Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 7 pdf

The series about z = ∞ for 1/(z + 2) is 1 2 + z = 1/z 1 + 2/z = 1 z ∞  n=0 (−2/z) n , for |2/z| < 1 = ∞  n=0 (−1) n 2 n z −n−1 , for |z| > 2 = −1  n=−∞ (−1) n+1 2 n+1 z n , for |z| > 2 To find the expansions in the three regions, we just choose the appropriate series. 1. f(z) = 1 1 + z − 1 2 + z = ∞  n=0 (−1) n z n − ∞  n=0 (−1) n 2 n+1 z n , for |z| < 1 = ∞  n=0 (−1) n  1 − 1 2 n+1  z n , for |z| < 1 f(z) = ∞  n=0 (−1) n 2 n+1 − 1 2 n+1 z n , for |z| < 1 614 2. f(z) = 1 1 + z − 1 2 + z f(z) = −1  n=−∞ (−1) n+1 z n − ∞  n=0 (−1) n 2 n+1 z n , for 1 < |z| < 2 3. f(z) = 1 1 + z − 1 2 + z = −1  n=−∞ (−1) n+1 z n − −1  n=−∞ (−1) n+1 2 n+1 z n , for 2 < |z| f(z) = −1  n=−∞ (−1) n+1 2 n+1 − 1 2 n+1 z n , for 2 < |z| Solution 12.27 Laurent Series. We assume that m is a non-negative integer and that n is an integer. The Laurent series about the point z = 0 of f(z) =  z + 1 z  m is f(z) = ∞  n=−∞ a n z n where a n = 1 ı2π  C f(z) z n+1 dz 615 and C is a contour going around the origin once in the positive direction. We manipulate the coefficient integral into the desired form. a n = 1 ı2π  C (z + 1/z) m z n+1 dz = 1 ı2π  2π 0 ( e ıθ + e −ıθ ) m e ı(n+1)θ ı e ıθ dθ = 1 2π  2π 0 2 m cos m θ e −ınθ dθ = 2 m−1 π  2π 0 cos m θ(cos(nθ) − ı sin(nθ)) dθ Note that cos m θ is even and sin(nθ) is odd about θ = π. = 2 m−1 π  2π 0 cos m θ cos(nθ) dθ Binomial Series. Now we find the binomial series expansion of f(z).  z + 1 z  m = m  n=0  m n  z m−n  1 z  n = m  n=0  m n  z m−2n = m  n=−m m−n even  m (m − n)/2  z n 616 The coefficients in the series f(z) =  ∞ n=−∞ a n z n are a n =   m (m−n)/2  −m ≤ n ≤ m and m −n even 0 otherwise By equating the coefficients found by the two methods, we evaluate the desired integral.  2π 0 (cos θ) m cos(nθ) dθ =  π 2 m−1  m (m−n)/2  −m ≤ n ≤ m and m −n even 0 otherwise Solution 12.28 First we write f(z) in the form f(z) = g(z) (z − ı/2)(z − 2) 2 . g(z) is an entire function which grows no faster that z 3 at infinity. By expanding g(z) in a Taylor series about the origin, we see that it is a polynomial of degree no greater than 3. f(z) = αz 3 + βz 2 + γz + δ (z − ı/2)(z − 2) 2 Since f(z) is a rational function we expand it in partial fractions to obtain a form that is convenient to integrate. f(z) = a z − ı/2 + b z − 2 + c (z − 2) 2 + d We use the value of the integrals of f(z) to determine the constants, a, b, c and d.  |z|=1  a z − ı/2 + b z − 2 + c (z − 2) 2 + d  dz = ı2π ı2πa = ı2π a = 1 617  |z|=3  1 z − ı/2 + b z − 2 + c (z − 2) 2 + d  dz = 0 ı2π(1 + b) = 0 b = −1 Note that by applying the second constraint, we can change the third constraint to  |z|=3 zf(z) dz = 0.  |z|=3 z  1 z − ı/2 − 1 z − 2 + c (z − 2) 2 + d  dz = 0  |z|=3  (z − ı/2) + ı/2 z − ı/2 − (z − 2) + 2 z − 2 + c(z − 2) + 2c (z − 2) 2  dz = 0 ı2π  ı 2 − 2 + c  = 0 c = 2 − ı 2 Thus we see that the function is f(z) = 1 z − ı/2 − 1 z − 2 + 2 − ı/2 (z − 2) 2 + d, where d is an arbitrary constant. We can also write the function in the form: f(z) = dz 3 + 15 − ı8 4(z − ı/2)(z − 2) 2 . Complete Laurent Series. We find the complete Laurent series about z = 0 for each of the terms in the partial 618 fraction expansion of f(z). 1 z − ı/2 = ı2 1 + ı2z = ı2 ∞  n=0 (−ı2z) n , for | − ı2z| < 1 = − ∞  n=0 (−ı2) n+1 z n , for |z| < 1/2 1 z − ı/2 = 1/z 1 − ı/(2z) = 1 z ∞  n=0  ı 2z  n , for |ı/(2z)| < 1 = ∞  n=0  ı 2  n z −n−1 , for |z| < 2 = −1  n=−∞  ı 2  −n−1 z n , for |z| < 2 = −1  n=−∞ (−ı2) n+1 z n , for |z| < 2 619 − 1 z − 2 = 1/2 1 − z/2 = 1 2 ∞  n=0  z 2  n , for |z/2| < 1 = ∞  n=0 z n 2 n+1 , for |z| < 2 − 1 z − 2 = − 1/z 1 − 2/z = − 1 z ∞  n=0  2 z  n , for |2/z| < 1 = − ∞  n=0 2 n z −n−1 , for |z| > 2 = − −1  n=−∞ 2 −n−1 z n , for |z| > 2 620 2 − ı/2 (z − 2) 2 = (2 − ı/2) 1 4 (1 − z/2) −2 = 4 − ı 8 ∞  n=0  −2 n   − z 2  n , for |z/2| < 1 = 4 − ı 8 ∞  n=0 (−1) n (n + 1)(−1) n 2 −n z n , for |z| < 2 = 4 − ı 8 ∞  n=0 n + 1 2 n z n , for |z| < 2 2 − ı/2 (z − 2) 2 = 2 − ı/2 z 2  1 − 2 z  −2 = 2 − ı/2 z 2 ∞  n=0  −2 n  − 2 z  n , for |2/z| < 1 = (2 − ı/2) ∞  n=0 (−1) n (n + 1)(−1) n 2 n z −n−2 , for |z| > 2 = (2 − ı/2) −2  n=−∞ (−n − 1)2 −n−2 z n , for |z| > 2 = −(2 − ı/2) −2  n=−∞ n + 1 2 n+2 z n , for |z| > 2 We take the appropriate combination of these series to find the Laurent series expansions in the regions: |z| < 1/2, 621 1/2 < |z| < 2 and 2 < |z|. For |z| < 1/2, we have f(z) = − ∞  n=0 (−ı2) n+1 z n + ∞  n=0 z n 2 n+1 + 4 − ı 8 ∞  n=0 n + 1 2 n z n + d f(z) = ∞  n=0  −(−ı2) n+1 + 1 2 n+1 + 4 − ı 8 n + 1 2 n  z n + d f(z) = ∞  n=0  −(−ı2) n+1 + 1 2 n+1  1 + 4 − ı 4 (n + 1)  z n + d, for |z| < 1/2 For 1/2 < |z| < 2, we have f(z) = −1  n=−∞ (−ı2) n+1 z n + ∞  n=0 z n 2 n+1 + 4 − ı 8 ∞  n=0 n + 1 2 n z n + d f(z) = −1  n=−∞ (−ı2) n+1 z n + ∞  n=0  1 2 n+1  1 + 4 − ı 4 (n + 1)  z n + d, for 1/2 < |z| < 2 For 2 < |z|, we have f(z) = −1  n=−∞ (−ı2) n+1 z n − −1  n=−∞ 2 −n−1 z n − (2 − ı/2) −2  n=−∞ n + 1 2 n+2 z n + d f(z) = −2  n=−∞  (−ı2) n+1 − 1 2 n+1 (1 + (1 − ı/4)(n + 1))  z n + d, for 2 < |z| Solution 12.29 The radius of convergence of the series for f(z) is R = lim n→∞     k 3 /3 k (k + 1) 3 /3 k+1     = 3 lim n→∞     k 3 (k + 1) 3     = 3. 