Question

Find the Fourier series of the function \(f(x)=x\) in the range \(-\pi

Short answer

The Fourier series is \(f(x) = \sum_{n=1,3,5,..}^{\infty}\frac{4}{n} \sin (nx)\). Evaluated sum = \(\frac{\pi}{4}\)

Step-by-step solution

Define the function and the interval

The given function is \(f(x) = x\) for \(-\pi < x \leq \pi\). Determine this function’s Fourier Series.

Find the Fourier coefficients

For a function \(f(x)\) defined over the interval \(-L < x \leq L\), the Fourier series is given by \[f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos \frac{n\pi x}{L} + b_n \sin \frac{n\pi x}{L} \right)\]Since \(L = \pi\) here, rewrite with appropriate limits: \[f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos nx + b_n \sin nx \right)\]where \[a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx, \quad a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx \, dx, \quad b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx \, dx\].

Calculate the \(a_0\) coefficient

Evaluate \[a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} x \, dx\ = 0 \] since the integrand is an odd function over a symmetric interval.

Calculate the \(a_n\) coefficients

Evaluate \[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \cos nx \, dx\]. Use integration by parts where \u = x\ and \dv = \cos nx \, dx\. Notice that the boundary terms cancel out because \cos(nx) = 0\ at the boundaries for continuous x.

Simplify the integration by parts results

Continue integrating by parts: \[\int_{-\pi}^{\pi} x \cos(nx) \, dx = 0\], leading to \[a_n = 0\].

Calculate the \(b_n\) coefficients

Evaluate \[b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx) \, dx\]. The result is \[b_n = \frac{2}{n}(1 - (-1)^n) = \begin{cases} \frac{4}{n}, & \text{if } n \text{ is odd}\ 0, & \text{if } n \text{ is even} \end{cases}\].

Construct the Fourier Series

Combining the coefficients results in \[f(x) = x = \sum_{n=1,3,5,...}^{\infty} \frac{4}{n} \sin(nx)\].

Evaluate at a specific point

Set \(x = \frac{\pi}{2}\), \(\sum_{n=1,3,5,...} \frac{4}{n} \sin(\frac{n\pi}{2}) = \frac{\pi}{2} . \sin(\frac{\pi}{2})=1\). Notice the sines of multiples of \(\pi\) oscillate. Values reduce.

Derive series result

From \(\sum_{n=1}^{\infty} \frac{1-n^{^2}}{4n}=. . . =\frac{\pi}{2} . \sum=\frac{\pi}{4}\). Summation of form does confirm \(\sum_{n=1,3,.}(results) = \pi/4\).

Key concepts

Fourier coefficients

A Fourier series represents a periodic function as a sum of sine and cosine functions. To find the Fourier series of a function, we first need to calculate the Fourier coefficients, which include the coefficients \(a_0\), \(a_n\), and \(b_n\). These coefficients capture the function's behavior and are calculated using integrals over one period of the function.

• The coefficient \(a_0\) is the average value of the function over one period and is given by: \[a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx\].
• The coefficients \(a_n\) and \(b_n\) encode the amplitude of the cosine and sine terms in the series, respectively. They are defined as: \[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx\] and \[b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx\].

In our case, since \(f(x) = x\), an odd function, the coefficients \(a_0\) and \(a_n\) turn out to be zero due to symmetry. This leaves us with the \(b_n\) coefficients to determine the Fourier series representation.

Integration by parts

Integration by parts is a technique used to evaluate integrals where the integrand is a product of two functions. The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du. \]
To use this method effectively, identify parts of the integrand to set as \(u\) and \(dv\). For the given problem:

• Set \(u = x\) and \(dv = \cos(nx)dx\). This means that \(du = dx\) and \(v = \frac{\sin(nx)}{n}\).

Applying these settings to integrate \( \int_{-\pi}^{\pi} x \cos(nx)dx \), we use integration by parts:

• Compute the boundary terms: These evaluate to zero because sine of multiples of \(\pi\) is zero.
• Integrate the remaining term and simplify: The integral simplifies to zero as the boundary terms negate each other.

Thus, \(a_n\) coefficients are zero due to the properties of the definite integral of odd functions and symmetry over the interval from \(-\pi\) to \(\pi\).
Next, for the \(b_n\) coefficients, we set \(u = x\), \(dv = \sin(nx)dx\) and proceed similarly to evaluate the integral. The calculated value for \(b_n\) using this integration by parts method will help complete the Fourier series.

Trigonometric series

A trigonometric series is an infinite series that consists of sine and cosine terms. In a Fourier series, these terms help represent a function as a series of periodic components. The general form of a Fourier series is:
\[ f(x) = a_0 + \sum_{n=1}^{\infty}(a_n \cos(nx) + b_n \sin(nx)). \]
For the function \(f(x) = x\), we have identified that:

• \(a_0 = 0\) and \(a_n = 0\) for all \(n\).
• The non-zero coefficients are \(b_n = \frac{4}{n}\) for odd values of \(n\).

This simplifies the series to:

• \[ f(x) = \sum_{n=1,3,5,...}^\infty \frac{4}{n} \sin(nx). \]

To check the Fourier series, substitute values into the series formula and verify if the series converges to the function values. For example, at \(x = \frac{\pi}{2}\), the sine terms alternate, leading to a series form:
\[ 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ... = \frac{\pi}{4}. \] This alternating series illustrates the richness of using trigonometric series to represent and analyze functions, showing converging results reflecting the original function's values at specific points.