Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 1 pps

... 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→...

Ngày tải lên: 06/08/2014, 01:21

... equal. 516 upper bound for the series.  ∞ 1 1 x 2 dx ≤ ∞  n =1 1 n 2 ≤ 1 +  ∞ 1 1 x 2 dx 1 ≤ ∞  n =1 1 n 2 ≤ 2 1 2 3 4 1 1 2 3 4 1 Figure 12 . 1: Upper and Lower bounds to  ∞ n =1 1/n 2 . In general, ... 0 517 11 .5 Hints Hint 11 .1 Use the argument theorem. Hint 11 .2 Hint 11 .3 To evaluate the integral, consider the circle at infinity. Hint...

Ngày tải lên: 06/08/2014, 01:21

... Annulus. Example 12 . 6 .1 Find the Laurent series expansions of 1/ (1 + z). For |z| < 1, 1 1 + z = 1 +  1 1  z +  1 2  z 2 +  1 3  z 3 + ··· = 1 + ( 1) 1 z + ( 1) 2 z 2 + ( 1) 3 z 3 + ··· = 1 − z ... 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)...

Ngày tải lên: 06/08/2014, 01:21

... < 1 1 4 (1 + t + x) 2 1 < x + t < 0 1 4 (1 − t 2 + 2t (1 − x) + x (2 − x)) x −t < 1, 0 < x + t 1 2 (2t − t 2 − x 2 ) 1 < x − t, x + t < 1 1 4 (1 − t 2 + 2t (1 + x) − x (2 + ... − 1) πx 2L  − ∞  m =1 n =1 g mn   (2m − 1) π 2L  2 +  nπy H  2  2 √ LH sin  (2m − 1) πx 2L  cos  nπy H  = ∞  m =1  2 LH sin  (2m − 1) πξ...

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 = ... an arc and that f 1 (z) = f 2 (z) for all z ∈ D 1 ∩ D 2 . (See Figure 9.4.) Then the function f(z) =  f 1 (z) for z ∈ D 1 , f 2 (z) for z ∈ D 2...

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 ... 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 = πr R 2 + 1 R...

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 5 81 Solution 12 . 22 cos z = −cos(z − π) = − ∞  n=0 ( 1) n (z −π) 2n (2n)! = ∞  n=0 ( 1) n +1 (z ... convergent, so is the series. Solution 12 . 4 ∞  n =1 ( 1) n +1 n = ∞  n =1  1 2n − 1 − 1 2n  = ∞  n =1 1 (2n − 1) (2n) < ∞  n =1...

Ngày tải lên: 06/08/2014, 01:21

... lim →0 +  2 −  1 e ıθ − 1 ıe ıθ dθ = lim →0 +  log(e ıθ − 1)  2 −  639 − 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 , ... negative direction. 648 (b) 1 z (1 −z) = 1 z + 1 1 − z = 1 z − 1 z 1 1 − 1/ z = 1 z...

Ngày tải lên: 06/08/2014, 01:21

... − 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 n 1  k=0 (1 + 2k) e 2 k/n = 2 e ıπ/n n 2 n 1  k =1 k e 2 k/n = 2 e ıπ/n n 2 n e 2 /n 1 = π n ... lim z→0 cos z 1 = 1 3. 1 + z 2 z(z 1) 2 There is a first order pole at z = 0 and a sec ond order pole at...

Ngày tải lên: 06/08/2014, 01:21

... c  n k =1 (z −z k ). Let m 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  = ... 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 ...

Ngày tải lên: 06/08/2014, 01:21