Problem 46 - Thomas Calculus

Question

Fourier series A finite Fourier series is given by the sum $$\begin{aligned} f(x) &=\sum_{n=1}^{N} a_{n} \sin n x \\ &=a_{1} \sin x+a_{2} \sin 2 x+\cdots+a_{N} \sin N x \end{aligned}$$ Show that the \(m\) th coefficient \(a_{m}\) is given by the formula $$a_{m}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin m x d x$$

Step-by-Step Solution

Answer

The coefficient \( a_m \) is given by \( a_m = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin mx \; dx \).

Step 1: Understand the Fourier Series

A Fourier series is a way to represent a function as a sum of sinusoidal components. This particular Fourier series represents a function \( f(x) \) as a sum of sine terms with increasing frequencies: \( f(x) = a_1 \sin x + a_2 \sin 2x + \dots + a_N \sin Nx \). Each term in the series has a coefficient \( a_n \) corresponding to the amplitude of the sine wave.

Step 2: Define the Goal

We need to find the formula for the coefficient \( a_m \) given the finite Fourier series representation: \( f(x) = \sum_{n=1}^{N} a_n \sin nx \). We are to show that \( a_m = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin mx \; dx \).

Step 3: Use Orthogonality of Sine Functions

Sine functions have orthogonal properties over the interval \([-\pi, \pi]\). Specifically, the integral of the product of two different sine functions over \([-\pi, \pi]\) is zero: \( \int_{-\pi}^{\pi} \sin nx \sin mx \; dx = 0 \) if \( n eq m \). This property allows us to isolate individual coefficients in the Fourier series.

Step 4: Apply Orthogonality to Isolate \(a_m\)

Multiply both sides of the Fourier series by \( \sin mx \) and integrate over \([-\pi, \pi]\): \[ \int_{-\pi}^{\pi} f(x) \sin mx \; dx = \int_{-\pi}^{\pi} \left( \sum_{n=1}^{N} a_n \sin nx \right) \sin mx \; dx \]. Due to orthogonality, all terms for which \(n eq m\) vanish, leaving \( a_m \int_{-\pi}^{\pi} \sin^2 mx \; dx \).

Step 5: Evaluate the Integral of \( \sin^2 mx \)

The integral \( \int_{-\pi}^{\pi} \sin^2 mx \; dx = \pi \) can be found using a trigonometric identity or by recognizing that the average value of \( \sin^2 x \) over any full period is \(\frac{1}{2} \), yielding \( \pi \) when integrated over \([-\pi, \pi]\).

Step 6: Solve for \(a_m\)

From Step 4, we have \( a_m \cdot \pi = \int_{-\pi}^{\pi} f(x) \sin mx \; dx \). Solving for \( a_m \), we get: \[ a_m = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin mx \; dx \]. This confirms the given formula for \( a_m \).

Key Concepts

Orthogonality of sine functions

In mathematics, orthogonality is an important concept that helps simplify the process of finding coefficients in Fourier series. It's a bit like saying that two functions are "independent" from one another.
The sine functions you encounter have this special property: over the interval \([-\pi, \pi]\), the integral of the product of two different sine functions is zero. Mathematically, this is expressed as: \[\int_{-\pi}^{\pi} \sin nx \sin mx \; dx = 0 \quad \text{for} \quad n eq m\]
This zero value means that different sine waves don't overlap in this interval.

• This property allows us to find any specific coefficient in the Fourier series. By isolating terms with one particular sine function, we can get rid of all others.
• The only function that affects the integral is the one matched frequency. This makes solving for coefficients like \(a_m\) easier, as demonstrated in the problem.

Understanding orthogonality is key to working effectively with Fourier series.

Integration

Integration is the process we use to sum up small parts to find the whole. In the context of Fourier series, we use integration to calculate coefficients like \(a_m\).
This process typically involves the integral over one period of the function, often from \(-\pi\) to \(\pi\) for trigonometric functions. In this exercise, the integral \[\int_{-\pi}^{\pi} f(x) \sin mx \; dx\] is crucial.

• This integral calculates the effect of each sine function component on the total function \(f(x)\).
• Orthogonality helps simplify this process, as only the component with matching frequency (i.e., \(m = n\)) remains significant.
• The result gives the coefficient \(a_m\), necessary for reconstructing the original function using the Fourier series.

Integration simplifies and encapsulates the complex interaction of different frequencies within the total function.

Trigonometric identities

Trigonometric identities are formulas that involve trigonometric functions, making calculations easier when working with them. In the context of Fourier series, such identities come in handy.
For example, regarding the integral of \(\sin^2(mx)\), an identity tells us: \[\sin^2 x = \frac{1 - \cos 2x}{2}\]
Using this can help in evaluating integrals. In our exercise, this identity aids in finding: \[\int_{-\pi}^{\pi} \sin^2(mx) \; dx = \pi\]

• Since the integral of sine squared over the interval yields \(\pi\), we use it to solve for \(a_m\).
• This simplifies calculations and helps verify the periodic nature of sine and cosine functions.

Recognizing and utilizing trigonometric identities are fundamental for solving Fourier series problems efficiently.