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October 12, 2014 1 (a) 1 A B = + 1 − x2 1+x 1−x 1 = A(1 − x) + B(1 + x) Thus A = B = 1/2. (b) Z 1 dx 1 − x2 Z 1 1 = ( + )dx 2(1 + x) 2(1 − x) 1 = (ln |1 + x| − ln |1 − x|) + C 2 r 1+x = ln | |+C 1−x (c) y −y e2y −1 2y = 1+x . Thus If y = tanh−1 x, then x = tanh y = eey −e −y = e2y +1 . This implies e 1−x +e q 1+x y = ln 1−x , −1 < x < 1. The two functions are the same up to a constant when −1 < x < 1. 1 2 We use the method of substitution. Set u = x2 + 1. Z 4 2 xe−x −2x −1 dx Z 1 2 = e−u du 2 1 = E(u) 2 1 = E(x2 + 1) 2 3 (1) √ dP =− P dt 1 − √ dP = dt P √ −2 P + C = t When t = 0, P = 676. Thus C = 52. When P = 0, t = C = 52. This means in 52 weeks they will all be dead. (2) Suppose his stake is S dollars. Then we have dS 1 =− S dt 3 3dS − = dt S −3 ln S + C = t Thus -3 ln S(0) + 3 ln S(1) = t(0) − t(1). We know S(1) = 21 S(0) and t(0) = 0. Thus t(1) = 3 ln 2 2 (3) dN = kN 1+ dt dN = dt kN 1+ 1 N − = t + C k(−) N − 0 If N = N0 when t = 0, then C = − k . Thus t = − N k will go to 0, hence N will go to ∞. N0− k − N − k . When t goes to N0− k , 4 (a) κ dv = − v2 + g dt m dv = dt κ 2 v +g −m dv 1 1 q q ( + ) = dt κ κ 2g 1 + 1 − v v gm gm q κ r 1 gm 1 + gm v q ln =t+C κ 2g κ 1 − gm v When t = 0, v = 0. Thus C = 0. So we have v = q √ gκ gm e2t√ m +1 κ 2t gκ m −1 e (b) q q gm The limit exists and is equal to gm . This result makes sense because when v = κ κ , the right hand side of the differential equation is 0, which means the speed does not change. (c) The answer will not change. If v0 is not 0, then the only change is that we will replace t with t + C where C is determined by v0 . When t goes to infinity, the limit of v is still the same as in (b). 3 5 (a) Take m = 1/2, n = k and θ = x. Then the formula gives us sin 21 x cos kx = 12 (sin(k + 1 1 1 1 1 2 )x + sin( 2 − k)x) = 2 (sin(k + 2 )x − sin(k − 2 )x). This is what we want. (b) 1 2 sin x(cos x + cos 2x + ... + cos nx) 2 1 1 1 1 1 1 =(sin(1 + )x − sin(1 − )x) + (sin(2 + )x − sin(2 − )x) + ... + (sin(n + )x − sin(n − )x) 2 2 2 2 2 2 1 1 = sin(n + )x − sin x 2 2 This proves the proposition. (c) n X cos kx = k=1 = = sin(n + 12 )x − sin 21 x 2 sin 12 x 2 sin 21 (n + 1 2 − 12 ) cos 12 (n + 2 sin 12 x 1 2 + 21 ) sin 21 nx cos 12 (n + 1)x sin 12 x (d) Z 0 π 2 n π X iπ cos n→∞ 2n 2n cos xdx = lim i=1 (n+1)π 2n ) π sin( 21 π2 ) cos( 12 = lim π n→∞ 2n sin 4n π π π = sin cos lim 2n π 4 4 n→∞ sin 4n 1 = ·2 2 =1 4 6 Ra 0 xn dx + R an 0 1 y n dy = an+1 , a > 0. Now let’s check the above equation. Z Z a xn dx + 0 an 1 y n dy 0 1 = xn+1 |a0 + n+1 1 n 1 1 n y n +1 |a0 +1 an+1 n n+1 + a n+1 n+1 =an+1 = 5