# composite rule; trapezium vs. Simpson

Suppose the interval $[a,b]$ has been divided into $2m$ intervals by the points $x_i=a+ih$, and $$h=\frac{b-a}{2m}.$$ The Composite Simpson rule says that $$\int_a^bf(x)dx\approx\frac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+\dots+2f(x_{2m-2})+4f(x_{2m-1}+f(x_{2m})]$$ So basically the quadrature weights are $1,4,2,4,2,\dots,2,4,1$, when multiplied by $h/3$. It can be shown that this approximation is better than the Composite Trapezium rule, which says that $$\int_a^bf(x)dx\approx h\left[\frac{1}{2}f(x_0)+f(x_1)+\dots+f(x_{m-1})+\frac{1}{2}f(x_m)\right],$$ where we've dividing our interval $[a,b]$ into $m$ intervals. Taking into account that we use the double amount of intervals for a given $m$ in the composite Simpson rule, the error for this rue is still much smaller than the error for the composite trapezium rule. I've seen the proof, where we use the Lagrange interpolation polynomials and what now, but I fail to see on the intuitive level why $1,4,2,4\dots,2,4,1$ would be a better distribution of the quadrature weights compared to $1/2,1,1,\dots,1,1/2$. Is there an intuitive explanation for the difference in the error?

## 1 Answer

The Simpson rule for $2n$ sub-intervals is the Richardson extrapolation of the trapezoidal rule for $n$ and $2n$ sub-intervals. If the error of the trapezium rule to the true integral value $I$ is $$T(n)=I+c_2n^{-2}+c_4n^{-4}+...$$ (note that it is an even function as the method is symmetric) then the extrapolation is \begin{align} S(n)=\frac{4T(2n)-T(n)}3&=I+c_2\frac{4(2n)^{-2}-n^{-2}}3+c_4\frac{4(2n)^{-4}-n^{-4}}3+...\\ &=I-\frac{c_4}4 n^{-4}+... \end{align} In other point of view, the trapezism rule approximates the function by a piecewise linear function while Simpson uses a piecewise quadratic approximation over the same sub-division. Even in the $2n$ subdivision, the quadratic approximation will in general be closer to the function than the line segments between the sample points.