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To avoid the $\log()$ function, I am looking for a good approximation of $\log(x)$ for very small $x$ (e.g. order $10^{-5}$).

I think Taylor series expansion is useless because around these small $x$, the first order derivative approachs $+\infty$.

I did try this approximation $\log_{10}(x) \approx 1 - \frac{1}{\sqrt{x}}$ but still don't have satisfactory results.

Could anyone suggest some better approximations?

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  • $\begingroup$ $\frac{\ln x}{\ln 10}$ $\endgroup$ – gammatester Oct 27 '18 at 8:32
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    $\begingroup$ Since $\log 1/x=-\log x$, an approximation for very small $x$ would be the same as an approximation for very large $x$. $\endgroup$ – Aaron Oct 27 '18 at 8:36
  • $\begingroup$ @gammatester very funny. $\endgroup$ – AlexTP Oct 27 '18 at 8:41
  • $\begingroup$ @Aaron thanks for quick comment. However, as I said, I am trying to avoid the $\log()$ function. Can you please tell me the good approximation of $\log(x)$ for very large $x$? $\endgroup$ – AlexTP Oct 27 '18 at 8:42
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    $\begingroup$ Please see here: math.stackexchange.com/a/977657/471884. $\endgroup$ – TheSimpliFire Oct 27 '18 at 8:47
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Since you seem to allow square roots, then the sequence of functions $\, f_n(x) := 2^n(\sqrt[2^n]{x}-1)\,$ give better and better results. In fact, $\, f_n(x) \to \ln(x)\,$ as $\, n \to \infty\,$ for all $\,x>0.\,$ Once you have $\,\ln(x)\,$ you can use $\, \log_{10}(x) = \frac{\ln(x)}{\ln(10)}.$

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$$\ln x\sim\frac{1-x^{-x}}x\qquad{(x\to 0)}$$

or

$$\ln x\sim \frac{x^x-1}x$$


Since $x^x\approx 1$ for small $x$,

$$\ln x=\frac1x\ln x^x=\frac1x\ln(1+(x^x-1))\sim\frac{x^x-1}x$$

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