Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability / 6 / a

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

i

Solution

h = 1/ N

and nodes

x_{j} = j / N

, the composite trapezoidal rule gives

T_{N} = \frac{1}{N} [\frac{1}{2} + j = 1 \sum N - 1 \frac{1}{1 + ( j / N ) ^{2}} + \frac{1}{4}] = \frac{3}{4 N} + j = 1 \sum N - 1 \frac{N}{N ^{2} + j ^{2}}, I \approx T_{N} .

The endpoint weights are half the interior weights, and the endpoint values are

1

and

1/2

.

ii

Solution

Write the sum in the question as the right Riemann sum

S_{N} = \sum_{j = 1}^{N} \frac{N}{N ^{2} + j ^{2}} = \frac{1}{N} \sum_{j = 1}^{N} \frac{1}{1 + ( j / N ) ^{2}} .

Relative to the composite trapezoidal rule, the only change is in the two endpoint weights:

S_{N} - T_{N} = \frac{1}{2 N} (f (1) - f (0)) = - \frac{1}{4 N} .

f (x) = 1/ (1 + x^{2})

is continuous on

[0, 1]

, so

T_{N} \to I

. Since an antiderivative is

arctan x

,

I = arctan 1 - arctan 0 = \frac{π}{4} .

Thus

N \to \infty lim S_{N} = \frac{π}{4} .

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