Problem 31 - Calculus

Question

Suppose that \(f(x)\) is a periodic function with period \(b\) Show that (a) \(g(t)=f(b t / 2 \pi)\) is periodic with period \(2 \pi\) and \(f(x)=g(2 \pi x / b)\) (b) The Fourier series for \(g\) is given by $$ g(t)=a_{0}+\sum_{k=1}^{\infty}\left(a_{k} \cos (k t)+b_{k} \sin (k t)\right) $$ where the coefficients \(a_{0}, a_{k}, b_{k}\) are given in the box on page 572 (c) The Fourier series for \(f\) is given by$$ f(x)=a_{0}+\sum_{k=1}^{\infty}\left(a_{k} \cos \left(\frac{2 \pi k x}{b}\right)+b_{k} \sin \left(\frac{2 \pi k x}{b}\right)\right)$$where the coefficients are the same as in part (b).

Step-by-Step Solution

Answer

The function \(g(t)\) is periodic with period \(2\pi\), and \(f(x) = g\left(\frac{2\pi x}{b}\right)\). Both functions share the same Fourier coefficients.

Step 1: Understanding the period of g(t)

We start by analyzing the function \(g(t) = f\left(\frac{b t}{2\pi}\right)\). Since \(f(x)\) is assumed to be periodic with period \(b\), this means that \(f(x + b) = f(x)\) for all \(x\). Substituting \(x\) with \(\frac{bt}{2\pi}\), we have \(f\left(\frac{bt}{2\pi} + b\right) = f\left(\frac{bt}{2\pi}\right)\). Set \(t' = t + 2\pi\), then \(g(t') = f\left(\frac{b(t + 2\pi)}{2\pi}\right) = f\left(\frac{bt}{2\pi} + b\right) = f\left(\frac{bt}{2\pi}\right) = g(t)\). Thus, \(g(t)\) is indeed periodic with period \(2\pi\).

Step 2: Verifying f(x) = g(2πx/b)

For part (a), we need to verify that \(f(x) = g\left(\frac{2\pi x}{b}\right)\). Substitute \(t = \frac{2\pi x}{b}\) into \(g(t)\), we find \(g\left(\frac{2\pi x}{b}\right) = f\left(\frac{b}{2\pi} \cdot \frac{2\pi x}{b}\right) = f(x)\). Therefore, the statement \(f(x) = g\left(\frac{2\pi x}{b}\right)\) holds true.

Step 3: The Fourier Series for g(t)

Given \(g(t)\) is periodic with period \(2\pi\), it has a Fourier series of the form: \[ g(t) = a_0 + \sum_{k=1}^{\infty}\left( a_k \cos(kt) + b_k \sin(kt) \right) \], where the coefficients \(a_0, a_k, b_k\) are specific for \(g(t)\) and determined by integration over one period \([0, 2\pi)\).

Step 4: Deriving the Fourier series for f(x) from g(t)

Since \(f(x) = g\left(\frac{2\pi x}{b}\right)\), we can transform each term of the Fourier series for \(g(t)\) to match \(f(x)\): \(\cos(kt)\) becomes \(\cos\left(k\cdot\frac{2\pi x}{b}\right)\) and \(\sin(kt)\) becomes \(\sin\left(k\cdot\frac{2\pi x}{b}\right)\). Thus, the Fourier series for \(f(x)\) becomes: \[ f(x) = a_0 + \sum_{k=1}^{\infty}\left(a_k \cos\left(\frac{2\pi k x}{b}\right) + b_k \sin\left(\frac{2\pi k x}{b}\right)\right) \]. The coefficients \(a_0, a_k, b_k\) remain the same as those obtained for \(g(t)\).

Key Concepts

Periodic functions

Periodic functions are fascinating because they repeat at regular intervals. Imagine something like a wave or the ticking of a clock. If a function is periodic, this means that there is a specific interval, called the period, after which the function's values start repeating. For example, with the function \( f(x) = \sin(x) \), it has a period of \( 2\pi \). This means that \( \sin(x + 2\pi) = \sin(x) \) for any value of \( x \). This repetition creates a predictable pattern.

When we take a periodic function like \( f(x) \) with a given period \( b \), and modify it as in this exercise with \( g(t) = f\left(\frac{b t}{2\pi}\right) \), we end up transforming its period to \( 2\pi \). Essentially, by scaling the input \( x \) in a proportional manner, the period adapts, which is quite useful for analyzing effects through a simpler interval.

Trigonometric series

A trigonometric series is a sum of sine and cosine functions. These series are particularly useful in representing periodic functions. A well-known example of a trigonometric series is the Fourier series, which expresses a periodic function as an infinite sum of sines and cosines. Each sine and cosine term oscillates at a different frequency, adding more detail to our approximation.

For instance, the Fourier series for a function \( g(t) \) with a period of \( 2\pi \) can be written as:

• \( a_0 + \sum_{k=1}^{\infty} \left( a_k \cos(kt) + b_k \sin(kt) \right) \)

In this case, \( a_0 \) is the average value of the function over one period, while \( a_k \) and \( b_k \) are coefficients that determine the amplitude of the cosine and sine terms, respectively.

The beauty of such a series lies in its ability to recreate complex wave patterns with just a few repetitive components. This not only simplifies analysis but also reveals interesting patterns within the original function.

Fourier coefficients

Fourier coefficients are numbers that weigh the importance of sine and cosine functions in the Fourier series representation of a periodic function. Determining these coefficients effectively "captures" the shape of the entire function. In simple terms, these coefficients tell us how much of each sine or cosine wave we need to recreate the original function.

For the Fourier series of \( g(t) \), the coefficients \( a_0 \), \( a_k \), and \( b_k \) are computed through integration over one period of the function, \([0, 2\pi)\). Specifically:

• \( a_0 = \frac{1}{2\pi} \int_{0}^{2\pi} g(t) \, dt \)
• \( a_k = \frac{1}{\pi} \int_{0}^{2\pi} g(t) \cos(kt) \, dt \)
• \( b_k = \frac{1}{\pi} \int_{0}^{2\pi} g(t) \sin(kt) \, dt \)

These formulas measure the correlation between the periodic function and the corresponding sine or cosine waves. By changing scales, like transforming \( g(t) \) to \( f(x) \), the same coefficients \( a_0 \), \( a_k \), and \( b_k \) can be used, keeping the transition consistent and allowing us to simplify such transformations in mathematical analysis.