Blericker74

Answered

2022-07-09

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

Answer & Explanation

gozaderaradiox5

Expert

2022-07-10Added 19 answers

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

$a{x}^{\frac{1}{a}}-a=\mathrm{log}x+O\left(\frac{(\mathrm{log}x{)}^{2}}{a}\right)$

so the approximation is accurate once $a$ is large relative to $(\mathrm{log}x{)}^{2}$. 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 $a$.

${x}^{\frac{1}{a}}={e}^{\frac{\mathrm{log}x}{a}}=1+\frac{\mathrm{log}x}{a}+O\left({\left(\frac{\mathrm{log}x}{a}\right)}^{2}\right)$

(by Taylor's theorem) which gives

$a{x}^{\frac{1}{a}}-a=\mathrm{log}x+O\left(\frac{(\mathrm{log}x{)}^{2}}{a}\right)$

so the approximation is accurate once $a$ is large relative to $(\mathrm{log}x{)}^{2}$. 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

2022-07-11Added 7 answers

This is the same thing as asserting that, if $a>0$ and a is very small, then

$\mathrm{log}(x)\approx \frac{{x}^{a}-1}{a},$

which is true, since

$\begin{array}{rl}\underset{a\to {0}^{+}}{lim}\frac{{x}^{a}-1}{a}& =\underset{a\to {0}^{+}}{lim}\frac{{e}^{a\mathrm{log}x}-1}{a}\\ & =\mathrm{log}(x);\end{array}$

in other words, the derivative at $0$ of $a\mapsto {x}^{a}$ is $\mathrm{log}x$.

$\mathrm{log}(x)\approx \frac{{x}^{a}-1}{a},$

which is true, since

$\begin{array}{rl}\underset{a\to {0}^{+}}{lim}\frac{{x}^{a}-1}{a}& =\underset{a\to {0}^{+}}{lim}\frac{{e}^{a\mathrm{log}x}-1}{a}\\ & =\mathrm{log}(x);\end{array}$

in other words, the derivative at $0$ of $a\mapsto {x}^{a}$ is $\mathrm{log}x$.

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