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Interpreting Matlab function simp |
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| Feb27-13, 06:15 AM | #1 |
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Interpreting Matlab function simp
Hello, guys
Hey guys, Got stuck with function integration using Simpson's rule and need your help. Please first refer to picture attached for full idea of my question: The Matlab command related to it is: for i=1:nr u1d(i)=4.0*pi*r(i)^2*u(it,i) end I1=simp(0.0,r0,nr,u1d)/(4.0/3.0*pi*r0^3) I1 is nr=21 r0=1.0; Does it mean that I1 is integrated 21 times between boundaries 0 and r0? |
| Feb27-13, 08:12 AM | #2 |
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Can you post more information? The code you posted references variables that you never define. Make it so that your code block can be copy/pasted into matlab.
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| Feb27-13, 08:27 AM | #3 |
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Yes. Sure
I have attached all commands with order from Pic.1 to Pic.3 |
| Feb27-13, 08:46 AM | #4 |
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Interpreting Matlab function simp
simp() is not a matlab function, so the information about the input arguments is not available in the documentation. I suggest looking at the function file for simp() to find this info.
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| Feb27-13, 09:04 AM | #5 |
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EDIT: I found information about this function in the MATLAB file exchange.
Code:
function s = simp(f, a, b, h)
x1 = a + 2 * h : 2 * h : b - 2 * h;
sum1 = sum(feval(f, x1));
x2 = a + h : 2 * h : b - h;
sum2 = sum(feval(f, x2));
s = h / 3 * (feval(f, a) + feval(f, b) + ...
2 * sum1 + 4 * sum2);
f=function, a=initial value, b=end value, h=interval size |
| Feb27-13, 04:27 PM | #6 |
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I have found it as separate m.file. Here are the commands:
function uint=simp(xl,xu,n,u) h=(xu-xl)/(n-1); uint(1)=u(1)-u(n); for i=3:2:n uint(1)=uint(1)+4.0*u(i-1)+2.0*u(i); end uint=h/3.0*uint; But why here different variables are used such as xl and xu? It seems to me that we use r in integration? |
| Feb27-13, 09:55 PM | #7 |
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At a glance it looks like
xl = beginning of interval xu = end of interval n = number of slices u = function So in your case of Code:
I1=simp(0.0,r0,nr,u1d) |
| Feb28-13, 03:50 AM | #8 |
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| integration, matlab, simpson's rule |
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