Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 4 / c

Take $n$ to be a positive even integer, as required by composite Simpson's rule. There are $n/2$ two-strip panels, each of width $2/n$. Applying the supplied panel bound with the global fourth-derivative minimum $0$ and maximum $1$, and summing, givesHere $E_n$ is the nonnegative absolute error. In this example Simpson's rule overestimates the integral, so $E_n=S_n-I$. Each panel has that same sign because its fourth derivative is nonnegative; cancellation is not being assumed.^[ref]

On the panel $[2j/n,2(j+1)/n]$, the maximum fourth derivative is at the left endpoint because it is decreasing. ThusThe geometric series and the sum of consecutive integers giveSubstitution provesFor $n>2$, at least one panel's maximum is smaller than the global maximum $1$, so this bound is strictly tighter than the global bound. For $n=2$, the two bounds coincide.