## Tuesday, October 20, 2015

### Deriving the Product Formula: The Easy Way

Recall from this post that: $\sum_{n=1}^{\infty} \frac{1}{x^2+n^2}=\frac{\pi}{2x} \coth(\pi x)-\frac{1}{2x^2}$ We then substitute $x=i z$: $\sum_{n=1}^{\infty} \frac{1}{n^2-z^2}=-\frac{\pi}{2z} \cot(\pi z)+\frac{1}{2z^2}$ We then go down the following line of calculation: $\sum_{n=1}^{\infty} \frac{2z}{n^2-z^2}=\frac{1}{z}-\pi\cot(\pi z)$ $\int\sum_{n=1}^{\infty} \frac{2z}{n^2-z^2}dz=C+\int \frac{1}{z}-\pi\cot(\pi z) dz$ $\sum_{n=1}^{\infty} -\ln \left (1-\frac{z^2}{n^2} \right )=C+\ln (z) - \ln (\sin (\pi z) )$ $\sin(\pi z)=C' z\prod_{n=1}^{\infty}\left ( 1-\frac{z^2}{n^2} \right )$ We can find $C'$ by looking at the behavior near zero, and so find that: $\sin(\pi z)=\pi z\prod_{n=1}^{\infty}\left ( 1-\frac{z^2}{n^2} \right )$ Therefore: $\sin(z)=z\prod_{n=1}^{\infty}\left ( 1-\frac{z^2}{\pi^2 n^2} \right )$

### Deriving the Product Formula: The Overkill Way, by Weierstrass' Factorization Theorem

Suppose a function can be expressed as $f(x)=A\frac{\prod_{n=1}^{M}\left ( x-z_n \right )}{\prod_{n=1}^{N}\left ( x-p_n \right )}$ Where $M \leq N$ and $N$ can be arbitrarily large, even tending to infinity. Assuming there are no poles of degree >1 (all poles are simple), we can rewrite this as $f(x)=K+\sum_{n=1}^{\infty} \frac{b_n}{x-p_n}$ Where some of the $b_n$ may be zero. We can also write this as $f(x)=f(0)+\sum_{n=1}^{\infty} b_n \cdot \left ( \frac{1}{x-p_n}+\frac{1}{p_n} \right )$ Suppose $f(0) \neq 0$, and that $f$ is an integral function (i.e. an entire function). In that case, the logarithmic derivative $f'(x)/f(x)$ has poles of degree 1. Moreover, $\lim_{x \rightarrow z_n} (x-z_n)\frac{f'(x)}{f(x)}=d_n$ Where $d_n$ is the degree of the zero at $z_n$. Thus: $\frac{f'(x)}{f(x)}=\frac{f'(0)}{f(0)}+\sum_{n=1}^{\infty} d_n \cdot \left ( \frac{1}{x-z_n}+\frac{1}{z_n} \right )$ Integrating: $\ln(f(x))=\ln(f(0))+x \frac{f'(0)}{f(0)}+\sum_{n=1}^{\infty} d_n \cdot \left ( \ln \left (1-\frac{x}{z_n} \right ) +\frac{x}{z_n} \right )$ $f(x)=f(0) e^{x \frac{f'(0)}{f(0)}} \prod_{n=1}^{\infty} \left (1-\frac{x}{z_n} \right )^{d_n} e^{x\frac{d_n}{z_n}}$ This is our main result, called the Weierstrass factorization theorem. In particular, for the function $f(x)=\sin(x)/x$ $\frac{\sin(x)}{x}=\prod_{n=-\infty, n \neq 0}^{\infty} \left (1-\frac{x}{n \pi} \right ) e^{x\frac{1}{n \pi}}=\prod_{n=1}^{\infty} \left (1-\frac{x^2}{n^2 \pi^2} \right )$ Thus $\sin(x)=x\prod_{n=1}^{\infty} \left (1-\frac{x^2}{\pi^2 n^2 } \right )$

