Given $f(x) = \ln(1+x)$, we can write $f(x)$ as $f(x) = p_1(x)+R_2(x)$ where $p_1(x)$ is the first taylor polynomial and $R_2(x)$ is the remainder.

Therefore, $f(x) = x-\frac{x^2}{2(1+c)^2}$ for some $0\leq c \leq x$.

Given that $x \in [-0.1, 0.1]$, and considering the approximation $\ln(1+x)\approx x$, show that the upper bound for the absolute error is $\frac{1}{162}$


$$|error| = |f(x)-p_1(x)| = |R_2(x)| = \left|\frac{-x^2}{2(1+c)^2}\right|$$

I'm a bit confused from here. Thanks.


1 Answer 1


You are almost there, just take $c=-0.1$ and $x=\pm 0.1^2$, hence $$ |R_2(x)| = \left|\frac{0.1^2}{2(1-0.1)^2)}\right| = \frac{1}{162}. $$


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