Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 2 docx

5 10 15 20 25 30 10000 20000 30000 40000 Figure 33.2: Plot of the integrand for Γ(10) We see that the ”important” part of the integrand is the hump centered around x = 9. If we find where the integrand of Γ(x) has its maximum d dx  e −t t x−1  = 0 − e −t t x−1 + (x − 1) e −t t x−2 = 0 (x − 1) − t = 0 t = x − 1, we see that the maximum varies with x. This could complicate our analysis. To take care of this problem we introduce 1614 the change of variables t = xs. Γ(x) =  ∞ 0 e −xs (xs) x−1 x ds = x x  ∞ 0 e −xs s x s −1 ds = x x  ∞ 0 e −x(s−log s) s −1 ds The integrands, ( e −x(s−log s) s −1 ), for Γ(5) and Γ(20) are plotted in Figure 33.3. 1 2 3 4 0.001 0.002 0.003 0.004 0.005 0.006 0.007 1 2 3 4 5·10 -10 1·10 -9 1.5·10 -9 2·10 -9 Figure 33.3: Plot of the integrand for Γ(5) and Γ(20). We see that the important part of the integrand is the hump that seems to be centered about s = 1. Also note that the the hump becomes narrower with increasing x. This makes sense as the e −x(s−log s) term is the most rapidly varying term. Instead of integrating from zero to infinity, we could get a good approximation to the integral by just integrating over some small n eighborhood centered at s = 1. Since s − log s has a minimum at s = 1, e −x(s−log s) has a maximum there. Because the important part of the integrand is the small area around s = 1, it makes sense to 1615 approximate s − log s with its Taylor series about that point. s − log s = 1 + 1 2 (s − 1) 2 + O  (s − 1) 3  Since the hump becomes increasingly narrow with increasi ng x, we will approximate the 1/s term in the i ntegrand with its value at s = 1. Substituting these approximations into the integral, we obtain Γ(x) ∼ x x  1+ 1− e −x(1+(s−1) 2 /2) ds = x x e −x  1+ 1− e −x(s−1) 2 /2 ds As x → ∞ both of the integrals  1− −∞ e −x(s−1) 2 /2 ds and  ∞ 1+ e −x(s−1) 2 /2 ds are exponentially small. Thus instead of integrating from 1 −  to 1 +  we can integrate from −∞ to ∞. Γ(x) ∼ x x e −x  ∞ −∞ e −x(s−1) 2 /2 ds = x x e −x  ∞ −∞ e −xs 2 /2 ds = x x e −x  2π x Γ(x) ∼ √ 2πx x−1/2 e −x as x → ∞. This is known as Stirling’s approximation to the Gamma function. In the table below, we see that the approximation is pretty good eve n for relatively small argument. 1616 n Γ(n) √ 2πx x−1/2 e −x relative error 5 24 23.6038 0.0165 15 8.71783 · 10 10 8.66954 · 10 10 0.0055 25 6.20448 · 10 23 6.18384 · 10 23 0.0033 35 2.95233 · 10 38 2.94531 · 10 38 0.0024 45 2.65827 · 10 54 2.65335 · 10 54 0.0019 In deriving Stirling’s approximation to the Gamma function we did a lot of hand waving. However, all of the steps can be justified and better approximations can be obtained by using Laplace’s method for finding the asymptotic behavior of integrals. 1617 33.6 Exercises Exercise 33.1 Given that  ∞ −∞ e −x 2 dx = √ π, deduce the value of Γ(1/2). Now find the value of Γ(n + 1/2). Exercise 33.2 Evaluate  ∞ 0 e −x 3 dx in terms of the gamma function. Exercise 33.3 Show that  ∞ 0 e −x sin(log x) dx = Γ(ı) + Γ(−ı) 2 . 1618 33.7 Hints Hint 33.1 Use the change of variables, ξ = x 2 in the integral. To find the value of Γ(n + 1/2) use the difference relation. Hint 33.2 Make the change of variable ξ = x 3 . Hint 33.3 1619 33.8 Solutions Solution 33.1  ∞ −∞ e −x 2 dx = √ π  ∞ 0 e −x 2 dx = √ π 2 Make the change of variables ξ = x 2 .  ∞ 0 e −ξ 1 2 ξ −1/2 dξ = √ π 2 Γ(1/2) = √ π Recall the difference relation for the Gamma function Γ(z + 1) = zΓ(z). Γ(n + 1/2) = (n − 1/2)Γ(n − 1/2) = 2n − 1 2 Γ(n − 1/2) = (2n − 3)(2n − 1) 2 2 Γ(n − 3/2) = (1)(3)(5) ···(2n − 1) 2 n Γ(1/2) Γ(n + 1/2) = (1)(3)(5) ···(2n − 1) 2 n √ π 1620 Solution 33.2 We make the change of variable ξ = x 3 , x = ξ 1/3 , dx = 1 3 ξ −2/3 dξ.  ∞ 0 e −x 3 dx =  ∞ 0 e −ξ 1 3 ξ −2/3 dξ = 1 3 Γ  1 3  Solution 33.3  ∞ 0 e −x sin(log x) dx =  ∞ 0 e −x 1 ı2  e ı log x − e −ı log x  dx = 1 ı2  ∞ 0 e −x  x ı − x −ı  dx = 1 ı2 (Γ(1 + ı) − Γ(1 − ı)) = 1 ı2 (ıΓ(ı) − (−ı)Γ(−ı)) = Γ(ı) + Γ(−ı) 2 1621 Chapter 34 Bessel Functions Ideas are angels. Implementations are a bitch. 34.1 Bessel’s Equation A commonly encountered differential equation in applied mathematics is Bessel’s equation y  + 1 z y  +  1 − ν 2 z 2  y = 0. For our purposes, we will consi der ν ∈ R 0+ . This equation arises when solving certain partial differential equations with the method of s eparation of variables in cylindrical coordinates. For this reason, the solutions of this equation are sometimes called cylindrical functions. This equation cannot be solved directly. However, we can find series representations of the solutions. There is a regular singular point at z = 0, so the Frobenius method is applicable there. The point at infinity is an irregular singularity, so we will look for asymptotic series about that point. Additionally, we will use Laplace’s method to find definite integral representations of the solutions. 1622 [...]... 2 πz 1 /2 sin z Using the recurrence relations, Jν+1 = ν Jν − Jν z and Jν−1 = we can find Jn+1 /2 for any integer n 1639 ν Jν + J ν , z 1 /2+ m z 1 /2+ 2m Example 34.3.1 To find J3 /2 (z), 1 /2 J1 /2 (z) − J1 /2 (z) z 1 /2 1 /2 2 1 = z −1 /2 sin z − − z π 2 J3 /2 (z) = 2 π 1 /2 z −3 /2 sin z − 2 π 1 /2 z −1 /2 cos z = 2 1 /2 π −1 /2 z −3 /2 sin z + 2 1 /2 π −1 /2 z −3 /2 sin z − 2 1 /2 π −1 /2 cos z 2 π 1 /2 = 2 π 1 /2 = z −3 /2. .. 1 −1 = 2 2 ζ ζ 1 1 2 4 c2 = − 2 3 = + 3 ζ ζ ζ ζ 1 −1 12 3 24 c3 = 2 − 2 − 4 = 2 + 4 2 ζ ζ ζ ζ ζ c1 = − We see that cn is a polynomial of degree n + 1 in 1/ζ One can show that  ζ4 ζn  2n−1 n! 1 + ζ 2 + + · · · + 2 4···n·(2n 2) ···(2n−n) ζ n+1 2( 2n 2) 2 4·(2n 2) (2n−4) cn (ζ) = ζ4 ζ n−1  2n−1 n! 1 + ζ 2 + + · · · + 2 4···(n−1)·(2n 2) ···(2n−(n−1)) ζ n+1 2( 2n 2) 2 4·(2n 2) (2n−4) 16 42 for even n for odd... nz n−1 2 2n 1 z n = 2 2 Jn (z) = 1 2 1 = 2 1 = 2 Jn (z) = z 2 z 2 z 2 n n n t−n−1 et−z 2 /4t dt + 1 2 z 2 n t−n−1 −2z 4t et−z 2 /4t dt n z −n−1 t−z2 /4t e − t dt z 2t z n 1 z n z n n 2 − + − 2 − − t−n−1 et−z /4t dt z z 2t z 2t 2t z 2t 2 2 n nz n 1 nz z 2 − − 2 − + 2 t−n−1 et−z /4t dt 2 z 2zt z 2t 2zt 4t n(n − 1) 2n + 1 z2 2 − + 2 t−n−1 et−z /4t dt 2 z 2t 4t 1 629 We substitute Jn (z) into... (Jν−1 − Jν+1 ) 2 1638 34.3 .5 Bessel Functions of Half-Integer Order Consider J1 /2 (z) Start with the series expansion ∞ J1 /2 (z) = Use the identity Γ(n + 1 /2) = (−1)m z m!Γ(1 /2 + m + 1) 2 m=0 1 /2+ 2m (1)(3)···(2n−1) √ π 2n ∞ (−1)m 2m+1 √ = m!