# Chaitanya's Random Pages

## August 9, 2011

### A Collection of Infinite Products – II

Filed under: mathematics — ckrao @ 11:57 am

Following on from my previous post on infinite products, here are some more to enjoy! Like last time, proof outlines are at the end.

(1) Seidel: $\displaystyle \prod_{n=1}^{\infty} \frac{2}{1 + x^{1/2^{n}}} = \frac{\log x}{x-1} = \frac{2}{1 + \sqrt{x}}\cdot \frac{2}{1 + \sqrt{\sqrt{x}}}\cdot \frac{2}{1 + \sqrt{\sqrt{\sqrt{x}}}} \cdot \ldots$

In particular x=2 gives von Seidel‘s product: $\displaystyle \log 2 = \prod_{n=1}^{\infty} \frac{2}{1 + 2^{1/2^{n}}} = \frac{2}{1 + \sqrt{2}}\cdot \frac{2}{1 + \sqrt{\sqrt{2}}}\cdot \frac{2}{1 + \sqrt{\sqrt{\sqrt{2}}}} \cdot \ldots$

(2) $\displaystyle \frac{\sin x}{x} = \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} \ldots = \prod_{n=1}^{\infty} \cos \frac{x}{2^n}$

(3) Viète: $\displaystyle \frac{2}{\pi} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2 + \sqrt{2}}}{2}\cdot \frac{\sqrt{2 + \sqrt{2+\sqrt{2}}}}{2}\cdots$

(4) $\displaystyle x = x^{1/2}x^{1/4}x^{1/8}\ldots = \prod_{n=1}^{\infty} x^{1/2^n}$

(5) $\displaystyle 1-x = \left(\frac{1-x}{1+x}\right)^{1/2}\left(\frac{1-x^2}{1+x^2}\right)^{1/4}\left(\frac{1-x^4}{1+x^4}\right)^{1/8}\ldots$

(6) $\displaystyle\prod_{k=0}^{\infty} \left(1 + x^{2^k}\right) = \frac{1}{1-x}$ for |x| < 1

(7) $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p \text{ prime}} \frac{p^s}{p^s-1}$

(8) $\displaystyle \prod_{p \text{ prime}} \frac{p^2}{p^2 - 1} = \frac{\pi^2}{6}$

(9) $\displaystyle \prod_{p \text{ prime}} \frac{p^2}{p^2 +1} = \frac{\pi^2}{15}$

(10) $\displaystyle \prod_{p \text{ prime}} \frac{p^2+1}{p^2 -1} = \frac{5}{2}$

(11) $\displaystyle \frac{\pi}{4} = \frac{3}{4} \cdot \frac{5}{4} \cdot \frac{7}{8} \cdot \frac{11}{12} \cdot \frac{13}{12} \cdot \frac{17}{16} \cdot \frac{19}{20} \cdot \frac{23}{24} \cdot \ldots =\prod_{p \text{ odd prime}} \frac{p}{p - (-1)^{(p-1)/2}}$
(in each fraction the numerator is an odd prime, the denominator is the nearest multiple of 4 to the numerator)

(12) $\displaystyle \frac{\pi}{2} = \frac{3}{2} \cdot \frac{5}{6} \cdot \frac{7}{6} \cdot \frac{11}{10} \cdot \frac{13}{14} \cdot \frac{17}{18} \cdot \frac{19}{18} \cdot \frac{23}{22} \cdot \ldots = \prod_{p \text{ odd prime}} \frac{p}{p + (-1)^{(p-1)/2}}$
(in each fraction the numerator is an odd prime, the denominator is the nearest even non-multiple of 4 to the numerator)

(13) $\displaystyle 2 = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{9} \cdot \frac{10}{9} \cdot \frac{12}{11} \cdot \ldots$
(in each fraction the numerator and denominator differ by 1, sum to the odd primes, and numerator is even)

(14) $\displaystyle \frac{\pi}{4} = \frac{4}{5} \cdot \frac{8}{7} \cdot \frac{10}{11} \cdot \frac{12}{13} \cdot \frac{14}{13} \cdot \frac{16}{17} \cdot \frac{17}{18} \cdot \frac{20}{19} \cdot \ldots$
(in each fraction the numerator and denominator differ by 1, sum to odd numbers that are not prime, and numerator is even)

