 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2017/1/b/solution

For

δ \in / 2 π Z

, the finite geometric series gives

\sum_{n = 1}^{N} e^{in δ} = \frac{e ^{i δ} ( 1 - e ^{i N δ} )}{1 - e ^{i δ}} = e^{i (N + 1) δ /2} \frac{s i n ( N δ /2 )}{s i n ( δ /2 )} .

The exponential prefactor has modulus one. Thus

n = 1 \sum N e^{in δ} = {∣ sin (N δ /2) / sin (δ /2) ∣, N, δ \in / 2 π Z, δ \in 2 π Z .

At an excluded phase every summand is one, which supplies the removable limiting value directly. Here

N

is a positive integer.

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