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

... n) n 11. ∞  n =2 (−1) n ln  1 n  12. ∞  n =2 (n!) 2 (2n)! 13. ∞  n =2 3 n + 4 n + 5 5 n − 4 n − 3 5 62 Im(z) Re(z) R R 2 1 Im(z) Re(z) R R 2 1 C r 1 r 2 z C C C 1 2 z Figure 12. 5: Contours for a Laurent ... closed form. (See Exercise 12. 9.) N−1  n=1 sin(nx) =  0 for x = 2 k cos(x /2) −cos((N−1 /2) x) 2 sin(x /2) for x = 2 k The partial sums have infinit...

Ngày tải lên: 06/08/2014, 01:21

... 1) 2 + y 2 ) 2 = −(x − 1) 2 + y 2 ((x − 1) 2 + y 2 ) 2 and 2( x − 1)y ((x − 1) 2 + y 2 ) 2 = 2( x − 1)y ((x − 1) 2 + y 2 ) 2 The Cauchy-Riemann equations are each identities. The first partial derivatives ... (z 2 ) has a second order pole at z = 0 and first order poles at z = (nπ) 1 /2 , n ∈ Z ± . lim z→0 z 2 sin (z 2 ) = lim z→0 2z 2z cos (z 2 ) = li...

Ngày tải lên: 06/08/2014, 01:21

... equation p(z) = z 6 −5z 2 +10 = 0 lie in the annulus 1 < |z| < 2. Exercise 11 .5 Evaluate as a function of t ω = 1 2  C e zt z 2 (z 2 + a 2 ) dz, 50 5 Integral Test. Result 12. 1 .2 If the coefficients ... 1) dz. There are singularities at z = 0 and z = −1. Let C 1 and C 2 be contours around z = 0 and z = −1. See Figure 11.6. We deform C onto C 1 and C 2 .  C...

Ngày tải lên: 06/08/2014, 01:21

... dish. 1 62 Let u = x 2 and dv = e 2x dx. Then du = 2x dx and v = 1 2 e 2x .  x 3 e 2x dx = 1 2 x 3 e 2x − 3 2  1 2 x 2 e 2x −  x e 2x dx   x 3 e 2x dx = 1 2 x 3 e 2x − 3 4 x 2 e 2x + 3 2  x e 2x dx Let ... = x and dv = e 2x dx. Then du = dx and v = 1 2 e 2x .  x 3 e 2x dx = 1 2 x 3 e 2x − 3 4 x 2 e 2x + 3 2  1 2 x e 2x − 1 2  e 2x dx   x 3 e 2...

Ngày tải lên: 06/08/2014, 01:21

... 4  x 2 + y 2 +  (x − 2) 2 + y 2 = 4 x 2 + y 2 = 16 − 8  (x − 2) 2 + y 2 + x 2 − 4x + 4 + y 2 x − 5 = 2  (x − 2) 2 + y 2 x 2 − 10x + 25 = 4x 2 − 16x + 16 + 4y 2 1 4 (x − 1) 2 + 1 3 y 2 = 1 Thus ... (1 2 ) 1 /2 = 1 1 /2 = ±1 and  1 1 /2  2 = (±1) 2 = 1. Example 6.6 .2 Consider 2 1 /5 , (1 + ı) 1/3 and (2 + ı) 5/ 6 . 2 1 /5 =...

Ngày tải lên: 06/08/2014, 01:21

... = 1 2 Log  x 2 + y 2  + ı Arctan(x, y). 4 52 2. We calculate the first partial derivatives of u and v. u x = 2 e x 2 −y 2 (x cos(2xy) − y sin(2xy)) u y = 2 e x 2 −y 2 (y cos(2xy) + x sin(2xy)) v x = ... x direction. f  (z) = u x + ıv x f  (z) = 2 e x 2 −y 2 (x cos(2xy) − y sin(2xy)) + 2 e x 2 −y 2 (y cos(2xy) + x sin(2xy)) f  (z) = 2 e x 2 −y 2 ((x +...

Ngày tải lên: 06/08/2014, 01:21

... =  2 0 e ınθ ı e ıθ dθ =     e ı(n+1)θ n+1  2 0 for n = −1 [ıθ] 2 0 for n = −1 =  0 for n = −1 2 for n = −1 2. We parameterize the contour and do the integration. z − z 0 = 2 + e ıθ , θ ∈ [0 . . . 2 )  C (z − z 0 ) n dz =  2 0  2 ... axis and is defined continuously on the real axis.) Hint, Solution 481      C Log z dz     ≤  C |Log z||dz| =  π...

Ngày tải lên: 06/08/2014, 01:21

... π) Hint 12. 23 CONTINUE Hint 12. 24 CONTINUE Hint 12. 25 Hint 12. 26 Hint 12. 27 Hint 12. 28 Hint 12. 29 Hint 12. 30 CONTINUE 58 1 Solution 12. 22 cos z = −cos(z − π) = − ∞  n=0 (−1) n (z −π) 2n (2n)! = ∞  n=0 (−1) n+1 (z ... 5| 2 lim k→∞     (k + 2) 2 (k + 1) 2     < 1 |z + 5| 2 lim k→∞ 2( k + 2) 2( k + 1) < 1 |z + 5| 2 lim k→∞ 2 2 < 1 |z + 5| 2 < 1...

Ngày tải lên: 06/08/2014, 01:21

... 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 ... − 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 1...

Ngày tải lên: 06/08/2014, 01:21

... the residue by expanding the function in a Laurent series. (1 − cos z) 2 z 7 = z −7  1 −  1 − z 2 2 + z 4 24 + O  z 6   2 = z −7  z 2 2 − z 4 24 + O  z 6   2 = z −7  z 4 4 − z 6 24 + O  z 8   = 1 4z 3 − 1 24 z + ... − n−1  k=0 lim z→ e ıπ(1+2k)/n  log z + (z − e ıπ(1+2k)/n )/z nz n−1  = − n−1  k=0  ıπ(1 + 2k)/n n e ıπ(1+2k)(n−1)/n  = − ıπ n 2 e ıπ(n−1)/n...

Ngày tải lên: 06/08/2014, 01:21