Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 8 ppt

... 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...

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 ... (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) and (1, 2, 5)?...

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 Exercises 3 .8 .1 ... 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 Fig...

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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  = ... [ 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. U...

Ngày tải lên: 06/08/2014, 01:21

... ı2 √ 3  8 − 8 √ 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 20 48 + ı √ 3 20 48 214 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...

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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

... . . . 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 Form...

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

... 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 ... w  (s) m|w  (s)| . 17 8 Example 6.5.2 Consider (5 + ı7) 11 . We will do the exponentiation in polar form and write the result in Cartesian form. (5 + ı7) 11...

Ngày tải lên: 06/08/2014, 01:21

... 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 defined for z = x, x ∈ [ 1 . . . 1] . Hence we introdu ... = 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...

Ngày tải lên: 06/08/2014, 01:21