"다이감마 함수(digamma function)"의 두 판 사이의 차이

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(피타고라스님이 이 페이지의 이름을 Digamma 함수로 바꾸었습니다.)
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px;">간단한 소개</h5>
  
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<math>\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}</math>
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<h5>반사공식</h5>
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<math>\psi(1 - x) - \psi(x) = \pi\,\!\cot{ \left ( \pi x \right ) }</math>
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px;">차분방정식</h5>
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<math>\psi(x + 1) = \psi(x) + \frac{1}{x}</math>
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px;">덧셈공식</h5>
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<math>\Gamma(z) \; \Gamma\left(z + \frac{1}{2}\right) = 2^{\frac{1}{2}-2z} \; \sqrt{2\pi} \; \Gamma(2z) \,\!</math> 의 로그미분을 통해서 다음을 얻을 수 있음.
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<math>\psi(2x)=\psi(x)+\psi(x+{1\over2})+\ln 2</math>
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px;">가우스의 Digamma 정리</h5>
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<math>\psi\left(\frac{m}{k}\right) = -\gamma -\ln(2k)  -\frac{\pi}{2}\cot\left(\frac{m\pi}{k}\right) +2\sum_{n=1}^{\lceil (k-1)/2\rceil} \cos\left(\frac{2\pi nm}{k} \right) \ln\left(\sin\left(\frac{n\pi}{k}\right)\right)</math>
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<h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px;">special values</h5>
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<math>\psi(1) = -\gamma\,\!</math>
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<math>\psi\left(\frac{1}{2}\right) = -2\ln{2} - \gamma</math>
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<math>\psi\left(\frac{1}{3}\right) = -\frac{\pi}{2\sqrt{3}} -\frac{3}{2}\ln{3} - \gamma</math>
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<math>\psi\left(\frac{1}{4}\right) = -\frac{\pi}{2} - 3\ln{2} - \gamma</math>
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<math>\psi\left(\frac{1}{6}\right) = -\frac{\pi}{2}\sqrt{3} -2\ln{2} -\frac{3}{2}\ln(3) - \gamma</math>
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<math>\psi\left(\frac{1}{8}\right) = -\frac{\pi}{2} - 4\ln{2} - \frac{1}{\sqrt{2}} \left\{\pi + \ln(2 + \sqrt{2}) - \ln(2 - \sqrt{2})\right\} - \gamma</math>
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2009년 7월 5일 (일) 01:10 판

간단한 소개

 

\(\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}\)

 

 

반사공식

\(\psi(1 - x) - \psi(x) = \pi\,\!\cot{ \left ( \pi x \right ) }\)

 

차분방정식

\(\psi(x + 1) = \psi(x) + \frac{1}{x}\)

 

 

덧셈공식

 

\(\Gamma(z) \; \Gamma\left(z + \frac{1}{2}\right) = 2^{\frac{1}{2}-2z} \; \sqrt{2\pi} \; \Gamma(2z) \,\!\) 의 로그미분을 통해서 다음을 얻을 수 있음.

\(\psi(2x)=\psi(x)+\psi(x+{1\over2})+\ln 2\)

 

 

가우스의 Digamma 정리

 

\(\psi\left(\frac{m}{k}\right) = -\gamma -\ln(2k) -\frac{\pi}{2}\cot\left(\frac{m\pi}{k}\right) +2\sum_{n=1}^{\lceil (k-1)/2\rceil} \cos\left(\frac{2\pi nm}{k} \right) \ln\left(\sin\left(\frac{n\pi}{k}\right)\right)\)

 

 

special values

 

\(\psi(1) = -\gamma\,\!\)

\(\psi\left(\frac{1}{2}\right) = -2\ln{2} - \gamma\)

\(\psi\left(\frac{1}{3}\right) = -\frac{\pi}{2\sqrt{3}} -\frac{3}{2}\ln{3} - \gamma\)

\(\psi\left(\frac{1}{4}\right) = -\frac{\pi}{2} - 3\ln{2} - \gamma\)

\(\psi\left(\frac{1}{6}\right) = -\frac{\pi}{2}\sqrt{3} -2\ln{2} -\frac{3}{2}\ln(3) - \gamma\)

\(\psi\left(\frac{1}{8}\right) = -\frac{\pi}{2} - 4\ln{2} - \frac{1}{\sqrt{2}} \left\{\pi + \ln(2 + \sqrt{2}) - \ln(2 - \sqrt{2})\right\} - \gamma\)

 

 

상위 주제

 

 

 

하위페이지

 

 

재미있는 사실

 

 

역사

 

많이 나오는 질문과 답변

 

관련된 고교수학 또는 대학수학

 

 

관련된 다른 주제들


 

관련도서 및 추천도서

 

참고할만한 자료

 

관련기사

 

 

블로그

 

이미지 검색

 

동영상