"감마함수"의 두 판 사이의 차이

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<math>\int_0^{\frac{\pi}{2}}\sin^{2n}\theta{d\theta}= \frac{\sqrt{\pi}\Gamma(n+\frac{1}{2})}{2\Gamma(n+1)}=\frac{\pi}{2}\frac{(\frac{1}{2})_n}{(1)_n}</math>
 
  
 
 
 
 
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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>
 
* <math>\Gamma(z) \; \Gamma\left(z + \frac{1}{2}\right) = 2^{\frac{1}{2}-2z} \; \sqrt{2\pi} \; \Gamma(2z) \,\!</math>
 
* <math>\Gamma(z) \; \Gamma\left(z + \frac{1}{m}\right) \; \Gamma\left(z + \frac{2}{m}\right) \cdots \Gamma\left(z + \frac{m-1}{m}\right) = (2 \pi)^{(m-1)/2} \; m^{1/2 - mz} \; \Gamma(mz). \,\!</math>
 
* <math>\Gamma(z) \; \Gamma\left(z + \frac{1}{m}\right) \; \Gamma\left(z + \frac{2}{m}\right) \cdots \Gamma\left(z + \frac{m-1}{m}\right) = (2 \pi)^{(m-1)/2} \; m^{1/2 - mz} \; \Gamma(mz). \,\!</math>
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39번째 줄: 33번째 줄:
 
<math>\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}</math>
 
<math>\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}</math>
  
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* 반사공식
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<math>\psi(1 - x) - \psi(x) = \pi\,\!\cot{ \left ( \pi x \right ) }</math>
 
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*  차분방정식<br><math>\psi(x + 1) = \psi(x) + \frac{1}{x}</math><br>
 
  
 
 
 
 
  
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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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<math>\psi(1) = -\gamma\,\!</math>
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<h5>삼각함수의 적분과 감마함수</h5>
  
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<math>\psi\left(\frac{1}{2}\right) = -2\ln{2} - \gamma</math>
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<math>\int_0^{\frac{\pi}{2}}\sin^{p}\theta{d\theta}= \frac{\sqrt{\pi}}{2} \frac{\Gamma(\frac{p}{2}+\frac{1}{2})}{\Gamma(\frac{p}{2}+1)}</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>\int_0^{\frac{\pi}{2}}\sin^{2n}\theta{d\theta}= \frac{\sqrt{\pi}\Gamma(n+\frac{1}{2})}{2\Gamma(n+1)}=\frac{\pi}{2}\frac{(\frac{1}{2})_n}{(1)_n}</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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<h5>삼각함수의 적분과 감마함수</h5>
 
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<math>\int_0^{\frac{\pi}{2}}\sin^{p}\theta{d\theta}= \frac{\sqrt{\pi}}{2} \frac{\Gamma(\frac{p}{2}+\frac{1}{2})}{\Gamma(\frac{p}{2}+1)}</math>
 
  
 
 
 
 

2009년 7월 5일 (일) 01:03 판

정의
  • \(\Gamma(s) = \int_0^\infty e^{-t} t^{s} \frac{dt}{t}\)
  • \(\Gamma(s+1) =s\Gamma(s)\)

 

 

반사공식
  • \(\Gamma(1-z) \; \Gamma(z) = {\pi \over \sin{(\pi z)}} \,\!\)
  • \(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\)
  • 일반적으로 
    \(\Gamma(n+\frac{1}{2})=(\frac{1}{2})_n\sqrt{\pi}\)
    (증명)

\(\Gamma(n+\frac{1}{2})=\Gamma(\frac{2n+1}{2})=(\frac{2n-1}{2})\Gamma(\frac{2n-1}{2})=(\frac{2n-1}{2})(\frac{2n-3}{2})\Gamma(\frac{2n-3}{2})=(\frac{2n-1}{2})\cdots(\frac{1}{2})\Gamma(\frac{1}{2})=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2n-1}{2}\sqrt{\pi}=(\frac{1}{2})_n\sqrt{\pi}\)

 

 

곱셈공식
  • \(\Gamma(z) \; \Gamma\left(z + \frac{1}{2}\right) = 2^{\frac{1}{2}-2z} \; \sqrt{2\pi} \; \Gamma(2z) \,\!\)
  • \(\Gamma(z) \; \Gamma\left(z + \frac{1}{m}\right) \; \Gamma\left(z + \frac{2}{m}\right) \cdots \Gamma\left(z + \frac{m-1}{m}\right) = (2 \pi)^{(m-1)/2} \; m^{1/2 - mz} \; \Gamma(mz). \,\!\)

 

 

Digamma  함수

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


 


 

삼각함수의 적분과 감마함수

\(\int_0^{\frac{\pi}{2}}\sin^{p}\theta{d\theta}= \frac{\sqrt{\pi}}{2} \frac{\Gamma(\frac{p}{2}+\frac{1}{2})}{\Gamma(\frac{p}{2}+1)}\)

\(\int_0^{\frac{\pi}{2}}\sin^{2n}\theta{d\theta}= \frac{\sqrt{\pi}\Gamma(n+\frac{1}{2})}{2\Gamma(n+1)}=\frac{\pi}{2}\frac{(\frac{1}{2})_n}{(1)_n}\)

 

 

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