Problem 14 - Advanced Engineering Mathematics

Question

Expand the given function in an appropriate cosine or sine series. $$ f(x)=x,-\pi

Step-by-Step Solution

Answer

The function \(f(x) = x\) is expanded as a sine series: \(f(x) = \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n} \sin(nx)\).

Step 1: Identify Function Type

The function given is an odd function, as \( f(-x) = -f(x) \). This means it can be expanded in terms of a sine series, since sine functions are also odd.

Step 2: Define the Interval and Function

We are given the function \( f(x)=x \) defined on the interval \( -\pi < x < \pi \). Analyzing it as a half-range sine series involves focusing on its properties from 0 to \( \pi \) due to symmetry.

Step 3: Determine the Coefficients

For a sine series, the coefficients \( b_n \) are found using the formula: \[b_n = \frac{2}{\pi} \int_0^{\pi} f(x) \sin(nx) \, dx \]Substitute \( f(x) = x \) into the formula:\[b_n = \frac{2}{\pi} \int_0^{\pi} x \sin(nx) \, dx\]This integral must be calculated for each coefficient.

Step 4: Integrate to Find Coefficients

To solve for \( b_n \), we integrate by parts:Let \( u = x \) and \( dv = \sin(nx) \, dx \), which gives us \( du = dx \) and \( v = -\frac{1}{n} \cos(nx) \):\[b_n = \frac{2}{\pi} \left[-\frac{x}{n} \cos(nx) \Big|_0^{\pi} + \frac{1}{n} \int_0^{\pi} \cos(nx) \, dx \right] \]Evaluate the boundary terms and remaining integral.

Step 5: Evaluate the Boundary Terms and Solve Integral

Substitute the limits into \([-\frac{x}{n} \cos(nx)]_0^{\pi}\):\[-\frac{\pi}{n} \cos(n\pi) + 0 = \frac{\pi(-1)^{n+1}}{n}\]And the integral of \( \frac{1}{n} \int_0^{\pi} \cos(nx) \, dx \) is zero because it's an integral over a full period of cosine, which cancels out. Therefore:\[b_n = \frac{2(-1)^{n+1}}{n}\]

Step 6: Compose the Sine Series

The sine series representation of \( f(x) = x \) over the interval \( -\pi < x < \pi \) is:\[f(x) = \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n} \sin(nx)\]

Key Concepts

Sine Series

A sine series is a type of Fourier series specifically used to represent functions through sine terms. This expansion is helpful when dealing with odd functions because sine, like any odd function, satisfies the property \( f(-x) = -f(x) \). By using sines, we avoid introducing mismatched symmetry. This property is crucial because it ensures that the original function's structure remains intact when expanded. To create a sine series, you express the function as a sum of sine terms, each involving a sine function of a multiple of \( x \), like \( \sin(nx) \):

• Each term is weighted by a coefficient, \( b_n \).
• The coefficients \( b_n \) themselves are calculated through integration, ensuring they fit the original function's shape.

Employing a sine series takes advantage of the symmetry that comes from focusing only on the sine components in the Fourier expansion, valuable for keeping the integrity of odd functions.

Odd Function

An odd function is one with the characteristic property that \( f(-x) = -f(x) \). This means that the function has a particular type of symmetry about the origin; if you were to rotate its graph 180 degrees about the origin, it would look the same. Odd functions are important in cases of sine series because each sine function itself is odd.

• This directly relates to symmetry properties utilized in Fourier expansions and integrations.
• For example, the sine function \( \sin(x) \) reflects this odd nature as \( \sin(-x) = -\sin(x) \).

Such symmetry considerations simplify the process of finding the Fourier series, allowing us to focus solely on sine functions, reducing complexity and maintaining fidelity to the function's inherent characteristics.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is particularly useful in deriving the coefficients for sine series when the function involves a product of \( x \) and \( \sin(nx) \).To perform integration by parts, follow these steps:

• Choose \( u = x \) and \( dv = \sin(nx) \, dx \).
• This gives \( du = dx \) and \( v = -\frac{1}{n} \cos(nx) \).
• Integrate using the formula \( \int u \, dv = uv - \int v \, du \).

This technique helps by breaking down a complex integral into simpler parts, allowing for more straightforward evaluation of the coefficients in a Fourier series. It is essential for handling the integration needed to calculate \( b_n \), as seen in the solution's step-by-step process.

Half-Range Expansion

Half-range expansion is a technique used in Fourier series to represent a function only over a portion of its usual range. Specifically, it focuses on either the positive or negative half of the function's domain, which can be advantageous when the function is symmetric, such as an odd function.When dealing with the function \( f(x) = x \) over the interval \( -\pi < x < \pi \):

• We restrict our integration for \( b_n \) from 0 to \( \pi \), utilizing only the positive half-range.
• This corresponds to the specific use of sine functions that match the odd symmetry in half-range expansions.
• The result is a more efficient calculation of coefficients, focusing only on the range required for the symmetric properties.

By focusing on a half-range, the series leverages the symmetric properties of sines and cosine functions to simplify derivations and computations, reducing the complexity of expansion steps.