Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 8 potx

... − t 3 6 + O(t 5 ). Thus the Laplace transform of sin t has the behavior L[sin t] ∼ 1 s 2 − 1 s 4 + O(s −6 ) as s → +∞. 1 48 8 We use integration by parts and utilize the homogeneous boundary conditions. [v(pv  )  ] b a − ... ≤ √ 6 Thus the smallest zero of J 0 (x) is less than or equal to √ 6 ≈ 2 .44 94. (The smallest zero of J 0 (x) is approximately 2 .40 483 .) Solution 29.9 We as...

Ngày tải lên: 06/08/2014, 01:21

... take the Laplace transform of the differential equation and solve for ˆy(s). s 2 ˆy −sy(0) − y  (0) + 4sˆy −4y(0) + 4 y = 4 s + 1 s 2 ˆy −2s + 3 + 4sˆy 8 + 4 y = 4 s + 1 ˆy = 4 (s + 1)(s + 2) 2 + 2s ... define the Fourier transform. Define the two functions f + (x) =      1 for x > 0 1/2 for x = 0 0 for x < 0 , f − (x) =      0 for x > 0 1/2 for x = 0 1...

Ngày tải lên: 06/08/2014, 01:21

... Z z = 4m 1 + 4n , m, n ∈ Z There are an infinite set of rational numbers for which ı z has 1 as one of its values. For example, ı 4/ 5 = 1 1/5 =  1, e ı2π/5 , e 4 /5 , e ı6π/5 , e 8 /5  7 .8 Riemann ... See Figure 7. 18 and Figure 7.19 for plots of the real and imaginary parts of the cosine and sine, respectively. Figure 7.20 shows the modulus of the cosine and the sine. T...

Ngày tải lên: 06/08/2014, 01:21

... by expanding the function in a Laurent series. (1 − cos z) 2 z 7 = z −7  1 −  1 − z 2 2 + z 4 24 + O  z 6   2 = z −7  z 2 2 − z 4 24 + O  z 6   2 = z −7  z 4 4 − z 6 24 + O  z 8   = 1 4z 3 − 1 24z + ... 13.39 Hint 13 .40 Definite Integrals Involving Sine and Cosine Hint 13 .41 Hint 13 .42 Hint 13 .43 Hint 13 .44 Make the changes of variables x = sin ξ and the...

Ngày tải lên: 06/08/2014, 01:21

... doubles every hour. For the continuous problem, we assume that this i s true for y(t). We write this as an equation: y  (t) = αy(t). 775 1 2 3 4 4 8 12 16 1 2 3 4 4 8 12 16 Figure 14. 1: The p opulation ... integrating factor for the Equation 14. 4. The proof of this is left as an exercise for the reader. (See Exercise 14. 2.) Result 14. 4 .4 Homogeneous Coefficient Differentia...

Ngày tải lên: 06/08/2014, 01:21

... 0, 0 for x > 0. This behavior is shown in Figure 14. 10. The first graph plots the solutions for α = 1/1 28, 1/ 64, . . . , 1. The second graph plots the solutions for α = 1, 2, . . . , 1 28. 83 0 Solution ... x 2 =   1 1 0   , x 3 =   2 2 1   . 84 9 Result 14. 8 .4 The behavior at infinity of dw dz + p(z)w = 0 is the same as the behavior at ζ = 0 of du dζ − p(1/ζ) ζ 2 u...

Ngày tải lên: 06/08/2014, 01:21

... identity,  ∞ −∞ f(x)δ (n) (x) dx = (−1) n f (n) (0), 1055 -4 -2 2 4 -0.6 -0 .4 -0.2 0.2 0 .4 0.6 Figure 21 .4: Plot of G(x|0). 21.7.1 Green Functions for Sturm-Liouville Problems Consider the problem L[y] ... the form of a particular solution. This form will contain some unknown parameters. We substitute this form into the differential equation to determine the parameters and thus d...

Ngày tải lên: 06/08/2014, 01:21

... q N−j  a j (α 2 ) and d n = lim α→α 2  d dα  (α − α 2 )a n (α)   for n ≥ 0. 12 08 0.2 0 .4 0.6 0 .8 1 1.2 1 .4 0.7 0 .8 0.9 1.1 1.2 Figure 23.1: Plot of the Numerical Solution and the First Three ... 1)c n+2 + c n−2 = 0, for n ≥ 2 c n +4 = − c n (n + 4) (n + 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 +...

Ngày tải lên: 06/08/2014, 01:21

... equation is α 2 + 1 4 α − 1 8 = 0  α + 1 2  α − 1 4  = 0. Since the roots are distinct and do not differ by an integer, there will be two solutions in the Frobenius form. w 1 = z 1 /4 ∞  n=0 a n (α 1 )z n , ... =  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 (x) dx =  1 −1 1 4...

Ngày tải lên: 06/08/2014, 01:21

... 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 see ... 21) π 4 P 4 (x) = 105 8 4 [(315 − 30π 2 )x 4 + ( 24 2 − 270)x 2 + (27 − 2π 2 )] The cosine and this polynomial are plotted in the second graph in Figure 25.1. The le ast squar...

Ngày tải lên: 06/08/2014, 01:21