[Math] Richardson extrapolation Simpson’s rule

numerical methodssimpsons rule

"Assume that S(h) is equivalent to the (composite) Simpson's rule where h is the size of the step. Correct use of Richardson's extrapolation gives the formula: $R(h) = \frac {16S(h)-S(a)} {b}$. What's a and b?"

I'm completely lost here. Can't find any good resources online so don't really know where to start or how to look at the question. Some hints/tips would be greatly appreciated.

Best Answer

Given a method $A(h) = \alpha_0 + \alpha_1 h^k +\mathcal{O}(h^{k+1})$ that approximates $\alpha_0$, the idea is to combine $A$ at varying step sizes to eliminate the lowest power of $h$, say $h^k$.

Since the error of Simpson's rule is $\mathcal{O}(h^4)$, we have $S(h) = \alpha_0 + \alpha_1 h^4 + \mathcal{O}(h^5)$ and \begin{align*} 16 S(h) &= 16 \alpha_0 + 16 \alpha_1 h^4 + \mathcal{O}(h^5) \\ S(2 h) &= \alpha_0 + 16 \alpha_1 h^4 + \mathcal{O}(h^5) \end{align*} which implies $$16 S(h) - S(2h) = 15\alpha_0 + \mathcal{O}(h^5).$$

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[Math] Length of bezier curve with Simpson’s rule

This page gives the formula for arclength calculation. It covers the 3D case, but you can recover the 2D case just by ignoring $z$, of course. The formula involves an integral that can't be represented analytically. You can use any numerical integration technique you like, of course. The same page has a simple implementation of evaluation using Simpson's rule.

You should stick with the parametric form. Converting to $y = f(x)$ form is impossible, in general.

[Math] Numerical integration in Matlab (Simpson’s rule)

A couple of notes. Since the integrand is never less than $1$, you know that $x_l<170$. In fact, when $x=170$, the integrand will be about $1.087$, so the eventual answer is bounded below by $\frac{170}{1.087}=156.5$. That gives you a pretty good idea about the starting point.

Bisection is useful when you don't have the derivative available, but here the you have the derivative in the form of the integrand, so it seems to be simpler to implement Newton's method in this case unless bisection is the point of the assignment.

Also you know the integral up to a point near to the actual root at each step after the first, so you need not integrate all the way from zero to the next guess for the root, just from the previous guess. Even if you integrated from zero every time, this program would zip through pretty fast anyway.

EDIT: Newton's method

% simp.m

f = @(x) sqrt(1+(x^2/68000)^2);

err = 1;
tol = 1.0e-8;
N = 10000;
a = 0;
b = 163;
while abs(err) > tol,
    h = (b-a)/N;
    y = f(a);
    for i = 1:N/2,
        y = y+4*f((2*i-1)*h)+2*f(2*i*h);
    end
    y = y+4*f(b-h)+f(b);
    y = h/3*y-170;
    yp = f(b);
    err = y/yp;
    b = b-err;
end
b

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