Using 1/3 simpson's rule for evaluating an erf function up to a specific error
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So a homework problem I am doing asks for me to use the 1/3 Simpson's rule to evaluate the erf function given an X with an absolute true error of 5E-8 with the minimum number of segments (n). However when I make a function that evaluates the erf function for each n it never makes it to the absolute true error of 5E-8 and instead just runs infinitely and at most giving me an error of around 5E-6. So how do I get to such a high accuracy with the 1/3 Simpson's rule?
My code so far.
clear,clc
n = 0;
x = 1;
t = @(x) exp(-x.^2);
format long
error = 1;
while error > 5E-8
n = n+1;
w = (1:1:n);
h = x/n ;
r = linspace(0,x,n);
odd = w(mod(w,2)~=0);
even = w(mod(w,2)==0);
E = r(even);
O = r(odd);
if n == 2
I = h/3*(t(r(1))+t(r(2))+t(r(n)));
else
I = h/3*(t(r(1))+2*sum(t(O))+4*sum(t(E))+t(r(n)));
end
Int = 2/sqrt(pi)*I
error = abs(erf(x) - Int);
end
result = [Int,error,length(w)]
回答（1 个）
David Hill
2020-6-9
a=0;
b=1;
n=2.1603e7;%limit the loop based on testing
iT=sqrt(pi)/2*erf(b);
error=1;
tol=5e-8;
t = @(x) exp(-x.^2);
while error>tol
n=n+2;
x=linspace(0,1,n);
it = (b-a)/(3*n)*sum(t(x).*([1,repmat([2 4],1,(n-2)/2),1]));
error=abs(iT-it);
end
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2020-6-9
2020-6-9
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