Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 4 / a

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

i

Solution

By the fundamental theorem of calculus,

\int_{0}^{1} 3 x^{2} d x = [x^{3}]_{0}^{1} = 1.

ii

Solution

For the composite trapezoidal rule, the mesh width is

h = 1/ N

and the nodes are

x_{n} = n / N

. The endpoint weights are halves and the interior weights are ones, so

T_{N} = \frac{1}{N} [\frac{3 ( 0 ) ^{2}}{2} + n = 1 \sum N - 1 3 (\frac{n}{N})^{2} + \frac{3 ( 1 ) ^{2}}{2}] = \frac{3}{N ^{3}} n = 1 \sum N - 1 n^{2} + \frac{3}{2 N} .

This formula is an approximation to the definite integral, not an exact equality with it for finite

N

.

iii

Solution

Let

S_{N} = N^{- 3} \sum_{n = 1}^{N} n^{2}

. The expression in part (ii) gives

T_{N} = 3 (S_{N} - \frac{1}{N}) + \frac{3}{2 N} = 3 S_{N} - \frac{3}{2 N} .

Because

3 x^{2}

is continuous on

[0, 1]

, the composite trapezoidal rule converges to its definite integral, which is

1

by part (i). In fact its standard composite trapezoidal error bound is

∣ T_{N} - 1∣ \leq 1/ (2 N^{2})

, since

f^{''} = 6

. Therefore

N \to \infty lim S_{N} = N \to \infty lim (\frac{T _{N}}{3} + \frac{1}{2 N}) = \frac{1}{3} .

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