Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 1 pps

... 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

... rule. 2. lim z→ı 1 + z 2 2 + 2z 6 =  2z 12 z 5  z=ı = 1 6 lim z→ıπ sinh(z) e z +1 =  cosh(z) e z  z=ıπ = 1 399 8.7 Hints Complex Derivatives Hint 8 .1 Hint 8.2 Start with the Cauchy-Riemann equation 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...

Ngày tải lên: 06/08/2014, 01:21

... Condition. ˆ F y  y  = 12 (ˆy  ) 2 − 12 = 12   β α  2 − 1       < 0 for |β/α| < 1 = 0 for |β/α| = 1 > 0 for |β/α| > 1 Thus we see that β α x may be a minimum for |β/α| ≥ 1 and may be ... =  sin(xy) for x ≥ 1 and y ≤ 1, 0 otherwise In both cases state for which values of λ the solution obtained is valid. Exercise 48.24 1. Suppose tha...

Ngày tải lên: 06/08/2014, 01:21

... 11 .4 Hint 11 .5 Hint 11 .6 Hint 11 .7 Hint 11 .8 Hint 11 .9 Hint 11 .10 509 The series clearly diverges for |z| ≥ 1 since the terms do not vanish as n → ∞. Consider the partial sum, S N (z) ≡  N 1 n=0 z n , ... integral  C e zt z 2 (z + 1) dz. There are singularities at z = 0 and z = 1. Let C 1 and C 2 be contours around z = 0 and z = 1. See Figure 11 .6. We defor...

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 ... ··· 558 For |z| > 1, 1 1 + z = 1/ z 1 + 1/ z = 1 z  1 +  1 1  z 1 +  1 2  z −2 + ···  = z 1 − z −2 + z −3 − ··· 559 in the annulus...

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 ... doubles every hour. For the continuous problem, we assume that this i s true for y(t). We write this as an equation: y  (t) = αy(t). 775 1 2 3 4 4 8 12 16 1 2 3 4 4 8...

Ngày tải lên: 06/08/2014, 01:21

... Equations 10 17 19 .6 Hints The Constant Coefficient Equation Normal Form Hint 19 .1 Transform the equation to normal form. Transformations of the Independent Variable Integral Equations Hint 19 .2 Transform ... olutions, u 1 and u 2 . u = c 1 u 1 (x) + c 2 u 2 (x) The solution of the Ricatti equation is then y = − c 1 u  1 (x) + c 2 u  2 (x) a(x)(c 1 u 1 (x) + c 2 u 2 (...

Ngày tải lên: 06/08/2014, 01:21

... problem. Example 21. 4.2 Again conside r y  − y = e −α|x| , y(±∞) = 0, α > 0, α = 1. 10 72 0.5 1 -0.3 -0.2 -0 .1 0 .1 0.5 1 -0.3 -0.2 -0 .1 0 .1 0.5 1 -0.3 -0.2 -0 .1 0 .1 0.5 1 -0.3 -0.2 -0 .1 0 .1 Figure 21. 3: ... equation. u  1 y 1 + 2u  1 y  1 + u 1 y  1 + u  2 y 2 + 2u  2 y  2 + u 2 y  2 + p(u  1 y 1 + u 1 y  1 + u  2 y 2 + u 2...

Ngày tải lên: 06/08/2014, 01:21

... 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 11 81 0.2 0.4 0 .6 0.8 1 1.2 0.3 0.4 0.5 0 .6 0.7 0.8 0.9 Figure 23.2: Plot of the solution and approximations. Recall ... 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...

Ngày tải lên: 06/08/2014, 01:21

... =  1 1 1 dx = 2  1 1 P 1 (x)P 1 (x) dx =  1 1 x 2 dx =  x 3 3  1 1 = 2 3  1 1 P 2 (x)P 2 (x) dx =  1 1 1 4  9x 4 − 6x 2 + 1  dx =  1 4  9x 5 5 − 2x 3 + x  1 1 = 2 5  1 1 P 3 (x)P 3 (x) ... 1) a n +1 + 1 2 (n + 1) a n +1 + a n  z n 1 + 1 2z a 1 + 1 z a 0 = 0. Equating powers of z, z 1 : a 1 2 + a 0 = 0 → a 1 = −2a 0 z n 1 :  n...

Ngày tải lên: 06/08/2014, 01:21