Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 3 ppt

... are u x = 3x 2 − 3y 2 − 2y, u y = −6xy − 2x + 1. The derivative of f(z) is f  (z) = u x − ıu y = 3x 2 − 2y 2 − 2y + ı(6xy − 2x + 1). On the real axis we have f  (z = x) = 3x 2 − ı2x + ı. Using ... max a≤x≤b |f(x)|. 466 with a, b and c complex-valued constants and d a real constant. Substituting z = x + ıy and expanding products yields, a  x 3 + ı3x 2 y − 3xy 2 − ıy 3...

Ngày tải lên: 06/08/2014, 01:21

... =     x 20 20 ! cos ξ     ≤ |x| 20 20 ! . x 20 /20 ! is plotted in Figure 3. 13. 2 4 6 8 10 0 .2 0.4 0.6 0.8 1 Figure 3. 13: Plot of x 20 /20 !. Note that the error is very small for x < 6, ... 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...

Ngày tải lên: 06/08/2014, 01:21

... 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 2( −1) 2 − 1 2( 1) 2 + 1 2 2  = 0 3. ... C.  C log 2 z 1 + z 2 dz = − π 3 4  ∞ 0 ln 2 r 1 + r 2 dr +  0 ∞ (ln r + ıπ) 2 1 + r 2 dr = − π 3 4 2  ∞ 0 ln 2 x 1 + x 2 dx + 2  ∞ 0 ln...

Ngày tải lên: 06/08/2014, 01:21

... 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 13. 46 From ... .      C R dz z 3 + 1     ≤ 2 R 3 1 R 3 − 1 → 0 as R → ∞ We take R → ∞ and solve for the desired integral.  1 + e −ıπ /3   ∞ 0 dx x 3 +...

Ngày tải lên: 06/08/2014, 01:21

... = 3 x y(x)−2x 2y(x)−x − y(x) (2y(x) −x) 2 , y  (x) = 3 x(y(x) −2x) − y(x)(2y(x) − x) (2y(x) −x) 3 , y  (x) = −6 x 2 − xy(x) + [y(x)] 2 (2y(x) −x) 3 , y  (x) = −18 (2y(x) −x) 3 , 105 The ... cos x  = 0 2 = 0 lim x→0  csc x − 1 x  = 0 109 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 po...

Ngày tải lên: 06/08/2014, 01:21

... ı √ 3  8  −1 =   2 + 2 √ 3  2 + 2 √ 3  4  −1 =   2 + 2 √ 3  −8 − ı8 √ 3  2  −1 =  2 + 2 √ 3  − 128 + ı 128 √ 3  −1 =  −5 12 − ı5 12 √ 3  −1 = 1 5 12 −1 1 + ı √ 3 = 1 5 12 −1 1 + ı √ 3 1 ... 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 + ı...

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...

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 − 1) ... 7 .21 1. For each value of z, f(z) = z 1 /3 has three values. f(z) = z 1 /3 = 3 √ z e ık2π /3 , k = 0, 1, 2 2. g(w) = w 3 = |w| 3 e 3 arg(w) 32 8 3...

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 ... 8.4 .3 1/ sin (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 co...

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