Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 5 pdf

... = 1 (0 − 1) (0 − 2)(0 − 3) = − 1 6 b = 1 (1) (1 − 2) (1 − 3) = 1 2 c = 1 (2)(2 − 1) (2 − 3) = − 1 2 d = 1 (3)(3 − 1) (3 − 2) = 1 6 1 x(x − 1) (x − 2)(x − 3) = − 1 6x + 1 2(x − 1) − 1 2(x − 2) + 1 6(x ... 5 .10 . -1 -0 .5 0 0 .5 1 -1 -0 .5 0 0 .5 1 0 0 .5 1 1 .5 2 -1 -0 .5 0 0 .5 1 0 0 .5 1 1 .5 2 Figure 5 .10 : Paraboloid, Tangent...

Ngày tải lên: 06/08/2014, 01:21

... Equations Hint 17 .11 Hint 17 .12 Hint 17 .13 Hint 17 .14 Substitute y = x λ into the differential equation. Consider the three cases: a 2 > b, a 2 < b and a 2 = b. Hint 17 . 15 Hint 17 .16 Exact Equations Hint ... conditions, y 1 (0) = 1 and y  1 (0) = 0, for the first solution become, c 1 = 1, −ac 1 + √ a 2 − b c 2 = 0, c 1 = 1, c 2 = a √ a 2 − b . The c...

Ngày tải lên: 06/08/2014, 01:21

... ∼ 4 ∞  n =1 oddn cos(nx). For x = nπ, this implies 0 = 4 ∞  n =1 oddn cos(nx), 13 61 -1 -0 .5 0 .5 1 -0.2 -0 .1 0 .1 0.2 1 0. 25 0 .1 0 0 .1 1 1 1 0 .5 Figure 28.8: Three Term Approximation for a Function ... =      −x − 1 for − 1 < x < 1/ 2 x for − 1/ 2 < x < 1/ 2 −x + 1 for 1/ 2 < x < 1. 13 55 1. Let S N be the sum of the fi...

Ngày tải lên: 06/08/2014, 01:21

... with u = α (1 − v) for special values of α and r. Find v(x) for this special case. 216 8 L 1 L 2 (L 1 x) = L 1 (L 1 l 2 − I)x = L 1 (λx − x) = (λ − 1) (L 1 x) L 1 L 2 (L 2 x) = (L 2 L 1 + I)L 2 x = ... 2 √ 1 − t 2 log (1 + t), 1 2  G + (t) + G − (t)  = ıt log (1 + t). For t ∈ (−∞, 1) , G + (t) = ı  √ 1 − t 2 + t  (log(−t − 1) + ıπ) , G − (t) = ı  − √...

Ngày tải lên: 06/08/2014, 01:21

... . . . 19 10 43 .10 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 11 44 Transform Methods 19 18 44 .1 Fourier Transform for Partial ... . . . . . 10 04 18 .10 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 06 19 Transformations and Canonical Forms...

Ngày tải lên: 06/08/2014, 01:21

... a and b are orthogonal, (perpendicular), or one of a and b are zero. 26 1 2 1 2 2 4 6 8 10 1 2 Figure 1. 11: Plots of f(x) = p(x)/q(x). 1. 6 Hints Hint 1. 1 area = constant ×diameter 2 . Hint 1. 2 A ... with vertices at (1, 1) , (4, 2), (3, 7) and (2, 3)? Hint, Solution Exercise 2.7 What is the volume of the tetrahedron with vertices at (1, 1, 0), (3, 2, 1) , (2, 4, 1) a...

Ngày tải lên: 06/08/2014, 01:21

... are f 1 (x) = 1 f 2 (x) = 1 + x f 3 (x) = 1 + x + x 2 2 f 4 (x) = 1 + x + x 2 2 + x 3 6 The four approximations are graphed in Figure 3 .11 . -1 -0 .5 0 .5 1 0 .5 1 1 .5 2 2 .5 -1 -0 .5 0 .5 1 0 .5 1 1 .5 2 2 .5 -1 -0 .5 ... 0 .5 1 0 .5 1 1 .5 2 2 .5 -1 -0 .5 0 .5 1 0 .5 1 1 .5 2 2 .5 -1 -0 .5 0 .5 1 0 .5 1 1 .5 2 2 .5 -1 -0 .5 0...

Ngày tải lên: 06/08/2014, 01:21

... [ 1, 1] and 1 ≤ cos(x 0 ) ≤ 1, the approximation sin x ≈ x − x 3 6 + x 5 12 0 has a maximum error of 1 50 40 ≈ 0.00 019 8. Using this polynomial to approximate sin (1) , 1 − 1 3 6 + 1 5 12 0 ≈ 0.8 416 67. 10 7 2. f  (0) ... thumb: 12 1 c. ln  lim x→+∞  1 + 1 x  x  = lim x→+∞  ln  1 + 1 x  x  = lim x→+∞  x ln  1 + 1 x  = lim x→+∞  ln  1 +...

Ngày tải lên: 06/08/2014, 01:21

... For instance, (1 2 ) 1/ 2 = 1 1/2 = 1 and  1 1/2  2 = ( 1) 2 = 1. Example 6.6.2 Consider 2 1/ 5 , (1 + ı) 1/ 3 and (2 + ı) 5/ 6 . 2 1/ 5 = 5 √ 2 e ı2πk /5 , for k = 0, 1, 2, 3, 4 19 9 Example 6 .5 .1 ... 2 √ 1 − z 2 . The volume of the intersection of the cylinders is  1 1 4  1 − z 2  dz. We compute the volume of the intersecting cylinders. -1 -...

Ngày tải lên: 06/08/2014, 01:21

... ı2 √ 3  −8 − ı8 √ 3  2  1 =  −2 + ı2 √ 3  12 8 + 12 8 √ 3  1 =  − 51 2 − ı 51 2 √ 3  1 = 1 51 2 1 1 + ı √ 3 = 1 51 2 1 1 + ı √ 3 1 − ı √ 3 1 − ı √ 3 = − 1 2048 + ı √ 3 2048 214 exists a unique ... y 2 . 250 -2 -1 0 1 2 x -2 -1 0 1 2 y -5 0 5 -2 -1 0 1 2 x -2 -1 0 1 2 y Figure 7 .10 : A few branches of arg(z). -2 -1 0 1 2 x -2 -1...

Ngày tải lên: 06/08/2014, 01:21