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

조회 수: 176(최근 30일)
Drew Levis 16 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

로그인 to comment.

### 채택된 답변

John D'Errico 17 Sep 2019
편집: John D'Errico 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

#### 댓글 수: 1

Drew Levis 17 Sep 2019
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.

로그인 to comment.

### 추가 답변(2개)

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

#### 댓글 수: 1

Drew Levis 17 Sep 2019
thank you

로그인 to comment.

Navadeep Ganesh U 30 Nov 2019
f=@(x)sin(x.^2);
x=linspace(-pi,pi,100);
plot(x,f(x))
grid

#### 댓글 수: 0

로그인 to comment.

이 질문에 답변하려면 로그인을(를) 수행하십시오.

Translated by