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Infinite product

We proposed

(1)

$$\left(\frac{e^{\pi}+1}{e^{\pi}-1}\cdot\frac{e^{3\pi}+1}{e^{3\pi}-1}\cdot\frac{e^{5\pi}+1}{e^{5\pi}-1}\cdots\right)^8=2$$

How one go about proving (1) ?

take the ln

$$\frac{1}{8}\ln(2)=\ln\left(\frac{e^{\pi}+1}{e^{\pi}-1}\right)+\ln\left(\frac{e^{3\pi}+1}{e^{3\pi}-1}\right)+\ln\left(\frac{e^{5\pi}+1}{e^{5\pi}-1}\right)\cdots $$

May be I have to use

$$\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)=x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\cdots$$

This series is the closest I see to the product, but it seem too complicate when substitute in...

I am stuck, can't go another further please help!

  • That power series expansion won't even work since $e^{(2k+1)\pi} > 1$, that is, the power series expansion is not valid for $x>1$. – Fimpellizzeri May 20 '16 at 20:51
  • What is the motivation for this claim? (Mathematica does seem to support it numerically, but I'm curious as to the source.) – Semiclassical May 20 '16 at 20:51
  • Perhaps one can use that for $a=e^{(2k+1)\pi}$, we may write $$\frac{a+1}{a-1}=1+\frac{2}{a-1}$$and use the power series expansion for $\ln(1+x)$. – Fimpellizzeri May 20 '16 at 20:55
  • @Fimpellizieri The power series is a valid approach if one writes $\dfrac{e^{(2k+1)\pi}+1}{e^{(2k+1)\pi}-1}=\dfrac{1+e^{-(2k+1)\pi}}{1-e^{-(2k+1)\pi}}$ and uses $x=e^{-(2k+1)\pi}$. – Semiclassical May 20 '16 at 20:58
  • @Semiclassical That's a great idea actually. – Fimpellizzeri May 20 '16 at 21:01
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    Have a look at Marko's solution here; your product is essentially $\prod\limits_{k=1}^\infty \mathrm{coth}\left(\pi\left(k-\frac12\right)\right)$ – J. M. ain't a mathematician May 20 '16 at 21:24
  • $$ \sum_{n\text{ odd}}\text{arctanh}\left(e^{-n\pi}\right)=\sum_{m\text{ odd}}e^{-m \pi}\sum_{\substack{d\mid m \ m\text{ odd}}}\frac{1}{d}=\sum_{m\text{ odd}}\sigma(m)\frac{e^{-m\pi}}{m}$$ but is this really related with an Eisenstein series (https://en.wikipedia.org/wiki/Eisenstein_series) ? – Jack D'Aurizio May 20 '16 at 21:44
  • We prolly don't need to bring something that complicated; theta functions should suffice. – J. M. ain't a mathematician May 20 '16 at 21:51
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    @J.M.: you're right, it should be enough to exploit Jacobi's triple product (https://en.wikipedia.org/wiki/Jacobi_triple_product) for $x=e^{-\pi}$, then Jacobi theta functions. – Jack D'Aurizio May 20 '16 at 22:32

1 Answers1

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Ok, this is simply Ramanujan's class invariants $g_{n}, G_{n}$ in disguise. We have \begin{align}G_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 + e^{-\pi\sqrt{n}})(1 + e^{-3\pi\sqrt{n}})(1 + e^{-5\pi\sqrt{n}})\cdots\tag{1}\\ g_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 - e^{-\pi\sqrt{n}})(1 - e^{-3\pi\sqrt{n}})(1 - e^{-5\pi\sqrt{n}})\cdots\tag{2}\end{align} Putting $n = 1$ in both the formulas and dividing them we get the expression in question as $$(G_{1}/g_{1})^{8}$$ We have $G_{1} = 1, g_{1} = 2^{-1/8}$ so the desired value $2$ is obtained easily.

It is known from the theory of elliptic integrals/functions and theta functions that $G_{n}, g_{n}$ are algebraic numbers for positive rational number $n$. Calculating these values for general rational $n$ is difficult (except for guys like Ramanujan who hand calculated a host of such values for many integers $n$).

For the current question it is sufficient to use the formulas $$G_{n} = (2kk')^{-1/12},\,g_{n} = \left(\frac{2k}{k'^{2}}\right)^{-1/12}\tag{3}$$ where $k$ is a real number with $0 < k < 1$ and $$k = k(e^{-\pi\sqrt{n}}), k' = \sqrt{1 - k^{2}}\tag{4}$$ and in general $$k = k(q) = \frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)}\tag{5}$$ in terms of Jacobi's Theta functions. The important thing to note here is that if $n = 1$ then $q = e^{-\pi}$ and $k = k' = 1/\sqrt{2}$ and then values of $G_{1}, g_{1}$ are easily obtained from $(3)$.

Paramanand Singh
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  • Thank you for nicely answer this question and ease my search for this Gn and g values. I was looking all over the site and couldn't find it until now, you point me to it, thank you. Nearly over 8 years now that why i couldn't Remember when I last saw it. –  May 21 '16 at 04:01
  • @mahdi: Glad to know that my answer was of some help in finding exactly what you needed. – Paramanand Singh May 21 '16 at 08:27
  • I have another infinite product Ramanujan type involving g function can you have a look at it and help to prove it. I will let you know when I post it. –  May 23 '16 at 15:11
  • @mahdishafici: I will definitely have a look at it. – Paramanand Singh May 23 '16 at 18:51
  • Thank you @paramanand singh –  May 23 '16 at 21:07