Register to reply

Interpreting Matlab function simp

by Ein Krieger
Tags: integration, matlab, simpson's rule
Share this thread:
Ein Krieger
#1
Feb27-13, 06:15 AM
P: 34
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?
Attached Thumbnails
2.jpg  
Phys.Org News Partner Science news on Phys.org
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
kreil
#2
Feb27-13, 08:12 AM
kreil's Avatar
P: 545
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.
Ein Krieger
#3
Feb27-13, 08:27 AM
P: 34
Yes. Sure

I have attached all commands with order from Pic.1 to Pic.3
Attached Thumbnails
Pic 1.jpg   Pic 2.jpg.png   pic 3.jpg  

kreil
#4
Feb27-13, 08:46 AM
kreil's Avatar
P: 545
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.
kreil
#5
Feb27-13, 09:04 AM
kreil's Avatar
P: 545
EDIT: I found information about this function in the MATLAB file exchange.

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);
It appears that the inputs are:

f=function, a=initial value, b=end value, h=interval size
Ein Krieger
#6
Feb27-13, 04:27 PM
P: 34
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?
kreil
#7
Feb27-13, 09:55 PM
kreil's Avatar
P: 545
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
I1=simp(0.0,r0,nr,u1d)
You are integrating u1d from 0 to r0 with nr intervals
Ein Krieger
#8
Feb28-13, 03:50 AM
P: 34
Quote Quote by kreil View Post
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
I1=simp(0.0,r0,nr,u1d)
You are integrating u1d from 0 to r0 with nr intervals
So you mean simp can be uniformly used for every type of variable assuming their correct order?


Register to reply

Related Discussions
Ramp Function in MATLAB? also window adjustments (MATLAB Help) Introductory Physics Homework 2
Simple structure function - interpreting answer [Probability & Reliability Theory] Calculus & Beyond Homework 0
Interpreting a function based on it's equation. General Math 28
Simp,e Harmonic Motion with Damping Classical Physics 4
Simp Simultaneous Equation Precalculus Mathematics Homework 7