Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $a \to Inf$?

share|cite|improve this question
Here is the technique of the solution. – Mhenni Benghorbal Nov 19 '12 at 7:10
up vote 0 down vote accepted

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.

share|cite|improve this answer
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? – PartiallyCovered Nov 14 '12 at 3:32
Try any example value for 0 < r < 1 – PartiallyCovered Nov 14 '12 at 3:35
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. – Antonio Vargas Nov 14 '12 at 3:40
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 $W_{-1}(x)$. – Antonio Vargas Nov 14 '12 at 3:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.