"중심이항계수 (central binomial coefficient)"의 두 판 사이의 차이
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Pythagoras0 (토론 | 기여) |
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| − | + | ==개요== | |
| − | + | * 다음과 같은 [[이항계수와 조합|이항계수]]로 정의<br>'''<math>{2n \choose n}=\frac{(2n)!}{(n!)^2}</math>'''<br> 1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, ...<br> | |
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| − | * 다음과 | ||
* 동전을 2n 번 던질때, 앞뒷면이 각각 n 번 나올 확률을 표현할 때 등장<br> | * 동전을 2n 번 던질때, 앞뒷면이 각각 n 번 나올 확률을 표현할 때 등장<br> | ||
| − | * | + | * 아페리가 [[Ζ(3)는 무리수이다(아페리의 정리)]]를 증명하는데 활용됨<br> |
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| − | + | ==중심이항계수의 근사식== | |
* http://planetmath.org/encyclopedia/AsymptoticsOfCentralBinomialCoefficient.html<br> | * http://planetmath.org/encyclopedia/AsymptoticsOfCentralBinomialCoefficient.html<br> | ||
| 40번째 줄: | 32번째 줄: | ||
<math>\frac{1}{2^{2n}}{{(2n)!} \over {n!n!}}= \frac{1}{2^{2n}}{2n\choose n} \approx \frac{1}{\sqrt{\pi n}}</math> | <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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| − | + | ==급수와 중심이항계수== | |
* [[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> | ||
* [[역삼각함수]]<br><math>2(\arcsin x)^2=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}{n}}</math><br><math>\frac{2x \arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n\binom{2n}{n}}</math><br><math>\frac{4 \left(\sqrt{4-z}+\sqrt{z} \sin ^{-1}\left(\frac{\sqrt{z}}{2}\right)\right)}{(4-z)^{3/2}}=\sum_{n=0}^{\infty}\frac{z^n}{ {{2n}\choose {n}}}</math><br> | * [[역삼각함수]]<br><math>2(\arcsin x)^2=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}{n}}</math><br><math>\frac{2x \arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n\binom{2n}{n}}</math><br><math>\frac{4 \left(\sqrt{4-z}+\sqrt{z} \sin ^{-1}\left(\frac{\sqrt{z}}{2}\right)\right)}{(4-z)^{3/2}}=\sum_{n=0}^{\infty}\frac{z^n}{ {{2n}\choose {n}}}</math><br> | ||
| − | * [[카탈란 수열(Catalan numbers)]] | + | * [[카탈란 수열(Catalan numbers)]] 의 생성함수<br><math>G(x)= \frac{1-\sqrt{1-4x}}{2x}=\sum_{n=0}^{\infty}\frac{1}{n+1}{2n\choose n}x^n</math><br> |
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| − | + | == 중심이항계수가 나타나는 급수== | |
* '''[Lehmer1985]''' 참조<br> | * '''[Lehmer1985]''' 참조<br> | ||
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<math>\sum_{n=1}^{\infty}\frac{1}{n\binom{2n}{n}}=\frac{\pi\sqrt{3}}{9}</math> | <math>\sum_{n=1}^{\infty}\frac{1}{n\binom{2n}{n}}=\frac{\pi\sqrt{3}}{9}</math> | ||
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(증명) | (증명) | ||
| − | + | <math>\frac{2x \arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n\binom{2n}{n}}</math> 에서 <math>x=\frac{1}{2}</math>인 경우, <math>\sum_{n=1}^{\infty}\frac{1}{n\binom{2n}{n}}=\frac{\pi\sqrt{3}}{9}</math> 를 얻는다. \[FilledSquare] | |
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<math>\sum_{n=1}^{\infty}\frac{1}{n^{2}\binom{2n}{n}}=\frac{\pi^2}{18}</math> | <math>\sum_{n=1}^{\infty}\frac{1}{n^{2}\binom{2n}{n}}=\frac{\pi^2}{18}</math> | ||
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(증명) | (증명) | ||
| − | + | <math>2(\arcsin x)^2=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}{n}}</math>에서 <math>x=\frac{1}{2}</math>인 경우, <math>\sum_{n=1}^{\infty}\frac{1}{n^{2}\binom{2n}{n}}=\frac{\pi^2}{18}</math> 를 얻는다. \[FilledSquare] | |
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| − | <math>\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}}=\frac{2\pi}{3}\operatorname{Cl} | + | <math>\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)</math> |
| − | + | 여기서 <math>\operatorname{Cl}_ 2(\theta)</math> 는 [[로바체프스키 함수|로바체프스키와 클라우센 함수]], <math>\psi^{(1)}</math>는 [[트리감마 함수(trigamma function)]]. | |
