# 4 randall d knight physics for scientists and engineers a strategic approach with modern physics 17

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## Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 4 pptx

... approximate sin (1) , 13 15 1 + ≈ 0.8 41 6 67 12 0 10 7 To see that this has the required accuracy, sin (1) ≈ 0.8 41 4 71 Solution 3 .19 Expanding the terms in the approximation in Taylor series, ∆x3 ∆x4 ∆x2 f ... Example 4. 4 .1 Consider the partial fraction expansion of + x + x2 (x − 1) 3 The expansion has the form a0 a1 a2 + + (x − 1) (x − 1) x 1 127 The coeﬃcients are (1 + x + x2 )|x =1 = 3, 0! d a1 = (1 + ... − x→0 10 9 x =0 c ln lim x→+∞ 1+ x x = lim x→+∞ = lim x→+∞ = lim x→+∞ = lim ln + 1/ x 1+ x→+∞ 1+ =1 Thus we have lim x→+∞ 1+ 11 0 x x 1 x − x2 1/ x2 x→+∞ = lim x ln 1+ x x ln + x x = e x 1 d It...
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## Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx

... continuously deformed to C2 on the domain where the integrand is analytic Thus the integrals have the same value 5 14 -4 -2 -2 -4 Figure 11 .2: The contours and the singularities of 3z +1 -6 -4 -2 -2 -4 -6 ... , C2 and C2 C z dz = z3 − C1 z− + √ C2 z− C3 z− + = 2 + 2 + 2 z− √ √ √ z− z− z √ dz e 2 /3 z − e− 2 /3 z √ √ dz 2 /3 z − 9e z − e− 2 /3 z √ √ dz z − e 2 /3 z − e− 2 /3 z− 9 √ z e 2 /3 ... deform C onto C1 and C2 = C + C1 520 C2 -4 C1 C2 -2 C -2 -4 Figure 11.5: The contours for (z +z+ı) sin z z +ız We use the Cauchy Integral Formula to evaluate the integrals along C1 and C2 ...
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## Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 4 pot

... 4x3 To check the theorem, d x x2 d ∆[A(x)] = dx dx x2 x4 2x x x2 = + 2x 4x3 x x = x4 − 2x3 + 4x4 − 2x3 = 5x4 − 4x3 16 .4. 2 The Wronskian of a Set of Functions A set of functions {y1 , y2 , ... Example 16 .4. 1 Consider the the matrix A(x) = x x2 x2 x4 The determinant is x5 − x4 thus the derivative of the determinant is 5x4 − 4x3 To check the theorem, d x x2 d ∆[A(x)] = dx dx x2 x4 2x x ... [¯] = y 9 03 For the same reason, if yp is a particular solution, then yp is a particular solution as well Since the real and imaginary parts of a function y are linear combinations of y and y ,...
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## Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 1 docx

... diﬀerential equation as ∞ w = c1 z 3 /4 1+ n =1 n 16 ∞ zn n!(n + 1) ! + c2 z 1 /4 n =1 ∞ c1 z 3 /4 + c2 z 1 /4 1+ n =1 23.2 .1 1+ 16 n 16 n zn n!(n + 1) ! zn n!(n + 1) ! Indicial Equation Now let’s ... c2 = (1 − c1 r1 ) r2 We substitute this into the second equation (1 − c1 r1 )r2 = r2 c1 (r1 − r1 r2 ) = − r2 c r1 + 11 81 n c1 = = = − r2 − r1 r2 r1 √ 1 √ √ 1+ 5 √ 1+ √ √ 1+ 5 1 =√ Substitute ... 31 an = 2(n 1) /2 (n − 2)(n − 4) · · · (1) For the even terms, a2 = a4 = 22 a6 = 42 an = 2(n−2)/2 (n − 2)(n − 4) · · · (2) Thus an = 2(n 1) /2 (n−2)(n 4) ··· (1) 2(n−2)/2 (n−2)(n 4) ···(2) 11 83 for...
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## Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 2 pps

... + x + x2 /2 + O(x3 ))(x + a2 x2 + a3 x3 + O(x4 )) = (2a2 x + 6a3 x2 ) + (2x + 4a2 x2 ) + (6x + 6(1 + a2 )x2 ) = O(x3 ) = 17 a2 = 4, a3 = 17 y1 = x − 4x2 + x3 + O(x4 ) Now we see if the second ... yields z= t dz = − dt t d d = −t2 dz dt w + d2 d d = −t2 −t2 dz dt dt d d = t4 + 2t3 dt dt 1 24 0 The equation for u is then t4 u + 2t3 u + (2t + 3t2 )(−t2 )u + t2 u = u + −3u + u = t We see that ... − a0 (2n)(2n − 2) · · · · a0 = (−1)n n , n≥0 m=1 2m = (−1)n a2n−1 2n + a2n−3 = (2n + 1)(2n − 1) a2n+1 = − a1 (2n + 1)(2n − 1) · · · · a1 = (−1)n n , n≥0 m=1 (2m + 1) = (−1)n If {w1 , w2 } is...
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## Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 3 pdf

