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A simple fsolve problem from a MATLAB beginner

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Peter J
Peter J on 23 Jan 2015
Commented: Peter J on 23 Jan 2015
Hi. I would like to solve a nonlinear equation using fsolve. The variable is denoted as phi. It should have three roots. Below is the mfile that I built.
function out = underwoodroot(phi)
alphaD = 1.00; alphaC = 2.5; alphaB = 6.25; alphaA = 15.625; xAfeed = 0.25; xBfeed = 0.25; xCfeed = 0.25; xDfeed = 0.25;
out = alphaA*xAfeed/(alphaA-phi)+alphaB*xBfeed/(alphaB-phi)+alphaC*xCfeed/(alphaC-phi)+alphaD*xDfeed/(alphaD-phi);
phi1 = fsolve(@underwoodroot,(alphaB+alphaA)/2)
phi2 = fsolve(@underwoodroot,(alphaC+alphaB)/2)
phi3 = fsolve(@underwoodroot,(alphaD+alphaC)/2)
end
However, when I tried to run it, MATLAB returned a "Not enough input arguments" comment. I wonder what happened to my code. As a MATLAB beginner, I would really appreciate any help you have. Thank you so much in advance!

Accepted Answer

Shoaibur Rahman
Shoaibur Rahman on 23 Jan 2015
Define underwoodroot function separately, and call them from another m-file.
In one m-file, write the following code, and save and run:
alphaD = 1.00; alphaC = 2.5; alphaB = 6.25; alphaA = 15.625;
phi1 = fsolve(@underwoodroot,(alphaB+alphaA)/2)
phi2 = fsolve(@underwoodroot,(alphaC+alphaB)/2)
phi3 = fsolve(@underwoodroot,(alphaD+alphaC)/2)
In another m-file, write the underwoodroot function, and save as underwoodroot.m
function out = underwoodroot(phi)
alphaD = 1.00; alphaC = 2.5; alphaB = 6.25; alphaA = 15.625; xAfeed = 0.25; xBfeed = 0.25; xCfeed = 0.25; xDfeed = 0.25;
out = alphaA*xAfeed/(alphaA-phi)+alphaB*xBfeed/(alphaB-phi)+alphaC*xCfeed/(alphaC-phi)+alphaD*xDfeed/(alphaD-phi);
However, if you want to keep them in a single m-file, then use two functions names, like below. Save the m-file as myFunction.m and run that.
function myFunction
alphaD = 1.00; alphaC = 2.5; alphaB = 6.25; alphaA = 15.625;
phi1 = fsolve(@underwoodroot,(alphaB+alphaA)/2)
phi2 = fsolve(@underwoodroot,(alphaC+alphaB)/2)
phi3 = fsolve(@underwoodroot,(alphaD+alphaC)/2)
function out = underwoodroot(phi)
alphaD = 1.00; alphaC = 2.5; alphaB = 6.25; alphaA = 15.625; xAfeed = 0.25; xBfeed = 0.25; xCfeed = 0.25; xDfeed = 0.25;
out = alphaA*xAfeed/(alphaA-phi)+alphaB*xBfeed/(alphaB-phi)+alphaC*xCfeed/(alphaC-phi)+alphaD*xDfeed/(alphaD-phi);
  1 Comment
Peter J
Peter J on 23 Jan 2015
Thank you so much for your help. That solves the problem.

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