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伽瑪函式


GammaFunction

(完全)伽瑪函式 Gamma(n) 被定義為 階乘複數實數 變數的擴充套件。它與 階乘 的關係為

 Gamma(n)=(n-1)!,
(1)

勒讓德引入了一個稍微不幸的符號,現在被普遍使用,而不是高斯更簡單的 Pi(n)=n! (Gauss 1812; Edwards 2001, p. 8)。

除了在 z=0, -1, -2, ... 之外,它在所有地方都是 解析 的,並且在 z=-k 的留數為

 Res_(z=-k)Gamma(z)=((-1)^k)/(k!).
(2)

沒有 z 點使得 Gamma(z)=0

伽瑪函式在 Wolfram 語言 中實現為Gamma[z]。

對於伽瑪函式的冪,有許多常用的符號慣例。雖然像 Watson (1939) 這樣的作者使用 Gamma^n(z) (即,使用類似於三角函式的慣例),但通常也寫成 [Gamma(z)]^n

伽瑪函式可以定義為 R[z]>0定積分 (尤拉積分形式)

Gamma(z)=int_0^inftyt^(z-1)e^(-t)dt
(3)
=2int_0^inftye^(-t^2)t^(2z-1)dt,
(4)

 Gamma(z)=int_0^1[ln(1/t)]^(z-1)dt.
(5)

完全伽瑪函式 Gamma(x) 可以推廣到上 不完全伽瑪函式 Gamma(a,x) 和下 不完全伽瑪函式 gamma(a,x)

GammaReImAbs
最小值 最大值
實部
虛部 Powered by webMathematica

上面展示了複平面中 Gamma(z) 的實部和虛部的圖。

對於 實數 變數,透過對公式 (3) 進行分部積分,可以看出

Gamma(x)=int_0^inftyt^(x-1)e^(-t)dt
(6)
=[-t^(x-1)e^(-t)]_0^infty+int_0^infty(x-1)t^(x-2)e^(-t)dt
(7)
=(x-1)int_0^inftyt^(x-2)e^(-t)dt
(8)
=(x-1)Gamma(x-1).
(9)

如果 x 是一個 整數 n=1, 2, 3, ...,則

Gamma(n)=(n-1)Gamma(n-1)
(10)
=(n-1)(n-2)Gamma(n-2)
(11)
=(n-1)(n-2)...1
(12)
=(n-1)!,
(13)

因此,對於 正整數 變數,伽瑪函式簡化為 階乘

Gamma(z)黎曼 zeta 函式 zeta(z) 之間存在一個優美的關係,由下式給出

 zeta(z)Gamma(z)=int_0^infty(u^(z-1))/(e^u-1)du
(14)

對於 R[z]>1 (Havil 2003, p. 60)。

伽瑪函式也可以透過 無窮乘積 形式定義 (Weierstrass 形式)

 Gamma(z)=[ze^(gammaz)product_(r=1)^infty(1+z/r)e^(-z/r)]^(-1),
(15)

其中 gamma尤拉-馬歇羅尼常數 (Krantz 1999, p. 157; Havil 2003, p. 57)。對 (◇) 兩邊取對數,

 -ln[Gamma(z)]=lnz+gammaz+sum_(n=1)^infty[ln(1+z/n)-z/n].
(16)

求導,

-(Gamma^'(z))/(Gamma(z))=1/z+gamma+sum_(n=1)^(infty)((1/n)/(1+z/n)-1/n)
(17)
=1/z+gamma+sum_(n=1)^(infty)(1/(n+z)-1/n)
(18)
Gamma^'(z)=-Gamma(z)[1/z+gamma+sum_(n=1)^(infty)(1/(n+z)-1/n)]
(19)
=Gamma(z)Psi(z)
(20)
=Gamma(z)psi_0(z)
(21)
Gamma^'(1)=-Gamma(1){1+gamma+[(1/2-1)+(1/3-1/2)+...+(1/(n+1)-1/n)+...]}
(22)
=-(1+gamma-1)
(23)
=-gamma
(24)
Gamma^'(n)=-Gamma(n){1/n+gamma+[(1/(1+n)-1)+(1/(2+n)-1/2)+(1/(3+n)-1/3)+...]}
(25)
=-(n-1)!(1/n+gamma-sum_(k=1)^(n)1/k),
(26)

