tapia jat4858 HW10 clark 52990 This print out should have 21 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points 1 Consequently on the interval 4 4 the function f defined by f x x 2x3 3x5 6 42 44 4 can be identified with Determine the value of f 2 when f x x 2x3 3x5 6 42 44 4 f x As x 2 lies in 4 4 we thus see that Hint differentiate the power series expansion of x2 42 1 1 f 2 42 x x2 42 2 f 2 1 50 2 correct 2 f 2 25 3 f 2 4 5 4 f 2 1 10 4 5 f 2 25 Explanation The geometric series 1 1 1 42 x 42 1 x 42 1 x x2 x3 2 1 2 4 6 4 4 4 4 has interval of convergence 16 16 But if we now restrict x to the interval 4 4 and replace x by x2 we see that 2 25 keywords 002 10 0 points Find a power series representation for the function f y ln 4 y 1 f y ln 4 X n 0 2 f y yn n 4n X yn n4n n 1 3 f y ln 4 X yn 4n n 1 4 f y X n 0 yn n 4n X yn 5 f y ln 4 4n n 0 1 2 4 x2 x2 x4 x6 1 2 1 2 4 6 4 4 4 4 6 f y ln 4 X n 1 yn correct n 4n on the interval 4 4 In addition in this interval the series expansion of the derivative of the left hand side is the term by term derivative of the series on the right Explanation We can either use the known power series representation 1 2x 4x3 6x5 2x 2 2 4 6 2 x 42 2 4 4 4 4 X xn ln 1 x n n 1 tapia jat4858 HW10 clark 52990 or the fact that ln 1 x Z x 0 x Z X s n n 0 X n 0 x Z 0 Explanation When lim 1 ds 1 s 0 2 cn 1 cn n 2 the Ratio Test ensures that the series ds sn ds X n 1 xn n X n 0 is 1 i convergent when x and 2 1 ii divergent when x 2 For then by properties of logs 1 1 f y ln 4 1 y ln 4 ln 1 y 4 4 so that cn xn On the other hand since f y ln 4 003 X n 1 yn n 4n 10 0 points Compare the radius of convergence R1 of the series X cn xn n 0 with the radius of convergence R2 of the series X n cn xn 1 n 1 when lim n 1 R1 R2 cn 1 cn lim n n 1 cn 1 ncn 1 2 3 2R1 R2 1 2 4 R1 R2 2 X n cn xn 1 n 1 is 1 i convergent when x and 2 1 ii divergent when x 2 Consequently 2 R1 R2 004 1 2 10 0 points Find a power series representation for the function y f y 9y 1 1 f y X 1 n 3n y n 1 n 0 5 R1 2R2 2 6 2R1 R2 2 n cn 1 cn the Ratio Test ensures also that the series 1 correct 2 2 R1 2R2 lim 2 f y X n 0 3n y n tapia jat4858 HW10 clark 52990 3 f y X h 1 5 h 1 2 interval of cgce 5 1 interval of cgce 2n n 3 y n 0 4 f y X 32n y n 1 n 0 5 f y X 6 f y 4 interval of cgce 5 5 correct 1 n 32n y n 1 correct Explanation After simplification 5 interval of cgce Explanation Since 1 1 x x2 x3 1 x On the other hand X 1 xn 1 x we see that n 0 1 1 2 1 x 1 x 2 Thus X 9y n 0 y X n 1 x2 x2 2 x2 3 Z n 0 while Z x X 0 keywords 005 x 1 dt tan 1 x 2 1 t 0 1 n 32n y n 1 10 0 points Determine the interval of convergence for the power series representation of x f x tan 1 5 centered at the origin obtained by integrating the power series expansion for 1 1 x2 1 n x2n Now Consequently f y X n 0 1 n 32n y n n 0 X 1 1i 5 5 6 interval of cgce 5 5 y y 9y 1 1 9y f y y 5 1 n 3n y n n 0 f y 1 5 1i 3 interval of cgce 5 5 n 0 X 3 n 0 X 1 n 2n 1 x 1 n x2n dt 2n 1 n 0 Thus 1 tan X 1 n 2n 1 x x 2n 1 n 0 from which it follows that 1 f x tan x 5 X 1 n x 2n 1 2n 1 5 n 0 tapia jat4858 HW10 clark 52990 To determine the interval of convergence of the power series set 1 n x 2n 1 an 2n 1 5 Then 2n 1 x 2 2n 3 5 an 1 an and so x 2 an 1 n an 5 By the Ratio Test therefore the power series converges when x 5 and diverges when x 5 On the other hand at x 5 the series reduces to X 1 n 2n 1 lim n 0 which converges by the Alternating series Test while at x 5 the series reduces to 3 f t 4 f t 5 f t X 32n 2n t 2n n 1 X n 1 1 n 32n 1 2n 1 t 2n 1 Explanation We know that x2 x3 2 3 X 1 n 1 n x n ln 1 x x n 1 while x2 x3 ln 1 x x 2 3 X 1 n x n n 0 interval of convergence 5 5 X 32n 1 2n 1 t 2n 1 n 1 X 1 n 1 2n 1 which converges again by the Alternating Series Test Consequently the power series representation for f x obtained from the series expansion for 1 1 x has 4 n 1 Thus x3 x5 ln 1 x ln 1 x 2 x 3 5 X 1 x2n 1 2 2n 1 n 1 keywords 006 10 0 points Find a power series representation for the function r 1 3t f t ln 1 3t Hint remember properties of logs X 32n 2n t 1 f t 2n Now by properties of logs r 1 1 3t 1 3t ln ln 1 3t 2 1 3t 1 ln 1 3t ln 1 3t 2 Thus 2 f t 2 X n 1 1 3t …