Fourier Series Representation Of A Piecewise Function

In the realm of mathematical analysis, the Fourier series stands as a powerful tool for representing periodic functions as an infinite sum of sines and cosines. This decomposition allows us to analyze complex waveforms by breaking them down into simpler, more manageable components. Understanding Fourier series is crucial in various fields, including signal processing, image analysis, and physics. In this article, we will delve into the process of finding the Fourier series representation for a given piecewise function, providing a step-by-step guide and addressing common challenges.

Understanding Fourier Series

Before diving into the specifics of our problem, it's essential to grasp the fundamental concepts behind Fourier series. A Fourier series is an expansion of a periodic function f(x) in terms of an infinite sum of sines and cosines. The general form of the Fourier series is given by:

f(x) = a₀/2 + Σ[an * cos(nx) + bn * sin(nx)], where the summation is taken from n = 1 to infinity.

Here,

• a₀, an, and bn are the Fourier coefficients, which determine the amplitude of each sine and cosine component.
• n represents the harmonic number, indicating the frequency of the component.

The coefficients a₀, an, and bn are calculated using the following integral formulas:

• a₀ = (1/π) ∫[-π, π] f(x) dx
• an = (1/π) ∫[-π, π] f(x) cos(nx) dx
• bn = (1/π) ∫[-π, π] f(x) sin(nx) dx

These formulas essentially project the function f(x) onto the basis functions (cosines and sines) to determine the contribution of each frequency component. Calculating these coefficients is the heart of finding the Fourier series representation.

Problem Statement: Finding the Fourier Series

Our objective is to determine the Fourier series for the piecewise function f(x) defined as follows:

• f(x) = -π, for -π < x ≤ 0
• f(x) = x, for 0 < x < π

This function is defined over the interval (-π, π), which is the standard interval for Fourier series representation. Piecewise functions, like the one given, are commonly encountered in applications and require careful treatment when computing the Fourier coefficients.

Step-by-Step Solution

To find the Fourier series, we need to calculate the coefficients a₀, an, and bn using the integral formulas mentioned earlier. Let's proceed step by step.

1. Calculate a₀

The coefficient a₀ represents the average value of the function over the interval (-π, π). It is calculated as follows:

a₀ = (1/π) ∫[-π, π] f(x) dx

Since f(x) is a piecewise function, we need to split the integral into two parts:

a₀ = (1/π) [∫[-π, 0] (-π) dx + ∫[0, π] (x) dx]

Now, we evaluate the integrals:

• ∫[-π, 0] (-π) dx = -π[x] from -π to 0 = -π(0 - (-π)) = -π²*
• ∫[0, π] (x) dx = [x²/2] from 0 to π = (π²/2) - (0²/2) = π²/2*

Substituting these results back into the expression for a₀:

a₀ = (1/π) [-π² + π²/2] = (1/π) [-π²/2] = -π/2

Therefore, the coefficient a₀ is equal to -π/2. This value contributes to the constant term in the Fourier series.

2. Calculate an

The coefficients an represent the amplitudes of the cosine terms in the Fourier series. They are calculated using the formula:

an = (1/π) ∫[-π, π] f(x) cos(nx) dx

Again, we split the integral into two parts based on the piecewise definition of f(x):

an = (1/π) [∫[-π, 0] (-π) cos(nx) dx + ∫[0, π] (x) cos(nx) dx]

Let's evaluate the integrals separately.

Integral 1: ∫[-π, 0] (-π) cos(nx) dx

This integral can be solved directly:

∫[-π, 0] (-π) cos(nx) dx = -π ∫[-π, 0] cos(nx) dx = -π [sin(nx)/n] from -π to 0

= -π [sin(0)/n - sin(-nπ)/n] = -π [0 - 0] = 0

Since sin(0) and sin(-nπ) are both zero, this integral evaluates to zero.

Integral 2: ∫[0, π] (x) cos(nx) dx

This integral requires integration by parts. Let u = x and dv = cos(nx) dx. Then, du = dx and v = sin(nx)/n.

Using integration by parts formula, ∫u dv = uv - ∫v du:

∫[0, π] (x) cos(nx) dx = [x sin(nx)/n] from 0 to π - ∫[0, π] (sin(nx)/n) dx

= [π sin(nπ)/n - 0 sin(0)/n] - (1/n) ∫[0, π] sin(nx) dx

Since sin(nπ) = 0 and sin(0) = 0, the first term becomes zero.

Now, we evaluate the remaining integral:

• (1/n) ∫[0, π] sin(nx) dx = -(1/n) [-cos(nx)/n] from 0 to π

= (1/n²) [cos(nπ) - cos(0)] = (1/n²) [cos(nπ) - 1]

Now, we know that cos(nπ) = (-1)^n, so the expression becomes:

(1/n²) [(-1)^n - 1]

Substituting the results of both integrals back into the expression for an:

an = (1/π) [0 + (1/n²) [(-1)^n - 1]] = (1/(πn²)) [(-1)^n - 1]

Therefore, the coefficients an are given by:

an = (1/(πn²)) [(-1)^n - 1]

This expression tells us that:

• If n is even, then (-1)^n = 1, so an = 0.
• If n is odd, then (-1)^n = -1, so an = -2/(πn²).