其中 Psi(z)雙伽瑪函式psi_0(z)多伽瑪函式n 階導數以 多伽瑪函式 psi_n, psi_(n-1), ..., psi_0 的形式給出。

實數 x=x_0Gamma(x) 的最小值 x_0 在以下情況下達到

 Gamma^'(x_0)=Gamma(x_0)psi_0(x_0)=0
(27)
 psi_0(x_0)=0.
(28)

這可以透過數值求解得到 x_0=1.46163... (OEIS A030169; Wrench 1968),其 連分數 為 [1, 2, 6, 63, 135, 1, 1, 1, 1, 4, 1, 38, ...] (OEIS A030170)。在 x_0 處,Gamma(x_0) 達到值 0.8856031944... (OEIS A030171),其 連分數 為 [0, 1, 7, 1, 2, 1, 6, 1, 1, ...] (OEIS A030172)。

尤拉極限形式為

 Gamma(z)=1/zproduct_(n=1)^infty[(1+1/n)^z(1+z/n)^(-1)],
(29)

所以

Gamma(z)=lim_(n->infty)((n+1)^z)/(z(1+z)(1+z/2)(1+z/3)...(1+z/n))
(30)
=lim_(n->infty)((n+1)^zn!)/(z(z+1)(z+2)(z+3)...(z+n))
(31)
=lim_(n->infty)(n!)/((z)_(n+1))(n+1)^z
(32)
=lim_(n->infty)(n!)/((z)_(n+1))n^z
(33)

(Krantz 1999, p. 156)。

伽瑪函式的倒數 1/Gamma(z) 是一個 整函式,可以表示為

 1/(Gamma(z))=zexp[gammaz-sum_(k=2)^infty((-1)^kzeta(k)z^k)/k],
(34)

其中 gamma尤拉-馬歇羅尼常數zeta(z)黎曼 zeta 函式 (Wrench 1968)。 1/Gamma(z)漸近級數 由下式給出

 1/(Gamma(z))∼z+gammaz^2+1/(12)(6gamma^2-pi^2)z^3+1/(12)[2gamma^3-gammapi^2+4zeta(3)]z^4+....
(35)

寫作

 1/(Gamma(z))=sum_(k=1)^inftya_kz^k,
(36)

a_k 滿足

 a_n=na_1a_n-a_2a_(n-1)+sum_(k=2)^n(-1)^kzeta(k)a_(n-k)
(37)

(Bourguet 1883, Davis 1933, Isaacson and Salzer 1943, Wrench 1968)。Wrench (1968) 數值計算了以下函式在 0 附近的級數展開的係數

 1/(z(1+z)Gamma(z))=1+(gamma-1)z+[1+1/2(gamma-2)gamma-1/(12)pi^2]z^2+....
(38)

Lanczos 近似 給出了 Gamma(z+1) 對於 z>0 的級數展開,以任意常數 sigma 表示,使得 R[z+sigma+1/2]>0

伽瑪函式滿足以下 函式方程

Gamma(1+z)=zGamma(z)
(39)
Gamma(1-z)=-zGamma(-z).
(40)

其他恆等式為

Gamma(x)Gamma(-x)=-pi/(xsin(pix))
(41)
Gamma(x)Gamma(1-x)=pi/(sin(pix))
(42)
|(ix)!|^2=(pix)/(sinh(pix))
(43)
|(n+ix)!|=sqrt((pix)/(sinh(pix)))product_(s=1)^(n)sqrt(s^2+x^2).
(44)

使用 (41),有理數 r 的伽瑪函式 Gamma(r) 可以簡化為一個常數乘以 Gamma(frac(r))1/Gamma(frac(r))。例如,

Gamma(2/3)=(2pi)/(sqrt(3)Gamma(1/3))
(45)
Gamma(3/4)=(sqrt(2)pi)/(Gamma(1/4))
(46)
Gamma(3/5)=sqrt(2-2/(sqrt(5)))pi/(Gamma(2/5))
(47)
Gamma(4/5)=sqrt(2+2/(sqrt(5)))pi/(Gamma(1/5)).
(48)

