MATLAB Answers

Defining function handles in MATLAB

조회 수: 1,788(최근 30일)
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.

  댓글 수: 2

Leia Sofia Mendez
Leia Sofia Mendez 26 Jul 2017
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
Walter Roberson 26 Jul 2017
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.

Sign in to comment.

채택된 답변

Chandra Kurniawan
Chandra Kurniawan 8 Jan 2012
편집: MathWorks Support Team 22 May 2019
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:

  댓글 수: 3

Chandra Kurniawan
Chandra Kurniawan 8 Jan 2012
to evaluate this function just type :
y = feval(@f,x)
Richard
Richard 8 Jan 2012
Thanks, Chandra! :)
Salaheddin Hosseinzadeh
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

Sign in to comment.

추가 답변(5개)

Walter Roberson
Walter Roberson 8 Jan 2012
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.

  댓글 수: 1

Salaheddin Hosseinzadeh
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.

Sign in to comment.


Junaid
Junaid 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.

  댓글 수: 1

Richard
Richard 8 Jan 2012
Thanks, Junaid! :)

Sign in to comment.


cyril
cyril 21 Mar 2014
편집: cyril 21 Mar 2014
> f = @(x) 2*x^3+7*x^2+x;
> f(0)
0
surprising no one mentioned anonymous functions...

  댓글 수: 1

Salaheddin Hosseinzadeh
@ Cyril
Walter did, just make sure you checked the other answers and comments!

Sign in to comment.


samy youssef
samy youssef 11 Mar 2015
편집: Walter Roberson 26 Sep 2016
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

  댓글 수: 1

Walter Roberson
Walter Roberson 26 Sep 2016
Okay... but irrelevant to the original question.

Sign in to comment.


Nikitha Challa
Nikitha Challa 26 Sep 2016
x = x + a/x 2 in matlab code

  댓글 수: 2

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

Sign in to comment.

제품


Translated by