Solving exponential equations with different powers

Question

What is the best approach in solving such an exponential equation:

$-50 + 0.1k^{20} + 50k^{23} = 0$

Does it include estimation methods like bisection etc?

I am only concerned with finding the real number solution

Answer 1

Actually, this is called a polynomial equation. Exponential equations have the variable in the exponent. Yes, you will probably have to solve it numerically. By Descartes' rule of signs, we know that it has exactly one positive solution and either two or zero negative solutions.

The bisection method is rather slow, but it always works. There are faster methods.

The python script

import numpy as np

x=24*[0]
x[0] = -50
x[20] = .1
x[23] = 50

y = np.array(x)
roots = np.polynomial.polynomial.polyroots(y)
print(roots)

returned 22 complex roots and the one real root $.99991311$

Answer 2

If you consider that you look for the zero of function $$f(k)=50 k^{23}+\frac{1}{10}k^{20}-50$$ by inspection, you should notice that the solution is quite close to $k=1$.

For a quick approximation, use Taylor series around $k=1$ to get $$f(k)=\frac{1}{10}+1152 (k-1)+O\left((k-1)^2\right)\implies k=\frac{11519}{11520}\approx 0.9999131944$$ This is equivalent to the first step of Newton method.

For more accuracy, continue the expansion to get $$f(k)=\frac{1}{10}+1152 (k-1)+12669 (k-1)^2+O\left((k-1)^3\right)$$ Solve the quadratic in $(k-1)$ to get $$k=\frac{120930+\sqrt{33050910}}{126690}\approx 0.9999131114$$

If you still want more accuracy, use Newton method starting using $k_0=1$. The iterates will be given by $$k_{n+1}=\frac{11000 k_n^{23}+19 k_n^{20}+500}{20 \left(575 k_n^{22}+k_n^{19}\right)}$$ $$\left( \begin{array}{cc} n & k_n \\ 0 & 1.000000000000000000000000000000000000000 \\ 1 & \color{red}{0.9999131}944444444444444444444444444444444 \\ 2 &\color{red}{0.999913111468}8605891558531445789506996528 \\ 3 &\color{red}{0.9999131114687848659999735}811337718381139 \\ 4 &\color{red}{ 0.9999131114687848659999735180692687244066} \end{array} \right)$$ which is solution for fourty significant figures.

Edit

If you perfer approximations in the form of rational numbers, we can build $[1,n]$ Padé approximants at $k=1$. These would give $$\left( \begin{array}{ccc} n & \text{fraction} & \text{decimal}\\ 0 & \frac{11519}{11520} & 0.99991319444444444444 \\ 1 & \frac{4419073}{4419457} & 0.99991311149763421162 \\ 2 & \frac{21796759619}{21798653672} & 0.99991311146878612605 \\ 3 & \frac{3512024318956863}{3512329500108271} & 0.99991311146878486415 \\ 4 & \frac{8786931220058840463}{8787694769950168348} & 0.99991311146878486600 \end{array} \right)$$