How to solve this nonlinear equation in Python, Octave, or Matlab?

Question

How do I solve this equation using Python, Octave, or Matlab?

Need to be able to find the solution of x for different values of c.

Answer 1

Use Solvers

# this is an example
>>> from sympy.solvers import solve
>>> from sympy import Symbol
>>> x = Symbol('x')
>>> solve(x ** 2 - 1, x)
[-1, 1]

Answer 2

As the equation has some float constants, it is probably best to use nsolve to solve it numerically:

from sympy import Eq, log, exp, nsolve
from sympy.abc import x

c = 267
sol = nsolve(Eq(x ** 0.22 * (c * exp(-c * x ** 0.22) - 1), c * log(0.7657)), x, 1)
print(sol)  # 264587674.024352

Answer 3

No Toolbox Required:

In MATLAB, you can use fminsearch() to do this numerically.

The key idea is to rewrite the equation y = w as y - w = 0. Then, construct a convex function to guarantee you the local minima returned is indeed globally optimal, e.g. abs(y-w).

% Rewrite equation y = w as y - w = 0.
% Then minimize abs(y-w).
fh =@(x,c) abs((x.^(0.22)).*(c*exp(-c*(x.^(0.22)))-1) - c*log(0.7657));

c = 267;
x0 = 5;                                      % initial guess
[x, abserr] = fminsearch(@(x) fh(x,c),x0)    % x = 264587674.0243530

Note that abserr = 4.2633e-14 which is pretty darn close to zero (equality).

Similarly, you could minimize the sum of squared error to achieve the same answer.

% minimize sum((y-w).^2)
gh =@(x,c) sum(((x.^(0.22)).*(c*exp(-c*(x.^(0.22)))-1) - c*log(0.7657)).^2);
[x, sse] = fminsearch(@(x) fh(x,c),x0) 

---

Rather than using arrayfun(), to get the different x values for each c value, it is probably easier just to loop through.

C = [100 267 300].';
Xval = zeros(size(C));
for ii = 1:length(C)
    Xval(ii) = fminsearch(@(x) fh(x,C(ii)),x0);
end

---

Not Recommended:
While fzero() will also work to solve y - w = 0, notice that it aborts searching for an interval containing a sign change (if x is negative). So you either have to add a penalty for that direction somehow, or try from a different starting value.

% solve y-w==0
zh =@(x,c) (x.^(0.22)).*(c*exp(-c*(x.^(0.22)))-1) - c*log(0.7657)

Using the same starting point,

x = fzero(@(x) zh(x,c),x0)

returns the error

Exiting fzero: aborting search for an interval containing a sign change because complex function value encountered during search. (Function value at -1.4 is 70.4499-0.686399i.) Check function or try again with a different starting value. x = NaN

while

x = fzero(@(x) zh(x,c),2e8)       % adjusted starting guess

works just fine.

_{Tested with MATLAB R2019a}