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Orthogonal Polynomials and Special Functions Comments Euler's formula (1) gives Γ(z + 1) z (m + 1)(z + 1) = lim = z Γ(z) z + 1 m!1 m + 1 + z Hence we obtain the most remarkable property for the Gamma function: Γ(z + 1) = zΓ(z) with Γ(1) = 1 = 0! : (3) Orthogonal Polynomials and Special Functions Comments LTCC course, 19 Feb - 20 March, 2018 Ana F. Loureiro ([email protected]) Outline Comments I Part 1. Special Functions I Gamma, Digamma and Beta Function I Hypergeometric Functions and Hypergeometric Series I Confluent Hypergeometric Function I Part 2. Orthogonal Polynomials I Main properties Recurrence relations, zeros, distribution of the zeros and so on and on.... I Classical Orthogonal Polynomials Hermite, Laguerre, Bessel and Jacobi!! I Other notions of "classical orthogonal polynomials" How to identify this on the Askey Scheme? I Semiclassical Orthogonal Polynomials How do these link to Random Matrix Theory, Painlev´eequations and so on? I Part 3. Multiple Orthogonal Polynomials When the orthogonality measure is spread across a vector of measures? Special Functions Comments 1. Gamma function. Introduced by Euler in 1729 in a letter to Golbach, the Gamma function arose to answer the question of finding a function mapping any nonnegative integer n to its factorial n!, that is Γ : N0 −! N0 N0 8 n Y <> n! = n(n − 1) ::: 3 · 2 · 1 = j if n ≥ 1; n 7−! j=1 :> 0! = 1 if n = 0: Gamma Function - Euler's definition Comments +1 z −1! 1 Y 1 z Γ(z) = 1 + 1 + (1) z j j j=1 valid for any z such that z 6= 0; −1; −2;:::. Observe that n−1 −1 n−1 z 1 Y z (n − 1)! Y 1 z 1 + = and 1 + = n z j (z)n j j=1 j=1 Therefore, we can conclude that the function Γ(z) can be equivalently given by (n − 1)!nz Γ(z) = lim : (2) n!+1 (z)n Gamma Function - Euler's definition Comments +1 z −1! 1 Y 1 z Γ(z) = 1 + 1 + (1) z j j j=1 valid for any z such that z 6= 0; −1; −2;:::. Observe that n−1 −1 n−1 z 1 Y z (n − 1)! Y 1 z 1 + = and 1 + = n z j (z)n j j=1 j=1 Therefore, we can conclude that the function Γ(z) can be equivalently given by (n − 1)!nz Γ(z) = lim : (2) n!+1 (z)n Euler's formula (1) gives Γ(z + 1) z (m + 1)(z + 1) = lim = z Γ(z) z + 1 m!1 m + 1 + z Hence we obtain the most remarkable property for the Gamma function: Γ(z + 1) = zΓ(z) with Γ(1) = 1 = 0! : (3) Gamma Function - integral representation Comments Theorem For z > 0, we have Z +1 Γ(z) = e−t tz−1dt: 0 Proof. For z > 0 and for any positive integer n, let Z n t n Z 1 Π(z; n) = 1 − tz−1dt = nz (1 − τ)n τ z−1dτ 0 n 0 Repeated integration by parts gives n(n − 1) ::: 2 · 1 Z 1 Π(z; n) = nz τ z+n−1dτ z(z + 1) ::: (z + n − 1) 0 n(n − 1) ::: 2 · 1 n!nz = nz = z(z + 1) ::: (z + n) (z)n so that Γ(z) = lim Π(z; n): n!+1 Comments On the other hand, observe that lim 1 − t n = e−t ; and n!+1 n t n t2e−t 0 ≤ e−t − 1 − ≤ : n n Since ! ! Z +1 Z n t n e−t tz−1dt − Γ(z) = lim e−t − 1 − tz−1dt 0 n!+1 0 n 1 Z n ≤ lim tz+1e−t dt n!+1 n 0 1 Z +1 < tz+1e−t dt −! 0 n 0 n!1 then Z +1 Γ(z) = e−t tz−1dt: 0 Gamma Function - properties Comments Integration by parts Z +1 +1 Z +1 −t z −t z −t z−1 Γ(z + 1) = e t dt = −e t − −e zt dt 0 0 0 Z +1 = z e−t tz−1dt = z Γ(z) 0 gives Γ(z + 1) = z Γ(z) Remark. The Gamma function does not satisfy any differential equation with rational coefficients (H¨older,1887). Gamma Function: a plot Comments 6 4 2 Γ(x) �������� 1 -4 -2 2 4 Γ(x) -2 -4 Weierstrass form of the Gamma function Comments The reciprocal of the Gamma function has the product representation +1 1 zγ Y z z = ze 1 + exp − (4) Γ(z) n n n=1 where n−1 ! X 1 γ := lim − log(n) = 0:5772156649::: (5) n!