Is there an approximation to $\frac{1}{-\log(1-e^{-\beta})}$

Question

Is there a way to simplify $$\frac{1}{-\log(1-e^{-\beta})}$$ where $\beta>0$ on the real line

I considered the taylor series for $log(1-t)$ but since it is divided by 1, I'm not sure how to proceed

Thank you

Answer 1

Approximations (by their nature) work for some special cases only, in other words we can't find an approximation for the whole $\beta>0$, we need to consider separate cases.

The most natural ones are:

1) $\beta \ll 1$

Then we expand the exponential function and get:

$$-\frac{1}{\log \left(1-1+\beta-\frac{\beta^2}{2}+\frac{\beta^3}{6}-\cdots \right)} \approx - \frac{1}{\log \beta}$$

2) $\beta \gg 1$

Then we have $e^{-\beta} \ll 1$ so we can expand the logarithm:

$$\frac{1}{e^{-\beta}+\frac{1}{2}e^{-2\beta}+\frac{1}{3}e^{-3\beta}+\cdots} \approx e^\beta$$

Here's both approximations plotted along the exact function to see where they fit:

Answer 2

You can use the Taylor series for 1/(1-u) = 1+u+u² ...