(증명) | (증명) | ||
| 92번째 줄: | 84번째 줄: | ||
[http://www.research.att.com/%7Enjas/sequences/A145438 http://www.research.att.com/~njas/sequences/A145438] | [http://www.research.att.com/%7Enjas/sequences/A145438 http://www.research.att.com/~njas/sequences/A145438] | ||
| − | <math>\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}}=4\int_{0}^{\frac{1}{2}}(\arcsin x)^2 | + | <math>\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</math> |
[http://www.wolframalpha.com/input/?i=integrate+%28arcsin+x%29%5E2/x+dx+from+x%3D0+to+1/2 http://www.wolframalpha.com/input/?i=integrate+(arcsin+x)^2/x+dx+from+x%3D0+to+1/2] | [http://www.wolframalpha.com/input/?i=integrate+%28arcsin+x%29%5E2/x+dx+from+x%3D0+to+1/2 http://www.wolframalpha.com/input/?i=integrate+(arcsin+x)^2/x+dx+from+x%3D0+to+1/2] | ||
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| − | + | 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%283%29/3%2Bpi*sqrt%283%29*%28trigamma%281/3%29-trigamma%282/3%29%29/18 http://www.wolframalpha.com/input/?i=-4*zeta(3)/3%2Bpi*sqrt(3)*(trigamma(1/3)-trigamma(2/3))/18] | |
| − | + | \[FilledSquare] | |
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(Comtet의 공식) | (Comtet의 공식) | ||
| 112번째 줄: | 104번째 줄: | ||
<math>\sum_{n=1}^\infty \frac{1}{n^4\binom{2n}{n}}=\frac{17\pi^4}{3240}</math> | <math>\sum_{n=1}^\infty \frac{1}{n^4\binom{2n}{n}}=\frac{17\pi^4}{3240}</math> | ||
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(증명) | (증명) | ||
| − | <math>2(\arcsin x)^2=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}{n}}</math> | + | <math>2(\arcsin x)^2=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}{n}}</math> 의 양변을 <math>2x</math>로 나눈뒤, 다음과 같은 적분을 구하자. |
<math>\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}</math> | <math>\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}</math> | ||
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우변으로부터 <math>\sum_{n=1}^{\infty}\frac{1}{8n^4\binom{2n}{n}}</math>을 얻는다. | 우변으로부터 <math>\sum_{n=1}^{\infty}\frac{1}{8n^4\binom{2n}{n}}</math>을 얻는다. | ||
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한편 | 한편 | ||
| − | <math>\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</math> | + | <math>\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</math> 이므로, |
| − | + | <math>x=\sin\frac{t}{2}</math>로 치환하면, | |
| − | <math>\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</math> | + | <math>\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</math> 를 얻는다. |
따라서, | 따라서, | ||
| − | <math>\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</math> | + | <math>\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</math> 이다. |
| − | 이제 [[로그 사인 적분 (log sine integrals)]] | + | 이제 [[로그 사인 적분 (log sine integrals)]] 에서 얻은 다음 결과를 사용하자. |
<math>\int_{0}^{\pi/3}x\log^2(2\sin \frac{x}{2})\,dx=\frac{17\pi^4}{6480}</math> | <math>\int_{0}^{\pi/3}x\log^2(2\sin \frac{x}{2})\,dx=\frac{17\pi^4}{6480}</math> | ||
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| − | + | <math>\sum_{n=1}^\infty \frac{1}{n^4\binom{2n}{n}}=\frac{17\pi^4}{3240}</math> 를 얻는다. \[FilledSquare] | |
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| − | + | ==원주율의 유리수 근사와 중심이항계수== | |
| − | + | <math>\sum_{n=1}^{\infty}\frac{2^{n}}{\binom{2n}{n}}=\frac{\pi}{2}+1</math> | |
<math>\sum_{n=1}^{\infty}\frac{n2^{n}}{\binom{2n}{n}}=\pi+3</math> | <math>\sum_{n=1}^{\infty}\frac{n2^{n}}{\binom{2n}{n}}=\pi+3</math> | ||
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<math>\sum_{n=1}^{\infty}\frac{n^{10} 2^{n}}{\binom{2n}{n}}=229093376\pi+719718067</math> | <math>\sum_{n=1}^{\infty}\frac{n^{10} 2^{n}}{\binom{2n}{n}}=229093376\pi+719718067</math> | ||
