Exponential Sum Approximation

Question

Is it possible to show mathematically that, for a short portion, sum of two decaying exponential can be approximated by a single decaying exponential? i.e. $Ae^{-ax}+Be^{-bx}\approx Ce^{-cx}$?

Answer 1

Assuming that the two exponents $(a,b)$ are not "too different", what you could do, for the range $\alpha \leq x \leq \beta$, is to minimize $$\Phi(C,c)=\int_\alpha^\beta \Big[C e^{-c x}-A e^{-a x}-B e^{-b x} \Big]^2\, dx$$ which is equivalent to an exponential regression based on an infinite number of data points.

The antiderivative is $$-\frac{A^2 e^{-2 a x}}{a}-\frac{4 A B e^{-(a+b)x}}{a+b}+\frac{4 A C e^{ -(a+c)x}}{a+c}-\frac{B^2 e^{-2 b x}}{b}+\frac{4 B C e^{ -(b+c)x}}{b+c}-\frac{C^2 e^{-2 c x}}{c}$$

Apply the bounds to get $\Phi(C,c)$, compute the partial derivatives and set them equal to $0$. $$\frac{\partial \Phi(C,c)}{\partial C}=0 \implies C=f(c)\qquad \text{(which is an explicit function)}$$ and you are left with $$\frac{\partial \Phi(C,c)}{\partial c}=\frac{\partial \Phi(f(c),c)}{\partial c}=0$$ which will require some numerical method (a quite nasty nonlinear equation in $c$ but not difficult to solve using Newton method with $c_0=\frac{a+b}2$).

Probably, generating data points and using nonlinear regression could be easier since the exponential fitting is quite trivial. Generate $n$ data points $(x_i,y_i)$ with $y_i=A e^{-a x_i}+B e^{-b x_i}$ to face the model $$y=C e^{-c x}$$ In a first step, take logarithms and a linear regression will give estimates of $\log(C)$ and $c$ which will be good starting values for the nonlinear regression.

For illustration, I used $A=123$, $a=0.8$, $B=234$, $b=1.1$, $\alpha=3$, $beta=5$ and generated $100$ data points. The nonlinear regression gives $(R^2 > 0.9999)$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ A & 306.804 & 0.60903 & \{305.596,308.013\} \\ a & 0.91475 & 0.00057 & \{0.91363,0.91587\} \\ \end{array}$$ showing a maximum absolute error of $0.06$ while the sum of the two exponentials vary between $20$ and $3$.