I found an interesting series

$$\sum_{k=1}^\infty \log \left(\tanh \frac{\pi k}{2} \right)=\log(\vartheta_4(e^{-\pi}))=\log \left(\frac{\pi^{\frac{1}{4}}}{2^{\frac{1}{4}}\Gamma \left( \frac{3}{4}\right)} \right)$$

  • Does anybody know how to approach this series using Jacobi Theta Function?
  • Also, can any one suggest any good papers/books on Jacobi theta functions and Jacobi Elliptic functions?

Thank you very much!

  • $\begingroup$ What do you mean, "how to approach this series using Jacobi Theta Function"? Do you mean, how to prove the equalities in the display? Where did you "find" this series? Was there no information there? $\endgroup$ – Gerry Myerson Apr 16 '13 at 9:11
  • 1
    $\begingroup$ I have added some references for learning the theory of elliptic and theta functions in an update to my answer. You may find it useful. $\endgroup$ – Paramanand Singh May 22 '16 at 10:42

We have $$\vartheta_{4}(q) = \prod_{n = 1}^{\infty}(1 - q^{2n})(1 - q^{2n - 1})^{2}\tag{1}$$ It is easily seen that the above product can be written as $$\prod_{n = 1}^{\infty}(1 - q^{2n})\cdot\frac{(1 - q^{n})^{2}}{(1 - q^{2n})^{2}} = \prod_{n = 1}^{\infty}\frac{(1 - q^{n})^{2}}{1 - q^{2n}} = \prod_{n = 1}^{\infty}\frac{1 - q^{n}}{1 + q^{n}}\tag{2}$$ Putting $q = e^{-\pi}$ and noting that $$\tanh(n\pi/2) = \frac{e^{n\pi/2} - e^{n\pi/2}}{e^{n\pi/2} + e^{n\pi/2}} = \frac{1 - e^{-n\pi}}{1 + e^{-n\pi}} = \frac{1 - q^{n}}{1 + q^{n}}\tag{3}$$ we have via virtue of equations $(1), (2), (3)$ $$\vartheta_{4}(e^{-\pi}) = \prod_{n = 1}^{\infty}\tanh\left(\frac{n\pi}{2}\right)$$ so that the first equality is proved.

Now to calculate the value of $\vartheta_{4}(q)$ for $q = e^{-\pi}$ note that $$\vartheta_{4}(q) = \frac{\vartheta_{4}(q)}{\vartheta_{3}(q)}\cdot \vartheta_{3}(q) = \sqrt{k'}\sqrt{\frac{2K}{\pi}}\tag{4}$$ For $q = e^{-\pi}$ we have $k = k' = 1/\sqrt{2}$ and $$K = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - (1/2)\sin^{2}x}} = \frac{1}{4\sqrt{\pi}}\Gamma^{2}\left(\frac{1}{4}\right)\tag{5}$$ and hence $$\vartheta_{4}(e^{-\pi}) = 2^{-3/4}\pi^{-3/4}\Gamma(1/4)$$ and noting that $$\Gamma(1/4)\Gamma(3/4) = \frac{\pi}{\sin(\pi/4)} = \sqrt{2}\pi$$ we get $$\vartheta_{4}(e^{-\pi}) = \frac{\pi^{1/4}}{2^{1/4}\Gamma(3/4)}$$ so that the second equality of the question is also proved.

The integral in $(5)$ is easily evaluated via the use of beta and gamma functions. Thus \begin{align} K &= K(1/\sqrt{2}) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - (1/2)\sin^{2}x}}\notag\\ &= \sqrt{2}\int_{0}^{\pi/2}\frac{dx}{\sqrt{2 - \sin^{2}x}}\notag\\ &= \sqrt{2}\int_{0}^{\pi/2}\frac{dx}{\sqrt{1 + \cos^{2}x}}\notag\\ &= \sqrt{2}\int_{0}^{1}\frac{dt}{\sqrt{1 - t^{4}}}\text{ (by putting }t = \cos x)\notag\\ &= \frac{\sqrt{2}}{4}\int_{0}^{1}x^{-3/4}(1 - x)^{-1/2}\,dx\text{ (by putting }t^{4} = x)\notag\\ &= \frac{\sqrt{2}}{4}B\left(\frac{1}{4}, \frac{1}{2}\right)\notag\\ &= \frac{\sqrt{2}}{4}\cdot\dfrac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{4}\right)}\notag\\ &= \frac{\sqrt{2\pi}}{4}\cdot\dfrac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}\notag\\ &= \frac{\sqrt{2\pi}}{4}\cdot\dfrac{\Gamma^{2}\left(\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)}\notag\\ &= \frac{\sqrt{2\pi}}{4}\cdot\dfrac{\Gamma^{2}\left(\frac{1}{4}\right)}{\dfrac{\pi}{\sin(\pi/4)}}\notag\\ &= \frac{1}{4\sqrt{\pi}}\Gamma^{2}\left(\frac{1}{4}\right)\notag \end{align}

Update: Apparently I forgot to shed some light on the second question asked by OP namely any references/books on the theory of Jacobi Elliptic and Theta functions. This I deal with now.

