"Gieseking's constant"의 두 판 사이의 차이

수학노트
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http://mathworld.wolfram.com/GiesekingsConstant.html
 
http://mathworld.wolfram.com/GiesekingsConstant.html
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http://www.research.att.com/~njas/sequences/A143298
  
 
http://www.wolframalpha.com/input/?i=Gieseking's+constant.
 
http://www.wolframalpha.com/input/?i=Gieseking's+constant.
  
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http://www.research.att.com/~njas/sequences/A143298
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http://www.wolframalpha.com/input/?i=sqrt(3)*(trigamma(1/3)-trigamma(2/3))/12
  
 
 
 
 
  
 
define similarly
 
define similarly
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<h5 style="line-height: 3.42em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.16em; background-image: ; background-color: initial; background-position: 0px 100%;">이 항목의 스프링노트 원문주소</h5>
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* [http://pythagoras0.springnote.com/pages/5883171 중심이항계수(central binomial coefficient)]<br>
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<h5 style="line-height: 3.42em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.16em; background-image: ; background-color: initial; background-position: 0px 100%;">개요</h5>
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*  다음과 같은 [http://pythagoras0.springnote.com/pages/4783943 이항계수]로 정의<br>'''<math>{2n \choose n}=\frac{(2n)!}{(n!)^2}</math>'''<br>
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*  아페리가 [http://pythagoras0.springnote.com/pages/5169545 ζ(3)는 무리수] 임을 증명하는데 활용됨<br>
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">급수와 중심이항계수</h5>
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* [http://pythagoras0.springnote.com/pages/4874531 이항급수와 이항정리]<br><math>\frac{1}{\sqrt{1-4z}}=\sum_{n=0}^{\infty} {{2n}\choose {n}} z^n</math><br>
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* [http://pythagoras0.springnote.com/pages/5900383 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>
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* [http://pythagoras0.springnote.com/pages/4784349 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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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;"> 중심이항계수가 나타나는 급수</h5>
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* '''[Lehmer1985]''' 참조<br>
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<math>\sum_{n=1}^{\infty}\frac{1}{n\binom{2n}{n}}=\frac{\pi\sqrt{3}}{9}</math>
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(증명)
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 <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. ■
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<math>\sum_{n=1}^{\infty}\frac{1}{n^{2}\binom{2n}{n}}=\frac{\pi^2}{18}</math>
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(증명)
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 <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. ■
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<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)=\frac{\pi\sqrt{3}}{18}(\psi^{(1)}(\frac{1}{3})-\psi^{(1)}(\frac{2}{3}))-\frac{4}{3}\zeta(3)</math>
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여기서 <math>\psi^{(1)}</math>는 트리감마(trigamma)함수. [http://pythagoras0.springnote.com/pages/6091635 트리감마 함수(trigamma function)]항목 참조
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(증명)
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http://www.research.att.com/~njas/sequences/A145438
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<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> 
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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
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우변 http://www.wolframalpha.com/input/?i=-4*zeta(3)/3%2Bpi*sqrt(3)*(trigamma(1/3)-trigamma(2/3))/18
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■
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(Comtet의 공식)
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<math>\sum_{n=1}^\infty \frac{1}{n^4\binom{2n}{n}}=\frac{17\pi^4}{3240}</math>
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(증명)
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<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.
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<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.
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한편
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<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,
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 <math>x=\sin\frac{t}{2}</math>로 치환하면,
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<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.
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따라서,
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<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.
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이제 [http://pythagoras0.springnote.com/pages/5896529 로그 사인 적분 (log sine integrals)] 에서 얻은 다음 결과를 사용하자.
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<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. ■
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">원주율의 유리수 근사와 중심이항계수</h5>
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 <math>\sum_{n=1}^{\infty}\frac{2^{n}}{\binom{2n}{n}}=\frac{\pi}{2}+1</math>
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<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^2 2^{n}}{\binom{2n}{n}}=\frac{7\pi}{2}+11</math>
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<math>\sum_{n=1}^{\infty}\frac{n^3 2^{n}}{\binom{2n}{n}}=\frac{35\pi}{2}+55</math>
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<math>\sum_{n=1}^{\infty}\frac{n^4 2^{n}}{\binom{2n}{n}}=113\pi+355</math>
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<math>\sum_{n=1}^{\infty}\frac{n^{5} 2^{n}}{\binom{2n}{n}} = \frac{1787\pi}{2}+2807</math>
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<math>\sum_{n=1}^{\infty}\frac{n^{6} 2^{n}}{\binom{2n}{n}} = \frac{16717\pi}{2}+26259</math>
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<math>\sum_{n=1}^{\infty}\frac{n^{10} 2^{n}}{\binom{2n}{n}}=229093376\pi+719718067</math>
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 http://www.wolframalpha.com/input/?i=sum+1%2F%28m%5E3*binom%282m%2Cm%29%29+from+1+to+infinity
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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,
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<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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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">리만제타함수</h5>
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<math>\zeta(2)=3\sum_{n=1}^{\infty}\frac{1}{n^{2}\binom{2n}{n}}</math>
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<math>\zeta(3) = \frac{5}{2} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3\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>
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<h5 style="line-height: 3.42em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.16em; background-image: ; background-color: initial; background-position: 0px 100%;">재미있는 사실</h5>
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* Math Overflow http://mathoverflow.net/search?q=
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* 네이버 지식인 http://kin.search.naver.com/search.naver?where=kin_qna&query=
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<h5 style="line-height: 3.42em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.16em; background-image: ; background-color: initial; background-position: 0px 100%;">역사</h5>
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* http://www.google.com/search?hl=en&tbs=tl:1&q=
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* [http://pythagoras0.springnote.com/pages/3304643 수학사연표]
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*  
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<h5 style="line-height: 3.42em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.16em; background-image: ; background-color: initial; background-position: 0px 100%;">메모</h5>
