Discussion
Let \(P(x) = 1 + e^{2\pi ix} + 2e^{3\pi ix}\), \(x \in \mathbb{R}\), \(i = \sqrt{-1}\). Then
\[ \lim_{N \to \infty}\frac{1}{N}\sum_{k=0}^{N-1}P(k\sqrt{2}) \]
is equal to
Answer: B
The limit is 1.
Average the three terms of \(P(k\sqrt{2})\) separately. The constant 1 averages to 1. For the others write them as \(e^{2\pi i k\theta}\):
\[ \begin{aligned} e^{2\pi i k\sqrt{2}}&:\ \theta = \sqrt{2} \\ e^{3\pi i k\sqrt{2}}&:\ \theta = \tfrac{3}{2}\sqrt{2} \end{aligned} \]
Both values of \(\theta\) are irrational. By Weyl's equidistribution theorem, for irrational \(\theta\) the sequence \(k\theta\) mod 1 is equidistributed, so the averages of \(e^{2\pi ik\theta}\) tend to \(\int_{0}^{1}e^{2\pi it}dt = 0\). (Directly, the geometric sum is bounded while N grows.)
So the limit is 1 + 0 + 2 × 0 = 1. 4 would be P(0), the value if the exponentials stayed at 1, and 3 or 0 drop or keep the wrong terms.
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