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Γ函數

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Γ函數
定義無窮乘積Gamma積分遞歸公式推導遞歸公式重要性質特殊值斯特靈公式解析延拓睇埋出面網頁

Γ 函數，亦叫做伽瑪函數（英文：Gamma function），係一個將階乘推廣到複數上嘅方法。Γ 函數係一個亞純函數，喺複平面上面除咗0同埋負整數，其他地方都係有定義嘅。佢喺理論研究同應用上都有好重要嘅意義。對任何嘅正整數，都有

Γ ( n ) = ( n − 1 ) ! {\displaystyle \Gamma (n)=(n-1)!} {\displaystyle \Gamma (n)=(n-1)!}

Γ ( z ) {\displaystyle \Gamma (z)} {\displaystyle \Gamma (z)}呢個符號係勒壤得揀嘅，個函數嘅定義係：

Γ ( z ) = ∫ 0 ∞ e − t t z − 1 d t ( t > 0 ) {\displaystyle \Gamma (z)=\int _{0}^{\infty }e^{-t}t^{z-1}\mathrm {d} t\quad (t>0)} {\displaystyle \Gamma (z)=\int _{0}^{\infty }e^{-t}t^{z-1}\mathrm {d} t\quad (t>0)}

呢個積分喺實數 z > 0 {\displaystyle z>0} {\displaystyle z>0} 時係絕對收斂，亦可以考慮 z {\displaystyle z} {\displaystyle z} 係複數嘅情形，呢個時候要求 z {\displaystyle z} {\displaystyle z} 嘅實部 R e ( z ) > 0 {\displaystyle \mathrm {Re} (z)>0} {\displaystyle \mathrm {Re} (z)>0}。

Γ 函數可以用無窮乘積嚟表示：

Γ ( z ) = lim n → + ∞ n ! n z z ( z + 1 ) ⋯ ( z + n ) {\displaystyle \Gamma (z)=\lim _{n\to {+\infty }}{\frac {n!\;n^{z}}{z\;(z+1)\cdots (z+n)}}} {\displaystyle \Gamma (z)=\lim _{n\to {+\infty }}{\frac {n!\;n^{z}}{z\;(z+1)\cdots (z+n)}}}
Γ ( z ) = e − γ z z ∏ n = 1 + ∞ ( 1 + z n ) − 1 e z / n {\displaystyle \Gamma (z)={\frac {e^{-\gamma z}}{z}}\prod _{n=1}^{+\infty }\left(1+{\frac {z}{n}}\right)^{-1}e^{z/n}} {\displaystyle \Gamma (z)={\frac {e^{-\gamma z}}{z}}\prod _{n=1}^{+\infty }\left(1+{\frac {z}{n}}\right)^{-1}e^{z/n}}

其中 γ {\displaystyle \gamma } {\displaystyle \gamma } 就係歐拉常數。

1 = ∫ 0 ∞ x ( α − 1 ) λ α e ( − λ x ) Γ ( α ) d x {\displaystyle 1=\int _{0}^{\infty }{\frac {x^{\left(\alpha -1\right)}\lambda ^{\alpha }e^{\left(-\lambda x\right)}}{\Gamma \left(\alpha \right)}}dx} {\displaystyle 1=\int _{0}^{\infty }{\frac {x^{\left(\alpha -1\right)}\lambda ^{\alpha }e^{\left(-\lambda x\right)}}{\Gamma \left(\alpha \right)}}dx}

⇒ Γ ( α ) λ α = ∫ 0 ∞ x α − 1 e − λ x d x {\displaystyle \Rightarrow {\frac {\Gamma \left(\alpha \right)}{\lambda ^{\alpha }}}=\int _{0}^{\infty }x^{\alpha -1}e^{-\lambda x}dx} {\displaystyle \Rightarrow {\frac {\Gamma \left(\alpha \right)}{\lambda ^{\alpha }}}=\int _{0}^{\infty }x^{\alpha -1}e^{-\lambda x}dx}

Γ 函數嘅遞歸公式係：

Γ ( x + 1 ) = x Γ ( x ) {\displaystyle \Gamma (x+1)=x\Gamma (x)} {\displaystyle \Gamma (x+1)=x\Gamma (x)}

對於正整數 n，有

Γ ( n + 1 ) = n ! {\displaystyle \Gamma (n+1)=n!} {\displaystyle \Gamma (n+1)=n!}

可以話Γ 函數係階乘嘅推廣。

推導遞歸公式

Γ ( n + 1 ) = ∫ 0 ∞ e − x x n + 1 − 1 d x = ∫ 0 ∞ e − x x n d x {\displaystyle \Gamma (n+1)=\int _{0}^{\infty }e^{-x}x^{n+1-1}dx=\int _{0}^{\infty }e^{-x}x^{n}dx} {\displaystyle \Gamma (n+1)=\int _{0}^{\infty }e^{-x}x^{n+1-1}dx=\int _{0}^{\infty }e^{-x}x^{n}dx}

