# 5 randall d knight physics for scientists and engineers a strategic approach with modern physics 12

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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...
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## 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...
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## Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 4 ppt

... φ(x, 0) = for < x < l, with boundary conditions φ(0, t) = for t > 0, and φ(l, t) + φx (l, t) = for t > Obtain two series solutions for this problem, one which is useful for large t and the other ... Solution 1696 36 .4 Hints Hint 36.1 Hint 36.2 Hint 36.3 1697 36 .5 Solutions Solution 36.1 For y = −1, the equation is parabolic For this case it is already in the canonical form, uxx = For y = −1, ... σ and τ σx = σy = 2y τx = 2x τy = Then we calculate the derivatives of u ux = 2xuτ uy = 2yuσ uxx = 4x2 uτ τ + 2uτ uyy = 4y uσσ + 2uσ Finally we transform the equation to canonical form σ (4 ...
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## 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...
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## 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...
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## 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...
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## 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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