Thursday, 19 June 2014

`Convergence' of Some Divergent Series

‘Convergence’ of Some Divergent Series!

The topic of this article, the idea of attaching a finite value to divergent series, is no longer a purely mathematical exercise. These finite values of divergent series have found application in String theory and quantum field theory (Casimir effect).

The finite sum of real/complex numbers is always finite. The infinite sum of real/complex numbers can be either finite or infinite. For instance,

k=1nk = 1 + 2 + + n = n(n + 1) 2

is finite and

k=1k = 1 + 2 + 3 + = +.

If an infinite sum has finite value it is said to converge, otherwise it is said to diverge. Divergence do not always mean it grows to ±. For instance,

k=1k, k=11 k and  k=0(1)k

are diverging, while

k=1 1 kss  and (s) > 1,

is converging. The convergence of a series is defined by the convergence of its partial sum. For instance,

k=1k

diverges because its partial sum

sn := k=1nk = n(n + 1) 2

diverges, as n . The geometric series

k=0zk = 1 1 z, (1)

for all |z| < 1, because its partial sum

sn := zn 1 z 1 n 1 z 1.

Note that the map T(z) = 1 1z is well-defined between T : {1} . In fact, T is a holomorphic (complex differentiable) functions and, hence, analytic. The analytic function T exists for all complex numbers except z = 1 and, inside the unit disk {z |z| < 1}, T is same as the power series in (1). Thus, T may be seen as an analytic continuation of

k=0zk

outside the unit disk, except at z = 1. A famous unique continuation result from complex analysis says that the analytic continuation is unique. Thus, one may consider the value of T(z), outside the unit disk, as an ‘extension’ of the divergent series

k=0zk.

In this sense, setting z = 2 in T(z), the divergent series may be seen as taking the finite value

k=02k = 1 + 2 + 4 + 8 + 16 + = 1.

Similarly, setting z = 1 in T(z), the oscillating divergent series may be seen as taking the finite value

k=0(1)k = 1 1 (1) = 1 2.

This sum coincides with the Cesàro sum which is a special kind of convergence for series which do not diverge to ±. For instance,

k=1k

diverges in Cesàro sum too. The Cesàro sum of a series

k=1a k

is defined as the

lim n1 n i=1n k=1ia k .

Observe that the sequence

1 n i=1n k=0i1(1)k 1 2.

Since T is not defined on 1, we have not assigned any finite value to the divergent series

1 + 1 + 1 + .

The Riemann zeta function

ζ(z) := k=1 1 kz

converges for z with (z) > 1. Note that for z , kz = ez ln(k), where ln is the real logarithm. If z = σ + it then kz = kσeit ln(k) and |kz| = kσ since |eit ln(n)| = 1. Therefore, the infinite series converges for all z such that (z) > 1 (i.e. σ > 1) because it is known that k=1kσ converges for all σ > 1 and diverges for all σ 1.

The Riemann zeta function is a special case of the Dirichlet series

D(z) := k=1ak kz

with ak = 1 for all k. Note that for z = 2, we get the Basel series

ζ(2) := k=1 1 k2 = π2 6

and ζ(3) = 1.202056903159594 called the Apéry’s constant. In fact, the following result of Euler gives the value of Riemann zeta function for all even positive integers.

Theorem 1 (Euler). For all k ,

ζ(2k) = (1)k+1(2π)2k 2(2k)!B2k

where B2k is the 2k-th Bernoulli number.

If ζ(z) is well-defined for (z) > 1 then 21zζ(z) is also well-defined for (z) > 1. Therefore,

ζ(z)(1 21z) = k=1(1)k1nz =: η(z).

Observe that η(z) is also a Dirichlet series. It can be shown that (1 21z)η(z) is analytic for all (z) > 0 and (z)1. Thus, we have analytically continued ζ(z) for all (z) > 0 and (z)1. There is a little work to be done on the zeroes of 1 21z but is fixable. In the strip 0 < (z) < 1, called the critical strip, the Riemann zeta function satisfies the relation

ζ(z) = 2z(π)z1 sin zπ 2 Γ(1 z)ζ(1 z).

This relation is used to extend the Riemann zeta function to z with non-positive real part, thus, extending to all complex number z1. Setting z = 1 in the relation yields

ζ(1) = 1 2π2 sin π 2 Γ(2)ζ(2) = 1 2π2(1)π2 6 = 1 12 .

Since ζ(1) is extension of the series

k=1 1 kz

one may think of ζ(1) as the finite value corresponding to 1 + 2 + 3 + . The value of ζ(0) is obtained limiting process because it involves ζ(1) which is not defined. However, there is an equivalent formulation of Riemann zeta function for all z {1} as

ζ(z) = 1 1 21z k=0 1 2k+1 i=0k(1)ik i (i + 1)z

with a simple pole and residue 1 at z = 1.

Recall that the analytic continuation T of the geometric series is not defined for z = 1 and, hence, we could not assign a finite value to

T(1) = 1 + 1 + 1 + .

But, setting z = 0 above,

T(1) = ζ(0) = k=0 1 2k+1 i=0k(1)ik i = k=0 δ0k 2k+1 = 1 2

where δ0k is the Kronecker delta defined as

δ0k = 1k = 00 k 0.

The harmonic series

n=11 n

corresponds to ζ(1) which is not defined. The closest one can conclude about ζ(1) is that

lim n k=1n1 k ln(n) = γ

where γ is the Euler-Mascheroni constant which has the value γ = 0.57721566.

