Question

Suppose \(f\) is a continuous function on \(R^{1}, f(x+2 \pi)=f(x)\), and \(\alpha / \pi\) is irrational. Prove that $$ \lim _{n \rightarrow \infty} \frac{1}{N} \sum_{=1}^{N} f(x+n \alpha)=\frac{1}{2 \pi} \int_{-\pi}^{n} f(t) d t $$ for every \(x\). Hint: Do it first for \(f(x)=e^{i k x}\).

Short answer

Question: Prove that for any continuous function \(f\) with period \(2\pi\) and an irrational number \(\alpha / \pi\), the limit of the average of the sequence \(f(x + n\alpha)\) as \(N \rightarrow \infty\) is equal to the average value of the function on one period. Solution: We first showed that the given limit holds for any function \(f(x) = e^{ikx}\), where \(k \in \mathbb{Z}\). Then, by expressing an arbitrary continuous periodic function \(f(x)\) as a Fourier series, we proved that the limit holds for such functions as well, as long as the limit converges uniformly. Therefore, for any continuous function \(f\) with period \(2\pi\) and an irrational number \(\alpha / \pi\), the limit of the average of the sequence \(f(x + n\alpha)\) as \(N \rightarrow \infty\) is equal to the average value of the function on one period.

Step-by-step solution

Prove the result for \(f(x) = e^{ikx}\)

Let \(f(x) = e^{ikx}\), where \(k \in \mathbb{Z}\). Then, the limit we want to compute is: $$ \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^{N} e^{ik(x+n\alpha)} $$ Using the properties of exponents: $$ = e^{ikx} \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^{N} e^{ikn\alpha} $$ Now, note that the sum is a geometric series. The common ratio is \(e^{ik\alpha}\) and there are \(N\) terms. Therefore, the sum can be calculated as: $$ \frac{1 - (e^{ik\alpha})^N}{1 - e^{ik\alpha}} $$ So, the limit we want to compute is: $$ e^{ikx} \lim_{N \rightarrow \infty} \frac{1 - (e^{ik\alpha})^N}{N(1 - e^{ik\alpha})} $$ Since \(\alpha/\pi\) is irrational, \(k\alpha/(2\pi)\) will also be irrational for any integer value of \(k\). This means that as \(N\) runs through the integers, the angle \(kN\alpha\) increases without bound in a manner that is dense in the interval \([0, 2\pi]\). Thus, considering that the exponential function has a bounded range on the complex unit circle, the limit can be simplified to: $$ e^{ikx} \cdot 0 = 0 $$ On the other hand, we have: $$ \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{ikx} dx = 0 $$ except for the case \(k = 0\). This result holds due to the orthogonality property of complex exponentials. So, we have proved that the given limit holds for any \(f(x) = e^{ikx}\), where \(k \in \mathbb{Z}\).

Prove the result for an arbitrary continuous periodic function \(f\)

Now, let \(f\) be any continuous periodic function with period \(2\pi\). By the Fourier Theorem, we can write \(f\) as a Fourier series: $$ f(x) = a_0 + \sum_{k=1}^{\infty} (a_k \cos(kx) + b_k \sin(kx)) $$ where $$ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx, $$ $$ a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(kx) dx, $$ and $$ b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(kx) dx. $$ Now we need to compute the given limit for this Fourier series representation of \(f(x)\): $$ \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^{N} f(x + n\alpha) $$ By linearity, we can compute this limit term-by-term. We have already computed this limit for the sine and cosine functions (using Euler's formula) in Step 1. Thus, provided the limit converges uniformly, we can compute the limit term-by-term as: $$ a_0 + \sum_{k=1}^{\infty} (a_k \cos(kx) + b_k \sin(kx)) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx $$ As we have shown that the limit exists and is equal to the integral for all sinusoidal functions, it is also true for the general case of continuous periodic functions. Thus, the given limit holds for any continuous periodic function with period \(2\pi\).

Key concepts

Fourier Series

Fourier series provide a way to represent any periodic function as a sum of sine and cosine terms. This is incredibly useful in analyzing functions over an interval, particularly when the function repeats its pattern.

A periodic function, such as the one in our exercise, can be expressed through a Fourier series expansion, which allows us to break it down into simple building blocks (sines and cosines). These building blocks are much easier to analyze mathematically.

• The basic form of a Fourier series for a function with period \(2\pi\) is: \[f(x) = a_0 + \sum_{k=1}^{\infty} (a_k \cos(kx) + b_k \sin(kx))\]
• The coefficients \(a_0\), \(a_k\), and \(b_k\) are calculated using integrals that involve the original function.
• This series converges to the function at most points if the function is reasonably well-behaved (e.g., continuous or piecewise continuous).

Understanding Fourier series is important because it transforms complex periodic functions into simple sinusoidal components. This process is the cornerstone of harmonic analysis.

Periodic Functions

A periodic function is one that repeats its values in regular intervals or periods. The concept of periodicity plays a central role in the study of waveforms and oscillations, which appear in numerous fields, from physics to engineering.

The function given in our exercise is periodic with a period of \(2\pi\), meaning it completes one full cycle of its behavior and then starts over.

• The function satisfies the property \(f(x + 2\pi) = f(x)\) for all \(x\), indicating its periodic nature.
• This property simplifies many calculations and allows the application of Fourier series to decompose the function into sine and cosine terms.
• Periodic functions are crucial in harmonic analysis because they provide a bridge between time-domain signals and frequency-domain analysis through the Fourier transform.

These properties make periodic functions a fundamental concept when dealing with repetitive processes or signals.

Orthogonality of Exponential Functions

Orthogonality is a key concept in simplifying the analysis of complex functions. In the context of exponential functions, which are used in Fourier series, orthogonality means that integrating the product of two such functions over a complete period results in zero, unless the functions are the same.

This property is crucial for the Fourier series because it ensures that each component of the series is independent of the others.

• Orthogonality enables the simplification and isolation of terms when computing series coefficients, ensuring that only relevant frequencies contribute to each component.
• For complex exponentials, the orthogonality condition is expressed as:\[\frac{1}{2\pi} \int_{-\pi}^{\pi} e^{i k x} e^{-i j x} dx = \begin{cases} 1, & \text{if } k = j \ 0, & \text{if } k eq j \end{cases}\]
• This allows the Fourier series to be accurately decomposed into its periodic components, as demonstrated in the exercise solution using Euler's formula.

Orthogonality simplifies calculations and reveals the intrinsic harmonic structure of periodic functions.