Quotes around the name of the function

Asked by Antonis V. on 15 Dec 2012

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When do we put quotes around the name of the function and when not ? For example, I used fsolve(myfun,x0) and I got no results , but when I used fsolve('myfun',x0) I got the root I was looking for ...

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Answer by Matt Fig on 15 Dec 2012

Edited by Matt Fig on 15 Dec 2012

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Typically, you want to pass a handle to a function, not a string.

fsovle(@myfun,x0) % Notice the @ symbol --> a function handle.

The string argument gets evaluated the same way, probably for backward compatibility. The modern way is to use function handles.

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Antonis V. on 15 Dec 2012

Thanks !! But I still don't get why fsolve(f,0) worked with f(x) = exp(x) - cos(x) - 1 and didn't work with f(x) = exp(x) - x^2 - 68 where I had to use the quotes or the function handle .

Matt Fig on 15 Dec 2012

You would have to show exactly what you did in order for me to know why one code failed.

f = @(x) exp(x) - cos(x) - 1;  % f is a function handle.
fsolve(f,0)   % Seems to work....
f = @(x) exp(x) - x^2 - 68 ;  % Another function handle.
fsolve(f,0)   % Also works....

Antonis V. on 15 Dec 2012

Ok i found out why it failed ! I had created and saved this function so I guess I had to use fsolve(@f,0) in first place ! Thanks a lot ! And one last thing :P I am a new Matlab user , and I haven't yet figured out how initial guesses work !For instance, I want to find out the roots of the equation F(X) = exp(x) - x^2 - 68 . What values should I put to x0 ? I 've tried several values close to 0 and I got the same result but I noticed I can't use very big numbers , whereas I can use very small numbers. If I want to find all the roots at F's domain range should I put more than one values at x0 ?

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