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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?

Star Strider
on 24 Dec 2014

Edited: Star Strider
on 24 Dec 2014

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.

Star Strider
on 24 Dec 2014

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')

Star Strider
on 24 Dec 2014

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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