Comega%2C+k%29%29%5Ccdots%281+-+k%5E%7B2%7Dx%5E%7B2%7D%5Ctext%7Bsn%7D%5E%7B2%7D%282%28p+-+1%29%5Comega%2C+k%29%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='V = (1 - k^{2}x^{2}&#92;text{sn}^{2}(4&#92;omega, k))(1 - k^{2}x^{2}&#92;text{sn}^{2}(8&#92;omega, k))&#92;cdots(1 - k^{2}x^{2}&#92;text{sn}^{2}(2(p - 1)&#92;omega, k))' title='V = (1 - k^{2}x^{2}&#92;text{sn}^{2}(4&#92;omega, k))(1 - k^{2}x^{2}&#92;text{sn}^{2}(8&#92;omega, k))&#92;cdots(1 - k^{2}x^{2}&#92;text{sn}^{2}(2(p - 1)&#92;omega, k))' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%281+-+k%5E%7B2%7Dx%5E%7B2%7D%5Ctext%7Bsn%7D%5E%7B2%7D%284s%5Comega%2C+k%29%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = &#92;prod_{s = 1}^{(p - 1)/2}(1 - k^{2}x^{2}&#92;text{sn}^{2}(4s&#92;omega, k))' title='&#92;displaystyle = &#92;prod_{s = 1}^{(p - 1)/2}(1 - k^{2}x^{2}&#92;text{sn}^{2}(4s&#92;omega, k))' class='latex' /></p>
<p>and the polynomials <img src='http://s0.wp.com/latex.php?latex=A%2C+B%2C+C%2C+D&amp;bg=fff&amp;fg=222&amp;s=0' alt='A, B, C, D' title='A, B, C, D' class='latex' /> are given by</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+A+%3D+%5Cleft%281+%2B+%5Cfrac%7Bx%7D%7B%5Ctext%7Bsn%7D%28K+-+4%5Comega%2C+k%29%7D%5Cright%29%5Cleft%281+%2B+%5Cfrac%7Bx%7D%7B%5Ctext%7Bsn%7D%28K+-+8%5Comega%2C+k%29%7D%5Cright%29%5Ccdots%5Cleft%281+%2B+%5Cfrac%7Bx%7D%7B%5Ctext%7Bsn%7D%28K+-+2%28p+-+1%29%5Comega%2C+k%29%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle A = &#92;left(1 + &#92;frac{x}{&#92;text{sn}(K - 4&#92;omega, k)}&#92;right)&#92;left(1 + &#92;frac{x}{&#92;text{sn}(K - 8&#92;omega, k)}&#92;right)&#92;cdots&#92;left(1 + &#92;frac{x}{&#92;text{sn}(K - 2(p - 1)&#92;omega, k)}&#92;right)' title='&#92;displaystyle A = &#92;left(1 + &#92;frac{x}{&#92;text{sn}(K - 4&#92;omega, k)}&#92;right)&#92;left(1 + &#92;frac{x}{&#92;text{sn}(K - 8&#92;omega, k)}&#92;right)&#92;cdots&#92;left(1 + &#92;frac{x}{&#92;text{sn}(K - 2(p - 1)&#92;omega, k)}&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%281+%2B+%5Cfrac%7Bx%7D%7B%5Ctext%7Bsn%7D%28K+-+4s%5Comega%2C+k%29%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(1 + &#92;frac{x}{&#92;text{sn}(K - 4s&#92;omega, k)}&#92;right)' title='&#92;displaystyle = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(1 + &#92;frac{x}{&#92;text{sn}(K - 4s&#92;omega, k)}&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+B+%3D+%5Cleft%281+-+%5Cfrac%7Bx%7D%7B%5Ctext%7Bsn%7D%28K+-+4%5Comega%2C+k%29%7D%5Cright%29%5Cleft%281+-+%5Cfrac%7Bx%7D%7B%5Ctext%7Bsn%7D%28K+-+8%5Comega%2C+k%29%7D%5Cright%29%5Ccdots%5Cleft%281+-+%5Cfrac%7Bx%7D%7B%5Ctext%7Bsn%7D%28K+-+2%28p+-+1%29%5Comega%2C+k%29%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle B = &#92;left(1 - &#92;frac{x}{&#92;text{sn}(K - 4&#92;omega, k)}&#92;right)&#92;left(1 - &#92;frac{x}{&#92;text{sn}(K - 8&#92;omega, k)}&#92;right)&#92;cdots&#92;left(1 - &#92;frac{x}{&#92;text{sn}(K - 2(p - 1)&#92;omega, k)}&#92;right)' title='&#92;displaystyle B = &#92;left(1 - &#92;frac{x}{&#92;text{sn}(K - 4&#92;omega, k)}&#92;right)&#92;left(1 - &#92;frac{x}{&#92;text{sn}(K - 8&#92;omega, k)}&#92;right)&#92;cdots&#92;left(1 - &#92;frac{x}{&#92;text{sn}(K - 2(p - 1)&#92;omega, k)}&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%281+-+%5Cfrac%7Bx%7D%7B%5Ctext%7Bsn%7D%28K+-+4s%5Comega%2C+k%29%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(1 - &#92;frac{x}{&#92;text{sn}(K - 4s&#92;omega, k)}&#92;right)' title='&#92;displaystyle = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(1 - &#92;frac{x}{&#92;text{sn}(K - 4s&#92;omega, k)}&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+C+%3D+%281+%2B+kx%5C%2C%5Ctext%7Bsn%7D%28K+-+4%5Comega%2C+k%29%29%281+%2B+kx%5C%2C%5Ctext%7Bsn%7D%28K+-+8%5Comega%2C+k%29%29%5Ccdots%281+%2B+kx%5C%2C%5Ctext%7Bsn%7D%28K+-+2%28p+-+1%29%5Comega%2C+k%29%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle C = (1 + kx&#92;,&#92;text{sn}(K - 4&#92;omega, k))(1 + kx&#92;,&#92;text{sn}(K - 8&#92;omega, k))&#92;cdots(1 + kx&#92;,&#92;text{sn}(K - 2(p - 1)&#92;omega, k))' title='&#92;displaystyle C = (1 + kx&#92;,&#92;text{sn}(K - 4&#92;omega, k))(1 + kx&#92;,&#92;text{sn}(K - 8&#92;omega, k))&#92;cdots(1 + kx&#92;,&#92;text{sn}(K - 2(p - 1)&#92;omega, k))' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%281+%2B+kx%5C%2C%5Ctext%7Bsn%7D%28K+-+4s%5Comega%2C+k%29%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = &#92;prod_{s = 1}^{(p - 1)/2}(1 + kx&#92;,&#92;text{sn}(K - 4s&#92;omega, k))' title='&#92;displaystyle = &#92;prod_{s = 1}^{(p - 1)/2}(1 + kx&#92;,&#92;text{sn}(K - 4s&#92;omega, k))' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+D+%3D+%281+-+kx%5C%2C%5Ctext%7Bsn%7D%28K+-+4%5Comega%2C+k%29%29%281+-+kx%5C%2C%5Ctext%7Bsn%7D%28K+-+8%5Comega%2C+k%29%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle D = (1 - kx&#92;,&#92;text{sn}(K - 4&#92;omega, k))(1 - kx&#92;,&#92;text{sn}(K - 8&#92;omega, k))' title='&#92;displaystyle D = (1 - kx&#92;,&#92;text{sn}(K - 4&#92;omega, k))(1 - kx&#92;,&#92;text{sn}(K - 8&#92;omega, k))' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ccdots%281+-+kx%5C%2C%5Ctext%7Bsn%7D%28K+-+2%28p+-+1%29%5Comega%2C+k%29%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;cdots(1 - kx&#92;,&#92;text{sn}(K - 2(p - 1)&#92;omega, k))' title='&#92;displaystyle &#92;cdots(1 - kx&#92;,&#92;text{sn}(K - 2(p - 1)&#92;omega, k))' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%281+-+kx%5C%2C%5Ctext%7Bsn%7D%28K+-+4s%5Comega%2C+k%29%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = &#92;prod_{s = 1}^{(p - 1)/2}(1 - kx&#92;,&#92;text{sn}(K - 4s&#92;omega, k))' title='&#92;displaystyle = &#92;prod_{s = 1}^{(p - 1)/2}(1 - kx&#92;,&#92;text{sn}(K - 4s&#92;omega, k))' class='latex' /></p>
<p>The invariance of relation between <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> under the transformation <img src='http://s0.wp.com/latex.php?latex=%28x%2C+y%29+%5Cto+%281%2Fkx%2C+1%2Fly%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(x, y) &#92;to (1/kx, 1/ly)' title='(x, y) &#92;to (1/kx, 1/ly)' class='latex' /> is satisfied if we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l+%3D+k%5E%7Bp%7D%5C%7B%5Ctext%7Bsn%7D%28K+-+4%5Comega%2C+k%29%5C%2C%5Ctext%7Bsn%7D%28K+-+8%5Comega%2C+k%29%5Ccdots%5C%2C%5Ctext%7Bsn%7D%28K+-+2%28p+-1%29%5Comega%2C+k%29%5C%7D%5E%7B4%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l = k^{p}&#92;{&#92;text{sn}(K - 4&#92;omega, k)&#92;,&#92;text{sn}(K - 8&#92;omega, k)&#92;cdots&#92;,&#92;text{sn}(K - 2(p -1)&#92;omega, k)&#92;}^{4}' title='&#92;displaystyle l = k^{p}&#92;{&#92;text{sn}(K - 4&#92;omega, k)&#92;,&#92;text{sn}(K - 8&#92;omega, k)&#92;cdots&#92;,&#92;text{sn}(K - 2(p -1)&#92;omega, k)&#92;}^{4}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+k%5E%7Bp%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%28%5Ctext%7Bsn%7D%28K+-+4s%5Comega%2C+k%29%29%5E%7B4%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = k^{p}&#92;prod_{s = 1}^{(p - 1)/2}(&#92;text{sn}(K - 4s&#92;omega, k))^{4}' title='&#92;displaystyle = k^{p}&#92;prod_{s = 1}^{(p - 1)/2}(&#92;text{sn}(K - 4s&#92;omega, k))^{4}' class='latex' /></p>
<p>and</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+M+%3D+%28-1%29%5E%7B%28p+-1%29%2F2%7D%5Cleft%5C%7B%5Cfrac%7B%5Ctext%7Bsn%7D%28K+-+4%5Comega%2C+k%29%7D%7B%5Ctext%7Bsn%7D%284%5Comega%2C+k%29%7D%5C%2C%5Cfrac%7B%5Ctext%7Bsn%7D%28K+-+8%5Comega%2C+k%29%7D%7B%5Ctext%7Bsn%7D%288%5Comega%2C+k%29%7D%5C%2C%5Ccdots%5C%2C%5Cfrac%7B%5Ctext%7Bsn%7D%28K+-+2%28p+-+1%29%5Comega%2C+k%29%7D%7B%5Ctext%7Bsn%7D%282%28p+-+1%29%5Comega%2C+k%29%7D%5Cright%5C%7D%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle M = (-1)^{(p -1)/2}&#92;left&#92;{&#92;frac{&#92;text{sn}(K - 4&#92;omega, k)}{&#92;text{sn}(4&#92;omega, k)}&#92;,&#92;frac{&#92;text{sn}(K - 8&#92;omega, k)}{&#92;text{sn}(8&#92;omega, k)}&#92;,&#92;cdots&#92;,&#92;frac{&#92;text{sn}(K - 2(p - 1)&#92;omega, k)}{&#92;text{sn}(2(p - 1)&#92;omega, k)}&#92;right&#92;}^{2}' title='&#92;displaystyle M = (-1)^{(p -1)/2}&#92;left&#92;{&#92;frac{&#92;text{sn}(K - 4&#92;omega, k)}{&#92;text{sn}(4&#92;omega, k)}&#92;,&#92;frac{&#92;text{sn}(K - 8&#92;omega, k)}{&#92;text{sn}(8&#92;omega, k)}&#92;,&#92;cdots&#92;,&#92;frac{&#92;text{sn}(K - 2(p - 1)&#92;omega, k)}{&#92;text{sn}(2(p - 1)&#92;omega, k)}&#92;right&#92;}^{2}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%28-1%29%5E%7B%28p+-+1%29%2F2%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cfrac%7B%5Ctext%7Bsn%7D%28K+-+4s%5Comega%2C+k%29%7D%7B%5Ctext%7Bsn%7D%284s%5Comega%2C+k%29%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = (-1)^{(p - 1)/2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{&#92;text{sn}(K - 4s&#92;omega, k)}{&#92;text{sn}(4s&#92;omega, k)}&#92;right)^{2}' title='&#92;displaystyle = (-1)^{(p - 1)/2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{&#92;text{sn}(K - 4s&#92;omega, k)}{&#92;text{sn}(4s&#92;omega, k)}&#92;right)^{2}' class='latex' /></p>
<p>In order to prove that the above mentioned transformation indeed has all the properties desired takes some reasonable amount of algebraic manipulations and some analysis of the various equations involved. In what follows elliptic functions will be used heavily and therefore we will drop the modulus <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' />. If any other modulus is used for these elliptic elliptic functions it will written explicitly.</p>
<p>We start with the equation</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1+-+y%7D%7B1+%2B+y%7D+%3D+%5Cfrac%7B1+-+x%7D%7B1+%2B+x%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cdfrac%7B1+-+%5Cdfrac%7Bx%7D%7B%5Ctext%7Bsn%7D%28K+-+4s%5Comega%29%7D%7D%7B1+%2B+%5Cdfrac%7Bx%7D%7B%5Ctext%7Bsn%7D%28K+-+4s%5Comega%29%7D%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{1 - y}{1 + y} = &#92;frac{1 - x}{1 + x}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{1 - &#92;dfrac{x}{&#92;text{sn}(K - 4s&#92;omega)}}{1 + &#92;dfrac{x}{&#92;text{sn}(K - 4s&#92;omega)}}&#92;right)^{2}' title='&#92;displaystyle &#92;frac{1 - y}{1 + y} = &#92;frac{1 - x}{1 + x}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{1 - &#92;dfrac{x}{&#92;text{sn}(K - 4s&#92;omega)}}{1 + &#92;dfrac{x}{&#92;text{sn}(K - 4s&#92;omega)}}&#92;right)^{2}' class='latex' /></p>
<p>and will show that this leads to all the other equations. Moreover from the form of the equations given earlier it follows that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7BMdy%7D%7B%5Csqrt%7B%281+-+y%5E%7B2%7D%29%281+-+l%5E%7B2%7Dy%5E%7B2%7D%29%7D%7D+%3D+%5Cfrac%7Bdx%7D%7B%5Csqrt%7B%281+-+x%5E%7B2%7D%29%281+-+k%5E%7B2%7Dx%5E%7B2%7D%29%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{Mdy}{&#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = &#92;frac{dx}{&#92;sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}' title='&#92;displaystyle &#92;frac{Mdy}{&#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = &#92;frac{dx}{&#92;sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}' class='latex' /></p>
<p>Upon solving for <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> we will obtain an expression <img src='http://s0.wp.com/latex.php?latex=y+%3D+xN%281%2C+x%5E%7B2%7D%29%2FD%281%2C+x%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = xN(1, x^{2})/D(1, x^{2})' title='y = xN(1, x^{2})/D(1, x^{2})' class='latex' />, where <img src='http://s0.wp.com/latex.php?latex=N%2C+D&amp;bg=fff&amp;fg=222&amp;s=0' alt='N, D' title='N, D' class='latex' /> are homogeneous polynomials of degree <img src='http://s0.wp.com/latex.php?latex=%28p+-+1%29%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='(p - 1)/2' title='(p - 1)/2' class='latex' /> (this is because <img src='http://s0.wp.com/latex.php?latex=x%2C+y&amp;bg=fff&amp;fg=222&amp;s=0' alt='x, y' title='x, y' class='latex' /> change their sign together and thus <img src='http://s0.wp.com/latex.php?latex=y%2Fx&amp;bg=fff&amp;fg=222&amp;s=0' alt='y/x' title='y/x' class='latex' /> is a rational and even function).</p>
<p>Thus if we are able to show that <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> vanishes for <img src='http://s0.wp.com/latex.php?latex=x+%3D+%5Cpm%5C%2C%5Ctext%7Bsn%7D%5C%2C4t%5Comega%2C+t+%3D+0%2C+1%2C+2%2C%5Cldots%2C+%28p+-+1%29%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = &#92;pm&#92;,&#92;text{sn}&#92;,4t&#92;omega, t = 0, 1, 2,&#92;ldots, (p - 1)/2' title='x = &#92;pm&#92;,&#92;text{sn}&#92;,4t&#92;omega, t = 0, 1, 2,&#92;ldots, (p - 1)/2' class='latex' />and that <img src='http://s0.wp.com/latex.php?latex=y+%5Cto+%5Cpm%5Cinfty&amp;bg=fff&amp;fg=222&amp;s=0' alt='y &#92;to &#92;pm&#92;infty' title='y &#92;to &#92;pm&#92;infty' class='latex' /> as <img src='http://s0.wp.com/latex.php?latex=x+%5Cto+%5Cpm+1%2F%28k%5C%2C%5Ctext%7Bsn%7D%5C%2C4t%5Comega%29%2C+t+%3D1%2C+2%2C+%5Cldots%2C+%28p+-+1%29%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='x &#92;to &#92;pm 1/(k&#92;,&#92;text{sn}&#92;,4t&#92;omega), t =1, 2, &#92;ldots, (p - 1)/2' title='x &#92;to &#92;pm 1/(k&#92;,&#92;text{sn}&#92;,4t&#92;omega), t =1, 2, &#92;ldots, (p - 1)/2' class='latex' /> then it will be obvious that <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> has to be of the form:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+y+%3D+%5Cfrac%7Bx%7D%7BC%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cdfrac%7B1+-+%5Cdfrac%7Bx%5E%7B2%7D%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D4s%5Comega%7D%7D%7B1+-+k%5E%7B2%7Dx%5E%7B2%7D%5Ctext%7Bsn%7D%5E%7B2%7D4s%5Comega%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle y = &#92;frac{x}{C}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{1 - &#92;dfrac{x^{2}}{&#92;text{sn}^{2}4s&#92;omega}}{1 - k^{2}x^{2}&#92;text{sn}^{2}4s&#92;omega}&#92;right)' title='&#92;displaystyle y = &#92;frac{x}{C}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{1 - &#92;dfrac{x^{2}}{&#92;text{sn}^{2}4s&#92;omega}}{1 - k^{2}x^{2}&#92;text{sn}^{2}4s&#92;omega}&#92;right)' class='latex' /></p>
<p>Also from the relation between <img src='http://s0.wp.com/latex.php?latex=x%2C+y&amp;bg=fff&amp;fg=222&amp;s=0' alt='x, y' title='x, y' class='latex' /> it is clear that <img src='http://s0.wp.com/latex.php?latex=y+%3D+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = 1' title='y = 1' class='latex' /> when <img src='http://s0.wp.com/latex.php?latex=x+%3D+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = 1' title='x = 1' class='latex' /> and therefore we must have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+C+%3D+%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cdfrac%7B1+-+%5Cdfrac%7B1%7D%7B%5Ctext%7Bsn%7D%5E%7B2%7D4s%5Comega%7D%7D%7B1+-+k%5E%7B2%7D%5Ctext%7Bsn%7D%5E%7B2%7D4s%5Comega%7D%5Cright%29+%3D+%28-1%29%5E%7B%28p+-+1%29%2F2%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cfrac%7B%5Ctext%7Bcn%7D%5C%2C4s%5Comega%7D%7B%5Ctext%7Bsn%7D%5C%2C4s%5Comega%5C%2C%5Ctext%7Bdn%7D%5C%2C4s%5Comega%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle C = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{1 - &#92;dfrac{1}{&#92;text{sn}^{2}4s&#92;omega}}{1 - k^{2}&#92;text{sn}^{2}4s&#92;omega}&#92;right) = (-1)^{(p - 1)/2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{&#92;text{cn}&#92;,4s&#92;omega}{&#92;text{sn}&#92;,4s&#92;omega&#92;,&#92;text{dn}&#92;,4s&#92;omega}&#92;right)^{2}' title='&#92;displaystyle C = &#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{1 - &#92;dfrac{1}{&#92;text{sn}^{2}4s&#92;omega}}{1 - k^{2}&#92;text{sn}^{2}4s&#92;omega}&#92;right) = (-1)^{(p - 1)/2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{&#92;text{cn}&#92;,4s&#92;omega}{&#92;text{sn}&#92;,4s&#92;omega&#92;,&#92;text{dn}&#92;,4s&#92;omega}&#92;right)^{2}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%28-1%29%5E%7B%28p+-+1%29%2F2%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cfrac%7B%5Ctext%7Bsn%7D%28K+-+4s%5Comega%29%7D%7B%5Ctext%7Bsn%7D%5C%2C4s%5Comega%7D%5Cright%29%5E%7B2%7D+%3D+M&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = (-1)^{(p - 1)/2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{&#92;text{sn}(K - 4s&#92;omega)}{&#92;text{sn}&#92;,4s&#92;omega}&#92;right)^{2} = M' title='&#92;displaystyle = (-1)^{(p - 1)/2}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;frac{&#92;text{sn}(K - 4s&#92;omega)}{&#92;text{sn}&#92;,4s&#92;omega}&#92;right)^{2} = M' class='latex' /></p>
<p>With this expression of <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> we at once see that <img src='http://s0.wp.com/latex.php?latex=y+%3D+U%2FV&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = U/V' title='y = U/V' class='latex' /> where <img src='http://s0.wp.com/latex.php?latex=U%2C+V&amp;bg=fff&amp;fg=222&amp;s=0' alt='U, V' title='U, V' class='latex' /> are as mentioned above. Also if assume the invariance of this relation under the transformation <img src='http://s0.wp.com/latex.php?latex=%28x%2C+y%29+%5Cto+%281%2Fkx%2C+1%2Fly%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(x, y) &#92;to (1/kx, 1/ly)' title='(x, y) &#92;to (1/kx, 1/ly)' class='latex' /> we at once arrive at the value of <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> described above. Again from equation <img src='http://s0.wp.com/latex.php?latex=%281+-+y%29%2F%281+%2B+y%29+%3D+%28%281+-+x%29B%5E%7B2%7D%29%2F%28%281+%2B+x%29A%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(1 - y)/(1 + y) = ((1 - x)B^{2})/((1 + x)A^{2})' title='(1 - y)/(1 + y) = ((1 - x)B^{2})/((1 + x)A^{2})' class='latex' /> we obtain</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+y+%3D+%5Cfrac%7BU%7D%7BV%7D+%3D+%5Cfrac%7B%281+%2B+x%29A%5E%7B2%7D+-+%281+-+x%29B%5E%7B2%7D%7D%7B%281+%2B+x%29A%5E%7B2%7D+%2B+%281+-+x%29B%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle y = &#92;frac{U}{V} = &#92;frac{(1 + x)A^{2} - (1 - x)B^{2}}{(1 + x)A^{2} + (1 - x)B^{2}}' title='&#92;displaystyle y = &#92;frac{U}{V} = &#92;frac{(1 + x)A^{2} - (1 - x)B^{2}}{(1 + x)A^{2} + (1 - x)B^{2}}' class='latex' /></p>