The theory of Elliptic functions and theta functions is a very fascinating one. My own sources of study in this field are the following books (order of the books listed is not important here):

  • A Course of Modern Analysis by Whittaker and Watson: A definitive resource for many many topics apart from theta functions. The focus here is on using the methods of complex analysis to develop a theory of elliptic and theta functions.
  • An Elementary Treatise on Elliptic Functions by Cayley: This book develops the theory of elliptic functions in a very elementary fashion and I learnt most of the topic from this book. The presentation is easy to follow and does not require any deep skills apart from a good knowledge of calculus.
  • Elliptic Functions by Armitage and Eberlein: This is another good book from Cambridge University Press which I read. It uses both elementary techniques as well as complex analysis to develop the theory of elliptic functions.
  • Pi and the AGM by Borwein and Borwein: This is a truly singular book which connects elliptic integrals with Arithmetic Geometric Mean and is an interesting but somewhat difficult read.
  • Ramanujan Notebooks Vol 3 by Berndt: Ramanujan developed his own theory of theta functions independently of Jacobi and went far ahead, but his techniques are mostly unknown and Berndt has tried to discern his methods (to some extent) and establish most of his formulas in the theory of theta functions.
  • Collected Papers of Ramanujan : There are some papers dealing with some applications of theory of theta functions. The value of this book is to get an insight into some of the techniques Ramanujan used. In particular one must read his monumental paper "Modular Equations and Approximations to $\pi$".
  • Fundamenta Nova by Jacobi: This is a very good book but unfortunately written in Latin. I did manage to study some parts with the help of google translate and you can give it a try if you are willing to put extra effort of translation. Fortunately this is now available in English and of course freely.

All the above books combined together cover most of the elementary and some advanced topics related to elliptic and theta functions except its link with the algebraic number theory (mainly the link with imaginary quadratic extensions of $\mathbb{Q}$). This is perhaps the most important and deep topic which is now famous by the name of modular forms. Unfortunately I don't have much knowledge on this topic.

The references listed above can be found online for free (if you search enough). I have tried to extract material from these references and present a coherent theory of elliptic integrals/functions and theta functions in my blog posts (in fact I started blogging only to document whatever I had learnt about these mysterious theta functions). The blog also contains the Ramanujan's theory of theta functions and the Borwein's approach via Arithmetic Geometric Mean. The advantage of the blog posts is that they are concise and are written in a particular order such that pre-requisites for proving a result are discussed before presenting the result.

  • $\begingroup$ That was quite nice. $\endgroup$ – Felix Marin May 21 '16 at 22:57
  • $\begingroup$ @Paramanand Singh: It's very nice to see an answer to an old forgotten question. I really appreciate it. Your blog is amazing. I will surely go through it. $\endgroup$ – Shobhit Bhatnagar May 23 '16 at 3:49
  • $\begingroup$ Where did the product representation (1) come from? Would you mind including a proof of that, derived from the series definition of $\theta_4$? $\endgroup$ – Franklin Pezzuti Dyer Apr 2 '18 at 15:27
  • $\begingroup$ @Frpzzd: that's a corollary of Jacobi triple product $$\sum_{n\in\mathbb{Z}} z^nq^{n^2}=\prod_{n=1}^{\infty}(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1})$$ obtained by putting $z=-1$. For a proof of Jacobi triple product see paramanands.blogspot.com/2011/02/… $\endgroup$ – Paramanand Singh Apr 3 '18 at 4:05
  • $\begingroup$ Awesome, thank you! $\endgroup$ – Franklin Pezzuti Dyer Apr 3 '18 at 23:07

It may be of interest to note that the sum $$ g(x) = \sum_{k=1}^\infty \log\tanh (kx)$$ is harmonic and may be evaluated using Mellin transforms, yielding an asymptotic expansion about zero.

