Estimating value of $h$ needed to cap the approximation error of $\ln{(1+h)}$

Question

We know that

$$\ln{(1+h)} = h - \frac{h^2}{2} +...$$

What is the method for finding how much we should cap $h$ at, to cap the approximation error of $\ln{(1+h)}=h$ at a certain value, like $10^{-3}$ or something like this? I want to solve this using Taylor remainder theorem.

Answer 1

If you approximate $\ln(1+h)$ to $h$, then Taylor's remainder theorem yields that there exists $t\in(0,h)$ such that $\ln(1+h)=h-\dfrac{1}{(1+t)^2}\dfrac{h^2}{2}$. Then the error $\left\lvert\ln(1+h)-h\right\rvert\le\dfrac{h^2}{2}$, since $\dfrac{1}{(1+t)^2}$ is decreasing. If you want to cap the approximation error at $10^{-3}$, then $\dfrac{h^2}{2}\le 10^{-3}$, and you can do the rest of the calculation.

Answer 2

Alternatives to Taylor's Remainder Theorem

We can use $$ \frac{h}{1+h}\le\overbrace{\int_0^h\frac{\mathrm{d}x}{1+x}}^{\log(1+h)}\le h $$ we get the maximum error is $$ \left|\,\log(1+h)-h\,\right|\le\frac{h^2}{1+h} $$

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However, if $h\ge0$, we can use Jensen's inequality and the convex function $\frac1{1+x}$ to get $$ \frac1h\int_0^h\frac{\mathrm{d}x}{1+x}\ge\frac1{1+\frac1h\int_0^hx\,\mathrm{d}x}=\frac1{1+\frac12h} $$ which gives a better bound of $$ \left|\,\log(1+h)-h\,\right|\le\frac{h^2}{2+h} $$