Composite Simpson's rule ($S_n$)

Composite Simpson's rule applies Simpson's rule to consecutive pairs of equal subintervals. Choose a positive even integer $n\ge 2$, put $h=(b-a)/n$, and take quadrature nodes $x_j=a+jh$ for $j=0,\dots ,n$. There are $n$ subintervals, $n/2$ two-subinterval panels, and $n+1$ quadrature nodes. ThenThe second sum is empty when $n=2$, recovering one panel. The quadrature weights, before multiplication by $h/3$, are $1,4,2,4,\dots ,2,4,1$. 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.^[ref]

For the exponential function on the interval $[0,1]$, the exact integral is $I=e-1$. With $n=2$, the formula gives $S_2\approx 1.718861151876593$, an overestimate of about $5.793234175\times 10^{-4}$. With $n=128$, it gives $S_{128}\approx 1.718281828494607$, an overestimate of about $3.556140064\times 10^{-11}$. The figure shows the intermediate refinements and compares every computed absolute error with the error bound for composite Simpson's rule.

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 yieldsA Taylor series at $h=0$ givesThe 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.^[ref]

Suppose $f$ has a fourth derivative that is a continuous function on the interval $[a,b]$, and $|f^{(4)}(x)|\le M$ there. With $I={\int }_a^{b}f(x)dx$, the signed quadrature error here is $I-S_n$. Then, for some point $\xi \in (a,b)$,In particular, a strictly positive fourth derivative makes $I-S_n<0$: composite Simpson's rule overestimates the integral.^[ref]

The bound's scale follows by adding the panel errors. A panel of width $2h$ has signed quadrature error $-h^5{f}^{(4)}({\xi }_j)/90$ and absolute error at most $M{h}^5/90$. There are $(b-a)/(2h)$ panels, so the triangle inequality gives $M(b-a)h^4/180$. The intermediate value theorem for the continuous fourth derivative replaces the average of the panel values $f^{(4)}({\xi }_j)$ by a single value $f^{(4)}(\xi )$, giving the signed formula.

The underlying single-panel calculation can be seen through the Peano kernel theorem. On the interval $[0,2]$, define the raw Peano kernel $K(t)=L[(x-t{)}_{+}^3]$, where $L$ is integral minus Simpson's rule and $(x-t{)}_{+}=\max (x-t,0)$ is the positive part. Then $K(t)=-t^3(4-3t)/12$ for $0\le t\le 1$ and $K(t)=-(2-t{)}^3(3t-2)/12$ for $1\le t\le 2$. It is strictly negative inside the interval and has integral $-1/15$. ThusThe uniform norm $\Vert f^{(4)}{\Vert }_{\infty }$ bounds the fourth derivative. The polynomial $f(x)=x^4/24$ attains equality. Scaling to endpoint spacing $h$ multiplies the coefficient by $h^5$, giving the sharp single-panel bound $h^5\Vert f^{(4)}{\Vert }_{\infty }/90$. The normalized Peano kernel is $K/3!$, so the external factorial must be omitted when using that normalization.

For the exponential function on $[0,1]$, $f^{(4)}(x)=e^x$, so $M=e$ and the bound is $e/(180n^4)$. Its leading actual absolute error is $(e-1)/(180n^4)$, smaller than the bound because a uniform upper bound on the fourth derivative does not use its variation across the interval. The theorem bounds truncation error with exact function evaluations; sample errors and roundoff error are additional contributions, as described by noise sensitivity of a quadrature rule.