The Maclaurin series for is for

MATHEMATICS · Multi-part · AP · Calculus BC · Infinite Sequences and Series · Infinite Sequences and Series

Problem

The Maclaurin series for arctan⁡x\arctan x is ∑k=0∞(−1)kx2k+12k+1\displaystyle\sum_{k=0}^{\infty}(-1)^{k}\dfrac{x^{2k+1}}{2k+1} for ∣x∣≤1|x|\le 1. A fifth–degree polynomial is used to approximate arctan⁡(0.7)\arctan(0.7). (a) Write the fifth–degree Maclaurin polynomial P5(x)P_5(x) for arctan⁡x\arctan x and compute P5(0.7)P_5(0.7), giving your answer to five decimal places. (b) Using the alternating nature of the series, obtain an upper bound for ∣arctan⁡(0.7)−P5(0.7)∣|\arctan(0.7)-P_5(0.7)| that involves the first neglected term. Provide a numerical value accurate to six decimal places. (c) Establish a separate upper bound for the same error by considering a suitable higher derivative of arctan⁡x\arctan x on [0,0.7][0,0.7]. Evaluate this second bound numerically to six decimal places. (d) Compare the bounds found in parts (b) and (c). Which bound is tighter, and why? Support your answer with both the numerical evidence and a brief theoretical justification.

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