Question

Find the Fourier sine series. $$ f(x)=x \sin x ; \quad[0, \pi] $$

Short answer

Answer: The Fourier sine series representation of the function f(x) = x * sin(x) in the interval [0, π] is given by: $$ f(x) = \sum_{n=1}^{\infty} \left(-\frac{1}{\pi}\left[\frac{2}{(n-1)^2} - \frac{2}{(n+1)^2}\right]\right) \sin(nx) $$

Step-by-step solution

Understand the Fourier sine series formula

The Fourier sine series of a function f(x) is represented as: $$ f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right) $$ where L is half the length of the interval (in this case, L = π) and the coefficients \(b_n\) are given by: $$ b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right)dx $$

Find the bn coefficients

Using the formula for \(b_n\) and substituting the given function f(x) = x * sin(x), we have: $$ b_n = \frac{2}{\pi} \int_{0}^{\pi} x \sin x \sin\left(\frac{n\pi x}{\pi}\right)dx $$ Simplify the expression: $$ b_n = \frac{2}{\pi} \int_{0}^{\pi} x \sin x \sin(nx) dx $$

Compute the bn integral

To find the integral, we will use integration by parts. Let: $$ u = x \Rightarrow du = dx \\ dv = \sin x \sin nx \Rightarrow v = \int \sin x \sin nx dx $$ To compute the integral for v, use the product-to-sum identity: $$ \sin x \sin nx = \frac{1}{2}\left[\cos\left((n-1)x\right) - \cos\left((n+1)x\right)\right] $$ Now, integrate with respect to x: $$ v = \frac{1}{2}\int\left[\cos\left((n-1)x\right) - \cos\left((n+1)x\right)\right]dx = \frac{1}{2}\left[\frac{1}{n-1}\sin\left((n-1)x\right) - \frac{1}{n+1}\sin\left((n+1)x\right)\right] $$ Now apply integration by parts formula: $$ b_n = \frac{2}{\pi} \left[u\cdot v \bigg|_0^{\pi} - \int_{0}^{\pi}v\cdot du\right] $$ Substitute the values of u, v, and du: $$ b_n = \frac{2}{\pi} \left[\left(\frac{1}{2}\pi\left[\frac{1}{n-1}\sin\left((n-1)\pi\right) - \frac{1}{n+1}\sin\left((n+1)\pi\right)\right] - 0\right) - \int_{0}^{\pi}\frac{1}{2}\left[\frac{1}{n-1}\sin\left((n-1)x\right) - \frac{1}{n+1}\sin\left((n+1)x\right)\right]dx\right] $$ The sine terms in the first part of the expression will be zero since sine of any multiple of π is zero. Therefore: $$ b_n = -\frac{1}{\pi} \int_{0}^{\pi}\left[\frac{1}{n-1}\sin\left((n-1)x\right) - \frac{1}{n+1}\sin\left((n+1)x\right)\right]dx $$ Integrate with respect to x: $$ b_n = -\frac{1}{\pi} \left[\frac{1}{(n-1)^2}\left[\cos\left((n-1)x\right) - 1\right] - \frac{1}{(n+1)^2}\left[\cos\left((n+1)x\right) - 1\right]\right]_0^{\pi} $$ Simplify the expression: $$ b_n = -\frac{1}{\pi}\left[\frac{2}{(n-1)^2} - \frac{2}{(n+1)^2}\right] $$

Write the Fourier sine series

Now that we have found the \(b_n\) coefficients, we can write the Fourier sine series for f(x) = x * sin(x): $$ f(x) = \sum_{n=1}^{\infty} \left(-\frac{1}{\pi}\left[\frac{2}{(n-1)^2} - \frac{2}{(n+1)^2}\right]\right) \sin(nx) $$ This is the Fourier sine series representation of the function f(x) = x * sin(x) in the interval [0, π].

Key concepts

Integration by Parts

Integration by parts is a technique ideal for solving integrals that are products of two functions. It simplifies the integral by allowing you to differentiate one function while integrating the other. The core formula used in integration by parts is: \[\int u\, dv = uv - \int v\, du\]where:

• \( u \) and \( dv \) are parts chosen from the integral.
• \( du \) is the derivative of \( u \).
• \( v \) is the integral of \( dv \).

This method is very useful in handling integrals where one term reduces upon differentiation, thus simplifying the integrand. In our exercise, we used it on \( x \sin x \cdot \sin(nx) \) by setting \( u = x \) and \( dv = \sin x \sin(nx) dx \). This way, we successfully reduced the complexity of the integral.

Product-to-Sum Identity

The product-to-sum identity is an identity in trigonometry that helps simplify the product of two trigonometric functions into a sum of simpler ones. For sine functions, it is expressed as:\[\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]\]This identity allows us to express the product of sines, \( \sin x \sin(nx) \), as a difference of cosines:

• \( \sin x \sin(nx) = \frac{1}{2} [\cos((n-1)x) - \cos((n+1)x)] \)

By doing this transformation, integrals that are initially complex products of trigonometric functions become manageable sums. This is particularly useful when computing Fourier coefficients because it simplifies the integration process.

Trigonometric Integrals

Trigonometric integrals are integrals that involve trigonometric functions. They often require special techniques like substitution, integration by parts, or using identities to solve. In our exercise, dealing with integrals like \( \int \sin x \sin(nx) dx \), we first used the product-to-sum identity to break the product into a sum of cosines.The typical approach with these integrals involves:

• Applying trigonometric identities to simplify the expression.
• Using integration by parts for products involving linear terms and trigonometric functions.
• Evaluating definite integrals by calculating boundary terms and using limits.

In the context of Fourier series problems, tackling trigonometric integrals allows us to determine the coefficients \( b_n \) necessary for constructing the series.

Fourier Coefficients

Fourier coefficients are essential in breaking down functions into their sinusoidal components. In a Fourier sine series, the coefficient \( b_n \) represents the contribution of the sine function at frequency \( n \). Calculating these coefficients involves solving integrals with the formula:\[b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx\]For the function \( f(x) = x \sin x \), and over an interval \([0, \pi]\), we set \( L = \pi \) and calculated \( b_n \) by evaluating the integral:

• Involves using techniques like substitution and integration by parts.
• Allows decomposition into simpler periodic components representing the original function.

These coefficients are the building blocks of the series, ensuring that the infinite sum accurately approximates the target function within the given interval.