+1 k k=1 and called the Euler's constant. (Proof due to Schl¨omilchand Newman in 1848.) Proof of the Weierstrass form of the Gamma Function Comments We have the identity (n−1 )! n−1 ! (z)n X 1 Y z z = z exp z − log(n) 1 + exp − (n − 1)!nz k k k k=1 k=1 z z −2 Observe that log 1 + k exp − k = O(k ) and therefore the product n−1 Y z z 1 + exp − converges uniformly in bounded sets as n ! +1. k k k=1 Furthermore, n−1 n−1 n−1 X 1 X Z k+1 1 1 X Z k+1 t − k − log(n) = − dt = dt k k t kt k=1 k=1 k k=1 k Z k+1 t − k and dt = O(k−2) so the sum converges as n ! +1. Hence, k kt the result now follows due to n−1 ! (n − 1)!nz X 1 Γ(z) = lim and γ := lim − log(n) n!+1 (z)n n!+1 k k=1 The asymptotic behaviour of the Gamma function for large argument. Comments It can be shown that for large values of x, p −x x− 1 Γ(x) = e x 2 2π(1 + O(1=x)); and this is known as the Stirling's asymptotic formula for the Gamma function. Reflection formula for the Gamma function Comments π Γ(z)Γ(1 − z) = ; z 2= : (6) sin(πz) Z After a multiplication by z, we obtain a symmetric version of (6): πz Γ(1 + z)Γ(1 − z) = ; sin(πz) which can be generalised to n 2 2 πz Y z Γ(n + 1 + z)Γ(n + 1 − z) = (n!) 1 − ; n = 1; 2;:::: sin(πz) k2 k=1 1 As an immediate consequence, the choice of z = 2 brings 1 p Γ = π: (7) 2 Quiz: What is the value for +1 Z 2 e−x dx ? −∞ Beta function Comments The Beta function or Beta integral is a function of two variables a and b and is only defined for a > 0 and b > 0 by Z 1 B(a; b) = sa−1(1 − s)b−1ds: (8) 0 Theorem The Beta function satisfies the following identities Z 1 B(a; b) = sa−1(1 − s)b−1ds 0 Z +1 a+b a−1 1 = u du (9) 0 1 + u Γ(a)Γ(b) = Γ(a + b) which are valid for a; b > 0. proof: the Beta integral in terms of Gamma function Comments s The first equal sign is a consequence of the c.o.v. u = 1−s . For the 2nd equal sign, consider the product of Γ(a)Γ(b): Z +1 Z +1 Γ(a)Γ(b) = e−(s+t)sa−1tb−1dsdt: 0 0 Take the c.o.v. s = xu and t = x(1 − u), whose Jacobian is @(t; s) u x = det = −x @(x; u) 1 − u −x so that Z 1 Z +1 Γ(a)Γ(b) = e−x x a−1ua−1x b−1(1 − u)b−1x dxdu 0 0 Z 1 Z +1 = ua−1(1 − u)b−1du e−x x a+b−1dx 0 0 = Γ(a + b)B(a; b) The Legendre's duplication formula and the Gauss multiplication formula Comments We start by observing that for any integer n > 1, we have 1 n−1 Γ(2n) = (2n − 1)! = Q (2k + 1)(2k + 2) = 22n−1 1 (n − 1)! 2 n 2n k=0 2n−1 2p 1 = π Γ(n)Γ n + 2 : More generally, for z > 0, 2 1 1 Γ(z) Z Z 2 = B(z; z) = sz−1(1 − s)z−1ds = 2 sz−1(1 − s)z−1ds Γ(2z) 0 0 and, with the c.o.v. t = 4s(1 − s), it follows 1 1 Γ(z)2 Z t z 1 1 Γ(z)Γ = 2 p dt = 21−2z B z; = 21−2z 2 : Γ(2z) 4 2 1 0 t 1 − t Γ z + 2 By analytic continuation, we obtain the Legendre's duplication formula: 22z−1 1 Γ(2z) = p Γ(z)Γ z + for 2z 6= 0; −1; −2;:::: (10) π 2 1.2. Hypergeometric Series and Functions Comments X Definition. A series cn is called an hypergeometric series when c0 = 1 and n≥0 c n+1 is a rational function in n (possibly complex valued). cn Examples. n X z z X n 1 = e ; x = (for jxj < 1) n! 1 − x n≥0 n≥0 cn+1 The property c0 = 1 and is a rational function in n is satisfied if cn n (a1)n ··· (ap)n x cn = (b1)n ··· (bq)n n! where the symbol (α)n is the Pochhammer symbol: n−1 Y (α)n := α(α + 1) ··· (α + n − 1) = (α + σ); n ≥ 1; σ=0 (α)0 := 1 Comments Hence, we may represent an hypergeometric series by n a1;:::; ap X (a1)n ··· (ap)n x pFq ; x := (11) b1;:::; bq (b1)n ··· (bq)n n! n≥0 The previous examples..
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