用分部積分法嚟計呢個積分：

∫ 0 ∞ e − x x n d x = [ − x n e x ] 0 ∞ + n ∫ 0 ∞ e − x x n − 1 d x {\displaystyle \int _{0}^{\infty }e^{-x}x^{n}dx=\left[{\frac {-x^{n}}{e^{x}}}\right]_{0}^{\infty }+n\int _{0}^{\infty }e^{-x}x^{n-1}dx} {\displaystyle \int _{0}^{\infty }e^{-x}x^{n}dx=\left[{\frac {-x^{n}}{e^{x}}}\right]_{0}^{\infty }+n\int _{0}^{\infty }e^{-x}x^{n-1}dx}

當 x = 0 時， − 0 n e 0 = 0 1 = 0 {\displaystyle {\frac {-0^{n}}{e^{0}}}={\frac {0}{1}}=0} {\displaystyle {\frac {-0^{n}}{e^{0}}}={\frac {0}{1}}=0}。當 x 趨於無窮大時，根據洛必達法則，有：

lim x → ∞ − x n e x = lim x → ∞ − n ! ⋅ 0 e x = 0 {\displaystyle \lim _{x\rightarrow \infty }{\frac {-x^{n}}{e^{x}}}=\lim _{x\rightarrow \infty }{\frac {-n!\cdot 0}{e^{x}}}=0} {\displaystyle \lim _{x\rightarrow \infty }{\frac {-x^{n}}{e^{x}}}=\lim _{x\rightarrow \infty }{\frac {-n!\cdot 0}{e^{x}}}=0}.

因此第一項 [ − x n e x ] 0 ∞ {\displaystyle \left[{\frac {-x^{n}}{e^{x}}}\right]_{0}^{\infty }} {\displaystyle \left[{\frac {-x^{n}}{e^{x}}}\right]_{0}^{\infty }}變咗零，所以：

Γ ( n + 1 ) = n ∫ 0 ∞ e − x x n − 1 d x {\displaystyle \Gamma (n+1)=n\int _{0}^{\infty }e^{-x}x^{n-1}dx} {\displaystyle \Gamma (n+1)=n\int _{0}^{\infty }e^{-x}x^{n-1}dx}

等式嘅右面啱啱就係n Γ ( n ) {\displaystyle \Gamma (n)} {\displaystyle \Gamma (n)}。所以遞歸公式係：

