Expanding X Sin(x) Into A Fourier Sine Series A Comprehensive Guide

Jul 16, 2025 by ADMIN 68 views

Expanding x * sin(x) into a Fourier Sine Series Over (0, π)

Expanding a function into a Fourier series is a powerful technique in mathematics and engineering, allowing us to represent complex functions as a sum of simpler trigonometric functions. In this article, we'll delve into the process of expanding the function f(x) = x * sin(x) into a Fourier sine series over the interval (0, π). This means we'll express x * sin(x) as an infinite sum of sine functions with different frequencies and amplitudes. Understanding this process requires a grasp of Fourier series, integration techniques, and some algebraic manipulation. Let's embark on this mathematical journey to see how this expansion unfolds.

Understanding Fourier Series

Before we dive into the specifics of expanding x * sin(x), let's first establish a solid understanding of Fourier series in general. Fourier series are a way to represent periodic functions as an infinite sum of sines and cosines. The fundamental idea behind Fourier series is that any periodic function, no matter how complex, can be decomposed into a sum of simple harmonic oscillations. This decomposition allows us to analyze and manipulate complex functions more easily by working with their constituent sine and cosine components.

The General Form of a Fourier Series

The general form of a Fourier series for a function f(x) defined on the interval (-L, L) is given by:

f(x) = a₀/2 + Σ[n=1 to ∞] (aₙ * cos(nπx/L) + bₙ * sin(nπx/L))

Where a₀, aₙ, and bₙ are the Fourier coefficients, which determine the amplitudes of the cosine and sine terms. These coefficients are calculated using the following formulas:

a₀ = (1/L) ∫[-L to L] f(x) dx
aₙ = (1/L) ∫[-L to L] f(x) cos(nπx/L) dx
bₙ = (1/L) ∫[-L to L] f(x) sin(nπx/L) dx

These integrals essentially project the function f(x) onto the cosine and sine basis functions, giving us the coefficients that represent the contribution of each frequency component.

Fourier Sine Series

In our case, we are interested in a Fourier sine series, which is a special case of the Fourier series where we only use sine terms. This is particularly useful for functions that are odd or that we want to represent as odd extensions over a symmetric interval. The Fourier sine series is given by:

f(x) = Σ[n=1 to ∞] bₙ * sin(nπx/L)

Here, the coefficients bₙ are calculated as:

bₙ = (2/L) ∫[0 to L] f(x) sin(nπx/L) dx

Notice that we are integrating over the interval (0, L) instead of (-L, L), and the factor in front of the integral is 2/L instead of 1/L. This is because we are exploiting the symmetry of the sine function and only considering the positive half of the interval. In the specific problem we are addressing, the interval is (0, π), so L = π. This simplifies our coefficient calculation to:

bₙ = (2/π) ∫[0 to π] f(x) sin(nx) dx

Now that we have a solid understanding of Fourier series and Fourier sine series, we can proceed to apply this knowledge to our specific function, f(x) = x * sin(x).

Calculating the Fourier Sine Coefficients for x * sin(x)

To expand f(x) = x * sin(x) into a Fourier sine series over the interval (0, π), the crucial step is to calculate the Fourier sine coefficients bₙ. As we established earlier, these coefficients are given by:

bₙ = (2/π) ∫[0 to π] x * sin(x) * sin(nx) dx

This integral represents the projection of our function x * sin(x) onto the sine basis functions. Evaluating this integral requires a bit of trigonometric manipulation and integration techniques. Let's break down the process step by step.

Trigonometric Identity

First, we can use the trigonometric identity:

sin(a) * sin(b) = (1/2) [cos(a - b) - cos(a + b)]

Applying this identity to our integral, we get:

bₙ = (2/π) ∫[0 to π] x * (1/2) [cos((n - 1)x) - cos((n + 1)x)] dx

Simplifying, we have:

bₙ = (1/π) ∫[0 to π] x * [cos((n - 1)x) - cos((n + 1)x)] dx

Now we have two integrals to evaluate:

bₙ = (1/π) [∫[0 to π] x * cos((n - 1)x) dx - ∫[0 to π] x * cos((n + 1)x) dx]

Integration by Parts

To evaluate these integrals, we'll use integration by parts. Recall the formula for integration by parts:

∫ u dv = uv - ∫ v du

For the first integral, ∫[0 to π] x * cos((n - 1)x) dx, let:

u = x => du = dx
dv = cos((n - 1)x) dx => v = sin((n - 1)x) / (n - 1) (for n ≠ 1)

Applying integration by parts, we get:

∫[0 to π] x * cos((n - 1)x) dx = [x * sin((n - 1)x) / (n - 1)][0 to π] - ∫[0 to π] sin((n - 1)x) / (n - 1) dx

The first term evaluates to zero because sin((n - 1)π) = 0 and sin(0) = 0. The second integral evaluates to:

∫[0 to π] sin((n - 1)x) / (n - 1) dx = [cos((n - 1)x) / (n - 1)²][0 to π] = [cos((n - 1)π) - 1] / (n - 1)²

Similarly, for the second integral, ∫[0 to π] x * cos((n + 1)x) dx, let:

u = x => du = dx
dv = cos((n + 1)x) dx => v = sin((n + 1)x) / (n + 1)

Applying integration by parts, we get:

∫[0 to π] x * cos((n + 1)x) dx = [x * sin((n + 1)x) / (n + 1)][0 to π] - ∫[0 to π] sin((n + 1)x) / (n + 1) dx