| − | + | 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%5E6*2%5Em/%28binom%282m,m%29%29+from+1+to+infinity http://www.wolframalpha.com/input/?i=sum+m^6*2^m/(binom(2m,m))+from+1+to+infinity] | [http://www.wolframalpha.com/input/?i=sum+m%5E6*2%5Em/%28binom%282m,m%29%29+from+1+to+infinity http://www.wolframalpha.com/input/?i=sum+m^6*2^m/(binom(2m,m))+from+1+to+infinity] | ||
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일반적으로 <math>k\in\mathbb{N}</math>에 대하여, | 일반적으로 <math>k\in\mathbb{N}</math>에 대하여, | ||
| − | <math>\sum_{n=1}^{\infty}\frac{n^{k} 2^{n}}{\binom{2n}{n}}=a\pi+b</math> | + | <math>\sum_{n=1}^{\infty}\frac{n^{k} 2^{n}}{\binom{2n}{n}}=a\pi+b</math> , (a와 b는 유리수) 형태로 주어진다. '''[Lehmer1985] '''참조 |
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| − | + | ==리만제타함수== | |
<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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<math>\zeta(4) = \frac{36}{17} \sum_{n=1}^\infty \frac{1}{n^4\binom{2n}{n}}</math> | <math>\zeta(4) = \frac{36}{17} \sum_{n=1}^\infty \frac{1}{n^4\binom{2n}{n}}</math> | ||
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| − | + | ==메모== | |
* http://math.stackexchange.com/questions/153504/the-fermat-prime-257-and-binomial-sum-sum-n-0-infty-frac-1n-binom-8<br> | * http://math.stackexchange.com/questions/153504/the-fermat-prime-257-and-binomial-sum-sum-n-0-infty-frac-1n-binom-8<br> | ||
* [http://mathoverflow.net/questions/98897/a-coincidence-concerning-fermat-primes-binomial-sums-and-eta-quotients ]http://mathoverflow.net/questions/98897/a-coincidence-concerning-fermat-primes-binomial-sums-and-eta-quotients<br> | * [http://mathoverflow.net/questions/98897/a-coincidence-concerning-fermat-primes-binomial-sums-and-eta-quotients ]http://mathoverflow.net/questions/98897/a-coincidence-concerning-fermat-primes-binomial-sums-and-eta-quotients<br> | ||
* [http://tpiezas.wordpress.com/2012/06/03/fermat-primes-and-binomial-sums/ ]http://tpiezas.wordpress.com/2012/06/03/fermat-primes-and-binomial-sums/<br> | * [http://tpiezas.wordpress.com/2012/06/03/fermat-primes-and-binomial-sums/ ]http://tpiezas.wordpress.com/2012/06/03/fermat-primes-and-binomial-sums/<br> | ||
| − | * http://www1.au.edu.tw/ox_view/edu/tojms/j_paper/Full_text/Vol-25/No-2/25%282%297-2%28141-151%29.pdf<br> | + | * http://www1.au.edu.tw/ox_view/edu/tojms/j_paper/Full_text/Vol-25/No-2/25%282 %297-2%28141-151%29.pdf<br> |
* Math Overflow http://mathoverflow.net/search?q= | * Math Overflow http://mathoverflow.net/search?q= | ||
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| − | + | ==역사== | |
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* http://www.google.com/search?hl=en&tbs=tl:1&q= | * http://www.google.com/search?hl=en&tbs=tl:1&q= | ||
* [[수학사연표 (역사)|수학사연표]] | * [[수학사연표 (역사)|수학사연표]] | ||
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| − | + | ==메모== | |
'''[Lehmer1985]''' | '''[Lehmer1985]''' | ||
| 228번째 줄: | 220번째 줄: | ||
<math>\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)</math> | <math>\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)</math> | ||
| − | + | 여기서 <math>\psi^{(1)}</math>는 트리감마(trigamma)함수. [[트리감마 함수(trigamma function)]]항목 참조 | |
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| − | + | ==관련된 항목들== | |
* [[카탈란 수열(Catalan numbers)]]<br> | * [[카탈란 수열(Catalan numbers)]]<br> | ||
| − | * [[ | + | * [[Z(3)는 무리수이다(아페리의 정리)]]<br> |
* [[폴리로그 함수(polylogarithm)]]<br> | * [[폴리로그 함수(polylogarithm)]]<br> | ||
* [[이항급수와 이항정리]]<br> | * [[이항급수와 이항정리]]<br> | ||
| 243번째 줄: | 235번째 줄: | ||
