Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 3 pdf

... x 2 , x 3 , x 4 } is independent, but not orthogonal in the interval [−1, 1]. Using Gramm-Schmidt orthogo- 12 84 1 2 3 4 5 6 -60 -40 -20 Figure 24. 3: log(error in approximation) In Figure 24. 4 we ... = 1 √ π x −1 e −x 2 − 1 √ π  ∞ x t −2 e −t 2 dt = 1 √ π x −1 e −x 2 − 1 √ π  − 1 2 t 3 e −t 2  ∞ x + 1 √ π  ∞ x 3 2 t 4 e −t 2 dt = 1 √ π e −x 2  x −1 − 1 2 x 3  + 1 √...

Ngày tải lên: 06/08/2014, 01:21

... solution and discuss in as much detail as possible what goes wrong. 13 64 Note that this formula is valid for m = 0, 1, 2, . . Similarly, we can multiply by sin(mx) and integrate to solve for b m . ... + n 2 x 2 − 2 n 3 sin(nx) 137 0 -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 Figure 28.2: Graph of ˆ f(x). would give a better approximation? Least squared error fit. The most common crit...

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... . . . . . . . . . . . . . . 1 630 34 . 3. 3 Bessel Functions of Non-Integer Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 633 34 . 3 .4 Recursion Formulas . . . . . . . . . . . ... . . 1910 43 .10Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1911 44 Transform Methods 1918 44 .1 Fourier Transfor...

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... + ∆x 3 6 f  (x) + ∆x 4 24 f  (x 1 ), f(x −∆x) = f(x) −∆xf  (x) + ∆x 2 2 f  (x) − ∆x 3 6 f  (x) + ∆x 4 24 f  (x 2 ), where x ≤ x 1 ≤ x + ∆x and x − ∆x ≤ x 2 ≤ x. Hint 3. 20 Hint 3. 21 a. ... positive for −π < x < 0 and negative for 0 < x < π. Since the sign of y  goes from positive to negative, x = 0 is a relative maxima. See Figure 3. 7. Exam...

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... = 1 2 x 3 e 2x − 3 4 x 2 e 2x + 3 2  1 2 x e 2x − 1 2  e 2x dx   x 3 e 2x dx = 1 2 x 3 e 2x − 3 4 x 2 e 2x + 3 4 x e 2x − 3 8 e 2x +C Solution 4. 15 Expanding the integrand in partial fractions, 1 x 2 − 4 = 1 (x ... +  2 0 √ x dx =  2 3 x 3/ 2  1 0 +  2 3 x 3/ 2  2 0 = 2 3 + 2 3 2 3/ 2 = 2 3 (1 + 2 √ 2) Solution 4. 3 d dx  x 2 x f(ξ) dξ = f(x...

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... =  ız 4 4 − 1 2z 2  ı 1+ı = 1 2 + ı In this example, the anti-derivative is single-valued. 2.  C sin 2 z cos z dz =  sin 3 z 3  ıπ π = 1 3  sin 3 (ıπ) −sin 3 (π)  = −ı sinh 3 (π) 3 Again ... max a≤x≤b |f(x)|. 46 6 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 2 − ıy...

Ngày tải lên: 06/08/2014, 01:21

... cosh z z 3 sin z sinh z =  1 − z 2 2 + z 4 24 − ···  1 + z 2 2 + z 4 24 + ···  z 3  z − z 3 6 + z 5 120 − ···  z + z 3 6 + z 5 120 + ···  = 1 − z 4 6 + ··· z 3  z 2 + z 6  −1 36 + 1 60  + ... ı /4) (n + 1))  z n + d, for 2 < |z| Solution 12.29 The radius of convergence of the series for f(z) is R = lim n→∞     k 3 /3 k (k + 1) 3 /3 k+1    ...

Ngày tải lên: 06/08/2014, 01:21

... polynomial are α 1 = 1 +  1 − 3 /4 2 = 3 4 , α 2 = 1 −  1 − 3 /4 2 = 1 4 . Thus our two series solutions will be of the form w 1 = z 3 /4 ∞  n=0 a n z n , w 2 = z 1 /4 ∞  n=0 b n z n . Substituting ... + 3) For our first solution we have the difference equation a 0 = 1, a 1 = 0, a 2 = 0, a 3 = 0, a n +4 = − a n (n + 4) (n + 3) . For our second solution, b 0 = 0, b...

Ngày tải lên: 06/08/2014, 01:21

... relation for the coefficients. (k + 1)kc k+1 + b(k + 1)c k+1 − kc k − ac k = 0 c k+1 = k + a (k + 1)(k + b) c k 1 244 23. 6 Hints Hint 23. 1 Hint 23. 2 Hint 23. 3 Hint 23 .4 Hint 23. 5 Hint 23. 6 Hint 23. 7 Hint ... =  1 −1 x 2 dx =  x 3 3  1 −1 = 2 3  1 −1 P 2 (x)P 2 (x) dx =  1 −1 1 4  9x 4 − 6x 2 + 1  dx =  1 4  9x 5 5 − 2x 3 + x  1 −1 = 2 5  1 −1 P 3 (x)P 3...

Ngày tải lên: 06/08/2014, 01:21

... the formula to obtain information about the eigenvalues before we solve a problem. Example 27 .4. 2 Consider the self-adjoint eigenvalue problem −y  = λy, y(0) = y(π) = 0. 132 2 Example 27 .4. 1 ... eigenfunction φ. Green’s formula states φ|L[φ] − L[φ]|φ = 0 φ|λφ − λφ|φ = 0 (λ − λ)φ|φ = 0 Since φ ≡ 0, φ|φ > 0. Thus λ = λ and λ is real. 131 9 27.7 Hints Hint 27.1 132 7 2...

Ngày tải lên: 06/08/2014, 01:21