# hw11_solutions - Carnegie Mellon University Department of...

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Unformatted text preview: Carnegie Mellon University Department of Mathematical Sciences 21-122, Spring 2017 Homework 11 Solutions Instructions: Please show all your work, following the instructions posted on Canvas. 1. Find the Taylor series for each of the following (you may use known series for ex ; sin x; cos x; k arctan x; and (1 + x) ) (a) f (x) = x cos x2 ; a = 0 Solution: 1 n X ( 1) x2 x cos x2 = x (2n)! n=0 2n 1 n X ( 1) 4n = x x (2n)! n=0 = 1 n X ( 1) 4n+1 x (2n)! n=0 x (b) f (x) = 1 xe ; x 6= 0; f (0) = 1; a = 0 Solution: ! 1 X 1 n 1 ex 1 = 1 x x x n! n=0 = 1 x = 1 x 1 X = n=1 1 X sin x ;a= 4 4 1 n x n! 1 = 4 1 n X ( 1) x (2n + 1)! n=0 x (d) f (x) = arctan ; x 6= 0; f (0) = 1; a = 0 x Solution: 1 2n+1 arctan x 1X n x = ( 1) x x n=0 2n + 1 = (e) f (x) = x 1=3 !! 1 xm (m + 1)! m=0 = (c) f (x) = sin x Solution: 1 X 1 n 1 1+ x n! n=1 ! 1 X 1 n x n! n=1 ;a=8 1 n X ( 1) 2n x 2n + 1 n=0 2n+1 4 1 1 x; Solution: 1=3 x = = = = = = = 1=3 (8 + (x 8)) x 1 1+ 1 2 8 1 4 1 X 31 3 2 n=0 1 1X 2 n=0 1 3 1=3 1 3 n+1 1 3 n+1 n! 4 3 8n n! 1 n 1 n 1 X ( 1) + 2 n=1 1 X ( 1) + 2 n=1 1 4 3 8n n! 1 3 4 3 1 3 3n 2 3 1 X ( 1) (1) (4) (3n + 2 n=1 2 (24n n!) For example, T2 (x) = n 1 2 1 48 (x 8) + 1 576 (x 2) 1 n (x 2 3n 3 2 (8n n!) n x 8 8) n (x 8) n (x 8) (x 8) n 2 8) (black below) and f (x) = x 1=3 (red below) 2 y 1 0 2 4 6 8 10 12 -1 2. Evaluate the integral as an in…nite series. (a) R sin(x3 ) x dx Solution: Z Z 1 n sin x3 1 X ( 1) 2n+1 dx = x3 dx x x n=0 (2n + 1)! Z 1 n 1 X ( 1) = x6n+3 dx x n=0 (2n + 1)! 1 n Z X ( 1) = x6n+2 dx (2n + 1)! n=0 = 1 X n ( 1) x6n+3 + C (6n + 3) (2n + 1)! n=0 14 16 18 20 x (b) R 4 e x dx Solution: Z Z X 1 4 1 n e x dx = x4 dx n! n=0 Z X 1 n ( 1) 4n = x dx n! n=0 1 n Z X ( 1) x4n dx = n! n=0 1 X (c) n ( 1) x4n+1 + C (4n + 1) n! n=0 = Rp x3 + 8 dx Solution: Z Z p 3 x + 8 dx = = = = = = 1=2 dx x3 + 8 Z p 3 2 2 1 + (x=2) 1 p Z X 2 2 p Z 2 2 1 2 1=2 1 2 1 2 n+1 n! n=0 1+ dx 1 X 1 2 1 2 1 2 x 2 n+1 8n n! n=1 Z 1 X p Z 1 3 2 2 1 dx + x dx + 16 n=2 1 X p 1 2 2 x + x2 + 4 n=2 1 2 1 2 2n 8n 1 2 3n dx 3n x 1 2 = = 1 n X ( 1) x2n+1 n! 2n + 1 n=0 1 X n ( 1) n! (2n + 1) n=0 x=1 x=0 dx 1 2 n+1 8n n! 3 2n 2 (3n + 1) n! 3n+1 x 1 n 1 X p ( 1) (1) (3) (2n 1 = 2 2 x + x2 + 4n (3n + 1) n! 4 2 n=2 p 1 n 1 p X 2 2 ( 1) (1) (3) (2n = x+ x +2 2 4n 2 2 (3n + 1) n! n=2 3. Use a series to approximate the integral to within ": R1 2 1 (a) 0 e x dx; " < 1000 Z 1 Z 1X 1 2 1 n e x dx = x2 dx 0 0 n=0 n! Z 1X 1 n ( 1) 2n = x dx n! 0 n=0 1 n Z 1 X ( 1) = x2n dx n! 0 n=0 ! ! Z +C ! 3) x3n+1 3) x3n+1 + C +C ! x3n dx 1 We have 5!(2(5)+1) = Z 1 2 e x dx 1 1320 1 1000 ; < 1 1000 ; so with n = 4; bn+1 < and we approximate s4 0 = = (b) R1 0 4 X n ( 1) n! (2n + 1) n=0 5651 7560 0:74748::: 1 cos x2 dx; " < 1000 Z Z 1 cos x2 dx = 0 0 Z 1 1X n ( 1) x2 (2n)! n=0 1 1X 2n dx n ( 1) 4n x dx 0 n=0 (2n)! 1 n Z 1 X ( 1) x4n dx = (2n)! 0 n=0 = = = 1 n X ( 1) x4n+1 (2n)! 