The estimation error is strangely obtained from Simpson's 1/3 rule

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재훈 2024-5-22

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https://ww2.mathworks.cn/matlabcentral/answers/2121361-the-estimation-error-is-strangely-obtained-from-simpson-s-1-3-rule

评论： 재훈 2024-5-24

Hello, I am looking for the estimation error of Simpson's rule of thirds. When dx is 1.01, the integral value is 2908800 and the error is 0.01, but the estimation error is 5.5567e-09. Where did it go wrong?

clc; clear all; close all;
a = -1.93317553181561E-24;
b = 3.788630291530091e-21;
c = -2.3447910280083294e-18
d = -0.019531249999999518;
e = 18.74999999999999
fun = @(t) a.*t.^5 + b.*t.^4 + c.*t.^3 + d.*t.^2+e.*t
x = 0:0.1:960;
fx =fun(x);
n=960;
dx=1.01
int=0;
for i =1:n
    plot(x, fx,'k','linewidth',2);
    mid=((i-1)+i)/2;
    fx_mid = fun(mid);
    fx_left = fun(i-1);
    fx_right = fun(i); 
    area_temp = dx/6*(fx_left +4*fx_mid+fx_right);
    
    int = int + area_temp; 
    
    x_segment = linspace(i-1, i,100);
    Px = fx_left * ((x_segment-mid).*(x_segment-i))/((i-1-mid)*(i-1-i))...
        + fx_mid*((x_segment-i+1)).*(x_segment-i)/((mid-i+1)*(mid-i))...
        + fx_right * ((x_segment-i+1).*(x_segment-mid))/((i-i+1)*(i-mid));
  
   area(x_segment,Px); hold on;
end
C=480;
E_a = -((960.^5)/(2880.*1.01.^4)).*(a.*120.*C+24.*b);%Is there a problem here?
disp('E_a');
disp(E_a);
disp(int);
int_true = 2880000
rel_error=norm(int_true-int)/norm(int_true);
disp('rel_error');
disp(rel_error);

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Steven Lord 2024-5-23

The user posted about this at least three times. While I closed one of them as duplicate, both this post and https://www.mathworks.com/matlabcentral/answers/2121401-the-estimation-error-is-strangely-obtained-from-simpson-s-1-3-rule?s_tid=prof_contriblnk have useful comments or answers.

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Answer 1

Ayush Anand 2024-5-22

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2121361-the-estimation-error-is-strangely-obtained-from-simpson-s-1-3-rule#answer_1461491

编辑：Ayush Anand 2024-5-22

There is a mistake with your formulae for estimation of

. The formula for the error estimate in Simpson's rule is generally given by:

where ξ is some number in the interval [a, b], n is the number of intervals, and

is the fourth derivative of the function being integrated.

Given your function:

; the fourth derivative of

would be:

So, given the interval

and

with

, the formulae for

looks like:

Replacing formulae for

in your code with this should give you the correct results.

You can check this out for reading more on Simpson's rule: https://www.mathworks.com/matlabcentral/fileexchange/72526-simpson-s-1-3-rule-composite

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Ayush Anand 2024-5-23

If the fourth derivative coefficient is very small, the error estimate will indeed be very small. The estimation error provided by Simpson's rule is not always an exact reflection of the true error; it's an estimate that depends on several factors, including the behavior of the function being integrated and the specifics of the numerical method being used. It assumes that the derivative in consideration is relatively constant or at least does not exhibit extreme variations within the interval. For well-behaved functions, where these assumptions hold true, the error estimate can be quite accurate or at least provide a good approximation of the actual error. For functions with significant variations in higher-order derivatives or with singularities, the estimate might not be as reliable.

재훈 2024-5-24