對於 R[z]=-1/2,

 |(-1/2+iy)!|^2=pi/(cosh(piy)).
(49)

變數為 2z 的伽瑪函式可以使用 Legendre 倍乘公式 表示

 Gamma(2z)=(2pi)^(-1/2)2^(2z-1/2)Gamma(z)Gamma(z+1/2).
(50)

變數為 3z 的伽瑪函式可以使用三倍公式表示

 Gamma(3z)=(2pi)^(-1)3^(3z-1/2)Gamma(z)Gamma(z+1/3)Gamma(z+2/3).
(51)

一般結果是 Gauss 乘法公式

 Gamma(z)Gamma(z+1/n)...Gamma(z+(n-1)/n)=(2pi)^((n-1)/2)n^(1/2-nz)Gamma(nz).
(52)

伽瑪函式也透過下式與 黎曼 zeta 函式 zeta(z) 相關

 Gamma(s/2)pi^(-s/2)zeta(s)=Gamma((1-s)/2)pi^(-(1-s)/2)zeta(1-s).
(53)

對於整數 n=1, 2, ...,Gamma(n) 的前幾個值為 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, ... (OEIS A000142)。對於半整數變數,Gamma(n/2) 具有特殊形式

 Gamma(1/2n)=((n-2)!!sqrt(pi))/(2^((n-1)/2)),
(54)

其中 n!!雙階乘n=1, 3, 5, ... 的前幾個值因此為

Gamma(1/2)=sqrt(pi)
(55)
Gamma(3/2)=1/2sqrt(pi)
(56)
Gamma(5/2)=3/4sqrt(pi),
(57)

15sqrt(pi)/8, 105sqrt(pi)/16, ... (OEIS A001147A000079; Wells 1986, p. 40)。一般來說,對於 n正整數 n=1, 2, ...

Gamma(1/2+n)=(1·3·5...(2n-1))/(2^n)sqrt(pi)
(58)
=((2n-1)!!)/(2^n)sqrt(pi)
(59)
Gamma(1/2-n)=((-1)^n2^n)/(1·3·5...(2n-1))sqrt(pi)
(60)
=((-1)^n2^n)/((2n-1)!!)sqrt(pi).
(61)

對於 Gamma(1/n),當 n 為正整數 n>2 時,似乎不存在這種型別的簡單閉式表示式。然而,Borwein 和 Zucker (1992) 給出了各種恆等式,將伽瑪函式與平方根和 橢圓積分奇異值 k_n 相關聯,即,橢圓模量 k_n 使得

 (K^'(k_n))/(K(k_n))=sqrt(n),
(62)

其中 K(k)第一類完全橢圓積分K^'(k)=K(k^')=K(sqrt(1-k^2)) 是互補積分。M. Trott(私人通訊)開發了一種演算法,用於自動生成數百個這樣的恆等式。

