## Thomas' Calculus 13th Edition

Recall the Taylor series for: $\ln (1+x)=x-\dfrac{ x^2}{2}+\dfrac{x^3}{3}-....$ ; $-1 \leq x \leq 1$ and $\ln (1-x)=-x-\dfrac{ x^2}{2}-\dfrac{x^3}{3}-....$ ; $-1 \leq x \leq 1$ Now, we have $| Error|=|\dfrac{(-1)^n x^n}{n}|$ Set $x=0.1$; then we have $| Error|=|\dfrac{(-1)^n x^n}{n}|=\dfrac{1}{n (10^n)}$ But $\dfrac{1}{n (10^n)} \lt \dfrac{1}{10^8}$ This implies that $n \geq 8$ Thus, we need $7$ terms for the accuracy.