(15) $\displaystyle e = \frac{2}{1} \cdot \frac{4}{3} \cdot \frac{15}{14} \cdot \frac{64}{63} \cdot \ldots = \prod_{n=1}^{\infty} \frac{e_n+1}{e_n}$,

where $e_1 = 1, e_{n+1} = (n+1)(e_n + 1)$

(16) Pippenger: $\displaystyle \frac{e}{2} = \left(\frac{2}{1}\right)^{1/2}\left(\frac{2}{3} \cdot \frac{4}{3}\right)^{1/4}\left(\frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7}\right)^{1/8} \ldots$

(17) Catalan: $\displaystyle e = \frac{2}{1} \left( \frac{4}{3}\right)^{1/2} \left( \frac{6}{5} \cdot \frac{8}{7} \right)^{1/4} \left( \frac{10}{9} \cdot \frac{12}{11} \cdot \frac{14}{13} \cdot \frac{16}{15}\right)^{1/8}\ldots$

(18) Pentagonal number theorem: $\displaystyle \prod_{n=1}^{\infty} (1-x^n) = \sum_{k=1}^{\infty} (-1)^k x^{k(3k-1)/2}$

#### Proof Outlines

A. Products based on geometric series

The following proof of (1) is based on . We start with repeated application of the difference of perfect squares identity: $\begin{array}{lcl} x-y &=& (x^{1/2} - y^{1/2})(x^{1/2} + y^{1/2})\\&=& (x^{1/4} - y^{1/4})(x^{1/4} + y^{1/4})(x^{1/2} + y^{1/2})\\&=& (x^{1/8} - y^{1/8})(x^{1/8} + y^{1/8})(x^{1/4} + y^{1/4})(x^{1/2} + y^{1/2})\\&=&\ldots\\&=& (x^{1/2^N} - y^{1/2^N})\prod_{n=1}^N (x^{1/2^n} + y^{1/2^n})\\&=& 2^N (x^{1/2^N} - y^{1/2^N}) \prod_{n=1}^N \left(\frac{x^{1/2^n} + y^{1/2^n}}{2}\right) \quad \quad (*) \end{array}$

In this expression let y=1. After rearrangement this gives $\displaystyle 2^N \left( x^{1/2^N} - 1 \right) = (x-1) \prod_{n=1}^N \left( \frac{2}{x^{1/2^n}+ 1} \right).$

Next we take the limit of both sides as $N\rightarrow \infty$. The left side becomes $\lim_{\epsilon \rightarrow 0+} \frac{x^{\epsilon}-1}{\epsilon} = \log x$ and so we have $\displaystyle \prod_{n=1}^{\infty} \frac{2}{1 + x^{1/2^{k}}} = \frac{\log x}{x-1} = \frac{2}{1 + \sqrt{x}}\cdot \frac{2}{1 + \sqrt{\sqrt{x}}}\cdot \frac{2}{1 + \sqrt{\sqrt{\sqrt{x}}}} \cdot \ldots$

As noted in  one can also set $x = e^{i\theta}, y = e^{-i\theta}$ in (*) and arrive at $\displaystyle \prod_{n=1}^{N} \cos \left(\frac{x}{2^n}\right) = \frac{\sin x}{2^N \sin \frac{x}{2^n}}$ so letting $N\rightarrow \infty$ gives us (2). It need not be arrived at via the double angle formula!

Setting $x = \pi/2$ in (2) automatically gives (3), which also can be derived by considering regular (2^n)-gons being approximated by a circle in the limit.

Next (4) is based on the infinite series $1 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots$ while (5) is apparent from the similar equality $\displaystyle a_0 = \left(\frac{a_0}{a_1}\right)^{1/2}\left(\frac{a_0 a_1}{a_2}\right)^{1/4}\left(\frac{a_0 a_1 a_2}{a_3}\right)^{1/8}\ldots$

with $\displaystyle a_0 = 1-x, a_1 = 1+x, a_2 = 1+x^2, a_3 = 1+x^4, \ldots$ .

Finally, (6) is based on the fact that every positive integer has a unique representation in binary, so every term $x^n$ is represented in exactly one product on the left side (e.g. $x^{13} = x^8.x^4.x^1$), and so $\displaystyle\prod_{k=0}^{\infty} \left(1 + x^{2^k}\right) = 1 + x + x^2 + x^3 + x^4 + \ldots,$

which is summed as a geometric series for |x| < 1.

B. Euler products

The next products over the primes [(7)-(14)] are due to Euler and come from his expansion of the Riemann zeta function in (7): $\displaystyle \zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p \text{ prime}} \left(\sum_{k=0}^{\infty} \frac{1}{p^{ks}}\right) = \prod_{p \text{ prime}} \frac{p^s}{p^s-1}$

This is a essentially a statement of unique factorisation, that every positive integer n can be written uniquely as a product of primes.

