Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 3 potx
... of t and t , −2t − 1 x ∼ 0 t ∼ − 1 2x t ∼ − 1 2 ln x as x → ∞. 16 75 We can transform Equation 36 .3 to this form with the change of variables σ = ξ + ψ, τ = ξ − ψ. Equation 36 .3 becomes u σσ − ... defined h i = ∂x 1 ∂ξ i 2 + ∂x 2 ∂ξ i 2 + ∂x 3 ∂ξ i 2 . The gradient, divergence, etc., follow. ∇u = a 1 h 1 ∂u ∂ξ 1 + a 2 h 2 ∂u ∂ξ 2 + a 3 h 3 ∂u ∂ξ 3 ∇ · v...
Ngày tải lên: 06/08/2014, 01:21
... + x 2 2 + x 3 6 The four approximations are graphed in Figure 3. 11. -1 -0 .5 0 .5 1 0 .5 1 1 .5 2 2 .5 -1 -0 .5 0 .5 1 0 .5 1 1 .5 2 2 .5 -1 -0 .5 0 .5 1 0 .5 1 1 .5 2 2 .5 -1 -0 .5 0 .5 1 0 .5 1 1 .5 2 2 .5 Figure 3. 11: ... through the points (x, y(x)) and (x + ∆x, y(x + ∆x)). See Figure 3 .5. y x ∆ y ∆ x Figure 3 .5: The increments ∆x and ∆y. 56 -10 -5...
Ngày tải lên: 06/08/2014, 01:21
... dz = sin 3 z 3 ıπ π = 1 3 sin 3 (ıπ) −sin 3 (π) = −ı sinh 3 (π) 3 Again the anti-derivative is single-valued. 491 3. e x (sin x cos y cosh y −cos x sin y sinh y) Exercise 9 .3 For an analytic ... max a≤x≤b |f(x)|. 466 with a, b and c complex-valued constants and d a real constant. Substituting z = x + ıy and expanding products yields, a x 3 + ı3x 2 y − 3xy...
Ngày tải lên: 06/08/2014, 01:21
Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 3 pdf
... 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). 1286 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 it’s ... assumptions, u x −2 → u 5 16 x −2 u x −1 → (u ) 2 5 16 x −2 . Thus we obtain −2x 1/2 u + 5 16 x −2 ∼ 0 u ∼ 5 32 x 5/...
Ngày tải lên: 06/08/2014, 01:21
Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 8 potx
... ∼ f(0) s + f (0) s 2 + f (0) s 3 + ··· as s → +∞. Example 31 .2 .5 The Taylor series expans ion of sin t about the origin is sin t = t − t 3 6 + O(t 5 ). Thus the Laplace transform of sin t has the behavior L[sin ... − e −sT . Example 31 .3. 1 Consider the inverse Laplace transform of 1 s 3 −s 2 . First we factor the denominator. 1 s 3 − s 2 = 1 s 2 1 s − 1 We know the inver...
Ngày tải lên: 06/08/2014, 01:21
Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 10 potx
... converges for |(ω)| < a. This domain is shown in Figure 32 .1. Re(z) Im(z) Figure 32 .1: The Domain of Convergence 154 6 32 .6 The Fourier Cosine and Sine Transform 32 .6.1 The Fourier Cosine Transform Suppose ... Fourier transform. Define the two functions f + (x) = 1 for x > 0 1/2 for x = 0 0 for x < 0 , f − (x) = 0 for x > 0 1/2 for x = 0...
Ngày tải lên: 06/08/2014, 01:21
Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 1 pdf
... < b. 159 7 Hint 32 .10 Hint 32 .11 The left side is the convolution of u(x) and e −ax 2 . Hint 32 .12 Hint 32 . 13 Hint 32 .14 Hint 32 . 15 Hint 32 .16 Hint 32 .17 Hint 32 .18 Hint 32 .19 Hint 32 .20 157 9 Solution ... integration with a semi-circle in the lower or upper half plane. Hint 32 .3 Hint 32 .4 Hint 32 .5 Hint 32 .6 Hint 32 .7 Hint 32 .8 Hint 32 .9 157...
Ngày tải lên: 06/08/2014, 01:21
Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 2 docx
... error 5 24 23. 6 038 0.01 65 15 8.717 83 · 10 10 8.66 954 · 10 10 0.0 055 25 6.20448 · 10 23 6.1 838 4 · 10 23 0.0 033 35 2. 952 33 · 10 38 2.9 4 53 1 · 10 38 0.0024 45 2. 658 27 · 10 54 2. 6 53 35 · 10 54 0.0019 In ... 33 .2 We make the change of variable ξ = x 3 , x = ξ 1 /3 , dx = 1 3 ξ −2 /3 dξ. ∞ 0 e −x 3 dx = ∞ 0 e −ξ 1 3 ξ −2 /3 dξ = 1 3 Γ ...
Ngày tải lên: 06/08/2014, 01:21
Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 4 ppt
... it. 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. ... = 2 3 x √ −x − y dy dx = − √ −x, y = − 2 3 x √ −x + c, ψ = 2 3 x √ −x + y Next we determine x and y in terms of ξ and ψ. x = − 3 4 (ξ + ψ) 1 /3 , y = ψ −ξ 2 We c...
Ngày tải lên: 06/08/2014, 01:21
Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 5 ppt
... to 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 = ν(X Y T + XY T ) T νT = X X + Y Y = ... solution of the partial differential equation and is thus twice continuously differentiable, (u ∈ C 2 ). In particular, this implies that R and Θ are boun...
Ngày tải lên: 06/08/2014, 01:21