Blericker74

2022-07-09

$\mathrm{ln}\left(x\right)\approx a{x}^{1/a}-a$ , which is good for large value of $a$. Where does it come from?

Expert

For large $a$ we have
${x}^{\frac{1}{a}}={e}^{\frac{\mathrm{log}x}{a}}=1+\frac{\mathrm{log}x}{a}+O\left({\left(\frac{\mathrm{log}x}{a}\right)}^{2}\right)$
(by Taylor's theorem) which gives
$a{x}^{\frac{1}{a}}-a=\mathrm{log}x+O\left(\frac{\left(\mathrm{log}x{\right)}^{2}}{a}\right)$
so the approximation is accurate once $a$ is large relative to $\left(\mathrm{log}x{\right)}^{2}$. By the way, I would describe this in exactly the opposite way: that $\mathrm{log}x$ is a good approximation to $a{x}^{\frac{1}{a}}-a$! It's not as if $a{x}^{\frac{1}{a}}-a$ is particularly easy to compute for large $a$.

Raul Walker

Expert

This is the same thing as asserting that, if $a>0$ and a is very small, then
$\mathrm{log}\left(x\right)\approx \frac{{x}^{a}-1}{a},$
which is true, since
$\begin{array}{rl}\underset{a\to {0}^{+}}{lim}\frac{{x}^{a}-1}{a}& =\underset{a\to {0}^{+}}{lim}\frac{{e}^{a\mathrm{log}x}-1}{a}\\ & =\mathrm{log}\left(x\right);\end{array}$
in other words, the derivative at $0$ of $a↦{x}^{a}$ is $\mathrm{log}x$.

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