<p>Looking at the degrees of <img src='http://s0.wp.com/latex.php?latex=V&amp;bg=fff&amp;fg=222&amp;s=0' alt='V' title='V' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%281+%2B+x%29A%5E%7B2%7D+%2B+%281+-+x%29B%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='(1 + x)A^{2} + (1 - x)B^{2}' title='(1 + x)A^{2} + (1 - x)B^{2}' class='latex' /> we can see that <img src='http://s0.wp.com/latex.php?latex=V&amp;bg=fff&amp;fg=222&amp;s=0' alt='V' title='V' class='latex' /> must be a constant multiple of <img src='http://s0.wp.com/latex.php?latex=%281+%2B+x%29A%5E%7B2%7D+%2B+%281+-+x%29B%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='(1 + x)A^{2} + (1 - x)B^{2}' title='(1 + x)A^{2} + (1 - x)B^{2}' class='latex' /> and on putting <img src='http://s0.wp.com/latex.php?latex=x+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = 0' title='x = 0' class='latex' /> we see that <img src='http://s0.wp.com/latex.php?latex=V+%3D+%281%2F2%29%28%281+%2B+x%29A%5E%7B2%7D+%2B+%281+-+x%29B%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='V = (1/2)((1 + x)A^{2} + (1 - x)B^{2})' title='V = (1/2)((1 + x)A^{2} + (1 - x)B^{2})' class='latex' /> and so <img src='http://s0.wp.com/latex.php?latex=U+%3D+%281%2F2%29%28%281+%2B+x%29A%5E%7B2%7D+-+%281+-+x%29B%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='U = (1/2)((1 + x)A^{2} - (1 - x)B^{2})' title='U = (1/2)((1 + x)A^{2} - (1 - x)B^{2})' class='latex' /> and hence it follows that <img src='http://s0.wp.com/latex.php?latex=1+%2B+y+%3D+%28V+%2B+U%29%2FV+%3D+%28%281+%2B+x%29A%5E%7B2%7D%29%2FV&amp;bg=fff&amp;fg=222&amp;s=0' alt='1 + y = (V + U)/V = ((1 + x)A^{2})/V' title='1 + y = (V + U)/V = ((1 + x)A^{2})/V' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=1+-+y+%3D+%28V+-+U%29%2FV+%3D+%28%281+-+x%29B%5E%7B2%7D%29%2FV&amp;bg=fff&amp;fg=222&amp;s=0' alt='1 - y = (V - U)/V = ((1 - x)B^{2})/V' title='1 - y = (V - U)/V = ((1 - x)B^{2})/V' class='latex' />. Thus the expressions of <img src='http://s0.wp.com/latex.php?latex=1+-+y%2C+1+%2B+y&amp;bg=fff&amp;fg=222&amp;s=0' alt='1 - y, 1 + y' title='1 - y, 1 + y' class='latex' /> are also verified. Using the invariance under <img src='http://s0.wp.com/latex.php?latex=%28x%2C+y%29+%5Cto+%281%2Fkx%2C+1%2Fly%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(x, y) &#92;to (1/kx, 1/ly)' title='(x, y) &#92;to (1/kx, 1/ly)' class='latex' /> we can then verify the expressions for <img src='http://s0.wp.com/latex.php?latex=1+-+ly%2C+1+%2B+ly&amp;bg=fff&amp;fg=222&amp;s=0' alt='1 - ly, 1 + ly' title='1 - ly, 1 + ly' class='latex' />.</p>
<p>Hence all the above expressions relating <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> are verified provided we establish that zeros and poles (points where a function tends to <img src='http://s0.wp.com/latex.php?latex=%5Cpm%5Cinfty&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;pm&#92;infty' title='&#92;pm&#92;infty' class='latex' />) of <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> are as described above. To that end we put <img src='http://s0.wp.com/latex.php?latex=x+%3D+%5Ctext%7Bsn%7D%5C%2Cu&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = &#92;text{sn}&#92;,u' title='x = &#92;text{sn}&#92;,u' class='latex' /> in the expession for <img src='http://s0.wp.com/latex.php?latex=%281+-+y%29%2F%281+%2B+y%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(1 - y)/(1 + y)' title='(1 - y)/(1 + y)' class='latex' /> and obtain</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1+-+y%7D%7B1+%2B+y%7D+%3D+%5Cfrac%7B1+-+%5Ctext%7Bsn%7D%5C%2Cu%7D%7B1+%2B+%5Ctext%7Bsn%7D%5C%2Cu%7D%5Cprod_%7Bs+%3D+1%7D%5E%7B%28p+-+1%29%2F2%7D%5Cleft%28%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%28K+-+4s%5Comega%29%7D%7D%7B1+%2B+%5Cdfrac%7B%5Ctext%7Bsn%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%28K+-+4s%5Comega%29%7D%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{1 - y}{1 + y} = &#92;frac{1 - &#92;text{sn}&#92;,u}{1 + &#92;text{sn}&#92;,u}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}&#92;,u}{&#92;text{sn}(K - 4s&#92;omega)}}{1 + &#92;dfrac{&#92;text{sn}&#92;,u}{&#92;text{sn}(K - 4s&#92;omega)}}&#92;right)^{2}' title='&#92;displaystyle &#92;frac{1 - y}{1 + y} = &#92;frac{1 - &#92;text{sn}&#92;,u}{1 + &#92;text{sn}&#92;,u}&#92;prod_{s = 1}^{(p - 1)/2}&#92;left(&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}&#92;,u}{&#92;text{sn}(K - 4s&#92;omega)}}{1 + &#92;dfrac{&#92;text{sn}&#92;,u}{&#92;text{sn}(K - 4s&#92;omega)}}&#92;right)^{2}' class='latex' /></p>
<p>Now we require some identities related to the elliptic functions which can be verified at once using addition formulas or otherwise:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5C%7B%281+%2B+%5Ctext%7Bsn%7D%28u+%2B+v%29%5C%7D%5C%7B1+%2B+%5Ctext%7Bsn%7D%28u+-+v%29%5C%7D+%3D+%5Cdfrac%7B%5Ctext%7Bcn%7D%5E%7B2%7Dv%5Cleft%281+%2B+%5Cdfrac%7B%5Ctext%7Bsn%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%28K+-+v%29%7D%5Cright%29%5E%7B2%7D%7D%7B1+-+k%5E%7B2%7D%5Ctext%7Bsn%7D%5E%7B2%7Du%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7Dv%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;{(1 + &#92;text{sn}(u + v)&#92;}&#92;{1 + &#92;text{sn}(u - v)&#92;} = &#92;dfrac{&#92;text{cn}^{2}v&#92;left(1 + &#92;dfrac{&#92;text{sn}&#92;,u}{&#92;text{sn}(K - v)}&#92;right)^{2}}{1 - k^{2}&#92;text{sn}^{2}u&#92;,&#92;text{sn}^{2}v}' title='&#92;displaystyle &#92;{(1 + &#92;text{sn}(u + v)&#92;}&#92;{1 + &#92;text{sn}(u - v)&#92;} = &#92;dfrac{&#92;text{cn}^{2}v&#92;left(1 + &#92;dfrac{&#92;text{sn}&#92;,u}{&#92;text{sn}(K - v)}&#92;right)^{2}}{1 - k^{2}&#92;text{sn}^{2}u&#92;,&#92;text{sn}^{2}v}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5C%7B%281+-+%5Ctext%7Bsn%7D%28u+%2B+v%29%5C%7D%5C%7B1+-+%5Ctext%7Bsn%7D%28u+-+v%29%5C%7D+%3D+%5Cdfrac%7B%5Ctext%7Bcn%7D%5E%7B2%7Dv%5Cleft%281+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%28K+-+v%29%7D%5Cright%29%5E%7B2%7D%7D%7B1+-+k%5E%7B2%7D%5Ctext%7Bsn%7D%5E%7B2%7Du%5C%2C%5Ctext%7Bsn%7D%5E%7B2%7Dv%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;{(1 - &#92;text{sn}(u + v)&#92;}&#92;{1 - &#92;text{sn}(u - v)&#92;} = &#92;dfrac{&#92;text{cn}^{2}v&#92;left(1 - &#92;dfrac{&#92;text{sn}&#92;,u}{&#92;text{sn}(K - v)}&#92;right)^{2}}{1 - k^{2}&#92;text{sn}^{2}u&#92;,&#92;text{sn}^{2}v}' title='&#92;displaystyle &#92;{(1 - &#92;text{sn}(u + v)&#92;}&#92;{1 - &#92;text{sn}(u - v)&#92;} = &#92;dfrac{&#92;text{cn}^{2}v&#92;left(1 - &#92;dfrac{&#92;text{sn}&#92;,u}{&#92;text{sn}(K - v)}&#92;right)^{2}}{1 - k^{2}&#92;text{sn}^{2}u&#92;,&#92;text{sn}^{2}v}' class='latex' /></p>
<p>and therefore</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B%5C%7B%281+-+%5Ctext%7Bsn%7D%28u+%2B+v%29%5C%7D%5C%7B1+-+%5Ctext%7Bsn%7D%28u+-+v%29%5C%7D%7D%7B%5C%7B%281+%2B+%5Ctext%7Bsn%7D%28u+%2B+v%29%5C%7D%5C%7B1+%2B+%5Ctext%7Bsn%7D%28u+-+v%29%5C%7D%7D+%3D+%5Cleft%28%5Cdfrac%7B1+-+%5Cdfrac%7B%5Ctext%7Bsn%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%28K+-+v%29%7D%7D%7B1+%2B+%5Cdfrac%7B%5Ctext%7Bsn%7D%5C%2Cu%7D%7B%5Ctext%7Bsn%7D%28K+-+v%29%7D%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{&#92;{(1 - &#92;text{sn}(u + v)&#92;}&#92;{1 - &#92;text{sn}(u - v)&#92;}}{&#92;{(1 + &#92;text{sn}(u + v)&#92;}&#92;{1 + &#92;text{sn}(u - v)&#92;}} = &#92;left(&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}&#92;,u}{&#92;text{sn}(K - v)}}{1 + &#92;dfrac{&#92;text{sn}&#92;,u}{&#92;text{sn}(K - v)}}&#92;right)^{2}' title='&#92;displaystyle &#92;frac{&#92;{(1 - &#92;text{sn}(u + v)&#92;}&#92;{1 - &#92;text{sn}(u - v)&#92;}}{&#92;{(1 + &#92;text{sn}(u + v)&#92;}&#92;{1 + &#92;text{sn}(u - v)&#92;}} = &#92;left(&#92;dfrac{1 - &#92;dfrac{&#92;text{sn}&#92;,u}{&#92;text{sn}(K - v)}}{1 + &#92;dfrac{&#92;text{sn}&#92;,u}{&#92;text{sn}(K - v)}}&#92;right)^{2}' class='latex' /></p>
<p>In the above relation we put <img src='http://s0.wp.com/latex.php?latex=v+%3D+4%5Comega%2C+8%5Comega%2C+%5Cldots%2C+2%28p+-+1%29%5Comega&amp;bg=fff&amp;fg=222&amp;s=0' alt='v = 4&#92;omega, 8&#92;omega, &#92;ldots, 2(p - 1)&#92;omega' title='v = 4&#92;omega, 8&#92;omega, &#92;ldots, 2(p - 1)&#92;omega' class='latex' /> successively and multiply the resulting equations and further multiply both sides by <img src='http://s0.wp.com/latex.php?latex=%281+-+%5Ctext%7Bsn%7D%5C%2Cu%29%2F%281+%2B+%5Ctext%7Bsn%7D%5C%2Cu%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(1 - &#92;text{sn}&#92;,u)/(1 + &#92;text{sn}&#92;,u)' title='(1 - &#92;text{sn}&#92;,u)/(1 + &#92;text{sn}&#92;,u)' class='latex' />. Also while doing so we need to observe that <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsn%7D%28u+-+4s%5Comega%29+%3D+%5Ctext%7Bsn%7D%28u+%2B+4%28p+-+s%29%5Comega%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{sn}(u - 4s&#92;omega) = &#92;text{sn}(u + 4(p - s)&#92;omega)' title='&#92;text{sn}(u - 4s&#92;omega) = &#92;text{sn}(u + 4(p - s)&#92;omega)' class='latex' />. Thus we obtain the following:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1+-+y%7D%7B1+%2B+y%7D+%3D+%5Cprod_%7Bs+%3D+0%7D%5E%7Bp+-+1%7D%5Cfrac%7B1+-+%5Ctext%7Bsn%7D%28u+%2B+4s%5Comega%29%7D%7B1+%2B+%5Ctext%7Bsn%7D%28u+%2B+4s%5Comega%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{1 - y}{1 + y} = &#92;prod_{s = 0}^{p - 1}&#92;frac{1 - &#92;text{sn}(u + 4s&#92;omega)}{1 + &#92;text{sn}(u + 4s&#92;omega)}' title='&#92;displaystyle &#92;frac{1 - y}{1 + y} = &#92;prod_{s = 0}^{p - 1}&#92;frac{1 - &#92;text{sn}(u + 4s&#92;omega)}{1 + &#92;text{sn}(u + 4s&#92;omega)}' class='latex' /></p>
<p>From the above relation it is easy to see that the <img src='http://s0.wp.com/latex.php?latex=%281+-+y%29%2F%281+%2B+y%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(1 - y)/(1 + y)' title='(1 - y)/(1 + y)' class='latex' /> and hence <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> is invariant when <img src='http://s0.wp.com/latex.php?latex=u&amp;bg=fff&amp;fg=222&amp;s=0' alt='u' title='u' class='latex' /> is replaced <img src='http://s0.wp.com/latex.php?latex=u+%2B+4%5Comega&amp;bg=fff&amp;fg=222&amp;s=0' alt='u + 4&#92;omega' title='u + 4&#92;omega' class='latex' /> (in doing so each factor is changed into next factor in the product and the last factor is changed into the first so that the product itself remains invariant). If we put <img src='http://s0.wp.com/latex.php?latex=u+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = 0' title='u = 0' class='latex' /> we have <img src='http://s0.wp.com/latex.php?latex=x+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = 0' title='x = 0' class='latex' /> and hence <img src='http://s0.wp.com/latex.php?latex=y+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = 0' title='y = 0' class='latex' />. Therefore <img src='http://s0.wp.com/latex.php?latex=y+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = 0' title='y = 0' class='latex' /> whenenver <img src='http://s0.wp.com/latex.php?latex=u+%3D+0%2C+4%5Comega%2C+8%5Comega%2C+%5Cldots%2C+4%28p+-+1%29%5Comega&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = 0, 4&#92;omega, 8&#92;omega, &#92;ldots, 4(p - 1)&#92;omega' title='u = 0, 4&#92;omega, 8&#92;omega, &#92;ldots, 4(p - 1)&#92;omega' class='latex' /> i.e. whenever <img src='http://s0.wp.com/latex.php?latex=x+%3D+%5Ctext%7Bsn%7D%5C%2C4t%5Comega%2C+t+%3D+0%2C+1%2C+2%2C+%5Cldots%2C+%28p+-+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = &#92;text{sn}&#92;,4t&#92;omega, t = 0, 1, 2, &#92;ldots, (p - 1)' title='x = &#92;text{sn}&#92;,4t&#92;omega, t = 0, 1, 2, &#92;ldots, (p - 1)' class='latex' />. Since <img src='http://s0.wp.com/latex.php?latex=%5Ctext%7Bsn%7D%5C%2C4%28p+-+t%29%5Comega+%3D+-%5Ctext%7Bsn%7D%5C%2C4t%5Comega&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;text{sn}&#92;,4(p - t)&#92;omega = -&#92;text{sn}&#92;,4t&#92;omega' title='&#92;text{sn}&#92;,4(p - t)&#92;omega = -&#92;text{sn}&#92;,4t&#92;omega' class='latex' /> it follows that <img src='http://s0.wp.com/latex.php?latex=y+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = 0' title='y = 0' class='latex' /> whenever <img src='http://s0.wp.com/latex.php?latex=x+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = 0' title='x = 0' class='latex' /> or <img src='http://s0.wp.com/latex.php?latex=x+%3D+%5Cpm%5C%2C%5Ctext%7Bsn%7D%5C%2C4t%5Comega%2C+t+%3D+0%2C+1%2C+2%2C+%5Cldots%2C+%28p+-+1%29%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = &#92;pm&#92;,&#92;text{sn}&#92;,4t&#92;omega, t = 0, 1, 2, &#92;ldots, (p - 1)/2' title='x = &#92;pm&#92;,&#92;text{sn}&#92;,4t&#92;omega, t = 0, 1, 2, &#92;ldots, (p - 1)/2' class='latex' />. Thus the zeroes of <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> are verified.</p>
<p>It is easily seen that <img src='http://s0.wp.com/latex.php?latex=u+%3D+iK%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = iK&#039;' title='u = iK&#039;' class='latex' /> is a pole for <img src='http://s0.wp.com/latex.php?latex=x+%3D+%5Ctext%7Bsn%7D%5C%2Cu&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = &#92;text{sn}&#92;,u' title='x = &#92;text{sn}&#92;,u' class='latex' /> and hence a pole for <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> (clearly <img src='http://s0.wp.com/latex.php?latex=y+%5Cto+%5Cinfty&amp;bg=fff&amp;fg=222&amp;s=0' alt='y &#92;to &#92;infty' title='y &#92;to &#92;infty' class='latex' /> as <img src='http://s0.wp.com/latex.php?latex=x+%5Cto+%5Cinfty&amp;bg=fff&amp;fg=222&amp;s=0' alt='x &#92;to &#92;infty' title='x &#92;to &#92;infty' class='latex' />). Thus <img src='http://s0.wp.com/latex.php?latex=y+%5Cto+%5Cinfty&amp;bg=fff&amp;fg=222&amp;s=0' alt='y &#92;to &#92;infty' title='y &#92;to &#92;infty' class='latex' /> whenever <img src='http://s0.wp.com/latex.php?latex=x+%5Cto+%5Ctext%7Bsn%7D%28iK%27+%2B+4t%5Comega%29%2C+t+%3D+1%2C+2%2C+%5Cldots%2C+%28p+-+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='x &#92;to &#92;text{sn}(iK&#039; + 4t&#92;omega), t = 1, 2, &#92;ldots, (p - 1)' title='x &#92;to &#92;text{sn}(iK&#039; + 4t&#92;omega), t = 1, 2, &#92;ldots, (p - 1)' class='latex' /> i.e whenever <img src='http://s0.wp.com/latex.php?latex=x+%5Cto+%5Ctext%7Bsn%7D%28iK%27+%5Cpm+4t%5Comega%29+%3D+%5Cpm+1%2F%28k%5C%2C%5Ctext%7Bsn%7D%5C%2C4t%5Comega%29%2C+t+%3D+1%2C+2%2C+%5Cldots%2C+%28p+-+1%29%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='x &#92;to &#92;text{sn}(iK&#039; &#92;pm 4t&#92;omega) = &#92;pm 1/(k&#92;,&#92;text{sn}&#92;,4t&#92;omega), t = 1, 2, &#92;ldots, (p - 1)/2' title='x &#92;to &#92;text{sn}(iK&#039; &#92;pm 4t&#92;omega) = &#92;pm 1/(k&#92;,&#92;text{sn}&#92;,4t&#92;omega), t = 1, 2, &#92;ldots, (p - 1)/2' class='latex' /> and so the position of poles of <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> are also verified.</p>
<p>Thus the tranformations of order <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> described in the beginning of the post using complicated product expansions containing elliptic functions are verified. We will discuss the repercussions of these transformations in the next post.</p>
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		<title>Elementary Approach to Modular Equations: Jacobi&#8217;s Transformation Theory 2</title>
		<link>http://paramanands.wordpress.com/2011/10/27/elementary-approach-to-modular-equations-jacobis-transformation-theory-2/</link>
		<comments>http://paramanands.wordpress.com/2011/10/27/elementary-approach-to-modular-equations-jacobis-transformation-theory-2/#comments</comments>
		<pubDate>Thu, 27 Oct 2011 06:25:30 +0000</pubDate>
		<dc:creator>paramanands</dc:creator>
				<category><![CDATA[Elliptic Functions]]></category>
		<category><![CDATA[Elliptic Integrals]]></category>
		<category><![CDATA[Mathematical Analysis]]></category>

		<guid isPermaLink="false">http://paramanands.wordpress.com/?p=2211</guid>
		<description><![CDATA[In this post we will apply the technique described in previous post to obtain modular equations of degree and . Cubic Transformation Since we have , the degrees of and are both and hence they both are constants. Since it is clear that the function depends only on ratio and hence we can take and [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=paramanands.wordpress.com&amp;blog=8565079&amp;post=2211&amp;subd=paramanands&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>In this post we will apply the technique described in <a title="Elementary Approach to Modular Equations: Jacobi’s Transformation Theory 1" href="http://paramanands.wordpress.com/2011/10/24/elementary-approach-to-modular-equations-jacobis-transformation-theory-1/" target="_blank">previous post</a> to obtain modular equations of degree <img src='http://s0.wp.com/latex.php?latex=3&amp;bg=fff&amp;fg=222&amp;s=0' alt='3' title='3' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=5&amp;bg=fff&amp;fg=222&amp;s=0' alt='5' title='5' class='latex' />.</p>
<p><strong>Cubic Transformation <img src='http://s0.wp.com/latex.php?latex=p+%3D+3&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = 3' title='p = 3' class='latex' /></strong></p>
<p>Since we have <img src='http://s0.wp.com/latex.php?latex=p+%3D+4%5Ccdot+1+-+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = 4&#92;cdot 1 - 1' title='p = 4&#92;cdot 1 - 1' class='latex' />, the degrees of <img src='http://s0.wp.com/latex.php?latex=P&amp;bg=fff&amp;fg=222&amp;s=0' alt='P' title='P' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=Q&amp;bg=fff&amp;fg=222&amp;s=0' alt='Q' title='Q' class='latex' /> are both <img src='http://s0.wp.com/latex.php?latex=2%5Ccdot+1+-+2+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='2&#92;cdot 1 - 2 = 0' title='2&#92;cdot 1 - 2 = 0' class='latex' /> and hence they both are constants. Since</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+y+%3D+%5Cfrac%7Bx%28P%5E%7B2%7D+%2B+2PQ+%2B+Q%5E%7B2%7Dx%5E%7B2%7D%29%7D%7BP%5E%7B2%7D+%2B+2PQx%5E%7B2%7D+%2B+Q%5E%7B2%7Dx%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle y = &#92;frac{x(P^{2} + 2PQ + Q^{2}x^{2})}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' title='&#92;displaystyle y = &#92;frac{x(P^{2} + 2PQ + Q^{2}x^{2})}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' class='latex' /></p>
<p>it is clear that the function <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> depends only on ratio <img src='http://s0.wp.com/latex.php?latex=Q%2FP&amp;bg=fff&amp;fg=222&amp;s=0' alt='Q/P' title='Q/P' class='latex' /> and hence we can take <img src='http://s0.wp.com/latex.php?latex=P+%3D+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='P = 1' title='P = 1' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=Q+%3D+%5Calpha&amp;bg=fff&amp;fg=222&amp;s=0' alt='Q = &#92;alpha' title='Q = &#92;alpha' class='latex' />. Thus we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+y+%3D+%5Cfrac%7Bx%281+%2B+2%5Calpha+%2B+%5Calpha%5E%7B2%7Dx%5E%7B2%7D%29%7D%7B1+%2B+2%5Calpha+x%5E%7B2%7D+%2B+%5Calpha%5E%7B2%7Dx%5E%7B2%7D%7D+%3D+x%5Ccdot+%5Cfrac%7B%281+%2B+2%5Calpha%29+%2B+%5Calpha%5E%7B2%7Dx%5E%7B2%7D%7D%7B1+%2B+%5Calpha%28%5Calpha+%2B+2%29x%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle y = &#92;frac{x(1 + 2&#92;alpha + &#92;alpha^{2}x^{2})}{1 + 2&#92;alpha x^{2} + &#92;alpha^{2}x^{2}} = x&#92;cdot &#92;frac{(1 + 2&#92;alpha) + &#92;alpha^{2}x^{2}}{1 + &#92;alpha(&#92;alpha + 2)x^{2}}' title='&#92;displaystyle y = &#92;frac{x(1 + 2&#92;alpha + &#92;alpha^{2}x^{2})}{1 + 2&#92;alpha x^{2} + &#92;alpha^{2}x^{2}} = x&#92;cdot &#92;frac{(1 + 2&#92;alpha) + &#92;alpha^{2}x^{2}}{1 + &#92;alpha(&#92;alpha + 2)x^{2}}' class='latex' /></p>
