Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 10 ppt

... u  3 y  3 + u 3 y  3 = u 1 y  1 + u 2 y  2 + u 3 y  3 y  p = u  1 y  1 + u 1 y  1 + u  2 y  2 + u 2 y  2 + u  3 y  3 + u 3 y  3 Substituting the expressions for ... u 3 y  3 ) + p 0 (u 1 y 1 + u 2 y 2 + u 3 y 3 ) = f(x) u  1 y  1 + u  2 y  2 + u  3 y  3 + u 1 L[y 1 ] + u 2 L[y 2 ] + u 3 L[y 3 ] = f(x) u  1 y  1...

Ngày tải lên: 06/08/2014, 01:21

... .      C R dz z 3 + 1     ≤ 2πR 3 1 R 3 − 1 → 0 as R → ∞ We take R → ∞ and solve for the desired integral.  1 + e −ıπ /3   ∞ 0 dx x 3 + 1 = − ı2π e ıπ /3 3  ∞ 0 dx x 3 + 1 = 2π 3 √ 3 746 2  ∞ 0 x 1/2 log ... z k ) −ı2π e ı2πa  735 We calculate the residue at that point. Res  1 z 3 + 1 , z = e ıπ /3  = lim z→ e ıπ /3  (z − e ıπ /3 ) 1 z 3 + 1  =...

Ngày tải lên: 06/08/2014, 01:21

... 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 1 i + 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 3 b 1 )j + (a...

Ngày tải lên: 06/08/2014, 01:21

... minimum. b. f  (x) = 2 3 (x −2) −1 /3 The first derivative exists and is nonzero for x = 2. At x = 2, the derivative does not exist and thus x = 2 is a critical point. For x < 2, f  (x) < 0 and for x > ... for r > 0, it must be a global minimum. The cup has a radius of 4 3 √ π cm and a height of 4 3 √ π . 106 a x x x x x x∆ 1 2 3 i n-2 n-1 b f( ) ξ 1 F...

Ngày tải lên: 06/08/2014, 01:21

... ı √ 3  10 =  2 e ıπ /3  10 = 2 10 e − 10 /3 = 1 102 4  cos  − 10 3  + ı sin  − 10 3  = 1 102 4  cos  4π 3  − ı sin  4π 3  = 1 102 4  − 1 2 + ı √ 3 2  = − 1 2048 + ı √ 3 2048 2. (11 + ı4) 2 = 105 + ı88 Solution ... ı2 √ 3  −2 + ı2 √ 3  4  −1 =   −2 + ı2 √ 3  −8 − ı8 √ 3  2  −1 =  −2 + ı2 √ 3  −128 + ı128 √ 3  −1 =  −512...

Ngày tải lên: 06/08/2014, 01:21

... sin 3 z = 1. sin 3 z = 1 sin z = 1 1 /3 e ız − e −ız ı2 = 1 1 /3 e ız −ı2(1) 1 /3 − e −ız = 0 e ı2z −ı2(1) 1 /3 e ız −1 = 0 e ız = ı2(1) 1 /3 ±  −4(1) 2 /3 + 4 2 e ız = ı(1) 1 /3 ±  1 − (1) 2 /3 z ... See Figure 7.18 and Figure 7.19 for plots of the real and imaginary parts of the cosine and sine, respectively. Figure 7.20 shows the modulus of the cosine and the sine....

Ngày tải lên: 06/08/2014, 01:21

... direction. Hint 7 .30 Hint 7 .31 To define the branch, define angles from each of the branch points in the finite complex plane. Hint 7 .32 Hint 7 .33 Hint 7 .34 Hint 7 .35 Hint 7 .36 Hint 7 .37 30 0 Now to define ... z 1 /3 = 3 √ z e ık2π /3 , 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 ∈...

Ngày tải lên: 06/08/2014, 01:21

... =  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 the ... 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 −ıψ /3 = 3  r st e ı(θ−φ−ψ) /3...

Ngày tải lên: 06/08/2014, 01:21

... = ı and an isolated essential singularity at z = 1. (b) sin(z − 3) (z − 3) (z + ı) 6 has a removable singularity at z = 3, a pole of order 6 at z = −ı and an essential singularity at z ∞ . 436 Result ... = 1 r ∂ ∂r  r ∂u ∂r  + 1 r 2 ∂ 2 u ∂θ 2 = 0. Therefore u is harmonic and is the real part of some analytic function. Example 9 .3. 9 Find an analytic function f(z) whose real...

Ngày tải lên: 06/08/2014, 01:21

... integral converges for (a) < 1. 725 Figure 13. 8: The real and imaginary part of the integrand for several values of α. Note that the integral exists for all nonzero real α and that lim α→0 +  1 −1 1 x ... = ı  = ıπ 3 lim z→ı 1 z + ı = π 3 2 Now we return to Equation 13. 1. 2  ∞ 0 ln 2 x 1 + x 2 dx + ı2π  ∞ 0 ln x 1 + x 2 dx = π 3 4 We equate the real and imaginary...

Ngày tải lên: 06/08/2014, 01:21