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Defining function handles in MATLAB

Asked by Richard on 8 Jan 2012
Latest activity Edited by MathWorks Support Team on 22 May 2019
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.

  2 Comments

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 年 7 月 26 日
Leia Sofia Mendez:
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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6 Answers

回答者: Chandra Kurniawan 2012 年 1 月 8 日
編集済み: MathWorks Support Team 2019 年 5 月 22 日
 採用された回答

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:
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:

  3 Comments

Chandra Kurniawan 2012 年 1 月 8 日
to evaluate this function just type :
y = feval(@f,x)
Thanks, Chandra! :)
The problem or lets say the dificulty of this code is that this couple of lines should be saved as a function and I personally don't wanna do that for a simple function like this! or even complicated ones!
Not my favorite way of defining a function

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回答者: Walter Roberson
2012 年 1 月 8 日

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.

  1 Comment

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 by Junaid
on 8 Jan 2012

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.

  1 Comment

Thanks, Junaid! :)

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Answer by cyril
on 21 Mar 2014
Edited by cyril
on 21 Mar 2014

> f = @(x) 2*x^3+7*x^2+x;
> f(0)
0
surprising no one mentioned anonymous functions...

  1 Comment

Salaheddin Hosseinzadeh 2014 年 3 月 21 日
@ Cyril
Walter did, just make sure you checked the other answers and comments!

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回答者: samy youssef 2015 年 3 月 11 日
編集済み: Walter Roberson
2016 年 9 月 26 日

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

  1 Comment

Walter Roberson
2016 年 9 月 26 日
Okay... but irrelevant to the original question.

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回答者: Nikitha Challa 2016 年 9 月 26 日

x = x + a/x 2 in matlab code

  2 Comments

John D'Errico
2016 年 9 月 26 日
Not a function at all. This is not even valid MATLAB code as written.
Walter Roberson
2016 年 9 月 26 日
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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