"오일러-가우스 초기하함수2F1"의 두 판 사이의 차이

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(피타고라스님이 이 페이지의 이름을 가우스의 초기하급수로 바꾸었습니다.)
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<h5 style="line-height: 3.428em; 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%;">이 항목의 스프링노트 원문주소</h5>
  
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<h5 style="line-height: 3.428em; 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%;">개요</h5>
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<math>\,_2F_1(a,b;c;z)=\sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(c)_nn!}z^n</math>
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미분방정식
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<math>z(1-z)\frac{d^2w}{dz^2}+(c-(a+b+1)z)\frac{dw}{dz}-abw = 0</math>
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을 만족시킴.
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[[타원적분]]
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<math>K(k) =\frac{\pi}{2}\,_2F_1(\frac{1}{2},\frac{1}{2};1;k^2)</math>
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<math>E(k) =\frac{\pi}{2}\,_2F_1(\frac{1}{2},-\frac{1}{2};1;k^2)</math>
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<h5 style="line-height: 3.428em; 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%;">타원적분과 초기하급수</h5>
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* [[제1종타원적분 K (complete elliptic integral of the first kind)]]<br>
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<math>K(k) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1-k^2 \sin^2\theta}} = \int_0^{\frac{\pi}{2}}\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_n}{n!} k^{2n}\sin^{2n}\theta{d\theta}  </math>
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<math>\int_0^{\frac{\pi}{2}}\sin^{2n}\theta{d\theta}=\frac{\pi}{2}\frac{(\frac{1}{2})_n}{(1)_n}</math> ([[#|감마함수]]) 이므로
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<math>K(k) = \frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_n(\frac{1}{2})_n}{n!(1)_n}k^{2n} = \frac{\pi}{2}\,_2F_1(\frac{1}{2},\frac{1}{2};1;k^2)</math>
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<h5 style="line-height: 3.428em; 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%;">special values</h5>
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<math>\frac{\pi}{2}\,_2F_1(\frac{1}{2},\frac{1}{2};1;\frac{1}{2})=K(\frac{1}{\sqrt{2}})=\frac{1}{4}B(1/4,1/4)=\frac{\Gamma(\frac{1}{4})^2}{4\sqrt{\pi}}=1.8540746773\cdots</math>
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<h5 style="line-height: 3.428em; 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%;">재미있는 사실</h5>
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* [[periods]]<br>
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<h5 style="line-height: 3.428em; 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%;">수학용어번역</h5>
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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: 3.428em; 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%;">관련논문</h5>
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* [http://people.math.jussieu.fr/~miw/articles/pdf/TranscendencePeriods.pdf Transcendence of periods: the state of the art.]<br>
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**  M. Waldschmidt., Pure Appl. Math. Q. 2 (2006), 435-463.<br>
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*   <br>[http://www.jstor.org/stable/2975319 On the Kummer Solutions of the Hypergeometric Equation]<br>
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** Reese T. Prosser, <cite style="line-height: 2em;">The American Mathematical Monthly</cite>, Vol. 101, No. 6 (Jun. - Jul., 1994), pp. 535-543
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* [http://math.dongascience.com/ 수학동아]
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* [http://www.ams.org/mathmoments/ Mathematical Moments from the AMS]
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* [http://betterexplained.com/ BetterExplained]

2009년 12월 2일 (수) 15:47 판

이 항목의 스프링노트 원문주소

 

 

 

개요

 

\(\,_2F_1(a,b;c;z)=\sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(c)_nn!}z^n\)

미분방정식

\(z(1-z)\frac{d^2w}{dz^2}+(c-(a+b+1)z)\frac{dw}{dz}-abw = 0\)

을 만족시킴.

 

타원적분

\(K(k) =\frac{\pi}{2}\,_2F_1(\frac{1}{2},\frac{1}{2};1;k^2)\)

\(E(k) =\frac{\pi}{2}\,_2F_1(\frac{1}{2},-\frac{1}{2};1;k^2)\)

 

 

 

타원적분과 초기하급수

\(K(k) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1-k^2 \sin^2\theta}} = \int_0^{\frac{\pi}{2}}\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_n}{n!} k^{2n}\sin^{2n}\theta{d\theta} \)

\(\int_0^{\frac{\pi}{2}}\sin^{2n}\theta{d\theta}=\frac{\pi}{2}\frac{(\frac{1}{2})_n}{(1)_n}\) (감마함수) 이므로

\(K(k) = \frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_n(\frac{1}{2})_n}{n!(1)_n}k^{2n} = \frac{\pi}{2}\,_2F_1(\frac{1}{2},\frac{1}{2};1;k^2)\)

 

 

 

 

special values

\(\frac{\pi}{2}\,_2F_1(\frac{1}{2},\frac{1}{2};1;\frac{1}{2})=K(\frac{1}{\sqrt{2}})=\frac{1}{4}B(1/4,1/4)=\frac{\Gamma(\frac{1}{4})^2}{4\sqrt{\pi}}=1.8540746773\cdots\)

 

 

재미있는 사실

 

 

 

역사

 

 

메모

 

 

관련된 항목들

 

 

수학용어번역

 

 

사전 형태의 자료

 

 

관련논문

 

관련도서 및 추천도서

 

 

관련기사

 

 

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