# Solving $\sum_i^d a_i \exp(-q_i k)=b_0$ for $k$

Suppose $$a_i,q_i,b_0$$ are positive real numbers. I need to solve the following equation for $$k$$

$$\sum_i^d a_i \exp(-q_i k)=b_0$$

Is this a well-known problem? One my special cases has $$a_i=q_i$$

In my application $$d> 20000$$, $$q_i\approx 0$$. Wondering if equation structure makes it possible to solve faster than Newton's method.

• If the $q$ are all small, and if you expect $k$ to be of moderate size, then $e^{-q_ik}\approx 1-q_ik$ and the whole problem becomes one simple linear equation in $k$. At the very least, this will provide for a good starting guess for a Newton method for solving the "real" problem. Mar 1 at 22:31
• I think it becomes $(1-q_i)^k$. This is actually the equation I started with, before switching to $e$ approximation, assuming some transform or series expansion of $e$ might be useful here Mar 1 at 23:03
• Cross-posted on dsp.SE Mar 2 at 17:52
• The first order Taylor expansions of $e^{-qk}$ and $(1-q)^k$ are both $1-qk$ (as you can see by multiplying out what $(1-q)^k$ actually is). If $qk\ll 1$, you can choose whichever you want, and $1-qk$ is definitely the easier one. Mar 2 at 22:13
• Newton should converge in very few steps and without numerical problems, since the function and all its derivatives are monotonic; and each step costs 2x as much as evaluating the function itself. It seems hard to beat if you consider these points. Mar 4 at 21:46

If $$q_i$$ are small, you might get away with solving for k with only one summand where q_i is smallest, as that will be the dominant term. The other terms where q_i are larger will be small due to the exponential function. As the comments suggest, that might just serve as a good starting point for a newton iteration.
• Good point, that approach gives a lower bound on true value. And I can get upper bound by averaging $q_i$'s together and using Jensen's Mar 3 at 8:18