I'd like to implement a sum of exponentials regression program, and found what looks to be a reasonable algorithm from @JJacquelin in his article on using integral equations to carry out nonlinear regressions without having to resort to numerical optimization or monte carlo simulation.


In the article he takes the biexponential case:

$$ y = b\ \text{exp}(p\ x) + c\ \text{exp}(q\ x) $$

and derives the integral equation:

linear expression of biexponential function derived by @JJacquelin

from which ordinary least squares regression can be used (on the first and second cummulative numerical integrations) to estimate $p$ and $q$, followed by another ordinary least squares regression on the original formula to determine the remaining coefficients.

I'd like to understand how he derived this equation. He says he took two successive integrations and then used the resulting formulas to do a substitution which eliminated the exponential terms from the final equation shown above. While I can compute the antiderivatives, I'm lost as to how to actually derive this equation. Any guidance would be appreciated.


$$y(x) = b\ e^{px}+c\ e^{qx}$$ $$\begin{cases}\int y\ dx=\frac{b}{p}e^{px}+\frac{c}{q}e^{qx}+c_1 \\ \int\int y\ dx\ dx=\frac{b}{p^2}e^{px}+\frac{c}{q^2}e^{qx}+c_1x+c_2\end{cases}$$ Solve this linear system for the two unknowns $e^{px}$ and $e^{qx}$.

Then, put them into $y = b\ e^{px}+c\ e^{qx}$

You obtain : $y=-pq\int\int y\ dx\ dx+(p+q)\int y\ dx+Cx+D$

Going from indefinite integrals to definite integrals with chosen lower bounds introduces some constants which are then included in $C$ and $D$.

  • $\begingroup$ Is your full name Speedy J. Jacquelin ? $\endgroup$ – Claude Leibovici Dec 11 '17 at 11:28
  • $\begingroup$ Ha-ha ! Not always so speedy... $\endgroup$ – JJacquelin Dec 11 '17 at 11:44
  • $\begingroup$ Thanks, can you explain why $x$ in the system of equations ($c_1 x$) is not an 'unknown'? $\endgroup$ – wdkrnls Dec 11 '17 at 15:03
  • $\begingroup$ $x$ is a set of experimental values as well as $y$.. The unknowns are $p$ , $q$ , $C$ and $D$. $\endgroup$ – JJacquelin Dec 11 '17 at 15:11
  • $\begingroup$ but in the second integration equation the left hand side is a scalar value, so mustn't $x$ be also a scalar? Thanks for your patience. I guess then I move the $c_1 x$ and $c_2$ parts to the left hand side, and then invert the matrix. It feels alien to me to do that with $x$. $\endgroup$ – wdkrnls Dec 11 '17 at 15:15

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