"월리스 곱 (Wallis product formula)"의 두 판 사이의 차이

수학노트
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1번째 줄: 1번째 줄:
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<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">이 항목의 스프링노트 원문주소</h5>
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==개요==
  
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* [[월리스 곱 (Wallis product formula)]]
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* 1655년, 영국 수학자 월리스([http://en.wikipedia.org/wiki/John_Wallis John Wallis])는 월리스 곱이라 불려지는 다음과 같은 공식을 남긴다
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:<math>\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}</math>
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:<math>\prod_{k=1}^{\infty}\frac{4k^2-1}{4k^2}=\frac{2}{\pi}</math>
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* [http://bomber0.byus.net/index.php/2008/07/12/686 스털링이 드무아브르가 남긴 문제를 해결]할때 이 월리스의 공식을 사용 :<math>\frac{\pi}{2}=\lim_{n\to\infty}{1\over{2n}}\cdot{{2^{4n}\,(n!)^4}\over{((2n)!)^2}}</math>
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* 이는 다음을 말해준다
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:<math>\frac{1}{2^{2n}}{{(2n)!}\over{n!n!}}=\frac{1}{2^{2n}}{2n\choose n}\approx\frac{1}{\sqrt{\pi n}}</math>
  
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==증명==
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* 다음과 같이 수열 <math>\{a_n\}</math>을 정의하자 :<math>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)}</math> 여기서 <math>B(x,y)</math>는 [[오일러 베타적분(베타함수)|오일러 베타적분]].
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* 수열 <math>\{a_n\}</math>은 다음 점화식을 만족시킨다 :<math>a_0=\pi,a_1=2,</math> :<math>a_{n}=\frac{n-1}{n}a_{n-2} \label{rec}</math>
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;보조정리1
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:<math>\frac{\frac{a_{2n}}{a_{2n+1}}}{\pi /2}=\prod _{k=1}^n \frac{4 k^2-1}{4 k^2}\label{prod}</math>
  
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<h5>개요</h5>
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;증명
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\ref{rec}로부터 다음을 얻는다
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:<math>\frac{a_{2k}}{a_{2k-2}}\frac{a_{2k-1}}{a_{2k+1}}=\frac{4 k^2-1}{4 k^2}</math> 으로부터
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:<math>\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}</math>을 얻는다.
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한편,
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:<math>\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}</math>
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로부터 \ref{prod}을 얻는다. ■
  
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* 1655년, 영국 수학자 월리스([http://en.wikipedia.org/wiki/John_Wallis John Wallis])는 월리스 곱이라 불려지는 다음과 같은 공식을 남긴다.
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;보조정리2
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:<math>\lim_{n\to \infty } \, \frac{a_{2 n}}{a_{2 n+1}}=1 \label{lim}</math>
  
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<math>\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}</math>
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;증명
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<math>a_{n}</math>은 단조감소수열이므로, 다음 부등식이 성립한다
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:<math>1 \le \frac{a_{2n}}{a_{2n+1}} \le \frac{a_{2n-1}}{a_{2n+1}}=\frac{2n+1}{2n}</math>
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우변에서는 \ref{rec}이 사용되었다.
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따라서 [[샌드위치 정리]]에 의해 :<math>\lim_{n\to \infty } \, \frac{a_{2 n}}{a_{2 n+1}}=1</math> ■
  
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* [http://bomber0.byus.net/index.php/2008/07/12/686 스털링이 드무아브르가 남긴 문제를 해결]할때 이 월리스의 공식을 사용
 
  
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<math>\frac{\pi}{2}=\lim_{n\to\infty}{1\over{2n}}\cdot{{2^{4n}\,(n!)^4}\over{((2n)!)^2}}</math>
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보조정리1과 보조정리2로부터 월리스 곱을 얻는다  ■
  
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* 이는 다음을 말해준다.
 
