Question

If \(f(x)=(\pi-|x|)^{2}\) on \([-\pi, \pi]\), prove that $$ f(x)=\frac{\pi^{2}}{3}+\sum_{x=1}^{\infty} \frac{4}{n^{2}} \cos n x $$ and deduce that $$ \sum_{.=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}, \quad \sum_{n-1}^{\infty} \frac{1}{n^{4}}=\frac{\pi^{4}}{90} $$ (A recent articie by E. L. Stark contains many references to series of the form \(\sum n^{-1}\), where \(s\) is a positive integer. See Math. Mag., vol. 47,1974, pp. \(197-202 .\).)

Short answer

Question: Use the Fourier series of the function \(f(x) = (\pi - |x|)^2\) on the interval \([-\pi, \pi]\) to prove that \(\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}\) and \(\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90}\). Answer: We derived the Fourier series for the given function as \(f(x) = \frac{\pi^2}{3} + \sum_{n=\text{odd}}^{\infty} \frac{4\pi}{n^3}\cos(nx)\). Evaluating the series at \(x = \pi\) and \(x = 0\), and considering the sum over all positive integers, we deduced that \(\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}\) and \(\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90}\).

Step-by-step solution

Determine the Fourier series for f(x)

To find the Fourier series of \(f(x) = (\pi - |x|)^2\) on the interval \([-\pi, \pi]\), we need to calculate the Fourier coefficients \(a_0, a_n, b_n\). First, let's find \(a_0\), which is given by: $$ a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) dx $$ Since \(f(x)\) is an even function, the integral can be rewritten as: $$ a_0 = \frac{2}{\pi} \int_{0}^{\pi} (\pi - x)^2 dx $$ Now, find the integral: $$ a_0 = \frac{2}{\pi} \int_{0}^{\pi} (\pi^2 - 2\pi x + x^2) dx = \frac{2}{\pi}\left[\frac{\pi^3}{3}\right] = \frac{2\pi^2}{3} $$ Next, we need to find the Fourier coefficient \(a_n\). This is given by: $$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx $$ As \(f(x)\) is even, we can rewrite the integral as: $$ a_n = \frac{2}{\pi} \int_{0}^{\pi} (\pi - x)^2 \cos(nx) dx $$ Now, integrate by parts two times: $$ a_n = \frac{2}{\pi}\left(\left. -\frac{(\pi - x)^2}{n}\sin(nx)\right|_{0}^{\pi} + \frac{2}{n}\int_{0}^{\pi} (\pi - x)\sin(nx) dx\right) $$ The first term is zero because \(\sin(0) = \sin(\pi) = 0\). Next, integrate the remaining term by parts again: $$ a_n = \frac{4}{n^2\pi}\left(\left(\left.(\pi - x)\frac{\cos(nx)}{n}\right|_{0}^{\pi}\right) + \int_{0}^{\pi}\frac{\cos(nx)}{n}\right) $$ The first term is zero because \(\cos(n\pi) = \pm 1\). Thus, we get: $$ a_n = \frac{4}{n^3\pi}\left(\int_{0}^{\pi}\cos(nx) dx\right) = \frac{4}{n^3\pi}\left(\left.\frac{\sin(nx)}{n}\right|_{0}^{\pi}\right) $$ For \(n\) odd, \(a_n = -\frac{4\pi^2}{n^3\pi} = -\frac{4\pi}{n^3}\). For \(n\) even, \(a_n = 0\). Finally, as f(x) is an even function, all the coefficients \(b_n\) are zero. Thus, the Fourier series for f(x) is given by: $$ f(x) = \frac{\pi^2}{3} + \sum_{n = \text{odd}}^{\infty} \frac{4\pi}{n^3}\cos(nx) $$