<p>The requirement of invariance under the transformation <img src='http://s0.wp.com/latex.php?latex=%28x%2C+y%29+%5Cto+%281%2Fkx%2C+1%2Fly%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(x, y) &#92;to (1/kx, 1/ly)' title='(x, y) &#92;to (1/kx, 1/ly)' class='latex' /> leads to the following conditions</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l+%3D+%5Cfrac%7Bk%5E%7B3%7D%7D%7B%5CDelta%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l = &#92;frac{k^{3}}{&#92;Delta^{2}}' title='&#92;displaystyle l = &#92;frac{k^{3}}{&#92;Delta^{2}}' class='latex' /></p>
<p>and</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%281+%2B+2%5Calpha%29k%5E%7B2%7Dx%5E%7B2%7D+%2B+%5Calpha%5E%7B2%7D+%3D+%5CDelta%5C%7B1+%2B+%5Calpha%28%5Calpha+%2B+2%29x%5E%7B2%7D%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle (1 + 2&#92;alpha)k^{2}x^{2} + &#92;alpha^{2} = &#92;Delta&#92;{1 + &#92;alpha(&#92;alpha + 2)x^{2}&#92;}' title='&#92;displaystyle (1 + 2&#92;alpha)k^{2}x^{2} + &#92;alpha^{2} = &#92;Delta&#92;{1 + &#92;alpha(&#92;alpha + 2)x^{2}&#92;}' class='latex' /></p>
<p>Thus we have the following equations</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l+%3D+%5Cfrac%7Bk%5E%7B3%7D%7D%7B%5CDelta%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l = &#92;frac{k^{3}}{&#92;Delta^{2}}' title='&#92;displaystyle l = &#92;frac{k^{3}}{&#92;Delta^{2}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Calpha%5E%7B2%7D+%3D+%5CDelta&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;alpha^{2} = &#92;Delta' title='&#92;alpha^{2} = &#92;Delta' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=k%5E%7B2%7D%282%5Calpha+%2B+1%29+%3D+%5CDelta+%5Calpha%28%5Calpha+%2B+2%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='k^{2}(2&#92;alpha + 1) = &#92;Delta &#92;alpha(&#92;alpha + 2)' title='k^{2}(2&#92;alpha + 1) = &#92;Delta &#92;alpha(&#92;alpha + 2)' class='latex' /></p>
<p>and since <img src='http://s0.wp.com/latex.php?latex=1%2FM+%3D+dy%2Fdx&amp;bg=fff&amp;fg=222&amp;s=0' alt='1/M = dy/dx' title='1/M = dy/dx' class='latex' /> at <img src='http://s0.wp.com/latex.php?latex=x+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='x = 0' title='x = 0' class='latex' /> we have <img src='http://s0.wp.com/latex.php?latex=M+%3D+1%2F%282%5Calpha+%2B+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='M = 1/(2&#92;alpha + 1)' title='M = 1/(2&#92;alpha + 1)' class='latex' />. Clearly we have then</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+k%5E%7B2%7D+%3D+%5Cfrac%7B%5Calpha%5E%7B3%7D%282+%2B+%5Calpha%29%7D%7B2%5Calpha+%2B+1%7D%2C%5C%2C%5C%2C+l%5E%7B2%7D+%3D+%5Calpha+%5Cleft%28%5Cfrac%7B2+%2B+%5Calpha%7D%7B2%5Calpha+%2B+1%7D%5Cright%29%5E%7B3%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle k^{2} = &#92;frac{&#92;alpha^{3}(2 + &#92;alpha)}{2&#92;alpha + 1},&#92;,&#92;, l^{2} = &#92;alpha &#92;left(&#92;frac{2 + &#92;alpha}{2&#92;alpha + 1}&#92;right)^{3}' title='&#92;displaystyle k^{2} = &#92;frac{&#92;alpha^{3}(2 + &#92;alpha)}{2&#92;alpha + 1},&#92;,&#92;, l^{2} = &#92;alpha &#92;left(&#92;frac{2 + &#92;alpha}{2&#92;alpha + 1}&#92;right)^{3}' class='latex' /></p>
<p>Eliminating <img src='http://s0.wp.com/latex.php?latex=%5Calpha&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;alpha' title='&#92;alpha' class='latex' /> from the above equation is bit tricky. After some manipulations we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+k%27%5E%7B2%7D+%3D+%5Cfrac%7B%281+-+%5Calpha%29%281+%2B+%5Calpha%29%5E%7B3%7D%7D%7B2%5Calpha+%2B+1%7D%2C%5C%2C%5C%2C+l%27%5E%7B2%7D+%3D+%5Cfrac%7B%281+%2B+%5Calpha%29%281+-+%5Calpha%29%5E%7B3%7D%7D%7B%282%5Calpha+%2B+1%29%5E%7B3%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle k&#039;^{2} = &#92;frac{(1 - &#92;alpha)(1 + &#92;alpha)^{3}}{2&#92;alpha + 1},&#92;,&#92;, l&#039;^{2} = &#92;frac{(1 + &#92;alpha)(1 - &#92;alpha)^{3}}{(2&#92;alpha + 1)^{3}}' title='&#92;displaystyle k&#039;^{2} = &#92;frac{(1 - &#92;alpha)(1 + &#92;alpha)^{3}}{2&#92;alpha + 1},&#92;,&#92;, l&#039;^{2} = &#92;frac{(1 + &#92;alpha)(1 - &#92;alpha)^{3}}{(2&#92;alpha + 1)^{3}}' class='latex' /></p>
<p>Clearly the pattern is now obvious if we multiply the expressions for <img src='http://s0.wp.com/latex.php?latex=k%2C+l&amp;bg=fff&amp;fg=222&amp;s=0' alt='k, l' title='k, l' class='latex' /> and (similarly for <img src='http://s0.wp.com/latex.php?latex=k%27%2C+l%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='k&#039;, l&#039;' title='k&#039;, l&#039;' class='latex' />)</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csqrt%7Bkl%7D+%3D+%5Cfrac%7B%5Calpha%282+%2B+%5Calpha%29%7D%7B2%5Calpha+%2B+1%7D%2C%5C%2C%5C%2C+%5Csqrt%7Bk%27l%27%7D+%3D+%5Cfrac%7B1+-+%5Calpha%5E%7B2%7D%7D%7B2%5Calpha+%2B+1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;sqrt{kl} = &#92;frac{&#92;alpha(2 + &#92;alpha)}{2&#92;alpha + 1},&#92;,&#92;, &#92;sqrt{k&#039;l&#039;} = &#92;frac{1 - &#92;alpha^{2}}{2&#92;alpha + 1}' title='&#92;displaystyle &#92;sqrt{kl} = &#92;frac{&#92;alpha(2 + &#92;alpha)}{2&#92;alpha + 1},&#92;,&#92;, &#92;sqrt{k&#039;l&#039;} = &#92;frac{1 - &#92;alpha^{2}}{2&#92;alpha + 1}' class='latex' /></p>
<p>and thus we obtain the modular equation of degree <img src='http://s0.wp.com/latex.php?latex=3&amp;bg=fff&amp;fg=222&amp;s=0' alt='3' title='3' class='latex' />:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Csqrt%7Bkl%7D+%2B+%5Csqrt%7Bk%27l%27%7D+%3D+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;sqrt{kl} + &#92;sqrt{k&#039;l&#039;} = 1' title='&#92;sqrt{kl} + &#92;sqrt{k&#039;l&#039;} = 1' class='latex' />.</p>
<p>Using the variable <img src='http://s0.wp.com/latex.php?latex=u+%3D+k%5E%7B1%2F4%7D%2C+v+%3D+l%5E%7B1%2F4%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = k^{1/4}, v = l^{1/4}' title='u = k^{1/4}, v = l^{1/4}' class='latex' /> we can obtain an expression devoid of algebraic irrationalities. Clearly we have <img src='http://s0.wp.com/latex.php?latex=%5Calpha%5E%7B4%7D+%3D+k%5E%7B3%7D%2Fl&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;alpha^{4} = k^{3}/l' title='&#92;alpha^{4} = k^{3}/l' class='latex' /> so that <img src='http://s0.wp.com/latex.php?latex=%5Calpha+%3D+u%5E%7B3%7D%2Fv&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;alpha = u^{3}/v' title='&#92;alpha = u^{3}/v' class='latex' /> and since <img src='http://s0.wp.com/latex.php?latex=%5Csqrt%7Bkl%7D+%3D+%5Calpha%282+%2B+%5Calpha%29%2F%282%5Calpha+%2B+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;sqrt{kl} = &#92;alpha(2 + &#92;alpha)/(2&#92;alpha + 1)' title='&#92;sqrt{kl} = &#92;alpha(2 + &#92;alpha)/(2&#92;alpha + 1)' class='latex' /> it follows that <img src='http://s0.wp.com/latex.php?latex=u%5E%7B2%7Dv%5E%7B2%7D+%3D+%5Calpha%282+%2B+%5Calpha%29%2F%282%5Calpha+%2B+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='u^{2}v^{2} = &#92;alpha(2 + &#92;alpha)/(2&#92;alpha + 1)' title='u^{2}v^{2} = &#92;alpha(2 + &#92;alpha)/(2&#92;alpha + 1)' class='latex' />. Putting the value of <img src='http://s0.wp.com/latex.php?latex=%5Calpha&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;alpha' title='&#92;alpha' class='latex' /> in this equation we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+u%5E%7B2%7Dv%5E%7B2%7D+%3D+%5Cdfrac%7B%5Cdfrac%7Bu%5E%7B3%7D%7D%7Bv%7D%5Cleft%282+%2B+%5Cdfrac%7Bu%5E%7B3%7D%7D%7Bv%7D%5Cright%29%7D%7B%5Cdfrac%7B2u%5E%7B3%7D%7D%7Bv%7D+%2B+1%7D+%3D+%5Cfrac%7Bu%5E%7B3%7D%282v+%2B+u%5E%7B3%7D%29%7D%7Bv%282u%5E%7B3%7D+%2B+v%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle u^{2}v^{2} = &#92;dfrac{&#92;dfrac{u^{3}}{v}&#92;left(2 + &#92;dfrac{u^{3}}{v}&#92;right)}{&#92;dfrac{2u^{3}}{v} + 1} = &#92;frac{u^{3}(2v + u^{3})}{v(2u^{3} + v)}' title='&#92;displaystyle u^{2}v^{2} = &#92;dfrac{&#92;dfrac{u^{3}}{v}&#92;left(2 + &#92;dfrac{u^{3}}{v}&#92;right)}{&#92;dfrac{2u^{3}}{v} + 1} = &#92;frac{u^{3}(2v + u^{3})}{v(2u^{3} + v)}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5CRightarrow+u%28u%5E%7B3%7D+%2B+2v%29+%3D+v%5E%7B3%7D%282u%5E%7B3%7D+%2B+v%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;Rightarrow u(u^{3} + 2v) = v^{3}(2u^{3} + v)' title='&#92;Rightarrow u(u^{3} + 2v) = v^{3}(2u^{3} + v)' class='latex' /></p>
<p>and thus finally</p>
<p><img src='http://s0.wp.com/latex.php?latex=u%5E%7B4%7D+-+v%5E%7B4%7D+%2B+2uv%281+-+u%5E%7B2%7Dv%5E%7B2%7D%29+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='u^{4} - v^{4} + 2uv(1 - u^{2}v^{2}) = 0' title='u^{4} - v^{4} + 2uv(1 - u^{2}v^{2}) = 0' class='latex' /></p>
<p><strong>Quintic Transformation <img src='http://s0.wp.com/latex.php?latex=p+%3D+5&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = 5' title='p = 5' class='latex' /></strong></p>
<p>Here we have <img src='http://s0.wp.com/latex.php?latex=p+%3D+5+%3D+4%5Ccdot+1+%2B+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = 5 = 4&#92;cdot 1 + 1' title='p = 5 = 4&#92;cdot 1 + 1' class='latex' />, hence the degree of <img src='http://s0.wp.com/latex.php?latex=P&amp;bg=fff&amp;fg=222&amp;s=0' alt='P' title='P' class='latex' /> is <img src='http://s0.wp.com/latex.php?latex=1&amp;bg=fff&amp;fg=222&amp;s=0' alt='1' title='1' class='latex' /> and that of <img src='http://s0.wp.com/latex.php?latex=Q&amp;bg=fff&amp;fg=222&amp;s=0' alt='Q' title='Q' class='latex' /> is zero. Thus we can take <img src='http://s0.wp.com/latex.php?latex=P+%3D+1+%2B+%5Cbeta+x%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='P = 1 + &#92;beta x^{2}' title='P = 1 + &#92;beta x^{2}' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=Q+%3D+%5Calpha&amp;bg=fff&amp;fg=222&amp;s=0' alt='Q = &#92;alpha' title='Q = &#92;alpha' class='latex' /> (note that the degrees being talked of are degrees when <img src='http://s0.wp.com/latex.php?latex=P%2C+Q&amp;bg=fff&amp;fg=222&amp;s=0' alt='P, Q' title='P, Q' class='latex' /> are expressed as polynomials in <img src='http://s0.wp.com/latex.php?latex=x%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='x^{2}' title='x^{2}' class='latex' />).</p>
<p>We have now</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+y+%3D+%5Cfrac%7Bx%28P%5E%7B2%7D+%2B+2PQ+%2B+Q%5E%7B2%7Dx%5E%7B2%7D%29%7D%7BP%5E%7B2%7D+%2B+2PQx%5E%7B2%7D+%2B+Q%5E%7B2%7Dx%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle y = &#92;frac{x(P^{2} + 2PQ + Q^{2}x^{2})}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' title='&#92;displaystyle y = &#92;frac{x(P^{2} + 2PQ + Q^{2}x^{2})}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' class='latex' /></p>
<p>so that in this case</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+y+%3D+%5Cfrac%7Bx%5C%7B%282%5Calpha+%2B+1%29+%2B+%282%5Calpha%5Cbeta+%2B+2%5Cbeta+%2B+%5Calpha%5E%7B2%7D%29x%5E%7B2%7D+%2B+%5Cbeta%5E%7B2%7Dx%5E%7B4%7D%5C%7D%7D%7B1+%2B+%282%5Cbeta+%2B+2%5Calpha+%2B+%5Calpha%5E%7B2%7D%29x%5E%7B2%7D+%2B+%28%5Cbeta%5E%7B2%7D+%2B+2%5Calpha%5Cbeta%29x%5E%7B4%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle y = &#92;frac{x&#92;{(2&#92;alpha + 1) + (2&#92;alpha&#92;beta + 2&#92;beta + &#92;alpha^{2})x^{2} + &#92;beta^{2}x^{4}&#92;}}{1 + (2&#92;beta + 2&#92;alpha + &#92;alpha^{2})x^{2} + (&#92;beta^{2} + 2&#92;alpha&#92;beta)x^{4}}' title='&#92;displaystyle y = &#92;frac{x&#92;{(2&#92;alpha + 1) + (2&#92;alpha&#92;beta + 2&#92;beta + &#92;alpha^{2})x^{2} + &#92;beta^{2}x^{4}&#92;}}{1 + (2&#92;beta + 2&#92;alpha + &#92;alpha^{2})x^{2} + (&#92;beta^{2} + 2&#92;alpha&#92;beta)x^{4}}' class='latex' /></p>
<p>This relation is invariant under the transformation <img src='http://s0.wp.com/latex.php?latex=%28x%2C+y%29+%5Cto+%281%2Fkx%2C+1%2Fly%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(x, y) &#92;to (1/kx, 1/ly)' title='(x, y) &#92;to (1/kx, 1/ly)' class='latex' /> so we have<br />
<img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+l+%3D+%5Cfrac%7Bk%5E%7B5%7D%7D%7B%5CDelta%5E%7B2%7D%7D%5C%2C%5C%2C%5C%2C%5Ccdots+%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle l = &#92;frac{k^{5}}{&#92;Delta^{2}}&#92;,&#92;,&#92;,&#92;cdots (1)' title='&#92;displaystyle l = &#92;frac{k^{5}}{&#92;Delta^{2}}&#92;,&#92;,&#92;,&#92;cdots (1)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cbeta%5E%7B2%7D+%3D+%5CDelta%5C%2C%5C%2C%5C%2C%5Ccdots+%282%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;beta^{2} = &#92;Delta&#92;,&#92;,&#92;,&#92;cdots (2)' title='&#92;beta^{2} = &#92;Delta&#92;,&#92;,&#92;,&#92;cdots (2)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=k%5E%7B2%7D%282%5Calpha%5Cbeta+%2B+2%5Cbeta+%2B+%5Calpha%5E%7B2%7D%29+%3D+%5CDelta%282%5Cbeta+%2B+2%5Calpha+%2B+%5Calpha%5E%7B2%7D%29%5C%2C%5C%2C%5C%2C%5Ccdots+%283%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='k^{2}(2&#92;alpha&#92;beta + 2&#92;beta + &#92;alpha^{2}) = &#92;Delta(2&#92;beta + 2&#92;alpha + &#92;alpha^{2})&#92;,&#92;,&#92;,&#92;cdots (3)' title='k^{2}(2&#92;alpha&#92;beta + 2&#92;beta + &#92;alpha^{2}) = &#92;Delta(2&#92;beta + 2&#92;alpha + &#92;alpha^{2})&#92;,&#92;,&#92;,&#92;cdots (3)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=k%5E%7B4%7D%282%5Calpha+%2B+1%29+%3D+%5CDelta%28%5Cbeta%5E%7B2%7D+%2B+2%5Calpha%5Cbeta%29%5C%2C%5C%2C%5C%2C%5Ccdots+%284%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='k^{4}(2&#92;alpha + 1) = &#92;Delta(&#92;beta^{2} + 2&#92;alpha&#92;beta)&#92;,&#92;,&#92;,&#92;cdots (4)' title='k^{4}(2&#92;alpha + 1) = &#92;Delta(&#92;beta^{2} + 2&#92;alpha&#92;beta)&#92;,&#92;,&#92;,&#92;cdots (4)' class='latex' /></p>
<p>Like in the case of cubic transformation we have <img src='http://s0.wp.com/latex.php?latex=M+%3D+1%2F%282%5Calpha+%2B+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='M = 1/(2&#92;alpha + 1)' title='M = 1/(2&#92;alpha + 1)' class='latex' />.</p>
<p>Obtaining a relation between <img src='http://s0.wp.com/latex.php?latex=k%2C+l&amp;bg=fff&amp;fg=222&amp;s=0' alt='k, l' title='k, l' class='latex' /> by eliminating <img src='http://s0.wp.com/latex.php?latex=%5Calpha%2C+%5Cbeta%2C+%5CDelta&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;alpha, &#92;beta, &#92;Delta' title='&#92;alpha, &#92;beta, &#92;Delta' class='latex' /> is reasonably cumbersome and it becomes manageable when we use the associated variables <img src='http://s0.wp.com/latex.php?latex=u+%3D+k%5E%7B1%2F4%7D%2C+v+%3D+l%5E%7B1%2F4%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='u = k^{1/4}, v = l^{1/4}' title='u = k^{1/4}, v = l^{1/4}' class='latex' />. Using these variables and equations (1) and (2) we have <img src='http://s0.wp.com/latex.php?latex=%5Cbeta+%3D+%5Csqrt%7B%5CDelta%7D+%3D+u%5E%7B5%7D%2Fv&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;beta = &#92;sqrt{&#92;Delta} = u^{5}/v' title='&#92;beta = &#92;sqrt{&#92;Delta} = u^{5}/v' class='latex' /> and putting these values in equation (4) we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+u%5E%7B16%7D%282%5Calpha+%2B+1%29+%3D+%5Cfrac%7Bu%5E%7B10%7D%7D%7Bv%5E%7B2%7D%7D%5Cleft%28%5Cfrac%7Bu%5E%7B10%7D%7D%7Bv%5E%7B2%7D%7D+%2B+2%5Calpha%5Ccdot+%5Cfrac%7Bu%5E%7B5%7D%7D%7Bv%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle u^{16}(2&#92;alpha + 1) = &#92;frac{u^{10}}{v^{2}}&#92;left(&#92;frac{u^{10}}{v^{2}} + 2&#92;alpha&#92;cdot &#92;frac{u^{5}}{v}&#92;right)' title='&#92;displaystyle u^{16}(2&#92;alpha + 1) = &#92;frac{u^{10}}{v^{2}}&#92;left(&#92;frac{u^{10}}{v^{2}} + 2&#92;alpha&#92;cdot &#92;frac{u^{5}}{v}&#92;right)' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%282%5Calpha+%2B+1%29uv%5E%7B4%7D+%3D+u%5E%7B5%7D+%2B+2%5Calpha+v+%5CRightarrow+2%5Calpha+v%281+-+uv%5E%7B3%7D%29+%3D+u%28v%5E%7B4%7D+-+u%5E%7B4%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(2&#92;alpha + 1)uv^{4} = u^{5} + 2&#92;alpha v &#92;Rightarrow 2&#92;alpha v(1 - uv^{3}) = u(v^{4} - u^{4})' title='(2&#92;alpha + 1)uv^{4} = u^{5} + 2&#92;alpha v &#92;Rightarrow 2&#92;alpha v(1 - uv^{3}) = u(v^{4} - u^{4})' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+2%5Calpha+%3D+%5Cfrac%7Bu%28v%5E%7B4%7D+-+u%5E%7B4%7D%29%7D%7Bv%281+-+uv%5E%7B3%7D%29%7D%5C%2C%5C%2C%5C%2C%5Ccdots+%285%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;Rightarrow 2&#92;alpha = &#92;frac{u(v^{4} - u^{4})}{v(1 - uv^{3})}&#92;,&#92;,&#92;,&#92;cdots (5)' title='&#92;displaystyle &#92;Rightarrow 2&#92;alpha = &#92;frac{u(v^{4} - u^{4})}{v(1 - uv^{3})}&#92;,&#92;,&#92;,&#92;cdots (5)' class='latex' /></p>
<p>Again using equation (3) we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%28v%5E%7B2%7D+-+u%5E%7B2%7D%29%282%5Cbeta+%2B+%5Calpha%5E%7B2%7D%29+%3D+u%5E%7B2%7D%281+-+u%5E%7B3%7Dv%29%282%5Calpha%29+%3D+%5Cfrac%7Bu%5E%7B3%7D%7D%7Bv%7D%5Cfrac%7B%28v%5E%7B4%7D+-+u%5E%7B4%7D%29%281+-+u%5E%7B3%7Dv%29%7D%7B1+-+uv%5E%7B3%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle (v^{2} - u^{2})(2&#92;beta + &#92;alpha^{2}) = u^{2}(1 - u^{3}v)(2&#92;alpha) = &#92;frac{u^{3}}{v}&#92;frac{(v^{4} - u^{4})(1 - u^{3}v)}{1 - uv^{3}}' title='&#92;displaystyle (v^{2} - u^{2})(2&#92;beta + &#92;alpha^{2}) = u^{2}(1 - u^{3}v)(2&#92;alpha) = &#92;frac{u^{3}}{v}&#92;frac{(v^{4} - u^{4})(1 - u^{3}v)}{1 - uv^{3}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+2%5Cbeta+%2B+%5Calpha%5E%7B2%7D+%3D+%5Cfrac%7Bu%5E%7B3%7D%7D%7Bv%7D%5Cfrac%7B%28v%5E%7B2%7D+%2B+u%5E%7B2%7D%29%281+-+u%5E%7B3%7Dv%29%7D%7B1+-+uv%5E%7B3%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;Rightarrow 2&#92;beta + &#92;alpha^{2} = &#92;frac{u^{3}}{v}&#92;frac{(v^{2} + u^{2})(1 - u^{3}v)}{1 - uv^{3}}' title='&#92;displaystyle &#92;Rightarrow 2&#92;beta + &#92;alpha^{2} = &#92;frac{u^{3}}{v}&#92;frac{(v^{2} + u^{2})(1 - u^{3}v)}{1 - uv^{3}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+%5Calpha%5E%7B2%7D+%3D+%5Cfrac%7Bu%5E%7B3%7D%7D%7Bv%7D%5Cleft%5C%7B%5Cfrac%7B%28v%5E%7B2%7D+%2B+u%5E%7B2%7D%29%281+-+u%5E%7B3%7Dv%29%7D%7B1+-+uv%5E%7B3%7D%7D+-+2u%5E%7B2%7D%5Cright%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;Rightarrow &#92;alpha^{2} = &#92;frac{u^{3}}{v}&#92;left&#92;{&#92;frac{(v^{2} + u^{2})(1 - u^{3}v)}{1 - uv^{3}} - 2u^{2}&#92;right&#92;}' title='&#92;displaystyle &#92;Rightarrow &#92;alpha^{2} = &#92;frac{u^{3}}{v}&#92;left&#92;{&#92;frac{(v^{2} + u^{2})(1 - u^{3}v)}{1 - uv^{3}} - 2u^{2}&#92;right&#92;}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5CRightarrow+%5Calpha%5E%7B2%7D+%3D+%5Cfrac%7Bu%5E%7B3%7D%7D%7Bv%7D%5Cfrac%7B%28v%5E%7B2%7D+-+u%5E%7B2%7D%29%281+%2B+u%5E%7B3%7Dv%29%7D%7B1+-+uv%5E%7B3%7D%7D%5C%2C%5C%2C%5C%2C%5Ccdots+%286%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;Rightarrow &#92;alpha^{2} = &#92;frac{u^{3}}{v}&#92;frac{(v^{2} - u^{2})(1 + u^{3}v)}{1 - uv^{3}}&#92;,&#92;,&#92;,&#92;cdots (6)' title='&#92;displaystyle &#92;Rightarrow &#92;alpha^{2} = &#92;frac{u^{3}}{v}&#92;frac{(v^{2} - u^{2})(1 + u^{3}v)}{1 - uv^{3}}&#92;,&#92;,&#92;,&#92;cdots (6)' class='latex' /></p>