  
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<math>\frac{1}{2^{2n}}{{(2n)!}\over{n!n!}}=\frac{1}{2^{2n}}{2n\choose n}\approx\frac{1}{\sqrt{\pi n}}</math>
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==사인함수의 무한곱 표현을 이용한 증명==
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* 다음 사인함수의 무한곱 표현에서 <math>x=1/2</math> 일 때, 월리스 곱을 얻는다
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:<math>\sin{\pi x} = \pi x \prod _{n=1}^{\infty } \left(1-\frac{x^2}{n^2}\right)\label{sinpro}</math>
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* [[삼각함수의 무한곱 표현]] 항목 참조
  
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==역사==
  
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* [[수학사 연표]]
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<h5>월리스의 증명</h5>
 
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* [[오일러 베타적분(베타함수)|오일러 베타적분]]<br><math>\int_0^{\pi}\sin^{p}\theta{d\theta}= B(\frac{p+1}{2},\frac{1}{2})=\frac{\sqrt{\pi}\Gamma(\frac{p}{2}+\frac{1}{2})}{\Gamma(\frac{p}{2}+1)}</math><br>
 
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<h5>역사</h5>
 
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* [[수학사연표 (역사)|수학사연표]]
 
 
* 1655 - 존 월리스가 <em style="">Arithmetica Infinitorum</em>를 저술
 
* 1655 - 존 월리스가 <em style="">Arithmetica Infinitorum</em>를 저술
  
46번째 줄: 52번째 줄:
 
* 뉴턴(1643년 1월-1727년 3월)
 
* 뉴턴(1643년 1월-1727년 3월)
  
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<h5>메모</h5>
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==메모==
  
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<h5>관련된 항목들</h5>
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==관련된 항목들==
  
 
* [[스털링 공식]]
 
* [[스털링 공식]]
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* [[search?q=%EC%A4%91%EC%8B%AC%EA%B7%B9%ED%95%9C%EC%A0%95%EB%A6%AC&parent id=2681644|중심극한정리]]
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* [[드무아브르-라플라스 중심극한정리]]
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* [[오일러 베타적분(베타함수)|오일러 베타적분]]
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* [[오일러 베타적분(베타함수)]]
 
* [[삼각함수의 적분]]
 
* [[삼각함수의 적분]]
  
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==매스매티카 파일 및 계산 리소스==
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* https://docs.google.com/file/d/0B8XXo8Tve1cxQ1pCSTA2YzVhWG8/edit
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<h5>사전형태의 자료</h5>
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==사전형태의 자료==
  
 
* http://en.wikipedia.org/wiki/Wallis_product
 
* http://en.wikipedia.org/wiki/Wallis_product
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* [http://en.wikipedia.org/wiki/Wallis_product ]http://www22.wolframalpha.com/input/?i=wallis+product
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* http://www22.wolframalpha.com/input/?i=wallis+product
  
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<h5>블로그</h5>
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==블로그==
  
 
* [http://bomber0.byus.net/index.php/2008/07/12/687 드무아브르의 중심극한정리(iii) : 숫자 파이와 동전던지기] 피타고라스의 창, 2008-7-12
 
