Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 5 / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii

i

Solution

The power rule for a definite integral gives

I = [\frac{x ^{3}}{3}]_{0}^{1} = \frac{1}{3} .

ii

Solution

h = 1/4

, the composite trapezoidal rule gives half weight to the two endpoints:

T_{4} = \frac{1}{4} [\frac{0 ^{2} + 1 ^{2}}{2} + (\frac{1}{4})^{2} + (\frac{1}{2})^{2} + (\frac{3}{4})^{2}] = \frac{1}{4} (\frac{1}{2} + \frac{1 + 4 + 9}{16}) .

Therefore

T_{4} = \frac{11}{32} = 0.34375.

It exceeds the exact definite integral by

1/96

, as expected because

x^{2}

is a convex function and each trapezoid lies above the curve.

iii

Solution

For

N

equal intervals, the composite trapezoidal rule is

T_{N} = \frac{1}{N} [\frac{1}{2} + \sum_{n = 1}^{N - 1} (\frac{n}{N})^{2}] = \frac{1}{N ^{3}} \sum_{n = 1}^{N - 1} n^{2} + \frac{1}{2 N} .

Adding the missing

N^{2} / N^{3}

term therefore gives the exact relation

\frac{1}{N ^{3}} \sum_{n = 1}^{N} n^{2} = T_{N} + \frac{1}{2 N} .

Since

x^{2}

is continuous, its Riemann sums, and hence its endpoint-averaged trapezoidal sums, converge to its definite integral. Thus

T_{N} \to 1/3

and

1/ (2 N) \to 0

N \to \infty lim \frac{1}{N ^{3}} n = 1 \sum N n^{2} = \frac{1}{3} .

No pre-existing formula for the sum of squares is needed.

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