Joy of Mathematics

An eta Dedekind type product

March 16, 2023 Tolaso

Prove that

$\prod_{n=1}^{\infty} \left (1 + 2e^{-n \pi \sqrt{3}} \cosh \frac{n \pi}{\sqrt{3}} \right ) = \frac{\exp \left ( \frac{\pi \sqrt{3}}{18} \right )}{\sqrt[4]{3}}$

Solution

Let $\tau = \frac{i}{\sqrt{3}}$ and $q = \exp \left ( 2 \pi i \tau \right )$ . Hence,

$\begin{align*} \prod_{n=1}^{\infty} \left (1 + 2e^{-n \pi \sqrt{3}} \cosh \frac{n \pi}{\sqrt{3}} \right ) &= \prod_{n=1}^{\infty} \left ( 1 + q + q^{2n} \right ) \\ &= \frac{\prod_{n=1}^{\infty} \left ( 1 - q^{3n} \right )}{\prod_{n=1}^{\infty} \left ( 1 - q^n \right )} \\ &= \exp\left ( - \frac{i \pi \tau}{6}\right ) \frac{\eta(3\tau)}{\eta(\tau)} \\ &= \exp\left ( - \frac{i \pi \tau}{6}\right ) \frac{\eta \left ( - 1/\tau \right )}{\eta(\tau)}\\ &= \exp\left ( - \frac{i \pi \tau}{6}\right ) \sqrt{-i \tau} \\ &= frac{\exp \left ( \frac{\pi \sqrt{3}}{18} \right )}{\sqrt[4]{3}} \end{align*}$

A special function sum

December 11, 2022 Tolaso

Let $\varphi$ denote the Euler’s totient function and $\Lambda$ denote the von Mangoldt function. Evaluate the sum

$\sum_{a \mid n} \varphi(a) \log a + \sum_{a \mid n} \sum_{b \mid \frac{n}{a}} \varphi(a) \Lambda (b)$

where $n$ is a positive integer.

Solution

We have successively:

$\begin{align*} \sum_{a \mid n} \varphi(a) \log a + \sum_{a \mid n} \sum_{b \mid \frac{n}{a}} \varphi(a) \Lambda (b) &= \sum_{a \mid n} \varphi(a) \log a +\sum_{a \mid n} \varphi(a) \sum_{b \mid \frac{n}{a}} \Lambda(b) \\ &= \sum_{a \mid n} \varphi(a) \log a + \sum_{a \mid n} \varphi (a) \log \frac{n}{a} \\ &= \sum_{a \mid n} \varphi(a) \left ( \log a + \log \frac{n}{a} \right ) \\ &= \log n \sum_{a \mid n} \varphi(a) \\ &= n \log n \end{align*}$

A Möbius inversion sum

November 29, 2022November 29, 2022 Tolaso

Let $q \in \mathbb{R}$ such that $|q|<1$ . Prove that

$\sum_{n=1}^{\infty} \mu(n) \cdot \frac{q^n}{1-q^n} = q$

Solution

We have successively:

$\begin{align*} \sum_{n=1}^{\infty} \mu(n) \cdot \frac{q^n}{1-q^n} &= \sum_{n=1}^{\infty} \mu(n) \sum_{m=0}^{\infty} q^{nm} \\ &= \sum_{n=1}^{\infty} \sum_{d \mid n} \mu(d)q^n \\ &= q + \sum_{n=2}^{\infty} \sum_{d \mid n} \mu(d)q^n \end{align*}$

since $\sum \limits_{d \mid n} \mu(d) =0$ whenever $n \geq 2.$

A Möbius function product

November 29, 2022November 29, 2022 Tolaso

Let $\mu$ denote the Möbius function. Prove that

$\prod_{n=1}^{\infty} \left ( 1 - x^n \right )^{-\mu(n)/n} = e^x \quad , \quad |x|<1$

Solution

Taking logarithms we have:

$\begin{align*} \log \prod_{n=1}^{\infty} \left ( 1 - x^n \right )^{-\mu(n)/n} &= -\sum_{n=1}^{\infty} \frac{\mu(n)}{n} \log \left ( 1 - x^n \right ) \\ &= \sum_{n=1}^{\infty} \frac{\mu(n)}{n} \sum_{m=1}^{\infty} \frac{x^{nm}}{m} \\ &= \sum_{n=1}^{\infty} \mu(n) \sum_{m=1}^{\infty} \frac{x^{nm}}{nm} \\ &= \sum_{m=1}^{\infty} \frac{x^m}{m} \sum_{d \mid m} \mu(d) \\ &= x + \sum_{m=2}^{\infty} \frac{x^m}{m} \sum_{d \mid m} \mu(d) \\ &= x \end{align*}$

since $\sum \limits_{d \mid m} \mu(d) =0$ whenever $m \geq 2$ .

Identity with Beta and Gamma function

November 10, 2022November 10, 2022 Tolaso

Let $\mathrm{B}$ and $\Gamma$ denote the Euler’s Beta and Gamma functions. Prove that

$\frac{\Gamma(a_1) \Gamma(a_2) \cdots \Gamma(a_n)}{\Gamma \left ( a_1 + a_2 + \cdots + a_n \right )} = \mathrm{B} \left ( a_1, a_2 \right ) \mathrm{B} \left ( a_1+a_2, a_3 \right ) \cdots \mathrm{B} \left ( a_1+a_2 + \cdots + a_{n-1}, a_n \right )$

Solution

We have successively:

$\begin{align*} \frac{\Gamma(a_1) \Gamma(a_2) \cdots \Gamma(a_n)}{\Gamma \left ( a_1 + a_2 + \cdots + a_n \right )} &= \frac{\Gamma(a_1) \Gamma(a_2)}{\Gamma(a_1+a_2)} \cdot \frac{\Gamma(a_1+a_2) \Gamma(a_3)}{\Gamma(a_1+a_2+a_3)} \cdots \\ &\quad \quad \quad \cdots \frac{\Gamma\left ( a_1+a_2 + \cdots + a_{n-1}\right ) \Gamma(a_n)}{\Gamma \left ( a_1+a_2 + \cdots + a_{n} \right )} \\ &=\mathrm{B} \left ( a_1, a_2 \right ) \mathrm{B} \left ( a_1+a_2, a_3 \right ) \cdots \\ &\quad \quad \quad \cdots \mathrm{B} \left ( a_1+a_2 + \cdots + a_{n-1}, a_n \right ) \end{align*}$

The result follows.