Problem 11 - Advanced Engineering Mathematics

Question

$$\begin{array}{l} a_{0}=\frac{1}{2} \int_{-2}^{2} f(x) d x=\frac{1}{2}\left(\int_{-1}^{0}-2 d x+\int_{0}^{1} 1 d x\right)=-\frac{1}{2} \\ a_{n}=\frac{1}{2} \int_{-2}^{2} f(x) \cos \frac{n \pi}{2} x d x=\frac{1}{2}\left(\int_{-1}^{0}(-2) \cos \frac{n \pi}{2} x d x+\int_{0}^{1} \cos \frac{n \pi}{2} x d x\right)=-\frac{1}{n \pi} \sin \frac{n \pi}{2} \\ b_{n}=\frac{1}{2} \int_{-2}^{2} f(x) \sin \frac{n \pi}{2} x d x=\frac{1}{2}\left(\int_{-1}^{0}(-2) \sin \frac{n \pi}{2} x d x+\int_{0}^{1} \sin \frac{n \pi}{2} x d x\right)=\frac{3}{n \pi}\left(1-\cos \frac{n \pi}{2}\right) \\ f(x)=-\frac{1}{4}+\sum_{n=1}^{\infty}\left[-\frac{1}{n \pi} \sin \frac{n \pi}{2} \cos \frac{n \pi}{2} x+\frac{3}{n \pi}\left(1-\cos \frac{n \pi}{2}\right) \sin \frac{n \pi}{2} x\right] \end{array}$$

Step-by-Step Solution

Answer

The function is a Fourier series with terms calculated for average value, cosine, and sine components over the specified range.

Step 1: Evaluate the Constant Term

The first given term is \( a_0 = -\frac{1}{2} \). This term represents the average value of the function \( f(x) \) over the interval \(-2\) to \(2\). It is calculated by finding the integral of \( f(x) \) from \(-2\) to \(2\), and then dividing by \(2\). Thus, the function seems to have an average value of \(-\frac{1}{2} \) in this interval.

Step 2: Compute the Coefficient \( a_n \)

The coefficient \( a_n \) is provided as \(-\frac{1}{n \pi} \sin\left(\frac{n \pi}{2}\right)\). This term involves an integral of the function multiplied by \( \cos\left(\frac{n \pi}{2} x\right) \). This part suggests that it accounts for the cosine terms in the Fourier series, and only the sine component of the integral contributes, as delineated that the cosine factors cancel out due to the intervals chosen.

Step 3: Compute the Coefficient \( b_n \)

Here, \( b_n \) is calculated as the expression \( \frac{3}{n \pi}\left(1 - \cos\left(\frac{n \pi}{2}\right)\right) \). This indicates the sine terms in the Fourier series expansion. The term \(1 - \cos\left(\frac{n \pi}{2}\right)\) enhances the sine components depending on \( n \), suggesting it accounts for odd harmonics due to its value behavior (becoming zero sometimes).

Step 4: Analyze the Function Representation

The function \( f(x) \) is represented as: \( f(x) = -\frac{1}{4} + \sum_{n=1}^{\infty}\left[-\frac{1}{n \pi} \sin\left(\frac{n \pi}{2}\right) \cos\left(\frac{n \pi}{2} x\right) + \frac{3}{n \pi}\left(1 - \cos\left(\frac{n \pi}{2}\right)\right) \sin\left(\frac{n \pi}{2} x\right)\right] \). This sum combines the effect of both sine and cosine components to fully describe \( f(x) \) over the interval \(-2\) to \(2\), making it a periodic function approximated via Fourier series.

Key Concepts

Fourier Coefficients

Fourier coefficients are essential in expressing a periodic function as a sum of sine and cosine functions. In Fourier series, you typically find three types of coefficients: the constant term (often labeled as \(a_0\)), and the coefficients for cosine and sine terms (\(a_n\) and \(b_n\), respectively). These coefficients are calculated using integral calculus over a specified interval.

The constant term \(a_0\) gives the average value of the function across the interval. For our problem, this is calculated to be \(-\frac{1}{2}\), indicating the overall shift of the function downward in this range. The formula for \(a_0\) is:

• \( a_0 = \frac{1}{2L} \int_{-L}^{L} f(x)\, dx \), where \(L\) is half the period of the function.

For non-zero coefficients \(a_n\) and \(b_n\), you find these by integrating the product of the function with cosine or sine functions over the same range.

• \(a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n \pi}{L}x\right) dx\)
• \(b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n \pi}{L}x\right) dx\)

These coefficients are crucial for accurately constructing the original function using its Fourier series.

Sine and Cosine Functions

Sine and cosine functions play a critical role in Fourier series, which allows periodic functions to be represented in terms of these trigonometric functions. In such series, sine functions are used to model the odd properties and cosine functions are for the even properties of the periodic function.

Each term in the Fourier series specifically utilizes these functions to capture distinct harmonics of the original function. The cosine term, \(a_n \cos\left(\frac{n \pi}{L}x\right)\), and the sine term, \(b_n \sin\left(\frac{n \pi}{L}x\right)\), contribute differently:

• Cosine terms (\(a_n\)) describe variations in phase or amplitude that are symmetric around some vertical line.
• Sine terms (\(b_n\)) capture asymmetries or changes that are antisymmetric.

Understanding the behavior of these functions helps in predicting how each contributes to the series, essentially reconstructing the periodic pattern defined by the function over the given interval.

Integral Calculus

Integral calculus is the backbone of finding Fourier coefficients. Each coefficient \(a_0\), \(a_n\), and \(b_n\) involves integrating the product of the function with sine or cosine terms over a complete interval. This process leverages the powerful principles of calculus to "weigh" parts of the function with respect to sine and cosine waves.

When computing integrals for these coefficients, the properties of sine and cosine functions are exploited, notably their orthogonality over a full period. This means that the integral of a sine function multiplied by a different sine or cosine function over its period results in zero. Only when they are the same type (either sine with sine of the same frequency or cosine with cosine) do these products yield non-zero results.

This is why the integration steps in the Fourier analysis are significant—they selectively extract (or "amplify") those parts of the function that match the specific wave patterns of sine and cosine. Understanding this principle through the use of integral calculus deepens your insight into how Fourier series dissect and rebuild functions from basic trigonometric components.