The Maclaurin series for is for

MATHEMATICS · Single correct · AP · Calculus BC · Infinite Sequences and Series · Infinite Sequences and Series

Problem

The Maclaurin series for

\arctan x

\displaystyle \sum_{n=0}^{\infty}(-1)^{n}\dfrac{x^{2n+1}}{2n+1}

for

|x|\le 1

. A student approximates

\arctan(0.5)

with the degree-3 Maclaurin polynomial and claims that the magnitude of the error is less than

\dfrac{0.5^{5}}{5}

. Which of the following statements best justifies the student’s claim? (a) Because

|0.5|<1

, the Maclaurin series converges absolutely; therefore the absolute error is less than the next term,

\dfrac{0.5^{5}}{5}

. (b) The Lagrange remainder gives

R_3(0.5)=\dfrac{f^{(4)}(\xi)}{4!}(0.5)^4

. Since

|f^{(4)}(\xi)|\le 1

[0,0.5]

, we have

|R_3(0.5)|<\dfrac{0.5^{5}}{5}

. (c) For

x=0.5

, the terms

\dfrac{0.5^{2n+1}}{2n+1}

decrease in magnitude to

0

. By the Alternating Series Estimation Theorem, the truncation error is at most the first neglected term,

\dfrac{0.5^{5}}{5}

. (d) Because the ratio of successive terms at

x=0.5

0.25<1

, the remaining terms form a geometric‐type tail whose entire sum is less than its first term,

\dfrac{0.5^{5}}{5}