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Solving a simple equation

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James
James on 25 May 2011
I have a function that can not be simplified further that goes something like this
0 = (1+exp(x))/(1+exp(-x))
(Simplified, my function is much longer and convoluted. The point is that I can't just write it as 'x = ...')
How can Matlab approximate this equation by choosing a good value for x? What is the command for this?

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Accepted Answer

John D'Errico
John D'Errico on 25 May 2011
If you are looking for a numerical solution, then this is a rootfinding problem. Use fzero, a tool designed to solve exactly that problem.
xfinal = fzero(@(x) (1+exp(x))./(1+exp(-x)),xstart);
You must supply a value for xstart. Better yet is if you can supply a pair of points that are known to bracket a solution.
Of course, this function has no solution, so it will always fail, but I assume that your true function does have one.
If you have truly tried to confuse things, and your real function is multi-dimensional, then you can use tools from the optimization toolbox. Here one would use fsolve.
If you are looking for a symbolic solution then solve (which requires the symbolic toolbox) is the answer.

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More Answers (1)

Ben Mitch
Ben Mitch on 25 May 2011
You can solve an equation algebraically using the symbolic math toolbox, e.g. to solve x+3=0 use:
solve('x+3')
You can solve it numerically using various techniques, including the optimization toolbox, e.g. to solve x+3=0 use:
fminsearch(@(x) (x+3)^2, 0)
However, the equation you've posted has no finite solutions, which is why both of these commands generate trash:
solve('(1+exp(x))/(1+exp(-x))')
fminsearch(@(x) ((1+exp(x))/(1+exp(-x)))^2, 0)
Note that the numerical approach (the second command) does find an answer, but it's only a value of x where the equation is approximately equal to zero, and there's an infinity of them so the particular answer returned is not meaningful.

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