Γ ( n + 1 ) = n Γ ( n ) {\displaystyle \Gamma (n+1)=n\Gamma (n)} {\displaystyle \Gamma (n+1)=n\Gamma (n)}。
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Γ 函數喺實軸上嘅函數圖形
  • 當 z → 0 + {\displaystyle z\to 0^{+}} {\displaystyle z\to 0^{+}}時， Γ ( z ) → + ∞ {\displaystyle \Gamma (z)\to +\infty } {\displaystyle \Gamma (z)\to +\infty }
  • 歐拉反射公式：
Γ ( z ) Γ ( 1 − z ) = π sin ⁡ π z ( 0 < R e ( z ) < 1 ) {\displaystyle \Gamma (z)\Gamma (1-z)={\frac {\pi }{\sin {\pi z}}}\quad (0<\mathrm {Re} (z)<1)} {\displaystyle \Gamma (z)\Gamma (1-z)={\frac {\pi }{\sin {\pi z}}}\quad (0<\mathrm {Re} (z)<1)}
由上面條式可以知道當 z = 1/2 時， Γ ( 1 2 ) = π {\displaystyle \Gamma \left({\frac {1}{2}}\right)={\sqrt {\pi }}} {\displaystyle \Gamma \left({\frac {1}{2}}\right)={\sqrt {\pi }}}。
  • 乘法定理：
Γ ( z ) Γ ( z + 1 2 ) = 2 1 − 2 z π Γ ( 2 z ) . {\displaystyle \Gamma (z)\;\Gamma \left(z+{\frac {1}{2}}\right)=2^{1-2z}\;{\sqrt {\pi }}\;\Gamma (2z).} {\displaystyle \Gamma (z)\;\Gamma \left(z+{\frac {1}{2}}\right)=2^{1-2z}\;{\sqrt {\pi }}\;\Gamma (2z).}
Γ ( z ) Γ ( z + 1 m ) Γ ( z + 2 m ) ⋯ Γ ( z + m − 1 m ) = ( 2 π ) ( m − 1 ) / 2 m 1 / 2 − m z Γ ( m z ) . {\displaystyle \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).} {\displaystyle \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).}
  • 補充：
Γ ( n + 1 2 ) = ( 2 n ) ! π n ! 4 n {\displaystyle \Gamma \left(n+{\frac {1}{2}}\right)={\frac {(2n)!{\sqrt {\pi }}}{n!4^{n}}}} {\displaystyle \Gamma \left(n+{\frac {1}{2}}\right)={\frac {(2n)!{\sqrt {\pi }}}{n!4^{n}}}}
呢條式可以用嚟協助計算 t 分布機率密度函數、卡方分布機率密度函數、F 分布機率密度函數等嘅累計機率。
  • 其他用乘法定理計到嘅數：
Γ ( 1 / 6 ) = Γ ( 1 / 3 ) 2 / π ∗ 2 2 / 3 ∗ sin ⁡ ( π / 3 ) {\displaystyle \Gamma (1/6)=\Gamma (1/3)^{2}/{\sqrt {\pi }}*2^{2/3}*\sin({\pi /3})} {\displaystyle \Gamma (1/6)=\Gamma (1/3)^{2}/{\sqrt {\pi }}*2^{2/3}*\sin({\pi /3})}
Γ ( 5 / 6 ) = 1 / Γ ( 1 / 3 ) 2 ∗ π 3 ∗ 2 4 / 3 / 3 {\displaystyle \Gamma (5/6)=1/\Gamma (1/3)^{2}*{\sqrt {\pi }}^{3}*2^{4/3}/{\sqrt {3}}} {\displaystyle \Gamma (5/6)=1/\Gamma (1/3)^{2}*{\sqrt {\pi }}^{3}*2^{4/3}/{\sqrt {3}}}
Γ ( 1 / 10 ) = Γ ( 1 / 5 ) ∗ Γ ( 2 / 5 ) / π ∗ 2 4 / 5 ∗ sin ⁡ ( 2 ∗ π / 5 ) {\displaystyle \Gamma (1/10)=\Gamma (1/5)*\Gamma (2/5)/{\sqrt {\pi }}*2^{4/5}*\sin({2*\pi /5})} {\displaystyle \Gamma (1/10)=\Gamma (1/5)*\Gamma (2/5)/{\sqrt {\pi }}*2^{4/5}*\sin({2*\pi /5})}
Γ ( 3 / 10 ) = Γ ( 1 / 5 ) / Γ ( 2 / 5 ) ∗ π / 2 3 / 5 / sin ⁡ ( 3 ∗ π / 10 ) {\displaystyle \Gamma (3/10)=\Gamma (1/5)/\Gamma (2/5)*{\sqrt {\pi }}/2^{3/5}/\sin({3*\pi /10})} {\displaystyle \Gamma (3/10)=\Gamma (1/5)/\Gamma (2/5)*{\sqrt {\pi }}/2^{3/5}/\sin({3*\pi /10})}
Γ ( 7 / 10 ) = Γ ( 2 / 5 ) / Γ ( 1 / 5 ) ∗ π ∗ 2 3 / 5 {\displaystyle \Gamma (7/10)=\Gamma (2/5)/\Gamma (1/5)*{\sqrt {\pi }}*2^{3/5}} {\displaystyle \Gamma (7/10)=\Gamma (2/5)/\Gamma (1/5)*{\sqrt {\pi }}*2^{3/5}}
Γ ( 9 / 10 ) = 1 / ( Γ ( 1 / 5 ) ∗ Γ ( 2 / 5 ) ) ∗ π 3 / 2 4 / 5 / ( sin ⁡ ( π / 10 ) ∗ sin ⁡ ( 2 ∗ π / 5 ) ) {\displaystyle \Gamma (9/10)=1/(\Gamma (1/5)*\Gamma (2/5))*{\sqrt {\pi }}^{3}/2^{4/5}/(\sin(\pi /10)*\sin({2*\pi /5}))} {\displaystyle \Gamma (9/10)=1/(\Gamma (1/5)*\Gamma (2/5))*{\sqrt {\pi }}^{3}/2^{4/5}/(\sin(\pi /10)*\sin({2*\pi /5}))}

[1]

