Question

Let \(P(x) = 1 + e^{2\pi i x} + 2 e^{3\pi i x}\), \(x \in \mathbb{R}\), \(i = \sqrt{-1}\). Then
\[ \lim_{N \to \infty} \frac{1}{N} \sum_{k=0}^{N-1} P(k\sqrt{2}) \] is equal to

• A. 0
• B. 1
• C. 3
• D. 4

Hint:

For limits of the form \(\lim_{N\to\infty} \frac{1}{N} \sum_{k=0}^{N-1} f(k\alpha)\) where \(\alpha\) is irrational, the limit often equals the integral of the function over its period, \(\int_0^1 f(x)dx\). For \(f(x) = e^{2\pi i n x}\) with integer \(n \neq 0\), this integral is 0. For \(n=0\) (i.e., a constant), the integral is the constant itself. This provides a fast way to solve such problems.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The question asks for the limit of the Cesàro mean (arithmetic average) of a sequence. The sequence is generated by evaluating a function \(P(x)\) at points \(x_k = k\alpha\), where \(\alpha = \sqrt{2}\) is an irrational number. This relates to a key result in ergodic theory or Fourier analysis concerning uniformly distributed sequences.

Step 2: Key Formula or Approach:
A fundamental result states that for any irrational number \(\alpha\) and a complex exponential function \(f(x) = e^{i\omega x}\), the time average is:
\[ \lim_{N\to\infty} \frac{1}{N} \sum_{k=0}^{N-1} e^{i\omega (k\alpha)} = \begin{cases} 1 & \text{if } \omega\alpha \text{ is a multiple of } 2\pi, \\ 0 & \text{otherwise} \end{cases} \]
Since \(\alpha = \sqrt{2}\) is irrational, \(\omega \alpha\) can only be a multiple of \(2\pi\) if \(\omega=0\). For any \(\omega \neq 0\), the limit is 0. We can apply this property to each term of \(P(x)\) by linearity of limits and sums.

Step 3: Detailed Calculation:
Let \(L\) be the limit we want to compute.
\[ L = \lim_{N\to\infty} \frac{1}{N} \sum_{k=0}^{N-1} P(k\sqrt{2}) \]
Substitute the expression for \(P(x)\):
\[ L = \lim_{N\to\infty} \frac{1}{N} \sum_{k=0}^{N-1} \left( 1 + e^{2\pi i (k\sqrt{2})} + 2e^{3\pi i (k\sqrt{2})} \right) \]
By linearity, we can split the limit into three parts:
\[ L = \lim_{N\to\infty} \frac{1}{N} \sum_{k=0}^{N-1} 1 + \lim_{N\to\infty} \frac{1}{N} \sum_{k=0}^{N-1} e^{i(2\pi\sqrt{2})k} + 2 \lim_{N\to\infty} \frac{1}{N} \sum_{k=0}^{N-1} e^{i(3\pi\sqrt{2})k} \]

1. First Term: The average of the constant 1.
\[ \lim_{N\to\infty} \frac{1}{N} \sum_{k=0}^{N-1} 1 = \lim_{N\to\infty} \frac{N}{N} = 1 \]

2. Second Term: This is of the form \(\lim_{N\to\infty} \frac{1}{N} \sum_{k=0}^{N-1} e^{i\theta k}\) with \(\theta = 2\pi\sqrt{2}\). Since \(\sqrt{2}\) is irrational, \(\theta\) is not a multiple of \(2\pi\). Therefore, the limit is 0.

3. Third Term: This is of the form \(\lim_{N\to\infty} \frac{1}{N} \sum_{k=0}^{N-1} e^{i\theta k}\) with \(\theta = 3\pi\sqrt{2}\). Since \(\sqrt{2}\) is irrational, \(\theta\) is not a multiple of \(2\pi\). Therefore, the limit is 0.

Combining the results:
\[ L = 1 + 0 + 2(0) = 1 \]

Step 4: Final Answer:
The value of the limit is 1.

Step 5: Why This is Correct:
The calculation correctly applies the principle that the time average of a non-constant complex exponential \(e^{i\theta k}\) is zero, provided \(\theta\) is not a multiple of \(2\pi\). Since the step size \(\sqrt{2}\) is irrational, this condition holds for both exponential terms in \(P(x)\), leaving only the average of the constant term.