# Help with For-Loop question?

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I keep getting the same error, about not being any real or logical answer.

here is the question; Create a time vector x = 0:2*pi. Using a for-loop, plot ten different curves on the same graph where the first curve is x1 = sin(x), the second curve is x2 = sin(x1), the third curve is x3 = sin(x2) and so on. Note that by using a for-loop it is not necessary to create ten separate variables to solve this problem.

Here is what I've been entering

x=0

>> for i=[pi/5:pi/5:2*pi]

x(i)=sin(x(i-(pi/5)))

end

??? Attempted to access x(0); index must be a positive integer or logical.

Any help would be appreciated

##### 1 Comment

Walter Roberson
on 5 Oct 2011

Duplicate is at http://www.mathworks.com/matlabcentral/answers/17480-help-with-for-loop-question

### Accepted Answer

Robert
on 5 Oct 2011

The answer above doesn't achieve what you were asking for. Try this:

N=10;

x=zeros(N,numel(pi/5:pi/5:2*pi));

x(1,:)=pi/5:pi/5:2*pi;

for j=2:N

x(j,:)=sin(x(j-1,:));

end

plot(x(1,:),x')

##### 4 Comments

Fangjun Jiang
on 5 Oct 2011

x(1,:) is all the data in the 1st row of the matrix x.

x(j,:) is all the data in the jth row of the matrix x.

This link might be helpful for you. It's also available in your local copy of the MATLAB documentation. Type doc to start browsing.

http://www.mathworks.com/help/techdoc/math/f1-85462.html

Walter Roberson
on 5 Oct 2011

### More Answers (1)

Fangjun Jiang
on 5 Oct 2011

When you use i=[pi/5:pi/5:2*pi], i is a vector with increment as pi/5, hardly any of the element is integer, you can not use i as the index of a vector as x(i). Try something like,

N=10;

x=zeros(N,1);

for k=1:N

a=k*pi/5;

x(k)=sin(a);

end

i is the square root of -1 so avoid using i. Try this:

clear i

i

##### 4 Comments

Sean de Wolski
on 5 Oct 2011

for your own knowledge, you should just type it at the command line and see what it creates

Fangjun Jiang
on 5 Oct 2011

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