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Let

$$P(q)=1+240\sum_{n=1}^{\infty}\frac{n^3q^n}{1-q^n}$$

We have a closed form for $q=e^{-\frac{\pi}{x}\cdot2^{2^{-k}\cdot{n}}}$

$$P\left(e^{-\frac{\pi}{x}\cdot2^{2^{-k}\cdot{n}}}\right)=\left[(2x)^{2^k}\cdot{2^{-n}}\right]^{2^{2-k}}$$

We believe it valids for $n\le 2^k-1$ and $x\ge 1$

We was able to find the closed form for it! Can anyone kindly give us a bit of details what all this roughly mean and where is the $e^{\pi}$ disappeared to and this $e^{\pi}$ either gives an almost integer or sometimes an integer. why? probably this $e^\pi$ has properties of cancellation and also produces a fast convergences to an integer.

Can anyone verify its correctness

For example,

Setting $k=3$, $n=5$ and $x=4$

$$P\left(e^{\frac{-\pi}{4}\cdot2^{ \frac{5}{8}}}\right)=\left[8^8\cdot2^{-5}\right]^{\frac{1}{2}}=724.0773439...$$

  • ... And where did you get this? More context could help. – Simply Beautiful Art May 15 '16 at 15:44
  • Don't get it. There is a 'n' which is both a dummy index of the summation, and it appears in square brackets on the r.h.s. of the final formula....can you clarify? – Pierpaolo Vivo May 15 '16 at 17:41
  • Just a partial annotation. The series can be evaluated in closed form. Write $$ \sum_{n=1}^{\infty}\frac{n^3q^n}{1-q^n}=\sum_{n=1}^{\infty}n^3\sum_{k=1}^\infty q^{nk}=\frac{1}{\log^3 q}\sum_{k=1}^\infty \frac{d^3}{ds^3}\sum_{n=1}^\infty q^{nk+ns}\Big|{s\to 0}= $$ $$ =\frac{1}{\log^3 q}\sum{k=1}^\infty \frac{d^3}{ds^3}\frac{q^{k+s}}{1-q^{k+s}}\Big|_{s\to 0}=\boxed{\frac{\psi _q^{(3)}(1)}{\log ^4(q)}}\ , $$ where $\psi _q^{(3)}(x)$ is the q-Polygamma function [http://mathworld.wolfram.com/q-PolygammaFunction.html]. – Pierpaolo Vivo May 15 '16 at 18:12

1 Answers1

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Ok this one appears tough. The function you have mentioned is actually denoted by $Q$ and not by $P$. Let $$Q(q) = 1 + 240\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{n}}$$ This is a kind of Eisentein series in modern language and is a modular form of weight $4$. The function is deeply connected with the theory of elliptic integrals/functions and theta functions.

Let $0 < k < 1$ and define $$k' = \sqrt{1 - k^{2}}, K = K(k) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - k^{2}\sin^{2}x}}, K' = K(k'), q = \exp\left(-\pi\frac{K'}{K}\right)$$ then it can be proved that $$Q(q) = \left(\frac{2K}{\pi}\right)^{4}\left(1 + 14k^{2} + k^{4}\right)$$ It is further known that the value of $k$ is algebraic if $q = e^{-\pi\sqrt{n}}$ where $n$ is a positive rational number.

You mention a specific value of $q$ such that $$\log q = -\frac{\pi}{x}\cdot 2^{2^{-k}\cdot n}$$ so that $$\frac{K'}{K} = \frac{2^{2^{-k}\cdot n}}{x}$$ and it appears that $x$ is real and $n$ is a positive integer in your question. Note that if $q = 0$ then $k = 0$ and $K = \pi/2$ so that $Q(q) = 1$. If we let $x \to 0^{+}$ in your formula the LHS becomes $Q(0) = 1$ and RHS is $0$. So definitely we can't have the formula valid for $x \to 0^{+}$. The evaluation of $Q(q)$ for values of $q$ other than $\exp(-\pi\sqrt{n})$ is really difficult. I wonder if there is a simple way to evaluate $Q(q)$ for some arbitrary value of $q$.

I am highly doubtful of a formula for $Q(q)$ which gives its value in closed form for all values of $q$ in a certain sub-interval of $(0, 1)$. Can you let us know how you arrived at this formula (or at least give some specific values of $k, n, x$ for which you think it is true)?

Also the values of $e^{\pi\sqrt{n}}$ being close to integer is not so much of a mystery as people think it is. See this question Approximation of $e^{\pi\sqrt{n}}$ using Ramanujan's Class Invariants

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Paramanand Singh
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  • I have given an example. I am so sorry, foolish of me to post thing I don't at ALL understand. The reason I posted it, because it quite bother me, I wanted to know whether it is corrected or not. You won't see these nonsense questions from me again. Thank you. –  May 16 '16 at 21:21
  • @pisquare: You don't need to feel sorry for asking any question here. It is very difficult to say whether your formula is correct or not because I don't have a way to do numerical calculations to check your example. I was trying to figure out if there is some proof theoretically and I am unable to do so. – Paramanand Singh May 17 '16 at 02:00