Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 4 pptx

... 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  = ... [ 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 5 040 ≈ 0.00 019 8....

Ngày tải lên: 06/08/2014, 01:21

... ( 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...

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 Fo...

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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 ... 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, 2, 3), (2, 3...

Ngày tải lên: 06/08/2014, 01:21

... 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 x 2 + a 2 dx = 1 a arctan  x a  14 0 Solution 4 .18  1 0 1 √ x dx ... = 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) =...

Ngày tải lên: 06/08/2014, 01:21

... 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 ... 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 + y 2 x 2 − 10 x + 25 = 4x 2 − 16 x + 16 + 4y 2 1 4 (x − 1) 2...

Ngày tải lên: 06/08/2014, 01:21

... ı2 √ 3  −8 − ı8 √ 3  2  1 =  −2 + ı2 √ 3  12 8 + 12 8 √ 3  1 =  − 512 − ı 512 √ 3  1 = 1 512 1 1 + ı √ 3 = 1 512 1 1 + ı √ 3 1 − ı √ 3 1 − ı √ 3 = − 1 2 048 + ı √ 3 2 048 2 14 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 0...

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 ... 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...

Ngày tải lên: 06/08/2014, 01:21

... ı  1/ 4 =  2 e −ıπ/6  1/ 4 = 4 √ 2 e −ıπ/ 24 1 1 /4 = 4 √ 2 e ı(πn/2−π/ 24) , n = 0, 1, 2, 3 1 ı /4 = e (ı /4) log 1 = e (ı /4) (ı2πn) = e −πn/2 , n ∈ Z 308 7 .11 Hints Cartesian and Modulus-Argument Form Hint ... coordinates.    e z 2    =    e (x+ıy) 2    =    e x 2 −y 2 +ı2xy    = e x 2 −y 2 3 04 25 50 75 10 0 -1 1 Figure 7 .43 : The values o...

Ngày tải lên: 06/08/2014, 01:21

... −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 ... branch point where z 1/ 2 − 1 = 0. This occurs at z = 1 on the branch of z 1/ 2 on which 1 1/2 = 1. (1 1/2 has the value 1 on one branch of z 1/ 2...

Ngày tải lên: 06/08/2014, 01:21