"월리스 곱 (Wallis product formula)"의 두 판 사이의 차이
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Pythagoras0 (토론 | 기여) (→메타데이터) |
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| (사용자 2명의 중간 판 47개는 보이지 않습니다) | |||
| 1번째 줄: | 1번째 줄: | ||
| − | + | ==개요== | |
| − | <math | + | * 1655년, 영국 수학자 월리스([http://en.wikipedia.org/wiki/John_Wallis John Wallis])는 월리스 곱이라 불려지는 다음과 같은 공식을 남긴다 |
| + | :<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> | ||
| + | :<math>\prod_{k=1}^{\infty}\frac{4k^2-1}{4k^2}=\frac{2}{\pi}</math> | ||
| + | * [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> | ||
| + | * 이는 다음을 말해준다 | ||
| + | :<math>\frac{1}{2^{2n}}{{(2n)!}\over{n!n!}}=\frac{1}{2^{2n}}{2n\choose n}\approx\frac{1}{\sqrt{\pi n}}</math> | ||
| − | |||
| − | + | ==증명== | |
| + | * 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]]. | ||
| + | * 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> | ||
| + | ;보조정리1 | ||
| + | :<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> | ||
| − | + | ;증명 | |
| + | \ref{rec}로부터 다음을 얻는다 | ||
| + | :<math>\frac{a_{2k}}{a_{2k-2}}\frac{a_{2k-1}}{a_{2k+1}}=\frac{4 k^2-1}{4 k^2}</math> 으로부터 | ||
| + | :<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. | ||
| + | 한편, | ||
| + | :<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> | ||
| + | 로부터 \ref{prod}을 얻는다. ■ | ||
| − | + | ;보조정리2 | |
| + | :<math>\lim_{n\to \infty } \, \frac{a_{2 n}}{a_{2 n+1}}=1 \label{lim}</math> | ||
| − | + | ;증명 | |
| + | <math>a_{n}</math>은 단조감소수열이므로, 다음 부등식이 성립한다 | ||
| + | :<math>1 \le \frac{a_{2n}}{a_{2n+1}} \le \frac{a_{2n-1}}{a_{2n+1}}=\frac{2n+1}{2n}</math> | ||
| + | 우변에서는 \ref{rec}이 사용되었다. | ||
| + | 따라서 [[샌드위치 정리]]에 의해 :<math>\lim_{n\to \infty } \, \frac{a_{2 n}}{a_{2 n+1}}=1</math> ■ | ||
| − | |||
| − | + | 보조정리1과 보조정리2로부터 월리스 곱을 얻는다 ■ | |
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| − | + | ==사인함수의 무한곱 표현을 이용한 증명== | |
| + | * 다음 사인함수의 무한곱 표현에서 <math>x=1/2</math> 일 때, 월리스 곱을 얻는다 | ||
| + | :<math>\sin{\pi x} = \pi x \prod _{n=1}^{\infty } \left(1-\frac{x^2}{n^2}\right)\label{sinpro}</math> | ||
| + | * [[삼각함수의 무한곱 표현]] 항목 참조 | ||
| − | |||
| − | |||
| − | + | ==역사== | |
| − | < | + | * [[수학사 연표]] |
| + | * 1655 - 존 월리스가 <em style="">Arithmetica Infinitorum</em>를 저술 | ||
| − | + | * 드무아브르의 발견은 대략 1730년대 즈음 | |
| + | * 데카르트(1596년 3월-1650년 2월) | ||
| + | * 뉴턴(1643년 1월-1727년 3월) | ||
| − | + | ||
| − | + | ||
| − | + | ==메모== | |
| − | + | ||
| − | + | ||
| + | |||
| + | ==관련된 항목들== | ||
* [[스털링 공식]] | * [[스털링 공식]] | ||
| − | * [[ | + | * [[드무아브르-라플라스 중심극한정리]] |
| + | * [[오일러 베타적분(베타함수)]] | ||
| + | * [[삼각함수의 적분]] | ||
| + | |||
| + | |||
| + | ==매스매티카 파일 및 계산 리소스== | ||
| + | * https://docs.google.com/file/d/0B8XXo8Tve1cxQ1pCSTA2YzVhWG8/edit | ||
| + | |||
| + | |||
| + | ==사전형태의 자료== | ||
| + | |||
| + | * http://en.wikipedia.org/wiki/Wallis_product | ||
| + | * http://www22.wolframalpha.com/input/?i=wallis+product | ||
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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 | ||
| − | |||
| − | < | + | ==관련논문== |
| + | * 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. | ||
| − | |||
| − | + | [[분류:원주율]] | |
| + | [[분류:미적분학]] | ||
| − | + | == 노트 == | |
| − | + | ===말뭉치=== | |
| + | # 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> | ||
| + | # 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> | ||
| + | # There are some interesting details in the historical calculation of what is now called the Wallis product.<ref name="ref_22afbcdd" /> | ||
| + | # 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" /> | ||