... 10−29 4. 14 × 10 37 2.09 × 10 45 One Term Relative Error 0 .32 03 0.1 044 0.0507 0.0296 0.0192 0.0 135 0.0100 0.0077 0.0061 0.0 049 Three Term Relative Error 0. 649 7 0.0182 0.0020 3. 9 · 10 4 1.1 · 10 4 3. 7 ... = x 3x2 − P2 (x) = 5x − 3x P3 (x) = 35 x4 − 30 x2 + P4 (x) = Expanding cos(πx) in Legendre polynomials cos(πx) ≈ cn Pn (x), n=0 and calculating the generalized Fourier coeﬃcients with the formula ... even for fairly small values of x 24. 3 Integration by Parts Example 24. 3. 1 The complementary error function erfc(x) = √ π 12 63 ∞ e−t dt x 1.75 1.5 1.25 0.75 0.5 0.25 Figure 24. 1: Plot of K0 (x) and...
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## Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 4 docx

... the eigenvalues as λn and the eigenfunctions as φn for n ∈ Z+ For the moment we assume that λ = is not an eigenvalue and that the eigenfunctions are real-valued We expand the function f (x) ... compute the eigenvalues However, we can often use the formula to obtain information about the eigenvalues before we solve a problem Example 27 .4. 2 Consider the self-adjoint eigenvalue problem −y ... equation formally self-adjoint xy + y + xy = d (xy ) + xy = dx Result 27.2.1 If L = L∗ then the linear operator L is formally self-adjoint Second order formally self-adjoint operators have the form...
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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 4 Part 6 pps

... Parseval’s theorem for this series to ﬁnd the value of ∞ 16 n=1 π 1 = n6 π −π ∞ 16 n=1 ∞ n=1 x3 π − x 3 16 = n6 945 6 = n6 945 13 76 ∞ n=1 n6 dx Solution 28.2 We diﬀerentiate the partial sum of ... 16 = n4 π n=1 ∞ π x4 dx −π 2π 2π + 16 = n4 n=1 ∞ n=1 4 = n4 90 1375 Now we integrate the series for f (x) = x2 x ξ2 − ∞ π2 3 dξ = n=1 ∞ (−1)n n2 x cos(nξ) dξ x π (−1)n − x =4 sin(nx) 3 n3 n=1 ... = π 4( −1)n = n2 a0 = Thus the Fourier series is ∞ π2 (−1)n x = +4 cos(nx) for x ∈ (−π π) n2 n=1 ∞ n=1 n4 We apply Parseval’s theorem for this series to ﬁnd the value of ∞ 2π 1 + 16 = n4 π...
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## Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 7 docx

... eigenvalues and eigenfunctions for: d4 φ = λφ, dx4 φ(0) = φ (0) = 0, φ(1) = φ (1) = Hint, Solution 144 2 29.5 Hints Hint 29.1 Hint 29.2 Hint 29.3 Hint 29 .4 Write the problem in Sturm-Liouville form to ... = 0, λ = 1 /4 is not an eigenvalue 144 9 Now consider the case λ = 1 /4 A set of solutions is √ (x + 1)(1+ 1 4 )/2 , (x + 1)(1− √ 1 4 )/2 We can write this in terms of the exponential and the logarithm ... p0 y = µy, for a ≤ x ≤ b, α1 y(a) + α2 y (a) = 0, β1 y(b) + β2 y (b) = 0, where the pj are real and continuous and p2 > on [a, b], and the αj and βj are real can be written in the form of a regular...
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## Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 8 potx

... s 146 8 From part (a) we know that there are only positive eigenvalues The general solution of the diﬀerential equation is φ = c1 cos(λ1 /4 x) + c2 cosh(λ1 /4 x) + c3 sin(λ1 /4 x) + c4 sinh(λ1 /4 x) ... conditions c1 sin(λ1 /4 ) + c2 sinh(λ1 /4 ) = −c1 λ1/2 sin(λ1 /4 ) + c2 λ1/2 sinh(λ1 /4 ) = We see that sin(λ1 /4 ) = The eigenvalues and eigenfunctions are λn = (nπ )4 , φn = sin(nπx), 146 9 n ∈ N Chapter ... r≤ √ Thus the smallest zero of J0 (x) is less than or equal to ≈ 2 .44 94 (The smallest zero of J0 (x) is approximately 2 .40 483 .) (1 Solution 29.9 We assume that < l < π Recall that the solution...
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## Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 9 docx

... the answer for ν = 0? Hint, Solution 1 49 9 Exercise 31.12 Find the inverse Laplace transform of ˆ f (s) = s3 − 2s2 +s−2 with the following methods ˆ Expand f (s) using partial fractions and then ... transform to ﬁnd y(t) We could expand the right side in partial fractions and then use a table of Laplace transforms Since the function is analytic except for 15 29 isolated singularities and vanishes ... Laplace transform of y (s) by ﬁrst ﬁnding its partial fraction expansion ˆ s/3 s/3 s − + +1 s +4 s +1 s/3 4s/3 + =− s +4 s +1 y(t) = − cos(2t) + cos(t) 3 y (s) = ˆ s2 Example 31 .4. 3 Consider...
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