3. Calculate bn

The coefficients bn represent the amplitudes of the sine terms in the Fourier series. They are calculated using the formula:

bn = (1/π) ∫[-π, π] f(x) sin(nx) dx

Again, we split the integral into two parts based on the piecewise definition of f(x):

bn = (1/π) [∫[-π, 0] (-π) sin(nx) dx + ∫[0, π] (x) sin(nx) dx]

Let's evaluate the integrals separately.

Integral 1: ∫[-π, 0] (-π) sin(nx) dx

This integral can be solved directly:

∫[-π, 0] (-π) sin(nx) dx = -π ∫[-π, 0] sin(nx) dx = -π [-cos(nx)/n] from -π to 0

= (π/n) [cos(nx)] from -π to 0 = (π/n) [cos(0) - cos(-nπ)]

= (π/n) [1 - cos(nπ)] = (π/n) [1 - (-1)^n]

Integral 2: ∫[0, π] (x) sin(nx) dx

This integral requires integration by parts. Let u = x and dv = sin(nx) dx. Then, du = dx and v = -cos(nx)/n.

Using integration by parts formula, ∫u dv = uv - ∫v du:

∫[0, π] (x) sin(nx) dx = [-x cos(nx)/n] from 0 to π - ∫[0, π] (-cos(nx)/n) dx

= [-π cos(nπ)/n - 0] + (1/n) ∫[0, π] cos(nx) dx

= [-π (-1)^n / n] + (1/n) [sin(nx)/n] from 0 to π

= [-π (-1)^n / n] + (1/n) [sin(nπ)/n - sin(0)/n]

Since sin(nπ) = 0 and sin(0) = 0, the second term becomes zero.

So, the integral simplifies to:

[-π (-1)^n / n]

Substituting the results of both integrals back into the expression for bn:

bn = (1/π) [(π/n) [1 - (-1)^n] + [-π (-1)^n / n]]

bn = (1/n) [1 - (-1)^n] + (-1)^n+1 / n

bn = (1/n) - ((-1)^n / n) + (-1)^n+1 / n

bn = 1/n - ((-1)^n / n) - ((-1)^n / n)

bn = (1 - (-1)^n - (-1)^n) / n

bn = (1 - 2(-1)^n) / n

Therefore, the coefficients bn are given by:

bn = (1 - 2(-1)^n) / n

4. Construct the Fourier Series

Now that we have calculated the coefficients a₀, an, and bn, we can construct the Fourier series representation of f(x).

The Fourier series is given by:

f(x) = a₀/2 + Σ[an * cos(nx) + bn * sin(nx)], where the summation is taken from n = 1 to infinity.

Substituting the values we found:

f(x) = (-π/2)/2 + Σ[((1/(πn²)) [(-1)^n - 1]) * cos(nx) + ((1 - 2(-1)^n) / n) * sin(nx)]

f(x) = -π/4 + Σ[((1/(πn²)) [(-1)^n - 1]) * cos(nx) + ((1 - 2(-1)^n) / n) * sin(nx)]

We know that an is zero for even n, so we can rewrite the summation for an considering only odd values of n.

For odd n, an = -2/(πn²).

So, the Fourier series becomes:

f(x) = -π/4 + Σ[-2/(πn²) * cos(nx)] + Σ[((1 - 2(-1)^n) / n) * sin(nx)], where the first summation is taken over odd n from 1 to infinity, and the second summation is taken over all n from 1 to infinity.

This is the Fourier series representation of the given piecewise function f(x).

Summary of the Fourier Series

In conclusion, we have successfully determined the Fourier series representation for the given piecewise function f(x). The steps involved calculating the Fourier coefficients a₀, an, and bn using integral formulas and then substituting these coefficients into the general Fourier series formula. The resulting Fourier series is:

f(x) = -π/4 + Σ[-2/(πn²) * cos(nx)] + Σ[((1 - 2(-1)^n) / n) * sin(nx)]

Where the first summation ranges over odd positive integers n, and the second summation ranges over all positive integers n.

Conclusion

The Fourier series is a powerful tool for representing periodic functions, and understanding how to calculate the Fourier series for piecewise functions is essential in many applications. This article has provided a detailed step-by-step solution for finding the Fourier series of a specific piecewise function. By following this approach, you can tackle similar problems and gain a deeper understanding of Fourier analysis.

Fourier Series, Piecewise Function, Fourier Coefficients, Mathematical Analysis, Signal Processing, Integral Formulas, Periodic Functions, Trigonometric Series, Cosine Series, Sine Series, Integration by Parts, Odd Functions, Even Functions, Harmonic Analysis, Frequency Domain

Mathematics, Calculus, Differential Equations, Engineering Mathematics, Signal Processing, Applied Mathematics