Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 3 pptx

... ( 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 ....

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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....

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... in Figure 1. 6. Figure 1. 6: y = Arcsin x Example 1. 3. 3 Consider 1 1 /3 . Since x 3 is a one-to-one function, x 1/ 3 is a single-valued function. (See Figure 1. 7.) 1 1 /3 = 1. Example 1. 3. 4 Consider ... . . . 19 10 43 .10 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 11 44 Transform Methods 19...

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... k a 1 a 2 a 3 b 1 b 2 b 3       = i     a 2 a 3 b 2 b 3     − j     a 1 a 3 b 1 b 3     + k     a 1 a 2 b 1 b 2     = (a 2 b 3 − a 3 b 2 )i − (a 1 b 3 − a 3 b 1 )j ... b 2 j + b 3 k) = a 1 b 2 k + a 1 b 3 (−j) + a 2 b 1 (−k) + a 2 b 3 i + a 3 b 1 j + a 3 b 2 (−i) = (a 2 b 3 − a 3 b 2 )i − (a 1 b 3 − a...

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... lim α→∞  1 2 arctan  x 2   α 0 = 1 2  π 2 − 0  = π 4 14 9 a = 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 ... − 2 15 .  x + 1 x 3 + x 2 − 6x dx =   − 1 6x + 3 10 (x − 2) − 2 15 (x + 3)  dx = − 1 6...

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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 ... form  √ 3 + ı  2 n by successive squaring.  √ 3 + ı  2 = 2 + ı2 √ 3  √ 3 + ı  4 = −8 + ı8 √ 3  √ 3 + ı  8 = 12 8 − 12 8 √ 3  √ 3 + ı  16...

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 2048 + ı √ 3 2048 214 exists a unique ... ı √ 3  10 =  2 e ıπ /3  10 = 2 10 e − 10 π /3 = 1 1024  cos  − 10 π 3  + ı sin  − 10 π 3  = 1 1024  cos  4π 3 ...

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 ... logarithm. sin 1 z = −ı log  ız ± √ 1 − z 2  Example 7.7.4 Consider the equation sin 3 z = 1. sin 3 z = 1 sin z = 1 1 /3 e ız − e −ız ı2 = 1 1 /3...

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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) + ... k = 0, 1, 2 2. g(w) = w 3 = |w| 3 e 3 arg(w) 32 8 3 1+ ı = e (1+ ı) log 3 = e (1+ ı)(ln 3+ ı2πn) = e ln 3 2πn e ı(ln 3+ 2πn) , n ∈ Z  √ 3 − ı  1/...

Ngày tải lên: 06/08/2014, 01:21

... result. 34 8 w( 3) =  3 − √ 2  1/ 3  3 + √ 2  1/ 3 ( 3 + 2) 1/ 3 = 3  3 + √ 2 e ıπ /3 3  3 − √ 2 e ıπ /3 3 √ 1 e ıπ /3 = 3 √ 7 e ıπ = − 3 √ 7 The value of the function is w = 3 √ abc e ı(α+β+γ) /3 . Consider ... real variable counterpart. 36 4 With f(z) = z 1/ 3 (z −ı) 1/ 3 (z + ı) 1/ 3 = 3 √ r e ıθ /3 1 3 √ s e −ıφ /3 1 3 √ t e −ı...

Ngày tải lên: 06/08/2014, 01:21