Γ ( − 3 / 2 ) = 4 π 3 ≈ 2.363 Γ ( − 1 / 2 ) = − 2 π ≈ − 3.545 Γ ( 1 / 2 ) = π ≈ 1.772 Γ ( 1 ) = 0 ! = 1 Γ ( 3 / 2 ) = π 2 ≈ 0.886 Γ ( 2 ) = 1 ! = 1 Γ ( 5 / 2 ) = 3 π 4 ≈ 1.329 Γ ( 3 ) = 2 ! = 2 Γ ( 7 / 2 ) = 15 π 8 ≈ 3.323 Γ ( 4 ) = 3 ! = 6 {\displaystyle {\begin{array}{lll}\Gamma (-3/2)&={\frac {4{\sqrt {\pi }}}{3}}&\approx 2.363\\\Gamma (-1/2)&=-2{\sqrt {\pi }}&\approx -3.545\\\Gamma (1/2)&={\sqrt {\pi }}&\approx 1.772\\\Gamma (1)&=0!&=1\\\Gamma (3/2)&={\frac {\sqrt {\pi }}{2}}&\approx 0.886\\\Gamma (2)&=1!&=1\\\Gamma (5/2)&={\frac {3{\sqrt {\pi }}}{4}}&\approx 1.329\\\Gamma (3)&=2!&=2\\\Gamma (7/2)&={\frac {15{\sqrt {\pi }}}{8}}&\approx 3.323\\\Gamma (4)&=3!&=6\\\end{array}}} {\displaystyle {\begin{array}{lll}\Gamma (-3/2)&={\frac {4{\sqrt {\pi }}}{3}}&\approx 2.363\\\Gamma (-1/2)&=-2{\sqrt {\pi }}&\approx -3.545\\\Gamma (1/2)&={\sqrt {\pi }}&\approx 1.772\\\Gamma (1)&=0!&=1\\\Gamma (3/2)&={\frac {\sqrt {\pi }}{2}}&\approx 0.886\\\Gamma (2)&=1!&=1\\\Gamma (5/2)&={\frac {3{\sqrt {\pi }}}{4}}&\approx 1.329\\\Gamma (3)&=2!&=2\\\Gamma (7/2)&={\frac {15{\sqrt {\pi }}}{8}}&\approx 3.323\\\Gamma (4)&=3!&=6\\\end{array}}}

斯特靈公式可以用嚟估計 Γ 函數嘅增長速度：

Γ ( x + 1 ) ∼ 2 π x ( x e ) x {\displaystyle \Gamma (x+1)\sim {\sqrt {2\pi x}}\left({\frac {x}{e}}\right)^{x}} {\displaystyle \Gamma (x+1)\sim {\sqrt {2\pi x}}\left({\frac {x}{e}}\right)^{x}}
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Γ 函數嘅絕對值函數圖形

注意到喺 Γ 函數的積分定義當中如果攞 z {\displaystyle z} {\displaystyle z} 嚟做實部大於零嘅複數、則積分存在，而且喺右半複平面上定義一個全純函數。利用函數方程

Γ ( z ) Γ ( 1 − z ) = π sin ⁡ π z ( 0 < R e ( z ) < 1 ) {\displaystyle \Gamma (z)\Gamma (1-z)={\frac {\pi }{\sin {\pi z}}}\quad (0<\mathrm {Re} (z)<1)} {\displaystyle \Gamma (z)\Gamma (1-z)={\frac {\pi }{\sin {\pi z}}}\quad (0<\mathrm {Re} (z)<1)}

並注意到函數 sin ⁡ ( π z ) {\displaystyle \sin(\pi z)} {\displaystyle \sin(\pi z)} 係成個複平面上有解析延拓，我地可以喺 R e ( z ) < 1 {\displaystyle \mathrm {Re} (z)<1} {\displaystyle \mathrm {Re} (z)<1} 時設

Γ ( z ) := π Γ ( 1 − z ) sin ⁡ π z {\displaystyle \Gamma (z):={\dfrac {\pi }{\Gamma (1-z)\sin {\pi z}}}} {\displaystyle \Gamma (z):={\dfrac {\pi }{\Gamma (1-z)\sin {\pi z}}}}

從而將 Γ 函數延拓為成個複平面上嘅亞純函數，佢喺 z = 0 , − 1 , − 2 , − 3 ⋯ {\displaystyle z=0,-1,-2,-3\cdots } {\displaystyle z=0,-1,-2,-3\cdots } 有單極點，留數係

R e s ( Γ , − n ) = ( − 1 ) n n ! {\displaystyle \mathrm {Res} (\Gamma ,-n)={\dfrac {(-1)^{n}}{n!}}} {\displaystyle \mathrm {Res} (\Gamma ,-n)={\dfrac {(-1)^{n}}{n!}}}
  • 雙伽瑪函數
  • 多伽瑪函數
  • 階乘
  • 機率密度函數
  • GAMMA 函數嘅性質
  1. [1]
    Mada, L. (2020-04-24). "Relations of the Gamma function". R code on Github. Code publicly available on Github [Personal Research]. 原先內容歸檔喺2021-04-02. 喺2020-04-24搵到. Relations of the Gamma function
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定義

無窮乘積

Gamma積分

遞歸公式

重要性質

特殊值

斯特靈公式

解析延拓

睇埋

出面網頁