중심이항계수 (central binomial coefficient)

이 항목의 스프링노트 원문주소

• 중심이항계수(central binomial coefficient)
  

 

 

개요

• 다음과 같은 이항계수로 정의
  \({2n \choose n}=\frac{(2n)!}{(n!)^2}\)
  
• 아페리가 ζ(3)는 무리수 임을 증명하는데 활용됨
  

 

 

 

중심이항계수의 근사식

• http://planetmath.org/encyclopedia/AsymptoticsOfCentralBinomialCoefficient.html
  

 

 

 

급수와 중심이항계수

• 이항급수와 이항정리
  \(\frac{1}{\sqrt{1-4z}}=\sum_{n=0}^{\infty} {{2n}\choose {n}} z^n\)
  
• 역삼각함수
  \(2(\arcsin x)^2=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}{n}}\)
  \(\frac{2x \arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n\binom{2n}{n}}\)
  
• 카탈란 수열(Catalan numbers) 의 생성함수
  \(G(x)= \frac{1-\sqrt{1-4x}}{2x}=\sum_{n=0}^{\infty}\frac{1}{n+1}{2n\choose n}x^n\)
  

 

 

 

 중심이항계수가 나타나는 급수

• [Lehmer1985] 참조
  

 

\(\sum_{n=1}^{\infty}\frac{1}{n\binom{2n}{n}}=\frac{\pi\sqrt{3}}{9}\)

(증명)

 \(\frac{2x \arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n\binom{2n}{n}}\) 에서 \(x=\frac{1}{2}\)인 경우, \(\sum_{n=1}^{\infty}\frac{1}{n\binom{2n}{n}}=\frac{\pi\sqrt{3}}{9}\) 를 얻는다. ■

 

 

\(\sum_{n=1}^{\infty}\frac{1}{n^{2}\binom{2n}{n}}=\frac{\pi^2}{18}\)

(증명)

 \(2(\arcsin x)^2=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}{n}}\)에서 \(x=\frac{1}{2}\)인 경우, \(\sum_{n=1}^{\infty}\frac{1}{n^{2}\binom{2n}{n}}=\frac{\pi^2}{18}\) 를 얻는다. ■

 

 

\(\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}}=\frac{2\pi}{3}\operatorname{Cl}_2(\frac{\pi}{3})-\frac{4}{3}\zeta(3)=\pi\operatorname{Cl}_2(\frac{2\pi}{3})-\frac{4}{3}\zeta(3)=\frac{\pi\sqrt{3}}{18}(\psi^{(1)}(\frac{1}{3})-\psi^{(1)}(\frac{2}{3}))-\frac{4}{3}\zeta(3)\)

여기서 \(\operatorname{Cl}_2(\theta)\) 는 로바체프스키와 클라우센 함수, \(\psi^{(1)}\)는 트리감마 함수(trigamma function).

(증명)

http://www.research.att.com/~njas/sequences/A145438

\(\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}}=4\int_{0}^{\frac{1}{2}}(\arcsin x)^2}\frac{dx}{x}=-2\int_{0}^{\pi/3}x\log(2\sin \frac{x}{2})\,dx\)

http://www.wolframalpha.com/input/?i=integrate+(arcsin+x)^2/x+dx+from+x%3D0+to+1/2

 

좌변 http://www.wolframalpha.com/input/?i=sum+1%2F%28m%5E3*binom%282m%2Cm%29%29+from+1+to+infinity

우변 http://www.wolframalpha.com/input/?i=-4*zeta(3)/3%2Bpi*sqrt(3)*(trigamma(1/3)-trigamma(2/3))/18

■

 

 

(Comtet의 공식)

\(\sum_{n=1}^\infty \frac{1}{n^4\binom{2n}{n}}=\frac{17\pi^4}{3240}\)

 

(증명)

\(2(\arcsin x)^2=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}{n}}\) 의 양변을 \(2x\)로 나눈뒤, 다음과 같은 적분을 구하자.

