91Ó°ÊÓ

An odd function is a type of mathematical function that has a specific symmetry. For a function to be classified as odd, it must satisfy the condition:

• For any value of x, the function fulfills the equation: \( f(-x) = -f(x) \).

This means if you were to graph an odd function, you would notice that it is symmetric about the origin. To visualize this, imagine folding the graph along the line \( y = 0 \) and the line \( x = 0 \). If the two halves match up perfectly, it's an odd function. This property is crucial when determining the type of Fourier series to apply for expansions. For the function \( f(x) = x \), which was given in our exercise, checking the condition \( f(-x) = -x = -f(x) \) shows it is indeed an odd function. As a result, it is best represented using a sine series, as sine functions are inherently odd.

Symmetry in functions plays a fundamental role in simplifying problems, particularly in Fourier analysis. There are generally two types of symmetry related to Fourier series – even and odd symmetry. Symmetry simplifies the process of finding the function's Fourier coefficients.

• Even functions: Symmetrical about the y-axis, fulfilling the condition \( f(x) = f(-x) \).
• Odd functions: Symmetrical about the origin, satisfying \( f(-x) = -f(x) \).

Leveraging these symmetries can help decide whether a function should be expanded as a cosine or sine series. Cosine functions are even, making them suitable for even functions, while sine functions are used for odd functions. By determining the symmetry of \( f(x) = x \), we can opt for the sine series, which neatly aligns with the function's inherent properties.

Integration by parts is a technique used in calculus, particularly useful for integrating products of two functions. The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

Choosing the appropriate \( u \) and \( dv \) is critical for simplifying the integration process. In our exercise, we set:

• \( u = x \) and \( dv = \sin(nx) \, dx \)
• then \( du = dx \), and \( v = -\frac{1}{n} \cos(nx) \).

Applying the integration by parts formula assists in solving the integral for the Fourier coefficient \( b_n \). In the case of the function \( x \), this technique simplifies the potentially complicated integration by reducing it into manageable parts, eventually leading to the Fourier series representation.

In computing a Fourier series, determining the coefficients is essential as they shape the series representation of the function. For a Fourier sine series, the coefficients \( b_n \) are calculated using the integral:

• \( b_n = \frac{2}{L} \int_{-L}^{L} f(x) \sin\left( \frac{n\pi x}{L} \right) dx \)

This integral gives a measure of how much each sine term contributes to the series. For our specific function \( f(x) = x \), the boundaries from \(-\pi\) to \(\pi\) define the interval, leading to

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

The integration by parts technique is applied, shown in the original solution. Interestingly, due to the properties of sine and cosine and the chosen interval's periodicity, certain integrals simplify further, resulting in coefficient values of \( b_n = \frac{2(-1)^{n+1}}{n} \). These coefficients are crucial for reconstructing the function \( f(x) = x \) using the Fourier sine series.