### Corollary 1: Wallis Product

Let us plug in $x=\pi/2$: $\sin(\pi/2)=1=\frac{\pi}{2}\prod_{n=1}^{\infty} \left (1-\frac{1}{4 n^2 } \right )$ $\pi=2\prod_{n=1}^{\infty} \left (\frac{4 n^2}{4 n^2-1 } \right )=2\frac{2 \cdot 2}{1 \cdot 3} \cdot \frac{4 \cdot 4}{3 \cdot 5} \cdot \frac{6 \cdot 6}{5 \cdot 7} \cdot \frac{8 \cdot 8}{7 \cdot 9} \cdots$ More generally: $\pi=\frac{N}{M} \sin(\pi M/N) \prod_{n=1}^{\infty} \left (\frac{N^2 n^2}{N^2 n^2 -M^2} \right )$ This is useful when $\sin(\pi M/N)$ is easily computable, such as when $\sin(\pi M/N)$ is algebraic (e.g. $M=1$, $N=2^m$ ). For example: $\pi=2 \sqrt{2} \prod_{n=1}^{\infty} \left (\frac{4^2 n^2}{4^2 n^2 -1^2} \right )$ $\pi=\frac{2}{3} \sqrt{2} \prod_{n=1}^{\infty} \left (\frac{4^2 n^2}{4^2 n^2 -3^2} \right )$ $\pi=\frac{3}{2} \sqrt{3} \prod_{n=1}^{\infty} \left (\frac{3^2 n^2}{3^2 n^2 -1^2} \right )$ $\pi=\frac{3}{4} \sqrt{3} \prod_{n=1}^{\infty} \left (\frac{3^2 n^2}{3^2 n^2 -2^2} \right )$ $\pi=3 \prod_{n=1}^{\infty} \left (\frac{6^2 n^2}{6^2 n^2 -1^2} \right )$ $\pi=\frac{3}{5} \prod_{n=1}^{\infty} \left (\frac{6^2 n^2}{6^2 n^2 -5^2} \right )$ $\pi=3\sqrt{2}(-1+\sqrt{3}) \prod_{n=1}^{\infty} \left (\frac{12^2 n^2}{12^2 n^2 -1^2} \right )$

### Corollary 2: Product Formula for Cosine

Let us evaluate the sine formula at $x+\pi/2$: $\sin(x+\pi/2)=\cos(x)=\left (x+\frac{\pi}{2} \right )\prod_{n=-\infty, n \neq 0}^{\infty} \left (1-\frac{x+\pi/2}{\pi n } \right )$ $\cos(x)=\frac{\sin(x+\pi/2)}{\sin(\pi/2)}=\left (1+\frac{x}{\pi/2} \right )\prod_{n=-\infty, n \neq 0}^{\infty} \frac{\left (1-\frac{x+\pi/2}{\pi n } \right )}{\left (1-\frac{\pi/2}{\pi n } \right )}$ $\cos(x)=\left (1+\frac{x}{\pi/2} \right )\prod_{n=-\infty, n \neq 0}^{\infty} \left (1-\frac{x}{\pi (n-1/2) } \right )=\prod_{n=-\infty}^{\infty} \left (1-\frac{x}{\pi (n-1/2) } \right )$ $\cos(x)=\prod_{n=1}^{\infty} \left (1-\frac{x^2}{\pi^2 (n-1/2)^2 } \right )$ Alternatively, we can derive this directly from the Weierstrass factorization theorem.
Additionally, by using imaginary arguments, we can derive the formulae: $\sinh(x)=x\prod_{n=1}^{\infty} \left (1+\frac{x^2}{\pi^2 n^2 } \right )$ $\cosh(x)=\prod_{n=1}^{\infty} \left (1+\frac{x^2}{\pi^2 (n-1/2)^2 } \right )$