Again, the first term evaluates to zero. The second integral evaluates to:

∫[0 to π] sin((n + 1)x) / (n + 1) dx = [-cos((n + 1)x) / (n + 1)²][0 to π] = [1 - cos((n + 1)π)] / (n + 1)²

Combining the Results

Now, substituting these results back into our expression for bₙ, we get (for n ≠ 1):

bₙ = (1/π) [[[cos((n - 1)π) - 1] / (n - 1)²] - [[1 - cos((n + 1)π)] / (n + 1)²]]

We can simplify this further by noting that cos(kπ) = (-1)ᵏ, where k is an integer. Therefore:

bₙ = (1/π) [[((-1)^(n-1) - 1) / (n - 1)²] - [(1 - (-1)^(n+1)) / (n + 1)²]]

Special Case: n = 1

Our integration by parts calculation had a condition that n ≠ 1. We need to calculate b₁ separately:

b₁ = (2/π) ∫[0 to π] x * sin(x) * sin(x) dx = (2/π) ∫[0 to π] x * sin²(x) dx

Using the identity sin²(x) = (1 - cos(2x)) / 2, we get:

b₁ = (1/π) ∫[0 to π] x * (1 - cos(2x)) dx = (1/π) [∫[0 to π] x dx - ∫[0 to π] x * cos(2x) dx]

The first integral is straightforward:

∫[0 to π] x dx = [x²/2][0 to π] = π²/2

For the second integral, we use integration by parts again:

∫[0 to π] x * cos(2x) dx

Let:

u = x => du = dx
dv = cos(2x) dx => v = sin(2x) / 2

Applying integration by parts:

∫[0 to π] x * cos(2x) dx = [x * sin(2x) / 2][0 to π] - ∫[0 to π] sin(2x) / 2 dx

The first term is zero, and the second integral evaluates to:

∫[0 to π] sin(2x) / 2 dx = [-cos(2x) / 4][0 to π] = 0

Therefore, b₁ = (1/π) [π²/2 - 0] = π/2.

Constructing the Fourier Sine Series

We have now calculated the Fourier sine coefficients bₙ for our function f(x) = x * sin(x). We have a general formula for bₙ when n ≠ 1 and a specific value for b₁. Let's summarize our findings:

b₁ = π/2
bₙ = (1/π) [[((-1)^(n-1) - 1) / (n - 1)²] - [(1 - (-1)^(n+1)) / (n + 1)²]] (for n ≠ 1)

Now we can construct the Fourier sine series:

x * sin(x) = Σ[n=1 to ∞] bₙ * sin(nx)

Substituting our values for bₙ, we get:

x * sin(x) = (π/2) * sin(x) + Σ[n=2 to ∞] (1/π) [[((-1)^(n-1) - 1) / (n - 1)²] - [(1 - (-1)^(n+1)) / (n + 1)²]] * sin(nx)

This is the Fourier sine series representation of x * sin(x) over the interval (0, π). It expresses the function as an infinite sum of sine waves with varying amplitudes and frequencies. This expansion is a powerful tool for analyzing and manipulating the function in various applications.

Further Simplification

We can further simplify the expression for bₙ by considering even and odd values of n separately.

If n is even, then (-1)^(n-1) = -1 and (-1)^(n+1) = -1, so: bₙ = (1/π) [(-2) / (n - 1)² - (2) / (n + 1)²] = (-2/π) [1/(n - 1)² + 1/(n + 1)²]
If n is odd and n > 1, then (-1)^(n-1) = 1 and (-1)^(n+1) = 1, so bₙ = 0.

Therefore, our Fourier sine series becomes:

x * sin(x) = (π/2) * sin(x) - (2/π) Σ[n=2,4,6,...] [1/(n - 1)² + 1/(n + 1)²] * sin(nx)

This form highlights that only the sine terms with even multiples of x contribute to the series (besides the first term).

Convergence and Applications

The Fourier sine series we derived converges to x * sin(x) over the interval (0, π). This means that as we add more and more terms to the series, the approximation gets closer and closer to the original function. However, it's important to note that the convergence behavior at the endpoints and points of discontinuity can be more complex, a phenomenon known as the Gibbs phenomenon.

Applications of Fourier Series

Fourier series have a wide range of applications in various fields, including:

Signal processing: Analyzing and synthesizing signals, such as audio and images.
Image compression: Representing images using a smaller number of Fourier coefficients.
Solving differential equations: Transforming differential equations into algebraic equations.
Heat transfer: Modeling the flow of heat in materials.
Quantum mechanics: Describing the behavior of particles.

In the case of x * sin(x), its Fourier sine series representation can be useful in applications where this function appears as a component of a larger system. For example, it might arise in the analysis of vibrations or oscillations. The Fourier series allows us to decompose the function into its constituent frequencies, which can be crucial for understanding the system's behavior.

Conclusion

Expanding f(x) = x * sin(x) into a Fourier sine series over the interval (0, π) is a challenging but rewarding exercise. It demonstrates the power of Fourier analysis in representing functions as a sum of simpler trigonometric functions. We have seen how to calculate the Fourier sine coefficients using integration by parts and trigonometric identities, and how to construct the Fourier sine series representation. This series provides a valuable tool for analyzing and manipulating the function x * sin(x) in various applications. Understanding Fourier series is fundamental for anyone working in mathematics, physics, engineering, or related fields, as it provides a powerful framework for analyzing and solving a wide range of problems involving periodic phenomena and functions.

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