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Правка en49, от bfsof123, 2024-06-02 18:04:42

In this blog we prove a Ramanujan-type identity:

$$$S(n_1, n_2, n_3) := \sum\limits_{n_1 \in \mathbb{Z}}(-1)^{n_1}\sum\limits_{(n_2, n_3) \in \mathbb{Z}^2}\frac{1}{\sqrt{n_1^2 + (n_2+0.5)^2+(n_3+0.5)^2}\sinh{(\pi\sqrt{n_1^2 + (n_2+0.5)^2+(n_3+0.5)^2})}} = 1$$$. (1)

Magic code

First, we consider a 4D lattice sum:

$$$U(n_1, n_2, n_3, n_4) := \sum\limits_{(n_1, n_2, n_3, n_4) \in \mathbb{Z}^4} \frac{(-1)^{n_1+n_4}}{n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2 + n_4^2}$$$. We will show later that $$$U(n_1, n_2, n_3, n_4)$$$ equals to $$$\pi$$$ (code shown below).

Pi code

Note that the above series converge super slow, and the Python code is also super slow. Be patient please!

For $$$\lambda > 0$$$, $$$\int_{0}^\infty e^{-\lambda x}dx = \frac{1}{\lambda}$$$, therefore $$$U(n_1, n_2, n_3, n_4) = \sum\limits_{(n_1, n_2, n_3, n_4) \in \mathbb{Z}^4} \int_{0}^\infty (-1)^{n_1+n_4} \exp((-n_1^2-(n_2+0.5)^2-(n_3+0.5)^2-n_4^2)x)dx$$$. (2)

By the Poisson summation formula of the theta function (see this),

$$$\sum\limits_{n \in \mathbb{Z}}(-1)^n \exp(-n^2 x) = \sqrt{\frac{\pi}{x}} \sum\limits_{n \in \mathbb{Z}}\exp(\frac{-(n+0.5)^2\pi^2}{x})$$$ (3)

By combining (2) and (3), $$$U$$$ can be written as a Mellin-transform-type integral:

$$$U(n_1, n_2, n_3, n_4) = \sum\limits_{(n_1, n_2, n_3, n_4) \in \mathbb{Z}^4} \sqrt{\pi}\int_{0}^\infty (-1)^{n_1} x^{-0.5} \exp(-(n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2)x - \frac{(n_4+0.5)^2\pi^2}{x})dx$$$. (4)

We then use a famous integral representation of the Modified Bessel functions of the second kind (i.e., $$$K$$$):

$$$M_s[\exp(-ax-\frac{b}{x})] := \int_{0}^\infty x^{s-1}\exp(-ax-\frac{b}{x})dx = 2(\frac{b}{a})^{\frac{s}{2}}K_s(2\sqrt{ab})$$$. (5)

In (5), $$$M_s$$$ denotes the Mellin transform with parameter $$$s$$$. And we note that when $$$s=0.5$$$, $$$K_{0.5}(z)$$$ has a closed-form: $$$K_{0.5}(z) = \sqrt\frac{\pi}{2} \frac{e^{-z}}{\sqrt{z}}$$$.

Then, $$$\int_{0}^\infty x^{-0.5}\exp(-ax-\frac{b}{x})dx = 2(\frac{b}{a})^{0.25} K_{0.5}(2\sqrt{ab}) = 2(\frac{b}{a})^{0.25}\sqrt{\frac{\pi}{2}}\frac{\exp(-2\sqrt{ab})}{\sqrt{2}(ab)^{0.25}} = \sqrt{\frac{\pi}{a}}\exp(-2\sqrt{ab})$$$. (6)

We set $$$a = n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2$$$ and $$$b = (n_4+0.5)^2\pi^2$$$, and

$$$U(n_1, n_2, n_3, n_4) = \sum\limits_{(n_1, n_2, n_3) \in \mathbb{Z}^3} (-1)^{n_1}\frac{\pi}{\sqrt{n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2}}\sum\limits_{n_4 \in \mathbb{Z}} \exp(-\pi |2n_4+1| \sqrt{n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2})$$$. (7)

In (7), the $$$\pi$$$ comes from two places. One $$$\sqrt{\pi}$$$ comes from the Poisson summation formula (Eq.(3)), and the other $$$\sqrt{\pi}$$$ comes from the Bessel function $$$K_{0.5}$$$. $$$\sum\limits_{n_4 \in \mathbb{Z}} \exp(-\pi |2n_4+1| \sqrt{n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2})$$$ is symmetric about $$$n_4 = -0.5$$$. $$$n_4 = 0,1,2,...,\infty$$$ and $$$n_4 = -1,-2,...,-\infty$$$, form two identical geometric series with initial term $$$a_1 := \exp(-\pi \sqrt{n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2})$$$ and common ratio $$$q := \exp(-2\pi \sqrt{n_1^2 + (n_2+0.5)^2 + (n_3+0.5)^2})$$$. Each one contributes $$$\frac{a_1}{1-q}$$$ and they two contribute $$$\frac{2a_1}{1-q} = \frac{1}{\sinh{(\pi\sqrt{n_1^2 + (n_2+0.5)^2+(n_3+0.5)^2})}}$$$ in total, amazing!

Therefore, $$$U(n_1, n_2, n_3, n_4) = \pi S(n_1, n_2, n_3)$$$. It is sufficient to show that $$$U = \pi$$$. Now we check the Zucker1974 paper Exact results for some lattice sums in 2, 4, 6 and 8 dimensions with a scihub link. $$$U(n_1, n_2, n_3, n_4)$$$ is equal to $$$U(2, 2)$$$ (Eq. (2.10)) with $$$s = 1$$$. By checking Table 1 in the paper, $$$U(n_1, n_2, n_3, n_4) = 8\beta(0)\beta(1)$$$, where $$$\beta(s) := \sum\limits_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$$$ is the Dirichlet beta function. It is well known that $$$\beta(0) = 0.5$$$ and $$$\beta(1) = \frac{\pi}{4}$$$. Finally, $$$U(n_1, n_2, n_3, n_4) = \pi$$$ and $$$S(n_1, n_2, n_3) = 1$$$.

Related words: Codeforces, lattice sum, Ramanujan, theta function, modular form, integral representation of the Gamma function, Bessel function, hyperbolic sine and hyperbolic cosine (sinh, cosh).

This work is inspired by the Madelung constant.

Теги math, modular form, ramanujan, lattice sum, sinh, codeforces

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