Basel problem

From Wikipedia, the free encyclopedia

The Basel problem is a famous problem in number theory, first posed by Pietro Mengoli in 1644, and solved by Leonhard Euler in 1735. Since the problem had withstood the attacks of the leading mathematicians of the day, Euler's solution brought him immediate fame when he was twenty-eight. Euler generalised the problem considerably, and his ideas were taken up years later by Bernhard Riemann in his seminal 1859 paper On the Number of Primes Less Than a Given Magnitude, in which he defined his zeta function and proved its basic properties. The problem is named after Basel, hometown of Euler as well as of the Bernoulli family, who unsuccessfully attacked the problem.

The Basel problem asks for the precise sum of the reciprocals of the squares of the positive integers, i.e. the precise sum of the infinite series:

$\sum_{n=1}^\infin \frac{1}{n^2} = \lim_{n \to \infty}\left(\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2}\right)$

The series is approximately equal to 1.644934. The Basel problem asks for the exact sum of this series (in closed form), as well as a proof that this sum is correct. Euler found the exact sum to be π²/6 and announced this discovery in 1735. His arguments were based on manipulations that were not justified at the time, and it was not until 1741 that he was able to produce a truly rigorous proof.

1 Euler attacks the problem
2 The Riemann zeta function
3 A rigorous proof
4 References
5 External links

[edit] Euler attacks the problem

Euler's original "derivation" of the value π²/6 is clever and original. He essentially extended observations about finite polynomials and assumed that these same properties hold true for infinite series. Of course, Euler's original reasoning requires justification, but even without justification, by simply obtaining the correct value, he was able to verify it numerically against partial sums of the series. The agreement he observed gave him sufficient confidence to announce his result to the mathematical community.

To follow Euler's argument, recall the Taylor series expansion of the sine function

$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots.$

Dividing through by x, we have

$\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots.$

Now, the roots (zeros) of sin(x)/x occur precisely at $x = n\cdot\pi$ where $n = \pm1, \pm2, \pm3, ...$ Let us assume we can express this infinite series as a product of linear factors given by its roots, just as we do for finite polynomials:

$\frac{\sin(x)}{x} = \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots$

$= \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots.$

If we formally multiply out this product and collect all the x² terms, we see that the x² coefficient of sin(x)/x is

$-\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots \right) = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.$

But from the original infinite series expansion of sin(x)/x, the coefficient of x² is −1/(3!) = −1/6. These two coefficients must be equal; thus,

$-\frac{1}{6} = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.$

Multiplying through both sides of this equation by −π² gives the sum of the reciprocals of the positive square integers.

[edit] The Riemann zeta function

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:

$\zeta(s) = \sum_{n=1}^\infin \frac{1}{n^s}.$

Taking s = 2, we see that ζ(2) is equal to the sum of the reciprocals of the squares of the positive integers:

$\zeta(2) = \sum_{n=1}^\infin \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots \approx 1.644934.$

Convergence can be proved with the following inequality:

$\sum_{n=1}^N \frac{1}{n^2} < 1 + \sum_{n=2}^N \frac{1}{n(n-1)} = 1 + \sum_{n=2}^N \left( \frac{1}{n-1} - \frac{1}{n} \right) = 1 + 1 - \frac{1}{N} \; \stackrel{N \to \infty}{\longrightarrow} \; 2.$

This gives us the upper bound ζ(2) < 2, and because the infinite sum has only positive terms, it must converge. It can be shown that ζ(s) has a nice expression in terms of the Bernoulli numbers whenever s is a positive even integer.

[edit] A rigorous proof

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).

[edit] History of this proof

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 1960s".

[edit] What you need to know

To understand the proof, you will need to understand the following facts:

De Moivre's formula states that for any real number x and any integer n,

$(\cos x + i\sin x)^n = \cos (nx) + i \sin (nx). \,\!$

Proof: This can be proved from Euler's formula; see the article for more details.

The binomial theorem, which states that for any real numbers x and y and any nonnegative integer n,

$(x+y)^n=\sum_{k=0}^n{n \choose k}x^ky^{n-k}$

where we have the binomial coefficients

${n \choose k}=\frac{n!}{k!(n-k)!}$ (See factorial)

Proof: This requires mathematical induction and some properties of the binomial coefficients.

The function cot² x is one-to-one on the interval (0, π/2).
- Proof: Suppose cot² x = cot² y for some x, y in the interval (0, π/2). Using the definition of cotangent cot x = (cos x)/(sin x) and the trigonometric identity cos² x = 1 − sin² x, we see that (sin² x)(1 − sin² y) = (sin² y)(1 − sin² x). Adding (sin² x)(sin² y) to each side, we have sin² x = sin² y. Since the sine function is always nonnegative on the interval (0, π/2), this means sin x = sin y, but it is geometrically evident (by looking at the unit circle, e.g.) that the sine function is 1-1 on the interval (0, π/2), so that x = y.

If p(t) is a polynomial of degree m, then p has no more than m distinct roots.
- Proof: This is a consequence of the fundamental theorem of algebra.

If p(t) = a_mt^m + a_{m − 1}t^{m − 1} + ... + a₁t + a₀, where a_m ≠ 0, then the sum of the roots of p (counting multiplicities) is −a_{m − 1}/a_m.
- Proof: If a_m = 1, then p(t) = product of all (t − s), where s ranges over all roots of p. Expanding this product, we see the coefficient of t^{m − 1} is minus the sum of all the roots. If a_m ≠ 1, then we can divide each term by it, obtaining a new polynomial with the same roots, whose leading coefficient is now 1; applying the preceding argument shows that the sum of the roots of the original p(t) = sum of the roots of this new polynomial = −a_{m − 1}/a_m.

