Question

Find the Fourier series of \(f(x)=(x-\pi) \sin x\) on \([-\pi, \pi]\).

Short answer

Answer: The Fourier series of the function $(x-\pi) \sin x$ on $[-\pi, \pi]$ is the zero function.

Step-by-step solution

1. Formula to compute Fourier coefficients

The Fourier coefficients \(a_n\) and \(b_n\) can be computed using the following formulas: $$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx$$ $$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx$$ We will compute these coefficients for our given function \(f(x) = (x-\pi) \sin x\).

2. Compute the Fourier coefficients

First, let's compute \(a_n\). We have $$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (x-\pi) \sin x \cos(nx) dx$$ We can observe that the integrand is an odd function since \((x-\pi)\sin x \cos(nx)\) is odd, so this integral is zero. Therefore, \(a_n = 0\) for all \(n\). Now, let's compute \(b_n\). We have $$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (x-\pi) \sin x \sin(nx) dx$$ We can break this integral into two smaller integrals and integrate by parts: $$b_n = \frac{1}{\pi} \left( \int_{-\pi}^{\pi} x \sin x \sin(nx) dx - \int_{-\pi}^{\pi} \pi \sin x \sin(nx) dx \right)$$ Let's compute the first integral using integration by parts: $$u = x, dv = \sin x \sin(nx) dx$$ $$du = dx, v = -\frac{1}{n} \cos x \sin(nx)$$ Hence, the first integral becomes $$\int_{-\pi}^{\pi} x \sin x \sin(nx) dx = \left[ -\frac{1}{n} x \cos x \sin(nx) \right]_{-\pi}^{\pi} + \frac{1}{n} \int_{-\pi}^{\pi} \cos x \sin(nx) dx$$ The second term of the right expression is zero since \(\cos x \sin(nx)\) is an odd function. Moreover, the first term is also zero since \(-\frac{1}{n} x \cos x \sin(nx)\) is zero at both \(x=\pi\) and \(x=-\pi\). Thus, we have $$\int_{-\pi}^{\pi} x \sin x \sin(nx) dx = 0$$ Now let's compute the second integral: $$\int_{-\pi}^{\pi} \pi \sin x \sin(nx) dx = \pi \int_{-\pi}^{\pi} \sin x \sin(nx) dx$$ Since \(\sin x \sin(nx)\) is odd, this integral is zero as well. Thus, we have \(b_n = 0\) for all \(n\).

3. Write down the Fourier series

Since all the Fourier coefficients \(a_n\) and \(b_n\) are zero, the Fourier series of the given function \((x-\pi) \sin x\) is simply the zero function: $$f(x) \approx \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)) = 0$$ So, the Fourier series of \(f(x) = (x-\pi) \sin x\) on \([-\pi, \pi]\) is the zero function.

Key concepts

Fourier coefficients

Fourier coefficients play a fundamental role in forming a Fourier series. They help us understand how a periodic function behaves by breaking it down into a sum of simpler trigonometric functions. Essentially, these coefficients determine the weights or amplitudes of the cosine and sine waves that compose the original function.
For any periodic function, the Fourier coefficients are calculated using specific integrals over a period. In this exercise:

• The coefficient \(a_n\) is derived from the integral of the function multiplied by \(\cos(nx)\).
• The coefficient \(b_n\) is found using the integral of the function times \(\sin(nx)\).

When computed, these coefficients dictate the Fourier series' formation. In our example, because our function is symmetric and has specific properties (odd), the coefficients \(a_n\) and \(b_n\) all result in zero, leading the Fourier series of the function to also be zero.

Odd function

Understanding the properties of odd functions is key in this exercise. An odd function is symmetric around the origin, meaning that if \(f(x)\) is an odd function, then \(f(-x) = -f(x)\). This property is quite useful when evaluating integrals over symmetric intervals like \([-\pi, \pi]\).
In this context, when we integrate a product involving an odd function over a symmetric interval, the result is zero. This is precisely why our integral for the Fourier coefficients \(a_n\) and the adjusted terms in the \(b_n\) computation turned out to be zero. Recognizing odd functions helps us simplify calculations significantly, as it allows us to skip manual integration in certain cases and directly conclude that the integral evaluates to zero.

Integration by parts

Integration by parts is a technique used to solve more complex integrals by breaking them down into simpler parts. It's particularly handy when dealing with products of functions, like in our situation of integrating \((x-\pi)\sin x \sin(nx)\).
The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \]In this exercise, by selecting \(u = x\) and \(dv = \sin x \sin(nx) \, dx\), we can transform the integral into a more manageable form. After differentiating \(u\) and integrating \(dv\), we apply the formula which simplifies the problem. In doing so, we’re able to reduce the original integral step by step. Any terms that further remain are analyzed based on the properties of the function, like symmetry, to conclude if they are zero. This systematic approach is integral in solving problems with complex trigonometric forms.

Trigonometric integrals

Trigonometric integrals involve integrating functions with sine and cosine. They often appear in the calculation of Fourier coefficients within the Fourier series. In many cases, these integrals can be efficiently solved by leveraging certain properties of sine and cosine functions, like orthogonality and symmetry.
In this exercise, the integrals involve products like \((x-\pi) \sin x \sin(nx)\) which can be tricky. The zero result for \(b_n\) confirmed that the expression was either inherently odd or effectively canceled out over the symmetric interval. This phenomenon is part of the 'orthogonality' concept, where sine and cosine functions over a full period return zero when combining in specific ways.
Understanding these underlying properties aids in solving such integrals more effectively, reducing the complexity of the problem-solving process. So when dealing with Fourier series, knowing how to handle and evaluate trigonometric integrals is crucial for determining the series’ comprehensive elements.