\(\int_{0}^{\frac{1}{2}}\int_{0}^{u}\frac{(\arcsin x)^2}{x}\,dx\,\frac{du}{u}=\sum_{n=1}^{\infty}\int_{0}^{\frac{1}{2}}\int_{0}^{u}\frac{(2x)^{2n-1}}{n^2\binom{2n}{n}}\,dx\,\frac{du}{u}=\sum_{n=1}^{\infty}\int_{0}^{\frac{1}{2}}\frac{(2u)^{2n}}{4n^3\binom{2n}{n}}\,\frac{du}{u}\)

우변으로부터 \(\sum_{n=1}^{\infty}\frac{1}{8n^4\binom{2n}{n}}\)을 얻는다.

 

한편

\(\int_{0}^{\frac{1}{2}}\int_{0}^{u}\frac{(\arcsin x)^2}{x}\,dx\,\frac{du}{u}=\int_{0}^{\frac{1}{2}}\int_{x}^{\frac{1}{2}}\frac{(\arcsin x)^2}{xu}\,du\,dx=\int_{0}^{\frac{1}{2}}\log 2x\frac{(\arcsin x)^2}{x}\,dx\) 이므로,

 \(x=\sin\frac{t}{2}\)로 치환하면,

\(\int_{0}^{\frac{1}{2}}\log 2x\frac{(\arcsin x)^2}{x}\,dx=\frac{1}{4}\int_{0}^{\pi/3}x\log^2(2\sin \frac{x}{2})\,dx\)  를 얻는다.

따라서,

\(\frac{1}{8}\sum_{n=1}^{\infty}\frac{1}{n^4\binom{2n}{n}}=\frac{1}{4}\int_{0}^{\pi/3}x\log^2(2\sin \frac{x}{2})\,dx\) 이다.

이제 로그 사인 적분 (log sine integrals) 에서 얻은 다음 결과를 사용하자.

\(\int_{0}^{\pi/3}x\log^2(2\sin \frac{x}{2})\,dx=\frac{17\pi^4}{6480}\)

 

 \(\sum_{n=1}^\infty \frac{1}{n^4\binom{2n}{n}}=\frac{17\pi^4}{3240}\) 를 얻는다. ■

 

원주율의 유리수 근사와 중심이항계수

 \(\sum_{n=1}^{\infty}\frac{2^{n}}{\binom{2n}{n}}=\frac{\pi}{2}+1\)

\(\sum_{n=1}^{\infty}\frac{n2^{n}}{\binom{2n}{n}}=\pi+3\)

\(\sum_{n=1}^{\infty}\frac{n^2 2^{n}}{\binom{2n}{n}}=\frac{7\pi}{2}+11\)

\(\sum_{n=1}^{\infty}\frac{n^3 2^{n}}{\binom{2n}{n}}=\frac{35\pi}{2}+55\)

\(\sum_{n=1}^{\infty}\frac{n^4 2^{n}}{\binom{2n}{n}}=113\pi+355\)

\(\sum_{n=1}^{\infty}\frac{n^{5} 2^{n}}{\binom{2n}{n}} = \frac{1787\pi}{2}+2807\)

\(\sum_{n=1}^{\infty}\frac{n^{6} 2^{n}}{\binom{2n}{n}} = \frac{16717\pi}{2}+26259\)

\(\sum_{n=1}^{\infty}\frac{n^{10} 2^{n}}{\binom{2n}{n}}=229093376\pi+719718067\)

 http://www.wolframalpha.com/input/?i=sum+1%2F%28m%5E3*binom%282m%2Cm%29%29+from+1+to+infinity

http://www.wolframalpha.com/input/?i=sum+m^6*2^m/(binom(2m,m))+from+1+to+infinity

