Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 6 pps

... values. 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 ... ellipse. 18 7 6. 8 Hints Complex Numbers Hint 6 .1 Hint 6. 2 Hint 6. 3 Hint 6. 4 Hint 6. 5 Hint 6. 6 Hint 6. 7 The Complex Plane Hint 6. 8 Hi...

Ngày tải lên: 06/08/2014, 01:21

... u − 1 1000 , u(0) = 1 y 0 . u  + u = 1 1000 u = 1 y 0 e −t + e −t 10 00  t 0 e τ dτ u = 1 1000 +  1 y 0 − 1 1000  e −t 10 10 This equation is separable. dx = dy  c 1 e 2y −y − 1 2  1/ 2 x ... −y 3 dy 1 2 u 2 = − 1 4 y 4 + c 1 u =  2c 1 − 1 2 y 4  1/ 2 y  =  2c 1 − 1 2 y 4  1/ 2 dy (2c 1 − 1 2 y 4 ) 1/ 2 = dx Integrating gives us the implici...

Ngày tải lên: 06/08/2014, 01:21

... = a 0 2 n  k =1 a k cos(kx), where a k = 1 + ( 1) n−k 2 1 2 n 1  n (n − k)/2  . 13 88 3. A n + ıB n = n  k =1 z k = 1 − z n +1 1 − z = 1 − r n +1 e ı(n +1) x 1 − r e ıx = 1 − r e −ıx −r n +1 e ı(n +1) x +r n+2 e ınx 1 ... 1) ξ) sin (ξ) dξ = 1 2 −  C   z 2N +1  (z 1/ z)/(ı2) dz ız =   −  C z 2N +1 (z 2 − 1) dz  =   ıπ Res  z 2N +1 (z + 1) (z 1) ,...

Ngày tải lên: 06/08/2014, 01:21

... θ) = ∞  n =1 a n r n sin(nθ). We determine the coefficients from the boundary condition at r = 1. u (1, θ) = ∞  n =1 a n sin(nθ) = 1 a n = 2 π  π 0 sin(nθ) dθ = 2 πn (1 − ( 1) n ) 17 89 For 0 < ... θ Θ  )  +  µ − ν sin 2 θ  Θ = 0  1 − x 2  P    +  µ − n 2 1 − x 2  P = 0 P (x) sh ould be bounded at the endpoints, x = 1 and x = 1. 18 10 The general solution...

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... . . . 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...

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... 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 ... from 1 to n and are called free indices. Example 2 .1. 1 Consider the matrix equation: A · x = b. We can write out the matrix and vectors explicitly.    a 11 ··...

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 ... ( 1) n 1 (n − 1) !, for n ≥ 1. By Taylor’s theorem of the mean we have, ln x = (x − 1) − (x − 1) 2 2 + (x − 1) 3 3 − (x − 1) 4 4 + ··· + ( 1)...

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 5040 ≈ 0.00 019 8. Using this polynomial to approximate sin (1) , 1 − 1 3 6 + 1 5 12 0 ≈ 0.8 4 16 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

... = 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 ... lim δ→0 +  1 δ 0 1 (x − 1) 2 dx + lim →0 +  4 1+  1 (x − 1) 2 dx Hint 4 .18  1 0 1 √ x dx = lim →0 +  1  1 √ x dx Hint 4 .19...

Ngày tải lên: 06/08/2014, 01:21

... ı88 Solution 6. 4 1.  2 + ı 6 − (1 − ı2)  2 =  2 + ı 1 + ı8  2 = 3 + ı4 63 − 16 = 3 + ı4 63 − 16 63 + 16 63 + 16 = − 253 4225 − ı 204 4225 215 2. (1 − ı) 7 =  (1 − ı) 2  2 (1 − ı) 2 (1 − ... 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 0 1 2 y 0...

Ngày tải lên: 06/08/2014, 01:21