Section 14.4 Examples and questions
Example 14.4.1. Comparing the accuracy of Simpson's, trapezoidal, and midpoint rules.
For the function \(f(x) = e^x \) on the interval \([0, 1]\text{,}\) use Matlab to compute the left-, right, and midpoint approximations with \(n=10\) and find the error of each approximation (that is, its difference with the actual integral).
Then repeat the same with \(f(x) = e^{\sqrt{x}} \text{.}\) What could explain the lackluster performance of Simpson's rule here?
f = @(x) exp(x); % or exp(sqrt(x)) a = 0; b = 1; exact = exp(1)-1; % or 2 n = 10; h = (b-a)/n; x = a:h:b; y = f(x); T = (h/2)*(y(1) + y(end) + 2*sum(y(2:end-1))); S = (h/3)*(y(1) + y(end) + 4*sum(y(2:2:end-1)) + 2*sum(y(3:2:end-2))); midpoints = (x(1:end-1) + x(2:end))/2; M = h*sum(f(midpoints)); er = abs([T M S] - exact); fprintf('Errors: Trapezoidal %g, Midpoint %g, Simpson %g\n', er);
The function \(f(x) = e^{\sqrt{x}} \) is not differentiable at \(0\text{.}\) This behavior affects higher order methods more because they are based on comparing the function to a smooth one (such as a parabola). The midpoint method has an advantage in that it does not use the problematic point \(0\text{.}\)
Question 14.4.2. Richardson extrapolation of the midpoint rule.
We derived Simpson's rule from trapezoidal rule using the Richardson extrapolation. Why not do the same starting from the midpoint rule instead of trapezoidal?