Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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- 3. Given the Maclaurin series I calculated for f(x)= sinx from the video on Maclaurin series, write down a power series for sin(x)arrow_forwardUse the power series 00 E(-1)^x" 1 + x n = 0 to determine a power series, centered at 0, for the function. 1 dx х+ 1 f(x) = In(x + 1) = f(x) = > n = 0 Identify the interval of convergence. (Enter your answer using interval notation.)arrow_forward(2n)! Find a Maclaurin series for f(x). (Use for 1:3.5. (2n - 3).) --. 2"n!(2n-1) f(x) = 1 + f(x) = x + 16 n = 2arrow_forward
- 1 g(x) = (7 +x)3 g(x) = 2 5 „n (-1)"(n+1)(n + 2) 2.(7)"+3 n=0 Part 3 Use your answers above to now express the function as a power series (centered at x = 0). x2 h(x) = (7 + x)3 h(x) = n=0arrow_forward(2n+ 1)x Please explan Determine whether the series sin- converges or diverg es. If it converges, find its sum. thish L and how it nelotes to the n-D0 (4r + 1)x Since sin =1 and sin =-1 and the angle may be incremente by any multiple of 2x, then sin 2 =1 and (4r+3)a sin =-1, (any r). How does this nelate to the answer. Examine the partial sums of the series. The expanded series is sin 3n 5x + sin + sin ,+ sin + sin+ .. ansver Notice that the angles in consecutive even terms (n is even) differ by 2x, and the angles in consecutive odd terms (n is odd) differ by 2x. Since sin 0 = sin (0+2mx), all the even terms are equal to 1, and all the odd terms are equal to -1. Evaluating each term, the series is 1+(-1)+1+(-1)+1+ The partial sums altermate between 1 and 0, tending to (2n + 1)a neither value. Therefore, lim sin- k00 n=0 does not exist and the series diverges. detail. Pelease explain n thanke gou.arrow_forwardUse a power series to represent the following function: f(x)=x^2/2-xarrow_forward
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