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

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재훈
재훈 2024-5-22
评论： Steven Lord 2024-5-23
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 7.0844e-21. Where did it go wrong? I think In this part, the 5th coefficient is verry verry small, so it seems that the value will always be low. Isn't the estimation error always accurate?
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.*(960/1.01).^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);
  5 个评论
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Torsten
Torsten 2024-5-22
编辑：Torsten 2024-5-22
I didn't know to think about all the errors that occur due to loss of precision when summing 960 values in variable "int"
Plus the error in evaluating a polynomial with such small coefficients. You will have to go the symbolic way (see below).

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采纳的回答

Torsten
Torsten 2024-5-22
编辑：Torsten 2024-5-23
Ran in:
syms a b c d e real
syms t real
f(t) = a*t^5 + b*t^4 + c*t^3 + d*t^2 + e*t;
s = 0;
i = -1/2;
while i < 959.5
i = i + 1;
tleft = i-1/2;
tmiddle = i;
tright = i+1/2;
fleft = f(tleft);
fmiddle = f(tmiddle);
fright = f(tright);
s = s + sym(1)/sym(6)*(fleft+sym(4)*fmiddle+fright);
end
s = simplify(s)
s = 
s_exact = int(f,t,0,960)
s_exact = 
error = s-s_exact
error = 
double(subs(error,[a,b],[-1.93317553181561E-24,3.788630291530091e-21]))
ans = -6.8079e-21
  2 个评论
재훈
재훈 2024-5-23
Thank you for helping me. However, since it does not reach the error value of 0.01 that I obtained, it seems that my calculation method is wrong or that I need to study. It must have been very difficult, but thank you for your help. have a good day!
Torsten
Torsten 2024-5-23
编辑：Torsten 2024-5-23
Your calculation method is correct (of course with dx = 1 instead of dx=1.01 - I don't know how you come up with 1.01 ?), but you have an accumulation of errors when evaluating the function and summing the contributions to the integral. For such small values for the coefficients of a polynomial, you have to use symbolic math - and you see that the "correct" error between analytical integral and Simpson's rule with a stepsize of 1/2 between 0 and 960 is only 6.8079e-21 instead of 0.01.

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更多回答（1 个）

recent works
recent works 2024-5-22
The error calculation in your code should be
C = 480;
f4_max = -1.324287168211725E-19; % This is an approximation
h = 1.01;
E_a = -(960 / 180) * (h^4) * f4_max;
disp('E_a');
disp(E_a);
  1 个评论
재훈
재훈 2024-5-23
Thank you for helping me. However, since it does not reach the error value of 0.01 that I obtained, it seems that my calculation method is wrong or that I need to study. It must have been very difficult, but thank you for your help. have a good day!

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