Problem 31 - Advanced Engineering Mathematics

Question

Find the half-range cosine and sine expansions of the given function. $$ f(x)=\left\\{\begin{array}{ll} x, & 0

Step-by-Step Solution

Answer

Calculate the Fourier coefficients for both series using integrals over function sections.

Step 1: Identify the Interval for Half-Range Expansion

First, we observe that the function is defined on the interval \(0 < x < 2\). For half-range expansions, we use either the cosine or sine expansion, which generally involve even or odd extensions of the function, respectively, across a determined interval, such as \([0, L]\) or \([-L, L]\). Here, the maximum value is 2, so we consider \(L = 2\).

Step 2: Setting Up the Half-Range Cosine Series

The half-range cosine series uses even extension, implying \(f(x) = f(-x)\). Each coefficient \(a_n\) is given by \(a_n = \frac{2}{L} \int_0^L f(x) \cos\left(\frac{n\pi x}{L}\right) \, dx\). For our function and \(L=2\), this leads to the integrals over two sections, from 0 to 1, where \(f(x)=x\), and from 1 to 2, where \(f(x)=1\).

Step 3: Calculate \(a_0\) for Cosine Expansion

The zeroth coefficient \(a_0\) is given by \(a_0 = \frac{2}{L} \int_0^L f(x) \, dx\). This becomes \(a_0 = \frac{1}{2} \int_0^1 x \, dx + \frac{1}{2} \int_1^2 1 \, dx\). Calculate these integrals separately: \[\frac{1}{2} \left(\frac{x^2}{2} \bigg|_0^1\right) + \frac{1}{2} [x]_1^2\], resulting in \(a_0 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\).

Step 4: Calculate \(a_n\) for Cosine Expansion

For \(n \geq 1\), \(a_n = \frac{2}{L} \left( \int_0^1 x \cos\left(\frac{n\pi x}{2}\right) \, dx + \int_1^2 \cos\left(\frac{n\pi x}{2}\right) \, dx \right)\). Solve each integral using integration by parts and computing specifically from these bounds.

Step 5: Setting Up the Half-Range Sine Series

The half-range sine series uses odd extension, implying \(f(x) = -f(-x)\). Each coefficient \(b_n\) is calculated with \(b_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx\). With \(L=2\), compute \(b_n = \int_0^1 x \sin\left(\frac{n\pi x}{2}\right) \, dx + \int_1^2 \sin\left(\frac{n\pi x}{2}\right) \, dx\).

Step 6: Calculate \(b_n\) for Sine Expansion

Use integration by parts for these integrals. Compute \(b_n = \frac{2}{2} \left( \int_0^1 x \sin\left(\frac{n\pi x}{2}\right) \, dx + \int_1^2 \sin\left(\frac{n\pi x}{2}\right) \, dx \right)\). Similar to \(a_n\), calculate each part and compile into series form.

Key Concepts

Half-Range Expansions

Half-range expansions are a technique used to simplify Fourier series by focusing on a specific half of the interval, either the positive or negative half. This is especially helpful for functions that display symmetry across an interval. When we deal with a given function defined on an interval from 0 to a particular value \(L\), rather than from \(-L\) to \(L\), we form either a cosine or sine series as a half-range expansion.

• Cosine Series: This involves an even extension where the function mirrors itself evenly across the interval.
• Sine Series: This is used for odd extensions and reflects the function in an odd manner across the interval.

In practice, we extend the function's definition to benefit from the simplicity of sine or cosine representations, reducing computational difficulty.

Cosine Series

A cosine series is leveraged when extending a function in an even manner, leading to a symmetry about the vertical axis. This is particularly useful for periodic functions observed in physics and engineering.

• Even Extension: In the cosine series, the assumption \(f(x) = f(-x)\) helps in simplifying the Fourier expansion by focusing solely on cosine terms.
• Coefficient Calculations: For a function defined over \([0, L]\), the coefficients \(a_n\) are calculated as: \[a_n = \frac{2}{L} \int_0^L f(x) \cos\left(\frac{n\pi x}{L}\right) \, dx\]

This formula splits the integral into convenient pieces based on the piecewise form of the function. These computational steps help create an accurate series representation for practical uses.

Sine Series

The sine series involves extending a function using an odd pattern, resulting in symmetry about the origin. Such extensions are crucial for odd periodic functions.

• Odd Extension: This series applies when \(f(x) = -f(-x)\), capturing symmetry around zero and simplifying representations solely using sine terms.
• Coefficient Formulation: Similar to the cosine series, the sine series uses: \[b_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx\]

In practical application, each integral in the formula is calculated by considering the function's specific formula on different intervals. Through precise computation, we achieve simplified sine series representations, enhancing many analyses in science and engineering.

Integration by Parts

Integration by parts is a fundamental technique used to integrate products of functions. It's especially relevant when solving Fourier series problems, as shown in the computation of coefficients for half-range expansions.
This method is based on the product rule for differentiation and is represented as:\[\int u \, dv = uv - \int v \, du\]

• Choosing Functions: Selecting \(u\) and \(dv\) from the given integral is crucial. Typically, you would let \(u\) be a polynomial and \(dv\) as trigonometric functions, like sine or cosine.
• Application in Series: Integration by parts allows simplification of more complex integrals in cosine or sine series, leading to manageable computations for determining series coefficients.

By systematically applying integration by parts, one can break down intricate integrals into simpler parts, effectively facilitating the completion of Fourier expansions.