Hello, I haven't figured out to find a way to create a new function out of multiplying 2 functions, for example:
fun1 = @(x) sin(x);
fun2 = @(x) cos(x);
*I want to create fun3 out of them so that -*
fun3 = @(x) sin(x)*cos(x)
The reason I'm asking this is that if I define fun3 as-
fun3 = @(x) fun1(x)*fun2(x)
and obviously it doesn't turn into-
fun3=@(x) sin(x)*cos(x)
thus I can't integrate fun3-
integral(fun3,0,5) ~= int(sin(x)*cos(x),x,0,5)
Matlab just says-
Error using *
Inner matrix dimensions must agree.
Thank you

 Akzeptierte Antwort

Star Strider
Star Strider am 19 Apr. 2017

5 Stimmen

You need to use element-wise operations here, using the dot (.) operator for element-wise multiplication, (.*).
This works:
fun1 = @(x) sin(x);
fun2 = @(x) cos(x);
fun3 = @(x) fun1(x).*fun2(x);
int_fun3 = integral(fun3, 0, 5)
int_fun3 =
459.7679e-003
See the documentation on Array vs. Matrix Operations for a full discussion.

6 Kommentare

yuval
yuval am 19 Apr. 2017
Thanks, didn't realize that was the issue.
Star Strider
Star Strider am 19 Apr. 2017
My pleasure.
Many people don’t, and I suspect everyone has encountered the problem themselves at least once. (I have.)
JacobsonRadical
JacobsonRadical am 24 Apr. 2022
Great Answer. I've also been confused by this for a while. You saved me!
Star Strider
Star Strider am 24 Apr. 2022
Thank you!
Grant Peel
Grant Peel am 14 Apr. 2024
Does this method work with 2 dimensional functions:
fun1 = @(x,y) sin(x)*sin(y);
fun2 = @(x,y) xy;
fun3 = @(x,y) fun1(x,y).*fun2(x,y);
integral2(fun3, 0,1,0,1)
The easiest way to find out is to first fully vectorise the constituent equations, and then try it —
fun1 = @(x,y) sin(x).*sin(y);
fun2 = @(x,y) x.*y;
fun3 = @(x,y) fun1(x,y).*fun2(x,y);
int_fun3 = integral2(fun3, 0,1,0,1)
int_fun3 = 0.0907
It seems that it does!
.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by