Setting s=2 gives (8). Setting s=4 gives $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^4} = \prod_{p \text{prime}} \frac{p^4}{p^4-1} = \prod_{p \text{prime}} \frac{p^2}{p^2-1} \cdot \frac{p^2}{p^2+1}. \quad \quad (**)$

The left side is equal to $\frac{\pi^4}{90}$.

One way of seeing this is by taking the $x^5$ term in the infinite product expansion $\sin \pi x = \pi x \prod_{n=1}^{\infty} (1-\frac{x^2}{n^2})$ (seen in my previous post). This gives $\displaystyle \frac{(\pi x)^5}{5!} = \pi x \sum_{m>n} \frac{1}{m^2n^2}.$

Hence $\begin{array}{lcl} \sum_{n=1}^{\infty} \frac{1}{n^4} &=& \left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right)^2 -2 \sum_{m\neq n}^{\infty} \frac{1}{m^2n^2}\\&=& \left( \frac{\pi^2}{6} \right)^2 - 2\frac{\pi^4}{5!}\\ &=& \pi^4 \left( \frac{1}{36} - \frac{2}{120} \right) \\ &=& \frac{\pi^4}{90}. \end{array}$

Finally one obtains (9) by dividing (**) by (8). (10) simply arises from dividing (8) by (9).

To prove (11) we proceed in a similar manner to proving (7), this time using unique factorisation of the odd natural numbers, and that an odd number that is 3 modulo 4 must have all prime factors that are 3 modulo 4 occurring an odd number of times: $\begin{array}{lcl} \sum_{n=0}^{\infty} \frac{(-1)^ n}{(2n+1)^s} &=& \left(1 - \frac{1}{3^s} + \frac{1}{3^{2s}} - \ldots\right)\left(1 + \frac{1}{5^s} + \frac{1}{5^{2s}} + \ldots\right)\left(1 - \frac{1}{7^s} + \frac{1}{7^{2s}} - \ldots\right) \ldots \\ &=&\prod_{p \text{ odd prime}} \sum_{k=0}^{\infty} \left(\frac{(-1)^{(p-1)/2}}{p^s}\right)^{k}\\ &=& \prod_{p \text{ odd prime}} \frac{p^s}{p^s - (-1)^{(p-1)/2}}\end{array}$

To be more precise, if m is the largest prime less than N, one can show  that the difference $\displaystyle \left| \sum_{n=0}^{\infty} \frac{(-1)^ n}{(2n+1)^s} - \prod_{p \text{ odd prime}}^m \frac{p^s}{p^s - (-1)^{(p-1)/2}}\right| \leq \sum_{n=N}^{\infty} \left|\frac{(-1)^ n}{(2n+1)^{\text{Re }s}} \right|\rightarrow 0$

as $n \rightarrow \infty$.

Setting s to 1 and using the sum $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^ n}{2n+1} = \frac{\pi}{4}$ gives (11). We obtain (12) by dividing 3/4 times (8) by (11).

Dividing 3/4 times (8) by the square of (11) leads to (13) after some cancellation.

The next formula uses result (9) (the Wallis product) of my previous post: $\displaystyle \frac{\pi}{2} = \prod_{n=1}^{\infty} \left(1- \frac{1}{(2n+1)^2}\right) = \frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdot \frac{6}{5}\cdot \frac{6}{7}\cdot\frac{8}{7}\cdot \ldots \quad \quad (***)$

Dividing this by (13) leads to (14).

C. Formulas involving e

We obtain (14) by using the following infinite sum form of e: $\displaystyle e = \sum_{n=1}^{\infty} \frac{1}{n!}$

We can see by the definition of $e_n$ that the product $\prod_{n=1}^N \frac{e_n + 1}{e_n} = \frac{e_{N+1}}{(N+1)!}$. Then it can be seen that $e_N$ is the numerator of the partial sum $\sum_{n=0}^{N} \frac{1}{n!}$. The result follows by taking the limit $N \rightarrow \infty$.

