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The Gamma Function

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The Gamma Function The Gamma Function EMILY GULLERUD, michael vaughan University of Wisconsin - Eau Claire May 10, 2017 EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 1 / 15 What is the Gamma Function? Motivation The gamma function (Γ) is an extension of the factorial function to − C n Z [ 0. It is a solution to the interpolation problem of connecting the discrete points of the factorial function. EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 2 / 15 What is the Gamma Function? Definitions Factorial function definition + If n 2 Z , then Γ(n) = (n − 1)! Improper integral definition R 1 z−1 −x If z 2 fx + iy j x > 0g, then Γ(z) = 0 x e dx. Analytic continuation is required to extend the integral definition to − z 2 fx + iy j y 6= 0 when x 2 Z [ 0g. This function has simple poles at all the non-positive integers. EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 3 / 15 What is the Gamma Function? More Definitions Definition as an infinite product 1 1 z 1 Y 1 + Γ(z) = n z 1 + z n=1 n Weierstrass's definition 1 e−γz Y z −1 Γ(z) = 1 + ez=n z n n=1 EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 4 / 15 Properties Identities and Formulas Requirements for Γ to be an extension of the factorial function: Γ(1) = 1 zΓ(z) = Γ(z + 1) for Re(z) > 0 Property of conjugation: Γ(z) = Γ(z) Complement Formula: π Γ(z)Γ(1 − z) = for z 2= sin(πz) Z Duplication Formula: 1 1−2z p Γ(z)Γ(z + 2 ) = Γ(2z) · 2 π EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 5 / 15 Properties Proof of zΓ(z) = Γ(z + 1) Let α 2 C such that Re(α) > 0 Z 1 Γ(α) = xα−1e−x dx 0 1 Z 1 α−1 −x −x α−2 = x (−e ) − −e (α − 1)x dx 0 0 α−1 Z 1 x −x α−2 = − lim x + e (α − 1)x dx x!1 e 0 Z 1 = 0 + (α − 1) e−x xα−1−1dx 0 Γ(α) = (α − 1)Γ(α − 1) Take z = α − 1. Γ(z + 1) = zΓ(z) EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 6 / 15 Properties Values of the Gamma Function 3 4p Γ − = π 2 3 1 p Γ − = −2 π 2 Γ(1) = 1 1 p Γ(2) = 1 Γ = π 2 Γ(3) = 2 3 1p Γ = π Γ(4) = 6 2 2 Γ(5) = 24 5 3p Γ = π 2 4 7 15p Γ = π 2 8 EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 7 / 15 Properties p 1 π Example: Proving 2 ! = 2 1 3 ! = Γ 2 2 1 1 = Γ 2 2 1=2 Z 1 Let u = t 1 −1=2 −t 1 −1=2 = t e dt Then du = 2 t dt 2 0 1 Z 2 = e−u p0 π = 2 EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 8 / 15 Applications and Connections to Other Functions The log-gamma function The Gamma function grows rapidly, so taking the natural logarithm yields a function which grows much more slowly: ln Γ(z) = ln Γ(z + 1) − ln z This function is used in many computing environments and in the context of wave propogation. EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 9 / 15 Applications and Connections to Other Functions Derivative of the log-gamma function: Digamma function The Digamma function is defined to be the logarithmic derivative of the Gamma function: d Γ0 (z) (z) = ln (Γ(z)) = dz Γ(z) A general form of the Digamma function: the Polygamma function defined to be the (m + 1)th logarithmic derivative of the Gamma function: dm (m)(z) = ln (Γ(z)) dzm Notice that (0)(z) = (z) The Polygamma function is meromorphic on C (holomorphic on − C n Z [ 0 The nonpositive integers have poles of order m + 1 EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 10 / 15 Applications and Connections to Other Functions Digamma and Polygamma functions EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 11 / 15 Applications and Connections to Other Functions Digamma and Polygamma functions Finding a property of the Digamma function: Recall that Γ(z + 1) = zΓ(z) Take the derivative: 0 0 Γ (z + 1) = zΓ (z) + Γ(z) Divide by Γ(z + 1) = zΓ(z): Γ0 (z + 1) Γ0 (z) 1 = + Γ(z + 1) Γ(z) z Substitute in function: 1 (z + 1) = (z) + z EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 12 / 15 Applications and Connections to Other Functions Incomplete Gamma Functions upper incomplete gamma function R 1 s−1 −t Γ(s; x) = x x e dt lower incomplete gamma function R x s−1 −t γ(s; x) = 0 x e dt EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 13 / 15 Applications and Connections to Other Functions Pi and Beta Functions Definition of the Beta Function R 1 x−1 y−1 B(x; y) = 0 t (1 − t) dt for Re(x); Re(y) > 0 Beta function in terms of the Gamma function: Γ(x)Γ(y) B(x; y) = Γ(x + y) Definition of the Pi Function Π(z) = Γ(z + 1) = zΓ(z) EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 14 / 15 Applications and Connections to Other Functions Riemann Zeta Function The functional equation: πs ζ(s) = 2s πs−1 sin Γ(1 − s)ζ(1 − s) 2 The functional equation in another form: − s s − 1−s 1 − s π 2 Γ ζ(s) = π 2 Γ ζ(1 − s) 2 2 Another relation: Z 1 uz−1 ζ(z)Γ(z) = u du for Re(z) > 1 0 e − 1 EMILY GULLERUD, michael vaughan The Gamma Function May 10, 2017 15 / 15.
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