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solve nonlinear equation

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Eman Ahmed Elsayed
Eman Ahmed Elsayed 2011-5-31
I write a method to solve any equation
function [ result ] = get( func )
%UNTITLED2 Summary of this function goes here
% Detailed explanation goes here
result=solve(func);
end
when I run it on command window,I get an error.
>> f=@(x) 2*x+1
f =
@(x)2*x+1
>> [ result ] = get( f )
??? Error using ==> error
Function is not defined for 'function_handle' inputs.
Error in ==> solve>getEqns at 182
error('symbolic:solve:errmsg1', ...
Error in ==> solve at 67
[eqns,vars] = getEqns(varargin{:});
Error in ==> get at 4
result=solve(func);
so how can I fix the error

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回答（3 个）

Oleg Komarov
Oleg Komarov 2011-5-31
You cannot solve a function handle.
You have to convert the anonymous function handle to a string function:
fh = @(x,y) 2*x+1-2*y;
% Convert to string
fs = func2str(fh);
% Extract only function part w/o @(.)
fs = regexprep(fs,'@\([\w,]+\)','');
% Solve
solve(fs)
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Eman Ahmed Elsayed
Eman Ahmed Elsayed 2011-6-1
func2str doesn't work with me
>> f=@(x) 2*x+1
f =
@(x)2*x+1
>> func2str(f)
ans =
@(x)2*x+1
It's return the same equation
Matt Fig
Matt Fig 2011-6-1
Read the rest of the post. FUNC2STR is working just like Oleg showed...

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Walter Roberson
Walter Roberson 2011-6-1
solve() only applies to symbolic expressions. If you are going to use symbolic expressions you might as well do so from the start and save the trouble about constructing function handles.
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Walter Roberson
Walter Roberson 2011-6-1
If this is for an assignment, then it would help us to read the assignment question, as then we would not suggest things you cannot use.
Walter Roberson
Walter Roberson 2011-6-1
If the assignment prohibits you from passing around symbolic variables, then why does it permit you to use solve(), the symbolic equation solver?

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Eman Ahmed Elsayed
I'm still have the same problem
  1 个评论
Oleg Komarov
Oleg Komarov 2011-6-1
What do you mean you have the same problem? Post the whole code you're using and the error message.

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