 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013/10/c/solution

q = e^{i θ} /2

\sum_{n = 1}^{N} q^{n} = \frac{q ( 1 - q ^{N} )}{1 - q} = \frac{( q - q ^{N + 1} ) ( 1 - q )}{∣1 - q ∣ ^{2}} .

(5 - 4 cos θ) /4

\frac{1}{2} sin θ - \frac{s i n (( N + 1 ) θ )}{2 ^{N + 1}} + \frac{s i n ( Nθ )}{2 ^{N + 2}} .

Dividing and multiplying numerator and denominator by

2^{N}

proves

n = 1 \sum N \frac{sin ( n θ )}{2 ^{n}} = \frac{2 ^{N + 1} sin θ + sin ( Nθ ) - 2 sin (( N + 1 ) θ )}{2 ^{N} ( 5 - 4 cos θ )} .

For real

θ

5 - 4 cos θ \geq 1

, so there is no exceptional angle requiring a limiting interpretation. At

N = 1

the expression reduces to

sin θ /2

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