* [http://bomber0.byus.net/index.php/2008/07/12/687 드무아브르의 중심극한정리(iii) : 숫자 파이와 동전던지기] 피타고라스의 창, 2008-7-12
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==관련논문==
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* Friedmann, Tamar, and C. R. Hagen. “Quantum Mechanical Derivation of the Wallis Formula for <math>\pi</math>.” arXiv:1510.07813 [math-Ph, Physics:quant-Ph], October 27, 2015. http://arxiv.org/abs/1510.07813.
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[[분류:원주율]]
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[[분류:미적분학]]
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== 노트 ==
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===말뭉치===
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# Comparison of the convergence of the Wallis product (purple asterisks) and several historical infinite series foris the approximation after takingterms.<ref name="ref_277b1850">[https://en.wikipedia.org/wiki/Wallis_product Wallis product]</ref>
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# The product, as n goes to infinity, is known as the Wallis product, and it is amazingly equal to π/2 ≈ 1.571.<ref name="ref_22afbcdd">[https://mindyourdecisions.com/blog/2016/10/12/the-wallis-product-formula-for-pi-and-its-proof/ The Wallis Product Formula For Pi And Its Proof – Mind Your Decisions]</ref>
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# There are some interesting details in the historical calculation of what is now called the Wallis product.<ref name="ref_22afbcdd" />
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# 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.<ref name="ref_22afbcdd" />
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# We can derive the Wallis product formula from these integrals.<ref name="ref_22afbcdd" />
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# I know how the value for the Wallis product was found, but none of these techniques seem to work on this seemingly related product.<ref name="ref_0ee8acf8">[https://math.stackexchange.com/questions/698176/infinite-product-related-to-the-wallis-product Infinite product related to the Wallis product]</ref>
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# Do you know or can you provide simple proofs of the above equation which do not use the Wallis product?<ref name="ref_44bab671">[https://mathoverflow.net/questions/257440/accelerated-wallis-product Accelerated Wallis' product]</ref>
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# I would be very curious to see if anyone has written on seeing the Wallis product this way in terms of sphere volumes before.<ref name="ref_485b18fb">[https://www.3blue1brown.com/sridhars-corner/2018/5/12/another-simple-geometric-proof-of-the-wallis-product Another Simple Geometric Proof of the Wallis Product — 3Blue1Brown]</ref>
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# Last week my cousin told me Wallis' product was quite his favorite formula in mathematics.<ref name="ref_1097d418">[https://www.hpmuseum.org/forum/thread-14476.html Wallis' product exploration]</ref>
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# 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...<ref name="ref_1097d418" />
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# In the previous post I mentioned about how to demonstrate the Wallis product for pi by starting from the powered sine integration.<ref name="ref_5740b247">[https://albertusk95.github.io/posts/2020/05/wallis-product-proof-with-euler-sine-infinite-product/ Proof of Wallis Product for Pi with Euler’s Infinite Product for Sine]</ref>
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# This time, we’re gonna see how to derive the Wallis product with Euler’s infinite product representation for sine function.<ref name="ref_5740b247" />
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# As a refresher, here’s the Wallis product for pi that we are trying to prove.<ref name="ref_5740b247" />
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===소스===
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<references />
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== 메타데이터 ==
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===위키데이터===
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* ID :  [https://www.wikidata.org/wiki/Q1501324 Q1501324]
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===Spacy 패턴 목록===
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* [{'LOWER': 'wallis'}, {'LEMMA': 'product'}]
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* [{'LOWER': 'wallis'}, {'LOWER': "'"}, {'LEMMA': 'product'}]

2021년 2월 26일 (금) 00:38 기준 최신판

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)



메모

관련된 항목들


매스매티카 파일 및 계산 리소스


사전형태의 자료



블로그


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]

소스

  1. ↑ Wallis product
  2. ↑ 2.0 2.1 2.2 2.3 The Wallis Product Formula For Pi And Its Proof – Mind Your Decisions
  3. ↑ Infinite product related to the Wallis product
  4. ↑ Accelerated Wallis' product
  5. ↑ Another Simple Geometric Proof of the Wallis Product — 3Blue1Brown
  6. ↑ 6.0 6.1 Wallis' product exploration
  7. ↑ 7.0 7.1 7.2 Proof of Wallis Product for Pi with Euler’s Infinite Product for Sine

메타데이터

위키데이터

Spacy 패턴 목록

  • [{'LOWER': 'wallis'}, {'LEMMA': 'product'}]
  • [{'LOWER': 'wallis'}, {'LOWER': "'"}, {'LEMMA': 'product'}]
원본 주소 "https://wiki.mathnt.net/index.php?title=월리스_곱_(Wallis_product_formula)&oldid=52691"