<p>Dividing (6) by (5) we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+2%5Calpha+%3D+%5Cfrac%7B4u%5E%7B2%7D%281+%2B+u%5E%7B3%7Dv%29%7D%7Bv%5E%7B2%7D+%2B+u%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 2&#92;alpha = &#92;frac{4u^{2}(1 + u^{3}v)}{v^{2} + u^{2}}' title='&#92;displaystyle 2&#92;alpha = &#92;frac{4u^{2}(1 + u^{3}v)}{v^{2} + u^{2}}' class='latex' /></p>
<p>Comparing this with (5) we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=4uv%281+-+uv%5E%7B3%7D%29%281+%2B+u%5E%7B3%7Dv%29+-+%28v%5E%7B2%7D+%2B+u%5E%7B2%7D%29%28v%5E%7B4%7D+-+u%5E%7B4%7D%29+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='4uv(1 - uv^{3})(1 + u^{3}v) - (v^{2} + u^{2})(v^{4} - u^{4}) = 0' title='4uv(1 - uv^{3})(1 + u^{3}v) - (v^{2} + u^{2})(v^{4} - u^{4}) = 0' class='latex' /></p>
<p>and on simplifying further we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=u%5E%7B6%7D+-+v%5E%7B6%7D+%2B+5u%5E%7B2%7Dv%5E%7B2%7D%28u%5E%7B2%7D+-+v%5E%7B2%7D%29+%2B+4uv%281+-+u%5E%7B4%7Dv%5E%7B4%7D%29+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='u^{6} - v^{6} + 5u^{2}v^{2}(u^{2} - v^{2}) + 4uv(1 - u^{4}v^{4}) = 0' title='u^{6} - v^{6} + 5u^{2}v^{2}(u^{2} - v^{2}) + 4uv(1 - u^{4}v^{4}) = 0' class='latex' /></p>
<p>The reader must have understood by now that this technique of finding modular equations is quite unsuitable for larger values of <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' />. In fact Jacobi treated only the cubic and quintic transformations in his <em>Fundamenta Nova</em> and Arthur Cayley extended this approach to <img src='http://s0.wp.com/latex.php?latex=p+%3D+7&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = 7' title='p = 7' class='latex' /> in his <em>An Elementary Treatise on Elliptic Functions</em>. Our exposition here is based on Cayley&#8217;s book. For higher values of <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> Jacobi provided the transformation in a form which contains elliptic functions of <img src='http://s0.wp.com/latex.php?latex=K%2Fp&amp;bg=fff&amp;fg=222&amp;s=0' alt='K/p' title='K/p' class='latex' /> and thereby obtained equations <img src='http://s0.wp.com/latex.php?latex=L+%3D+K%2F%28pM%29%2C+L%27+%3D+K%27%2FM&amp;bg=fff&amp;fg=222&amp;s=0' alt='L = K/(pM), L&#039; = K&#039;/M' title='L = K/(pM), L&#039; = K&#039;/M' class='latex' /> from which follows the relation <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+pK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = pK&#039;/K' title='L&#039;/L = pK&#039;/K' class='latex' />. This we study in the next post.</p>
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		<title>Elementary Approach to Modular Equations: Jacobi&#8217;s Transformation Theory 1</title>
		<link>http://paramanands.wordpress.com/2011/10/24/elementary-approach-to-modular-equations-jacobis-transformation-theory-1/</link>
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		<pubDate>Mon, 24 Oct 2011 09:11:39 +0000</pubDate>
		<dc:creator>paramanands</dc:creator>
				<category><![CDATA[Elliptic Functions]]></category>
		<category><![CDATA[Elliptic Integrals]]></category>
		<category><![CDATA[Mathematical Analysis]]></category>

		<guid isPermaLink="false">http://paramanands.wordpress.com/?p=2119</guid>
		<description><![CDATA[To recapitulate the basics of elliptic integral theory (details here) we have We have here and the complementary modulus or in other words . Also we denote and we have the Legendre&#8217;s Identity . The Function From the definition it is clear that is a strictly increasing function of and moreover as . Thus function [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=paramanands.wordpress.com&amp;blog=8565079&amp;post=2119&amp;subd=paramanands&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>To recapitulate the basics of elliptic integral theory (details <a title="π(PI) and the AGM: Introduction to Elliptic Integrals" href="http://paramanands.wordpress.com/2009/08/09/pi-and-the-agm-intro-elliptic-integrals/" target="_blank">here</a>) we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+K+%3D+K%28k%29+%3D+%5Cint_%7B0%7D%5E%7B%5Cpi%2F2%7D%5Cfrac%7Bd%5Ctheta%7D%7B%5Csqrt%7B1+-+k%5E%7B2%7D%5Csin%5E%7B2%7D%5Ctheta%7D%7D+%3D+%5Cint_%7B0%7D%5E%7B1%7D%5Cfrac%7Bdx%7D%7B%5Csqrt%7B%281+-+x%5E%7B2%7D%29%281+-+k%5E%7B2%7Dx%5E%7B2%7D%29%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle K = K(k) = &#92;int_{0}^{&#92;pi/2}&#92;frac{d&#92;theta}{&#92;sqrt{1 - k^{2}&#92;sin^{2}&#92;theta}} = &#92;int_{0}^{1}&#92;frac{dx}{&#92;sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}' title='&#92;displaystyle K = K(k) = &#92;int_{0}^{&#92;pi/2}&#92;frac{d&#92;theta}{&#92;sqrt{1 - k^{2}&#92;sin^{2}&#92;theta}} = &#92;int_{0}^{1}&#92;frac{dx}{&#92;sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+E+%3D+E%28k%29+%3D+%5Cint_%7B0%7D%5E%7B%5Cpi%2F2%7D%5Csqrt%7B1+-+k%5E%7B2%7D%5Csin%5E%7B2%7D%5Ctheta%7D%5C%2Cd%5Ctheta+%3D+%5Cint_%7B0%7D%5E%7B1%7D%5Cfrac%7B%5Csqrt%7B1+-+k%5E%7B2%7Dx%5E%7B2%7D%7D%7D%7B%5Csqrt%7B1+-+x%5E%7B2%7D%7D%7D%5C%2Cdx&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle E = E(k) = &#92;int_{0}^{&#92;pi/2}&#92;sqrt{1 - k^{2}&#92;sin^{2}&#92;theta}&#92;,d&#92;theta = &#92;int_{0}^{1}&#92;frac{&#92;sqrt{1 - k^{2}x^{2}}}{&#92;sqrt{1 - x^{2}}}&#92;,dx' title='&#92;displaystyle E = E(k) = &#92;int_{0}^{&#92;pi/2}&#92;sqrt{1 - k^{2}&#92;sin^{2}&#92;theta}&#92;,d&#92;theta = &#92;int_{0}^{1}&#92;frac{&#92;sqrt{1 - k^{2}x^{2}}}{&#92;sqrt{1 - x^{2}}}&#92;,dx' class='latex' /></p>
<p>We have here <img src='http://s0.wp.com/latex.php?latex=0+%5Cleq+k+%5Cleq+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='0 &#92;leq k &#92;leq 1' title='0 &#92;leq k &#92;leq 1' class='latex' /> and the complementary modulus <img src='http://s0.wp.com/latex.php?latex=k%27+%3D+%5Csqrt%7B1+-+k%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='k&#039; = &#92;sqrt{1 - k^{2}}' title='k&#039; = &#92;sqrt{1 - k^{2}}' class='latex' /> or in other words <img src='http://s0.wp.com/latex.php?latex=k%5E%7B2%7D+%2B+k%27%5E%7B2%7D+%3D+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='k^{2} + k&#039;^{2} = 1' title='k^{2} + k&#039;^{2} = 1' class='latex' />. Also we denote <img src='http://s0.wp.com/latex.php?latex=K%27+%3D+K%28k%27%29%2C+E%27+%3D+E%28k%27%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039; = K(k&#039;), E&#039; = E(k&#039;)' title='K&#039; = K(k&#039;), E&#039; = E(k&#039;)' class='latex' /> and we have the <a title="π(PI) and the AGM: Legendre’s Identity" href="http://paramanands.wordpress.com/2009/08/10/pi-and-the-agm-lengendre-identity/" target="_blank">Legendre&#8217;s Identity</a> <img src='http://s0.wp.com/latex.php?latex=KE%27+%2B+K%27E+-+KK%27+%3D+%5Cpi%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='KE&#039; + K&#039;E - KK&#039; = &#92;pi/2' title='KE&#039; + K&#039;E - KK&#039; = &#92;pi/2' class='latex' />.</p>
<p><strong>The Function <img src='http://s0.wp.com/latex.php?latex=K%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;/K' title='K&#039;/K' class='latex' /></strong></p>
<p>From the definition it is clear that <img src='http://s0.wp.com/latex.php?latex=K&amp;bg=fff&amp;fg=222&amp;s=0' alt='K' title='K' class='latex' /> is a strictly increasing function of <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> and moreover <img src='http://s0.wp.com/latex.php?latex=K+%5Cto+%5Cinfty&amp;bg=fff&amp;fg=222&amp;s=0' alt='K &#92;to &#92;infty' title='K &#92;to &#92;infty' class='latex' /> as <img src='http://s0.wp.com/latex.php?latex=k+%5Cto+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='k &#92;to 1' title='k &#92;to 1' class='latex' />. Thus function <img src='http://s0.wp.com/latex.php?latex=K%28k%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='K(k)' title='K(k)' class='latex' /> maps interval <img src='http://s0.wp.com/latex.php?latex=%5B0%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='[0, 1)' title='[0, 1)' class='latex' /> to <img src='http://s0.wp.com/latex.php?latex=%5B%5Cpi%2F2%2C+%5Cinfty%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='[&#92;pi/2, &#92;infty)' title='[&#92;pi/2, &#92;infty)' class='latex' />. Similarly <img src='http://s0.wp.com/latex.php?latex=E&amp;bg=fff&amp;fg=222&amp;s=0' alt='E' title='E' class='latex' /> is strictly decreasing and maps <img src='http://s0.wp.com/latex.php?latex=%5B0%2C+1%5D&amp;bg=fff&amp;fg=222&amp;s=0' alt='[0, 1]' title='[0, 1]' class='latex' /> to <img src='http://s0.wp.com/latex.php?latex=%5B1%2C+%5Cpi%2F2%5D&amp;bg=fff&amp;fg=222&amp;s=0' alt='[1, &#92;pi/2]' title='[1, &#92;pi/2]' class='latex' />. Therefore it follows that <img src='http://s0.wp.com/latex.php?latex=K%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;' title='K&#039;' class='latex' /> is a strictly decreasing function of <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=E%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='E&#039;' title='E&#039;' class='latex' /> is a strictly increasing function of <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' />.</p>
<p>From the above arguments it follows that the function <img src='http://s0.wp.com/latex.php?latex=K%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;/K' title='K&#039;/K' class='latex' /> is a strictly decreasing function of <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' />. When <img src='http://s0.wp.com/latex.php?latex=k+%5Cto+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='k &#92;to 0' title='k &#92;to 0' class='latex' />, <img src='http://s0.wp.com/latex.php?latex=K%27%2FK+%5Cto+%5Cinfty&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;/K &#92;to &#92;infty' title='K&#039;/K &#92;to &#92;infty' class='latex' /> and as <img src='http://s0.wp.com/latex.php?latex=k+%5Cto+1%2C+K%27%2FK+%5Cto+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='k &#92;to 1, K&#039;/K &#92;to 0' title='k &#92;to 1, K&#039;/K &#92;to 0' class='latex' />. Thus the function <img src='http://s0.wp.com/latex.php?latex=K%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;/K' title='K&#039;/K' class='latex' /> maps the interval <img src='http://s0.wp.com/latex.php?latex=%280%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(0, 1)' title='(0, 1)' class='latex' /> to the interval <img src='http://s0.wp.com/latex.php?latex=%280%2C+%5Cinfty%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(0, &#92;infty)' title='(0, &#92;infty)' class='latex' />. For any given positive number <img src='http://s0.wp.com/latex.php?latex=%5Calpha&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;alpha' title='&#92;alpha' class='latex' /> we have a unique positive <img src='http://s0.wp.com/latex.php?latex=k+%5Cin+%280%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='k &#92;in (0, 1)' title='k &#92;in (0, 1)' class='latex' /> such that <img src='http://s0.wp.com/latex.php?latex=K%27%2FK+%3D+%5Calpha&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;/K = &#92;alpha' title='K&#039;/K = &#92;alpha' class='latex' />. Let <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> be any positive number and then <img src='http://s0.wp.com/latex.php?latex=p%5Calpha&amp;bg=fff&amp;fg=222&amp;s=0' alt='p&#92;alpha' title='p&#92;alpha' class='latex' /> is also positive and therefore there is a unique number <img src='http://s0.wp.com/latex.php?latex=l+%5Cin+%280%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='l &#92;in (0, 1)' title='l &#92;in (0, 1)' class='latex' /> such that <img src='http://s0.wp.com/latex.php?latex=K%28l%27%29%2FK%28l%29+%3D+p%5Calpha&amp;bg=fff&amp;fg=222&amp;s=0' alt='K(l&#039;)/K(l) = p&#92;alpha' title='K(l&#039;)/K(l) = p&#92;alpha' class='latex' />. We traditionally denote <img src='http://s0.wp.com/latex.php?latex=L+%3D+K%28l%29%2C+L%27+%3D+K%28l%27%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='L = K(l), L&#039; = K(l&#039;)' title='L = K(l), L&#039; = K(l&#039;)' class='latex' />. Therefore we can write</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7BL%27%7D%7BL%7D+%3D+p%5Calpha+%3D+p%5C%2C%5Cfrac%7BK%27%7D%7BK%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{L&#039;}{L} = p&#92;alpha = p&#92;,&#92;frac{K&#039;}{K}' title='&#92;displaystyle &#92;frac{L&#039;}{L} = p&#92;alpha = p&#92;,&#92;frac{K&#039;}{K}' class='latex' /></p>
<p>Again let&#8217;s think in the following way. Suppose a number <img src='http://s0.wp.com/latex.php?latex=k+%5Cin+%280%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='k &#92;in (0, 1)' title='k &#92;in (0, 1)' class='latex' /> and a positive number <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> is given. From <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> we can calculate <img src='http://s0.wp.com/latex.php?latex=K%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;/K' title='K&#039;/K' class='latex' /> and hence <img src='http://s0.wp.com/latex.php?latex=pK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='pK&#039;/K' title='pK&#039;/K' class='latex' /> corresponding to which there is a unique number <img src='http://s0.wp.com/latex.php?latex=l+%5Cin+%280%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='l &#92;in (0, 1)' title='l &#92;in (0, 1)' class='latex' /> such that <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+pK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = pK&#039;/K' title='L&#039;/L = pK&#039;/K' class='latex' />. In this fashion <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> turns out to be a function of <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' />. Therefore the equation <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+pK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = pK&#039;/K' title='L&#039;/L = pK&#039;/K' class='latex' /> determines <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> as a function of <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> which is one-one and invertible. From the continuity of <img src='http://s0.wp.com/latex.php?latex=K&amp;bg=fff&amp;fg=222&amp;s=0' alt='K' title='K' class='latex' /> the function turns out to be continuous. To derive these results formally we need the fundamental formulas:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bdk%27%7D%7Bdk%7D+%3D+-%5Cfrac%7Bk%7D%7Bk%27%7D%2C%5C%2C%5C%2C+%5Cfrac%7BdK%7D%7Bdk%7D+%3D+%5Cfrac%7BE+-+k%27%5E%7B2%7DK%7D%7Bkk%27%5E%7B2%7D%7D%2C%5C%2C%5C%2C+%5Cfrac%7BdE%7D%7Bdk%7D+%3D+%5Cfrac%7BE+-+K%7D%7Bk%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{dk&#039;}{dk} = -&#92;frac{k}{k&#039;},&#92;,&#92;, &#92;frac{dK}{dk} = &#92;frac{E - k&#039;^{2}K}{kk&#039;^{2}},&#92;,&#92;, &#92;frac{dE}{dk} = &#92;frac{E - K}{k}' title='&#92;displaystyle &#92;frac{dk&#039;}{dk} = -&#92;frac{k}{k&#039;},&#92;,&#92;, &#92;frac{dK}{dk} = &#92;frac{E - k&#039;^{2}K}{kk&#039;^{2}},&#92;,&#92;, &#92;frac{dE}{dk} = &#92;frac{E - K}{k}' class='latex' /></p>
<p>(for the proofs of these results read <a title="π(PI) and the AGM: Legendre’s Identity" href="http://paramanands.wordpress.com/2009/08/10/pi-and-the-agm-lengendre-identity/" target="_blank">here</a>)</p>
<p>Using these we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7BdK%27%7D%7Bdk%7D+%3D+%5Cfrac%7Bdk%27%7D%7Bdk%7D%5Cfrac%7BdK%27%7D%7Bdk%27%7D+%3D+-%5Cfrac%7Bk%7D%7Bk%27%7D%5Cfrac%7BE%27+-+k%5E%7B2%7DK%27%7D%7Bk%5E%7B2%7Dk%27%7D+%3D+%5Cfrac%7Bk%5E%7B2%7DK%27+-+E%27%7D%7Bkk%27%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{dK&#039;}{dk} = &#92;frac{dk&#039;}{dk}&#92;frac{dK&#039;}{dk&#039;} = -&#92;frac{k}{k&#039;}&#92;frac{E&#039; - k^{2}K&#039;}{k^{2}k&#039;} = &#92;frac{k^{2}K&#039; - E&#039;}{kk&#039;^{2}}' title='&#92;displaystyle &#92;frac{dK&#039;}{dk} = &#92;frac{dk&#039;}{dk}&#92;frac{dK&#039;}{dk&#039;} = -&#92;frac{k}{k&#039;}&#92;frac{E&#039; - k^{2}K&#039;}{k^{2}k&#039;} = &#92;frac{k^{2}K&#039; - E&#039;}{kk&#039;^{2}}' class='latex' /></p>
<p>and therefore</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bd%7D%7Bdk%7D%5Cleft%28%5Cfrac%7BK%27%7D%7BK%7D%5Cright%29+%3D+%5Cdfrac%7BK%5Cdfrac%7BdK%27%7D%7Bdk%7D+-+K%27%5Cdfrac%7BdK%7D%7Bdk%7D%7D%7BK%5E%7B2%7D%7D+%3D+%5Cfrac%7B1%7D%7BK%5E%7B2%7D%7D%5Cleft%28%5Cfrac%7BK%28k%5E%7B2%7DK%27+-+E%27%29%7D%7Bkk%27%5E%7B2%7D%7D+-+%5Cfrac%7BK%27%28E+-+k%27%5E%7B2%7DK%29%7D%7Bkk%27%5E%7B2%7D%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{d}{dk}&#92;left(&#92;frac{K&#039;}{K}&#92;right) = &#92;dfrac{K&#92;dfrac{dK&#039;}{dk} - K&#039;&#92;dfrac{dK}{dk}}{K^{2}} = &#92;frac{1}{K^{2}}&#92;left(&#92;frac{K(k^{2}K&#039; - E&#039;)}{kk&#039;^{2}} - &#92;frac{K&#039;(E - k&#039;^{2}K)}{kk&#039;^{2}}&#92;right)' title='&#92;displaystyle &#92;frac{d}{dk}&#92;left(&#92;frac{K&#039;}{K}&#92;right) = &#92;dfrac{K&#92;dfrac{dK&#039;}{dk} - K&#039;&#92;dfrac{dK}{dk}}{K^{2}} = &#92;frac{1}{K^{2}}&#92;left(&#92;frac{K(k^{2}K&#039; - E&#039;)}{kk&#039;^{2}} - &#92;frac{K&#039;(E - k&#039;^{2}K)}{kk&#039;^{2}}&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%5Cfrac%7BKK%27+-+KE%27+-+K%27E%7D%7Bkk%27%5E%7B2%7DK%5E%7B2%7D%7D+%3D+-%5Cfrac%7B%5Cpi%7D%7B2kk%27%5E%7B2%7DK%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = &#92;frac{KK&#039; - KE&#039; - K&#039;E}{kk&#039;^{2}K^{2}} = -&#92;frac{&#92;pi}{2kk&#039;^{2}K^{2}}' title='&#92;displaystyle = &#92;frac{KK&#039; - KE&#039; - K&#039;E}{kk&#039;^{2}K^{2}} = -&#92;frac{&#92;pi}{2kk&#039;^{2}K^{2}}' class='latex' /></p>
<p>Thus we have finally another fundamental relation</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bd%7D%7Bdk%7D%5Cleft%28%5Cfrac%7BK%27%7D%7BK%7D%5Cright%29+%3D+-%5Cfrac%7B%5Cpi%7D%7B2kk%27%5E%7B2%7DK%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{d}{dk}&#92;left(&#92;frac{K&#039;}{K}&#92;right) = -&#92;frac{&#92;pi}{2kk&#039;^{2}K^{2}}' title='&#92;displaystyle &#92;frac{d}{dk}&#92;left(&#92;frac{K&#039;}{K}&#92;right) = -&#92;frac{&#92;pi}{2kk&#039;^{2}K^{2}}' class='latex' /></p>
<p>which shows that <img src='http://s0.wp.com/latex.php?latex=K%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='K&#039;/K' title='K&#039;/K' class='latex' /> is strictly decreasing. On differentiating the equation <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+pK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = pK&#039;/K' title='L&#039;/L = pK&#039;/K' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bdl%7D%7Bll%27%5E%7B2%7DL%5E%7B2%7D%7D+%3D+p%5Cfrac%7Bdk%7D%7Bkk%27%5E%7B2%7DK%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{dl}{ll&#039;^{2}L^{2}} = p&#92;frac{dk}{kk&#039;^{2}K^{2}}' title='&#92;displaystyle &#92;frac{dl}{ll&#039;^{2}L^{2}} = p&#92;frac{dk}{kk&#039;^{2}K^{2}}' class='latex' /> or <img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bdl%7D%7Bdk%7D+%3D+p%5C%2C%5Cfrac%7Bll%27%5E%7B2%7D%7D%7Bkk%27%5E%7B2%7D%7D%5Cleft%28%5Cfrac%7BL%7D%7BK%7D%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{dl}{dk} = p&#92;,&#92;frac{ll&#039;^{2}}{kk&#039;^{2}}&#92;left(&#92;frac{L}{K}&#92;right)^{2}' title='&#92;displaystyle &#92;frac{dl}{dk} = p&#92;,&#92;frac{ll&#039;^{2}}{kk&#039;^{2}}&#92;left(&#92;frac{L}{K}&#92;right)^{2}' class='latex' /></p>
