Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 1 pdf

... 8 .12 Consider the complex function f(z) = u + ıv =  x 3 (1+ ı)−y 3 (1 ı) x 2 +y 2 for z = 0, 0 for z = 0. Show that the partial derivatives of u and v with respect to x and y exist at z = 0 and ... 8 .1. 3 and use Result 8 .1. 1. Hint 8.4 Use Result 8 .1. 1. Hint 8.5 Take the logarithm of the equation to get a linear equation. Cauchy-Riemann Equations Hint 8.6 Hint 8.7 Hint...

Ngày tải lên: 06/08/2014, 01:21

... negative direction. 648 (b) 1 z (1 −z) = 1 z + 1 1 − z = 1 z − 1 z 1 1 − 1/ z = 1 z − 1 z ∞  n=0  1 z  n , for |z| > 1 = − 1 z ∞  n =1 z −n , for |z| > 1 = − −∞  n=−2 z n , for |z| > 1 624 Result 13 .5.2 ... Exercise 13 .10 .) 642 The series about z = ∞ for 1/ (z + 2) is 1 2 + z = 1/ z 1 + 2/z = 1 z ∞  n=0 (−2/z) n , for |2/z| &l...

Ngày tải lên: 06/08/2014, 01:21

... = y(x) x . P (1, u) + Q (1, u)  u + x du dx  = 0 This equation is separable. P (1, u) + uQ (1, u) + xQ (1, u) du dx = 0 1 x + Q (1, u) P (1, u) + uQ (1, u) du dx = 0 ln |x| +  1 u + P (1, u)/Q (1, u) du ... divide it into two equations on separate domains. y  1 − y 1 = 0, y 1 (0) = 1, for x < 1 y  2 − y 2 = 1, y 2 (1) = y 1 (1) , for x > 1 797 • y  + 3xy ...

Ngày tải lên: 06/08/2014, 01:21

... Equations Hint 17 .11 Hint 17 .12 Hint 17 .13 Hint 17 .14 Substitute y = x λ into the differential equation. Consider the three cases: a 2 > b, a 2 < b and a 2 = b. Hint 17 .15 Hint 17 .16 Exact Equations Hint ... conditions, y 1 (0) = 1 and y  1 (0) = 0, for the first solution become, c 1 = 1, −ac 1 + √ a 2 − b c 2 = 0, c 1 = 1, c 2 = a √ a 2 − b . The co...

Ngày tải lên: 06/08/2014, 01:21

... that a n =   n/2 j=0  − 4j 2 −2j +1 (2j+2)(2j +1)  for even n, 0 for odd n. 11 93 c 1 = 1 − r 2 r 2 1 − r 1 r 2 = 1 − 1 √ 5 2 1+ √ 5 2 √ 5 = 1+ √ 5 2 1+ √ 5 2 √ 5 = 1 √ 5 Substitute this result into the equation for c 2 . c 2 = 1 r 2  1 ... second equation. c 1 r 2 1 + 1 r 2 (1 − c 1 r 1 )r 2 2 = 1 c 1 (r 2 1 − r 1 r 2 ) = 1 − r 2 1...

Ngày tải lên: 06/08/2014, 01:21

... continuous. -1 -0.5 0.5 1 -1 -0.5 0.5 1 -1 -0.5 0.5 1 -1 -0.5 0.5 1 Figure 25 .1: Polynomial Approximations to cos(πx). 12 86 0 1 2 3 4 5 0.25 0.5 0.75 1 1.25 1. 5 1. 75 2 Figure 24 .1: Plot of K 0 (x) and ... (t  ) 2 − 2x 1/ 2 t  − 1 2 x 1/ 2 = 0 1 4 x −2 + u  +  − 1 4 x 1 + u   2 − 2x 1/ 2  − 1 4 x 1 + u   − 1 2 x 1/ 2 = 0 u  + (u  ) 2 +...

Ngày tải lên: 06/08/2014, 01:21

... ∼ 4 ∞  n =1 oddn cos(nx). For x = nπ, this implies 0 = 4 ∞  n =1 oddn cos(nx), 13 61 -1 -0.5 0.5 1 -0.2 -0 .1 0 .1 0.2 1 0.25 0 .1 0 0 .1 1 1 1 0.5 Figure 28.8: Three Term Approximation for a Function ... =      −x − 1 for − 1 < x < 1/ 2 x for − 1/ 2 < x < 1/ 2 −x + 1 for 1/ 2 < x < 1. 13 55 1. Let S N be the sum of the first N ter...

Ngày tải lên: 06/08/2014, 01:21

... integration by parts we can evaluate the integral.  1 0 (1 − τ) n τ z 1 dτ =  (1 − τ) n τ z z  1 0 −  1 0 −n (1 − τ) n 1 τ z z dτ = n z  1 0 (1 − τ) n 1 τ z dτ = n(n − 1) z(z + 1)  1 0 (1 − τ) n−2 τ z +1 dτ = n(n ... 1)  1 0 (1 − τ) n−2 τ z +1 dτ = n(n − 1) ··· (1) z(z + 1) ···(z + n 1)  1 0 τ z+n 1 dτ = n(n − 1) ··· (1) z(z + 1) ···(z + n 1...

Ngày tải lên: 06/08/2014, 01:21

... case 1 < t. u(x, t) =                0, x + t < 1 1 2  x+t 1 (1 − |ξ|) dξ, 1 < x + t < 1 1 2  1 1 (1 − |ξ|) dξ, x − t < 1, 1 < x + t 1 2  1 x−t (1 − |ξ|) ... =                0, x + t < 1 1 2  x+t 1 (1 − |ξ|) dξ, x − t < 1 < x + t 1 2  x+t x−t (1 − |ξ|) dξ, 1 < x − t, x + t < 1 1 2  1 x−t...

Ngày tải lên: 06/08/2014, 01:21

... with u = α (1 − v) for special values of α and r. Find v(x) for this special case. 216 8 L 1 L 2 (L 1 x) = L 1 (L 1 l 2 − I)x = L 1 (λx − x) = (λ − 1) (L 1 x) L 1 L 2 (L 2 x) = (L 2 L 1 + I)L 2 x = ... 2 √ 1 − t 2 log (1 + t), 1 2  G + (t) + G − (t)  = ıt log (1 + t). For t ∈ (−∞, 1) , G + (t) = ı  √ 1 − t 2 + t  (log(−t − 1) + ıπ) , G − (t) = ı  − √...

Ngày tải lên: 06/08/2014, 01:21