Fast methods to find zeros of a sum of exponentials? I'm looking for a fast method to find a zero within a region $[x_\text{low}, x_\text{high}]$ for the following function:
$$f(x) = a_1 e^{b_1 x} + a_2 e^{b_2 x} + a_3 e^{b_3 x} + \cdots$$
Right now I'm using bisection, but I'm not satisfied by its performance and I'm looking for a faster method. I could try some other general-purpose root finding methods (Netwton's method, ITP method, ...) but I'm wondering if there's any specialized algorithm that could help here.
 A: Newton method is the first choice (the only problem being that you need a reasonable starting guess $x_0$).
In the most genearl case where the $b_i$ could be anything (integer, rational or not) you cannot reduce the problem to a polynomial.
In any manner, in the real domain, to have a solution you need that some of the $a_i$ be positive and other be negative. So, write
$$f(x)=\sum_{i=1}^p \alpha_i e^{b_i x}-\sum_{i=1}^p \beta_i e^{b_i x}+c$$ where the $\alpha_i$ correspond to the positive $a_i$ and the $\beta_i$ correspond to the absolute values of the negative $a_i$.
Now, to make the problem more linear (this is always good for a root-finding method), instead of looking for the zero of $f(x)$, consider instead
$$g(x)=\log\Bigg[c+\sum_{i=1}^p \alpha_i e^{b_i x} \Bigg]-\log\Bigg[\sum_{i=1}^p \beta_i e^{b_i x} \Bigg]\qquad \text{if} \qquad c>0$$ or
$$g(x)=\log\Bigg[\sum_{i=1}^p \alpha_i e^{b_i x} \Bigg]-\log\Bigg[-c+\sum_{i=1}^p \beta_i e^{b_i x} \Bigg]\qquad \text{if} \qquad c<0$$
This will make the problem quite close to the intersection of two straight lines.
A: Let $x = \ln u$. Then we have:
$$f(x) = a_1 u^{b_1} + a_2 u^{b_2} + a_3 u^{b_3} + \cdots$$
and it is simpler to find the roots of this expression using Newton's method, which may be a polynomial. The bounds for $u$ are just $[e^{x_{\text{low}}}, e^{x_{\text{high}}}]$ .
For example, $2e^{3x} - 3e^{4x} = 0$ becomes $2u^3 - 3u^4 = 0 \implies 2u^3 \left(1 - \frac{3}{2} u \right) = 0$ or $u = 0, u = \frac{2}{3}$. Since $\ln 0$ is undefined, the only root must be $x = \ln \frac{2}{3} = \ln 2 - \ln 3$.