* [[중심이항계수가 등장하는 어떤 급수에 대한 문제]]<br> | * [[중심이항계수가 등장하는 어떤 급수에 대한 문제]]<br> | ||
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| − | + | ==사전 형태의 자료== | |
* http://ko.wikipedia.org/wiki/ | * http://ko.wikipedia.org/wiki/ | ||
| 271번째 줄: | 248번째 줄: | ||
** http://www.research.att.com/~njas/sequences/?q= | ** http://www.research.att.com/~njas/sequences/?q= | ||
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| − | + | ==관련논문== | |
* Renzo Sprugnoli, Sum of reciprocals of the Central Binomial Coefficients, Integers: electronic journal of combinatorial number theory, 6 (2006) A27, 1-18 | * Renzo Sprugnoli, Sum of reciprocals of the Central Binomial Coefficients, Integers: electronic journal of combinatorial number theory, 6 (2006) A27, 1-18 | ||
| − | * [http://escholarship.org/uc/item/7wd7j9nz Experimental Determination of Apéry-like Identities for | + | * [http://escholarship.org/uc/item/7wd7j9nz Experimental Determination of Apéry-like Identities for \[Zeta](2n + 2)]<br> |
** David H. Bailey, Jonathan M. Borwein, and David M. Bradley | ** David H. Bailey, Jonathan M. Borwein, and David M. Bradley | ||
* [http://dx.doi.org/10.1007/s00010-005-2774-x Evaluations of binomial series]<br> | * [http://dx.doi.org/10.1007/s00010-005-2774-x Evaluations of binomial series]<br> | ||
| − | ** Jonathan M. Borwein1 | + | ** Jonathan M. Borwein1 and Roland Girgensohn, 2004 |
* [http://arxiv.org/abs/hep-th/0004153 Central Binomial Sums, Multiple Clausen Values and Zeta Values]<br> | * [http://arxiv.org/abs/hep-th/0004153 Central Binomial Sums, Multiple Clausen Values and Zeta Values]<br> | ||
** J. M. Borwein, D. J. Broadhurst, J. Kamnitzer, 2000 | ** J. M. Borwein, D. J. Broadhurst, J. Kamnitzer, 2000 | ||
| − | * [http://dx.doi.org/10.1016/S0370-2693%2800%2900574-8 http://dx.doi.org/10.1016/S0370-2693(00)00574-8] | + | * [http://dx.doi.org/10.1016/S0370-2693%2800 %2900574-8 http://dx.doi.org/10.1016/S0370-2693(00)00574-8] |
* '''[Lehmer1985]'''[http://www.jstor.org/stable/2322496 Interesting Series Involving the Central Binomial Coefficient]<br> | * '''[Lehmer1985]'''[http://www.jstor.org/stable/2322496 Interesting Series Involving the Central Binomial Coefficient]<br> | ||
| − | ** D. | + | ** D. H. Lehmer, The American Mathematical Monthly, Vol. 92, No. 7 (Aug. - Sep., 1985), pp. 449-457 |
| − | * [http://dx.doi.org/10.1016/0022-314X%2885%2990019-8 On the series | + | * [http://dx.doi.org/10.1016/0022-314X %2885 %2990019-8 On the series \[CapitalSigma]k = 1\[Infinity](k2k)\[Minus]1 k\[Minus]n and related sums]<br> |
| − | ** I. J. Zucker, Journal of Number Theory, | + | ** I. J. Zucker, Journal of Number Theory, Volume 20, Issue 1, February 1985, Pages 92-102 |
* Some wonderful formulas ... an introduction to polylogarithms<br> | * Some wonderful formulas ... an introduction to polylogarithms<br> | ||
| − | ** A.J. Van der Poorten, | + | ** 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.jstor.org/action/doBasicSearch?Query= | ||
| 297번째 줄: | 274번째 줄: | ||
* http://dx.doi.org/ | * http://dx.doi.org/ | ||
| − | + | ||
| − | + | ||
| − | + | ==관련도서== | |
* http://books.google.co.kr/books?id=C0HPgWhEssYC<br> | * http://books.google.co.kr/books?id=C0HPgWhEssYC<br> | ||
2012년 9월 20일 (목) 14:18 판
개요
- 다음과 같은 이항계수로 정의
\({2n \choose n}=\frac{(2n)!}{(n!)^2}\)
1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, ... - 동전을 2n 번 던질때, 앞뒷면이 각각 n 번 나올 확률을 표현할 때 등장
- 아페리가 Ζ(3)는 무리수이다(아페리의 정리)를 증명하는데 활용됨
중심이항계수의 근사식
- http://planetmath.org/encyclopedia/AsymptoticsOfCentralBinomialCoefficient.html
- 드무아브르의 중심극한정리(iii) : 숫자 파이와 동전던지기 피타고라스의 창, 2008-7-12
동전을 2n 번 던질때, 앞뒷면이 각각 n 번 나올 확률은 수학적으로 다음과 같다.
