5 randall d knight physics for scientists and engineers a strategic approach with modern physics 07

Nonlinear physics with mathematica for scientists and engineers r ennis, g mcguire

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

... 87 − 6 1/ 3 1/ 3 6−2/3 + √ 2/3 87 √ + 87 , 0, − 6 1/ 3   ≈ (0 .58 9 755 , 0, 0.347 81) 1/ 3 The closest point is shown graphically in Figure 5 .10 1- 1 -0 .5 0 .5 -1 -0 .5 0 0 .5 1. 5 0 .5 Figure 5 .10 : Paraboloid, ... dx2 = (1 + 2x) x= 1 = 1 x= 1 = x= 1 (2) x= 1 =1 Then we can the integration + x + x2 dx = (x + 1) 3 1 − + (x + 1) (x + 1) x +1 1 + + ln |x + 1| =− 2(x + 1) 2 x + x + 1/ 2 + ln |x + 1| = (x + 1) 2 dx ... x3 x +1 dx = + x2 − 6x 10 Setting x = −3 yields C = − 15 + − dx 6x 10 (x − 2) 15 (x + 3) ln |x − 2| − ln |x + 3| + C = − ln |x| + 10 15 |x − 2|3 /10 = ln 1/ 6 +C |x| |x + 3|2 / 15 − Solution 4 .17 dx...

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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 5 pps

... (−1)n ln 11 n =2 ∞ 12 n =2 ∞ 13 n =2 n (n! )2 (2n)! 3n + 4n + 5n − 4n − 5 62 ∞ 14 n =2 ∞ 15 n =2 ∞ 16 n=1 ∞ 17 n=1 ∞ 18 n=1 ∞ 19 n=1 ∞ 20 n =2 n! (ln n)n en ln(n!) (n! )2 (n2 )! n8 + 4n4 + 3n9 − n5 + 9n 1 ... z +2 z + 5z + 1 √ =√ + z/3 + z /2 −1 /2 z −1 /2 z = √ 1+ + + ··· 3 z 3z z z2 = √ 1− + + ··· 1− + + ··· 24 32 17 = √ − z + z2 + · · · 12 96 12. 6 1+ −1 /2 z −1 /2 + 2 z 2 + ··· Laurent Series Result 12. 6.1 ... = 2 n=−∞ C f (ζ) dζ z n ζ n+1 For the case of arbitrary z0 , simply make the transformation z → z − z0 55 7 Im(z) Im(z) r2 R2 r1 R1 C1 C2 R2 Re(z) R1 z Re(z) C Cz Figure 12. 5: Contours for...

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Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 5 pdf

... + 13 eλx = λ2 − 4λ + 13 = λ = ± 3i Thus two linearly independent solutions are e(2+3i)x , and 966 e(2−3i)x Noting that e(2+3i)x = e2x [cos(3x) + ı sin(3x)] e(2−3i)x = e2x [cos(3x) − ı sin(3x)], ... an exact equation d 3 ex /3 y = c1 ex /3 dx ex /3 ex y = c1 y = c1 e−x /3 ex /3 /3 9 45 dx + c2 dx + c2 e−x /3 Result 17 .3. 1 If you can write a diﬀerential equation in the form d F (x, y, y , ... = The linearly independent solutions are x2+ 3 , x2− 3 We can put this in a more understandable form x2+ 3 = x2 e 3 ln x = x2 cos (3 ln x) + x2 sin (3 ln x) We can write the general solution in...

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Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 5 pdf

... not a good approximation 13 54 0 .5 -1 0 .4 0.2 -0 .5 0 .5 -1 -0 .5 0 .5 -0 .5 -0.2 -1 -0 .4 Figure 28.7: Three Term Approximation for a Function with Jump Discontinuities and a Continuous Function A ... -1 0 .5 -0 .5 0 .5 1 .5 -1 -0 .5 0 .5 -0 .5 -0 .5 -1 1 .5 -1 Figure 28.3: A Function Deﬁned on the range −1 ≤ x < and the Function to which the Fourier Series Converges bn = = = 3/2 3 f (x) sin −1 5/ 2 ... + for − < x < −1/2 for − 1/2 < x < 1/2 for 1/2 < x < 1 355 0 .5 0.2 0.1 -1 -0 .5 0 .5 0. 25 0.1 -0.1 -0.2 0.1 Figure 28.8: Three Term Approximation for a Function with Continuous First Derivative and...

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

... 32 .10 Hint 32 .11 The left side is the convolution of u(x) and e−ax 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 15 7 9 Hint 32. 21 15 8 0 32 .11 ... integral 1 (1 − τ )n τ z τz − −n (1 − τ )n 1 dτ z z 0 n (1 − τ )n 1 τ z dτ = z n(n − 1) = (1 − τ )n−2 τ z +1 dτ z(z + 1) n(n − 1) · · · (1) τ z+n 1 dτ = z(z + 1) · · · (z + n − 1) (1 − τ )n τ z 1 dτ ... factor = 1 2z 3z · · · (n + 1) z lim z n→∞ (1 + z) (1 + z/2) · · · (1 + z/n) 1z 2z · · · nz ∞ 1 (n + 1) z = z n =1 + z/n nz Thus we have Gauss’ formula for the Gamma function ∞ Γ(z) = z n =1 1+ n z 1+ z...

Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 2 docx

... 15 25 35 45 Γ(n) 24 8.71783 · 1010 6 .20 448 · 1 023 2. 9 52 33 · 1038 2. 658 27 · 1 054 √ 2 xx−1 /2 e−x relative error 23 .6038 0.01 65 8.66 954 · 1010 0.0 055 6.18384 · 1 023 0.0033 2. 9 453 1 · 1038 0.0 024 54 ... 2 4···n·(2n 2) ···(2n−n) ζ n+1 2( 2n 2) 2 4·(2n 2) (2n−4) cn (ζ) = ζ4 ζ n−1  2n−1 n! + ζ + + · · · + 2 4···(n−1)·(2n 2) ···(2n−(n−1)) ζ n+1 2( 2n 2) 2 4·(2n 2) (2n−4) 16 42 for even n for odd n Uniform Convergence ... J1 /2 (z) z 1 /2 1 /2 = z −1 /2 sin z − − z π J3 /2 (z) = π 1 /2 z −3 /2 sin z − π 1 /2 z −1 /2 cos z = 2 1 /2 π −1 /2 z −3 /2 sin z + 2 1 /2 π −1 /2 z −3 /2 sin z − 2 1 /2 π −1 /2 cos z π 1 /2 = π 1 /2 = z −3/2...

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Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 3 potx

... Hint 34 .9 Hint 34 .10 1 655 Hint 34 .11 Hint 34 .12 Hint 34 . 13 Hint 34 .14 1 656 34 .10 Solutions Solution 34 .1 Bessel’s equation is L[y] ≡ z y + zy + z − n2 y = We consider a solution of the form e ... Hint, Solution 1682 z = r cos φ 35 . 2 Hints Hint 35 . 1 Hint 35 . 2 16 83 35 . 3 Solutions Solution 35 . 1 h1 = (cos θ)2 + (sin θ)2 + = h2 = (−r sin θ)2 + (r cos θ)2 + = r h3 = u= r √ ∂ ∂r + + 12 = ∂u ∂ ... ∂ 3 h3 ∂ 3 u= h1 h2 h3 ∂ u= h1 h2 h3 ∂ξ1 ·v = 1681 35 . 1 Exercises Exercise 35 . 1 Find the Laplacian in cylindrical coordinates (r, θ, z) x = r cos θ, y = r sin θ, z Hint, Solution Exercise 35 . 2...

Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 5 ppt

... exponentially decaying Hint 37.22 For parts (a), (b) and (c) use separation of variables For part (b) the eigen-solutions will involve Bessel functions For part (c) the eigen-solutions will involve ... boundary value problems for X(x) and Y (y) and a differential equation for T (t) X + µX = 0, X (0) = X (1) = Y + (λ − µ)Y = 0, Y (0) = Y (1) = T = −λνT The solutions for X(x) form a cosine series ... solution of the partial differential equation and is thus twice continuously differentiable, (u ∈ C ) In particular, this implies that R and Θ are bounded and that Θ is continuous and has a continuous...

Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 6 pps

... solution for R(r) is Rn = j0 (λr) Applying the boundary condition at r = a, we see that the eigenvalues and eigenfunctions are λn = γn , a Rn = j0 γn r , a The problem for T becomes Tn = −κ 17 85 γn ... dx for d = nπξ L for d = L , 2n L 2n The solution for u(x, t) is, 8dL2 v u(x, t) = π2c u(x, t) = v π2c ∞ n=1 ∞ n=1 n2 n(L2 cos − 4d2 n2 ) 2nπd + L sin 2nπd L sin nπξ L sin nπx sin L nπct L for ... 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, ∞ u(x, t) = un (t) sin n=1 nπx...

Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 7 docx

... , a2 = k , cρ (39.2) so that it is valid for diffusion in a non-homogeneous medium for which c and k are functions of x and φ and so that it is valid for a geometry in which A is a function of ... not uniformly convergent and we are not allowed to differentiate it with respect to x We substitute the expansion into the partial 18 27 differential equation, multiply by the eigenfunction and integrate ... temperature and A is the cross-sectional area 1833 39.2 Hints Hint 39.1 Hint 39.2 Hint 39.3 Hint 39.4 Hint 39 .5 Check that the expression for w(x, t) satisfies the partial differential equation and initial...

Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 8 doc

... was correct 18 75 Direct Solution D’Alembert’s solution is valid for all x and t We formally substitute t − T for t in this solution to solve the problem with deﬂection u(x, T ) = φ(x) and velocity ... determine α(ξ) and β(ξ) for ξ > Then use the initial condition to determine β(ξ) for ξ < Hint 41.6 Hint 41.7 a) Substitute u(x, t) = (A eıωt−αx ) into the partial diﬀerential equation and solve for α ... solution to solve this problem ∞ u(x, t) = H n=−∞ − x − − 8n − t 2 +H +H − x − − 8n + t 2 13 − x− − 8n − t 2 186 9 +H 13 − x− − 8n + t 2 The solution for several times is plotted in Figure 41.2 Note that...

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Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 5 pdf

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

Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 3 potx