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How can I plot a sin (x^2) function

Asked by Drew Levis on 16 Sep 2019
Latest activity Commented on by Drew Levis on 17 Sep 2019
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

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2 Answers

Answer by John D'Errico
on 17 Sep 2019
Edited by John D'Errico
on 17 Sep 2019
 Accepted Answer

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

  1 Comment

I appericate the help. I started to think and come to a simlar answer when looking at the figure again. I am glad my thoughts were confirmed. THANK YOU again.

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Answer by madhan ravi
on 17 Sep 2019

fplot(@(x)sin(x.^2),[-pi,pi])

  1 Comment

thank you

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