Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 9 ppt

... = 2 Res  z 2 (z 2 + 1) 2 , z = ı  Res  z 2 (z 2 + 1) 2 , z = ı  = lim z→ı d dz  (z − ı) 2 z 2 (z 2 + 1) 2  = lim z→ı d dz  z 2 (z + ı) 2  = lim z→ı  (z + ı) 2 2z − z 2 2(z + ı) (z + ... be 696 2. −  1 −1 1 x 3 dx = lim →0 +   − −1 1 x 3 dx +  1  1 x 3 dx  = lim →0 +   − 1 2x 2  − −1 +  − 1 2x 2  1   = lim →0 +  − 1 2( −) 2 + 1...

Ngày tải lên: 06/08/2014, 01:21

... Hint 7. 19 Hint 7 .20 Hint 7 .21 Hint 7 .22 Hint 7 .23 Hint 7 .24 Hint 7 .25 1. (z 2 + 1) 1 /2 = (z − ı) 1 /2 (z + ı) 1 /2 2. (z 3 − z) 1 /2 = z 1 /2 (z − 1) 1 /2 (z + 1) 1 /2 3. log (z 2 − 1) = log(z ... ı47 ( e ız + e −ız ) /2 ( e ız − e −ız ) /( 2) = ı47 e ız + e −ız = 47  e ız − e −ız  46 e ı2z −48 = 0 ı2z = log 24 23 z = − ı 2 log 24 23 z = − ı 2  ln 24...

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 ... 1     |dz| ≤  θ 0 +π θ 0     R 2 e 2 −1 R 2 e 2 +1     |R dθ| ≤ R  θ 0 +π θ 0 R 2 + 1 R 2 − 1 dθ =...

Ngày tải lên: 06/08/2014, 01:21

... principal value exists. 766 π 2  lim z→0  (z −1) 2 (z −3 + 2 √ 2) (z −3 2 √ 2)  + lim z→3 2 √ 2  (z −1) 2 z(z − 3 − 2 √ 2)  .  1 0 x 2 (1 + x 2 ) √ 1 − x 2 dx = (2 − √ 2) π 4 Infinite Sums Solution ... z (z + 1) 2 , −1  . 2  ∞ 0 x 1 /2 log x (x + 1) 2 dx + 2  ∞ 0 x 1 /2 (x + 1) 2 dx = 2 lim z→−1 d dz (z 1 /2 log z) 2  ∞ 0 x 1 /2 log x (x +...

Ngày tải lên: 06/08/2014, 01:21

... y  ) − (2x − y)(1 + 2y  ) (x + 2y) 2 = 5(y −xy  ) (x + 2y) 2 Substitute in the expression for y  . = − 10(x 2 − xy −y 2 ) (x + 2y) 2 Use the original implicit equation. = − 10 (x + 2y) 2 3.5 ... rule to evaluate the limit. For the Taylor series expansion method, csc 2 x − 1 x 2 = x 2 − sin 2 x x 2 sin 2 x = x 2 − (x − x 3 /6 + O(x 5 )) 2 x 2 (x + O(x 3 ))...

Ngày tải lên: 06/08/2014, 01:21

... cos x  = 0 2 = 0 lim x→0  csc x − 1 x  = 0 1 09 Solution 3.15 a. f  (x) = ( 12 −2x) 2 + 2x( 12 − 2x)( 2) = 4(x −6) 2 + 8x(x − 6) = 12( x 2) (x − 6) There are critical points at x = 2 and x = 6. f  (x) ... derivative exists and is nonzero for x = 2. At x = 2, the derivative does not exist and thus x = 2 is a critical point. For x < 2, f  (x) < 0 and...

Ngày tải lên: 06/08/2014, 01:21

... mapping. 24 3 2. z + 2 2 − ız = x + ıy + 2 2 − ı(x − ıy) = x + ı(y + 2) 2 − y −ıx = x + ı(y + 2) 2 − y −ıx 2 − y + ıx 2 − y + ıx = x (2 − y) − (y + 2) x (2 − y) 2 + x 2 + ı x 2 + (y + 2) (2 − y) (2 − ... of z 2 is |z 2 | =  z 2 z 2 = zz = (x + ıy)(x −ıy) = x 2 + y 2 . 25 0 -2 -1 0 1 2 x -2 -1 0 1 2 y -5 0 5 -2 -1 0 1 2 x -2 -1 0 1 2 y...

Ngày tải lên: 06/08/2014, 01:21

... cut. 29 2 -2 0 2 x -2 -1 0 1 2 y 2 4 -2 0 2 x -2 0 2 x -2 -1 0 1 2 y 0 2 4 -2 0 2 x Figure 7 .20 : Plots of |cos(z)| and |sin(z)|. Result 7.6.1 e z = e x (cos y + ı sin y) cos z = e ız + e −ız 2 sin ... form. Denote any multi-valuedness explicitly. 2 2/5 , 3 1+ı ,  √ 3 − ı  1/4 , 1 ı/4 . Hint, Solution 28 7 -2 -1 0 1 2 x -2 -1 0 1 2 y 0 0.5 1 -2...

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

... = 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 = ... y). f(z) = 2u  z 2 , −ı z 2  = 2 e −z /2  z 2 sin  −ı z 2  + ı z 2 cos  −ı z 2  + c = ız e −z /2  ı sin  ı z 2  + cos  −ı z 2  + c = ız e...

Ngày tải lên: 06/08/2014, 01:21