To obtain (15) we use Stirling’s formula for the asymptotic form of the factorial function: $\displaystyle n! \sim \left(\frac{n}{e}\right)^n \sqrt{2 \pi n} \Rightarrow e = \lim_{n \rightarrow \infty} \frac{n}{(n!)^{1/n}}$

Next we observe that each term in the right side can be written in terms of factorials and powers of 2. For example, $\begin{array}{lcl} \frac{8}{9} \cdot \frac{10}{9} \cdot \frac{10}{11} \cdot \ldots \cdot \frac{14}{15} \cdot \frac{16}{15} &=&\frac{2^8.(4.5.5.6.6.7.8)}{(9.11.13.15)^2}\\ &=& 2^8\frac{(8!)^2}{(4!)^2}\frac{4}{8} \frac{(8!)^2.(10.12.14.16)^2}{(16!)^2}\\ &=& 2^{15}\frac{(8!)^2}{(4!)^2}\frac{(8!)^2 (8!)^2}{(16!)^2(4!)^2}\\ &=& \frac{2^{15}.(8!)^6}{(4!)^4.(16!)^2} \end{array}$

The general term is of the form $\displaystyle \left[\frac{2^{2^n-1} (2^{n-1}!)^6}{(2^{n-2}!)^4(2^n!)^2} \right]^{\frac{1}{2^n}}$.

When these terms are multiplied, the product telescopes and we end up with $\begin{array}{lcl} \left(\frac{2}{1}\right)^{1/2}\left(\frac{2}{3} \cdot \frac{4}{3}\right)^{1/4}\left(\frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7}\right)^{1/8} \ldots &=& 2\sqrt{2} \lim_{N \rightarrow \infty} \prod_{n=2}^N \left[\frac{2^{2^n-1} (2^{n-1}!)^6}{(2^{n-2}!)^4(2^n!)^2} \right]^{\frac{1}{2^n}}\\ &=& 2\sqrt{2} \lim_{N \rightarrow \infty} 2^{N-1} \frac{1}{2^{1/4 + 1/8 + \ldots + 1/2^N}}.2.\frac{\left[ 2^{N-1}! \right]^{6/2^N - 2/2^{N-1}}}{\left[ 2^N !\right]^{2/2^N}}\\&=& 2\sqrt{2} \lim_{N \rightarrow \infty} 2^N.2^{-\left(1/2 - 1/2^N \right)}. \left[\frac{2^{N-1}!}{2^N!} \right]^{\frac{1}{2^{N-1}}}\\ &=& 2 \lim_{N \rightarrow \infty} 2^{1/2^N}.\left(\frac{2^N}{\left(2^N! \right)^{1/2^N}}\right)^2.\left(\frac{\left(2^{N-1}! \right)^{1/2^{N-1}}}{2^{N-1}}\right).\frac{1}{2}\\&=& 2.1.e^2.\frac{1}{e}.\frac{1}{2}\\&=& e\end{array}$

We remark that the product converges rapidly to e, since even if we use the more precise form $\displaystyle e \sim \left(\frac{n^{n+1/2}}{n!}\sqrt{2 \pi} \right)^{1/n}$ and replace $\pi$ with the Wallis product expansion, the left and right sides of (15) can be shown to be similar for N large.

It is amazing to see that (15) becomes Wallis’s product (***) for $\displaystyle \frac{\pi}{2}$ when the exponents are set to 1 (while keeping the same fractions)!

(16) can be easily obtained from (15) by multiplying both sides by $\displaystyle 2 = 2^{1/2}2^{1/4}2^{1/8}\ldots$

(although Catalan’s result, obtained through more involved means, long preceded that of Pippenger).

More similar products are given in .

D. Pentagonal number theorem

This result, again due to Euler, shows how much cancellation there is when the product $(1-x)(1-x^2)(1-x^3)\ldots$ is expanded. The coefficient of $x^n$ is the number of partitions of n into an even number of parts minus the number of partitions of n into an odd number of parts. The theorem says that this number is zero unless n is a pentagonal number (i.e. of the form k(3k-1)/2), in which case it is 1 or -1. Refer to this Wikipedia entry of this theorem for a combinatorial proof.

#### References

 T.J. Osler, Interesting finite and infinite products from simple algebraic identities, The Mathematical Gazette, 90(2006), pp. 90-93. Available here.

 K. Brown, Infinite Products and a Tangent Fan. Mathpages link

 P. Loya, Amazing and Aesthetic Aspects of Analysis: On the incredible infinite, available at http://www.math.binghamton.edu/dennis/478.f07/EleAna.pdf

 J. Sondow and H. Yi, New Wallis- and Catalan-Type Infinite Products for π , e, and sqrt(2+ sqrt(2)), available at http://arxiv.org/ftp/arxiv/papers/1005/1005.2712.pdf

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