Applications of Series — Question 6

Question 6 – Solution

Step 1: Expand the logarithm.

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

Step 2: Determine the direction.

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 $\log(1+x)$.

Step 3: Bound the error.

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 $0.01768$.