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

... 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

... . . 22 00 50.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 01 VIII Appendices 22 20 A Greek Letters 22 21 xxi 0.3 Warnings and ... . . . . . . . . . . . . . 22 2. 1.1 Scalars and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2. 1 .2 The Kronecker Delta and Einstein...

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

... 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...

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... 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...

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... = 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 + ı...

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

... z z 3  z=−ı + 2 2!  d 2 dz 2 (z 3 + z + ı) sin z z + ı  z=0 = 2 (−ı sinh(1)) + ıπ  2  3z 2 + 1 z + ı − z 3 + z + ı (z + ı) 2  cos z +  6z z + ı − 2( 3z 2 + 1) (z + ı) 2 + 2( z 3 + z + ı) (z ... formula.  C z z 2 + 1 dz =  C 1 /2 z −ı dz +  C 1 /2 z + ı dz = 1 2 2 + 1 2 2 = 2 3.  C z 2 + 1 z dz =  C  z + 1 z  dz =  C z dz +  C 1 z dz = 0 + 2 = 2...

Ngày tải lên: 06/08/2014, 01:21

... 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 infinite discontinuities at x = 2 k, k ∈ Z. The partial ... n)) 5. ∞  n=1 ln (2 n ) ln (3 n ) + 1 6. ∞  n=0 1 ln(n + 20 ) 7. ∞  n=0 4 n + 1 3 n − 2 8. ∞  n=0 (Log π 2) n 9. ∞  n =2 n 2 − 1 n 4 − 1 10. ∞  n =2 n 2 (ln n) n...

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 581 Solution 12. 22 cos z = −cos(z − π) = − ∞  n=0 (−1) n (z −π) 2n (2n)! = ∞  n=0 (−1) n+1 (z ... 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 1 (2n − 1) 2 < 1 2 ∞  n=1 1 n 2 = π 2 12 Th...

Ngày tải lên: 06/08/2014, 01:21