Problem 14 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The Fourier coefficients are calculated, and the series is formed as shown.

Step 1: Calculate Constant Term

Let's first calculate the constant term \(a_0\) in the Fourier series. The given equation gives us:\[a_0 = \frac{1}{2} \int_{-2}^{2} f(x) \, dx = \frac{1}{2}\left(\int_{-2}^{0}(2+x) \, dx + \int_{0}^{2} 2 \, dx\right) = 3\]Hence, the constant term is 3.

Step 2: Calculate Coefficient \(a_n\)

Next, we use the formula for \(a_n\):\[a_n = \frac{1}{2} \int_{-2}^{2} f(x) \, \cos \frac{n \pi}{2} x \, dx\]Breaking it down, we have:\[a_n = \frac{1}{2}\left(\int_{-2}^{0}(2+x) \cos \frac{n \pi}{2} x \, dx + \int_{0}^{2} 2 \cos \frac{n \pi}{2} x \, dx\right) = \frac{2}{n^{2} \pi^{2}}\left[1-(-1)^{n}\right]\]This coefficient depends on \(n\) and reflects how much each cosine term contributes to the series.

Step 3: Calculate Coefficient \(b_n\)

Now, we'll determine the coefficients for the sine terms using:\[b_n = \frac{1}{2} \int_{-2}^{2} f(x) \, \sin \frac{n \pi}{2} x \, dx\]Calculating this, we find:\[b_n = \frac{1}{2}\left(\int_{-2}^{0}(2+x) \sin \frac{n \pi}{2} x \, dx + \int_{0}^{2} 2 \sin \frac{n \pi}{2} x \, dx\right) = \frac{2}{n \pi}(-1)^{n+1}\]This coefficient reflects the contribution from the sine terms based on \(n\).

Step 4: Construct the Fourier Series

Using the values calculated for \(a_0\), \(a_n\), and \(b_n\), we build the Fourier series representation for \(f(x)\):\[f(x) = \frac{3}{2} + \sum_{n=1}^{\infty}\left[\frac{2}{n^{2} \pi^{2}}\left[1-(-1)^{n}\right] \cos \frac{n \pi}{2} x + \frac{2}{n \pi}(-1)^{n+1} \sin \frac{n \pi}{2} x\right]\]This series representation expresses \(f(x)\) in terms of its cosine and sine components, demonstrating a way to reconstruct the function from its harmonics.

Key Concepts

Cosine Series

The cosine series is a type of Fourier series, which uses only cosine terms to describe a function. In our exercise, the coefficient \(a_n\) represents the contribution of each cosine term to the series. These coefficients help in reconstructing the even parts of a function. The formula used to calculate these coefficients is:

• \(a_n = \frac{1}{2} \int_{-2}^{2} f(x) \cos \frac{n \pi}{2} x \, dx\)

Here, \(n\) is an integer that indicates the frequency of the cosine term. Higher \(n\) leads to higher frequency components in the reconstructed function. The cosine series is especially useful for analyzing functions that are symmetrical around the vertical axis.
Breaking down the integral into two parts simplifies the calculation, taking advantage of the symmetry of the cosine function. This approach ensures that the coefficients are derived accurately, reflecting how much each frequency component contributes to the original function.

Sine Series

The sine series is another form of Fourier series focusing exclusively on sine functions. In our example, the coefficients \(b_n\) reflect how sine terms contribute to reconstructing the function.

• \(b_n = \frac{1}{2} \int_{-2}^{2} f(x) \sin \frac{n \pi}{2} x \, dx\)

These coefficients identify the influence of the sine waves, which often illustrate the odd parts of a function. The sine function, being an odd function itself, lends to describing asymmetrical features in a function.This fractional integral considers the function over symmetric limits, thus accounting for any shifts or peculiarities arising within the given range. The results, indicated by a change in sign with powers of \((-1)^n\), provide insight into phase shifts or oscillation characteristics deeply tied to the sine components.

Integration Techniques

Integration is crucial in calculating Fourier coefficients. The integrals in this exercise are split into manageable parts. For the constant term \(a_0\), and coefficients \(a_n\) and \(b_n\), breaking the interval \([-2,2]\) into \([-2,0]\) and \([0,2]\) makes the computations straightforward.

• For \(a_0\), the integral of a simple polynomial \((2+x)\) over these intervals yields the constant term.
• With \(a_n\) and \(b_n\), integration of products of the original function and trigonometric functions is required.

The use of symmetry and properties of sine and cosine simplifies these integrations. Techniques such as substitution and integration by parts are implicit in handling trigonometric integrals. Splitting intervals takes advantage of zero values outside non-overlapping regions owing to sine and cosine functions' properties. This structured approach mitigates errors and eases the solution process, ensuring each coefficient reflects the desired harmonic component.

Harmonic Analysis

Harmonic analysis involves breaking down a function into basic trigonometric components: sines and cosines. This approach allows for a deep understanding of a function’s behavior through its frequency components.
In our example, the function is expressed as a sum of cosine and sine terms with coefficients derived from integration.

• The constant term adds a base value to the function, accounting for its average over the interval.
• Cosine terms (explained by \(a_n\) ) detail the function's even frequency components.
• Sine terms (explained by \(b_n\)) depict the odd frequency components.

Harmonic analysis is foundational in fields like signal processing, vibration analysis, and acoustics. It aids in comprehending how oscillations and fluctuations manifest in physical and theoretical systems. By reconstructing a function from its harmonic components, one attains insights into periodicity and wave behavior, proving invaluable in diverse scientific and engineering applications.