일반적으로 \(k\in\mathbb{N}\)에 대하여,

\(\sum_{n=1}^{\infty}\frac{n^{k} 2^{n}}{\binom{2n}{n}}=a\pi+b\) , (a와 b는 유리수) 형태로 주어진다. [Lehmer1985] 참조

 

 

 

리만제타함수

\(\zeta(2)=3\sum_{n=1}^{\infty}\frac{1}{n^{2}\binom{2n}{n}}\)

\(\zeta(3) = \frac{5}{2} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3\binom{2n}{n}}\)

\(\zeta(4) = \frac{36}{17} \sum_{n=1}^\infty \frac{1}{n^4\binom{2n}{n}}\)

 

 

재미있는 사실

 

• Math Overflow http://mathoverflow.net/search?q=
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역사

 

• http://www.google.com/search?hl=en&tbs=tl:1&q=
• 수학사연표
•  

 

 

메모

[Lehmer1985]

에는 다음과 같은 공식이 나오지만, 잘못된 것이다.

\(\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}}=-\frac{\zeta(3)}{3}-\frac{\pi\sqrt{3}}{72}(\psi^{(1)}(\frac{1}{3})-\psi^{(1)}(\frac{2}{3}))\)

바른 공식은 다음과 같다.

\(\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}}=\frac{\pi\sqrt{3}}{18}(\psi^{(1)}(\frac{1}{3})-\psi^{(1)}(\frac{2}{3}))-\frac{4}{3}\zeta(3)\)

여기서 \(\psi^{(1)}\)는 트리감마(trigamma)함수. 트리감마 함수(trigamma function)항목 참조

 

 

관련된 항목들

• 카탈란 수열(Catalan numbers)
  
• ζ(3)는 무리수이다(아페리의 정리)
  
• 폴리로그 함수(polylogarithm)
  
• 이항급수와 이항정리
  

 

 

수학용어번역

• 단어사전 http://www.google.com/dictionary?langpair=en%7Cko&q=
• 발음사전 http://www.forvo.com/search/
• 대한수학회 수학 학술 용어집
  
  • http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=
• 남·북한수학용어비교
• 대한수학회 수학용어한글화 게시판

 

 

사전 형태의 자료

• http://ko.wikipedia.org/wiki/
• http://en.wikipedia.org/wiki/Central_binomial_coefficient
• http://math world.wolfram.com/CentralBinomialCoefficient.html
• http://mathworld.wolfram.com/BinomialSums.html
• http://www.wolframalpha.com/input/?i=
• NIST Digital Library of Mathematical Functions
• The On-Line Encyclopedia of Integer Sequences
  
  • http://www.research.att.com/~njas/sequences/?q=

 

 

관련논문

• Experimental Determination of Apéry-like Identities for ζ(2n + 2)
  
  • David H. Bailey, Jonathan M. Borwein, and David M. Bradley
• Evaluations of binomial series
  
  • Jonathan M. Borwein1  and Roland Girgensohn, 2004
• Central Binomial Sums, Multiple Clausen Values and Zeta Values
  
  • J. M. Borwein, D. J. Broadhurst, J. Kamnitzer, 2000
• http://dx.doi.org/10.1016/S0370-2693(00)00574-8
• [Lehmer1985]Interesting Series Involving the Central Binomial Coefficient
  
  • D. H. Lehmer, The American Mathematical Monthly, Vol. 92, No. 7 (Aug. - Sep., 1985), pp. 449-457

• On the series Σk = 1∞(k2k)−1 k−n and related sums
  
  • I. J. Zucker, Journal of Number Theory, Volume 20, Issue 1, February 1985, Pages 92-102   
• Some wonderful formulas ... an introduction to polylogarithms
  
  • A.J. Van der Poorten, Queen's papers in Pure and Applied Mathematics, 54 (1979), 269-286

• http://www.jstor.org/action/doBasicSearch?Query=
• http://www.ams.org/mathscinet
• http://dx.doi.org/

 

 

관련도서

• http://books.google.co.kr/books?id=C0HPgWhEssYC
  
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