Quadrature Error

One way we might evaluate the error is to use Taylor expansions with integral remainders for f. We illustrate for the trapezium rule using the Taylor results

$$\begin{array} {lcl}f(x) & = & f(0)+xf'(0)+\int _0^h(x-t)_+f'... ...,\\ f(h) & = & f(0)+hf'(0)+\int _0^h(h-t)_+f''(t)dt\end{array}$$

to obtain

$$\begin{array} {lcl}E(f) & = & \int _0^hf(x)dx-\frac{h}{2}\le... ...t,\quad 0<\xi <h \\ & = & -\frac{h^3}{12}f''(\xi ),\end{array}$$

where we were able to use the integral mean value theorem in the penultimate line assuming continuity of f'', since t(t-h) is of constant sign.

We investigate the error in more generality in the following theorem:

Theorem 3839 ((Peano))

If the linear functional $E:C^{m+1}[a,b]\to \hbox{{\sf I}\kern-.15em\hbox{\sf R}}$is defined by $E(f)=\int _a^bf(x)dx-\sum _{i=0}^nH_if(x_i)$, and if E(p)=0 for all polynomials p of degree not greater than m then

$$E(f)=\int _a^bf^{(m+1)}(t)K_m(t)dt,$$

where

$$K_m(t)=\frac{1}{m!}E\left( (x-t)_+^m \right),$$

the Peano kernel.

Corollary 3843

If further K_m is of constant sign on [a,b], then

$$E(f)=\frac{f^{(m+1)}(\xi )}{(m+1)!}E(x^{m+1}),\quad a<\xi <b.$$

Examples

• For the trapezium rule, $E(f)=\int _0^hf(x)dx-\frac{h}{2}\left( f(0)+f(h) \right)$, so E(1)=E(x)=0, while $E(x^2)=-\frac{h^3}{6}$, so m=1, provided that f'' is continuous. Now

  $$\begin{array} {lcl}K_1(t) & = & E\left( (x-t)_+ \right)\\ & ... ...(h-t)^2-\frac{h}{2}(h-t)\\ & = & -\frac{t}{2}(h-t).\end{array}$$

  This does not change sign in [0,h], so we may use the corollary to obtain

  $$E(f)=\frac{f''(\xi )}{2!}E(x^2)=-\frac{h^3}{12}f''(\xi ).$$

• For Simpson's rule,

  $$E(f)=\int _{-h}^hf(x)dx-\frac{h}{3}\left\{ f(-h)+4f(0)+f(h) \right\} .$$

  We have already seen from undetermined coefficients that E(1)=E(x)=E(x²)=E(x³)=0, while $E(x^4)=-\frac{4}{15}h^5$, so m=3, provided that f⁽⁴⁾ is continuous. Now

  $$\begin{array} {lcl}3!K_3(t) & = & E\left( (x-t)_+^3 \right)\... ... (h-t)^3(3t+h) & 0\leq t\leq h \end{array}\right..\end{array}$$

  Hence $K_3(t)\leq 0$ for all $t\in [-h,h]$ and we may apply the corollary to obtain

  $$\begin{array} {lcl}E(f) & = & \frac{f^{(4)}(\xi )}{4!}E(x^4)... ...f^{(4)}(\xi )\\ & = & -\frac{h^5}{90}f^{(4)}(\xi ).\end{array}$$

The difficulty is always the establishment of a constant sign for K_m. If this cannot be done, we have to settle for a bound on the error using the result of the theorem:

$$\begin{array} {lcl}\left\vert \kern.05em E(f) \kern.05em \ri... ...kern.05em \int _a^bK_m(t)dt \kern.05em \right\vert,\end{array}$$

where

$$M=\stackrel{\max }{t\in [a,b]}\left\vert \kern.05em f^{(m+1)}(t) \kern.05em \right\vert.$$

This can be quite messy to evaluate.

Example

Find the open quadrature formula

$$\int _0^{3h}f(x)dx\approx \alpha _1f(h)+\alpha _2f(2h)$$

using the method of undetermined coefficients. Find also an error estimate using Peano's Theorem assuming $f\in C^2[a,b]$.

Let $E(f)= \int _0^{3h}f(x)dx- \alpha _1f(h)-\alpha _2f(2h)$. Then

$$\begin{array} {lcl}0~=~E(1) & = & 3h-\alpha _1-\alpha _2\\ 0~=~E(x) & = & \frac{9}{2}h^2-\alpha _1h-2\alpha _2h,\end{array}$$

from which we find the values $\alpha _1=\alpha _2=\frac{3}{2}h$. Now

$$E(f)= \int _0^{3h}f(x)dx- \frac{3}{2}h\left\{ f(h)+f(2h) \right\}$$

and $E(x^2)=\frac{3}{2}h^2$. From Peano's Theorem we have

$$\begin{array} {lcl}K_1(t) & = & \frac{1}{1!}E\left( (x-t)_+ ... ...nd{array} \right.\\ & \geq & 0~\forall t\in [0,3h],\end{array}$$

so

$$E(f)=\frac{f''(\xi )}{2!}E(x^2)=\frac{3}{4}h^3f''(\xi ).$$

We can theoretically construct more and more accurate quadrature formulae by taking higher degree interpolating polynomials. However, the coefficients in these formulae become large and therefore liable to produce rounding errors. An alternative and much commoner approach is to use composite rules.