<p>and therefore <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> is a strictly increasing function of <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> and maps <img src='http://s0.wp.com/latex.php?latex=%280%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(0, 1)' title='(0, 1)' class='latex' /> to <img src='http://s0.wp.com/latex.php?latex=%280%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(0, 1)' title='(0, 1)' class='latex' />. Also <img src='http://s0.wp.com/latex.php?latex=p+%3C%28%3D%2C%3E%29+1+%5CRightarrow+l+%3E+%28%3D%2C%3C%29+k&amp;bg=fff&amp;fg=222&amp;s=0' alt='p &lt;(=,&gt;) 1 &#92;Rightarrow l &gt; (=,&lt;) k' title='p &lt;(=,&gt;) 1 &#92;Rightarrow l &gt; (=,&lt;) k' class='latex' />. We also see that the ratio <img src='http://s0.wp.com/latex.php?latex=L%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L/K' title='L/K' class='latex' /> is a function of <img src='http://s0.wp.com/latex.php?latex=k%2C+l&amp;bg=fff&amp;fg=222&amp;s=0' alt='k, l' title='k, l' class='latex' /> and we traditionally call it the <em>multiplier</em> and denote by <img src='http://s0.wp.com/latex.php?latex=M_%7Bp%7D%28l%2C+k%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='M_{p}(l, k)' title='M_{p}(l, k)' class='latex' />. Thus we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+pM_%7Bp%7D%5E%7B2%7D%28l%2C+k%29+%3D+p%5Cleft%28%5Cfrac%7BL%7D%7BK%7D%5Cright%29%5E%7B2%7D+%3D+%5Cfrac%7Bkk%27%5E%7B2%7D%7D%7Bll%27%5E%7B2%7D%7D%5Cfrac%7Bdl%7D%7Bdk%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle pM_{p}^{2}(l, k) = p&#92;left(&#92;frac{L}{K}&#92;right)^{2} = &#92;frac{kk&#039;^{2}}{ll&#039;^{2}}&#92;frac{dl}{dk}' title='&#92;displaystyle pM_{p}^{2}(l, k) = p&#92;left(&#92;frac{L}{K}&#92;right)^{2} = &#92;frac{kk&#039;^{2}}{ll&#039;^{2}}&#92;frac{dl}{dk}' class='latex' /></p>
<p>Also it should be noted that if <img src='http://s0.wp.com/latex.php?latex=p+%3C+%28%3D%2C+%3E%29+1+%5CRightarrow+M_%7Bp%7D%28l%2C+k%29+%3E+%28%3D%2C+%3C%29+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='p &lt; (=, &gt;) 1 &#92;Rightarrow M_{p}(l, k) &gt; (=, &lt;) 1' title='p &lt; (=, &gt;) 1 &#92;Rightarrow M_{p}(l, k) &gt; (=, &lt;) 1' class='latex' />.</p>
<p><strong>Jacobi&#8217;s Transformation Theory</strong></p>
<p>Jacobi shows that the relation between <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> is algebraic when <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> is rational. First of all as an example we can take <img src='http://s0.wp.com/latex.php?latex=p+%3D+2&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = 2' title='p = 2' class='latex' />. Then by Landen&#8217;s transformation <img src='http://s0.wp.com/latex.php?latex=l+%3D+%281+-+k%27%29%2F%281+%2B+k%27%29+%3C+k&amp;bg=fff&amp;fg=222&amp;s=0' alt='l = (1 - k&#039;)/(1 + k&#039;) &lt; k' title='l = (1 - k&#039;)/(1 + k&#039;) &lt; k' class='latex' /> we see that <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+2K%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = 2K&#039;/K' title='L&#039;/L = 2K&#039;/K' class='latex' />. Clearly here the relation between <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> is algebraic and is also given by <img src='http://s0.wp.com/latex.php?latex=k+%3D+2%5Csqrt%7Bl%7D%2F%281+%2B+l%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='k = 2&#92;sqrt{l}/(1 + l)' title='k = 2&#92;sqrt{l}/(1 + l)' class='latex' />. Also</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bdl%7D%7Bdk%7D+%3D+%5Cfrac%7B%281+%2B+k%27%29%28k%2Fk%27%29+%2B+%281+-+k%27%29%28k%2Fk%27%29%7D%7B%281+%2B+k%27%29%5E%7B2%7D%7D+%3D+%5Cfrac%7B2k%7D%7Bk%27%281+%2B+k%27%29%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{dl}{dk} = &#92;frac{(1 + k&#039;)(k/k&#039;) + (1 - k&#039;)(k/k&#039;)}{(1 + k&#039;)^{2}} = &#92;frac{2k}{k&#039;(1 + k&#039;)^{2}}' title='&#92;displaystyle &#92;frac{dl}{dk} = &#92;frac{(1 + k&#039;)(k/k&#039;) + (1 - k&#039;)(k/k&#039;)}{(1 + k&#039;)^{2}} = &#92;frac{2k}{k&#039;(1 + k&#039;)^{2}}' class='latex' /> and therefore</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+2M_%7B2%7D%5E%7B2%7D%28l%2C+k%29+%3D+%5Cfrac%7Bkk%27%5E%7B2%7D%7D%7Bll%27%5E%7B2%7D%7D%5Cfrac%7Bdl%7D%7Bdk%7D+%3D+kk%27%5E%7B2%7D%5Cfrac%7B1+%2B+k%27%7D%7B1+-+k%27%7D%5Cfrac%7B%281+%2B+k%27%29%5E%7B2%7D%7D%7B4k%27%7D%5Cfrac%7B2k%7D%7Bk%27%281+%2B+k%27%29%5E%7B2%7D%7D+%3D+%5Cfrac%7Bk%5E%7B2%7D%7D%7B2%7D%5Cfrac%7B%281+%2B+k%27%29%5E%7B2%7D%7D%7B1+-+k%27%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 2M_{2}^{2}(l, k) = &#92;frac{kk&#039;^{2}}{ll&#039;^{2}}&#92;frac{dl}{dk} = kk&#039;^{2}&#92;frac{1 + k&#039;}{1 - k&#039;}&#92;frac{(1 + k&#039;)^{2}}{4k&#039;}&#92;frac{2k}{k&#039;(1 + k&#039;)^{2}} = &#92;frac{k^{2}}{2}&#92;frac{(1 + k&#039;)^{2}}{1 - k&#039;^{2}}' title='&#92;displaystyle 2M_{2}^{2}(l, k) = &#92;frac{kk&#039;^{2}}{ll&#039;^{2}}&#92;frac{dl}{dk} = kk&#039;^{2}&#92;frac{1 + k&#039;}{1 - k&#039;}&#92;frac{(1 + k&#039;)^{2}}{4k&#039;}&#92;frac{2k}{k&#039;(1 + k&#039;)^{2}} = &#92;frac{k^{2}}{2}&#92;frac{(1 + k&#039;)^{2}}{1 - k&#039;^{2}}' class='latex' /></p>
<p>so that <img src='http://s0.wp.com/latex.php?latex=L%2FK+%3D+M_%7B2%7D%28l%2C+k%29+%3D+%281+%2B+k%27%29%2F2+%3D+1%2F%281+%2B+l%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='L/K = M_{2}(l, k) = (1 + k&#039;)/2 = 1/(1 + l)' title='L/K = M_{2}(l, k) = (1 + k&#039;)/2 = 1/(1 + l)' class='latex' /> as should have been the case.</p>
<p>Next let&#8217;s understand that it is necessary only to study the case when <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> is a positive prime. For, if for positive prime <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> the relation <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+pK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = pK&#039;/K' title='L&#039;/L = pK&#039;/K' class='latex' /> is leading to an algebraic relation between <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> then we can show that the same kind of relation holds when <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> is any positive rational number. First suppose <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> is composite positive integer and <img src='http://s0.wp.com/latex.php?latex=p+%3D+p_%7B1%7Dp_%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = p_{1}p_{2}' title='p = p_{1}p_{2}' class='latex' /> where both <img src='http://s0.wp.com/latex.php?latex=p_%7B1%7D%2C+p_%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='p_{1}, p_{2}' title='p_{1}, p_{2}' class='latex' /> are primes. Then <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+pK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = pK&#039;/K' title='L&#039;/L = pK&#039;/K' class='latex' /> can be written <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+p_%7B1%7D%5CGamma%27%2F%5CGamma+%3D+p_%7B1%7D%28p_%7B2%7DK%27%2FK%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = p_{1}&#92;Gamma&#039;/&#92;Gamma = p_{1}(p_{2}K&#039;/K)' title='L&#039;/L = p_{1}&#92;Gamma&#039;/&#92;Gamma = p_{1}(p_{2}K&#039;/K)' class='latex' />. Since the relation between <img src='http://s0.wp.com/latex.php?latex=l%2C+%5Cgamma&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, &#92;gamma' title='l, &#92;gamma' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%5Cgamma%2C+k&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;gamma, k' title='&#92;gamma, k' class='latex' /> is algebraic by assumption, the relation between <img src='http://s0.wp.com/latex.php?latex=l%2C+k&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, k' title='l, k' class='latex' /> is algebraic. The same reasoning can be extended to the case when <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> is a product of any number of primes. Thus the case of positive integers is handled. In case of <img src='http://s0.wp.com/latex.php?latex=p+%3D+a%2Fb&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = a/b' title='p = a/b' class='latex' /> where <img src='http://s0.wp.com/latex.php?latex=a%2C+b&amp;bg=fff&amp;fg=222&amp;s=0' alt='a, b' title='a, b' class='latex' /> are positive integers we can write <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+pK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = pK&#039;/K' title='L&#039;/L = pK&#039;/K' class='latex' /> as <img src='http://s0.wp.com/latex.php?latex=bL%27%2FL+%3D+aK%27%2FK+%3D+%5CGamma%27%2F%5CGamma&amp;bg=fff&amp;fg=222&amp;s=0' alt='bL&#039;/L = aK&#039;/K = &#92;Gamma&#039;/&#92;Gamma' title='bL&#039;/L = aK&#039;/K = &#92;Gamma&#039;/&#92;Gamma' class='latex' />. Since the relation between <img src='http://s0.wp.com/latex.php?latex=%5Cgamma%2C+k&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;gamma, k' title='&#92;gamma, k' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%5Cgamma%2C+l&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;gamma, l' title='&#92;gamma, l' class='latex' /> is algebraic therefore the relation between <img src='http://s0.wp.com/latex.php?latex=k%2C+l&amp;bg=fff&amp;fg=222&amp;s=0' alt='k, l' title='k, l' class='latex' /> is algebraic.</p>
<p>So from now on let&#8217;s assume that <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> is a positive prime. The algebraic relation between <img src='http://s0.wp.com/latex.php?latex=l%2C+k&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, k' title='l, k' class='latex' /> implied by the equation <img src='http://s0.wp.com/latex.php?latex=L%27%2FL+%3D+pK%27%2FK&amp;bg=fff&amp;fg=222&amp;s=0' alt='L&#039;/L = pK&#039;/K' title='L&#039;/L = pK&#039;/K' class='latex' /> is called a <em>modular equation of degree</em> <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' />. The way Jacobi obtained these modular equations is related to the transformation of elliptic integrals. Consider the differential</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bdy%7D%7B%5Csqrt%7B%281+-+y%5E%7B2%7D%29%281+-+l%5E%7B2%7Dy%5E%7B2%7D%29%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{dy}{&#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})}}' title='&#92;displaystyle &#92;frac{dy}{&#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})}}' class='latex' /></p>
<p>By suitable rational transformation <img src='http://s0.wp.com/latex.php?latex=y+%3D+f%28x%29%2Fg%28x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = f(x)/g(x)' title='y = f(x)/g(x)' class='latex' /> where <img src='http://s0.wp.com/latex.php?latex=f%28x%29%2C+g%28x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(x), g(x)' title='f(x), g(x)' class='latex' /> are polynomials it is expected to reduce the differential to the form</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bdx%7D%7BM%5Csqrt%7B%281+-+x%5E%7B2%7D%29%281+-+k%5E%7B2%7Dx%5E%7B2%7D%29%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{dx}{M&#92;sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}' title='&#92;displaystyle &#92;frac{dx}{M&#92;sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}' class='latex' /></p>
<p>Thus in effect we wish to establish a relation of the form</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7BMdy%7D%7B%5Csqrt%7B%281+-+y%5E%7B2%7D%29%281+-+l%5E%7B2%7Dy%5E%7B2%7D%29%7D%7D+%3D+%5Cfrac%7Bdx%7D%7B%5Csqrt%7B%281+-+x%5E%7B2%7D%29%281+-+k%5E%7B2%7Dx%5E%7B2%7D%29%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{Mdy}{&#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = &#92;frac{dx}{&#92;sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}' title='&#92;displaystyle &#92;frac{Mdy}{&#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = &#92;frac{dx}{&#92;sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}' class='latex' /></p>
<p>where <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> is a rational function of <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=M&amp;bg=fff&amp;fg=222&amp;s=0' alt='M' title='M' class='latex' /> is some constant. That such a transformation is possible is not so obvious. Jacobi started his <em>Fundamenta Nova</em> by describing this elegant theory of transformation and showed that such a transformation exists for every value of the positive prime <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> which is called the order of the transformation. In what follows we shall assume that <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> is an odd prime (<img src='http://s0.wp.com/latex.php?latex=p+%3D+2&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = 2' title='p = 2' class='latex' /> is covered by the famous Landen Transformations).</p>
<p>The constant <img src='http://s0.wp.com/latex.php?latex=M&amp;bg=fff&amp;fg=222&amp;s=0' alt='M' title='M' class='latex' /> is called <em>multiplier</em> and depends upon the values of <img src='http://s0.wp.com/latex.php?latex=l%2C+k&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, k' title='l, k' class='latex' /> and is related to the multiplier <img src='http://s0.wp.com/latex.php?latex=M_%7Bp%7D%28l%2C+k%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='M_{p}(l, k)' title='M_{p}(l, k)' class='latex' />. To simplify the process of transformation the rational function <img src='http://s0.wp.com/latex.php?latex=y+%3D+U%28x%29%2FV%28x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = U(x)/V(x)' title='y = U(x)/V(x)' class='latex' /> is chosen to be of very specific form. With the requirement that <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> vanishes with <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> we take <img src='http://s0.wp.com/latex.php?latex=y+%3D+xf%28x%29%2Fg%28x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = xf(x)/g(x)' title='y = xf(x)/g(x)' class='latex' /> where <img src='http://s0.wp.com/latex.php?latex=f%28x%29%2C+g%28x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(x), g(x)' title='f(x), g(x)' class='latex' /> are polynomials of degrees <img src='http://s0.wp.com/latex.php?latex=%28p+-+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(p - 1)' title='(p - 1)' class='latex' />. Also <img src='http://s0.wp.com/latex.php?latex=f%28x%29%2C+g%28x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='f(x), g(x)' title='f(x), g(x)' class='latex' /> are supposed to be even functions so that they are functions of <img src='http://s0.wp.com/latex.php?latex=x%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='x^{2}' title='x^{2}' class='latex' /> and then we can write</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+y+%3D+%5Cfrac%7BxN%281%2C+x%5E%7B2%7D%29%7D%7BD%281%2C+x%5E%7B2%7D%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle y = &#92;frac{xN(1, x^{2})}{D(1, x^{2})}' title='&#92;displaystyle y = &#92;frac{xN(1, x^{2})}{D(1, x^{2})}' class='latex' /></p>
<p>where <img src='http://s0.wp.com/latex.php?latex=N%2C+D&amp;bg=fff&amp;fg=222&amp;s=0' alt='N, D' title='N, D' class='latex' /> are homogeneous polynomials of degree <img src='http://s0.wp.com/latex.php?latex=%28p+-+1%29%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='(p - 1)/2' title='(p - 1)/2' class='latex' />. Another condition which we impose is that the relation between <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> does not change when <img src='http://s0.wp.com/latex.php?latex=%28x%2C+y%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(x, y)' title='(x, y)' class='latex' /> is replaced by <img src='http://s0.wp.com/latex.php?latex=%281%2Fkx%2C+1%2Fly%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(1/kx, 1/ly)' title='(1/kx, 1/ly)' class='latex' />. That this is possible needs to be demonstrated. First of all we can see that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+N%5Cleft%281%2C+%5Cfrac%7B1%7D%7Bk%5E%7B2%7Dx%5E%7B2%7D%7D%5Cright%29+%3D+%5Cleft%28%5Cfrac%7B1%7D%7Bkx%7D%5Cright%29%5E%7Bp+-+1%7DN%28k%5E%7B2%7Dx%5E%7B2%7D%2C+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle N&#92;left(1, &#92;frac{1}{k^{2}x^{2}}&#92;right) = &#92;left(&#92;frac{1}{kx}&#92;right)^{p - 1}N(k^{2}x^{2}, 1)' title='&#92;displaystyle N&#92;left(1, &#92;frac{1}{k^{2}x^{2}}&#92;right) = &#92;left(&#92;frac{1}{kx}&#92;right)^{p - 1}N(k^{2}x^{2}, 1)' class='latex' /></p>
<p>We can next choose <img src='http://s0.wp.com/latex.php?latex=D%281%2C+x%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='D(1, x^{2})' title='D(1, x^{2})' class='latex' /> such that <img src='http://s0.wp.com/latex.php?latex=N%28k%5E%7B2%7Dx%5E%7B2%7D%2C+1%29+%3D+%5CDelta+D%281%2C+x%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='N(k^{2}x^{2}, 1) = &#92;Delta D(1, x^{2})' title='N(k^{2}x^{2}, 1) = &#92;Delta D(1, x^{2})' class='latex' /> where <img src='http://s0.wp.com/latex.php?latex=%5CDelta&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;Delta' title='&#92;Delta' class='latex' /> is a constant. Thus the coefficients of <img src='http://s0.wp.com/latex.php?latex=D%281%2C+x%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='D(1, x^{2})' title='D(1, x^{2})' class='latex' /> are same as those of <img src='http://s0.wp.com/latex.php?latex=N%281%2C+x%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='N(1, x^{2})' title='N(1, x^{2})' class='latex' />, but in reverse order and multiplied by suitable powers of <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' />. It thus follows that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+N%5Cleft%281%2C+%5Cfrac%7B1%7D%7Bk%5E%7B2%7Dx%5E%7B2%7D%7D%5Cright%29+%3D+%5CDelta%5Cleft%28%5Cfrac%7B1%7D%7Bkx%7D%5Cright%29%5E%7Bp+-+1%7DD%281%2C+x%5E%7B2%7D%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle N&#92;left(1, &#92;frac{1}{k^{2}x^{2}}&#92;right) = &#92;Delta&#92;left(&#92;frac{1}{kx}&#92;right)^{p - 1}D(1, x^{2})' title='&#92;displaystyle N&#92;left(1, &#92;frac{1}{k^{2}x^{2}}&#92;right) = &#92;Delta&#92;left(&#92;frac{1}{kx}&#92;right)^{p - 1}D(1, x^{2})' class='latex' /></p>
<p>Replacing <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=1%2Fkx&amp;bg=fff&amp;fg=222&amp;s=0' alt='1/kx' title='1/kx' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+N%281%2C+x%5E%7B2%7D%29+%3D+%5CDelta+x%5E%7Bp+-+1%7DD%5Cleft%281%2C+%5Cfrac%7B1%7D%7Bk%5E%7B2%7Dx%5E%7B2%7D%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle N(1, x^{2}) = &#92;Delta x^{p - 1}D&#92;left(1, &#92;frac{1}{k^{2}x^{2}}&#92;right)' title='&#92;displaystyle N(1, x^{2}) = &#92;Delta x^{p - 1}D&#92;left(1, &#92;frac{1}{k^{2}x^{2}}&#92;right)' class='latex' /></p>
