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

수학노트
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(사용자 2명의 중간 판 62개는 보이지 않습니다)
1번째 줄: 1번째 줄:
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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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==개요==
  
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* [[다이감마 함수(digamma function)|digamma 함수]]<br>
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* 감마함수의 로그미분으로 정의
  
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* [[차분방정식(difference equation) 과 유한미적분학 (finite calculus)|차분방정식]]에서 자연스럽게 등장함.
  
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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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*  감마함수의 로그미분으로 정의<br>
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==정의와 급수표현==
  
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<math>\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}</math>
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*  정의:<math>\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}</math>
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*  급수표현:<math>\psi(z)=-\frac{1}{z} -\gamma +\sum_{n=1}^\infty \frac{z}{n(n+z)} , z \neq 0, -1, -2, -3, \cdots</math>
  
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* [[차분방정식(difference equation) 과 유한미적분학 (finite calculus)|차분방정식]]에서 자연스럽게 등장함.<br>
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(증명)
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[[감마함수]]의 무한곱표현
  
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<math>\Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}</math>
  
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위의 식에 로그미분을 취하여 얻는다. ■
  
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<h5>급수표현</h5>
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* [[#]]
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* <math>z = 0, -1, -2, -3, \cdots</math> 에서 pole을 가진다
  
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==함수의 그래프==
  
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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%;">차분방정식과의 관계</h5>
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* <math>-3<x<3</math>일 때, <math>\psi(x)</math>의 그래프
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[[파일:다이감마 함수(digamma function)1.gif]]
  
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<math>\psi(x + 1) = \psi(x) + \frac{1}{x}</math>
 
  
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==도함수와 polygamma 함수==
  
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<h5>반사공식</h5>
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* trigamma  <math>\psi^{(1)}(z)</math>
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* tetragamma <math>\psi^{(2)}(z)</math>
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* pentagamma <math>\psi^{(3)}(z)</math>
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* [[폴리감마함수(polygamma functions)]]
  
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* [[감마함수]]의 반사공식<br><math>\Gamma(1-z) \; \Gamma(z) = {\pi \over \sin{(\pi z)}} \,\!</math><br>
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* 위의 식을 로그미분하여 다음을 얻는다<br>
 
  
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<math>\psi(1 - x) - \psi(x) = \pi\,\!\cot{ \left ( \pi x \right ) }</math>
 
  
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여기서 <math>x</math>를 <math>-x</math>로 두면 다음을 얻는다
 
  
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<math>\psi(1 + x) = \psi(-x) -\pi\,\!\cot{ \left ( \pi x \right ) }</math>
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==차분방정식과의 관계==
  
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* [[차분방정식(difference equation) 과 유한미적분학 (finite calculus)|차분방정식]]:<math>\Delta \psi=\frac{1}{x}</math> 즉,
  
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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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*  차분방정식의 기본정리를 적용하면:<math>\sum_{n=a}^{b-1}\frac{1}{n}=\psi(b)-\psi(a)</math>
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* [[조화급수와 조화 평균에서 '조화'란?|조화급수]]와의 관계:<math>\sum_{n=1}^{N}\frac{1}{n}=\psi(N+1)-\psi(1)=\psi(N+1)-\gamma</math>
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*  일반화:<math>\psi^{(n)}(x+1)-\psi^{(n)}(x)=\frac{(-1)^n n!}{x^{n+1}}</math>
  
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* 이항 덧셈공식<br><math>\psi(2x)=\psi(x)+\psi(x+{1\over2})+\ln 2</math><br>
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(증명)
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[[감마함수]]의 곱셈공식 
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==점근 급수==
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* [[점근 급수(asymptotic series)]] <math>x\sim \infty</math> 일 때,
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:<math>
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\begin{align}
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\psi(x) - \log(x) &= - \frac{1}{2x} - \sum_{n=1}^\infty \frac{B_{2n}}{2n(x^{2n})}\\
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&=-\frac{1}{12 x^2}+\frac{1}{120 x^4}-\frac{1}{252 x^6}+\frac{1}{240 x^8}-\frac{1}{132 x^{10}}+\cdots
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\end{align}
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</math>
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또는
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:<math>
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\psi(x) = \log(x) - \frac{1}{2x} + \sum_{n=1}^\infty \frac{\zeta(1-2n)}{x^{2n}}
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</math> 여기서 <math>B_{n}</math>은 [[베르누이 수]]
  
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<math>2^{2z}\Gamma(z) \; \Gamma\left(z + \frac{1}{2}\right) = 2\sqrt{\pi}\;\Gamma(2z)</math>
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로그를 취하면
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==반사공식==
  
