4 randall d knight physics for scientists and engineers a strategic approach with modern physics 07
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...
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...
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 ... [...
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+...
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...
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 coe...
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...
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 Converge...
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...
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...
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 =...
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 singular...
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