Gamma(1/3)=2^(7/9)3^(-1/12)pi^(1/3)[K(k_3)]^(1/3)
(63)
Gamma(1/4)=2pi^(1/4)[K(k_1)]^(1/2)
(64)
Gamma(1/6)=2^(-1/3)3^(1/2)pi^(-1/2)[Gamma(1/3)]^2
(65)
Gamma(1/8)Gamma(3/8)=(sqrt(2)-1)^(1/2)2^(13/4)pi^(1/2)K(k_2)
(66)
(Gamma(1/8))/(Gamma(3/8))=2(sqrt(2)+1)^(1/2)pi^(-1/4)[K(k_1)]^(1/2)
(67)
Gamma(1/(12))=2^(-1/4)3^(3/8)(sqrt(3)+1)^(1/2)pi^(-1/2)Gamma(1/4)Gamma(1/3)
(68)
Gamma(5/(12))=2^(1/4)3^(-1/8)(sqrt(3)-1)^(1/2)pi^(1/2)(Gamma(1/4))/(Gamma(1/3))
(69)
(Gamma(1/(24))Gamma((11)/(24)))/(Gamma(5/(24))Gamma(7/(24)))=sqrt(3)sqrt(2+sqrt(3))
(70)
(Gamma(1/(24))Gamma(5/(24)))/(Gamma(7/(24))Gamma((11)/(24)))=4·3^(1/4)(sqrt(3)+sqrt(2))pi^(-1/2)K(k_1)
(71)
(Gamma(1/(24))Gamma(7/(24)))/(Gamma(5/(24))Gamma((11)/(24)))=2^(25/18)3^(1/3)(sqrt(2)+1)pi^(-1/3)[K(k_3)]^(2/3)
(72)
Gamma(1/(24))Gamma(5/(24))Gamma(7/(24))Gamma((11)/(24))=384(sqrt(2)+1)(sqrt(3)-sqrt(2))(2-sqrt(3))pi[K(k_6)]^2
(73)
Gamma(1/(10))=2^(-7/10)5^(1/4)(sqrt(5)+1)^(1/2)pi^(-1/2)Gamma(1/5)Gamma(2/5)
(74)
Gamma(3/(10))=2^(-3/5)(sqrt(5)-1)pi^(1/2)(Gamma(1/5))/(Gamma(2/5))
(75)
(Gamma(1/(15))Gamma(4/(15))Gamma(7/(15)))/(Gamma(2/(15)))=2·3^(1/2)5^(1/6)sin(2/(15)pi)[Gamma(1/3)]^2
(76)
(Gamma(1/(15))Gamma(2/(15))Gamma(7/(15)))/(Gamma(4/(15)))=2^2·3^(2/5)sin(1/5pi)sin(4/(15)pi)[Gamma(1/5)]^2
(77)
(Gamma(2/(15))Gamma(4/(15))Gamma(7/(15)))/(Gamma(1/(15)))=(2^(-3/2)3^(-1/5)5^(1/4)(sqrt(5)-1)^(1/2)[Gamma(2/5)]^2)/(sin(4/(15)pi))
(78)
(Gamma(1/(15))Gamma(2/(15))Gamma(4/(15)))/(Gamma(7/(15)))=60(sqrt(5)-1)sin(7/(15)pi)[K(k_(15))]^2
(79)
(Gamma(1/(20))Gamma(9/(20)))/(Gamma(3/(20))Gamma(7/(20)))=2^(-1)5^(1/4)(sqrt(5)+1)
(80)
(Gamma(1/(20))Gamma(3/(20)))/(Gamma(7/(20))Gamma(9/(20)))=2^(4/5)(10-2sqrt(5))^(1/2)pi^(-1)sin(7/(20)pi)sin(9/(20)pi)[Gamma(1/5)]^2
(81)
(Gamma(1/(20))Gamma(7/(20)))/(Gamma(3/(20))Gamma(9/(20)))=2^(3/5)(10+2sqrt(5))^(1/2)pi^(-1)sin(3/(20)pi)sin(9/(20)pi)[Gamma(2/5)]^2
(82)
Gamma(1/(20))Gamma(3/(20))Gamma(7/(20))Gamma(9/(20))=160(sqrt(5)-2)^(1/2)pi[K(k_5)]^2.
(83)

Campbell (1966, p. 31) 也給出了一些。

一些有趣的恆等式包括

product_(n=1)^(2)Gamma(1/3n)=(2pi)/(sqrt(3))
(84)
product_(n=1)^(3)Gamma(1/3n)=(2pi)/(sqrt(3))
(85)
product_(n=1)^(4)Gamma(1/3n)=(2piGamma(1/3))/(3sqrt(3))
(86)
product_(n=1)^(5)Gamma(1/3n)=8/(27)pi^2
(87)
product_(n=1)^(6)Gamma(1/3n)=8/(27)pi^2
(88)
product_(n=1)^(7)Gamma(1/3n)=(32)/(243)pi^2Gamma(1/3)
(89)
product_(n=1)^(8)Gamma(1/3n)=(640pi^3)/(2187sqrt(3)),
(90)

其中 Magnus 和 Oberhettinger (1949, p. 1) 僅給出了最後一種情況,以及

 (Gamma^'(1))/(Gamma(1))-(Gamma^'(1/2))/(Gamma(1/2))=2ln2
(91)