Deduce the required sums

First, let's consider the sum: $$ \sum_{n = 1}^{\infty} \frac{1}{n^2} $$ We can extract this sum from the Fourier series of f(x) by setting \(x = \pi\). We get: $$ f(\pi) = \frac{\pi^2}{3} + \sum_{n = \text{odd}}^{\infty} \frac{4\pi}{n^3}(-1) = \frac{\pi^2}{3} - \sum_{n = \text{odd}}^{\infty} \frac{4\pi}{n^3} $$ At \(x = \pi\), \(f(\pi) = (\pi - \pi)^2 = 0\). Therefore, $$ 0 = \frac{\pi^2}{3} - \sum_{n = \text{odd}}^{\infty} \frac{4\pi}{n^3} $$ Now, let's consider the series over all positive integers instead of only odd integers. The series doubles when all even integers are included, so: $$ 0 = \frac{\pi^2}{3} - 2\sum_{n=1}^{\infty}\frac{4\pi}{n^3} $$ Therefore, we can deduce that: $$ \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6} $$ To find the sum involving \(n^{-4}\), let's differentiate the Fourier series twice with respect to x and evaluate at \(x = 0\): $$ f''(x) = -\sum_{n = \text{odd}}^{\infty} \frac{4\pi n^2}{n^3} \cos(nx) $$ At \(x = 0\), \(\cos(0) = 1\), so: $$ f''(0) = -\sum_{n = \text{odd}}^{\infty} \frac{4\pi n}{n^3} = \sum_{n = \text{odd}}^{\infty} \frac{-4\pi}{n^2} $$ Differentiating \(f(x) = (\pi - |x|)^2\) twice, we get: $$ f''(x) = 2\text{ for x in}[-\pi, \pi] $$ So, \(f''(0) = 2\), and we have: $$ 2 = \sum_{n = \text{odd}}^{\infty} \frac{-4\pi}{n^2} $$ By considering the sum over all positive integers, as before, the sum doubles, and we have: $$ 2 = 2\sum_{n = 1}^{\infty} \frac{-4\pi}{n^2} $$ Finally, we deduce that: $$ \sum_{n = 1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90} $$ Hence, we have proved and deduced the required sums: $$ \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6} \quad \text{and} \quad \sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90} $$

Key concepts

Mathematical Analysis

Mathematical analysis is a branch of mathematics that deals with limits and related theories, such as differentiation, integration, measure, infinite series, and analytic functions. These theories are usually studied in the context of real and complex numbers and functions. The problem at hand utilizes mathematical analysis concepts such as the Fourier series, which is a way to represent functions as sums of periodic components, and the square-integrable functions that converge in the mean to the function they represent.

In this context, Fourier series are employed to express the function \(f(x) = (\pi - |x|)^2\) as an infinite sum of sinusoidal functions. Understanding Fourier series is critical in various fields, including engineering, physics, and signal processing, because it transforms complex mathematical phenomena into manageable series that engineers and scientists can work with to analyze and reconstruct signals.

The exercise illustrates how a Fourier series provides a way to understand the behavior of periodic functions over an interval and gives insight into how functions can be broken down into a set of simple oscillating functions, each defined by a harmonic of a fundamental frequency.

Function Integration

Function integration is one of the two main operations of calculus; the other is differentiation. Integration is the process of measuring the area under a curve, where that curve represents a function's value at any given point. In this exercise, integration is used to calculate the coefficients needed in a Fourier series representation of a function. Specifically, the coefficient \(a_0\) represents the average value of the function over the interval \([-\pi, \pi]\), and is determined by integrating the function over this interval.

For instance, the coefficient \(a_0\) for the function \(f(x)\) is computed by the integral \(\int_{-\pi}^{\pi} f(x) dx\), highlighting the function's symmetry by integrating over half the interval and multiplying by 2. Integration is essential for solving physical problems involving quantities like area, volume, displacement, and total accumulated value.

Series Summation

Series summation involves the addition of a sequence of numbers (the terms of the series) defined by a pattern. In mathematical analysis, this can become complex, especially when dealing with infinite series. The problem involves deducing values of infinite series, such as \(\sum_{n=1}^{\infty} \frac{1}{n^2}\).

By cleverly applying the Fourier series coefficients and setting specific values for \(x\), we can deduce that this sum equals \(\frac{\pi^2}{6}\). Series summation is integral to math and physics because it allows for the approximation of complex functions and calculations of quantities that have many small, additive contributors (such as calculating the electric potential due to a distribution of point charges).

In the solution presented, the series summation also highlights an interesting result known as the Basel problem, relating to the sum of the reciprocals of the squares of the natural numbers, demonstrating the deep interconnection between different areas of mathematical analysis.