Solve exponential equation with constant term.

Question

Is there an analytical solution to an equation of the form $a e^{(\alpha + i \beta) x} - b e^{(\alpha - i \beta) x} + c = 0$? How can we solve such an equation to $x$?

EDIT 1: Thanks for your patience with me. I tried it based on your hint in Matlab with the following code:

a = 5.6;
b = -12.6;
c = 18.9;
alpha = 2.5;
beta = 3.3;
r = (alpha+1jbeta)/(alpha-1jbeta);
A = a/b;
C = c/b;
y = C;
for n = 1:100
  derivative = (nr)C^((nr)-1);
  y = y + ((A^n)/(factorial(n)))derivative;
end;
y
y_correct = fsolve(@(y)(ay^r - by + c), 1)

However, this yields $y = -1.497762285457014e+00 + 1.248026585918367e-02i$ and $y_{correct} = -1.517484388747081e+00 + 8.391487021758234e-03i$. When I check with

A*y^r + C - y

the result is not 0, so clearly there is an error here somewhere.

EDIT 2: found the problem:

a = 5.6;
b = -12.6;
c = 18.9;
alpha = 2.5;
beta = 3.3;
r = (alpha+1jbeta)/(alpha-1jbeta);
A = a/b;
C = c/b;
y = C;
for n = 1:100
  if n == 1
    derivative = C^r;
  else
    derivative = (nr)C^((nr)-1);
  endif;
  y = y + ((A^n)/(factorial(n)))derivative;
end;

works good!

EDIT 3: There was still an error in the derivative. I fixed it and wrote an answer.

You can substitute $y=e^{(\alpha-i\beta)x}$ and use series solution for general trinomial — Тyma Gaidash, Jun 10 '23 at 14:13
Thanks for your hint. However I've try to substitute as you recommend, ending up in $a e^{(\alpha + j \beta) x} - b y + c = 0$. $x$ remains in the equation and I'm still lost. — Doran Strehnisch, Jun 10 '23 at 17:37
You should get $e^x=y^\frac1{\alpha-i\beta}$, so $a y^\frac{\alpha+i\beta}{\alpha-i\beta}-by+c=0$. Also, are you looking for all complex solutions or just one? — Тyma Gaidash, Jun 10 '23 at 17:48
Ok, I tried to apply the stuff from your link to my problem, however did not succeed. Might need a little primer to get things in the right way. — Doran Strehnisch, Jun 11 '23 at 15:51
Sorry, but maybe you can add more details, like your attempts to solve the equation. Then, I can write an answer. — Тyma Gaidash, Jun 11 '23 at 17:08
For one solution, you can try Lagrange reversion: $Ay^r+C=y\implies y=C+\sum\limits_{n=1}^\infty \frac{A^n}{n!} \left.\frac{d^{n-1}}{dw^{n-1}}w^{nr}\right|_{w=C}$. There is a method to get multiple complex solutions, but it is a little more work . — Тyma Gaidash, Jun 11 '23 at 17:57

score 1 · Accepted Answer · answered Jun 12 '23 at 02:26

$\def\sgn{\operatorname{sgn}}$ Let $r=\alpha+\beta i,A=\frac ab,B=\frac ca$:

$$a e^{rz}-be^{\bar rz}+c=0\iff A e^{(r-\bar r)z}+B e^{-\bar rz}=1$$

and substitute like in this MathOverflow question:

$$B(e^z)^{-\bar r}=w,\ln(e^z)=-\frac{\ln\left(\frac wB\right)}{\bar r}-\frac{2\pi i k}{\bar r}\implies e^z=e^{-\frac{2\pi i k}{\bar r}}\left(\frac w B\right)^{-\frac1{\bar r}}$$

to get:

$$w=1-Ae^{2\pi i k(1-\sgn^2(r))}B^{\sgn^2(r)-1}w^{1-\sgn^2(r)}=1-pw^v$$

Using Lagrange reversion and factorial power $u^{(v)}$:

$$e^\frac{2\pi i k}{\bar r}e^z=\left(\frac1B\right)^{-\frac1{\bar r}}-\frac{B^\frac 1{\bar r}}{\bar r}\sum_{n=1}^\infty\frac{(-p)^n}{n!}\frac{d^{n-1}}{dt^{n-1}}\left(w^{vn}\frac d{dt}\left(\frac tB\right)^{-\frac1{\bar r}}\right)=B^\frac1{\bar r}-\frac{B^{\frac 1{\bar r}}}{\bar r}\sum_{n=1}^\infty\frac{(-p)^n}{n!}\left(vn-\frac1{\bar r}-1\right)^{(n-1)}$$

shown here is a solution to $B y^{-\bar r}=1-p (B y^{-\bar r})^v$

To be updated with more details – Тyma Gaidash Jun 12 '23 at 02:26 — Тyma Gaidash, Jun 12 '23 at 02:26

score 0 · Answer 2 · answered Jun 11 '23 at 20:04

Based on the answer of Tyma Gaidash, I came up with this solution:

clear all; close all;
format long;
a = 5.6;
b = -12.6;
c = 18.9;
alpha = 2.5;
beta = 3.3;
r = (alpha+1jbeta)/(alpha-1jbeta);
A = a/b;
C = c/b;
y = C;
for n = 1:100
  derivative = C^((nr) - (n-1));
  for j = 1:(n-1)
    derivative = derivative  ((nr) - (1-j));
  end;
  y = y + ((A^n)/(factorial(n)))derivative;
end;
y
y_correct = fsolve(@(y)(ay^r - by + c), 1)
A*y^r + C - y

Solve exponential equation with constant term.

2 Answers2