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Could someone please explain how to transform the nome $q = e^{-\pi K'/K}$ from $q^2$ to $q$ and then to $-q$? In other words, how does changing $q^2$ to $q$ and then $q$ to $-q$ affect $k$ and $K$. Here, $k$ is the elliptic modulus, $K = K(k)$ is the complete elliptic integral of the first kind, and $K' = K(k')$, where $k' = \sqrt{1 - k^2}$ is the complementary modulus. I know that the transformation of $q$ is somehow connected to Landen's transformation, but I do not understand how these two concepts are related. Thank you in advance.

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I tend to look at this using Jacobi thetanull functions, as these are usually expressed in terms of the nome $q$: $$\begin{align} \vartheta_2(q) &= q^{1/4}\sum_{n\in\mathbb{Z}} q^{n(n+1)} & \vartheta_3(q) &= \sum_{n\in\mathbb{Z}} q^{n^2} & \vartheta_4(q) &= \sum_{n\in\mathbb{Z}} (-q)^{n^2} \end{align}$$

As to the connection of your question with Landen's transformation, let me just remark that Landen's transformation gives rise to the arithmetic-geometric mean iteration step, which in turn resembles the transformation of $\vartheta_3^2$ and $\vartheta_4^2$ under $q\mapsto q^2$. To answer your question, it suffices to do that step backwards. I will focus on the Jacobi thetanulls here, and thus get along without considering integrals and Landen's transformation as such. Expect some ambiguity from inverse powers.

The formulae we need are $$\begin{align} \vartheta_2(-q) &= \pm\sqrt{\pm\mathrm{i}}\,\vartheta_2(q) & \vartheta_2^2(q) &= 2\,\vartheta_3(q^2)\,\vartheta_2(q^2) \\ \vartheta_3(-q) &= \vartheta_4(q) & \vartheta_3^2(q) &= \vartheta_3^2(q^2) + \vartheta_2^2(q^2) \\ \vartheta_4(-q) &= \vartheta_3(q) & \vartheta_4^2(q) &= \vartheta_3^2(q^2) - \vartheta_2^2(q^2) \end{align}$$ The formulae for $q\mapsto-q$ are obvious, and the last one of those for $q^2\mapsto q$ has been proven elsewhere on this site. The remaining ones follow in a similar manner.

For the relations of the Jacobi thetanulls to the complete elliptic integral of the first kind, one only needs to keep in mind that $$\begin{align} k(q) &= \frac{\vartheta_2^2(q)}{\vartheta_3^2(q)} & k'(q) &= \frac{\vartheta_4^2(q)}{\vartheta_3^2(q)} & K\left(k(q)\right) &= \frac{\pi}{2}\vartheta_3^2(q) \end{align}$$

From those identities we can deduce $$\begin{align} k(-q) &= \pm\mathrm{i}\frac{k(q)}{k'(q)} & k'(-q) &= \frac{1}{k'(q)} \\ k(q) &= \pm\frac{2\sqrt{k(q^2)}}{1+k(q^2)} & k'(q) &= \frac{1-k(q^2)}{1+k(q^2)} \\\therefore\quad k(-q) &= \pm\mathrm{i}\frac{2\sqrt{k(q^2)}}{1-k(q^2)} & k'(-q) &= \frac{1+k(q^2)}{1-k(q^2)} \end{align}$$ and for the complete elliptic integral: $$\begin{align} K\left(k(-q)\right) &= k'(q)\,K\left(k(q)\right) \\ K\left(k(q)\right) &= \left(1+k(q^2)\right) K\left(k(q^2)\right) \\\therefore\quad K\left(k(-q)\right) &= \left(1-k(q^2)\right) K\left(k(q^2)\right) \end{align}$$ which is what you have asked for.

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  • $\begingroup$ Thank you. Great answer. I managed to derive these relations using a slightly different approach. Now I know my calculations must be correct. $\endgroup$
    – glebovg
    Commented Mar 3, 2014 at 1:46
  • $\begingroup$ Do you know how changing $q$ into $q^3$ affect $k$ and $K$ by any chance? I could not find this transformation anywhere. $\endgroup$
    – glebovg
    Commented Oct 2, 2014 at 1:43
  • $\begingroup$ The following identity might help you getting started: $\vartheta_3(q)\vartheta_3(q^3) = \vartheta_2(q)\vartheta_2(q^3) + \vartheta_4(q)\vartheta_4(q^3)$ (Legendre). $\endgroup$
    – ccorn
    Commented Oct 4, 2014 at 4:30
  • $\begingroup$ Thanks. By the way, are there any books on the classical theory of elliptic functions that you would recommend? The only good ones I know of are Whittaker and Watson, Greenhill, Hancock, and Cayley. Lang's book is also good, but it focuses mostly on the algebraic aspects of elliptic functions. $\endgroup$
    – glebovg
    Commented Oct 4, 2014 at 7:35
  • $\begingroup$ I'd propose to first read Borwein & Borwein's Pi and the AGM, Wiley 1987. This will teach you how to "do" such things systematically. After that, you will recognize how the references you named build up the more general theory. $\endgroup$
    – ccorn
    Commented Oct 11, 2014 at 20:52

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