Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 2 pptx

... = 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

... y  ) − (2x − y)(1 + 2y  ) (x + 2y) 2 = 5(y −xy  ) (x + 2y) 2 Substitute in the expression for y  . = − 10(x 2 − xy −y 2 ) (x + 2y) 2 Use the original implicit equation. = − 10 (x + 2y) 2 3.5 ... =     x 20 20 ! cos ξ     ≤ |x| 20 20 ! . x 20 /20 ! is plotted in Figure 3.13. 2 4 6 8 10 0 .2 0.4 0.6 0.8 1 Figure 3.13: Plot of x 20 /20 !. Note that the erro...

Ngày tải lên: 06/08/2014, 01:21

... cos x  = 0 2 = 0 lim x→0  csc x − 1 x  = 0 109 Solution 3.15 a. f  (x) = ( 12 −2x) 2 + 2x( 12 − 2x)( 2) = 4(x −6) 2 + 8x(x − 6) = 12( x 2) (x − 6) There are critical points at x = 2 and x = 6. f  (x) ... 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 fo...

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 ... are bounded at the origin are: y =      1 A + t 2 A +2 , A > 0 1 A + t 2 A +2 + ct −A , A < 0, A = 2 − 1 2 + t 2 ln t + ct 2 , A = 2 Equations in the...

Ngày tải lên: 06/08/2014, 01:21

... 3     e −2t 0 0 0 e (−1−ı √ 2) t 0 0 0 e (−1+ı √ 2) t   1 6   2 2 2 −1 − ı5 √ 2/ 2 1 − 2 √ 2 4 + ı √ 2 −1 + ı5 √ 2/ 2 1 + 2 √ 2 4 − ı √ 2     1 0 0   x = 1 3   2 2 1   e −2t + 1 6   2 ... λ           856 e At =   2 e −2t +3 e −t − e −2t + e −t − e −2t + e −t 5 e −2t −8 e −t +3 e t 2 15 e −2t −16 e −t +13 e t 12 15 e −2t −16 e −t + e...

Ngày tải lên: 06/08/2014, 01:21

... 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) = ... 2t 3 d dt . 124 0 Solution 23 .2 1. First we write the differential equation in the standard form.  1 − x 2  y  − 2xy  + α(α + 1)y = 0 (23 .2) y  − 2x 1 − x 2 y  + α(α...

Ngày tải lên: 06/08/2014, 01:21

... + ζ 2 2(2n 2) + ζ 4 2 4·(2n 2) (2n−4) + ··· + ζ n 2 4···n·(2n 2) ···(2n−n)  for even n 2 n−1 n! ζ n+1  1 + ζ 2 2(2n 2) + ζ 4 2 4·(2n 2) (2n−4) + ··· + ζ n−1 2 4···(n−1)·(2n 2) ···(2n−(n−1))  for ... J 3 /2 (z), J 3 /2 (z) = 1 /2 z J 1 /2 (z) − J  1 /2 (z) = 1 /2 z  2 π  1 /2 z −1 /2 sin z −  − 1 2  2 π  1 /2 z −3 /2 sin z −  2 π  1 /2 z −1 /2 co...

Ngày tải lên: 06/08/2014, 01:21

... = 1 4π ln  ((x 2 − y 2 − ξ 2 + η 2 ) 2 + (2xy − 2 η) 2 ) ((x 2 − y 2 − ξ 2 + η 2 ) 2 + (2xy + 2 η) 2 )  u = 1 4π ln  ((x − ξ) 2 + (y − η) 2 ) ((x + ξ) 2 + (y + η) 2 ) ((x + ξ) 2 + (y − η) 2 ) ((x − ξ) 2 + ... − ξ) 2 + (y −ψ) 2 (x − ξ) 2 + (y + ψ) 2  . 20 18 u = 1 4π ln  ((x − ξ) 2 + (y − η) 2 ) ((x + ξ) 2 + (y + η) 2 ) ((x + ξ) 2...

Ngày tải lên: 06/08/2014, 01:21

... . . 22 00 50.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 01 VIII Appendices 22 20 A Greek Letters 22 21 xxi 0.3 Warnings and ... . . . . . . . . . . . . . 22 2. 1.1 Scalars and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2. 1 .2 The Kronecker Delta and Einstein...

Ngày tải lên: 06/08/2014, 01:21

... dish. 1 62 Let u = x 2 and dv = e 2x dx. Then du = 2x dx and v = 1 2 e 2x .  x 3 e 2x dx = 1 2 x 3 e 2x − 3 2  1 2 x 2 e 2x −  x e 2x dx   x 3 e 2x dx = 1 2 x 3 e 2x − 3 4 x 2 e 2x + 3 2  x e 2x dx Let ... = x and dv = e 2x dx. Then du = dx and v = 1 2 e 2x .  x 3 e 2x dx = 1 2 x 3 e 2x − 3 4 x 2 e 2x + 3 2  1 2 x e 2x − 1 2  e 2x dx   x 3 e 2...

Ngày tải lên: 06/08/2014, 01:21