In financial mathematics, the inverse series of: $$b(x) = -\frac{\log(1-e^{-x})}{x}$$ is needed in order to perform fast calculation on swaptions for G2++ calibration model. (see [this post](https://mathoverflow.net/questions/247977/taylor-series-expansion-for-an-implicitely-defined-family-of-functions) for further references). 


The unique positive solution for equation $$e^{-x}+e^{-bx}=1$$ implicitly defines $x$ as a function of $b$. There is no possible explicit solution for $x(b)$, unless when $b \in {0,1/4, 1/3, 1/2, 1,2,3,4}$, as it becomes Galois solvable. 


Nevertheless, it is always possible to express $b(x)$ as above such that the inverse series $x(b)$ would provide a solution.


I managed to expand x(b) up to order 5 as follows:


$$x(b) = +\log(2) - \frac{\log(2)}{2}(b-1) + \left(\frac{1}{4} \log(2)+\frac{1}{8} \log(2)^2\right) (b-1)^2 - \left(\frac{1}{8} \log(2)+\frac{3}{16} \log(2)^2\right) (b-1)^3 + \left(\frac{1}{16} \log(2)+\frac{3}{16} \log(2)^2 + \frac{1}{32} \log(2)^3-\frac{1}{192} \log(2)^4\right) (b-1)^4 - \left(\frac{1}{32} \log(2)+\frac{5}{32} \log(2)^2 + \frac{5}{64} \log(2)^3-\frac{5}{384} \log(2)^4\right) (b-1)^5 + o\left((b-1)^5\right)$$ 


However, that is not nearly enough for practical applications since the convergence is really slow.


So here is my question. Can we find a general formula for the $n^{th}$ term? Alternatively, is there any other faster converging Series that could be derived in this case?


To be a bit more precise, I am absolutely convinced that the coefficients of the expansion are rational polynomials in $\log(2)$, i.e. $a\_n = P\_n(\log(2))$. And I am looking for a recursive or explicit formulation of these polynomials. Or alternatively any other Series type (Chebyshev?) the convergence of which would be faster.


Thanks a lot for your help! 



Compute inverse series for implicit equation $b=-\log(1-e^{-x})/x$
Writing $e^{-x} = t$, your equation is $t + t^b = 1$. I'll assume $0 < b < 1$ (for the case $b > 1$, write $s = t^b$ and the equation becomes $s + s^{1/b} = 1$).


Now the slightly more general equation $t + \epsilon t^b = 1$ has a nice series solution 


$$ t = 1 - \sum\_{n=1}^\infty \left( \prod\_{j=0}^{n-2} (j-nb)\right) \frac{\epsilon^n}{n!}$$
 With our assumption $0 < b < 1$, at $\epsilon = 1$ the series should converge to a solution for your equation.


