Help with loops calculating value

조회 수: 1 (최근 30일)
amateurintraining
amateurintraining 2017년 11월 20일
답변: Walter Roberson 2017년 11월 20일
I have a function:
function [ I, h ] = Simpson( f, a, b, tol )
n=2;
h=(b-a)/n;
x=h/3;
I=x*(f(a)+4*f((a+b)/2)+f(b));
Iold=I;
n=2*n;
if abs((I-Iold)/Iold)>=tol
n=2*n;
h=(b-a)/n;
for j=1:n/2-1
term1=2*f(2*xj);
xj=a+j*h;
end
for i=1:n/2
term2=4*f(2*xj-1);
xj=a+j*h;
end
x=h/3;
I=x*(f(a)+term1+term2+f(b));
n=n*2;
end
However, I want the function to loop to keep calculating I until
abs((I-Iold)/Iold) is less than the input tol.
How do I do this? Presently, it only calculates when n=2.
  댓글 수: 1
Walter Roberson
Walter Roberson 2017년 11월 20일
Why do you have
n=2*n;
twice in your loop? And just before you loop you also have it. Considering the lack of comments, it looks like you are doing too many of those changes to n.

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답변 (2개)

KL
KL 2017년 11월 20일
편집: KL 2017년 11월 20일
I don't clearly understand what you're trying to do but it looks like you may want to use a while loop,
while abs((I-Iold)/Iold)>=tol
%here goes the part to be repeated until the above condition fails
end
additionally, I have noticed few other things, when you say
for j=1:n/2-1
term1=2*f(2*xj);
xj=a+j*h;
end
for i=1:n/2
term2=4*f(2*xj-1);
xj=a+j*h;
end
  1. You haven't defined xj previously so it's gonna throw you an error.
  2. you're simply repeating the task until you calculate a final value, then why not directly calculate the final value?
j=floor(n/2);
xj = a+j*h;
term1 = 2*f(2*xj);
and similarly the other loop content.

Walter Roberson
Walter Roberson 2017년 11월 20일
You have
for j=1:n/2-1
term1=2*f(2*xj);
xj=a+j*h;
end
every iteration of j, that is going to overwrite all of term1 and xj. There is no point in calculating values that are just going to be overwritten.
Perhaps you should be recording all of the terms and totaling them ?

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