<p>Multiplying the above two equations we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+N%281%2C+x%5E%7B2%7D%29N%5Cleft%281%2C+%5Cfrac%7B1%7D%7Bk%5E%7B2%7Dx%5E%7B2%7D%7D%5Cright%29+%3D+%5Cfrac%7B%5CDelta%5E%7B2%7D%7D%7Bk%5E%7Bp+-+1%7D%7DD%281%2C+x%5E%7B2%7D%29D%5Cleft%281%2C+%5Cfrac%7B1%7D%7Bk%5E%7B2%7Dx%5E%7B2%7D%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle N(1, x^{2})N&#92;left(1, &#92;frac{1}{k^{2}x^{2}}&#92;right) = &#92;frac{&#92;Delta^{2}}{k^{p - 1}}D(1, x^{2})D&#92;left(1, &#92;frac{1}{k^{2}x^{2}}&#92;right)' title='&#92;displaystyle N(1, x^{2})N&#92;left(1, &#92;frac{1}{k^{2}x^{2}}&#92;right) = &#92;frac{&#92;Delta^{2}}{k^{p - 1}}D(1, x^{2})D&#92;left(1, &#92;frac{1}{k^{2}x^{2}}&#92;right)' class='latex' /></p>
<p>Now let&#8217;s suppose that in the defining equation if we put <img src='http://s0.wp.com/latex.php?latex=1%2Fkx&amp;bg=fff&amp;fg=222&amp;s=0' alt='1/kx' title='1/kx' class='latex' /> in place of <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> then the value of <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> changes to <img src='http://s0.wp.com/latex.php?latex=y_%7B1%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='y_{1}' title='y_{1}' class='latex' />. Thus</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+y_%7B1%7D+%3D+%5Cdfrac%7B%5Cdfrac%7B1%7D%7Bkx%7DN%5Cleft%281%2C+%5Cdfrac%7B1%7D%7Bk%5E%7B2%7Dx%5E%7B2%7D%7D%5Cright%29%7D%7BD%5Cleft%281%2C+%5Cdfrac%7B1%7D%7Bk%5E%7B2%7Dx%5E%7B2%7D%7D%5Cright%29%7D+%3D+%5Cfrac%7B1%7D%7Bkx%7D%5Cfrac%7B%5CDelta%5E%7B2%7D%7D%7Bk%5E%7Bp+-+1%7D%7D%5Cfrac%7BD%281%2C+x%5E%7B2%7D%29%7D%7BN%281%2C+x%5E%7B2%7D%29%7D+%3D+%5Cfrac%7B%5CDelta%5E%7B2%7D%7D%7Bk%5E%7Bp%7D%7D%5Cfrac%7BD%281%2C+x%5E%7B2%7D%29%7D%7BxN%281%2C+x%5E%7B2%7D%29%7D+%3D+%5Cfrac%7B%5CDelta%5E%7B2%7D%7D%7Bk%5E%7Bp%7D%7D%5Cfrac%7B1%7D%7By%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle y_{1} = &#92;dfrac{&#92;dfrac{1}{kx}N&#92;left(1, &#92;dfrac{1}{k^{2}x^{2}}&#92;right)}{D&#92;left(1, &#92;dfrac{1}{k^{2}x^{2}}&#92;right)} = &#92;frac{1}{kx}&#92;frac{&#92;Delta^{2}}{k^{p - 1}}&#92;frac{D(1, x^{2})}{N(1, x^{2})} = &#92;frac{&#92;Delta^{2}}{k^{p}}&#92;frac{D(1, x^{2})}{xN(1, x^{2})} = &#92;frac{&#92;Delta^{2}}{k^{p}}&#92;frac{1}{y}' title='&#92;displaystyle y_{1} = &#92;dfrac{&#92;dfrac{1}{kx}N&#92;left(1, &#92;dfrac{1}{k^{2}x^{2}}&#92;right)}{D&#92;left(1, &#92;dfrac{1}{k^{2}x^{2}}&#92;right)} = &#92;frac{1}{kx}&#92;frac{&#92;Delta^{2}}{k^{p - 1}}&#92;frac{D(1, x^{2})}{N(1, x^{2})} = &#92;frac{&#92;Delta^{2}}{k^{p}}&#92;frac{D(1, x^{2})}{xN(1, x^{2})} = &#92;frac{&#92;Delta^{2}}{k^{p}}&#92;frac{1}{y}' class='latex' /></p>
<p>Clearly this reduces to <img src='http://s0.wp.com/latex.php?latex=1%2Fly&amp;bg=fff&amp;fg=222&amp;s=0' alt='1/ly' title='1/ly' class='latex' /> provided <img src='http://s0.wp.com/latex.php?latex=l+%3D+k%5E%7Bp%7D%2F%5CDelta%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='l = k^{p}/&#92;Delta^{2}' title='l = k^{p}/&#92;Delta^{2}' class='latex' />.</p>
<p>Hence a transformation of the form <img src='http://s0.wp.com/latex.php?latex=y+%3D+U%28x%29%2FV%28x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = U(x)/V(x)' title='y = U(x)/V(x)' class='latex' /> with <img src='http://s0.wp.com/latex.php?latex=U&amp;bg=fff&amp;fg=222&amp;s=0' alt='U' title='U' class='latex' /> of degree <img src='http://s0.wp.com/latex.php?latex=p&amp;bg=fff&amp;fg=222&amp;s=0' alt='p' title='p' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=V&amp;bg=fff&amp;fg=222&amp;s=0' alt='V' title='V' class='latex' /> of degree <img src='http://s0.wp.com/latex.php?latex=%28p+-+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(p - 1)' title='(p - 1)' class='latex' /> with <img src='http://s0.wp.com/latex.php?latex=U&amp;bg=fff&amp;fg=222&amp;s=0' alt='U' title='U' class='latex' /> being odd function and <img src='http://s0.wp.com/latex.php?latex=V&amp;bg=fff&amp;fg=222&amp;s=0' alt='V' title='V' class='latex' /> being even function is possible which remains invariant under the change of variables <img src='http://s0.wp.com/latex.php?latex=%28x%2C+y%29+%5Cto+%281%2Fkx%2C+1%2Fly%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(x, y) &#92;to (1/kx, 1/ly)' title='(x, y) &#92;to (1/kx, 1/ly)' class='latex' />.</p>
<p>Next we would like to have further restrictions on the form of <img src='http://s0.wp.com/latex.php?latex=U&amp;bg=fff&amp;fg=222&amp;s=0' alt='U' title='U' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=V&amp;bg=fff&amp;fg=222&amp;s=0' alt='V' title='V' class='latex' />. Putting <img src='http://s0.wp.com/latex.php?latex=y+%3D+U%2FV&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = U/V' title='y = U/V' class='latex' /> leads to</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+dy+%3D+%5Cfrac%7BVU%27+-+UV%27%7D%7BV%5E%7B2%7D%7D%5C%2Cdx%2C%5C%2C%5C%2C+%5Csqrt%7B%281+-+y%5E%7B2%7D%29%281+-+l%5E%7B2%7Dy%5E%7B2%7D%29%7D+%3D+%5Cfrac%7B1%7D%7BV%5E%7B2%7D%7D%5Csqrt%7B%28V%5E%7B2%7D+-+U%5E%7B2%7D%29%28V%5E%7B2%7D+-+l%5E%7B2%7DU%5E%7B2%7D%29%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle dy = &#92;frac{VU&#039; - UV&#039;}{V^{2}}&#92;,dx,&#92;,&#92;, &#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})} = &#92;frac{1}{V^{2}}&#92;sqrt{(V^{2} - U^{2})(V^{2} - l^{2}U^{2})}' title='&#92;displaystyle dy = &#92;frac{VU&#039; - UV&#039;}{V^{2}}&#92;,dx,&#92;,&#92;, &#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})} = &#92;frac{1}{V^{2}}&#92;sqrt{(V^{2} - U^{2})(V^{2} - l^{2}U^{2})}' class='latex' /></p>
<p>and hence</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bdy%7D%7B%5Csqrt%7B%281+-+y%5E%7B2%7D%29%281+-+l%5E%7B2%7Dy%5E%7B2%7D%29%7D%7D+%3D+%5Cfrac%7BVU%27+-+UV%27%7D%7B%5Csqrt%7B%28V%5E%7B2%7D+-+U%5E%7B2%7D%29%28V%5E%7B2%7D+-+l%5E%7B2%7DU%5E%7B2%7D%29%7D%7D%5C%2Cdx&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{dy}{&#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = &#92;frac{VU&#039; - UV&#039;}{&#92;sqrt{(V^{2} - U^{2})(V^{2} - l^{2}U^{2})}}&#92;,dx' title='&#92;displaystyle &#92;frac{dy}{&#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = &#92;frac{VU&#039; - UV&#039;}{&#92;sqrt{(V^{2} - U^{2})(V^{2} - l^{2}U^{2})}}&#92;,dx' class='latex' /></p>
<p>If it is possible that <img src='http://s0.wp.com/latex.php?latex=V+%2B+U+%3D+%281+%2B+x%29A%5E%7B2%7D%2C+V+-+U+%3D+%281+-+x%29B%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='V + U = (1 + x)A^{2}, V - U = (1 - x)B^{2}' title='V + U = (1 + x)A^{2}, V - U = (1 - x)B^{2}' class='latex' /> for some polynomials <img src='http://s0.wp.com/latex.php?latex=A%2C+B&amp;bg=fff&amp;fg=222&amp;s=0' alt='A, B' title='A, B' class='latex' /> then we can see that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+1+%2B+y+%3D+%281+%2B+x%29A%5E%7B2%7D%2FV%2C%5C%2C+1+-+y+%3D+%281+-+x%29B%5E%7B2%7D%2FV&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 1 + y = (1 + x)A^{2}/V,&#92;, 1 - y = (1 - x)B^{2}/V' title='&#92;displaystyle 1 + y = (1 + x)A^{2}/V,&#92;, 1 - y = (1 - x)B^{2}/V' class='latex' /></p>
<p>and by invariance of relation between <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=y&amp;bg=fff&amp;fg=222&amp;s=0' alt='y' title='y' class='latex' /> under the transformation <img src='http://s0.wp.com/latex.php?latex=%28x%2C+y%29+%5Cto+%281%2Fkx%2C+1%2Fky%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(x, y) &#92;to (1/kx, 1/ky)' title='(x, y) &#92;to (1/kx, 1/ky)' class='latex' /> we can see that we must also have the relations</p>
<p><img src='http://s0.wp.com/latex.php?latex=1+%2B+ly+%3D+%281+%2B+kx%29C%5E%7B2%7D%2FV%2C%5C%2C+1+-+ly+%3D+%281+-+kx%29D%5E%7B2%7D%2FV&amp;bg=fff&amp;fg=222&amp;s=0' alt='1 + ly = (1 + kx)C^{2}/V,&#92;, 1 - ly = (1 - kx)D^{2}/V' title='1 + ly = (1 + kx)C^{2}/V,&#92;, 1 - ly = (1 - kx)D^{2}/V' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=V+%2BlU+%3D+%281+%2B+kx%29C%5E%7B2%7D%2C%5C%2C+V+-+lU+%3D+%281+-+kx%29D%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='V +lU = (1 + kx)C^{2},&#92;, V - lU = (1 - kx)D^{2}' title='V +lU = (1 + kx)C^{2},&#92;, V - lU = (1 - kx)D^{2}' class='latex' /></p>
<p>Now we can see that if <img src='http://s0.wp.com/latex.php?latex=%28x+-+%5Calpha%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(x - &#92;alpha)' title='(x - &#92;alpha)' class='latex' /> is a factor of <img src='http://s0.wp.com/latex.php?latex=A&amp;bg=fff&amp;fg=222&amp;s=0' alt='A' title='A' class='latex' /> then <img src='http://s0.wp.com/latex.php?latex=%28x+-+%5Calpha%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='(x - &#92;alpha)^{2}' title='(x - &#92;alpha)^{2}' class='latex' /> divides <img src='http://s0.wp.com/latex.php?latex=V+%2B+U&amp;bg=fff&amp;fg=222&amp;s=0' alt='V + U' title='V + U' class='latex' /> and hence <img src='http://s0.wp.com/latex.php?latex=%28x+-+%5Calpha%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(x - &#92;alpha)' title='(x - &#92;alpha)' class='latex' /> divides <img src='http://s0.wp.com/latex.php?latex=%28V+%2B+U%29U%27+-+U%28V+%2B+U%29%27+%3D+VU%27+-+UV%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='(V + U)U&#039; - U(V + U)&#039; = VU&#039; - UV&#039;' title='(V + U)U&#039; - U(V + U)&#039; = VU&#039; - UV&#039;' class='latex' />. From the same logic it follows that <img src='http://s0.wp.com/latex.php?latex=A%2C+B%2C+C%2C+D&amp;bg=fff&amp;fg=222&amp;s=0' alt='A, B, C, D' title='A, B, C, D' class='latex' /> each divide <img src='http://s0.wp.com/latex.php?latex=VU%27+-+UV%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='VU&#039; - UV&#039;' title='VU&#039; - UV&#039;' class='latex' /> and since the degrees of <img src='http://s0.wp.com/latex.php?latex=VU%27+-+UV%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='VU&#039; - UV&#039;' title='VU&#039; - UV&#039;' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=ABCD&amp;bg=fff&amp;fg=222&amp;s=0' alt='ABCD' title='ABCD' class='latex' /> are same (equal to <img src='http://s0.wp.com/latex.php?latex=%282p+-+2%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(2p - 2)' title='(2p - 2)' class='latex' />) it follows that the ratio <img src='http://s0.wp.com/latex.php?latex=ABCD%2F%28VU%27+-+UV%27%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='ABCD/(VU&#039; - UV&#039;)' title='ABCD/(VU&#039; - UV&#039;)' class='latex' /> is a constant which we denote by <img src='http://s0.wp.com/latex.php?latex=M&amp;bg=fff&amp;fg=222&amp;s=0' alt='M' title='M' class='latex' />. Thus we arrive at</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7BMdy%7D%7B%5Csqrt%7B%281+-+y%5E%7B2%7D%29%281+-+l%5E%7B2%7Dy%5E%7B2%7D%29%7D%7D+%3D+%5Cfrac%7Bdx%7D%7B%5Csqrt%7B%281+-+x%5E%7B2%7D%29%281+-+k%5E%7B2%7Dx%5E%7B2%7D%29%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{Mdy}{&#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = &#92;frac{dx}{&#92;sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}' title='&#92;displaystyle &#92;frac{Mdy}{&#92;sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = &#92;frac{dx}{&#92;sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}' class='latex' /></p>
<p>The idea is now to choose polynomials <img src='http://s0.wp.com/latex.php?latex=A%2C+B&amp;bg=fff&amp;fg=222&amp;s=0' alt='A, B' title='A, B' class='latex' /> in a suitable form. We keep <img src='http://s0.wp.com/latex.php?latex=A+%3D+P+%2B+Qx%2C+B+%3D+P+-+Qx&amp;bg=fff&amp;fg=222&amp;s=0' alt='A = P + Qx, B = P - Qx' title='A = P + Qx, B = P - Qx' class='latex' /> such that <img src='http://s0.wp.com/latex.php?latex=P%2C+Q&amp;bg=fff&amp;fg=222&amp;s=0' alt='P, Q' title='P, Q' class='latex' /> are even functions of <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> and the degree of <img src='http://s0.wp.com/latex.php?latex=P+%5Cpm+Qx&amp;bg=fff&amp;fg=222&amp;s=0' alt='P &#92;pm Qx' title='P &#92;pm Qx' class='latex' /> is <img src='http://s0.wp.com/latex.php?latex=%28p+-+1%29%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='(p - 1)/2' title='(p - 1)/2' class='latex' />. In general if <img src='http://s0.wp.com/latex.php?latex=p+%3D+%284n+-+1%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = (4n - 1)' title='p = (4n - 1)' class='latex' /> then the degrees of <img src='http://s0.wp.com/latex.php?latex=P%2C+Q&amp;bg=fff&amp;fg=222&amp;s=0' alt='P, Q' title='P, Q' class='latex' /> are <img src='http://s0.wp.com/latex.php?latex=%282n+-+2%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(2n - 2)' title='(2n - 2)' class='latex' /> and if <img src='http://s0.wp.com/latex.php?latex=p+%3D+4n+%2B+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = 4n + 1' title='p = 4n + 1' class='latex' />, degree of <img src='http://s0.wp.com/latex.php?latex=P&amp;bg=fff&amp;fg=222&amp;s=0' alt='P' title='P' class='latex' /> is <img src='http://s0.wp.com/latex.php?latex=2n&amp;bg=fff&amp;fg=222&amp;s=0' alt='2n' title='2n' class='latex' /> and that of <img src='http://s0.wp.com/latex.php?latex=Q&amp;bg=fff&amp;fg=222&amp;s=0' alt='Q' title='Q' class='latex' /> is <img src='http://s0.wp.com/latex.php?latex=%282n+-+2%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(2n - 2)' title='(2n - 2)' class='latex' />.</p>
<p>Thus we arrive at the following transformation</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B1+-+y%7D%7B1+%2B+y%7D+%3D+%5Cfrac%7B1+-+x%7D%7B1+%2B+x%7D%5Cfrac%7B%28P+-+Qx%29%5E%7B2%7D%7D%7B%28P+%2B+Qx%29%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{1 - y}{1 + y} = &#92;frac{1 - x}{1 + x}&#92;frac{(P - Qx)^{2}}{(P + Qx)^{2}}' title='&#92;displaystyle &#92;frac{1 - y}{1 + y} = &#92;frac{1 - x}{1 + x}&#92;frac{(P - Qx)^{2}}{(P + Qx)^{2}}' class='latex' /></p>
<p>so that</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+y+%3D+%5Cfrac%7Bx%28P%5E%7B2%7D+%2B+2PQ+%2B+Q%5E%7B2%7Dx%5E%7B2%7D%29%7D%7BP%5E%7B2%7D+%2B+2PQx%5E%7B2%7D+%2B+Q%5E%7B2%7Dx%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle y = &#92;frac{x(P^{2} + 2PQ + Q^{2}x^{2})}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' title='&#92;displaystyle y = &#92;frac{x(P^{2} + 2PQ + Q^{2}x^{2})}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' class='latex' /></p>
<p>and then we obtain</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+1+-+y+%3D+%5Cfrac%7B%281+-+x%29%28P+-+Qx%29%5E%7B2%7D%7D%7BP%5E%7B2%7D+%2B+2PQx%5E%7B2%7D+%2B+Q%5E%7B2%7Dx%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 1 - y = &#92;frac{(1 - x)(P - Qx)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' title='&#92;displaystyle 1 - y = &#92;frac{(1 - x)(P - Qx)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+1+%2B+y+%3D+%5Cfrac%7B%281+%2B+x%29%28P+%2B+Qx%29%5E%7B2%7D%7D%7BP%5E%7B2%7D+%2B+2PQx%5E%7B2%7D+%2B+Q%5E%7B2%7Dx%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 1 + y = &#92;frac{(1 + x)(P + Qx)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' title='&#92;displaystyle 1 + y = &#92;frac{(1 + x)(P + Qx)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+1+-+ly+%3D+%5Cfrac%7B%281+-+kx%29%28P%27+-+Q%27x%29%5E%7B2%7D%7D%7BP%5E%7B2%7D+%2B+2PQx%5E%7B2%7D+%2B+Q%5E%7B2%7Dx%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 1 - ly = &#92;frac{(1 - kx)(P&#039; - Q&#039;x)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' title='&#92;displaystyle 1 - ly = &#92;frac{(1 - kx)(P&#039; - Q&#039;x)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+1+%2B+ly+%3D+%5Cfrac%7B%281+%2B+kx%29%28P%27+%2B+Q%27x%29%5E%7B2%7D%7D%7BP%5E%7B2%7D+%2B+2PQx%5E%7B2%7D+%2B+Q%5E%7B2%7Dx%5E%7B2%7D%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle 1 + ly = &#92;frac{(1 + kx)(P&#039; + Q&#039;x)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' title='&#92;displaystyle 1 + ly = &#92;frac{(1 + kx)(P&#039; + Q&#039;x)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}' class='latex' /></p>
<p>The actual methodology of finding the polynomials <img src='http://s0.wp.com/latex.php?latex=U%2C+V&amp;bg=fff&amp;fg=222&amp;s=0' alt='U, V' title='U, V' class='latex' /> (or <img src='http://s0.wp.com/latex.php?latex=P%2C+Q&amp;bg=fff&amp;fg=222&amp;s=0' alt='P, Q' title='P, Q' class='latex' />) would be dealt with in next post by showing examples with <img src='http://s0.wp.com/latex.php?latex=p+%3D+3&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = 3' title='p = 3' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=p+%3D+5&amp;bg=fff&amp;fg=222&amp;s=0' alt='p = 5' title='p = 5' class='latex' />. From the theoretical investigation above it is clear that the process of finding these polynomials (i.e. their coefficients) is totally algebraical and hence the relation between <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> is algebraic (note that <img src='http://s0.wp.com/latex.php?latex=l+%3D+k%5E%7Bp%7D%2F%5CDelta%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='l = k^{p}/&#92;Delta^{2}' title='l = k^{p}/&#92;Delta^{2}' class='latex' />, and that <img src='http://s0.wp.com/latex.php?latex=%5CDelta&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;Delta' title='&#92;Delta' class='latex' /> itself would be an algebraic function of <img src='http://s0.wp.com/latex.php?latex=l%2C+k&amp;bg=fff&amp;fg=222&amp;s=0' alt='l, k' title='l, k' class='latex' />). It should also be observed that the multiplier <img src='http://s0.wp.com/latex.php?latex=M&amp;bg=fff&amp;fg=222&amp;s=0' alt='M' title='M' class='latex' /> is also an algebraical function of <img src='http://s0.wp.com/latex.php?latex=l&amp;bg=fff&amp;fg=222&amp;s=0' alt='l' title='l' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=fff&amp;fg=222&amp;s=0' alt='k' title='k' class='latex' /> (calculation of <img src='http://s0.wp.com/latex.php?latex=M&amp;bg=fff&amp;fg=222&amp;s=0' alt='M' title='M' class='latex' /> is easy if we observe that <img src='http://s0.wp.com/latex.php?latex=1%2FM+%3D+dy%2Fdx+%5Ctext%7B+at+%7D+x+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='1/M = dy/dx &#92;text{ at } x = 0' title='1/M = dy/dx &#92;text{ at } x = 0' class='latex' />).</p>
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			<media:title type="html">paramanands</media:title>
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		<title>Elementary Approach to Modular Equations: Hypergeometric Series 2</title>
		<link>http://paramanands.wordpress.com/2011/10/22/elementary-approach-to-modular-equations-hypergeometric-series_2/</link>
		<comments>http://paramanands.wordpress.com/2011/10/22/elementary-approach-to-modular-equations-hypergeometric-series_2/#comments</comments>
		<pubDate>Sat, 22 Oct 2011 08:54:55 +0000</pubDate>
		<dc:creator>paramanands</dc:creator>
				<category><![CDATA[Elliptic Functions]]></category>
		<category><![CDATA[Mathematical Analysis]]></category>