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'''[Lehmer1985]'''
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에는 다음과 같은 공식이 나오지만, 잘못된 것이다.
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<math>\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}))</math>
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바른 공식은 다음과 같다.
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<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>
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여기서 <math>\psi^{(1)}</math>는 트리감마(trigamma)함수. [http://pythagoras0.springnote.com/pages/6091635 트리감마 함수(trigamma function)]항목 참조
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<h5 style="line-height: 3.42em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.16em; background-image: ; background-color: initial; background-position: 0px 100%;">관련된 항목들</h5>
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* [http://pythagoras0.springnote.com/pages/4784349 카탈란 수열(Catalan numbers)]<br>
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* [http://pythagoras0.springnote.com/pages/5169545 ζ(3)는 무리수이다(아페리의 정리)]<br>
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* [http://pythagoras0.springnote.com/pages/5828671 폴리로그 함수(polylogarithm)]<br>
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* [http://pythagoras0.springnote.com/pages/4874531 이항급수와 이항정리]<br>
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<h5 style="line-height: 3.42em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.16em; background-image: ; background-color: initial; background-position: 0px 100%;">수학용어번역</h5>
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* 단어사전 http://www.google.com/dictionary?langpair=en|ko&q=
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* 발음사전 http://www.forvo.com/search/
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* [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집]<br>
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** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=
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* [http://www.nktech.net/science/term/term_l.jsp?l_mode=cate&s_code_cd=MA 남·북한수학용어비교]
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* [http://kms.or.kr/home/kor/board/bulletin_list_subject.asp?bulletinid=%7BD6048897-56F9-43D7-8BB6-50B362D1243A%7D&boardname=%BC%F6%C7%D0%BF%EB%BE%EE%C5%E4%B7%D0%B9%E6&globalmenu=7&localmenu=4 대한수학회 수학용어한글화 게시판]
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<h5 style="line-height: 3.42em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.16em; background-image: ; background-color: initial; background-position: 0px 100%;">사전 형태의 자료</h5>
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* http://ko.wikipedia.org/wiki/
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* http://en.wikipedia.org/wiki/Central_binomial_coefficient
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* [http://mathworld.wolfram.com/CentralBinomialCoefficient.html http://math world.wolfram.com/CentralBinomialCoefficient.html]
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* http://mathworld.wolfram.com/BinomialSums.html
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* http://www.wolframalpha.com/input/?i=
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* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
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* [http://www.research.att.com/~njas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
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** http://www.research.att.com/~njas/sequences/?q=
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<h5 style="line-height: 3.42em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.16em; background-image: ; background-color: initial; background-position: 0px 100%;">관련논문</h5>
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* [http://escholarship.org/uc/item/7wd7j9nz Experimental Determination of Apéry-like Identities&nbsp;for ζ(2n + 2)]<br>
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** David H. Bailey, Jonathan M. Borwein, and David M. Bradley
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* [http://dx.doi.org/10.1007/s00010-005-2774-x Evaluations of binomial series]<br>
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** Jonathan M. Borwein1  and Roland Girgensohn, 2004
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* [http://arxiv.org/abs/hep-th/0004153 Central Binomial Sums, Multiple Clausen Values and Zeta Values]<br>
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** J. M. Borwein, D. J. Broadhurst, J. Kamnitzer, 2000
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* http://dx.doi.org/10.1016/S0370-2693(00)00574-8
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* '''[Lehmer1985]'''[http://www.jstor.org/stable/2322496 Interesting Series Involving the Central Binomial Coefficient]<br>
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** D. H. Lehmer, The American Mathematical Monthly, Vol. 92, No. 7 (Aug. - Sep., 1985), pp. 449-457
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* [http://dx.doi.org/10.1016/0022-314X(85)90019-8 On the series Σk = 1∞(k2k)−1 k−n and related sums]<br>
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** I. J. Zucker, Journal of Number Theory, Volume 20, Issue 1, February 1985, Pages 92-102   
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*  Some wonderful formulas ... an introduction to polylogarithms<br>
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** A.J. Van der Poorten, Queen's papers in Pure and Applied Mathematics, 54 (1979), 269-286
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* http://www.jstor.org/action/doBasicSearch?Query=
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* http://www.ams.org/mathscinet
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* http://dx.doi.org/
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<h5 style="line-height: 3.42em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.16em; background-image: ; background-color: initial; background-position: 0px 100%;">관련도서</h5>
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* http://books.google.co.kr/books?id=C0HPgWhEssYC<br>
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*  도서내검색<br>
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** http://books.google.com/books?q=
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** http://book.daum.net/search/contentSearch.do?query=
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*  도서검색<br>
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** http://books.google.com/books?q=
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** http://book.daum.net/search/mainSearch.do?query=
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** http://book.daum.net/search/mainSearch.do?query=
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<h5 style="line-height: 3.42em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.16em; background-image: ; background-color: initial; background-position: 0px 100%;">관련기사</h5>
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*  네이버 뉴스 검색 (키워드 수정)<br>
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** http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
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** http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
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2010년 7월 17일 (토) 03:53 판

http://mathworld.wolfram.com/GiesekingsConstant.html

http://www.research.att.com/~njas/sequences/A143298

http://www.wolframalpha.com/input/?i=Gieseking's+constant.

http://www.wolframalpha.com/input/?i=sqrt(3)*(trigamma(1/3)-trigamma(2/3))/12

 

define similarly

 

 

이 항목의 스프링노트 원문주소

 

 

개요

 

 

 

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. ■

 

 

 

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

 

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

 

 

 

 

(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

 

 

 

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

 

 

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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