Question

Show that the mixed Fourier sine series of \(f\) on \([0, L]\) is the restriction to \([0, L]\) of the Fourier sine series of $$ f_{4}(x)=\left\\{\begin{array}{cc} f(x), & 0 \leq x \leq L \\ f(2 L-x), & L

Short answer

#Answer# To prove this, we computed the Fourier sine series coefficients of \(f_4(x)\) on the interval \([0, 2L]\), and showed that they are equal to the mixed Fourier sine series coefficients of \(f\) on \([0, L]\). This result helps in proving Theorem 11.3.4 because it allows us to express the mixed Fourier sine series of \(f\) as the restriction of the Fourier sine series of \(f_4(x)\) on \([0, 2L]\), thus completing the proof of the theorem.

Step-by-step solution

Compute the Fourier sine series of \(f_4(x)\)

Let's compute the Fourier sine series of \(f_4(x)\) in the interval \([0, 2L]\). The coefficients for the Fourier sine series of \(f_4(x)\) can be found by using the formula: $$ b_n^{(4)} = \frac{2}{2L} \int_0^{2L} f_4(x) \sin \frac{n \pi x}{2L} dx $$ We can split the integral into two parts: $$ b_n^{(4)} = \frac{2}{2L}( \int_0^L f(x) \sin \frac{n \pi x}{2L} dx + \int_L^{2L} f(2L - x) \sin \frac{n \pi x}{2L} dx) \\ b_n^{(4)} = \frac{1}{L}( \int_0^L f(x) \sin \frac{n \pi x}{2L} dx + \int_L^{2L} f(2L - x) \sin \frac{n \pi x}{2L} dx) $$

Make a substitution

Let \(u = 2L - x\). Then, we have \(\frac{du}{dx} = -1\), which means \(dx = - du\). We need to change the limits of the integral as well: - When \(x = L\), \(u = L\). - When \(x = 2L\), \(u = 0\). Now, substitute \(u\) into the second integral: $$ \int_L^{2L} f(2L - x) \sin \frac{n \pi x}{2L} dx = -\int_L^{0} f(u) \sin \frac{n \pi (2L-u)}{2L} du $$ Change the order of the integration limits, which will remove the negative sign: $$ -\int_L^{0} f(u) \sin \frac{n \pi (2L-u)}{2L} du = \int_0^L f(u) \sin \frac{n \pi (2L-u)}{2L} du $$

Compute the final Fourier sine series coefficients

Now, let's rewrite the expression for \(b_n^{(4)}\) with the substitution: $$ b_n^{(4)} = \frac{1}{L}(\int_0^L f(x) \sin \frac{n \pi x}{2L} dx + \int_0^L f(u) \sin \frac{n \pi (2L-u)}{2L} du) $$ Notice that \(\sin(2 \pi k - \theta) = \sin \theta\). Applying this property to the expression above: $$ b_n^{(4)} = \frac{1}{L}(\int_0^L f(x) \sin \frac{n \pi x}{2L} dx + \int_0^L f(u) \sin \frac{n \pi u}{2L} du) $$ Now, change the variable back to \(x\): $$ b_n^{(4)} = \frac{2}{L} \int_0^L f(x) \sin \frac{n \pi x}{2L} dx $$

Prove Theorem 11.3.4

Now we can prove Theorem 11.3.4. We have shown that the mixed Fourier sine series of a function, \(f\), on an interval \([0, L]\) can be written as the restriction of the Fourier sine series of \(f_4(x)\) on \([0, 2L]\). Since the mixed Fourier sine series of a function \(f\) on \([0, L]\) is given by: $$ f(x) \sim \sum_{n=1}^{\infty} \frac{2}{L} b_n \sin \frac{n \pi x}{L} $$ We have shown that: $$ f_{4}(x) \sim \sum_{n=1}^{\infty} \frac{2}{L} b_n^{(4)} \sin \frac{n \pi x}{2L} $$ Therefore: $$ f(x) \sim \sum_{n=1}^{\infty} \frac{2}{L} \left( \frac{2}{L} \int_0^L f(x) \sin \frac{n \pi x}{2L} dx \right) \sin \frac{n \pi x}{L} $$ which completes the proof of Theorem 11.3.4.

Key concepts

Fourier sine series

A Fourier sine series is a way to represent a function using sine functions only. It's a part of the larger family of Fourier series, which can break down even complex waveforms into simpler sinusoidal components. In mathematical terms, the Fourier sine series focuses on solving problems with specific types of boundary conditions.

For a function defined on an interval \([0, L]\), the Fourier sine series is expressed as \( f(x) \sim \sum_{n=1}^{\infty} b_n \sin \frac{n\pi x}{L}\).

This representation is particularly useful in problems involving odd functions or situations where the function must be zero at both ends of the interval, which is a common scenario in physics and engineering applications.

• It primarily applies to problems with Dirichlet boundary conditions.
• It utilizes only the sine terms, which makes it excellent for representing functions with certain symmetries and behaviors.

The sine series is crucial for analyzing oscillating systems and wave equations, where these conditions often naturally appear.

mixed Fourier sine series

A mixed Fourier sine series is an intriguing extension of the basic Fourier sine series. It is used when analyzing functions that have behavior extended or modified beyond the interval \([0, L]\). In the exercise, this concept is illustrated by extending the function as \( f_4(x) =\begin{cases} f(x), & 0 \leq x \leq L \ f(2L-x), & L < x \leq 2L \end{cases}
\).

A mixed Fourier sine series essentially involves constructing a new function, like \( f_4(x)\), and analyzing it over a doubled interval \([0, 2L]\). This technique is useful because it enables simplification in calculating the Fourier series by maintaining some symmetry properties of the initial function across the extended interval.

• Such series incorporate any potential symmetry or periodicity the function could naturally possess.
• This extension can simplify computation, especially when calculating boundary or initial conditions for physical problems.

By doing this, it becomes possible to express the original function over \([0, L]\) as part of this doubled interval solution, adding flexibility and ease in dealing with particular boundary conditions.

Fourier coefficients

Fourier coefficients are the building blocks of the Fourier series. They determine "how much" of each sine (or cosine in the general case) function contributes to the complete series that represents the original function. In the context of the exercise, the Fourier sine series coefficients are denoted by \(b_n\) and are computed using the integral:

\[ b_n \qq = \frac{2}{L} \int_0^L f(x) \sin \frac{n \pi x}{L} \text{ } dx \]

These coefficients play a vital role as they encapsulate the interaction between the function and the sine wave at each harmonic. Each coefficient \(b_n\) is essentially a measure of the amplitude for the corresponding sine term in the series.

• They are derived from projecting the function onto each sine component, via integration over the defined interval.
• The magnitude of each coefficient indicates the strength or weight of the corresponding sine term.

Understanding these coefficients helps in decoding how the Fourier series has been constructed, providing insight into the behavior of the overall system or function being studied.