Defining function handles in MATLAB

Question

Richard 2012-1-8

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评论： Voss 2023-2-4

采纳的回答： Chandra Kurniawan

How might I define a function handle?

For example, I want to define a function f(x)=2*x^3+7*x^2+x

I want MATLAB to evaluate f(x) at random values x. I have heard of feval and fhandles, but I don't know how to do it.

Thanks.

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Leia Sofia Mendez 2017-7-26

编辑：Walter Roberson 2017-7-26

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This was the code I was trying to write:

a= 0.0009
a= convtime([1],'samples','seconds')

This code gave an error saying that my function (convtime) was undefined. How you define a function in MATLAB?

Walter Roberson 2017-7-26

Leia Sofia Mendez:

convtime appears to be from the File Exchange https://www.mathworks.com/matlabcentral/fileexchange/28204-time---rate-unit-conversion-function .

You should go to that link, and click the download button, and download the .zip file. You should unzip to a directory that is not under your MATLAB installation directory. You would then use pathtool in MATLAB to add that directory to your MATLAB path.

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Answer 1

Chandra Kurniawan 2012-1-8

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编辑：MathWorks Support Team 2019-5-22

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Hi, Richard.

To evaluate f(x) at different values of x, you can create an .m file and write this code:

function y = f(x)
y = 2 * (x^3) + 7 * (x^2) + x;

If you save the file under the name 'f.m', you can run the function by typing this code in the Command Window or a separate .m file.

x =  randi(7);
y = f(x)

The randi function above generates a 1-by-5 row vector of random integers between 1 and 10. The values returned by f are stored in a 1-by-5 row vector y.

For more information about creating functions, see:

http://www.mathworks.com/help/matlab/matlab_prog/create-functions-in-files.html.

You can create a handle to the function f with an @ sign. For example, create a handle named myHandle as follows:

myHandle = @f;

Now you can run f indirectly by using its handle.

y = myHandle(x)

For more information about function handles, see:

https://www.mathworks.com/help/matlab/matlab_prog/creating-a-function-handle.html.

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Piyush Gade 2022-2-4

Whatever this is, this doesn't work.

Mark Jackson 2022-10-11

I recommend clicking 'Getting Started.'

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Answer 2

Walter Roberson 2012-1-8

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Function handle version:

f = @(x) 2*x^3+7*x^2+x;

Then f is already the function handle, and you can call f(3.7) (for example)

There is no need to use feval() for this, but you could.

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Salaheddin Hosseinzadeh 2014-3-13

I rather this anonymous way of defining a function! It's way easier. I also know another way of doing this, surprisingly nobudy mentioned that so far! lol I'm gonna put it in the answers.

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Answer 3

Junaid 2012-1-8

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Dear Richard,

To define a function in matlab you can do following syntax of given function:

function n = F(x)
n= 2*x^3+7*x^2+x;

that is it. You can put end at the end of function. But it is also acceptable not to put to various matlab versions. If you put end for one function then you have to put for all function in single m file.

then you can generate random numbers, either integer or double, and can get the values of this function.

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Richard 2012-1-8

Thanks, Junaid! :)

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Answer 4

cyril 2014-3-21

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编辑：cyril 2014-3-21

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> f = @(x) 2*x^3+7*x^2+x;
> f(0)
0

surprising no one mentioned anonymous functions...

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Salaheddin Hosseinzadeh 2014-3-21

@ Cyril

Walter did, just make sure you checked the other answers and comments!

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Answer 5

samy youssef 2015-3-11

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编辑：Walter Roberson 2016-9-26

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here is a function i developed to calculate the log of any number with different base:

function d =log_for_diff_base(myNumber,myBase)
x=log(myNumber);
y=log(myBase);
d=x/y;
end

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Walter Roberson 2016-9-26

Okay... but irrelevant to the original question.

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Answer 6

Nikitha Challa 2016-9-26

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x = x + a/x 2 in matlab code

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John D'Errico 2016-9-26

编辑：John D'Errico 2016-9-26

Not a function at all. This is not even valid MATLAB code as written.

Walter Roberson 2016-9-26

在 MATLAB Online 中打开

That does not appear to be a question, and it is not an Answer to what was asked here?

If the question is to solve the equation

x == x + a/(x^2)

then for finite a values, the solutions are -inf and +inf as a/(x^2) would be 0 for those values, leading to the equality -inf == -inf and +inf == +inf

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