Applications of Series — Question 6

*Question 6 For $0<x\le1$, compare the linear approximation $L(x)=x$ with $\log(1+x)$. Decide whether $L$ overestimates or underestimates, and use the alternating series to bound the absolute error.

*Question 6 – Solution

$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots,\qquad 0<x\le1.$$ The first neglected term after the linear term is negative.

Subtract the function from the linear approximation: $$x-\log(1+x)=\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}-\cdots.$$ This is positive by the Alternating Series Test. Therefore $x$ is an overestimate of $\log(1+x)$.

The alternating remainder after the first term is no larger than the first omitted magnitude: $$\boxed{0<x-\log(1+x)\le\frac{x^2}{2}}.$$ For example, at $x=0.2$, the error is at most $0.02$; the actual error is approximately $0.01768$.