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<math>(2\ln 2)x+\ln \Gamma(x) +\ln \Gamma\left(x + \frac{1}{2}\right) = \ln 2\sqrt{\pi}+\ln \Gamma(2x)</math>
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* [[감마함수]]의 반사공식:<math>\Gamma(1-z) \; \Gamma(z) = {\pi \over \sin{(\pi z)}} \,\!</math>
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*  위의 식을 로그미분하여 다음을 얻는다
  
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미분하면,
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<math>\psi(1 - x) - \psi(x) = \pi\,\!\cot{ \left ( \pi x \right ) }</math>
  
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<math>\psi(2x)=\psi(x)+\psi(x+{1\over2})+\ln 2</math> ■
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여기서 <math>x</math>를 <math>-x</math>로 두면 다음을 얻는다
  
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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;">가우스의 Digamma 정리</h5>
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==덧셈공식==
  
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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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* [[감마함수]]의 곱셈공식에 따른 성질:<math>m\ln m+\psi(z)+ \psi\left(z + \frac{1}{m}\right) + \cdots+ \psi\left(z + \frac{m-1}{m}\right) = m\psi(mz)</math>
  
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<math>\psi\left(1-\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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(증명)
  
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[[감마함수]]의 곱셈공식은 적당한 상수 c에 대하여 다음과 같이 쓸 수 있다.
  
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<math>m^{mz}\Gamma(z)\cdots \Gamma\left(z + \frac{m-1}{m}\right) = c\Gamma(mz)</math>
  
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변수를 x로 바꾸고, 로그를 취하면,
  
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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>(m\ln m)x+\ln \Gamma(x) +\ln \Gamma\left(x + \frac{m-1}{m}\right) =\ln c+\ln \Gamma(mx)</math>
  
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<math>\psi(1) = -\gamma\,\!</math>
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미분하면,
  
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<math>\psi\left(\frac{1}{2}\right) = -2\ln{2} - \gamma</math>
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<math>m\ln m+\psi(x)+\cdots+\psi(x+\frac{m-1}{m})=m\psi(mx)</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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*  이항 덧셈공식:<math>2\psi(2x)=\psi(x)+\psi(x+{1\over2})+2\ln 2</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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==가우스의 Digamma 정리==
  
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<math>\psi\left(\frac{m}{k}\right) = -\gamma -\ln(2k)
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-\frac{\pi}{2}\cot\left(\frac{m\pi}{k}\right)
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+2\sum_{n=1}^{\lfloor (k-1)/2\rfloor}
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\cos\left(\frac{2\pi nm}{k} \right)
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\ln\left(\sin\left(\frac{n\pi}{k}\right)\right)
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</math>
  
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<math>\psi\left(1-\frac{m}{k}\right) = -\gamma -\ln(2k) +\frac{\pi}{2}\cot\left(\frac{m\pi}{k}\right)
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+2\sum_{n=1}^{\lfloor (k-1)/2\rfloor}
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\cos\left(\frac{2\pi nm}{k} \right)
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\ln\left(\sin\left(\frac{n\pi}{k}\right)\right)
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</math>
  
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==special values==
  
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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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<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 Overflow http://mathoverflow.net/search?q=
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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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* 네이버 지식인 http://kin.search.naver.com/search.naver?where=kin_qna&query=
 
  
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<math>\psi\left(\frac{2}{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{3}{4}\right) = \frac{\pi}{2} - 3\ln{2} - \gamma</math>
  
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<math>\psi\left(\frac{1}{5}\right) =-\gamma -\frac{1}{2} \sqrt{1+\frac{2}{\sqrt{5}}} \pi -\log (10)+2 \left(\frac{1}{8} \left(\sqrt{5}-1\right) \log \left(\frac{5}{8}-\frac{\sqrt{5}}{8}\right)+\frac{1}{8} \left(-1-\sqrt{5}\right) \log \left(\frac{5}{8}+\frac{\sqrt{5}}{8}\right)\right)</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%;">역사</h5>
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<math>\psi\left(\frac{2}{5}\right) =-\gamma -\frac{1}{2} \sqrt{1-\frac{2}{\sqrt{5}}} \pi -\log (10)+2 \left(\frac{1}{8} \left(-1-\sqrt{5}\right) \log \left(\frac{5}{8}-\frac{\sqrt{5}}{8}\right)+\frac{1}{8} \left(\sqrt{5}-1\right) \log \left(\frac{5}{8}+\frac{\sqrt{5}}{8}\right)\right) </math>
  
