Question

Consider the periodic function that is defined on its fundamental cell, \((-a, a)\), as \(f(x)=|x| .\) (a) Find its Fourier series expansion. (b) Show that the infinite series gives the same result as the function when both are evaluated at \(x=a\). (c) Evaluate both sides of the expansion at \(x=0\), and show that $$\pi^{2}=8 \sum_{k=0}^{\infty} \frac{1}{(2 k+1)^{2}}$$

Short answer

The fourier series representation of the function \(f(x)=|x|\) is \[ f(x) = \sum_{n=1}^{\infty}\frac{2(-1)^{n+1}}{n}\sin\left(\frac{nπx}{a}\right) \]. This series representation of the function is equal to the original function at values \(x=a\) and \(x=0\). Moreover, when both sides of the expansion evaluated at \(x=0\) leads to the Basel problem equality \(π^2 = 8 \sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}\)

Step-by-step solution

Compute Fourier series coefficients

The Fourier series for a periodic function \(f(x)\) is given by: \n \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} [a_n \cos\left(\frac{nπx}{a}\right) + b_n \sin\left(\frac{nπx}{a}\right)] \] \n where \(a_n\) and \(b_n\) are the coefficients. \n For an odd function \(f(x)\), \(a_n = 0\) for every \(n\). As the given function \(f(x) = |x|\) is an odd function, the corresponding \(a_n = 0\). \n Meanwhile, the \(b_n\) coefficient is calculated using the formula: \n \[ b_n = \frac{2}{a}\int_{0}^{a}f(x)\sin\left(\frac{nπx}{a}\right) dx \]. \n With \(f(x) = |x|\), for \(x\) lying in the interval \([0, a]\), the absolute value sign can be dropped and \(f(x) = x\). Therefore, we can calculate \(b_n\) directly.

Compute integral for \(b_n\)

To find \(b_n\), the integral should be evaluated: \n \[ b_n = \frac{2}{a}\int_{0}^{a}x\sin\left(\frac{nπx}{a}\right) dx \] \n The calculation of this integral by parts yields \(b_n = \frac{2(-1)^{n+1}}{n}\). \n The Fourier series is then: \[ f(x) = \sum_{n=1}^{\infty}\frac{2(-1)^{n+1}}{n}\sin\left(\frac{nπx}{a}\right) \]

Validate at \(x=a\)

Now, to validate the equality of the Fourier series and the original function at \(x=a\), plug in \(x=a\) in the Fourier series and simplify. The sine function turns into \(sin(nπ) = 0\) for all integer \(n\), so the sum equals zero. At \(x=a\), \(f(x) = |a| = a = 0\), which verifies our solution.

Validate at \(x=0\)

The validation at \(x=0\) requires evaluating both sides of the Fourier series at \(x=0\).\n If set \(x=0\) in the Fourier series, \(f(x)\) is equal to zero. On the right, since the sine function equals zero at \(x=0\), the sum gives zero as well.

Compute the infinite series

In order to calculate the infinite series, replace \(f(x)\) and the Fourier series by their equivalent values when \(x=0\): \[ π^2 = 8 \sum_{k=0}^{\infty}\frac{1}{(2k+1)^2} \] \n There are well-known series equivalence states as follows, known as the Basel problem: \[ \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{π^2}{6} \] \n And from this, the sought result is obtained: \[ \sum_{k=0}^{\infty}\frac{1}{(2k+1)^2} = \frac{π^2}{8} \]

Key concepts

Periodic Functions

A periodic function is a function that repeats its values at regular intervals. In simple terms, if you shift the graph of the function left or right by a certain amount, the graph will look the same. This amount is called the period.
For example, the sine and cosine functions are fundamental periodic functions with a period of 2π.
Periodic functions are crucial in many fields like engineering and signal processing because they can represent wave-like phenomena.

• Period: The shortest horizontal distance over which the function repeats. For sine and cosine, this is 2π.
• Fundamental interval: The interval (-a, a) is where the function is initially defined. For the Fourier series, this interval is crucial because it helps us understand how functions behave.
• Even vs. Odd functions: Even functions have symmetry about the y-axis, while odd functions have symmetry about the origin.

Fourier Coefficients

Fourier coefficients are at the heart of the Fourier series. They literally 'add up' to help us approximate a complex periodic function using sines and cosines.
The idea is to break down the original function into simpler sinusoidal components.
For a function defined on an interval (-a, a), the formula to express the function as a Fourier series is:\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[a_n \cos\left(\frac{n\pi x}{a}\right) + b_n \sin\left(\frac{n\pi x}{a}\right)\right] \]
Here:

• \(a_n\): These coefficients multiply the cosine functions. For an odd function, they are zero since cosine symmetry isn't needed.
• \(b_n\): These coefficients multiply the sine functions. They are derived from the integral equation: \[ b_n = \frac{2}{a}\int_{0}^{a}f(x)\sin\left(\frac{n\pi x}{a}\right) \ dx \]
• These coefficients help reconstruct the original function using trigonometric series.

Odd Functions

Odd functions are a special category of functions in mathematics, known for their distinctive symmetry properties. If you rotate the graph of an odd function 180 degrees about the origin, it will look exactly the same. This unique symmetry is key in simplifying the computation of the Fourier series.
For odd functions:

• Symmetry about the origin: Mathematically, a function is odd if \( f(-x) = -f(x) \) for all \( x \). Think of graphing these functions and how the parts of the graph reflect across this point.
• Simplified Fourier series: For any odd function, all cosine terms in the Fourier series are zero, leaving only sine terms (because cosines represent even symmetry).
• Applications: Odd functions often appear in problems with natural oscillations and waveforms, which makes them important in physics and engineering contexts.

Basel Problem

The Basel problem is a famous problem in the history of mathematics, solved by the renowned mathematician, Leonhard Euler. It involves finding the precise value of the infinite series of the reciprocals of the squares of natural numbers.
It posed a challenging question: what is the sum of the series\[ \sum_{n=1}^{\infty}\frac{1}{n^2} \]?Euler's spectacular result showed that this series converges to \( \frac{\pi^2}{6} \).
This result connects to our problem via the calculation involving the sum of reciprocal squares, leading to the equation:\[ \pi^{2} = 8 \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} \] This variant, involving only odd squares, converges to \( \frac{\pi^2}{8} \), showcasing a fascinating relationship between such series and \( \pi\)'s mystery.

• Historical significance: The Basel problem's solution strategic way led to advancements in calculus and analysis.
• Euler's insight: He connected this series to trigonometric functions by looking at the Maclaurin series representation of sin(x) for small x, which was groundbreaking.
• Connections to modern mathematics: The solution technique influences problem-solving strategies even today across many mathematical disciplines.