Basel problem
In mathematics, the Basel problem is a famous problem in number theory, first posed by Pietro Mengoli in 1644, and solved by Leonhard Euler in 1735. The problem had withstood the attacks of the leading mathematicians of the day, so Euler's solution gained him immediate notoriety at the age of 28. Euler generalised the problem considerably, and his ideas were taken up years later by Bernhard Riemann in his seminal 1859 paper, in which he defined his zeta function and proved its basic properties.The Basel problem asks for the precise sum of the reciprocals of the squares of the positive integers, i.e. it asks for the precise sum of the infinite series
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2 The Riemann zeta function 3 A rigorous proof 4 References 5 External Links |
The Riemann zeta function ζ(s) is one of the most important functions in mathematics, because of its relationship to the distribution of the prime numbers. The function is defined for any complex number s with real part > 1 by the following formula:
The following argument proves the identity ζ(2) = π2/6, where ζ(s) is the Riemann zeta function. It is by far the simplest proof yet available; while most proofs utilise results from advanced mathematics, such as Fourier analysis, complex analysis, and multivariable calculus, the following does not even require single-variable calculus (although a single limit is taken at the end).
The origin of the proof is unclear. It appeared in the journal Eureka in 1982, attributed to John Scholes, but Scholes claims he learned the proof from Peter Swinnerton-Dyer, and in any case he maintains the proof was "common knowledge at Cambridge in the late 1960's".
To understand the proof, you will need to understand the following facts:
The main idea behind the proof is to bound the partial sums
Let x be a real number with 0 < x < π/2, and let n be a positive integer. Then from De Moivre's formula and the definition of the cotangent function, we have
Euler attacks the problem
The Riemann zeta function
Taking s = 2, we see that ζ(2) is equal to the sum of the reciprocals of the squares of the positive integers:
How do we know it converges at all? We can easily demonstrate this with the following inequality:
This gives us the upper bound ζ(2) < 2, but the exact value ζ(2) = &pi2/6 was unknown for some time, until Leonhard Euler computed it in 1734. It can be shown that ζ(s) has a nice expression in terms of the Bernoulli numbers whenever s is a positive even integer.A rigorous proof
History of this proof
What you need to know
The proof
between two expressions, each of which will tend to π2/6 as m approaches infinity. The two expressions are derived from identities involving the cotangent and cosecant functions. These identities are in turn derived from De Moivre's formula, and we now turn to establishing these identities.
From the binomial theorem, we have
Combining the two equations and equating imaginary parts gives the identity
We take this identity and set n = 2m + 1, where m is a positive integer, and x = rπ/(2m + 1), where r = 1, 2, ..., m. Then nx = rπ, so that sin(nx) = 0, and so
This equation holds for each of the values x = rπ/(2m + 1), where r = 1, 2, ..., m. These values of x are distinct numbers strictly between 0 and π/2. Since the function cot2(x) is 1-1 on the interval (0, π/2), the numbers cot2(x) = cot2(rπ/(2m + 1)) are therefore distinct for r = 1, 2, ..., m. But by the above equation, each of these m distinct numbers is a root of the mth degree polynomial
This means that the numbers x = cot2(rπ/(2m + 1)), for r = 1, 2, ..., m are precisely the roots of the polynomial p(t). But we can calculate the sum of the roots directly by examining the coefficients, and the comparison shows that
Substituting the identity csc2 x = cot2 x + 1, we have
Now consider the inequality cot2 x < 1/x2 < csc2 x. If we add up all these inequalities for each of the numbers x = rπ/(2m + 1), and if we use the two identities above, we get
Multiplying through by (π/(2m + 1))2, this becomes
As m approaches infinity, the left and right hand expressions each approach π2/6, so by the squeeze theorem,
and this completes the proof. Q.E.DReferences
External Links