Question

Find the mixed Fourier sine series. $$ f(x)=\left\\{\begin{array}{ll} 1, & 0 \leq x \leq \frac{L}{2} \\ 0, & \frac{L}{2}

Short answer

Answer: The mixed Fourier sine series representation of the given function f(x) is: $$ f(x) = \sum_{n=1}^{\infty} \frac{1}{\pi n}(1 - \cos(\pi n)) \sin\left(\frac{2\pi nx}{L}\right) $$

Step-by-step solution

Define the function f(x) and its Fourier sine series representation

The given function f(x) is: $$ f(x)=\left\\{\begin{array}{ll} 1, & 0 \leq x \leq \frac{L}{2} \\\ 0, & \frac{L}{2}<x<L ; \end{array}\quad[0, L]\right. $$ Its mixed Fourier sine series representation is given by: $$ f(x) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{2\pi nx}{L}\right) $$ Where the \(B_n\) coefficients are found using: $$ B_n = \frac{2}{L}\int_{0}^{L} f(x) \sin\left(\frac{2\pi nx}{L}\right)dx $$

Evaluate the integral to find B_n

Since the function f(x) is piecewise defined, we will integrate it over each interval. We have: $$ B_n = \frac{2}{L}\left(\int_{0}^{L/2} f(x) \sin\left(\frac{2\pi nx}{L}\right)dx + \int_{L/2}^{L} f(x) \sin\left(\frac{2\pi nx}{L}\right)dx\right) $$ From the definition of f(x), we know that: $$ f(x) = 1 \quad \text{for} \quad 0 \leq x \leq \frac{L}{2} $$ $$ f(x) = 0 \quad \text{for} \quad \frac{L}{2} < x < L $$ Plugging these values into both integrals: $$ B_n = \frac{2}{L}\left(\int_{0}^{L/2} 1 \cdot \sin\left(\frac{2\pi nx}{L}\right)dx + \int_{L/2}^{L} 0 \cdot \sin\left(\frac{2\pi nx}{L}\right)dx\right) $$ Since the second integral has a 0 in the integrand, its result will also be 0. $$ B_n = \frac{2}{L}\int_{0}^{L/2} \sin\left(\frac{2\pi nx}{L}\right)dx $$

Evaluate the integral to find the B_n coefficients

Let's evaluate the integral: $$ B_n = \frac{2}{L} \left[ -\frac{L}{2\pi n}\cos\left(\frac{2\pi nx}{L}\right)\right]_0^{L/2} $$ Now, substitute the values x = 0 and x = L/2 into the integral: $$ B_n = \frac{2}{L} \left[-\frac{L}{2\pi n}\cos\left(\frac{2\pi n(L/2)}{L}\right) + \frac{L}{2\pi n}\cos(0)\right] $$ Simplify the equation: $$ B_n = \frac{2}{L} \left[-\frac{L}{2\pi n}\cos(\pi n) + \frac{L}{2\pi n}\right] = \frac{1}{\pi n}(1 - \cos(\pi n)) $$

Write the mixed Fourier sine series of f(x)

Finally, we have the mixed Fourier sine series for the given function f(x): $$ f(x) = \sum_{n=1}^{\infty} \frac{1}{\pi n}(1 - \cos(\pi n)) \sin\left(\frac{2\pi nx}{L}\right) $$ This is the mixed Fourier sine series representation of the given function f(x) over the interval [0, L].

Key concepts

Mixed Sine Series

A Mixed Sine Series is a type of Fourier series that primarily uses sine functions to represent a known periodic function. In the case of a mixed sine series, we focus specifically on the sine part of the Fourier series. This is because the function is often odd and needs to be expressed through sine components exclusively.
The full representation of the function is given by:

• The sine functions oscillate between positive and negative values.
• They are used to account for the odd functions over a defined interval (often \( [0, L] \)).

In our example, the function \( f(x) \) is expressed in terms of sine functions only due to its specific nature. The mixed sine series allows us to effectively capture the behavior of the function by decomposing it into its sine component parts.

Piecewise Functions

Piecewise Functions are an essential concept in mathematics and particularly in calculus and Fourier series. These functions are defined by different expressions over distinct intervals of their domain.
In our problem, the function \( f(x) \) is piecewise defined over the interval \( [0, L] \):

• For \( 0 \leq x \leq \frac{L}{2} \), \( f(x) = 1 \).
• For \( \frac{L}{2} < x < L \), \( f(x) = 0 \).

The piecewise definition divides the problem into separate parts that are easier to solve individually. This allows us to integrate each section more effectively and apply the relevant boundary conditions to find the appropriate series representation.
Understanding how piecewise functions work enables solving more complex problems and evaluating integrals that arise in these contexts.

Integral Evaluation

Integral Evaluation is a crucial step in determining the coefficients of a Fourier series. It involves calculating the area under the curve defined by the function being represented. For sine series, integrals involve sine functions multiplied by the original function components.
In our exercise, we evaluate the integral for coefficients \( B_n \) using:

• The integral involves evaluating over each interval separately: \( [0, \frac{L}{2}] \) and \( [\frac{L}{2}, L] \).
• Since \( f(x) = 0 \) in one part, the integral from \( \frac{L}{2} \) to \( L \) is zero.

The result of the non-zero integral is simplified by substituting boundary values, thus focusing on the interval where \( f(x) = 1 \). By understanding integral evaluation, one can derive coefficients that properly reconstruct the function using its sine series.

Coefficients in Series

Coefficients in Series are the key parameters that configure how terms in a Fourier series combine to represent a target function. In sine series, these coefficients dictate the amplitude and phase of the sine terms employed.
We determine the \( B_n \) coefficients using an integral that involves both the given piecewise function and sine functions:

• The formula is \( B_n = \frac{2}{L}\displaystyle\int_{0}^{L} f(x) \sin\left(\frac{2\pi nx}{L}\right)dx \).
• Solutions reveal relationships between terms, such as \( B_n = \frac{1}{\pi n}(1 - \cos(\pi n)) \), elucidating the alternation and cancellation of terms.

Understanding these coefficients' role is essential, as they maintain the series' accuracy and alignment with the original periodic function. For learners, grasping this relation means mastering how series intricately map and replicate function behaviors.