Mathematics

Mathematics

Problem

For f(x)=1+xf(x)=\sqrt{1+x}, the second-degree Maclaurin polynomial is P2(x)=1+x2−x28P_2(x)=1+\dfrac{x}{2}-\dfrac{x^{2}}{8}. Find the greatest RR with 0<R<10<R<1 such that ∣f(x)−P2(x)∣≤5.0×10−4whenever ∣x∣≤R.|f(x)-P_2(x)|\le 5.0\times10^{-4}\quad\text{whenever }|x|\le R. Report RR to two decimal places.
Intuitive explanation

The error in a Taylor polynomial is controlled by the next unused term in the series—here, the third-degree term. You'll use the Lagrange remainder formula, which bounds ∣f(x)−P2(x)∣|f(x) - P_2(x)| by the maximum value of the third derivative over your interval, scaled by ∣x∣3/3!|x|^3/3!. Find where that error bound equals your tolerance 5.0×10−45.0 \times 10^{-4}, then solve for the largest RR satisfying the inequality.

Intuitive explanation

The error in a Taylor polynomial is controlled by the next unused term in the series—here, the third-degree term. You'll use the Lagrange remainder formula, which bounds ∣f(x)−P2(x)∣|f(x) - P_2(x)| by the maximum value of the third derivative over your interval, scaled by ∣x∣3/3!|x|^3/3!. Find where that error bound equals your tolerance 5.0×10−45.0 \times 10^{-4}, then solve for the largest RR satisfying the inequality.

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