Suppose that the interval of integration $[a,b]$ is divided into equal subintervals of length $h$ each such that $r = \frac{b-a}{h}$ is even. Denote by $R_1$ the result of applying the composite trapezoidal method with step size $2h$ and by $R_2$ the result of applying the same method with step size h. Show that one application of Richardson extrapolation, reading
$$S = \frac{4R_2-R_1}{3}$$
yields the composite Simpson method
So if we replace our step size with $2h$, that just means we will have half as many subintervals. So $R_1$ will look like
$$\sum_{i=0}^{\frac{r}{2}-1} \int_{t_i}^{t_{i+1}}f(x)dx \approx R_1 = \sum_{i=0}^{\frac{r}{2}-1} h[f(t_i)+f(t_{i+1})]$$
With regular step size $h$, $R_2$ will look like
$$\sum_{i=0}^{r-1} \int_{t_i}^{t_{i+1}}f(x)dx \approx R_2 = \sum_{i=0}^{r-1} \frac{h}{2}[f(t_i)+f(t_{i+1})]$$
Now plugging into $S$, we get
$$S = \frac{4R_2-R_1}{3} = \frac{h}{3}[f(t_0)+2f(t_1)+…+3f(t_{\frac{r}{2}})+…+4f(t_{r-1})+2f(t_r)]$$
I am confused how this is equivalent to composite simpson's method. Any help is appreciated. Thank you!
Best Answer
To compute the difference of the trapezium sums, you need to use the same sampling sequence in both expressions, that is the denser sequence of the second expression where the first expression uses the even-index sub-sequence. Then $$ R_1=\sum_{i=0}^{r-1}h[f(x_{2i})+f(x_{2i+2})] =h\left[f(0)+2\sum_{i=1}^{r-1}f(x_{2i})+f(x_{2r})\right] $$ and $$ R_2=\sum_{i=0}^{2r-1}\frac h2[f(x_{i})+f(x_{i+1})] =\frac h2\left[f(0)+2\sum_{i=0}^{r-1}f(x_{2i+1})+2\sum_{i=1}^{r-1}f(x_{2i})+f(x_{2r})\right]. $$ Then the extrapolation expression is \begin{align} \frac{4R_2-R_1}3&=\frac{h}3\left(\begin{aligned} &\left[2f(0)+4\sum_{i=0}^{r-1}f(x_{2i+1})+4\sum_{i=1}^{r-1}f(x_{2i})+2f(x_{2r})\right]… \\ …&- \left[f(0)+2\sum_{i=1}^{r-1}f(x_{2i})+f(x_{2r})\right] \end{aligned}\right) \\ &=\frac{h}3\left( f(0)+4\sum_{i=0}^{r-1}f(x_{2i+1})+2\sum_{i=1}^{r-1}f(x_{2i})+f(x_{2r}) \right) \\ &=\sum_{i=0}^{r-1}\frac{h}3[f(x_{2i}+4f(x_{2i+1})+f(x_{2i+2})] \end{align} which is the formula for the composite Simpson method.