I have a function that is "vectorized". I did that because I wanted to have one such function for each value of the parameter "a":
clc;
clear all;
tic;
a=[1/2,1/3,1/4];
p = @(x) 1/3.*(exp(x.^(a))-1)./(exp(x.^(a))+1) ;
I want to have the three plots (one for each value of "a") in the same figure. However, I am having trouble isolating each function for plotting. For instance, I expected that:
p1 = @(x) p(x,1,1)
would represent one of functions, but it doesn't. My other attempts have also failed. How can I get the desired result?

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Star Strider
Star Strider am 24 Dez. 2014
Bearbeitet: Star Strider am 24 Dez. 2014

1 Stimme

The easiest way is to pass ‘a’ as a parameter as well:
a=[1/2,1/3,1/4];
p = @(x,a) 1/3.*(exp(x.^(a))-1)./(exp(x.^(a))+1) ;
p1 = p(x,a(1));
Not that within the function, ‘a’ is whatever you want it to be. If you define it as a parameter, it will not pick up the value of ‘a’ from the workspace.

4 Kommentare

John
John am 24 Dez. 2014
Bearbeitet: John am 24 Dez. 2014
I suppose you meant:
p1 = @(x) p(x,a(1));
It does work. I could even get the plot without the auxiliary function "p1", like:
ezplot(@(x) p(x,a(1)),[0,1])
ETA: It seems that I would need the auxiliary "p1" if I use "plot" instead of "ezplot".
Actually, I meant it as I wrote it. It works with a vector ‘x’ and whatever value you want to give it for ‘a’. (I test all my code before I post it to be sure it works. If I can’t test it, I note that it is ‘untested code’.) Your version of it will work as well, with:
p2 = p1(x);
If you used plot instead of ezplot, you would plot it as:
x = linspace(0,1);
figure(1)
plot(x, p1(x))
So long as you define ‘a’ in ‘p1’ as you have, that will work.
If you want to be elaborate:
figure(1)
for k1 = 1:length(a)
plot(x, p(x,a(k1)))
hold on
ar{k1} = sprintf('a = %s',rats(a(k1),3));
end
grid
legend(ar, 'Location','SE')
John
John am 24 Dez. 2014
Bearbeitet: John am 24 Dez. 2014
You've already helped a lot. But if I don't put "@(x)"I get an error message:
clc;
clear all;
tic;
a=[1/2,1/3,1/4];
p = @(x,a) 1/3.*(exp(x.^(a))-1)./(exp(x.^(a))+1) ;
p1 = p(x,a(1));
%p2 = p1(x);
Gives me:
Undefined function or variable 'x'.
Error in Untitled2 (line 9) p1 = p(x,a(1));
Thank you. I do my best!
Now I understand what you’re doing. If you only want it as a function of ‘x’, you have to specify it as you did:
p1 = @(x) p(x,a(1));
It all depends on how you want to call your function. There are at least a few ways to do it correctly, so choose the one that works best for you in your application. I chose the way I usually do it in my two versions of the plot call.

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