### Corollary 3: Sine is Periodic

Let us evaluate the sine formula at $x+\pi$: $\sin(x+\pi)=\left (x+\pi \right )\prod_{n=-\infty, n \neq 0}^{\infty} \left (1-\frac{x+\pi}{\pi n } \right )$ $\sin(x+\pi)=\cdots \left (1+\frac{x+\pi}{3\pi} \right ) \left (1+\frac{x+\pi}{2\pi} \right )\left (1+\frac{x+\pi}{\pi} \right )\left (x+\pi \right ) \left (1-\frac{x+\pi}{\pi} \right )\left (1-\frac{x+\pi}{2\pi} \right ) \left (1-\frac{x+\pi}{3\pi} \right ) \cdots$ $\sin(x+\pi)=\cdots \left (\frac{4}{3}+\frac{x}{3\pi} \right ) \left (\frac{3}{2}+\frac{x}{2\pi} \right )\left (2+\frac{x}{\pi} \right ) \pi \left (1+\frac{x}{\pi}\right ) \left (\frac{-x}{\pi} \right )\left (\frac{1}{2}-\frac{x}{2\pi} \right ) \left (\frac{2}{3}-\frac{x}{3\pi} \right ) \cdots$ $\sin(x+\pi)=\cdots \frac{4}{3}\left (1+\frac{x}{4\pi} \right ) \frac{3}{2}\left (1+\frac{x}{3\pi} \right )2\left (1+\frac{x}{2\pi} \right ) \pi \left (1+\frac{x}{\pi}\right ) \left (\frac{-x}{\pi} \right ) \frac{1}{2}\left (1-\frac{x}{\pi} \right ) \frac{2}{3}\left (1-\frac{x}{2\pi} \right ) \cdots$ $\sin(x+\pi)=-2x\left ( \prod_{k=2}^{\infty} \frac{k^2-1}{k^2} \right ) \left ( \prod_{n=1}^{\infty} \left (1-\frac{x^2}{n^2 \pi^2} \right ) \right )=-\sin(x)$ As the first product easily telescopes. Thus $\sin(x+2\pi)=\sin((x+\pi)+\pi)=-\sin(x+\pi)=\sin(x)$. Therefore, sine is periodic with period $2\pi$.

### Corollary 3: Some Zeta Values

Let us begin expanding the product for sine in a power series $\sin(x)=x\prod_{n=1}^{\infty} \left (1-\frac{x^2}{\pi^2 n^2 } \right )=x-\frac{x^3}{\pi^2}\left (\frac{1}{1^2}+\frac{1}{2^2}+\cdots \right )+\frac{x^5}{\pi^4}\left (\frac{1}{1^2 \cdot2^2}+\frac{1}{1^2 \cdot3^2}+\cdots \frac{1}{2^2 \cdot3^2}+\frac{1}{2^2 \cdot4^2}+\cdots \right )+\cdots$ $\sin(x)=x-\frac{x^3}{\pi^2}\left (\sum_{k=1}^{\infty}\frac{1}{k^2} \right )+\frac{x^5}{\pi^4}\left (\sum_{m=1,n=1, m < n}^{\infty}\frac{1}{m^2n^2} \right )+\cdots$ $\sin(x)=x-\frac{x^3}{\pi^2}\left (\sum_{k=1}^{\infty}\frac{1}{k^2} \right )+\frac{x^5}{2\pi^4}\left (\left (\sum_{k=1}^{\infty}\frac{1}{k^2} \right )^2- \sum_{k=1}^{\infty}\frac{1}{k^4} \right )+\cdots$ By comparing this to the Taylor series for sine, we find: $\frac{1}{3!}=\frac{1}{\pi^2}\left (\sum_{k=1}^{\infty}\frac{1}{k^2} \right )$ $\frac{1}{5!}=\frac{1}{2\pi^4}\left (\left (\sum_{k=1}^{\infty}\frac{1}{k^2} \right )^2- \sum_{k=1}^{\infty}\frac{1}{k^4} \right )$ From which it follows that $\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$ $\sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{\pi^4}{90}$ In fact, for the fourth term, we find, similarly, that $\frac{1}{7!}=\frac{1}{6\pi^6}\left ( \left (\sum_{k=1}^{\infty}\frac{1}{k^2} \right )^3-3\left (\sum_{k=1}^{\infty}\frac{1}{k^2} \right )\left (\sum_{k=1}^{\infty}\frac{1}{k^4} \right )+2\left (\sum_{k=1}^{\infty}\frac{1}{k^6} \right ) \right )$ From which it follows that $\sum_{k=1}^{\infty}\frac{1}{k^6}=\frac{\pi^6}{945}$