"중심이항계수 (central binomial coefficient)"의 두 판 사이의 차이

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* [[중심이항계수(central binomial coefficient)]]<br>
 
* [[중심이항계수(central binomial coefficient)]]<br>
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*  다음과 같은 [[이항계수와 조합|이항계수]]로 정의<br>'''<math>{2n \choose n}=\frac{(2n)!}{(n!)^2}</math>'''<br>
 
*  다음과 같은 [[이항계수와 조합|이항계수]]로 정의<br>'''<math>{2n \choose n}=\frac{(2n)!}{(n!)^2}</math>'''<br>
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* [[이항급수와 이항정리]]<br><math>\frac{1}{\sqrt{1-4z}}=\sum_{n=0}^{\infty} {{2n}\choose {n}} z^n</math><br>
 
* [[이항급수와 이항정리]]<br><math>\frac{1}{\sqrt{1-4z}}=\sum_{n=0}^{\infty} {{2n}\choose {n}} z^n</math><br>
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<math>\zeta(2)=3\sum_{n=1}^{\infty}\frac{1}{n^{2}\binom{2n}{n}}</math>
 
<math>\zeta(2)=3\sum_{n=1}^{\infty}\frac{1}{n^{2}\binom{2n}{n}}</math>
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 http://www.wolframalpha.com/input/?i=integrate+log+%282x%29+*+%28arcsin+x%29%5E2%2Fx+dx+from+x%3D0+to+1%2F2
  
 
 
 
 
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* [[카탈란 수열(Catalan numbers)]]<br>
 
* [[카탈란 수열(Catalan numbers)]]<br>
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* 단어사전 http://www.google.com/dictionary?langpair=en|ko&q=
 
* 단어사전 http://www.google.com/dictionary?langpair=en|ko&q=
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* http://ko.wikipedia.org/wiki/
 
* http://ko.wikipedia.org/wiki/
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* [http://escholarship.org/uc/item/7wd7j9nz Experimental Determination of Apéry-like Identities&nbsp;for ζ(2n + 2)]<br>
 
* [http://escholarship.org/uc/item/7wd7j9nz Experimental Determination of Apéry-like Identities&nbsp;for ζ(2n + 2)]<br>
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*  도서내검색<br>
 
*  도서내검색<br>
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*  네이버 뉴스 검색 (키워드 수정)<br>
 
*  네이버 뉴스 검색 (키워드 수정)<br>
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*  구글 블로그 검색<br>
 
*  구글 블로그 검색<br>

2010년 6월 22일 (화) 17:25 판

이 항목의 스프링노트 원문주소

 

 

개요
  • 다음과 같은 이항계수로 정의
    \({2n \choose n}=\frac{(2n)!}{(n!)^2}\)

 

 

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
    \(G(x)= \frac{1-\sqrt{1-4x}}{2x}=\sum_{n=0}^{\infty}\frac{1}{n+1}{2n\choose n}x^n\)
     

 

 

 

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

 

 

 

역사

 

 

 

메모

 http://www.wolframalpha.com/input/?i=integrate+log+%282x%29+*+%28arcsin+x%29%5E2%2Fx+dx+from+x%3D0+to+1%2F2

 

 

관련된 항목들

 

 

수학용어번역

 

 

사전 형태의 자료

 

 

관련논문
  • 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

 

 

관련도서

 

 

관련기사

 

 

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