4n + 1 n=0 1 X x=1 x=0 n ( 1) (4n + 1) (2n)! n=0 1 1 = 9360 < We have (4(3)+1)((2(3))!) Z 1 cos x2 dx s2 1 1000 ; so with n = 2; bn+1 < 1 1000 ; and we approximate 0 = = = (c) R 1=2 p 0 Z 0 1=2 x4 + 1 dx; " < p x4 + 1 dx = 2 X n ( 1) (4n + 1) (2n)! n=0 977 1080 0:9046::: 1 1000 Z 0 1 1=2 X 1 2 1 2 1 2 n+1 n! n=0 x4 n dx ! 1 n 1 1 4 X ( 1) (1) (3) (2n 3) 4n = 1+ x + x dx 2 2n n! 0 n=2 Z 1=2 Z 1=2 Z 1 n 1 X 1 4 ( 1) (1) (3) (2n 3) 1=2 4n = 1 dx + x dx + x dx 2 2n n! 0 0 0 n=2 Z = = = = 1=2 1 X ( 1) 1 1 + + 2 320 n=2 n 1 (1) (3) 2n n! (2n 1 n 1 X 1 1 ( 1) (1) (3) (2n + + 2 320 n=2 24n+1 (4n + 1) 2n n! 1 X ( 1) 1 1 (1) (3) (2n + + 2 320 n=2 24n+1 (4n + 1) 2n (n!) 1 1 + 2 320 n 1 1 3 + 36 864 5111 808 3) 3) 3) 15 ::: 855 638 016 x4n+1 4n + 1 x=1=2 x=0 1 1 Because 36864 < 1000 we approximate Z 1=2 p x4 + 1 dx s1 0 1 1 + 2 320 161 320 0:503125 = = = 4. (a) Find the maximum possible error in approximating f (x) = x sin x by its Taylor polynomial T2 (x) on the interval 6 ; 6 : f 000 (x) d3 (x sin x) dx3 3 sin x x cos x = = So f 000 j3 sin x + x cos xj (x) 3 jsin xj + jxj jcos xj 3 (1) + (1) 6 3:523::: = M So 3:523 3 jxj 3! 3 3:523 3! 6 0:0842::: jR2 (x)j (b) Find n such that the Taylor polynomial Tn (x) approximates f (x) = ln 1 + x2 to within " = 1 : on the interval 101 ; 10 Solution: We test n = 0; then n = 1 and …nd: jf 00 (x)j = jR1 (x)j 2 2 2! x2 1 (x2 + 1) 1 10 2=M 2 2 = 1 100 y 2.0 1.5 1.0 0.5 -1.0 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1.0 x 1 100 So n = 1: (c) Find the interval [ d; d] such that the Taylor polynomial T2 (x) approximates f (x) = cos ( x) to 1 within " = 100 : Solution: f 000 3 (x) So 3 jR2 (x)j d 3 d 3! < 3 d3 < 1 100 3! (100) 3! 3 (100) < 1=3 0:124 614 107 so the interval is [ 0:124:::; 0:124:::] : 5. (a) Use Newton’s Method to …nd a solution of the equation cos x = x: (Show your work, and halt your sequence when you have 3 decimal places the same.) Solution: Let f (x) = cos x x; then f 0 (x) = sin x 1 x1 = x2 = x3 x4 1 cos 1 1 0:750 3::: sin 1 1 cos 0:750 3 0:750 3 = 0:750 3 sin 0:750 3 1 cos 0:739 1 0:739 1 = 0:739 1 sin 0:739 1 1 1 So the approximate solution is x 0:739 1::: 0:739 0::: 0:739::: (b) Determine the terms x1 through x11 using Newton’s Method with x1 = 1:5 and f (x) = x2 + 1 (x Graph this function and describe your result in terms of the graph. Solution: f 0 (x) = x2 + 1 + 2x (x 4:5) = 3x2 9x + 1 x1 x2 x3 x4 x5 x6 = 1:5 # x2 + 1 (x 4:5) = 1:5 = 0:195 652 174 3x2 9x + 1 x=1:5 " # x2 + 1 (x 4:5) = 1: 499 721 34 = 0:195 652 174 3x2 9x + 1 x= 0:195 652 174 " # x2 + 1 (x 4:5) = 0:195 652 243 = 1: 499 721 34 3x2 9x + 1 x=1: 499 721 34 " # x2 + 1 (x 4:5) = 1: 499 720 92 = 0:195 652 243 3x2 9x + 1 x= 0:195 652 243 " # x2 + 1 (x 4:5) = 1: 499 720 92 = 0:195 652 243 3x2 9x + 1 " x=1: 499 720 92 x7 x8 x9 x10 x11 = 1: 499 720 92 = 0:195 652 243 = 1: 499 720 92 = 0:195 652 243 = 1: 499 720 92 By considering the tangents, we see that we are pulled into a 2-cycle. 4:5). y -2 10 -1 1 -10 -20 2 3 4 5 6 7 8 9 10 x ...
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