Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 2 pps

... cases for even and odd n separately. a 2n = − a 2n 2 2n = a 2n 4 (2n)(2n − 2) = (−1) n a 0 (2n)(2n − 2) ·· 4 · 2 = (−1) n a 0  n m=1 2m , n ≥ 0 a 2n+1 = − a 2n−1 2n + 1 = a 2n−3 (2n + 1)(2n − 1) = ... =  x 3 3  1 −1 = 2 3  1 −1 P 2 (x)P 2 (x) dx =  1 −1 1 4  9x 4 − 6x 2 + 1  dx =  1 4  9x 5 5 − 2x 3 + x  1 −1 = 2 5  1 −1 P 3 (x)P 3 (x) dx =  1 −1 1 4...

Ngày tải lên: 06/08/2014, 01:21

... of  ∞ n=1 1 n 2 . ∞  n=1 4 n 2 = 1 π  π −π x 2 dx ∞  n=1 4 n 2 = 2 2 3 ∞  n=1 1 n 2 = π 2 6 13 74 a n = 2 π  π 0 f(x) cos(nx) dx = 2 π  π /2 0 cos(nx) dx = 2 πn sin(nπ /2) =  2 πn (−1) (n−1) /2 , for odd n 0 for even n The ... sum.  π −π (f(x)) 2 dx − πa 2 0 − 2 ∞  n=1  a 2 n + b 2 n  + πa 2 0 2 + π ∞  n=1  a 2 n + b 2 n  =...

Ngày tải lên: 06/08/2014, 01:21

... = 2x 2 x 2 + βx + χ Now we use the fact that f(x) is even to conclude that q(x) is even and thus β = 0. f(x) = 2x 2 x 2 + χ Finally, we use that f(1) = 1 to determine χ. f(x) = 2x 2 x 2 + 1 20 Einstein ... there and that χ = 0. f(x) = ax 2 x 2 + βx + χ We n ote that f (x) → 2 as x → ∞. This d etermine s the parameter a. lim x→∞ f(x) = lim x→∞ ax 2 x 2 + βx + χ = l...

Ngày tải lên: 06/08/2014, 01:21

... 4  x 2 + y 2 +  (x − 2) 2 + 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 − 10x + 25 = 4x 2 − 16x + 16 + 4y 2 1 4 (x − 1) 2 + 1 3 y 2 = 1 Thus ... exponential. cos 4 θ =  e ıθ + e −ıθ 2  4 = 1 16  e 4 +4 e 2 +6 + 4 e − 2 + e − 4  = 1 8  e 4 + e − 4 2  + 1 2  e 2...

Ngày tải lên: 06/08/2014, 01:21

... A = 0, 2 ln t + 1 2 t 2 + c, A = 0 − 1 2 t 2 + ln t + c, A = 2 y =      1 A + t 2 A +2 + ct −A , A = 2 ln t + 1 2 t 2 + c, A = 0 − 1 2 + t 2 ln t + ct 2 , A = 2 For positive A, the ... constants.   0 1 2 2 1 2 1 0 1     c 1 c 2 c 3   =   2 0 1   c 1 = 1, c 2 = 2, c 3 = 0 The solution subject to the initial condition is x =   0 2 1...

Ngày tải lên: 06/08/2014, 01:21

... 2 √ 2 c 1 e √ 2x −c 2 e − √ 2x c 1 e √ 2x +c 2 e − √ 2x u = 2  c 1 √ 2 e √ 2x −c 2 √ 2 e − √ 2x c 1 e √ 2x +c 2 e − √ 2x dx + c 3 u = 2 log  c 1 e √ 2x +c 2 e − √ 2x  + c 3 y =  c 1 e √ 2x +c 2 e − √ 2x  2 e c 3 The ... −y 3 dy 1 2 u 2 = − 1 4 y 4 + c 1 u =  2c 1 − 1 2 y 4  1 /2 y  =  2c 1 − 1 2 y 4  1 /2 dy (2c 1 − 1 2 y 4 ) 1 /2 = d...

Ngày tải lên: 06/08/2014, 01:21

... exp  − 2y 3 /2 3  = y 1 /2 exp  − 2y 3 /2 3  x exp  − 2y 3 /2 3  = −exp  − 2y 3 /2 3  + c 1 x = −1 + c 1 exp  2y 3 /2 3  x + 1 c 1 = exp  2y 3 /2 3  log  x + 1 c 1  = 2 3 y 3 /2 y =  3 2 log  x ... If a 2 = 4b then the solutions are y 1 = e (−a+ √ a 2 −4b)x /2 , y 2 = e (−a− √ a 2 −4b)x /2 If a 2 = 4b then we have y 1 = e −ax /2 , y 2 = x e −ax /2...

Ngày tải lên: 06/08/2014, 01:21

... equation. u  1 y 1 + 2u  1 y  1 + u 1 y  1 + u  2 y 2 + 2u  2 y  2 + u 2 y  2 + p(u  1 y 1 + u 1 y  1 + u  2 y 2 + u 2 y  2 ) + q(u 1 y 1 + u 2 y 2 ) = f u  1 y 1 + 2u  1 y  1 + u  2 y 2 + 2u  2 y  2 + ... y p = e t (at 2 + 4at + 2a) e t 2( at 2 + 2at) e t +at 2 e t = e t 2a e t = e t a = 1 2 A particular solution is y p = t 2...

Ngày tải lên: 06/08/2014, 01:21

... solution b n = n  j=0  − 4j 2 − 2j + 1 (2j + 2) (2j + 1)  . Thus we have that a n =   n /2 j=0  − 4j 2 −2j+1 (2j +2) (2j+1)  for even n, 0 for odd n. 1193 c 1 = 1 − r 2 r 2 1 − r 1 r 2 = 1 − 1− √ 5 2 1+ √ 5 2 √ 5 = 1+ √ 5 2 1+ √ 5 2 √ 5 = 1 √ 5 Substitute ... have c 1 r 1 +c 2 r 2 = 1 c 1 r 2 1 +c 2 r 2 2 = 1. Solving for c 2 in the first equation,...

Ngày tải lên: 06/08/2014, 01:21

... x 1 /2 → x −1  x 1 /2 t   x −1 /2 → x 2  x −1 /2 125 7 yields cos(πx) ≈ − 15 π 2 P 2 (x) + 45 (2 2 − 21 ) π 4 P 4 (x) = 105 8π 4 [(315 − 30π 2 )x 4 + ( 24 2 − 27 0)x 2 + (27 − 2 2 )] The cosine and ... u. t  + (t  ) 2 +  1 x − 2  t  −  1 x + ν 2 x 2  = 0 1 2x 2 + u  +  − 1 2x + u   2 +  1 x − 2  − 1 2x + u   −  1 x + ν...

Ngày tải lên: 06/08/2014, 01:21