		<guid isPermaLink="false">http://paramanands.wordpress.com/?p=2102</guid>
		<description><![CDATA[To continue our adventures (which started here) with the hypergeometric function we are going to establish the following identity If is neither zero nor a negative integer and if and , then A curious thing to observe here is that the term on the left does not change if we replace by , but the [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=paramanands.wordpress.com&amp;blog=8565079&amp;post=2102&amp;subd=paramanands&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>To continue our adventures (which started <a title="Elementary Approach to Modular Equations: Hypergeometric Series 1" href="http://paramanands.wordpress.com/2011/10/22/elementary-approach-to-modular-equations-hypergeometric-series-1/" target="_blank">here</a>) with the hypergeometric function we are going to establish the following identity</p>
<p><em>If <img src='http://s0.wp.com/latex.php?latex=a+%2B+b+%2B+%281%2F2%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='a + b + (1/2)' title='a + b + (1/2)' class='latex' /> is neither zero nor a negative integer and if <img src='http://s0.wp.com/latex.php?latex=%7Cx%7C+%3C+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='|x| &lt; 1' title='|x| &lt; 1' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%7C4x%281+-+x%29%7C+%3C+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='|4x(1 - x)| &lt; 1' title='|4x(1 - x)| &lt; 1' class='latex' />, then</em></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+F%5Cleft%28a%2C+b%3B+a+%2B+b+%2B+%5Cfrac%7B1%7D%7B2%7D%3B+4x%281+-+x%29%5Cright%29+%3D+F%5Cleft%282a%2C+2b%3B+a+%2B+b+%2B+%5Cfrac%7B1%7D%7B2%7D%3B+x%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle F&#92;left(a, b; a + b + &#92;frac{1}{2}; 4x(1 - x)&#92;right) = F&#92;left(2a, 2b; a + b + &#92;frac{1}{2}; x&#92;right)' title='&#92;displaystyle F&#92;left(a, b; a + b + &#92;frac{1}{2}; 4x(1 - x)&#92;right) = F&#92;left(2a, 2b; a + b + &#92;frac{1}{2}; x&#92;right)' class='latex' /><em></em></p>
<p>A curious thing to observe here is that the term on the left does not change if we replace <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> by <img src='http://s0.wp.com/latex.php?latex=%281+-+x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(1 - x)' title='(1 - x)' class='latex' />, but the right term does change its value. If we restrict ourselves to real positive values of <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> then we need to understand that in the right term the symbol <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> should actually be replaced by minimum of <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=fff&amp;fg=222&amp;s=0' alt='x' title='x' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=1+-+x&amp;bg=fff&amp;fg=222&amp;s=0' alt='1 - x' title='1 - x' class='latex' /> or we can say that the identity as it is holds only when <img src='http://s0.wp.com/latex.php?latex=0+%5Cleq+x+%5Cleq+1%2F2&amp;bg=fff&amp;fg=222&amp;s=0' alt='0 &#92;leq x &#92;leq 1/2' title='0 &#92;leq x &#92;leq 1/2' class='latex' />.</p>
<p>Again to establish the identity we start with function <img src='http://s0.wp.com/latex.php?latex=y+%3D+F%28a%2C+b%3B+a+%2B+b+%2B+1%2F2%3B+z%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = F(a, b; a + b + 1/2; z)' title='y = F(a, b; a + b + 1/2; z)' class='latex' /> which satisfies the differential equation</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+z%281+-+z%29%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdz%5E%7B2%7D%7D+%2B+%5Cleft%28a+%2B+b+%2B+%5Cfrac%7B1%7D%7B2%7D+-+%28a+%2B+b+%2B+1%29z%5Cright%29%5Cfrac%7Bdy%7D%7Bdz%7D+-+aby+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle z(1 - z)&#92;frac{d^{2}y}{dz^{2}} + &#92;left(a + b + &#92;frac{1}{2} - (a + b + 1)z&#92;right)&#92;frac{dy}{dz} - aby = 0' title='&#92;displaystyle z(1 - z)&#92;frac{d^{2}y}{dz^{2}} + &#92;left(a + b + &#92;frac{1}{2} - (a + b + 1)z&#92;right)&#92;frac{dy}{dz} - aby = 0' class='latex' /></p>
<p>Putting <img src='http://s0.wp.com/latex.php?latex=z+%3D+4x%281+-+x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='z = 4x(1 - x)' title='z = 4x(1 - x)' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bdy%7D%7Bdz%7D+%3D+%5Cdfrac%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%7D%7B%5Cdfrac%7Bdz%7D%7Bdx%7D%7D+%3D+%5Cfrac%7B1%7D%7B4%281+-+2x%29%7D%5Cfrac%7Bdy%7D%7Bdx%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{dy}{dz} = &#92;dfrac{&#92;dfrac{dy}{dx}}{&#92;dfrac{dz}{dx}} = &#92;frac{1}{4(1 - 2x)}&#92;frac{dy}{dx}' title='&#92;displaystyle &#92;frac{dy}{dz} = &#92;dfrac{&#92;dfrac{dy}{dx}}{&#92;dfrac{dz}{dx}} = &#92;frac{1}{4(1 - 2x)}&#92;frac{dy}{dx}' class='latex' /> and</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdz%5E%7B2%7D%7D+%3D+%5Cdfrac%7B%5Cdfrac%7Bd%7D%7Bdx%7D%5Cleft%28%5Cdfrac%7B1%7D%7B4%281+-+2x%29%7D%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%29%7D%7B%5Cdfrac%7Bdz%7D%7Bdx%7D%7D+%3D+%5Cfrac%7B1%7D%7B16%281+-+2x%29%7D%5Cleft%28%5Cfrac%7B1%7D%7B1+-+2x%7D%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdx%5E%7B2%7D%7D+%2B+%5Cfrac%7B2%7D%7B%281+-+2x%29%5E%7B2%7D%7D%5Cfrac%7Bdy%7D%7Bdx%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{d^{2}y}{dz^{2}} = &#92;dfrac{&#92;dfrac{d}{dx}&#92;left(&#92;dfrac{1}{4(1 - 2x)}&#92;dfrac{dy}{dx}&#92;right)}{&#92;dfrac{dz}{dx}} = &#92;frac{1}{16(1 - 2x)}&#92;left(&#92;frac{1}{1 - 2x}&#92;frac{d^{2}y}{dx^{2}} + &#92;frac{2}{(1 - 2x)^{2}}&#92;frac{dy}{dx}&#92;right)' title='&#92;displaystyle &#92;frac{d^{2}y}{dz^{2}} = &#92;dfrac{&#92;dfrac{d}{dx}&#92;left(&#92;dfrac{1}{4(1 - 2x)}&#92;dfrac{dy}{dx}&#92;right)}{&#92;dfrac{dz}{dx}} = &#92;frac{1}{16(1 - 2x)}&#92;left(&#92;frac{1}{1 - 2x}&#92;frac{d^{2}y}{dx^{2}} + &#92;frac{2}{(1 - 2x)^{2}}&#92;frac{dy}{dx}&#92;right)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+%5Cfrac%7B1%7D%7B16%281+-+2x%29%5E%7B2%7D%7D%5Cleft%28%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdx%5E%7B2%7D%7D+%2B+%5Cfrac%7B2%7D%7B1+-+2x%7D%5Cfrac%7Bdy%7D%7Bdx%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = &#92;frac{1}{16(1 - 2x)^{2}}&#92;left(&#92;frac{d^{2}y}{dx^{2}} + &#92;frac{2}{1 - 2x}&#92;frac{dy}{dx}&#92;right)' title='&#92;displaystyle = &#92;frac{1}{16(1 - 2x)^{2}}&#92;left(&#92;frac{d^{2}y}{dx^{2}} + &#92;frac{2}{1 - 2x}&#92;frac{dy}{dx}&#92;right)' class='latex' /></p>
<p>and after some symbolic manipulations we arrive at the following differential equation</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+x%281+-+x%29%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdx%5E%7B2%7D%7D+%2B+%5Cleft%28a+%2B+b+%2B+%5Cfrac%7B1%7D%7B2%7D+-+%282a+%2B+2b+%2B+1%29x%5Cright%29%5Cfrac%7Bdy%7D%7Bdx%7D+-+4aby+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle x(1 - x)&#92;frac{d^{2}y}{dx^{2}} + &#92;left(a + b + &#92;frac{1}{2} - (2a + 2b + 1)x&#92;right)&#92;frac{dy}{dx} - 4aby = 0' title='&#92;displaystyle x(1 - x)&#92;frac{d^{2}y}{dx^{2}} + &#92;left(a + b + &#92;frac{1}{2} - (2a + 2b + 1)x&#92;right)&#92;frac{dy}{dx} - 4aby = 0' class='latex' /></p>
<p>This is clearly satisfied by <img src='http://s0.wp.com/latex.php?latex=y+%3D+F%282a%2C+2b%3B+a+%2B+b+%2B+1%2F2%3B+x%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = F(2a, 2b; a + b + 1/2; x)' title='y = F(2a, 2b; a + b + 1/2; x)' class='latex' /> and hence the identity follows. On putting <img src='http://s0.wp.com/latex.php?latex=a+%3D+b+%3D+1%2F4%2C+x+%3D+k%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='a = b = 1/4, x = k^{2}' title='a = b = 1/4, x = k^{2}' class='latex' /> we see that <img src='http://s0.wp.com/latex.php?latex=4x%281+-+x%29+%3D+4k%5E%7B2%7D%281+-+k%5E%7B2%7D%29+%3D+%282kk%27%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='4x(1 - x) = 4k^{2}(1 - k^{2}) = (2kk&#039;)^{2}' title='4x(1 - x) = 4k^{2}(1 - k^{2}) = (2kk&#039;)^{2}' class='latex' /> and therefore we obtain the formula</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+K+%3D+%5Cfrac%7B%5Cpi%7D%7B2%7DF%5Cleft%28%5Cfrac%7B1%7D%7B4%7D%2C%5Cfrac%7B1%7D%7B4%7D%3B+1%3B+%282kk%27%29%5E%7B2%7D%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle K = &#92;frac{&#92;pi}{2}F&#92;left(&#92;frac{1}{4},&#92;frac{1}{4}; 1; (2kk&#039;)^{2}&#92;right)' title='&#92;displaystyle K = &#92;frac{&#92;pi}{2}F&#92;left(&#92;frac{1}{4},&#92;frac{1}{4}; 1; (2kk&#039;)^{2}&#92;right)' class='latex' /> or in simpler and expanded form</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7B2K%7D%7B%5Cpi%7D+%3D+1+%2B+%5Cleft%28%5Cfrac%7B1%7D%7B4%7D%5Cright%29%5E%7B2%7D%282kk%27%29%5E%7B2%7D+%2B+%5Cleft%28%5Cfrac%7B1%5Ccdot+5%7D%7B4%5Ccdot+8%7D%5Cright%29%5E%7B2%7D%282kk%27%29%5E%7B4%7D+%2B+%5Cleft%28%5Cfrac%7B1%5Ccdot+5%5Ccdot+9%7D%7B4%5Ccdot+8%5Ccdot+12%7D%5Cright%29%5E%7B2%7D%282kk%27%29%5E%7B6%7D+%2B+%5Ccdots&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;frac{2K}{&#92;pi} = 1 + &#92;left(&#92;frac{1}{4}&#92;right)^{2}(2kk&#039;)^{2} + &#92;left(&#92;frac{1&#92;cdot 5}{4&#92;cdot 8}&#92;right)^{2}(2kk&#039;)^{4} + &#92;left(&#92;frac{1&#92;cdot 5&#92;cdot 9}{4&#92;cdot 8&#92;cdot 12}&#92;right)^{2}(2kk&#039;)^{6} + &#92;cdots' title='&#92;displaystyle &#92;frac{2K}{&#92;pi} = 1 + &#92;left(&#92;frac{1}{4}&#92;right)^{2}(2kk&#039;)^{2} + &#92;left(&#92;frac{1&#92;cdot 5}{4&#92;cdot 8}&#92;right)^{2}(2kk&#039;)^{4} + &#92;left(&#92;frac{1&#92;cdot 5&#92;cdot 9}{4&#92;cdot 8&#92;cdot 12}&#92;right)^{2}(2kk&#039;)^{6} + &#92;cdots' class='latex' /></p>
<p>and this is valid for <img src='http://s0.wp.com/latex.php?latex=0+%5Cleq+k+%5Cleq+1%2F%5Csqrt%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='0 &#92;leq k &#92;leq 1/&#92;sqrt{2}' title='0 &#92;leq k &#92;leq 1/&#92;sqrt{2}' class='latex' />.</p>
<p><strong>Clausen&#8217;s Formula</strong></p>
<p>It turns out we can square the above identity on both sides and get another series which is like the hypergeometric series but slightly more complex. In order to proceed further we introduce the generalized hypergeometric series <img src='http://s0.wp.com/latex.php?latex=_%7B3%7DF_%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='_{3}F_{2}' title='_{3}F_{2}' class='latex' /> defined by</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+_%7B3%7DF_%7B2%7D%28a%2C+b%2C+c%3B+d%2C+e%3B+z%29+%3D+%5Csum_%7Bn+%3D+0%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%28a%29_%7Bn%7D%28b%29_%7Bn%7D%28c%29_%7Bn%7D%7D%7B%28d%29_%7Bn%7D%28e%29_%7Bn%7D%7D%5Cfrac%7Bz%5E%7Bn%7D%7D%7Bn%21%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle _{3}F_{2}(a, b, c; d, e; z) = &#92;sum_{n = 0}^{&#92;infty}&#92;frac{(a)_{n}(b)_{n}(c)_{n}}{(d)_{n}(e)_{n}}&#92;frac{z^{n}}{n!}' title='&#92;displaystyle _{3}F_{2}(a, b, c; d, e; z) = &#92;sum_{n = 0}^{&#92;infty}&#92;frac{(a)_{n}(b)_{n}(c)_{n}}{(d)_{n}(e)_{n}}&#92;frac{z^{n}}{n!}' class='latex' /></p>
<p>The differential equation satisfied by <img src='http://s0.wp.com/latex.php?latex=_%7B3%7DF_%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='_{3}F_{2}' title='_{3}F_{2}' class='latex' /> is obtained in the similar way as done for the hypergeometric function and we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5C%7B%5CTheta%28%5CTheta+%2B+d+-+1%29%28%5CTheta+%2B+e+-+1%29+-+z%28%5CTheta+%2B+a%29%28%5CTheta+%2B+b%29%28%5CTheta+%2B+c%29%5C%7D%5C%2C_%7B3%7DF_%7B2%7D+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;{&#92;Theta(&#92;Theta + d - 1)(&#92;Theta + e - 1) - z(&#92;Theta + a)(&#92;Theta + b)(&#92;Theta + c)&#92;}&#92;,_{3}F_{2} = 0' title='&#92;{&#92;Theta(&#92;Theta + d - 1)(&#92;Theta + e - 1) - z(&#92;Theta + a)(&#92;Theta + b)(&#92;Theta + c)&#92;}&#92;,_{3}F_{2} = 0' class='latex' /></p>
<p>This needs to be simplified by a laborious symbolic manipulation as follows:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5CTheta%28%5CTheta+%2B+d+-+1%29%28%5CTheta+%2B+e+-+1%29y+%3D+%5CTheta%28%5CTheta+%2B+d+-+1%29%28zy%27+%2B+%28e+-+1%29y%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;Theta(&#92;Theta + d - 1)(&#92;Theta + e - 1)y = &#92;Theta(&#92;Theta + d - 1)(zy&#039; + (e - 1)y)' title='&#92;Theta(&#92;Theta + d - 1)(&#92;Theta + e - 1)y = &#92;Theta(&#92;Theta + d - 1)(zy&#039; + (e - 1)y)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+%5CTheta%5C%7Bz%28zy%27+%2B+%28e+-+1%29y%29%27+%2B+%28d+-+1%29%28zy%27+%2B+%28e+-+1%29y%29%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='= &#92;Theta&#92;{z(zy&#039; + (e - 1)y)&#039; + (d - 1)(zy&#039; + (e - 1)y)&#92;}' title='= &#92;Theta&#92;{z(zy&#039; + (e - 1)y)&#039; + (d - 1)(zy&#039; + (e - 1)y)&#92;}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+%5CTheta%5C%7Bz%28zy%27%27+%2B+ey%27%29+%2B+z%28d+-+1%29y%27+%2B+%28d+-+1%29%28e+-+1%29y%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='= &#92;Theta&#92;{z(zy&#039;&#039; + ey&#039;) + z(d - 1)y&#039; + (d - 1)(e - 1)y&#92;}' title='= &#92;Theta&#92;{z(zy&#039;&#039; + ey&#039;) + z(d - 1)y&#039; + (d - 1)(e - 1)y&#92;}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D%5CTheta%5C%7Bz%5E%7B2%7Dy%27%27+%2B+%28d+%2B+e+-+1%29zy%27+%2B+%28d+-+1%29%28e+-+1%29y%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='=&#92;Theta&#92;{z^{2}y&#039;&#039; + (d + e - 1)zy&#039; + (d - 1)(e - 1)y&#92;}' title='=&#92;Theta&#92;{z^{2}y&#039;&#039; + (d + e - 1)zy&#039; + (d - 1)(e - 1)y&#92;}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+z%5C%7Bz%5E%7B2%7Dy%27%27+%2B+%28d+%2B+e+-+1%29zy%27+%2B+%28d+-+1%29%28e+-+1%29y%5C%7D%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='= z&#92;{z^{2}y&#039;&#039; + (d + e - 1)zy&#039; + (d - 1)(e - 1)y&#92;}&#039;' title='= z&#92;{z^{2}y&#039;&#039; + (d + e - 1)zy&#039; + (d - 1)(e - 1)y&#92;}&#039;' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+z%5C%7Bz%5E%7B2%7Dy%27%27%27+%2B+2zy%27%27+%2B+%28d+%2B+e+-+1%29zy%27%27+%2B+%28d+%2B+e+-+1%29y%27+%2B+%28d+-+1%29%28e+-+1%29y%27%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='= z&#92;{z^{2}y&#039;&#039;&#039; + 2zy&#039;&#039; + (d + e - 1)zy&#039;&#039; + (d + e - 1)y&#039; + (d - 1)(e - 1)y&#039;&#92;}' title='= z&#92;{z^{2}y&#039;&#039;&#039; + 2zy&#039;&#039; + (d + e - 1)zy&#039;&#039; + (d + e - 1)y&#039; + (d - 1)(e - 1)y&#039;&#92;}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+z%5C%7Bz%5E%7B2%7Dy%27%27%27+%2B+%28d+%2B+e+%2B+1%29zy%27%27+%2B+%28d+%2B+e+-+1+%2B+de+-+d+-+e+%2B+1%29y%27%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='= z&#92;{z^{2}y&#039;&#039;&#039; + (d + e + 1)zy&#039;&#039; + (d + e - 1 + de - d - e + 1)y&#039;&#92;}' title='= z&#92;{z^{2}y&#039;&#039;&#039; + (d + e + 1)zy&#039;&#039; + (d + e - 1 + de - d - e + 1)y&#039;&#92;}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+z%5C%7Bz%5E%7B2%7Dy%27%27%27+%2B+%28d+%2B+e+%2B+1%29zy%27%27+%2B+dey%27%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='= z&#92;{z^{2}y&#039;&#039;&#039; + (d + e + 1)zy&#039;&#039; + dey&#039;&#92;}' title='= z&#92;{z^{2}y&#039;&#039;&#039; + (d + e + 1)zy&#039;&#039; + dey&#039;&#92;}' class='latex' /></p>
<p>and</p>
<p><img src='http://s0.wp.com/latex.php?latex=z%28%5CTheta+%2B+a%29%28%5CTheta+%2B+b%29%28%5CTheta+%2B+c%29y+%3D+z%28%5CTheta+%2B+a%29%28%5CTheta+%2B+b%29%28zy%27+%2B+cy%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='z(&#92;Theta + a)(&#92;Theta + b)(&#92;Theta + c)y = z(&#92;Theta + a)(&#92;Theta + b)(zy&#039; + cy)' title='z(&#92;Theta + a)(&#92;Theta + b)(&#92;Theta + c)y = z(&#92;Theta + a)(&#92;Theta + b)(zy&#039; + cy)' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+z%28%5CTheta+%2B+a%29%5C%7Bz%28zy%27+%2B+cy%29%27+%2B+b%28zy%27+%2B+cy%29%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='= z(&#92;Theta + a)&#92;{z(zy&#039; + cy)&#039; + b(zy&#039; + cy)&#92;}' title='= z(&#92;Theta + a)&#92;{z(zy&#039; + cy)&#039; + b(zy&#039; + cy)&#92;}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+z%28%5CTheta+%2B+a%29%5C%7Bz%28zy%27%27+%2B+y%27+%2B+cy%27%29+%2B+bzy%27+%2B+bcy%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='= z(&#92;Theta + a)&#92;{z(zy&#039;&#039; + y&#039; + cy&#039;) + bzy&#039; + bcy&#92;}' title='= z(&#92;Theta + a)&#92;{z(zy&#039;&#039; + y&#039; + cy&#039;) + bzy&#039; + bcy&#92;}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+z%28%5CTheta+%2B+a%29%5C%7Bz%5E%7B2%7Dy%27%27+%2B+%28b+%2B+c+%2B+1%29zy%27+%2B+bcy%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='= z(&#92;Theta + a)&#92;{z^{2}y&#039;&#039; + (b + c + 1)zy&#039; + bcy&#92;}' title='= z(&#92;Theta + a)&#92;{z^{2}y&#039;&#039; + (b + c + 1)zy&#039; + bcy&#92;}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+z%5C%7Bz%28z%5E%7B2%7Dy%27%27+%2B+%28b+%2B+c+%2B+1%29zy%27+%2B+bcy%29%27+%2B+a%28z%5E%7B2%7Dy%27%27+%2B+%28b+%2B+c+%2B+1%29zy%27+%2B+bcy%29%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='= z&#92;{z(z^{2}y&#039;&#039; + (b + c + 1)zy&#039; + bcy)&#039; + a(z^{2}y&#039;&#039; + (b + c + 1)zy&#039; + bcy)&#92;}' title='= z&#92;{z(z^{2}y&#039;&#039; + (b + c + 1)zy&#039; + bcy)&#039; + a(z^{2}y&#039;&#039; + (b + c + 1)zy&#039; + bcy)&#92;}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+z%5C%7Bz%28z%5E%7B2%7Dy%27%27%27+%2B+2zy%27%27+%2B+%28b+%2B+c+%2B+1%29zy%27%27+%2B+%28b+%2B+c+%2B+1%29y%27+%2B+bcy%27%29+%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='= z&#92;{z(z^{2}y&#039;&#039;&#039; + 2zy&#039;&#039; + (b + c + 1)zy&#039;&#039; + (b + c + 1)y&#039; + bcy&#039;) +' title='= z&#92;{z(z^{2}y&#039;&#039;&#039; + 2zy&#039;&#039; + (b + c + 1)zy&#039;&#039; + (b + c + 1)y&#039; + bcy&#039;) +' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=a%28z%5E%7B2%7Dy%27%27+%2B+%28b+%2B+c+%2B+1%29zy%27+%2B+bcy%29%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='a(z^{2}y&#039;&#039; + (b + c + 1)zy&#039; + bcy)&#92;}' title='a(z^{2}y&#039;&#039; + (b + c + 1)zy&#039; + bcy)&#92;}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%3D+z%5C%7Bz%5E%7B3%7Dy%27%27%27+%2B+%28a+%2B+b+%2B+c+%2B+3%29z%5E%7B2%7Dy%27%27+%2B+%28b+%2B+c+%2B+1+%2B+bc+%2B+a%28b+%2B+c+%2B+1%29%29zy%27+%2B+abcy%5C%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='= z&#92;{z^{3}y&#039;&#039;&#039; + (a + b + c + 3)z^{2}y&#039;&#039; + (b + c + 1 + bc + a(b + c + 1))zy&#039; + abcy&#92;}' title='= z&#92;{z^{3}y&#039;&#039;&#039; + (a + b + c + 3)z^{2}y&#039;&#039; + (b + c + 1 + bc + a(b + c + 1))zy&#039; + abcy&#92;}' class='latex' /></p>
<p>Finally the differential equation turns out to be</p>