622 [...]... 4 1 = lim 24 cot(z) coth(z)csc(z )2 − 32z coth(z)csc(z)4 4! z→0 = lim − 16z cos(2z) coth(z)csc(z)4 + 22 z 2 cot(z) coth(z)csc(z)4 + 2z 2 cos(3z) coth(z)csc(z)5 + 24 cot(z) coth(z)csch(z )2 + 24 csc(z )2 csch(z )2 − 48z cot(z)csc(z )2 csch(z )2 − 48z coth(z)csc(z )2 csch(z )2 + 24 z 2 cot(z) coth(z)csc(z )2 csch(z )2 + 16z 2 csc(z)4 csch(z )2 + 8z 2 cos(2z)csc(z)4 csch(z )2 − 32z cot(z)csch(z)4 − 16z cosh(2z) cot(z)csch(z)4... cot(z)csch(z)4 + 22 z 2 cot(z) coth(z)csch(z)4 + 16z 2 csc(z )2 csch(z)4 + 8z 2 cosh(2z)csc(z )2 csch(z)4 + 2z 2 cosh(3z) cot(z)csch(z)5 = 1 4! =− − 56 15 7 45 633 Since taking the fourth derivative of z 2 cot z coth z really sucks, we would like a more elegant way of finding the residue We expand the functions in the integrand in Taylor series about the origin 2 4 2 4 z z 1 − z2 + 24 − · · · 1 + z2 + 24 + · ·... n=1 −∞ ∞ n=0 ∞ n=0 1 1 − n (z + 1) (z + 1) for |z + 1| > 1 and |z + 1| > 2 for |z + 1| > 2 for |z + 1| > 2 1 − 2 n−1 (z + 1)n , = n=0 2n , (z + 1)n n 1 2 , (z + 1)n 1 − 2n , (z + 1)n+1 ∞ for |z + 1| > 1 and |z + 1| > 2 for |z + 1| > 2 n= 2 2 First we factor the denominator of f (z) = 1/(z 4 + 4) z 4 + 4 = (z − 1 − ı)(z − 1 + ı)(z + 1 − ı)(z + 1 + ı) We look for an annulus about z = 1 containing the... integrals have the form ∞ eıωx f (x) dx −∞ We evaluate these integrals by closing the path of integration in the lower or upper half plane and using techniques of contour integration Consider the integral π /2 e−R sin θ dθ 0 Since 2 /π ≤ sin θ for 0 ≤ θ ≤ π /2, e−R sin θ ≤ e−R2θ/π for 0 ≤ θ ≤ π /2 π /2 π /2 e −R sin θ e−R2θ/π dθ dθ ≤ 0 0 π −R2θ/π π /2 e = − 2R 0 π −R = − (e −1) 2R π ≤ 2R → 0 as R → ∞ 6 47 We can use... dx = lim 2+ 1 a→+∞ x −a b 1 dx+ = lim 2+ 1 b→+∞ x 0 x2 1 dx +1 In the above example we instead computed R lim R→+∞ −R 1 dx x2 + 1 Note that for some integrands, the former and latter are not the same Consider the integral of ∞ −∞ x dx = lim 2+ 1 a→+∞ x = lim a→+∞ 0 −a x dx + lim 2+ 1 b→+∞ x b 0 1 log |a2 + 1| + lim b→+∞ 2 x x2 +1 x dx +1 1 − log |b2 + 1| 2 x2 Note that the limits do not exist and hence... i + O( 2 ) − 1 − log 1 + i + O( 2 ) − 1 = lim log −i + O( 2 ) − log i + O( 2 ) + →0 = lim Log + →0 + O( 2 ) + ı arg −ı + O( 2 ) − Log + O( 2 ) − ı arg ı + O( 2 ) π 3π −ı 2 2 = ıπ =ı Thus we obtain  0  1 − dz = ıπ  Cr z − 1  2 for r < 1, for r = 1, for r > 1 In the above example we evaluated the contour integral by parameterizing the contour This approach is only feasible when the integrand is... integrand is analytic Thus by Cauchy’s theorem the value of the integral is zero eız f (z) dz = 0 |z|=1 2 We use Cauchy’s integral formula to evaluate the integral |z|=1 2 (3) 2 3!33 f (z) dz = f (0) = = 2 z4 3! 3! 33 |z|=1 f (z) dz = 2 z4 3 We use Cauchy’s integral formula to evaluate the integral |z|=1 2 d f (z) ez dz = (f (z) ez ) 2 z 1! dz |z|=1 z=0 = 2 1!13 31 f (z) ez 2 dz = 2 z 3... integrals ∞ −∞ x dx = lim 2+ 1 a→−∞, b→∞ x b a x2 x dx +1 1 = lim ln(x2 + 1) a→−∞, b→∞ 2 1 b2 + 1 = lim ln 2 a→−∞, b→∞ a2 + 1 636 b a The integral diverges because a and b approach infinity independently Now consider what would happen if a and b were not independent If they approached zero symmetrically, a = −b, then the value of the integral would be zero b2 + 1 b2 + 1 1 lim ln 2 b→∞ =0 We could make the... integrand has first order poles at z = ±ı For 643 R > 1, we have C z2 1 dz = 2 Res +1 1 = 2 2 = π z2 1 ,ı +1 Now we examine the integral along CR We use the maximum modulus integral bound to show that the value of the integral vanishes as R → ∞ CR z2 1 1 dz ≤ πR max 2 z∈CR z + 1 +1 1 = πR 2 R −1 → 0 as R → ∞ Now we are prepared to evaluate the original real integral 1 dz = π +1 C 1 1 dx + dz = π 2+ 1... 2 z 3 Solution 12. 30 1 (a) 1 1 1 = + z(1 − z) z 1−z ∞ = 1 zn, + z n=0 for 0 < |z| < 1 ∞ 1 zn, = + z n=−1 623 for 0 < |z| < 1 (b) 1 1 1 = + z(1 − z) z 1−z 1 1 1 = − z z 1 − 1/z = 1 1 − z z =− 1 z ∞ n=0 1 z n , for |z| > 1 ∞ z −n , for |z| > 1 n=1 −∞ zn, =− n= 2 624 for |z| > 1 (c) 1 1 1 = + z(1 − z) z 1−z 1 1 + = (z + 1) − 1 2 − (z + 1) 1 1 1 1 = − , (z + 1) 1 − 1/(z + 1) (z + 1) 1 − 2/ (z + 1) 1 = (z . 1)(−1) n 2 −n z n , for |z| < 2 = 4 − ı 8 ∞  n=0 n + 1 2 n z n , for |z| < 2 2 − ı /2 (z − 2) 2 = 2 − ı /2 z 2  1 − 2 z  2 = 2 − ı /2 z 2 ∞  n=0  2 n  − 2 z  n , for |2/ z| < 1 = (2 −. 0.  |z|=3 z  1 z − ı /2 − 1 z − 2 + c (z − 2) 2 + d  dz = 0  |z|=3  (z − ı /2) + ı /2 z − ı /2 − (z − 2) + 2 z − 2 + c(z − 2) + 2c (z − 2) 2  dz = 0 2  ı 2 − 2 + c  = 0 c = 2 − ı 2 Thus we see that. 1 = (2 − ı /2) ∞  n=0 (−1) n (n + 1)(−1) n 2 n z −n 2 , for |z| > 2 = (2 − ı /2) 2  n=−∞ (−n − 1 )2 −n 2 z n , for |z| > 2 = − (2 − ı /2) 2  n=−∞ n + 1 2 n +2 z n , for |z| > 2 We take