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<math>\psi\left(\frac{3}{5}\right) =-\gamma +\frac{1}{2} \sqrt{1-\frac{2}{\sqrt{5}}} \pi -\log (10)+2 \left(\frac{1}{8} \left(-1-\sqrt{5}\right) \log \left(\frac{5}{8}-\frac{\sqrt{5}}{8}\right)+\frac{1}{8} \left(\sqrt{5}-1\right) \log \left(\frac{5}{8}+\frac{\sqrt{5}}{8}\right)\right) </math>
  
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* http://www.google.com/search?hl=en&tbs=tl:1&q=
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<math>\psi\left(\frac{4}{5}\right) =-\gamma +\frac{1}{2} \sqrt{1+\frac{2}{\sqrt{5}}} \pi -\log (10)+2 \left(\frac{1}{8} \left(\sqrt{5}-1\right) \log \left(\frac{5}{8}-\frac{\sqrt{5}}{8}\right)+\frac{1}{8} \left(-1-\sqrt{5}\right) \log \left(\frac{5}{8}+\frac{\sqrt{5}}{8}\right)\right)</math>
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* [[수학사연표 (역사)|수학사연표]]
 
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*  
 
  
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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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<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%;">메모</h5>
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<math>\psi\left(\frac{5}{6}\right) = \frac{\pi}{2}\sqrt{3} -2\ln{2} -\frac{3}{2}\ln(3) - \gamma</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%;">관련된 항목들</h5>
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==역사==
  
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* [[감마함수]]<br>
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* http://www.google.com/search?hl=en&tbs=tl:1&q=
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* [[수학사 연표]]
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* 
  
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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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==메모==
  
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* http://www.google.com/dictionary?langpair=en|ko&q=
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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://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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==관련된 항목들==
  
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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%;">사전 형태의 자료</h5>
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* [[감마함수]]
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* [[디리클레 L-함수]]
  
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* http://ko.wikipedia.org/wiki/
 
  
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* http://ko.wikipedia.org/wiki/
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==매스매티카 파일 및 계산 리소스==
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* http://en.wikipedia.org/wiki/Digamma_function
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* https://docs.google.com/file/d/0B8XXo8Tve1cxS2kzTUV2NTBHX1k/edit
 
* http://www76.wolframalpha.com/input/?i=Digamma+function
 
* http://www76.wolframalpha.com/input/?i=Digamma+function
  
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* http://en.wikipedia.org/wiki/
 
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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: 3.428em; 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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==수학용어번역==
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* http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=di&page=9
  
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* [http://dx.doi.org/10.1016/j.jnt.2009.02.007 Linear independence of digamma function and a variant of a conjecture of Rohrlich]<br>
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** Sanoli Gun, M. Ram Murty, and Purusottam Rath, Journal of Number Theory, Volume 129, Issue 8, August 2009, Pages 1858-1873
 
  
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* http://www.jstor.org/action/doBasicSearch?Query=
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==사전 형태의 자료==
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* http://dx.doi.org/
 
  
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* http://ko.wikipedia.org/wiki/
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* http://en.wikipedia.org/wiki/Digamma_function
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* http://mathworld.wolfram.com/PolygammaFunction.html
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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%;">관련도서</h5>
 
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* [http://books.google.com/books?id=yoGvQAAACAAJ Methods of Summation]<br>
 
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** Bertram Ross
 
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*  도서내검색<br>
 
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** http://books.google.com/books?q=Methods+of+Summation
 
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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: 3.428em; 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%;">관련기사</h5>
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==관련논문==
  
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* 네이버 뉴스 검색 (키워드 수정)<br>
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* [http://dx.doi.org/10.1016/j.jnt.2009.02.007 Linear independence of digamma function and a variant of a conjecture of Rohrlich] Sanoli Gun, M. Ram Murty, and Purusottam Rath, Journal of Number Theory, Volume 129, Issue 8, August 2009, Pages 1858-1873
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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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===Spacy 패턴 목록===
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2021년 2월 17일 (수) 04:01 기준 최신판

개요

  • 감마함수의 로그미분으로 정의



정의와 급수표현

  • 정의\[\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}\]
  • 급수표현\[\psi(z)=-\frac{1}{z} -\gamma +\sum_{n=1}^\infty \frac{z}{n(n+z)} , z \neq 0, -1, -2, -3, \cdots\]

(증명)

감마함수의 무한곱표현

\(\Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}\)

위의 식에 로그미분을 취하여 얻는다. ■


  • \(z = 0, -1, -2, -3, \cdots\) 에서 pole을 가진다



함수의 그래프

  • \(-3<x<3\)일 때, \(\psi(x)\)의 그래프

다이감마 함수(digamma function)1.gif


도함수와 polygamma 함수



차분방정식과의 관계

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

  • 차분방정식의 기본정리를 적용하면\[\sum_{n=a}^{b-1}\frac{1}{n}=\psi(b)-\psi(a)\]
  • 조화급수와의 관계\[\sum_{n=1}^{N}\frac{1}{n}=\psi(N+1)-\psi(1)=\psi(N+1)-\gamma\]
  • 일반화\[\psi^{(n)}(x+1)-\psi^{(n)}(x)=\frac{(-1)^n n!}{x^{n+1}}\]



