History and Introduction
The Basel problem involves finding the sum of the reciprocals of the squares of the natural numbers. In mathematical terms, it can be expressed as:
\[ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \ldots \]
Pietro Mengoli initially posed this problem in 1650. Many mathematicians attempted to solve it, but it wasn’t until 1734 that Leonhard Euler provided a remarkable solution. Euler found that the sum converges to a finite value, and he expressed this value as:
\[ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \]
This solution was groundbreaking and brought Euler considerable recognition at the age of 28. Euler’s method involved manipulating infinite series and introducing what later became known as the Basel identity. Euler’s solution not only solved the Basel problem but also laid the foundation for the study of convergent infinite series.
Euler’s work on the Basel problem had broader implications, leading to the development of complex analysis and inspiring future mathematicians. One notable example is Bernhard Riemann, who, in his 1859 paper “On the Number of Primes Less Than a Given Magnitude,” generalized Euler’s ideas by introducing the Riemann zeta function. The Riemann zeta function became a fundamental tool in number theory and has connections to various areas of mathematics, including the distribution of prime numbers.
Cauchy Proof – using almost elementary (high school) mathematics!
step 1. Setup of problem:
– The goal is to find the sum of the reciprocals of the squares of natural numbers, i.e., \( \sum_{k=1}^{\infty} \frac{1}{k^2} \).
– The proof focuses on bounding the partial sums \( \sum_{k=1}^{m} \frac{1}{k^2} \) between two expressions, each approaching \( \frac{\pi^2}{6} \) as \( m \) tends to infinity.
step 2. Use of Trigonometric Identities:
– De Moivre’s formula is employed to establish an identity involving the cotangent and cosecant functions.
– By manipulating this identity, expressions involving \( \cot^2 x \) and \( \csc^2 x \) are derived.
step 3. Polynomial and Vieta’s Formulas:
– The cotangent values at specific points \( x_r \) are considered, where \( x_r = r\pi / (2m + 1) \).
– Vieta’s formulas are used to relate the sum of cotangent squares to the coefficients of a polynomial.
step 4. Inequality Analysis:
– An inequality involving cotangent squares, reciprocal squares, and cosecant squares is established.
– This inequality is based on the geometric interpretation of the relationship between cotangent, cosecant, and reciprocal functions.
step 5. Bounding the Partial Sums:
– The inequality is applied to the partial sums \( \sum_{k=1}^{m} \frac{1}{k^2} \) for specific values of \( m \).
– The inequality is multiplied through by \( (\pi / (2m + 1))^2 \) to obtain bounds for the partial sums.
step 6. Squeeze Theorem:
– As \( m \) approaches infinity, both the upper and lower bounds of the partial sums approach \( \frac{\pi^2}{6} \).
– The squeeze theorem is invoked to conclude that \( \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6} \).
This proof method is based on cleverly using trigonometric identities, polynomial properties, and inequalities to establish the convergence of the series to the desired limit.
Infinite series as a Infinite product (Euler’s original method) via Fundamental theorem of Algebra
Euler’s derivation of the value \(\frac{\pi^2}{6}\) for the sum of the reciprocals of the squares involved an extension of observations about finite polynomials to infinite series. Despite lacking formal justification at the time (which Karl Weierstrass provided a century later through the Weierstrass factorization theorem), Euler’s numerical verification using partial sums instilled confidence in his result.
Starting with the Taylor series expansion of the sine function \( \sin x = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!} + \ldots \), Euler divided through by \(x\) to get \(\frac{\sin x}{x} = 1 – \frac{x^2}{3!} + \frac{x^4}{5!} – \frac{x^6}{7!} + \ldots\). Euler heuristically applied the concept of factorization as seen in finite polynomials to this infinite-degree polynomial. The Weierstrass factorization theorem later validated this approach.
The factorization yielded \(\frac{\sin x}{x} = \left(1 – \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 – \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 – \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \ldots = \left(1 – \frac{x^2}{\pi^2}\right)\left(1 – \frac{x^2}{4\pi^2}\right)\left(1 – \frac{x^2}{9\pi^2}\right) \ldots\).
By multiplying out and collecting the \(x^2\) terms, Euler showed, using Newton’s identities, that the coefficient of \(x^2\) in \(\frac{\sin x}{x}\) equates to \(-\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \ldots\right) = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}\).
Comparing this with the coefficient from the original infinite series expansion of \(\frac{\sin x}{x}\) (\(-\frac{1}{3!} = -\frac{1}{6}\)), Euler equated the two coefficients, leading to the fundamental identity \(-\frac{1}{6} = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}\). Multiplying both sides by \(-\pi^2\) then yielded the sum of the reciprocals of the positive square integers: \(\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}\).