\(\frac{1}{2^{2n}}{2n\choose n} = \frac{1}{2^{2n}}{{(2n)!} \over {n!n!}}\)
한편 월리스의 공식에서 일반항은 다음과 같은데,
\(p_n ={1\over{2n+1}}\prod_{k=1}^{n} \frac{(2k)^4 }{((2k)(2k-1))^2}={1\over{2n+1}}\cdot {{2^{4n}\,(n!)^4}\over {((2n)!)^2}}\)
따라서
\(p_n ={1\over{2n+1}}\cdot {{2^{4n}\,(n!)^4}\over {((2n)!)^2}} \approx {1\over{2n}}\cdot {{2^{4n}\,(n!)^4}\over {((2n)!)^2}}\)
이는 월리스의 공식을 다음과 같은 방식으로도 쓸 수 있다는 것을 말해준다.
\(\frac{\pi}{2} =\lim_{n \to \infty} {1\over{2n}}\cdot {{2^{4n}\,(n!)^4}\over {((2n)!)^2}}\)
그리고 이는 다음을 말해준다.
\(\frac{1}{2^{2n}}{{(2n)!} \over {n!n!}}= \frac{1}{2^{2n}}{2n\choose n} \approx \frac{1}{\sqrt{\pi n}}\)
급수와 중심이항계수
- 이항급수와 이항정리
\(\frac{1}{\sqrt{1-4z}}=\sum_{n=0}^{\infty} {{2n}\choose {n}} z^n\) - 역삼각함수
\(2(\arcsin x)^2=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}{n}}\)
\(\frac{2x \arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n\binom{2n}{n}}\)
\(\frac{4 \left(\sqrt{4-z}+\sqrt{z} \sin ^{-1}\left(\frac{\sqrt{z}}{2}\right)\right)}{(4-z)^{3/2}}=\sum_{n=0}^{\infty}\frac{z^n}{ {{2n}\choose {n}}}\) - 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. \[FilledSquare]
\(\sum_{n=1}^{\infty}\frac{1}{n^{2}\binom{2n}{n}}=\frac{\pi^2}{18}\)
(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. \[FilledSquare]
\(\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}}=\frac{2\pi}{3}\operatorname{Cl}_ 2(\frac{\pi}{3})-\frac{4}{3}\zeta(3)=\pi\operatorname{Cl}_ 2(\frac{2\pi}{3})-\frac{4}{3}\zeta(3)=\frac{\pi\sqrt{3}}{18}(\psi^{(1)}(\frac{1}{3})-\psi^{(1)}(\frac{2}{3}))-\frac{4}{3}\zeta(3)\)
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
\[FilledSquare]
(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. \[FilledSquare]
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
\(\zeta(2)=3\sum_{n=1}^{\infty}\frac{1}{n^{2}\binom{2n}{n}}\)
\(\zeta(3) = \frac{5}{2} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3\binom{2n}{n}}\)
\(\zeta(4) = \frac{36}{17} \sum_{n=1}^\infty \frac{1}{n^4\binom{2n}{n}}\)
메모
- http://math.stackexchange.com/questions/153504/the-fermat-prime-257-and-binomial-sum-sum-n-0-infty-frac-1n-binom-8
- [1]http://mathoverflow.net/questions/98897/a-coincidence-concerning-fermat-primes-binomial-sums-and-eta-quotients
- [2]http://tpiezas.wordpress.com/2012/06/03/fermat-primes-and-binomial-sums/
- http://www1.au.edu.tw/ox_view/edu/tojms/j_paper/Full_text/Vol-25/No-2/25%282 %297-2%28141-151%29.pdf
- Math Overflow http://mathoverflow.net/search?q=
역사
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)
- Z(3)는 무리수이다(아페리의 정리)
- 폴리로그 함수(polylogarithm)
- 이항급수와 이항정리
- 정규분포와 그 확률밀도함수
- 중심이항계수가 등장하는 어떤 급수에 대한 문제
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://planetmath.org/GeneratingFunctionForTheReciprocalCentralBinomialCoefficients.html
- NIST Digital Library of Mathematical Functions
- The On-Line Encyclopedia of Integer Sequences
관련논문
- Renzo Sprugnoli, Sum of reciprocals of the Central Binomial Coefficients, Integers: electronic journal of combinatorial number theory, 6 (2006) A27, 1-18
- Experimental Determination of Apéry-like Identities for \[Zeta(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
- %2900574-8 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
- %2885 %2990019-8 On the series \[CapitalSigmak = 1\[Infinity](k2k)\[Minus]1 k\[Minus]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