Could anyone point me in the right direction of finding the roots of equations of the form $$ \sum_{i=1}^n a_ie^{f_i(x)}, $$ where $a_i \in \mathbb{R}$ and the $f_i$ are each first degree polynomials $x$, $i = 1,2,\ldots, n$? For example, solve $$ a_1e^{f_1(x)} + a_2e^{f_2(x)} + a_3e^{f_3(x)}+ a_4e^{f_4(x)} = 0 $$ for $x$.

Note that if I can transform it into a linear combination of the $f_i$'s then it is simply root-finding for a first degree polynomial. E.g. for $n=2$, assuming $a_1>0$ and $a_2<0$ we get $$ \begin{align*} a_1e^{f_1(x)} + a_2e^{f_2(x)} = 0 & \iff a_1e^{f_1(x)} = -a_2e^{f_2(x)} \\ & \iff \log(a_1) + f_1(x) = \log(-a_2) + f_2(x) \\ &\iff f_1(x) - f_2(x) = \log(-a_2/a_1), \end{align*} $$ which is easily solved for $x$ since $f_1(x) - f_2(x) \in \mathbb{P}^1$.


Note that $\mathrm{e}^{a x+b} = \mathrm{e}^b \cdot \left(\mathrm{e}^x \right) ^a$. Making the change of variable $\mathrm{e}^x \rightarrow z$, we find that the equation you want to solve is actually a polynomial (if all the linear coefficients are positive integers) or a generalization to larger coefficient rings.

To be clear, $z^\pi - z^\mathrm{e}$ is not a polynomial because the powers are not positive integers. These powers come from the linear coefficients in the given $f_i$. So, depending on the constraints in the coefficients of the $f_i$ this may or may not be a polynomial.

If it is a polynomial, use usual root-finding techniques to find $z_i$ and then $x_i = \ln z_i$ are the roots of your original equation. If not, good luck: even the number of roots of an equation can be difficult to figure out. ($x^{\pi}-1=0$ has three roots, so you might think to round down to the nearest integer. But $x^{2\pi}-1=0$ has seven roots, so rounding down doesn't work.)

  • $\begingroup$ This is quite interesting, Eric. Where can we find a detailed discussion of equations like $x^{\pi} = 1$? $\endgroup$ – Manolito Pérez Aug 13 '15 at 15:08
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    $\begingroup$ @ManolitoPérez : I'm not aware of a reference on this specific topic. I approach it as described above. For $x^\pi - 1$, the roots are $\mathrm{e}^{0(\pi \mathrm{i})}$ and $\mathrm{e}^{\pm 2\pi \mathrm{i}}$, which you might recognize are just extracted from the $x = \ln z$ change of variables. $\endgroup$ – Eric Towers Aug 18 '15 at 3:58

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