The trigonometric identity csc² x = 1 + cot² x.
- Proof: This follows from the fundamental identity 1 = sin² x + cos² x after dividing through by sin² x.

For any real number x with 0 < x < π/2, we have the inequalities cot² x < 1/x² < csc² x.
- Proof: First note that 0 < sin x < x < tan x. This can be seen by considering the following picture:

To see that 0 < sin x < x, observe that in the picture, sin θ is the length of the line AC, and θ is the length of the circular arc AD.

To see that x < tan x, observe that the area of the triangle OAE is tan(θ)/2, the area of the sector OAD is θ/2, and that the sector is contained within the triangle.

Now, take the reciprocal of everything and square. Remember that the inequality switches direction.

Let a, b, and c be any real numbers, with a ≠ 0; then the limit of the function (am + b)/(am + c) as m approaches infinity is 1.
- Proof: Divide each term by m, and get (a + b/m)/(a + c/m). If we divide a fixed number by a larger and larger number, the quotient approaches zero; thus, both the numerator and the denominator above tend to a, and so their quotient tends to 1.

The squeeze theorem, which states that if a function is "squeezed" between two other functions, and each of those two functions approach a common limit, then the "squeezed" function also approaches that same limit.
- Proof: See the article for a thorough discussion and proof.

[edit] The proof

The main idea behind the proof is to bound the partial sums

$\sum_{k=1}^m \frac{1}{k^2} = \frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{m^2}$

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.

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

$\frac{\cos (nx) + i \sin (nx)}{(\sin x)^n} = \frac{(\cos x + i\sin x)^n}{(\sin x)^n} = \left(\frac{\cos x + i \sin x}{\sin x}\right)^n = (\cot x + i)^n.$

From the binomial theorem, we have

$(\cot x + i)^n = {n \choose 0} \cot^n x + {n \choose 1} (\cot^{n-1} x)i + \cdots + {n \choose {n-1}} (\cot x)i^{n-1} + {n \choose n} i^n$

$= \left[ {n \choose 0} \cot^n x - {n \choose 2} \cot^{n-2} x \pm \cdots \right] \; + \; i\left[ {n \choose 1} \cot^{n-1} x - {n \choose 3} \cot^{n-3} x \mp \cdots \right].$

Combining the two equations and equating imaginary parts gives the identity

$\frac{\sin (nx)}{(\sin x)^n} = \left[ {n \choose 1} \cot^{n-1} x - {n \choose 3} \cot^{n-3} x \mp \cdots \right].$

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

$0 = {{2m+1} \choose 1} \cot^{2m} x - {{2m+1} \choose 3} \cot^{2m-2} x \mp \cdots + (-1)^m{{2m+1} \choose {2m+1}}.$

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

$p(t) := {{2m+1} \choose 1}t^m - {{2m+1} \choose 3}t^{m-1} \mp \cdots + (-1)^m{{2m+1} \choose {2m+1}}.$

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

$\cot ^2 \left(\frac{\pi}{2m+1}\right) + \cot ^2 \left(\frac{2 \pi}{2m+1}\right) + \cdots + \cot ^2 \left(\frac{m \pi}{2m+1}\right)$

$= {{2m+1} \choose 3} / {{2m+1} \choose 1} = \frac{(2m)(2m-1)}{6}.$

Substituting the identity csc² x = cot² x + 1, we have

$\csc ^2 \left(\frac{\pi}{2m+1}\right) + \csc ^2 \left(\frac{2 \pi}{2m+1}\right) + \cdots + \csc ^2 \left(\frac{m \pi}{2m+1}\right)$

$= {{2m+1} \choose 3} / {{2m+1} \choose 1} + m = \frac{(2m)(2m+2)}{6}.$

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

$\frac{(2m)(2m-1)}{6} < \left( \frac{2m+1}{\pi} \right) ^2 + \left( \frac{2m+1}{2 \pi} \right) ^2 + \cdots + \left( \frac{2m+1}{m \pi} \right) ^2 < \frac{(2m)(2m+2)}{6}.$

Multiplying through by (π/(2m + 1))², this becomes

$\frac{\pi ^2}{6}\left(\frac{2m}{2m+1}\right)\left(\frac{2m-1}{2m+1}\right) < \frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{m^2} < \frac{\pi ^2}{6}\left(\frac{2m}{2m+1}\right)\left(\frac{2m+2}{2m+1}\right).$

As m approaches infinity, the left and right hand expressions each approach π²/6, so by the squeeze theorem,

$\zeta(2) = \sum_{k=1}^\infin \frac{1}{k^2} = \lim_{m \to \infty}\left(\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{m^2}\right) = \frac{\pi ^2}{6}$

and this completes the proof. Q.E.D.

[edit] References

Number Theory: An Approach Through History, Andre Weil, Springer, ISBN 0-8176-3141-0
Euler: The Master of Us All, William Dunham, MAA, ISBN 0-88385-328-0
Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics, John Derbyshire, Joseph Henry Press, ISBN 0-309-08549-7
Proofs From the Book, Martin Aigner, Gunter Ziegler, Springer, ISBN 3-540-67865-4
Riemann's Zeta Function, Harold M. Edwards, Dover, ISBN 0-486-41740-9