Problem with a variable not changing value in MATLAB

Question

So I have one problem with my code, which is that one variable is not working as it's supposed to do. Here are the codes I'm using:

format long
f = inline('-x.^2');
for i = 0:10
    [I(i+1) h(i+1) tid(i+1)] = trapets(f,0,1,2^i);
end

for i = 0:10
    trunk(i+1) = I(i+1) - log(2);
end

hold on
grid on
plot(log(h),log(trunk),'r+')

t = -7:0;
c = polyfit(log(h),log(trunk),1);
yy = polyval(c,t);
plot(t,yy)
grid off
hold off

koefficienter = real(c)

and also this:

function [ I,h,tid ] = trapets(f,a,b,n )
h=(b-a)/n;

tic;
I=(f(a)+f(b));
for k=2:2:n-2
    I = I+2*f(a+k*h);
end

for k = 1:2:n-1
    I = I + 4*f(a+k*h);
end

I = I * h/3;
tid = toc;
end

So the problem here is that the variable I is not changing value. It gets 11 values when I run the first code (I don't run the last code I wrote, it's only used by the first one), but the values are all the same. I don't know if the problem is that the variable n, which I use in the the second code, never change value, although I'm trying to do that with the "2^i" part in "trapets(f,0,1,2^i)". If the case is that n never change value, is there a solution to do that?

And if the problem is not the variable n, why doesn't the variable I change value in the code?

@nisse: Here's a general tip: Do not name your variables as l or l1, l2, etc. It's really hard to distinguish it from I or 1 in certain fonts. — abcd
– abcd, Commented Jun 4, 2011 at 18:49
Okay, thanks. As a clerification: k = 1:2:n-1 in the second code it says n-1, as the number one and not the variable 'I'. — nisse
– nisse, Commented Jun 4, 2011 at 19:06
@trutheality: I apologize, it was my mistake... I saw the comparison and this is what I saw everywhere: imgur.com/a/iD1OR (the top being the original and the bottom being your edit) and thought you had changed it from l to I, when it actually was the difference between normal font and monospaced font. This fumble on my part only further reinforces my first comment to the OP! — abcd
– abcd, Commented Jun 4, 2011 at 20:08

trutheality · Accepted Answer · 2011-06-04 20:43:02Z

1

After running your program I found out the I always equals -1/h after the for loops, which makes I = I * h/3; always give you the same result.

answered Jun 4, 2011 at 20:43

trutheality

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1 Comment

nisse Over a year ago

Okay, I just found out that I get different values for 'I' if i change the function to something with a higher power. Thanks for all the help anyway.

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