Can someone tell me why my error is so large for my composite simpsons rule?

2 次查看（过去 30 天）
Briyahna Deleon
Briyahna Deleon 2018-11-10
评论： Briyahna Deleon 2018-11-10
My assignment was to code a composite simpsons rule where the exact value of the integral of xsin(x) from 0 to 1 is 0.301168678 but I keep getting 0.862149054988026 as the approximation and that is not within my error bound. Am I entering the composite simpsons rule wrong???
if true
% EV = sin(1)-cos(1); %exact value
a=0; %starting point
b=1; %endpoint
f=@(x) x.*sin(x);
%d=f(a);
for j=0:6
n=10^(j);
h=(b-a)/n;
Err_CS= 1/(36*(n)^4); %error bound
CS=(1/3)*(f(a) + 2.*sum(f((a + 2*h):2*h:(b - 2*h)))+4.*sum(f((a + h):2*h:(b -h)) +f(b)))*h; %composite simpsons rule
fprintf('%8.2e %14.8e %14.8e %14.8e\n', n, CS, abs(CS-EV), Err_CS)
end
end

请先登录，再回答此问题。

采纳的回答

David Goodmanson
David Goodmanson 2018-11-10
编辑：David Goodmanson 2018-11-10
Hi Briyahna,
Yes the Simpson's rule expression is wrong, but only in the typo sense of having a misplaced parenthesis. Instead of
CS=(1/3)*(f(a) + 2.*sum(f((a + 2*h):2*h:(b - 2*h)))+4.*sum(f((a + h):2*h:(b -h)) +f(b)))*h;
it should be
CS=(1/3)*(f(a) + 2.*sum(f((a + 2*h):2*h:(b - 2*h)))+4.*sum(f((a + h):2*h:(b -h))) +f(b ))*h;
Even better would be to have (h/3) in front as per the usual Simpson expression rather than the h factor hiding all the way at the end.

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Numerical Integration and Differential Equations 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by