Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 10 doc

... −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 ... square root.  z 2 − 1  1/ 2 = (z + 1) 1/ 2 (z 1) 1/ 2 We see that there are branch points at z = 1 and z = 1. In particular we want the Arccos to be...

Ngày tải lên: 06/08/2014, 01:21

... . . . . . 10 04 18 .10 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 06 19 Transformations and Canonical Forms 10 18 19 .1 The Constant ... . . . 19 10 43 .10 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 11 44 Transform Methods 19 18 44 ....

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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 ··· a 1...

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... ( 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) n 1 (x − 1) n n + ( 1) n (x − 1) n +1 n + 1 1 ξ n +1 . 72 3.8 ... 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...

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... 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 + 1 x  1/ x  = lim x→+∞   1 + 1 x  1  − 1 x 2  1/ x 2  = lim x→+∞   1 + 1 x  1  = ... + 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...

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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 ... 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  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 ı2πk/5 , for k = 0, 1, 2, 3, 4 19 9 Example ... 4 x 2 + y 2 = 16 − 8  (x − 2) 2 + y 2 + x 2 − 4x + 4 + y 2 x − 5 = −2  (x − 2) 2 + y 2 x 2 − 10 x + 25 = 4x 2 − 16 x + 16 + 4y 2 1 4 (x − 1) 2 + 1 3 y 2 =...

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... + ı| = 1. 2 31 Now we do the multiplication in modulus-argument, (polar), form.  1 + ı √ 3  10 =  2 e ıπ/3  10 = 2 10 e − 10 π/3 = 1 102 4  cos  − 10 π 3  + ı sin  − 10 π 3  = 1 102 4  cos  4π 3  − ... 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...

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... 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 ... Cartesian form. Denote any multi-valuedness explicitly. 2 2/5 , 3 1+ ı ,  √ 3 − ı  1/ 4 , 1 ı/4 . Hint, Solution 287 -2 -1 0 1 2 x -2 -1 0 1 2 y 0 0.5...

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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) + ... z 1 = z 2 = 1. Arg(( 1) ( 1) ) = Arg (1) = 0, Arg( 1) + Arg( 1) = 2π Log(( 1) ( 1) ) = Log (1) = 0, Log( 1) + Log( 1) = ı2π 306 sign. This will change the value of arccos...

Ngày tải lên: 06/08/2014, 01:21