<p><img src='http://s0.wp.com/latex.php?latex=z%5E%7B2%7D%28z+-+1%29y%27%27%27+%2B+%5C%7B%28a+%2B+b+%2B+c+%2B+3%29z+-+%28d+%2B+e+%2B+1%29%5C%7Dzy%27%27%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='z^{2}(z - 1)y&#039;&#039;&#039; + &#92;{(a + b + c + 3)z - (d + e + 1)&#92;}zy&#039;&#039;&#92;,+' title='z^{2}(z - 1)y&#039;&#039;&#039; + &#92;{(a + b + c + 3)z - (d + e + 1)&#92;}zy&#039;&#039;&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5C%7B%28%28a+%2B+1%29%28b+%2B+c+%2B+1%29+%2B+bc%29z+-+de%5C%7Dy%27+%2B+abcy+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;{((a + 1)(b + c + 1) + bc)z - de&#92;}y&#039; + abcy = 0' title='&#92;{((a + 1)(b + c + 1) + bc)z - de&#92;}y&#039; + abcy = 0' class='latex' /></p>
<p>Using this differential equation Thomas Clausen in 1828 found a way to express the <img src='http://s0.wp.com/latex.php?latex=_%7B3%7DF_%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='_{3}F_{2}' title='_{3}F_{2}' class='latex' /> series as a square of the usual hypergeometric function and this lead to some sufficient conditions under which a generalized hypergeometric series could be guaranteed to be positive. His result is famously called the <em>Clausen&#8217;s Formula</em> which is as follows:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cleft%28F%5Cleft%28a%2C+b%3B+a+%2B+b+%2B+%5Cfrac%7B1%7D%7B2%7D%3B+z%5Cright%29%5Cright%29%5E%7B2%7D+%3D+%5C%2C_%7B3%7DF_%7B2%7D%5Cleft%282a%2C+2b%2C+a+%2B+b%3B+2a+%2B+2b%2C+a+%2B+b+%2B+%5Cfrac%7B1%7D%7B2%7D%3B+z%5Cright%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;left(F&#92;left(a, b; a + b + &#92;frac{1}{2}; z&#92;right)&#92;right)^{2} = &#92;,_{3}F_{2}&#92;left(2a, 2b, a + b; 2a + 2b, a + b + &#92;frac{1}{2}; z&#92;right)' title='&#92;displaystyle &#92;left(F&#92;left(a, b; a + b + &#92;frac{1}{2}; z&#92;right)&#92;right)^{2} = &#92;,_{3}F_{2}&#92;left(2a, 2b, a + b; 2a + 2b, a + b + &#92;frac{1}{2}; z&#92;right)' class='latex' /></p>
<p>The conditions for validity turn out to be <img src='http://s0.wp.com/latex.php?latex=%7Cz%7C+%3C+1&amp;bg=fff&amp;fg=222&amp;s=0' alt='|z| &lt; 1' title='|z| &lt; 1' class='latex' /> and both <img src='http://s0.wp.com/latex.php?latex=a+%2B+b+%2B+1%2F2%2C+2a+%2B+2b&amp;bg=fff&amp;fg=222&amp;s=0' alt='a + b + 1/2, 2a + 2b' title='a + b + 1/2, 2a + 2b' class='latex' /> should not be zero or a negative integer.</p>
<p>The equation satisfied by the series on the right side is</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+z%5E%7B2%7D%28z+-+1%29y%27%27%27+-+3z%5Cleft%28a+%2B+b+%2B+%5Cfrac%7B1%7D%7B2%7D+-+%28a+%2B+b+%2B+1%29z%5Cright%29y%27%27%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle z^{2}(z - 1)y&#039;&#039;&#039; - 3z&#92;left(a + b + &#92;frac{1}{2} - (a + b + 1)z&#92;right)y&#039;&#039;&#92;,+' title='&#92;displaystyle z^{2}(z - 1)y&#039;&#039;&#039; - 3z&#92;left(a + b + &#92;frac{1}{2} - (a + b + 1)z&#92;right)y&#039;&#039;&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5B%5C%7B2%28a%5E%7B2%7D+%2B+b%5E%7B2%7D+%2B+4ab%29+%2B+3%28a+%2B+b%29+%2B+1%5C%7Dz+-+%28a+%2B+b%29%282a+%2B+2b+%2B+1%29%5Dy%27+%2B+4ab%28a+%2B+b%29y+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='[&#92;{2(a^{2} + b^{2} + 4ab) + 3(a + b) + 1&#92;}z - (a + b)(2a + 2b + 1)]y&#039; + 4ab(a + b)y = 0' title='[&#92;{2(a^{2} + b^{2} + 4ab) + 3(a + b) + 1&#92;}z - (a + b)(2a + 2b + 1)]y&#039; + 4ab(a + b)y = 0' class='latex' /></p>
<p>On the other hand the function <img src='http://s0.wp.com/latex.php?latex=v+%3D+F%28a%2C+b%3B+a+%2B+b+%2B+1%2F2%3B+z%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='v = F(a, b; a + b + 1/2; z)' title='v = F(a, b; a + b + 1/2; z)' class='latex' /> satisfied the differential equation</p>
<p><img src='http://s0.wp.com/latex.php?latex=z%28z+-+1%29v%27%27+%2B+%5C%7B%28a+%2B+b+%2B+1%29z+-+a+-+b+-+1%2F2%5C%7Dv%27+%2B+abv+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='z(z - 1)v&#039;&#039; + &#92;{(a + b + 1)z - a - b - 1/2&#92;}v&#039; + abv = 0' title='z(z - 1)v&#039;&#039; + &#92;{(a + b + 1)z - a - b - 1/2&#92;}v&#039; + abv = 0' class='latex' /> or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%28z%5E%7B3%7D+-+z%5E%7B2%7D%29v%27%27+%2B+%5C%7B%28a+%2B+b+%2B+1%29z%5E%7B2%7D+-+%28a+%2B+b+%2B+1%2F2%29z%5C%7Dv%27+%2B+abzv+%3D+0%5C%2C%5C%2C%5C%2C+%5Ccdots+%5C%2C%5C%2C+%281%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='(z^{3} - z^{2})v&#039;&#039; + &#92;{(a + b + 1)z^{2} - (a + b + 1/2)z&#92;}v&#039; + abzv = 0&#92;,&#92;,&#92;, &#92;cdots &#92;,&#92;, (1)' title='(z^{3} - z^{2})v&#039;&#039; + &#92;{(a + b + 1)z^{2} - (a + b + 1/2)z&#92;}v&#039; + abzv = 0&#92;,&#92;,&#92;, &#92;cdots &#92;,&#92;, (1)' class='latex' /></p>
<p>Differentiating with respect to <img src='http://s0.wp.com/latex.php?latex=z&amp;bg=fff&amp;fg=222&amp;s=0' alt='z' title='z' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%28z%5E%7B3%7D+-+z%5E%7B2%7D%29v%27%27%27+%2B+%5C%7B%28a+%2B+b+%2B+4%29z%5E%7B2%7D+-+%28a+%2B+b+%2B+5%2F2%29z%5C%7Dv%27%27%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='(z^{3} - z^{2})v&#039;&#039;&#039; + &#92;{(a + b + 4)z^{2} - (a + b + 5/2)z&#92;}v&#039;&#039;&#92;,+' title='(z^{3} - z^{2})v&#039;&#039;&#039; + &#92;{(a + b + 4)z^{2} - (a + b + 5/2)z&#92;}v&#039;&#039;&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5C%7B%282a+%2B+2b+%2B+ab+%2B+2%29z+-+%28a+%2B+b+%2B+1%2F2%29%5C%7Dv%27+%2B+abv+%3D+0%5C%2C%5C%2C%5C%2C+%5Ccdots+%5C%2C%5C%2C+%282%29&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;{(2a + 2b + ab + 2)z - (a + b + 1/2)&#92;}v&#039; + abv = 0&#92;,&#92;,&#92;, &#92;cdots &#92;,&#92;, (2)' title='&#92;{(2a + 2b + ab + 2)z - (a + b + 1/2)&#92;}v&#039; + abv = 0&#92;,&#92;,&#92;, &#92;cdots &#92;,&#92;, (2)' class='latex' /></p>
<p>Let&#8217;s put <img src='http://s0.wp.com/latex.php?latex=y+%3D+v%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = v^{2}' title='y = v^{2}' class='latex' /> so that <img src='http://s0.wp.com/latex.php?latex=y%27+%3D+2vv%27%2C+y%27%27+%3D+2v%27%5E%7B2%7D+%2B+2vv%27%27%2C+y%27%27%27+%3D+6v%27v%27%27+%2B+2vv%27%27%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='y&#039; = 2vv&#039;, y&#039;&#039; = 2v&#039;^{2} + 2vv&#039;&#039;, y&#039;&#039;&#039; = 6v&#039;v&#039;&#039; + 2vv&#039;&#039;&#039;' title='y&#039; = 2vv&#039;, y&#039;&#039; = 2v&#039;^{2} + 2vv&#039;&#039;, y&#039;&#039;&#039; = 6v&#039;v&#039;&#039; + 2vv&#039;&#039;&#039;' class='latex' />.</p>
<p>Now mulplying (1) by <img src='http://s0.wp.com/latex.php?latex=6v%27&amp;bg=fff&amp;fg=222&amp;s=0' alt='6v&#039;' title='6v&#039;' class='latex' /> and (2) by <img src='http://s0.wp.com/latex.php?latex=2v&amp;bg=fff&amp;fg=222&amp;s=0' alt='2v' title='2v' class='latex' /> and adding the equations we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%28z%5E%7B3%7D+-+z%5E%7B2%7D%29%286v%27v%27%27+%2B+2vv%27%27%27%29+%2B+6%5C%7B%28a+%2B+b+%2B+1%29z%5E%7B2%7D+-+%28a+%2B+b+%2B+1%2F2%29z%5C%7Dv%27%5E%7B2%7D+%2B+6abzvv%27%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='(z^{3} - z^{2})(6v&#039;v&#039;&#039; + 2vv&#039;&#039;&#039;) + 6&#92;{(a + b + 1)z^{2} - (a + b + 1/2)z&#92;}v&#039;^{2} + 6abzvv&#039;&#92;,+' title='(z^{3} - z^{2})(6v&#039;v&#039;&#039; + 2vv&#039;&#039;&#039;) + 6&#92;{(a + b + 1)z^{2} - (a + b + 1/2)z&#92;}v&#039;^{2} + 6abzvv&#039;&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=2%5C%7B%28a+%2B+b+%2B+4%29z%5E%7B2%7D+-+%28a+%2B+b+%2B+5%2F2%29z%5C%7Dvv%27%27+%2B+2%5C%7B%282a+%2B+2b+%2B+ab+%2B+2%29z+-+%28a+%2B+b+%2B+1%2F2%29%5C%7Dvv%27+%2B+2abv%5E%7B2%7D+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='2&#92;{(a + b + 4)z^{2} - (a + b + 5/2)z&#92;}vv&#039;&#039; + 2&#92;{(2a + 2b + ab + 2)z - (a + b + 1/2)&#92;}vv&#039; + 2abv^{2} = 0' title='2&#92;{(a + b + 4)z^{2} - (a + b + 5/2)z&#92;}vv&#039;&#039; + 2&#92;{(2a + 2b + ab + 2)z - (a + b + 1/2)&#92;}vv&#039; + 2abv^{2} = 0' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%28z%5E%7B3%7D+-+z%5E%7B2%7D%29%286v%27v%27%27+%2B+2vv%27%27%27%29+%2B+3z%5C%7B%28a+%2B+b+%2B+1%29z+-+%28a+%2B+b+%2B+1%2F2%29%5C%7D%282v%27%5E%7B2%7D+%2B+2vv%27%27%29%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='(z^{3} - z^{2})(6v&#039;v&#039;&#039; + 2vv&#039;&#039;&#039;) + 3z&#92;{(a + b + 1)z - (a + b + 1/2)&#92;}(2v&#039;^{2} + 2vv&#039;&#039;)&#92;,+' title='(z^{3} - z^{2})(6v&#039;v&#039;&#039; + 2vv&#039;&#039;&#039;) + 3z&#92;{(a + b + 1)z - (a + b + 1/2)&#92;}(2v&#039;^{2} + 2vv&#039;&#039;)&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=2z%28z+-+1%29%28-2a+-+2b+%2B+1%29vv%27%27+%2B+%5C%7B2z%28a+%2B+b+%2B+2ab+%2B+1%29+-+%28a+%2B+b+%2B+1%2F2%29%5C%7D%282vv%27%29+%2B+2abv%5E%7B2%7D+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='2z(z - 1)(-2a - 2b + 1)vv&#039;&#039; + &#92;{2z(a + b + 2ab + 1) - (a + b + 1/2)&#92;}(2vv&#039;) + 2abv^{2} = 0' title='2z(z - 1)(-2a - 2b + 1)vv&#039;&#039; + &#92;{2z(a + b + 2ab + 1) - (a + b + 1/2)&#92;}(2vv&#039;) + 2abv^{2} = 0' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%28z%5E%7B3%7D+-+z%5E%7B2%7D%29%286v%27v%27%27+%2B+2vv%27%27%27%29+%2B+3z%5C%7B%28a+%2B+b+%2B+1%29z+-+%28a+%2B+b+%2B+1%2F2%29%5C%7D%282v%27%5E%7B2%7D+%2B+2vv%27%27%29%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='(z^{3} - z^{2})(6v&#039;v&#039;&#039; + 2vv&#039;&#039;&#039;) + 3z&#92;{(a + b + 1)z - (a + b + 1/2)&#92;}(2v&#039;^{2} + 2vv&#039;&#039;)&#92;,+' title='(z^{3} - z^{2})(6v&#039;v&#039;&#039; + 2vv&#039;&#039;&#039;) + 3z&#92;{(a + b + 1)z - (a + b + 1/2)&#92;}(2v&#039;^{2} + 2vv&#039;&#039;)&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=2%28-2a+-+2b+%2B+1%29%5Bz%28z+-+1%29vv%27%27+%2B+%5C%7B%28a+%2B+b+%2B+1%29z+-+%28a+%2B+b+%2B+1%2F2%29%5C%7Dvv%27+%2B+abv%5E%7B2%7D%5D%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='2(-2a - 2b + 1)[z(z - 1)vv&#039;&#039; + &#92;{(a + b + 1)z - (a + b + 1/2)&#92;}vv&#039; + abv^{2}]&#92;,+' title='2(-2a - 2b + 1)[z(z - 1)vv&#039;&#039; + &#92;{(a + b + 1)z - (a + b + 1/2)&#92;}vv&#039; + abv^{2}]&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=2%282a+%2B+2b+-+1%29%5B%5C%7B%28a+%2B+b+%2B+1%29z+-+%28a+%2B+b+%2B+1%2F2%29%5C%7Dvv%27+%2B+abv%5E%7B2%7D%5D%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='2(2a + 2b - 1)[&#92;{(a + b + 1)z - (a + b + 1/2)&#92;}vv&#039; + abv^{2}]&#92;,+' title='2(2a + 2b - 1)[&#92;{(a + b + 1)z - (a + b + 1/2)&#92;}vv&#039; + abv^{2}]&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5C%7B2z%28a+%2B+b+%2B+2ab+%2B+1%29+-+%28a+%2B+b+%2B+1%2F2%29%5C%7D%282vv%27%29+%2B+2abv%5E%7B2%7D+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;{2z(a + b + 2ab + 1) - (a + b + 1/2)&#92;}(2vv&#039;) + 2abv^{2} = 0' title='&#92;{2z(a + b + 2ab + 1) - (a + b + 1/2)&#92;}(2vv&#039;) + 2abv^{2} = 0' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%28z%5E%7B3%7D+-+z%5E%7B2%7D%29%286v%27v%27%27+%2B+2vv%27%27%27%29+%2B+3z%5C%7B%28a+%2B+b+%2B+1%29z+-+%28a+%2B+b+%2B+1%2F2%29%5C%7D%282v%27%5E%7B2%7D+%2B+2vv%27%27%29%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='(z^{3} - z^{2})(6v&#039;v&#039;&#039; + 2vv&#039;&#039;&#039;) + 3z&#92;{(a + b + 1)z - (a + b + 1/2)&#92;}(2v&#039;^{2} + 2vv&#039;&#039;)&#92;,+' title='(z^{3} - z^{2})(6v&#039;v&#039;&#039; + 2vv&#039;&#039;&#039;) + 3z&#92;{(a + b + 1)z - (a + b + 1/2)&#92;}(2v&#039;^{2} + 2vv&#039;&#039;)&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=2%28a+%2B+b+-+1%2F2%29%5C%7B%28a+%2B+b+%2B+1%29z+-+%28a+%2B+b+%2B+1%2F2%29%5C%7D%282vv%27%29%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='2(a + b - 1/2)&#92;{(a + b + 1)z - (a + b + 1/2)&#92;}(2vv&#039;)&#92;,+' title='2(a + b - 1/2)&#92;{(a + b + 1)z - (a + b + 1/2)&#92;}(2vv&#039;)&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5C%7B2z%28a+%2B+b+%2B+2ab+%2B+1%29+-+%28a+%2B+b+%2B+1%2F2%29%5C%7D%282vv%27%29+%2B+4ab%28a+%2B+b%29v%5E%7B2%7D%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;{2z(a + b + 2ab + 1) - (a + b + 1/2)&#92;}(2vv&#039;) + 4ab(a + b)v^{2}= 0' title='&#92;{2z(a + b + 2ab + 1) - (a + b + 1/2)&#92;}(2vv&#039;) + 4ab(a + b)v^{2}= 0' class='latex' /></p>
<p>and finally</p>
<p><img src='http://s0.wp.com/latex.php?latex=%28z%5E%7B3%7D+-+z%5E%7B2%7D%29%286v%27v%27%27+%2B+2vv%27%27%27%29+%2B+3z%5C%7B%28a+%2B+b+%2B+1%29z+-+%28a+%2B+b+%2B+1%2F2%29%5C%7D%282v%27%5E%7B2%7D+%2B+2vv%27%27%29%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='(z^{3} - z^{2})(6v&#039;v&#039;&#039; + 2vv&#039;&#039;&#039;) + 3z&#92;{(a + b + 1)z - (a + b + 1/2)&#92;}(2v&#039;^{2} + 2vv&#039;&#039;)&#92;,+' title='(z^{3} - z^{2})(6v&#039;v&#039;&#039; + 2vv&#039;&#039;&#039;) + 3z&#92;{(a + b + 1)z - (a + b + 1/2)&#92;}(2v&#039;^{2} + 2vv&#039;&#039;)&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5B%5C%7B2%28a%5E%7B2%7D+%2B+b%5E%7B2%7D+%2B+4ab%29+%2B+3%28a+%2B+b%29+%2B+1%5C%7Dz+-+%28a+%2B+b%29%282a+%2B+2b+%2B+1%29%5D%282vv%27%29%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='[&#92;{2(a^{2} + b^{2} + 4ab) + 3(a + b) + 1&#92;}z - (a + b)(2a + 2b + 1)](2vv&#039;)&#92;,+' title='[&#92;{2(a^{2} + b^{2} + 4ab) + 3(a + b) + 1&#92;}z - (a + b)(2a + 2b + 1)](2vv&#039;)&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=4ab%28a+%2B+b%29v%5E%7B2%7D%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='4ab(a + b)v^{2}= 0' title='4ab(a + b)v^{2}= 0' class='latex' /></p>
<p>Thus <img src='http://s0.wp.com/latex.php?latex=y+%3D+v%5E%7B2%7D+%3D+%28F%28a%2C+b%3B+a+%2B+b+%2B+1%2F2%3B+z%29%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='y = v^{2} = (F(a, b; a + b + 1/2; z))^{2}' title='y = v^{2} = (F(a, b; a + b + 1/2; z))^{2}' class='latex' /> satisfies the differential equation</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+z%5E%7B2%7D%28z+-+1%29y%27%27%27+-+3z%5Cleft%28a+%2B+b+%2B+%5Cfrac%7B1%7D%7B2%7D+-+%28a+%2B+b+%2B+1%29z%5Cright%29y%27%27%5C%2C%2B&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle z^{2}(z - 1)y&#039;&#039;&#039; - 3z&#92;left(a + b + &#92;frac{1}{2} - (a + b + 1)z&#92;right)y&#039;&#039;&#92;,+' title='&#92;displaystyle z^{2}(z - 1)y&#039;&#039;&#039; - 3z&#92;left(a + b + &#92;frac{1}{2} - (a + b + 1)z&#92;right)y&#039;&#039;&#92;,+' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5B%5C%7B2%28a%5E%7B2%7D+%2B+b%5E%7B2%7D+%2B+4ab%29+%2B+3%28a+%2B+b%29+%2B+1%5C%7Dz+-+%28a+%2B+b%29%282a+%2B+2b+%2B+1%29%5Dy%27+%2B+4ab%28a+%2B+b%29y+%3D+0&amp;bg=fff&amp;fg=222&amp;s=0' alt='[&#92;{2(a^{2} + b^{2} + 4ab) + 3(a + b) + 1&#92;}z - (a + b)(2a + 2b + 1)]y&#039; + 4ab(a + b)y = 0' title='[&#92;{2(a^{2} + b^{2} + 4ab) + 3(a + b) + 1&#92;}z - (a + b)(2a + 2b + 1)]y&#039; + 4ab(a + b)y = 0' class='latex' /></p>
<p>This establishes the Clausen&#8217;s Formula. Immediate application of this formula in the context of elliptic integrals is obtained by putting <img src='http://s0.wp.com/latex.php?latex=a+%3D+b+%3D+1%2F4%2C+z+%3D+%282kk%27%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='a = b = 1/4, z = (2kk&#039;)^{2}' title='a = b = 1/4, z = (2kk&#039;)^{2}' class='latex' /> as follows:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cleft%28%5Cfrac%7B2K%7D%7B%5Cpi%7D%5Cright%29%5E%7B2%7D+%3D+1+%2B+%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B3%7D%282kk%27%29%5E%7B2%7D+%2B+%5Cleft%28%5Cfrac%7B1%5Ccdot+3%7D%7B2%5Ccdot+4%7D%5Cright%29%5E%7B3%7D%282kk%27%29%5E%7B4%7D+%2B+%5Cleft%28%5Cfrac%7B1%5Ccdot+3%5Ccdot+5%7D%7B2%5Ccdot+4%5Ccdot+6%7D%5Cright%29%5E%7B3%7D%282kk%27%29%5E%7B6%7D+%2B+%5Ccdots&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;left(&#92;frac{2K}{&#92;pi}&#92;right)^{2} = 1 + &#92;left(&#92;frac{1}{2}&#92;right)^{3}(2kk&#039;)^{2} + &#92;left(&#92;frac{1&#92;cdot 3}{2&#92;cdot 4}&#92;right)^{3}(2kk&#039;)^{4} + &#92;left(&#92;frac{1&#92;cdot 3&#92;cdot 5}{2&#92;cdot 4&#92;cdot 6}&#92;right)^{3}(2kk&#039;)^{6} + &#92;cdots' title='&#92;displaystyle &#92;left(&#92;frac{2K}{&#92;pi}&#92;right)^{2} = 1 + &#92;left(&#92;frac{1}{2}&#92;right)^{3}(2kk&#039;)^{2} + &#92;left(&#92;frac{1&#92;cdot 3}{2&#92;cdot 4}&#92;right)^{3}(2kk&#039;)^{4} + &#92;left(&#92;frac{1&#92;cdot 3&#92;cdot 5}{2&#92;cdot 4&#92;cdot 6}&#92;right)^{3}(2kk&#039;)^{6} + &#92;cdots' class='latex' /></p>
<p>or in the more striking form as</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cleft%281+%2B+%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B2%7Dk%5E%7B2%7D+%2B+%5Cleft%28%5Cfrac%7B1%5Ccdot+3%7D%7B2%5Ccdot+4%7D%5Cright%29%5E%7B2%7Dk%5E%7B4%7D+%2B+%5Cleft%28%5Cfrac%7B1%5Ccdot+3%5Ccdot+5%7D%7B2%5Ccdot+4%5Ccdot+6%7D%5Cright%29%5E%7B2%7Dk%5E%7B6%7D+%2B+%5Ccdots%5Cright%29%5E%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle &#92;left(1 + &#92;left(&#92;frac{1}{2}&#92;right)^{2}k^{2} + &#92;left(&#92;frac{1&#92;cdot 3}{2&#92;cdot 4}&#92;right)^{2}k^{4} + &#92;left(&#92;frac{1&#92;cdot 3&#92;cdot 5}{2&#92;cdot 4&#92;cdot 6}&#92;right)^{2}k^{6} + &#92;cdots&#92;right)^{2}' title='&#92;displaystyle &#92;left(1 + &#92;left(&#92;frac{1}{2}&#92;right)^{2}k^{2} + &#92;left(&#92;frac{1&#92;cdot 3}{2&#92;cdot 4}&#92;right)^{2}k^{4} + &#92;left(&#92;frac{1&#92;cdot 3&#92;cdot 5}{2&#92;cdot 4&#92;cdot 6}&#92;right)^{2}k^{6} + &#92;cdots&#92;right)^{2}' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D+1+%2B+%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B3%7D%282kk%27%29%5E%7B2%7D+%2B+%5Cleft%28%5Cfrac%7B1%5Ccdot+3%7D%7B2%5Ccdot+4%7D%5Cright%29%5E%7B3%7D%282kk%27%29%5E%7B4%7D+%2B+%5Cleft%28%5Cfrac%7B1%5Ccdot+3%5Ccdot+5%7D%7B2%5Ccdot+4%5Ccdot+6%7D%5Cright%29%5E%7B3%7D%282kk%27%29%5E%7B6%7D+%2B+%5Ccdots&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;displaystyle = 1 + &#92;left(&#92;frac{1}{2}&#92;right)^{3}(2kk&#039;)^{2} + &#92;left(&#92;frac{1&#92;cdot 3}{2&#92;cdot 4}&#92;right)^{3}(2kk&#039;)^{4} + &#92;left(&#92;frac{1&#92;cdot 3&#92;cdot 5}{2&#92;cdot 4&#92;cdot 6}&#92;right)^{3}(2kk&#039;)^{6} + &#92;cdots' title='&#92;displaystyle = 1 + &#92;left(&#92;frac{1}{2}&#92;right)^{3}(2kk&#039;)^{2} + &#92;left(&#92;frac{1&#92;cdot 3}{2&#92;cdot 4}&#92;right)^{3}(2kk&#039;)^{4} + &#92;left(&#92;frac{1&#92;cdot 3&#92;cdot 5}{2&#92;cdot 4&#92;cdot 6}&#92;right)^{3}(2kk&#039;)^{6} + &#92;cdots' class='latex' /></p>
<p>and this is valid for <img src='http://s0.wp.com/latex.php?latex=0+%5Cleq+k+%5Cleq+1%2F%5Csqrt%7B2%7D&amp;bg=fff&amp;fg=222&amp;s=0' alt='0 &#92;leq k &#92;leq 1/&#92;sqrt{2}' title='0 &#92;leq k &#92;leq 1/&#92;sqrt{2}' class='latex' />.</p>
<p>This is a fundamental result which is used by Ramanujan to derive certain series for <img src='http://s0.wp.com/latex.php?latex=1%2F%5Cpi&amp;bg=fff&amp;fg=222&amp;s=0' alt='1/&#92;pi' title='1/&#92;pi' class='latex' />. With this we complete the background material in the theory of hypergeometric series which is needed to understand Ramanujan&#8217;s <em>Modular Equations and Approximations to <img src='http://s0.wp.com/latex.php?latex=%5Cpi&amp;bg=fff&amp;fg=222&amp;s=0' alt='&#92;pi' title='&#92;pi' class='latex' />.</em></p>
<p>In the next post we will define a modular equation and also derive such equations as done by Jacobi in his <em>Fundamenta Nova.</em></p>
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