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here is my code and it works but the figure is obviously not a sin(x^2) function from -pi to pi

syms x;

f=@(x)sin(x.^2);

x=[-pi pi];plot(x,f(x))

grid

John D'Errico
on 17 Sep 2019

Edited: John D'Errico
on 17 Sep 2019

First, there is ABSOLUTELY NO reason to predefine x as a sym. So this line is completely irrelevant:

syms x;

When you do create x, you overwrite the symbolic version of x that you created before anyway.

Next, you created a function

f=@(x)sin(x.^2);

Good there. But then what?

What do you think this does?

x=[-pi pi];

It does NOT create the interval [-pi,pi]. Instead, it creates a vector of length 2, so TWO values, -pi and pi. Then when you plotted, using

plot(x,f(x))

it plots TWO points, connected with a straight line. And since the two poiints each had function value of zero, the line is perfectly horizontal. Compare that to this version:

f=@(x)sin(x.^2);

x=linspace(-pi,pi,100);

plot(x,f(x))

grid

See that I never needed to pre-define x as symbolic. (Why would you want to do that anyway? Just because you don't know the value of something, does not mean it must automatically be symbolic. This is perhaps the one of most common mistakes I see made by new users.)

Simpler yet, you might have done just this:

f = @(x) sin(x.^2);

fplot(f,[-pi,pi])

grid

Navadeep Ganesh U
on 30 Nov 2019

f=@(x)sin(x.^2);

x=linspace(-pi,pi,100);

plot(x,f(x))

grid

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