Question

Let \(f\) first be extended into \((L, 2 L)\) so that it is symmetric about \(x=L ;\) that is, so as to satisfy \(f(2 L-x)=f(x)\) for \(0 \leq x

Short answer

Question: Show that the given function f(x), which is extended to be symmetric about x=L, odd about x=0, and periodic with period 4L, has a Fourier sine series representation of the form: $$ f(x) = \sum_{n=1}^{\infty} b_{n} \sin \frac{(2 n-1) \pi x}{2 L} $$ where $$ b_{n} = \frac{2}{L} \int_{0}^{L} f(x) \sin \frac{(2 n-1) \pi x}{2 L} d x $$ Also, show that the series converges to the original function on the interval (0, L]. Answer: To solve this problem, we followed these steps: 1. Recognize it as a Fourier sine series problem: Since we have an odd function f(x), the Fourier series representation would only involve sine terms. 2. Determine the Fourier coefficients b_n using the given formula. 3. Calculate the Fourier sine series using the coefficients b_n. 4. Acknowledge the convergence of the series: It is given in the problem statement that the sine series converges to the original function on (0, L]. Thus, we have shown that the given function f(x) has the desired Fourier sine series representation, and the series converges to the original function on the interval (0, L].

Step-by-step solution

Recognize it as a Fourier sine series problem

We are given an odd function f(x) and the objective is to find its Fourier sine series representation. Only sine terms are involved, because the function is odd.

Determine the Fourier coefficients b_n

According to the formula given in the problem, we can calculate the Fourier coefficients b_n as: $$ b_{n}=\frac{2}{L} \int_{0}^{L} f(x) \sin \frac{(2 n-1) \pi x}{2 L} d x $$

Calculate the sine series

Now, we can construct the Fourier sine series of the given function f(x): $$ f(x)=\sum_{n=1}^{\infty} b_{n} \sin \frac{(2 n-1) \pi x}{2 L} $$ where the coefficients b_n are calculated in Step 2.

Convergence of the series

It is given in the problem statement that the sine series converges to the original function on (0, L]. This means that the sine series representation of f(x) agrees with the original function f(x) on the interval (0, L]. In conclusion, we showed that the given function has a Fourier sine series representation of the given form with coefficients b_n obtained as in Step 2. It is provided in the exercise that the series converges to the original function on the interval (0, L].

Key concepts

Fourier Coefficients

To represent a function using a Fourier series, we need to determine the Fourier coefficients. In the case of a Fourier sine series, the coefficients are essential in expressing how each sine function contributes to the reconstruction of the original function. Specifically, the coefficients \(b_n\) in the given problem are calculated using the integral formula:

• \(b_{n}=\frac{2}{L} \int_{0}^{L} f(x) \sin \frac{(2 n-1) \pi x}{2 L} \, dx\)

This mathematical formula captures the essence of the Fourier sine coefficients, defining each term in the series by how the original function \(f(x)\) correlates with sine functions over the interval \([0, L]\).
The choice of sine functions, which are orthogonal over this interval, allows us to decompose \(f(x)\) uniquely in terms of these functions, ensuring that each coefficient specifically reflects the part of \(f(x)\) described by \(\sin((2n-1)\pi x / 2L)\).

Odd Function

An odd function is one that satisfies the condition \(f(-x) = -f(x)\). This property is crucial for constructing a Fourier sine series because it can only contain sine terms. Sine functions are inherently odd because \(\sin(-x) = -\sin(x)\).
In our problem, extending the function as an odd function over the interval \((-2L, 0)\) plays a pivotal role. This ensures that when creating the Fourier sine series, the resulting representation maintains the odd symmetry across the selected domain. Consequently, only sine terms appear in the series, effectively capturing the essence of odd symmetry in this function's periodic extension.
This reflects a fundamental characteristic of Fourier series: that we utilize odd functions (sines) to capture odd properties of functions when extended symmetrically in specified ways.

Series Convergence

The concept of series convergence is central in ensuring that a Fourier series accurately represents the original function. In this context, the problem states that the Fourier sine series converges to \(f(x)\) on the interval \((0, L]\).
Convergence means that as more terms \(n\) are included in the series sum, the approximation of \(f(x)\) becomes more precise, eventually matching \(f(x)\) for each point within this interval. This property ensures that, practically, the Fourier sine series can substitute \(f(x)\) without losing accuracy.
However, convergence often depends on the function's properties, including continuity and integrability. For our odd function, convergence assures us of equivalence with the original function's behavior across each considered interval within \((0, L]\). This fidelity underscores the effectiveness of Fourier series in replicating periodic functions.

Symmetric Extension

A symmetric extension involves expanding a function's domain such that its behavior is mirrored or repeated across a new interval. This concept is shown in the exercise by extending \(f(x)\) to \((L, 2L)\) while satisfying the symmetry condition \(f(2L - x) = f(x)\). This technique allows a periodic pattern to emerge naturally when considering Fourier series.
In practice, this means turning a non-periodic function originally defined over \([0, L]\) into a periodic one over \([0, 4L]\) by mirroring its values across the specified points within \((L, 2L)\). By ensuring this symmetry, all periodic aspects of the function can be leveraged when determining the sine series.
Such symmetric extensions are vital in constructing Fourier series, particularly because they create a smooth bridge between original and repeated function segments. These insights streamline both visual and analytical understanding of how the function behaves within the intended scope.