Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 6 pps

... = ∞  n=1 2(−1) n+1 n sin(nx) for x ∈ (−π . . . π). We apply Parseval’s theorem for this series to find the value of  ∞ n=1 1 n 2 . ∞  n=1 4 n 2 = 1 π  π −π x 2 dx ∞  n=1 4 n 2 = 2π 2 3 ∞  n=1 1 n 2 = π 2 6 13 74 a n = 2 π  π 0 f(x) ... no value of a for which both cos a and sin a vanish, the system is not orthogonal for any interval of length π. 2. First note that  π...

Ngày tải lên: 06/08/2014, 01:21

... ellipse. 187 6. 8 Hints Complex Numbers Hint 6. 1 Hint 6. 2 Hint 6. 3 Hint 6 .4 Hint 6. 5 Hint 6. 6 Hint 6. 7 The Complex Plane Hint 6. 8 Hint 6. 9 208 Part III Functions of a Complex Variable 179 6. 7 Exercises Complex ... y 2 = 4 x 2 + y 2 = 16 − 8  (x − 2) 2 + y 2 + x 2 − 4x + 4 + y 2 x − 5 = −2  (x − 2) 2 + y 2 x 2 − 10x + 25 = 4x 2 − 16x + 16 + 4y 2 1 4 (x − 1) 2...

Ngày tải lên: 06/08/2014, 01:21

... factor is I(t) = e R −4t −1 dt = e 4 log t = t 4 . We multiply by the integrating factor and integrate. d dt  t 4 u  = −2t 6 t 4 u = 2 5 t −5 + c u = 2 5 t −1 + ct 4 1011 18 .6 *Equidimensional-in-y ... − 1 4 y 4 + c 1 u =  2c 1 − 1 2 y 4  1/2 y  =  2c 1 − 1 2 y 4  1/2 dy (2c 1 − 1 2 y 4 ) 1/2 = dx Integrating gives us the implicit solution  1 (2c 1 − 1 2 y 4...

Ngày tải lên: 06/08/2014, 01:21

... =  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) dx =  1 −1 1 4  25x 6 − 30x 4 + 9x 2  dx =  1 4  25x 7 7 − 6x 5 + 3x 3  1 −1 = 2 7 Solution ... relation for the coefficients. (k + 1)kc k+1 + b(k + 1)c k+1 − kc k − ac k = 0 c k+1 = k + a (k + 1)(k + b) c k 1 244 23 .6 Hint...

Ngày tải lên: 06/08/2014, 01:21

... A 1 = 0, A 2 = 4 3 , A 3 = A 4 = ··· = 0 Ψ(θ, t) = − 1 3 P 0 (cos θ) + 4 3 P 2 (cos θ) exp  − 6a 2 R 2 t  Ψ(θ, t) = − 1 3 +  2 cos 2 θ − 2 3  exp  − 6a 2 R 2 t  Solution 37. 34 Since we have ... inhomogeneous partial differential equation, we will expand the solution in a series of eigenfunctions in x for which the coefficients are functions of t. The solution for u has the form...

Ngày tải lên: 06/08/2014, 01:21

... equations for µ and ν are satisfied if and only if the Cauchy-Riemann equations for u and v are satisfied. The continuity of the first partial derivatives of u and v implies the same of µ and ν. Thus ... =  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 that u x = v y and u y...

Ngày tải lên: 06/08/2014, 01:21

... , converges for α > 1 and diverges for α ≤ 1. Hint, Solution 5 64 Example 12.3.2 Convergence and Uniform Convergence. Consider the series log(1 − z) = − ∞  n=1 z n n . This series converges for |z| ... large. Thus this series is not uniformly convergent in the domain |z| ≤ 1, z = 1. The series is uniformly convergent for |z| ≤ r < 1. 545 12.2.2 Uniform Convergence and...

Ngày tải lên: 06/08/2014, 01:21

... 3) n 2. ∞  n=2 Log z ln n 3. ∞  n=1 z n 4. ∞  n=1 (z + 2) 2 n 2 5 94 Thus the series converges absolutely for |z| < 1 /4. 6. By the Cauchy-Hadamard formula, the radius of absolute convergence ... 1)!     = 0 The series is convergent. 17. ∞  n=1 n 8 + 4n 4 + 8 3n 9 − n 5 + 9n = ∞  n=1 1 n 1 + 4n 4 + 8n −8 3 − n 4 + 9n −8 > 1 4 ∞  n=1 1 n We s ee that the series is...

Ngày tải lên: 06/08/2014, 01:21

... Equations 1017 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 the ... equation to normal form an d then apply the scale transformation x = λξ + µ. Hint 19.3 Transform the equation to normal form an d then apply the scale transformation x = λξ. Hint 19 ....

Ngày tải lên: 06/08/2014, 01:21

... identity,  ∞ −∞ f(x)δ (n) (x) dx = (−1) n f (n) (0), 1055 -4 -2 2 4 -0 .6 -0 .4 -0.2 0.2 0 .4 0 .6 Figure 21 .4: Plot of G(x|0). 21.7.1 Green Functions for Sturm-Liouville Problems Consider the problem L[y] ... = 0 a − 1 = 0, b − 4a = 0, 2a − 2b + c = 0 a = 1, b = 4, c = 6 A particular solution is y p = t 2 + 4t + 6. 1 062 With the constraints, we have a system of linear equa...

Ngày tải lên: 06/08/2014, 01:21