Question

Find the Fourier series of $$ f(x)= \begin{cases}x-[x], & x \text { is not an integer, } \\ \frac{1}{2}, & x \text { is an integer. }\end{cases} $$ To what values does the Fourier series converge at the points \(x=5, x=3\), and \(x=1.5\) ?

Short answer

The Fourier series converges to 0 at \( x=5 \) and \( x=3 \), and converges to 0.5 at \( x=1.5 \).

Step-by-step solution

Identify the period of the function

The given function \( f(x) = x - [x] \), where \([x]\) is the floor function, represents the fractional part of \( x \). This function is periodic with period \( T = 1 \), since the fractional part repeats when an integer is added to \( x \).

Compute the Fourier coefficients

For a function \( f(x) \) with period \( T = 1 \), the Fourier series is given by \( f(x) = a_0 + \sum_{n=1}^{\infty} \left(a_n \cos(2\pi nx) + b_n \sin(2\pi nx) \right) \), where:- \( a_0 = \int_0^1 f(x) \, dx \)- \( a_n = 2\int_0^1 f(x) \cos(2\pi nx) \, dx \)- \( b_n = 2\int_0^1 f(x) \sin(2\pi nx) \, dx \)Calculate these integrals to find the coefficients.

Calculate the Fourier coefficient \( a_0 \)

Compute \( a_0 \):\[a_0 = \int_0^1 (x - [x]) \, dx = \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}.\]Thus, \( a_0 = \frac{1}{2} \).

Compute the sine coefficients \( b_n \)

Calculate \( b_n = 2\int_0^1 (x - [x]) \sin(2\pi nx) \, dx = 2\int_0^1 x \sin(2\pi nx) \, dx \).Using integration by parts, let \( u = x \) and \( dv = \sin(2\pi nx) \, dx \). Then \( du = dx \) and \( v = -\frac{1}{2\pi n}\cos(2\pi nx) \).Evaluating, \[b_n = \left[-x \frac{1}{2\pi n} \cos(2\pi nx) \right]_0^1 + \int_0^1 \frac{1}{2\pi n} \cos(2\pi nx) \, dx.\]This simplifies to \( b_n = \frac{1}{\pi n} \), for \( n eq 0 \).

Compute the cosine coefficients \( a_n \)

The coefficient \( a_n = 2 \int_0^1 (x - [x]) \cos(2\pi nx) \, dx = 2 \int_0^1 x \cos(2\pi nx) \, dx = 0 \).This integral is zero due to symmetry; hence, \( a_n = 0 \) for all \( n \).

Write the Fourier series

Substitute the coefficients into the Fourier series:\[f(x) = \frac{1}{2} + \sum_{n=1}^{\infty} \frac{1}{\pi n} \sin(2\pi nx) .\]

Evaluate the Fourier series at specific points

To determine the convergence at specific points:- At \( x = 5 \): The series converges to \( f(5) = 5 - [5] = 0 \).- At \( x = 3 \): The series converges to \( f(3) = 3 - [3] = 0 \).- At \( x = 1.5 \): The series converges to \( f(1.5) = 1.5 - [1.5] = 0.5 \).

Key concepts

Fourier Coefficients

In the study of Fourier series, the Fourier coefficients are fundamental components that allow us to express a periodic function as an infinite sum of sine and cosine terms. These coefficients, represented by \(a_0\), \(a_n\), and \(b_n\) for a given function \(f(x)\), determine the amplitude of each cosine and sine term in the series.

• \(a_0\) is the average value of the function over one period.
• \(a_n\) and \(b_n\) provide the amplitudes for the cosine and sine terms, respectively.

To compute each coefficient:

• \(a_0 = \int_0^1 f(x) \, dx\)
• \(a_n = 2 \int_0^1 f(x) \cos(2\pi nx) \, dx\)
• \(b_n = 2 \int_0^1 f(x) \sin(2\pi nx) \, dx\)

Understanding the calculation of these coefficients is crucial for constructing a Fourier series. Each coefficient reflects how much of a particular frequency is present in the original function. This information can be retrieved by integration over a complete period of the function.

Periodic Function

A periodic function is one that repeats its values at regular intervals, known as periods. The concept of periodicity is at the heart of Fourier series.
The function given in the original exercise, \(f(x) = x - [x]\), is periodic with a period of 1. This means that after every interval of 1 unit along the x-axis, the function values repeat themselves.

• Period: The interval at which a function repeats.
• Characterization: A periodic function can be characterized by its period \(T\) such that \(f(x + T) = f(x)\) for all \(x\).
• Importance: Identifying the period simplifies the calculation of Fourier coefficients and helps in constructing the series for representation.

Utilizing periodic functions allows us to apply Fourier series to approximate more complex functions efficiently over these repeating intervals.

Integration by Parts

Integration by parts is an important technique in calculus used to solve integrals that are products of two functions. It's often used when solving for Fourier coefficients involving trigonometric functions.
The formula for integration by parts is\[\int u \, dv = uv - \int v \, du\]
where \(u\) and \(dv\) are functions chosen to simplify the integral.
In the exercise, integration by parts is used to compute the sine coefficients \(b_n\). Choosing \(u = x\) and \(dv = \sin(2\pi nx) dx\) simplifies the integration process:

• \(du = dx\)
• \(v = -\frac{1}{2\pi n}\cos(2\pi nx)\)

This technique is valuable in the context of Fourier series as it helps evaluate terms where direct integration would be more challenging.

Floor Function

The floor function, often denoted as \([x]\), takes a real number \(x\) and returns the greatest integer less than or equal to \(x\). In the context of the exercise, the function \(f(x) = x - [x]\) essentially represents the fractional part of \(x\).

• Definition: \([x]\) is the greatest integer \(n\) such that \(n \leq x\).
• Properties: For any integer \(x\), \([x] = x\).
• Usage: Floor functions are used to manage periodicity and deal with integer and non-integer distinctions in problems.

Employing the floor function makes the resulting function periodic with a period of 1, which is crucial for calculating the Fourier series.
Understanding the behavior of the floor function in mathematical terms is essential for dealing with periodic functions and their Fourier series expansions.