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

12 vues (au cours des 30 derniers jours)
재훈 le 22 Mai 2024
Commenté : Steven Lord le 23 Mai 2024
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 commentairesAfficher 3 commentaires plus anciensMasquer 3 commentaires plus anciens
Torsten le 22 Mai 2024
Modifié(e) : Torsten le 22 Mai 2024
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).
Steven Lord le 23 Mai 2024

Connectez-vous pour commenter.

### Réponse acceptée

Torsten le 22 Mai 2024
Modifié(e) : Torsten le 23 Mai 2024
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 commentairesAfficher AucuneMasquer Aucune
재훈 le 23 Mai 2024
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 le 23 Mai 2024
Modifié(e) : Torsten le 23 Mai 2024
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.

Connectez-vous pour commenter.

### Plus de réponses (1)

recent works le 22 Mai 2024
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 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
재훈 le 23 Mai 2024
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!

Connectez-vous pour commenter.

### Catégories

En savoir plus sur Programming dans Help Center et File Exchange

R2024a

### Community Treasure Hunt

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

Start Hunting!

Translated by