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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 $a \to Inf$?

  • 0
    Here is [the technique](http://math.stackexchange.com/questions/185397/number-to-the-exponent-divided-by-exponent-value/185411#185411) of the solution.2012-11-19

1 Answers 1

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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.

  • 0
    If you'd like an asymptotic representation for this other solution, see the [wikipedia page for$W$under "Asymptotic expansions"](http://en.wikipedia.org/wiki/Lambert_W_function#Asymptotic_expansions) and use the one for $W_{-1}(x)$.2012-11-14