Math 132 Final Exam Fall 2016 20 multiple choice questions worth 5 points each Exam is comprehensive No calculators For the multiple choice questions mark your answer on the answer card Useful Formulas Pn i 1 Pn i 1 i3 Pn n n 1 2 i n n 1 2 i 1 2 i2 n n 1 2n 1 6 sin2 cos2 1 1 tan2 sec2 1 cot2 csc2 sin A B sin A cos B sin B cos A cos A B cos A cos B sin A sin B tan A B cos A cos B R 1 2 tan A tan B 1 tan A tan B cos A B cos A B sin A sin B 1 2 cos A B cos A B sin A cos B 1 2 sin A B cos A B sin2 x 12 1 cos 2x cos2 x 21 1 cos 2x sin 2 2 sin cos cos 2 cos2 sin2 csc x dx ln csc x cot x C R sec x dx ln sec x tan x C Math 132 Final Exam 1 Find a power series representation for f x A B C X n 1 X n 1 X Page 2 of 21 2x centered at 0 1 3x 2 2 3n n 1 xn 3n nxn 1 2 3n n 1 xn n 1 D X 2 3n 1nxn n 1 X 3n 1 nxn 1 E F n 1 X 2 3n 1 xn 1 n 1 G X 2 3n x2n n 1 Solution Observe that X 1 3n xn 1 3x n 0 and differentiating both sides gives X 3 3n n xn 1 1 3x 2 n 1 Then we can compute X 2x 2x X n 3 n 1 f x 3 n x 2 3n 1 n xn 3 1 3x 2 3 n 1 n 1 Math 132 Final Exam Page 3 of 21 2 Find the third degree Taylor polynomial T3 centered at x 2 for the function f x 1 1 x A f x 1 x x2 x3 1 B f x 13 19 x 2 27 x C f x 1 x 2 x 2 2 x 2 3 D f x 1 3 19 x 2 1 x 27 2 2 1 2 2 81 x 2 3 1 x 81 2 3 6 x 81 2 3 E f x 1 x x2 x3 F f x G f x 1 3 1 3 91 x 2 91 x 2 x 27 1 2 1 3 x 81 x 27 2 2 Solution Computing derivatives gives f 0 x 1 1 x 2 f 00 x 2 1 x 3 f 000 x 6 1 x 4 Then we can compute 1 f 2 3 Since an f 0 2 f n 2 n 1 a0 3 1 9 f 00 2 2 27 f 000 2 we can conclude a1 1 9 a2 1 27 a3 1 81 so the third degree Taylor polynomial is T3 x 1 1 1 1 x 2 x 2 2 x 2 3 3 9 27 81 6 2 81 27 Math 132 Final Exam Page 4 of 21 3 Find the Taylor series for f x e3x centered at 1 n 3 n X 3 e x 1 n A B C D E F G n 0 X 3n x 1 n n 0 X n 0 X n 0 X n 0 X n 0 X n 0 n xn n 3n e3 xn n x 1 n 3n 3e x 1 n n 3n 1 e3 x 1 n n Solution Observe that f n x 3n e3x so f n 1 3n e3 Thus we can conclude that the Taylor coefficients are 3n e3 f n 1 an n n and so the Taylor series for f x e3x centered at c 1 is X 3n e3 x 1 n n 0 n 4 Find the value of the series or conclude that it diverges 2n 2 2 4 6 8 X 1 n 2 2 2 2 2 2n 1 1 3 5 7 n 0 Math 132 Final Exam A B C D 2 4 1 2 3 2 2 2 E 2 4 F 2 G Page 5 of 21 3 2 H The series diverges Solution Recall that sin x equals its Maclaurin series for all x and in particular sin x X 1 n x2n 1 2n 1 n 0 Plugging in x 2 gives X 1 n 2n 1 2 1 sin 2 2n 1 n 0 and so X 1 n n 0 2n 2 2 2n 1 X 1 n 2 2n 1 2 n 0 2n 1 2 Math 132 Final Exam Z 5 The indefinite integral Page 6 of 21 arctan x2 dx has a power series centered at 0 that looks like C a1 x a2 x 2 a3 x 3 a4 x 4 a5 x 5 where C is the constant of integration Find a3 a4 A 0 B C 1 3 D E 1 7 1 7 F G 1 3 2 7 2 7 Solution Recall that arctan x X 1 n x2n 1 2n 1 n 0 for x 1 and also at x 1 which implies 2 arctan x X 1 n x2 2n 1 2n 1 n 0 X 1 n x4n 2 n 0 2n 1 for x 1 as well Then we can compute Z Z X X 1 n x4n 2 1 n x4n 3 dx C arctan x dx 2n 1 2n 1 4n 3 n 0 n 0 2 C So a3 1 3 and a4 0 x3 x7 x11 3 3 7 5 11 Math 132 Final Exam Page 7 of 21 6 Find the interval of convergence of the power series X 1 n x 4 n 2n 1 n 1 n 1 A B C D 2 2 29 72 21 12 6 2 E 29 72 F 6 2 G 2 2 H Solution First applying the Ratio Test gives lim n x 4 n 1 2n 1 n 1 x 4 n 1 x 4 an 1 lim n 2 lim n n 2 n an n 2 x 4 2 n 2 2 The Ratio Test implies that the series converges when this limit is less than 1 and diverges when the limit is greater than 1 Setting x 4 1 implies that x 4 2 2 and 2 x 4 2 which gives 6 x 2 Thus the series converges on 6 2 but we don t yet know what happens at the the endpoints Now we need to check the endpoints of the interval If x 6 we obtain X 1 n 6 4 n n 1 2n 1 n 1 X n 1 1 2 n 1 This series diverges by the Limit Comparison Test since p 1 and 1 2 n 1 lim 1 n n P 1 n 1 n diverges p series n 1 n 2n …
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