(1)(3) · · · (2m + 1) π m=0 z 2 1 /2+ 2m ∞ = (−1)m 2m+1 √ (2) (4) · · · (2m) · (1)(3) · · · (2m + 1) π m=0 = 2 πz 1 2 1 /2 ∞ (−1)m 2m+1 z (2m + 1)! m=0 We recognize... (z) into Bessel’s equation 1 Jn + J n + 1 − z 1 z n = 2 2 1 z n = 2 2 1 z n = 2 2 Since t−n−1 et−z 2 /4t n2 z2 Jn z2 1 n(n − 1) 2n + 1 n − + 2 + − z2 2t 4t z 2 2t 2 z n+1 2 + 2 t−n−1 et−z /4t dt 1− t 4t d −n−1 t−z2 /4t e t dt dt + 1− n2 z2 t−n−1 et−z 2 /4t dt is analytic in 0 < |t| < ∞ when n is an integer, the integral vanishes = 0 Thus for integer n, Jn (z) satisfies Bessel’s equation Jn (z)... t−z2 /4t t e dt = 0 dt Expanding the integral, z 2 −ν 2 2 t − νt−ν−1 et−z /4t dt = 0 4 z2 2 t−ν + t−ν 2 − νt−ν−1 et−z /4t dt = 0 4 t−ν + C 1 2 Since Jν (z) = 1 (z /2) ν 2 C z 2 t−ν−1 et−z ν 2 /4t C dt, 2 z −1 Jν−1 + 2 z z2 Jν+1 − νJν = 0 4 Jν−1 + Jν+1 = 1636 2 Jν z Differentiating the integral expression for Jν , 1 νz ν−1 2 2 ν 1 z Jν (z) = z 2 2 t−ν−1 et−z Jν (z) = C ν 2 /4t t−ν−1 e t−z 2 /4t... z 1 is the equation (1 + 2 )a1 = 0 a1 = 0 The coefficient of z k for k ≥ 2 gives us k 2 + 2kν ak + ak 2 = 0 ak 2 ak 2 ak = − 2 =− k + 2kν k(k + 2 ) From the recurrence relation we see that all the odd coefficients are zero, a2k+1 = 0 The even coefficients are a2k = − a2k 2 (−1)k a0 = 2k 4k(k + ν) 2 k!Γ(k + ν + 1) 1 624 Thus we have the series solution ∞ y(z) = a0 k=0 (−1)k z 2k 22 k k!Γ(k + ν + 1) −ν a0 is... + 1) 2 (−1)k+n z (k + n)!Γ(k + 1) 2 ∞ n J−n (z) = (−1) k=0 (−1)k z k!(k + n)! 2 2k−n 2k+n 2k+n J−n (z) = (−1)n Jn (z) Thus we see that J−n (z) and Jn (z) are not linearly independent for integer n 16 25 34 .2. 1 Behavior at Infinity With the change of variables z = 1/ζ, w(z) = u(ζ) Bessel’s equation becomes ζ 4 u + 2 3 u + ζ −ζ 2 u + 1 − ν 2 ζ 2 u = 0 1 1 2 u + u + − u = 0 ζ ζ4 2 The point ζ = 0 and. .. sin z − 2 1 /2 π −1 /2 cos z 2 π 1 /2 = 2 π 1 /2 = z −3 /2 sin z − 2 π 1 /2 z −1 /2 cos z z −3 /2 sin z − z −1 /2 cos z You can show that J−1 /2 (z) = 2 πz 1 /2 cos z Note that at a first glance it appears that J3 /2 ∼ z −1 /2 as z → 0 However, if you expand the sine and cosine you will see that the z −1 /2 and z 1 /2 terms vanish and thus J3 /2 (z) ∼ z 3 /2 as z → 0 as we showed previously Recall that we showed the... 0.3 0.6 0 .2 0.4 0.1 0 .2 5 2 4 6 8 10 14 12 -0 .2 10 15 20 -0.1 -0 .2 -0.4 Figure 34.1: Plots of J0 (x), J1 (x) and J5 (x) For the path of integration, we are free to choose any contour that encloses the origin Consider the circular path on |t| = 1 Since the integral is uniformly convergent, we can interchange the order of integration and summation Jn (z) = 1 2 z 2 n ∞ (−1)m z 2m 22 m m! m=0 1631 t−n−m−1 . n for relatively small argument. 1616 n Γ(n) √ 2 x x−1 /2 e −x relative error 5 24 23 .6038 0.01 65 15 8.71783 · 10 10 8.66 954 · 10 10 0.0 055 25 6 .20 448 · 10 23 6.18384 · 10 23 0.0033 35 2. 9 52 33. = (n − 1 /2) Γ(n − 1 /2) = 2n − 1 2 Γ(n − 1 /2) = (2n − 3)(2n − 1) 2 2 Γ(n − 3 /2) = (1)(3) (5) ···(2n − 1) 2 n Γ(1 /2) Γ(n + 1 /2) = (1)(3) (5) ···(2n − 1) 2 n √ π 1 620 Solution 33 .2 We make the change. equation. J  n + 1 z J  n +  1 − n 2 z 2  J n = 1 2  z 2  n   n(n − 1) z 2 − 2n + 1 2t + z 2 4t 2  +  n z 2 − 1 2t  +  1 − n 2 z 2  t −n−1 e t−z 2 /4t dt = 1 2  z 2  n   1 − n + 1 t + z 2 4t 2  t −n−1 e t−z 2 /4t dt = 1 2  z 2  n  d dt  t −n−1 e t−z 2 /4t  dt Since