(Magnus and Oberhettinger 1949, p. 1)。Ramanujan 也給出了一些引人入勝的恆等式

 (Gamma^2(n+1))/(Gamma(n+xi+1)Gamma(n-xi+1))=product_(k=1)^infty[1+(x^2)/((n+k)^2)]
(92)
 phi(m,n)phi(n,m)=(Gamma^3(m+1)Gamma^3(n+1))/(Gamma(2m+n+1)Gamma(2n+m+1)) 
 ×(cosh[pi(m+n)sqrt(3)]-cos[pi(m-n)])/(2pi^2(m^2+mn+n^2)),
(93)

其中

 phi(m,n)=product_(k=1)^infty[1+((m+n)/(k+m))^3],
(94)
 product_(k=1)^infty[1+(n/k)^3]product_(k=1)^infty[1+3(n/(n+2k))^2]=(Gamma(1/2n))/(Gamma[1/2(n+1)])(cosh(pinsqrt(3))-cos(pin))/(2^(n+2)pi^(3/2)n)
(95)

(Berndt 1994)。

Ramanujan 給出了無窮級數

 sum_(k=0)^infty(8k+1)[(Gamma(k+1/4))/(k!Gamma(1/4))]^4 
=1+9(1/4)^4+17((1·5)/(4·8))^4+25((1·5·9)/(4·8·12))^4+... 
=(2^(3/2))/(sqrt(pi)[Gamma(3/4)]^2)
(96)

 sum_(k=0)^infty(-1)^k(4k+1)[((2k-1)!!)/((2k)!!)]^5 
=1-5(1/2)^5+9((1·3)/(2·4))^5-13((1·3·5)/(2·4·6))^5+... 
=2/([Gamma(3/4)]^4)
(97)

(Hardy 1923; Hardy 1924; Whipple 1926; Watson 1931; Bailey 1935; Hardy 1999, p. 7)。

以下 漸近級數 在機率論中偶爾有用(例如,一維隨機遊走

 (Gamma(J+1/2))/(Gamma(J))=sqrt(J)(1-1/(8J)+1/(128J^2)+5/(1024J^3)-(21)/(32768J^4)+...)
(98)

(OEIS A143503A061549; Graham et al. 1994)。該級數還給出了 第一類斯特林數 到分數值的良好漸近推廣。

長期以來人們都知道 Gamma(1/4)pi^(-1/4)超越數 (Davis 1959),Gamma(1/3) 也是 (Le Lionnais 1983; Borwein and Bailey 2003, p. 138),並且 Chudnovsky 最近顯然證明了 Gamma(1/4) 本身是 超越數 (Borwein and Bailey 2003, p. 138)。

存在有效的迭代演算法來計算所有整數 kGamma(k/24) (Borwein and Bailey 2003, p. 137)。例如,Gamma(1/4)=3.6256099... (OEIS A068466) 的二次收斂迭代由定義給出

x_n=1/2(x_(n-1)^(1/2)+x_(n-1)^(-1/2))
(99)
y_n=(y_(n-1)x_(n-1)^(1/2)+x_(n-1)^(-1/2))/(y_(n-1)+1),
(100)

設定 x_0=sqrt(2)y_1=2^(1/4),然後

 Gamma(1/4)=2(1+sqrt(2))^(3/4)[product_(n=1)^inftyx_n^(-1)((1+x_n)/(1+y_n))^3]^(1/4)
(101)

(Borwein and Bailey 2003, pp. 137-138)。

對於 Gamma(1/5),尚不清楚是否存在這樣的迭代 (Borwein and Borwein 1987; Borwein and Zucker 1992; Borwein and Bailey 2003, p. 138)。


另請參閱

Bailey 定理, Barnes G 函式, Binet 的斐波那契數公式, Bohr-Mollerup 定理, 雙伽瑪函式, Fransén-Robinson 常數 Gauss 乘法公式, 不完全伽瑪函式, Knar 公式, Lambda 函式, Lanczos 近似, Legendre 倍乘公式, 對數伽瑪函式, Mellin 公式, Mu 函式, Nu 函式, Pearson 函式, 多伽瑪函式, 正則化伽瑪函式, Stirling 級數, 超階乘 在 課堂中探索此主題

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參考文獻

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