Consider the following expression:

$a*(1-\frac{1}{b})^{(a-1)} = r$

Provided some real number value for $b$, I need to find a positive real number $0 < a \leq b$ to satisfy the above equation, where $0 < r < 1$.

Must we appeal to an approximation for the above expression to solve for $a \leq b$? If so, what is a good approximation that becomes better as $a \to Inf$?


1 Answer 1


We have

$$ \begin{align*} a\left(1-\frac{1}{b}\right)^{a-1} &= r \\ a\left(1-\frac{1}{b}\right)^a &= \left(1-\frac{1}{b}\right)r \\ a e^{a \log\left(1-\frac{1}{b}\right)} &= \left(1-\frac{1}{b}\right)r \\ a \log\left(1-\frac{1}{b}\right) e^{a \log\left(1-\frac{1}{b}\right)} &= \left(1-\frac{1}{b}\right)r\log\left(1-\frac{1}{b}\right), \end{align*} $$

so that

$$ \begin{align*} a \log\left(1-\frac{1}{b}\right) &= W\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right) \\ a &= \frac{W\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right)}{\log\left(1-\frac{1}{b}\right)}, \end{align*} $$

where $W$ is the the Lambert W function. Note that $W(x)$ is double-valued when $x \in (-1/e,0)$, and the solution you want is given by the principal branch:

$$ a_0 = \frac{W_0\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right)}{\log\left(1-\frac{1}{b}\right)} $$

If you'd like, you can expand this in a series which converges for $b$ large:

$$ a_0 = r + \frac{r(r-1)}{b} - \frac{3r^2(r-1)}{2b^2} + O(b^{-3}). $$

Let us denote the other solution, given by the other branch of $W$, by

$$ a_{-1} = \frac{W_{-1}\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right)}{\log\left(1-\frac{1}{b}\right)}. $$

One can use the asymptotic series derived in this paper this paper (pp. 19-23) to calculate an expression for this solution as $b \to \infty$:

$$ a_{-1} = -\frac{1}{\log\!\left(1-\frac{1}{b}\right)}\left\{\log b + \log \log b - \log r + \frac{\log \log b}{\log b} - \frac{\log r}{\log b} + O\!\left(\frac{\log \log b}{\log b}\right)^2\right\}. $$

This is an okay approximation but note that the absolute error does not decrease to $0$ so it won't be very helpful for numerics.

  • $\begingroup$ The derivation seems to be correct, and I find the same thing using WolframAlpha (without your helpful derivation though). However, I seem to be finding $a \approx r$ using either the Lamber W function or power series approximation. Is there an obvious mistake I'm making? $\endgroup$ Nov 14, 2012 at 3:32
  • $\begingroup$ Try any example value for 0 < r < 1 $\endgroup$ Nov 14, 2012 at 3:35
  • $\begingroup$ It is true that $a \approx r$ when $b$ is large. This isn't the solution you're looking for? The other solution, given by the other branch of $W$, is larger than $b$. See this plot, with $b$ on the horizontal axis and the other solution, $a^*$, on the vertical axis. $\endgroup$ Nov 14, 2012 at 3:40
  • $\begingroup$ If you'd like an asymptotic representation for this other solution, see the wikipedia page for W under "Asymptotic expansions" and use the one for $W_{-1}(x)$. $\endgroup$ Nov 14, 2012 at 3:49

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