if number of steps is not multiple of 3 how to do simpsons 3/8 rule? why im getting more error in simpsons 3/8 rule than in simpsons 1/3 rule?

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Shamila 2024-4-4

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2102831-if-number-of-steps-is-not-multiple-of-3-how-to-do-simpsons-3-8-rule-why-im-getting-more-error-in-si

评论： Shamila 2024-4-8

Here, as dx is 0.1 and 0.01, n is not multiple of 3 for 3/8 rule.tried by making n same for all 3 rules. but error in 3/8 > error in 1/3. how to solve this issue?

% Define the function
f = @(x) 2-x+log(x);
% Define integration limits
a = 1;
b = 2;
% Exact integral value
og = integral(f,a,b);
%Step sizes
h1 =(b-a)/12;
h2 =(b-a)/102;
% Trapezoidal Rule
trph1 = h1/2 * (f(a) + f(b) + 2*sum(f(a+h1:h1:b-h1)));
trph2 = h2/2 * (f(a) + f(b) + 2*sum(f(a+h2:h2:b-h2)));
trperh1 = abs((trph1-og) / og) * 100;
trperh2 = abs((trph2-og) / og) * 100;
% Simpson's 1/3 Rule
simp1h1 = h1/3 * (f(a) + f(b) + 2*sum(f(a+2*h1:2*h1:b-2*h1)) + 4*sum(f(a+h1:2*h1:b-h1)));
simp1h2 = h2/3 * (f(a) + f(b) + 2*sum(f(a+2*h2:2*h2:b-2*h2)) + 4*sum(f(a+h2:2*h2:b-h2)));
simp1erh1 = abs((simp1h1 - og) / og) * 100;
simp1erh2 = abs((simp1h2 - og) / og) * 100;
% Simpson's 3/8 Rule
simp2h1 = (3*h1/8) * (f(a) +f(b) + 3*sum(f(a+h1:3*h1:b-2*h1)) + 3*sum(f(a+2*h1:3*h1:b-h1)) + 2*sum(f(a+3*h1:3*h1:b-3*h1)));
simp2h2 = (3*h2/8) * (f(a) +f(b) + 3*sum(f(a+h2:3*h2:b-2*h2)) + 3*sum(f(a+2*h2:3*h2:b-h2)) + 2*sum(f(a+3*h2:3*h2:b-3*h2)));
simp2erh1 = abs((simp2h1 - og) / og) * 100;
simp2erh2 = abs((simp2h2 - og) / og) * 100;
% Display results
disp('for h = 0.1 :-');
for h = 0.1 :-
disp(['Trapezoidal Rule: ',num2str(trph1), ', Error: ', num2str(trperh1)]);
Trapezoidal Rule: 0.88601, Error: 0.032634
disp(['Simpson''s 1/3 Rule: ', num2str(simp1h1), ', Error: ', num2str(simp1erh1)]);
Simpson's 1/3 Rule: 0.88629, Error: 5.2332e-05
disp(['Simpson''s 3/8 Rule: ', num2str(simp2h1), ',  Error: ', num2str(simp2erh1)]);
Simpson's 3/8 Rule: 0.88629,  Error: 0.00011653
% disp('for h = 0.01 :-');
disp(['Trapezoidal Rule: ', num2str(trph2), ',  Error: ', num2str(trperh2)]);
Trapezoidal Rule: 0.88629,  Error: 0.00045186
disp(['Simpson''s 1/3 Rule: ', num2str(simp1h2), ', Error: ', num2str(simp1erh2)]);
Simpson's 1/3 Rule: 0.88629, Error: 1.0133e-08
disp(['Simpson''s 3/8 Rule: ', num2str(simp2h2), ',  Error: ', num2str(simp2erh2)]);
Simpson's 3/8 Rule: 0.88629,  Error: 2.2795e-08

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Answer 1

Torsten 2024-4-4

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移动：Torsten 2024-4-4

if number of steps is not multiple of 3 how to do simpsons 3/8 rule?

The 1/3 rule can be used for the remaining subintervals without changing the order of the error term (conversely, the 3/8 rule can be used with a composite 1/3 rule for odd-numbered subintervals).

why im getting more error in simpsons 3/8 rule than in simpsons 1/3 rule?

Why do you think this should not be the case ? Both composite rules are of the same order as far as I know.