{title2=$S_n$}

<Composite Simpson's rule> applies <Simpson's rule> to consecutive pairs of equal subintervals. Choose a positive even <integer> $n\geq2$, put $h=(b-a)/n$, and take <quadrature nodes> $x_j=a+jh$ for $j=0,\ldots,n$. There are $n$ subintervals, $n/2$ two-subinterval panels, and $n+1$ <quadrature nodes>. Then
$$
S_n=\frac h3\left[f(x_0)+4\sum_{j=1}^{n/2}f(x_{2j-1})+2\sum_{j=1}^{n/2-1}f(x_{2j})+f(x_n)\right].
$$
The second <sum> is empty when $n=2$, recovering one panel. The <quadrature weights>, before multiplication by $h/3$, are $1,4,2,4,\ldots,2,4,1$. Every interior even-indexed <quadrature node> is an endpoint of two panels, so its two endpoint contributions add to two; every odd-indexed <quadrature node> is a <midpoint> with weight four. An odd $n$ cannot be covered by these pairs. The displayed formula requires equal spacing.<Fundamentals of Numerical Computation by Tobin A. Driscoll and others/5.6 Numerical integration>[Fundamentals of Numerical Computation by Tobin A. Driscoll and others; §5.6, equations 5.6.20–5.6.21: elementary and composite Simpson formulas]{ref}

For the <exponential function> on the <interval> $[0,1]$, the exact <integral> is $I=e-1$. With $n=2$, the formula gives $S_2\approx1.718861151876593$, an overestimate of about $5.793234175\times10^{-4}$. With $n=128$, it gives $S_{128}\approx1.718281828494607$, an overestimate of about $3.556140064\times10^{-11}$. The figure shows the intermediate refinements and compares every computed <absolute error> with the <error bound for composite Simpson's rule>.

\Image[/composite-simpson-exponential-convergence.png]
{title=Computed composite Simpson errors for the exponential function on zero to one approach fourth-order convergence; the dashed curve is the rigorous bound e over 180 n to the fourth power}

For this <exponential function>, a <geometric series> also gives a compact independent formula. With $q=e^h$ and $h=1/n$, summing the $n/2$ panel contributions yields
$$
S_n=(e-1)\frac{h(1+4q+q^2)}{3(q^2-1)}.
$$
A <Taylor series> at $h=0$ gives
$$
S_n-I=(e-1)\left(\frac{h^4}{180}-\frac{h^6}{1512}+O(h^8)\right).
$$
The <Big O notation> describes the remaining terms as $h$ tends to zero. Consequently $(S_n-I)/(S_{2n}-I)$ tends to sixteen: the computed ratios increase from about $15.65169$ to $15.99965$. Fourth-order <truncation error> is also obtained by <Richardson extrapolation> of the <composite trapezoidal rule>.<Fundamentals of Numerical Computation by Tobin A. Driscoll and others/5.6 Numerical integration>[Fundamentals of Numerical Computation by Tobin A. Driscoll and others; §5.6.2, equations 5.6.14–5.6.15: Simpson extrapolation and fourth-order error]{ref}


 Composite Simpson's rule (source code)