How to find exponent coefficients in a sum of exponents?

Question

It is easy to determine a coefficient 'c' of exp(c*x), just log it and find slope. Or if it's

exp(c1*x) + exp(c2*x)

then after log from 0 to the right of left we would find 'c1' and 'c2'. But what if we have more terms? For example, such a sum

exp(x) + exp(1.5*x) + exp(2.5*x) + exp(3*x)

if you loged it, almost would have no difference between exp(x) + exp(3*x) So, please, how could I find them in another way?

Answer 1

If you know, that $$y(x) = \exp(c_1 x)+\exp(c_2 x)+\cdots+\exp(c_n x),$$ then calculate $y(1), y(2), \ldots, y(n)\;$: $\;p_1 = y(1), p_2 = y(2), \ldots, p_n=y(n)$.
You'll get system of equations:

$$\left\{ \begin{array}{r} \exp(c_1)+\exp(c_2)+\cdots+\exp(c_n)=p_1; \\ \exp(2c_1)+\exp(2c_2)+\cdots+\exp(2c_n)=p_2; \\ \cdots \cdots \cdots \qquad \qquad \qquad \qquad \\ \exp(nc_1)+\exp(nc_2)+\cdots+\exp(nc_n)=p_n. \end{array} \right. $$

If denote $s_1=\exp(c_1), s_2=\exp(c_2), \ldots, s_n=\exp(c_n)$, then $$\left\{ \begin{array}{r} s_1+s_2+\cdots+s_n=p_1; \\ s_1^2+s_2^2+\cdots+s_n^2=p_2; \\ \cdots \cdots \cdots \qquad \qquad \\ s_1^n+s_2^n+\cdots+s_n^n=p_n. \end{array} \right. $$

With the help of Power sum symmetric polynomial you'll find values $e_1,e_2, \ldots, e_n$ $-$ elementary symmetric polynomials:

$$\left\{ \begin{array}{l} e_1 = s_1+s_2+\cdots+s_n = p_1; \\ e_2 = s_1s_2+s_1s_3+\cdots+s_{n-1}s_n = \dfrac{1}{2}(p_1^2-p_2); \\ \cdots \cdots \cdots\qquad \qquad \\ e_n = s_1s_2\cdots s_n=\dfrac{1}{n}\sum\limits_{j=1}^n (-1)^{j-1}e_{n-j}p_j; \end{array} \right. $$

so $s_1,s_2,\ldots,s_n$ are roots of equation

$$s^n-e_1\cdot s^{n-1}+\cdots+(-1)^{n-1}e_{n-1}\cdot s+(-1)^n e_n=0.$$

Finally, when you will find roots $s_1,s_2,\ldots,s_n$,

$$c_j=\ln(s_j), \; j=1,\ldots,n.$$

Answer 2

The answer depends much of the control you have over the studied function.

You may apply an integral transform to your function (with the advantage of additionnal constants in front of the exponentials possible) :

• Laplace transform for example : $e^{7\,x}+3\;e^{11\,x}\mapsto \frac 1{s-7}+\frac 3{s-11} $ see this Alpha result
• Fourier transform (multiplying $x$ by $-i$ first) to get for example : $e^{-5\,i\,x}+e^{-11\,xi\,}\mapsto \sqrt{2\pi}[\delta(\omega-5)+\delta(\omega-11)]$ with $\delta$ the Dirac distribution.
  Alpha result.
• wavelet transform and so on...

Another method is to use :

• the derivative at $0$ that will give you the sum of coefficients.
• the second derivative at 0 the sum of square of coefficients and so on.

You could too (as suggested by Peter Košinár) evaluate the largest exponential, subtract it from your function, estimate the next largest exponential and so on...