$\ln x \approx \frac{x-1}{\sqrt{x}}$ for $0.5<x<1.5$. Why?

Question

I've been looking at series expansions of data after different transformations and accidentally found that

$\ln x \approx \frac{x-1}{\sqrt{x}}$ when $0.5<x<1.5$

This is a particularly helpful approximation for me, but I don't want to use it without having some sort of derivation. Is there any trick I'm missing?

Answer 1

Use the Taylor expansion on both sides around $x=1$. Let $t=x-1$, then you get

$$\ln(1+t)\approx t-\frac{t^2}{2}+\frac{t^3}{3}-\cdots,$$

and the right-hand side becomes

$$\frac{t}{\sqrt{1+t}}=t-\frac{t^2}{2}+\frac{3t^3}{8}-\cdots.$$

So the first two Taylor terms of the functions agree. The leading order difference is $\,t^3/24$, which remains small if $|t|<0.5$, i.e., $0.5<x<1.5$.