Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 1 pdf

... 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

... imaginary part of some analytic function. Solution 8 .11 We write the real and imaginary parts of f(z) = u + ıv. u =  x 4/3 y 5/ 3 x 2 +y 2 for z = 0, 0 for z = 0. , v =  x 5/ 3 y 4/3 x 2 +y 2 for ... 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...

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 Fourier ... 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...

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

... 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 ... c 2 . c 2 = 1 r 2  1 − 1 √ 5 r 1  = 2 1 − √ 5  1 − 1 √ 5 1 + √ 5 2  = − 2...

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 ... = 2 √ π  ∞ x  1 2t   −2t e −t 2  dt = 2 √ π  1 2t e −t 2  ∞ x − 2 √ π  ∞ x 1 2 t −2 e −t 2 dt = 1 √ π x 1 e −x 2 −...

Ngày tải lên: 06/08/2014, 01:21

... 1 2t + z 2 4t 2  t −n 1 e t−z 2 /4t dt 16 29 n Γ(n) √ 2πx x 1/ 2 e −x relative error 5 24 23.6038 0. 01 65 15 8. 717 83 · 10 10 8.66 954 · 10 10 0.0 055 25 6.20448 · 10 23 6 .18 384 · 10 23 0.0033 35 2. 952 33 · 10 38 2.9 453 1 · 10 38 0.0024 45 ... ( 1) n J n (z) Thus we see that J −n (z) and J n (z) are not linearly independent for integer n. 16 25 2...

Ngày tải lên: 06/08/2014, 01:21

... variables t → 1/ t in the integral representation of y 1 (t). y 1 (t) =  C t n 1 e 1 2 z(t 1/ t) dt =  C ( 1/ t) n 1 e 1 2 z( 1/ t+t) 1 t 2 dt =  C ( 1) n t −n 1 e 1 2 z(t 1/ t) dt = ( 1) n y 2 (t) Thus ... equation.  C L  e 1 2 z(t 1/ t)  v(t) dt = 0  C  z 2 1 4  t + 1 t  2 + z 1 2  t − 1 t  2 +  z 2 − n 2   e 1 2 z(t 1/ t) v(t) dt = 0 (34 .1) By...

Ngày tải lên: 06/08/2014, 01:21

... I planned it that way. 17 05 36 .5 Solutions Solution 36 .1 For y = 1, the equation is parabolic. For this case it is already in the canonical form, u xx = 0. For y = 1, the equation is elliptic. ... and y in terms of ξ and ψ. x = −  3 4 (ξ + ψ)  1/ 3 , y = ψ −ξ 2 We calculate the derivatives of ξ and ψ. ξ x = √ −x =  3 4 (ξ + ψ)  1/ 6 , ξ y = 1 ψ x =  3 4 (ξ + ψ) ...

Ngày tải lên: 06/08/2014, 01:21

... 1) 2 R n = 0 α(α − 1) + α −(2n − 1) 2 = 0 α = ±(2n − 1) R n = c 1 r 2n 1 + c 2 r 1 2n The solution which is bounded in 0 ≤ r ≤ 1 is R n = r 2n 1 . 17 42 We substitute this form into the wave equation ... obtain a problem with homogeneous boundary conditions. Hint 37 .13 Hint 37 .14 Hint 37. 15 Hint 37 .16 Hint 37 .17 17 35 Solution 37.4 1. u t = ν(u xx + u yy ) XY T ...

Ngày tải lên: 06/08/2014, 01:21