Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 2 ppt

... 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 ... vertices at (1, 1, 0), (3, 2, 1) , (2, 4, 1) and (1, 2, 5)? Hint, Solution Exercise 2. 8 What is the equation of the plane that passes through the points (1,...

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 + ·...

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... + x cos x Solution 3 . 12 Let y(x) = sin x. Then y  (x) = cos x. d dy arcsin y = 1 y  (x) = 1 cos x = 1 (1 −sin 2 x) 1/ 2 = 1 (1 −y 2 ) 1/ 2 d dx arcsin x = 1 (1 −x 2 ) 1/ 2 10 3 is an improper integral. ... cos x  = 0 2 = 0 lim x→0  csc x − 1 x  = 0 10 9 Solution 3 .15 a. f  (x) = ( 12 −2x) 2 + 2x ( 12 − 2x)( 2) = 4(x −6) 2 + 8x(x − 6) = 12 (x...

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... arg(z). -2 -1 0 1 2 x -2 -1 0 1 2 y 0 1 2 -2 -1 0 1 2 x -2 -1 0 1 2 x -2 -1 0 1 2 y -2 0 2 -2 -1 0 1 2 x Figure 7 .11 : Plots of |z| and Arg(z). 2 51 -1 1 -0.4 -0 .2 0 .2 0.4 Figure 6 .20 : |(z)| + 5|(z)| = 1 -1 1 -1 1 Figure ... z 2 is |z 2 | =  z 2 z 2 = zz = (x + ıy)(x −ıy) = x 2 + y 2 . 25 0 -2 -1 0 1 2 x...

Ngày tải lên: 06/08/2014, 01:21

... arguments: 1. log( 1) = log  1 1  = log (1) − log( 1) = −log( 1) , therefore, log( 1) = 0. 2. 1 = 1 1 /2 = (( 1) ( 1) ) 1/ 2 = ( 1) 1/ 2 ( 1) 1/ 2 = ıı = 1, therefore, 1 = 1. Hint, Solution Exercise 7 .11 Write ... form. Denote any multi-valuedness explicitly. 2 2/5 , 3 1+ ı ,  √ 3 − ı  1/ 4 , 1 ı/4 . Hint, Solution 28 7 -2 -1 0 1 2 x -2 -1 0 1 2...

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... Hint 7 .19 Hint 7 .20 Hint 7. 21 Hint 7 .22 Hint 7 .23 Hint 7 .24 Hint 7 .25 1. (z 2 + 1) 1/ 2 = (z − ı) 1/ 2 (z + ı) 1/ 2 2. (z 3 − z) 1/ 2 = z 1/ 2 (z − 1) 1/ 2 (z + 1) 1/ 2 3. log (z 2 − 1) = log(z − 1) + ... the point z = 1. The point at infinity is not a branch point for (z 2 − 1) 1/ 2 . We factor the expression to verify this.  z 2 − 1 ...

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... . . . . . . . . . . . . . . . 11 17 21 . 11Hi nts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 23 21 . 12Solutions . . . . . . . . . . ... . . . . . . . . . 11 65 22 Difference Equations 11 66 22 .1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 66 22 .2...

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... = 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 ... dish. 16 2 Let u = x 2 and dv = e 2x dx. Then du = 2x dx and v = 1 2 e 2x .  x 3 e 2x dx = 1 2 x 3 e 2x − 3 2  1...

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... 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 2 k/5 , for k = 0, 1, 2, 3, 4 19 9 Example ... 4  x 2 + y 2 +  (x − 2) 2 + y 2 = 4 x 2 + y 2 = 16 − 8  (x − 2) 2 + y 2 + x 2 − 4x + 4 + y 2 x − 5 = 2  (x − 2) 2 +...

Ngày tải lên: 06/08/2014, 01:21

... 1) 1/ 2 (z 2) 1/ 2 (z −3) 1/ 2 There are branch points at z = 1, 2, 3. Now we examine the point at infinity. f  1 ζ  =  1 ζ − 1  1 ζ − 2  1 ζ − 3  1/ 2 = ζ −3 /2  1 − 1 ζ  1 − 2 ζ  1 − 3 ζ  1/ 2 Since ... = 1 and z = +∞ with z = r + 1, r ∈ [0 . . . ∞). w = r + 1 +  (r + 1) 2 − 1  1/ 2 = r + 1 ±  r(r + 2) = r  1...

Ngày tải lên: 06/08/2014, 01:21