NUMERICAL METHODS
The difference between the estimate of the integral and the exact answer is 1/12. Equation
(27.38) estimates this error as 2× 0. 25 ×〈f′′〉/12. Our (deliberately chosen!) integrand is
one for which〈f′′〉can be evaluated trivially. Becausef(x) is a quadratic function ofx,
its second derivative is constant, and equal to 2 in this case. Thus〈f′′〉has value 2 and
(27.38) estimates the error as 1/12; that the estimate is exactly right should be no surprise
since the Taylor expansion for a quadratic polynomial about any point always terminates
after three terms and so no higher-order terms inhhave been ignored in (27.38).
27.4.2 Simpson’s ruleWhereas the trapezium rule makes a linear interpolation off, Simpson’s rule
effectively mimics the local variation off(x) using parabolas. The strips are
treated two at a time (figure 27.4(c)) and therefore their number,N, should be
made even.
In the neighbourhood ofxi,foriodd, it is supposed thatf(x) can be adequatelyrepresented by a quadratic form,
f(xi+y)=fi+ay+by^2. (27.39)In particular, applying this toy=±hyields two expressions involvingb
fi+1=f(xi+h)=fi+ah+bh^2 ,fi− 1 =f(xi−h)=fi−ah+bh^2 ;thus
bh^2 =^12 (fi+1+fi− 1 − 2 fi).Now, in the representation (27.39), the area of the double strip fromxi− 1 toxi+1is given by
Ai(estim.) =∫h−h(fi+ay+by^2 )dy=2hfi+^23 bh^3.Substituting forbh^2 then yields, for the estimated area,
Ai(estim.) = 2hfi+^23 h×^12 (fi+1+fi− 1 − 2 fi)=^13 h(4fi+fi+1+fi− 1 ),an expression involving only given quantities. It should be noted that the values
of neitherbnoraneed be calculated.
For the full integral,I(estim.) =^13 h(f 0 +fN+4∑moddfm+2∑mevenfm). (27.40)
It can be shown, by following the same procedure as in the trapezium rule case,
that the error in the estimated area is approximately
∆I(estim.)≈(b−a)
180h^4 〈f(4)〉.