# 월리스 곱 (Wallis product formula)

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## 개요

• 1655년, 영국 수학자 월리스(John Wallis)는 월리스 곱이라 불려지는 다음과 같은 공식을 남긴다

$\lim_{n \rightarrow \infty}\big(\frac{2}{1}\cdot \frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdots \frac{2n}{2n - 1} \cdot\frac{2n}{2n+1}\big) = \frac{\pi}{2}$ $\prod_{k=1}^{\infty}\frac{4k^2-1}{4k^2}=\frac{2}{\pi}$

$\frac{1}{2^{2n}}{{(2n)!}\over{n!n!}}=\frac{1}{2^{2n}}{2n\choose n}\approx\frac{1}{\sqrt{\pi n}}$

## 증명

• 다음과 같이 수열 $$\{a_n\}$$을 정의하자 $a_n:=\int_0^{\pi}\sin^{n}\theta{d\theta}= B(\frac{n+1}{2},\frac{1}{2})=\frac{\sqrt{\pi}\Gamma(\frac{n}{2}+\frac{1}{2})}{\Gamma(\frac{n}{2}+1)}$ 여기서 $$B(x,y)$$는 오일러 베타적분.
• 수열 $$\{a_n\}$$은 다음 점화식을 만족시킨다 $a_0=\pi,a_1=2,$ $a_{n}=\frac{n-1}{n}a_{n-2} \label{rec}$
보조정리1

$\frac{\frac{a_{2n}}{a_{2n+1}}}{\pi /2}=\prod _{k=1}^n \frac{4 k^2-1}{4 k^2}\label{prod}$

증명

\ref{rec}로부터 다음을 얻는다 $\frac{a_{2k}}{a_{2k-2}}\frac{a_{2k-1}}{a_{2k+1}}=\frac{4 k^2-1}{4 k^2}$ 으로부터 $\prod _{k=1}^n \frac{a_{2k}}{a_{2k-2}}\frac{a_{2k-1}}{a_{2k+1}}=\prod _{k=1}^n \frac{4 k^2-1}{4 k^2}$을 얻는다. 한편, $\prod _{k=1}^n \frac{a_{2k}}{a_{2k-2}}\frac{a_{2k-1}}{a_{2k+1}}=\frac{a_{1}a_{2n}}{a_{0} a_{2n+1}}=\frac{\frac{a_{2n}}{a_{2n+1}}}{\pi /2}$ 로부터 \ref{prod}을 얻는다. ■

보조정리2

$\lim_{n\to \infty } \, \frac{a_{2 n}}{a_{2 n+1}}=1 \label{lim}$

증명

$$a_{n}$$은 단조감소수열이므로, 다음 부등식이 성립한다 $1 \le \frac{a_{2n}}{a_{2n+1}} \le \frac{a_{2n-1}}{a_{2n+1}}=\frac{2n+1}{2n}$ 우변에서는 \ref{rec}이 사용되었다. 따라서 샌드위치 정리에 의해 $\lim_{n\to \infty } \, \frac{a_{2 n}}{a_{2 n+1}}=1$ ■

보조정리1과 보조정리2로부터 월리스 곱을 얻는다 ■

## 사인함수의 무한곱 표현을 이용한 증명

• 다음 사인함수의 무한곱 표현에서 $$x=1/2$$ 일 때, 월리스 곱을 얻는다

$\sin{\pi x} = \pi x \prod _{n=1}^{\infty } \left(1-\frac{x^2}{n^2}\right)\label{sinpro}$

## 역사

• 드무아브르의 발견은 대략 1730년대 즈음
• 데카르트(1596년 3월-1650년 2월)
• 뉴턴(1643년 1월-1727년 3월)

## 관련논문

• Friedmann, Tamar, and C. R. Hagen. “Quantum Mechanical Derivation of the Wallis Formula for $$\pi$$.” arXiv:1510.07813 [math-Ph, Physics:quant-Ph], October 27, 2015. http://arxiv.org/abs/1510.07813.

## 노트

### 말뭉치

1. Comparison of the convergence of the Wallis product (purple asterisks) and several historical infinite series foris the approximation after takingterms.[1]
2. The product, as n goes to infinity, is known as the Wallis product, and it is amazingly equal to π/2 ≈ 1.571.[2]
3. There are some interesting details in the historical calculation of what is now called the Wallis product.[2]
4. The results involved π/4 as well as the fractions involved in the Wallis product, and Wallis could re-write the expressions to find π in terms of a fractional product.[2]
5. We can derive the Wallis product formula from these integrals.[2]
6. I know how the value for the Wallis product was found, but none of these techniques seem to work on this seemingly related product.[3]
7. Do you know or can you provide simple proofs of the above equation which do not use the Wallis product?[4]
8. I would be very curious to see if anyone has written on seeing the Wallis product this way in terms of sphere volumes before.[5]
9. Last week my cousin told me Wallis' product was quite his favorite formula in mathematics.[6]
10. I remembered the thread opened on this forum about PI approximations (like 355 / 113), and we decided to program Wallis product on Free42 while having a coffee...[6]
11. In the previous post I mentioned about how to demonstrate the Wallis product for pi by starting from the powered sine integration.[7]
12. This time, we’re gonna see how to derive the Wallis product with Euler’s infinite product representation for sine function.[7]
13. As a refresher, here’s the Wallis product for pi that we are trying to prove.[7]

## 메타데이터

### Spacy 패턴 목록

• [{'LOWER': 'wallis'}, {'LEMMA': 'product'}]
• [{'LOWER': 'wallis'}, {'LOWER': "'"}, {'LEMMA': 'product'}]