Proof by the Fourier Series Method
Parseval’s identity relates the coefficients of the Fourier series of a function to the integral of the squared function over its period. For a function \(f(x)\) with its Fourier series given by \(f(x) = \sum_{n=-\infty}^{\infty} c_n e^{inx}\), Parseval’s identity is stated as:
\[ \sum_{n=-\infty}^{\infty} |c_n|^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |f(x)|^2 \,dx \]
Now, consider the function \(f(x) = x\) over the interval \([- \pi, \pi]\). The coefficients \(c_n\) for this function are given by:
\[ c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} x e^{-inx} \,dx = \frac{1}{n}i \left((-1)^n – 1\right) \]
For \(n \neq 0\), \(|c_n|^2 = \frac{1}{n^2}\), and for \(n = 0\), \(c_0 = 0\). Therefore:
\[ |c_n|^2 = \begin{cases} \frac{1}{n^2}, & \text{for } n \neq 0 \\ 0, & \text{for } n = 0 \end{cases} \]
The sum of these squared coefficients over all \(n\) is twice the sum over positive \(n\):
\[ \sum_{n=-\infty}^{\infty} |c_n|^2 = 2 \sum_{n=1}^{\infty} \frac{1}{n^2} \]
Now, applying Parseval’s identity:
\[ \sum_{n=-\infty}^{\infty} |c_n|^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} x^2 \,dx \]
Solving the integral on the right-hand side:
\[ \frac{1}{2\pi} \int_{-\pi}^{\pi} x^2 \,dx = \frac{1}{2\pi} \left[\frac{x^3}{3}\right]_{-\pi}^{\pi} = \frac{\pi^2}{3} \]
Therefore:
\[ 2 \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{3} \]
Simplifying:
\[ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \]
This result aligns with the previously derived expression for the sum of the reciprocals of the positive square integers using Euler’s method.
Another Proof by using Parseval’s Identity
The given expression demonstrates the application of Parseval’s identity to a complete orthonormal basis in the space \(L^2_{\text{per}}(0,1)\) of L2 periodic functions over the interval \((0,1)\). The basis is denoted by \(\{e_i\}_{i=-\infty}^{\infty}\), where \(e_i(\theta) = \exp(2\pi i k \theta)\), and the inner product on this Hilbert space is defined as \(\langle f, g \rangle = \int_0^1 f(x) \overline{g(x)} \, dx\) for \(f, g \in L^2_{\text{per}}(0,1)\).
Parseval’s identity, in this context, is expressed as:
\[ \|x\|^2 = \sum_{i=-\infty}^{\infty} |\langle e_i, x \rangle|^2 \]
Now, considering the orthonormal basis \(\{e_k\}\) where \(\langle e_k, e_j \rangle = \int_0^1 e^{2\pi i (k-j)\theta} \, d\theta = \delta_{k,j}\), and taking \(f(\theta) = \theta\), the norm of \(f\) can be computed as:
\[ \|f\|^2 = \int_0^1 \theta^2 \, d\theta = \frac{1}{3} \]
Additionally, the inner product \(\langle f, e_k \rangle\) is calculated as:
\[ \langle f, e_k \rangle = \int_0^1 \theta e^{-2\pi i k \theta} \, d\theta \]
This leads to the following results:
\[ \langle f, e_k \rangle = \begin{cases} \frac{1}{2}, & \text{if } k = 0 \\ -\frac{1}{2\pi i k}, & \text{if } k \neq 0 \end{cases} \]
Now, applying Parseval’s identity, we have:
\[ \frac{1}{3} = \sum_{k \neq 0} \left| -\frac{1}{2\pi i k} \right|^2 + \frac{1}{4} = 2 \sum_{k=1}^{\infty} \frac{1}{(2\pi k)^2} + \frac{1}{4} \]
Simplifying further yields:
\[ \frac{\pi^2}{6} = \frac{2\pi^2}{3} – \frac{\pi^2}{2} = \zeta(2) \]
Thus, the expression concludes with the well-known result \(\zeta(2) = \frac{\pi^2}{6}\).
Proof by Lewin Argument using Dilogarithm function and some tricks
This above list of proofs just goes to say that a result in mathematics can be obtained in multitude of ways. It is also not to suggest that this is an exhaustive list of proofs for the said Basel problem. Each of these proofs show us that we may obtain the answer to a problem in the most elementary or elegant or elaborate way!
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