| + | # We can derive the Wallis product formula from these integrals.<ref name="ref_22afbcdd" /> | ||
| + | # 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> | ||
| + | # 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> | ||
| + | # 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> | ||
| + | # 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> | ||
| + | # 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" /> | ||
| + | # 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> | ||
| + | # 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" /> | ||
| + | # As a refresher, here’s the Wallis product for pi that we are trying to prove.<ref name="ref_5740b247" /> | ||
| + | ===소스=== | ||
| + | <references /> | ||
| − | + | == 메타데이터 == | |
| − | * | + | ===위키데이터=== |
| − | * | + | * ID : [https://www.wikidata.org/wiki/Q1501324 Q1501324] |
| + | ===Spacy 패턴 목록=== | ||
| + | * [{'LOWER': 'wallis'}, {'LEMMA': 'product'}] | ||
| + | * [{'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
\[\lim_{n \rightarrow \infty}\big(\frac{2}{1}\cdot \frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdots \frac{2n}{2n - 1} \cdot\frac{2n}{2n+1}\big) = \frac{\pi}{2}\] \[\prod_{k=1}^{\infty}\frac{4k^2-1}{4k^2}=\frac{2}{\pi}\]
- 스털링이 드무아브르가 남긴 문제를 해결할때 이 월리스의 공식을 사용 \[\frac{\pi}{2}=\lim_{n\to\infty}{1\over{2n}}\cdot{{2^{4n}\,(n!)^4}\over{((2n)!)^2}}\]
- 이는 다음을 말해준다
\[\frac{1}{2^{2n}}{{(2n)!}\over{n!n!}}=\frac{1}{2^{2n}}{2n\choose n}\approx\frac{1}{\sqrt{\pi n}}\]
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
\[\sin{\pi x} = \pi x \prod _{n=1}^{\infty } \left(1-\frac{x^2}{n^2}\right)\label{sinpro}\]
- 삼각함수의 무한곱 표현 항목 참조
역사
- 수학사 연표
- 1655 - 존 월리스가 Arithmetica Infinitorum를 저술
- 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
- 드무아브르의 중심극한정리(iii) : 숫자 파이와 동전던지기 피타고라스의 창, 2008-7-12
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
- Comparison of the convergence of the Wallis product (purple asterisks) and several historical infinite series foris the approximation after takingterms.[1]
- The product, as n goes to infinity, is known as the Wallis product, and it is amazingly equal to π/2 ≈ 1.571.[2]
- There are some interesting details in the historical calculation of what is now called the Wallis product.[2]
- 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]
- We can derive the Wallis product formula from these integrals.[2]
- 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]
- Do you know or can you provide simple proofs of the above equation which do not use the Wallis product?[4]
- 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]
- Last week my cousin told me Wallis' product was quite his favorite formula in mathematics.[6]
- 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]
- In the previous post I mentioned about how to demonstrate the Wallis product for pi by starting from the powered sine integration.[7]
- This time, we’re gonna see how to derive the Wallis product with Euler’s infinite product representation for sine function.[7]
- As a refresher, here’s the Wallis product for pi that we are trying to prove.[7]
소스
- ↑ Wallis product
- ↑ 2.0 2.1 2.2 2.3 The Wallis Product Formula For Pi And Its Proof – Mind Your Decisions
- ↑ Infinite product related to the Wallis product
- ↑ Accelerated Wallis' product
- ↑ Another Simple Geometric Proof of the Wallis Product — 3Blue1Brown
- ↑ 6.0 6.1 Wallis' product exploration
- ↑ 7.0 7.1 7.2 Proof of Wallis Product for Pi with Euler’s Infinite Product for Sine
메타데이터
위키데이터
- ID : Q1501324
Spacy 패턴 목록
- [{'LOWER': 'wallis'}, {'LEMMA': 'product'}]
- [{'LOWER': 'wallis'}, {'LOWER': "'"}, {'LEMMA': 'product'}]