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

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);

[Torsten]

Why do you insert C = 480 in the formula ?

And did you think about all the errors that arise because the function evaluation is imprecise and because you lose precision when summing the 960 values in the variable "int" ?

Maybe using advanced precision with vpa can help.

[재훈]

Oh I didn't know that.

[재훈]

I chose 480, which is the middle value between 0 and 960.

I didn't know to think about all the errors that occur due to loss of precision when summing 960 values ​​in variable "int".

I know this is really hard, but could you tell me about using advanced precision with vpa?

[Torsten]

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]

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/2121361-the-estimation-error-is-strangely-obtained-from-simpson-s-1-3-rule?s_tid=prof_contriblnk have useful comments or answers.

Answer 1

[Torsten]

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

[재훈]

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]

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.

Answer 2

[recent works]

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);

[재훈]

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!