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How can I solve an equation like $ x^a + bx + c = 0 $ ?

I figured that I can write this equation in a polynominal form as $ e^{wz} + b e^{z} + c $ or more generic as $ \sum{v_i e^{w_i z}} $ and that these are called exponential polynominals.

There are some papers about the zeros of these functions, but those are mostly concerned with problems where the exponent is imaginary and z is complex. In this case it seems, the exact positions of the zeros cannot be analytically calculated. (See "Polynomials" with non-integer exponents for a related question)

But I am only interested in the case of a real exponent ( $ z, w \in \mathbb{R} $ ).

(The underlying problem: I want to get the parameters of the equation $ f(x) = a + b e^{c x} $ defined by $ f(0) = y_0 $ , $ f(x_1) = y_1 $ , $ f(x_2) = y_2 $. I can solve this if $ x_1 = \frac{x_2}{2} $, because then the problem becomes a simple quadratic equation. But if $ x_1 \neq \frac{x_2}{2} $, then I get stuck with the problem at the top.)

Thanks for your help.

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There are no analytical solutions for the roots of such "polynomials", and usually the function is undefined for $x<0$.

Anyway, you can consider the intersections of the pseudo-parabola $y=x^a$ and the straight line $y=-bx-c$. If $b>0$, you must have $c<0$ for a real solution, which is unique.

Otherwise, you can easily obtain the position of the tangent of slope $-b$ to the curve, and from this check if the line lies below or above the tangent. This tells you the number of roots ($0, 1$ or $2$) and if there are two of them, it gives you a separation.

From gross initial values, you find the roots by Newton.

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  • $\begingroup$ What about something like the Budan's theorem on the number of roots in an interval - is there something like that for such "polynomial" with non-integer powers? Thank you. $\endgroup$
    – Confounded
    Jul 16, 2021 at 10:46

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