The Mellin transform $f^*(s)$ of the base function $$ f(x) = \log\tanh x$$ may be computed as follows

\begin{align} f^*(s) & = \mathfrak{M}(f(x); s) = \int_0^\infty \log \frac{e^x-e^{-x}}{e^x+e^{-x}} x^{s-1} \, dx = \int_0^\infty \log \left(1 - 2\frac{e^{-x}}{e^x+e^{-x}}\right) x^{s-1} dx \\[6pt] & = \int_0^\infty \log \left(1 - 2\frac{e^{-2x}}{1+e^{-2x}}\right) x^{s-1} dx = - \int_0^\infty \sum_{q\ge 1} \frac{1}{q} 2^q e^{-2qx}\left(\frac{1}{1+e^{-2x}}\right)^q x^{s-1} dx \\[6pt] & = -\sum_{q\ge 1} \frac{1}{q} 2^q \int_0^\infty e^{-2qx} \sum_{m\ge 0} (-1)^m \binom{m+q-1}{m} e^{-2mx} x^{s-1} dx \\[6pt] & = -\sum_{q\ge 1} \frac{1}{q} 2^q \sum_{m\ge 0} (-1)^m \binom{m+q-1}{m} \int_0^\infty e^{-2qx} e^{-2mx} x^{s-1} dx \\[6pt] & =-\Gamma(s)\sum_{q\ge 1} \frac{1}{q} 2^q \sum_{m\ge 0} (-1)^m \binom{m+q-1}{m} \frac{1}{(2m+2q)^s} \\[6pt] & = -\frac{\Gamma(s)}{2^s} \sum_{q\ge 1} \frac{1}{q} 2^q \sum_{m\ge 0} (-1)^m \binom{m+q-1}{m} \frac{1}{(m+q)^s}. \end{align}

To complete this calculation, ask about the coefficient of $$\frac{1}{n^s} = \frac{1}{(m+q)^s}.$$ It is given by $$\sum_{m=0}^{n-1} \frac{1}{n-m} 2^{n-m} (-1)^m \binom{n-1}{m} = - \frac{1}{n} (-1)^n + \frac{1}{n} \sum_{m=0}^n 2^{n-m} (-1)^m \binom{n}{m} = \frac{1}{n} \left(1-(-1)^n\right).$$ It follows that the double sum is $$\sum_{n\ge 1} \frac{1-(-1)^n}{n^{s+1}}= 2 \sum_{m\ge 0} \frac{1}{(2m+1)^{s+1}} = 2 \zeta(s+1) \left(1 - \frac{1}{2^{s+1}}\right) = \zeta(s+1) \left(2 - \frac{1}{2^s}\right).$$ This gives the following for $f^*(s):$ $$ f^*(s) = -\frac{\Gamma(s)}{2^s} \zeta(s+1) \left(2 - \frac{1}{2^s}\right).$$ Now introduce $g(x)$, the harmonic sum we are trying to calculate, so that $$ g(x) = \sum_{k=1}^\infty f(kx).$$ The Mellin transform of $g(x)$ is then given by $$ g^*(s) = \mathfrak{M}(g(x); s) = -\frac{\Gamma(s)}{2^s} \zeta(s) \zeta(s+1) \left(2 - \frac{1}{2^s}\right).$$ The zeros of the two zeta function terms cancel the poles of the gamma function, so that inverting $g^*(s)$ we only have two terms that contribute, namely $$\operatorname{Res}(g^*(s) x^{-s}; s=1) = -1/8\,{\frac {{\pi }^{2}}{x}}$$ and $$\operatorname{Res}(g^*(s) x^{-s}; s=0) = 1/2\,\log \left( 2\,\pi \right) -1/2\,\log \left( x \right).$$ This yields that in a neighborhood of zero $$ g(x) \sim 1/2\,\log \left( 2\,\pi \right) -1/2\,\log \left( x \right) -1/8\,{\frac {{\pi }^{2}}{x}}$$ Setting $x=\frac{\pi}{2}$, we obtain that $$g\left(\frac{\pi}{2}\right) \sim \log 2 - \frac{\pi}{4},$$ which produces only three good digits. On the other hand, for e.g. $x=\frac{1}{4}$, we get $$g\left(\frac{1}{4}\right) \sim 3/2\,\log \left( 2 \right) +1/2\,\log \left( \pi \right) -1/2\,{\pi }^{2} \sim -3.322716487,$$ which has nine good digits.

For e.g. $x=\frac{\pi}{16}$, we get $$g\left(\frac{\pi}{16}\right) \sim 5/2\,\log \left( 2 \right) -2\,\pi \sim -4.5503173557797232034$$ which has 20 good digits.

It seems quite intriguing to ask whether this expansion can also be derived directly from properties of the Jacobi theta function without using Mellin transforms.

  • $\begingroup$ @Michael Hardy. I usually rollback edits that are done for aesthetic reasons. As this one did no harm I will let it stand. Please remember this for the future. $\endgroup$ – Marko Riedel Dec 27 '15 at 23:38
  • 1
    $\begingroup$ Would you normally expect some harm to result from such edits? Usually they improve legibility. Is that harmful? ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 27 '15 at 23:41
  • $\begingroup$ If a user with 130K reputation makes an edit I would not expect any harm to come from it. With another user it may be different. Legibility is a somewhat subjective concept. I actually like the tree format for multiline equations. $\endgroup$ – Marko Riedel Dec 27 '15 at 23:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.