점근 급수

\[ \begin{align} \psi(x) - \log(x) &= - \frac{1}{2x} - \sum_{n=1}^\infty \frac{B_{2n}}{2n(x^{2n})}\\ &=-\frac{1}{12 x^2}+\frac{1}{120 x^4}-\frac{1}{252 x^6}+\frac{1}{240 x^8}-\frac{1}{132 x^{10}}+\cdots \end{align} \] 또는 \[ \psi(x) = \log(x) - \frac{1}{2x} + \sum_{n=1}^\infty \frac{\zeta(1-2n)}{x^{2n}} \] 여기서 \(B_{n}\)은 베르누이 수



반사공식

  • 감마함수의 반사공식\[\Gamma(1-z) \; \Gamma(z) = {\pi \over \sin{(\pi z)}} \,\!\]
  • 위의 식을 로그미분하여 다음을 얻는다

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

여기서 \(x\)를 \(-x\)로 두면 다음을 얻는다

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



덧셈공식

  • 감마함수의 곱셈공식에 따른 성질\[m\ln m+\psi(z)+ \psi\left(z + \frac{1}{m}\right) + \cdots+ \psi\left(z + \frac{m-1}{m}\right) = m\psi(mz)\]

(증명)

감마함수의 곱셈공식은 적당한 상수 c에 대하여 다음과 같이 쓸 수 있다.

\(m^{mz}\Gamma(z)\cdots \Gamma\left(z + \frac{m-1}{m}\right) = c\Gamma(mz)\)

변수를 x로 바꾸고, 로그를 취하면,

\((m\ln m)x+\ln \Gamma(x) +\ln \Gamma\left(x + \frac{m-1}{m}\right) =\ln c+\ln \Gamma(mx)\)

미분하면,

\(m\ln m+\psi(x)+\cdots+\psi(x+\frac{m-1}{m})=m\psi(mx)\) ■

  • 이항 덧셈공식\[2\psi(2x)=\psi(x)+\psi(x+{1\over2})+2\ln 2\]



가우스의 Digamma 정리

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

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


special values

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

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

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

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

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

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

\(\psi\left(\frac{1}{5}\right) =-\gamma -\frac{1}{2} \sqrt{1+\frac{2}{\sqrt{5}}} \pi -\log (10)+2 \left(\frac{1}{8} \left(\sqrt{5}-1\right) \log \left(\frac{5}{8}-\frac{\sqrt{5}}{8}\right)+\frac{1}{8} \left(-1-\sqrt{5}\right) \log \left(\frac{5}{8}+\frac{\sqrt{5}}{8}\right)\right)\)

\(\psi\left(\frac{2}{5}\right) =-\gamma -\frac{1}{2} \sqrt{1-\frac{2}{\sqrt{5}}} \pi -\log (10)+2 \left(\frac{1}{8} \left(-1-\sqrt{5}\right) \log \left(\frac{5}{8}-\frac{\sqrt{5}}{8}\right)+\frac{1}{8} \left(\sqrt{5}-1\right) \log \left(\frac{5}{8}+\frac{\sqrt{5}}{8}\right)\right) \)

\(\psi\left(\frac{3}{5}\right) =-\gamma +\frac{1}{2} \sqrt{1-\frac{2}{\sqrt{5}}} \pi -\log (10)+2 \left(\frac{1}{8} \left(-1-\sqrt{5}\right) \log \left(\frac{5}{8}-\frac{\sqrt{5}}{8}\right)+\frac{1}{8} \left(\sqrt{5}-1\right) \log \left(\frac{5}{8}+\frac{\sqrt{5}}{8}\right)\right) \)

\(\psi\left(\frac{4}{5}\right) =-\gamma +\frac{1}{2} \sqrt{1+\frac{2}{\sqrt{5}}} \pi -\log (10)+2 \left(\frac{1}{8} \left(\sqrt{5}-1\right) \log \left(\frac{5}{8}-\frac{\sqrt{5}}{8}\right)+\frac{1}{8} \left(-1-\sqrt{5}\right) \log \left(\frac{5}{8}+\frac{\sqrt{5}}{8}\right)\right)\)

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

\(\psi\left(\frac{5}{6}\right) = \frac{\pi}{2}\sqrt{3} -2\ln{2} -\frac{3}{2}\ln(3) - \gamma\)


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