Tuesday, 17 June 2014

Bernoulli Numbers and Polynomials

Bernoulli Numbers and Polynomials

The sum of first n natural numbers 1, 2, 3,,n is

S1(n) := m=1nm = n(n + 1) 2 = n2 2 + n 2.

This formula can be derived by noting that

S1(n) = 1 + 2 + + n S1(n) = n + (n 1) + + 1.

Therefore, summing term-by-term,

2S1(n) = (n + 1) + + (n + 1) ntimes = n(n + 1).

An alternate way of obtaining the above sum is by using the following two identity:

(i)
(m + 1)2 m2 = 2m + 1 and, hence,
2 m=1nm = m=1n (m + 1)2 m2 n.

(ii)
m=1n (m + 1)2 m2 = [22 12] + [32 22] + + +[n2 (n 1)2] + [(n + 1)2 n2] = (n + 1)2 1.

Thus,

S1(n) = (n + 1)2 (n + 1) 2 = n(n + 1) 2 .

More generally, the sum of k-th power of first n natural numbers is denoted as

Sk(n) := 1k + 2k + + nk.

Since a0 = 1, for any a, we have S0(n) = n. For k , one can compute Sk(n) using the identities:

(i)
(m + 1)k+1 mk+1 = i=0kk + 1 i mi

and, hence,

(k+1) m=1nmk = m=1n (m + 1)k+1 mk+1 i=0k1 m=1nk + 1 i mi.

(ii)
m=1n (m + 1)k+1 mk+1 = (n + 1)k+1 1.

Thus,

(k + 1) m=1nmk = (n + 1)k+1 1 i=0k1k + 1 i m=1nmi Sk(n) = (n + 1)k+1 k + 1 (n + 1) k + 1 1 k + 1 i=1k1k + 1 i Si(n).

The formula obtained in RHS is a (k + 1)-degree polynomial of n. Using the above formula, one can compute

S2(n) = n3 3 + n2 2 + n 6, S3(n) = n4 4 + n3 2 + n2 4 , S4(n) = n5 5 + n4 2 + n3 3 n 30, S5(n) = n6 6 + n5 2 + 5n4 12 n2 12, S6(n) = n7 7 + n6 2 + n5 2 n3 6 + n 42, S7(n) = n8 8 + n7 2 + 7n6 12 7n4 24 + n2 12,

James/Jacques/Jakob Bernoulli observed that the sum of first n whole numbers raised to the k-th power can be concisely written as,

Sk(n) = nk+1 k + 1 + 1 2nk + 1 12knk1 + 0 × nk2 + .

Note that the coefficients 1, 12, 112, 0, are independent of k. Jakob Bernoulli rewrote the above expression as

(k + 1)Sk(n) = nk+1 1 2(k + 1)nk + 1 6 k(k + 1) 2 nk1 + 0 1 30 (k 2)(k 1)k(k + 1) 4! +

and, hence,

Sk(n) = 1 k + 1 i=0kk + 1 i Bink+1i, (1)

where Bi are the i-th Bernoulli numbers and the formula is called Bernoulli formula. An easier way to grasp the above formula for Sk(n) is to rewrite1 it as

Sk(n) = (n + B)k+1 B k+1 k + 1

where B is a notation used to identify the i-th power of B with the i-th Bernoulli number Bi and

(n + B)k+1 := i=0k+1k + 1 i Bink+1i.

This notation also motivates the definition of Bernoulli polynomial of degree k as

Bk(t) := i=0kk i Bitki

where Bi are the Bernoulli numbers. In terms of Bernoulli polynomials, the k-th Bernoulli number Bk = Bk(0).

Two quick observation can be made from (1).

(i)
There is no constant term in Sk(n) because i does not take k + 1.
(ii)
The k-th Bernoulli number, Bk, is the coefficient of n in Sk(n). For instance, B0 is coefficient of n in S0(n) = n and, hence, B0 = 1. Similarly, B1 = 12,B2 = 16,B3 = 0,B4 = 130,B5 = 0,B6 = 142,B7 = 0,.

The beauty about the sequence of Bernoulli numbers is that one can compute them a priori and use it to calculate Sk(n), i.e., given n and k {0} it is enough to know Bi, for all 0 i k, to compute Sk(n). We already computed B0 = 1. Using n = 1 in the Bernoulli formula (1), we get

1 = 1 k + 1 i=0kk + 1 i Bi (2)

and, this implies that the k-th Bernoulli number, for k > 0, is defined as

Bk = 1 1 k + 1 i=0k1k + 1 i Bi.

Recall that B3,B5,B7 vanish. In fact, it turns out that Bk = 0 for all odd k > 1. The odd indexed Bernoulli numbers vanish because they have no n-term. Since there are no constant terms in Sk(n), the vanishing of Bernoulli numbers is equivalent to the fact that n2 is a factor of Sk(n). Some properties of Bernoulli numbers:

(i)
For odd k > 1, Bk = 0.
(ii)
For even k, Bk0.
(iii)
Bk for all k {0}.
(iv)
B0 = 1 is the only non-zero integer.
(v)
B4j is negative rational and B4j2 is positive rational, for all j .
(vi)
|B6 = 142| < |B2k|, for all k .

L. Euler gave a nice generating function for the Bernoulli numbers. He seeked a function f(x) such that f(k)(0) = B k where f(k) denotes the k-th derivative of f with the convention that f(0) = f. If such an f exists then it admits the Taylor series expansion, around 0,

f(x) = k=0f(k)(0)xk k! .

Therefore, for such a function

f(x) = k=0B kxk k! .

Recall the Taylor series of ex,

ex = k=0xk k!

defined for all x . Consider the product (discrete convolution/Cauchy product) of the infinite series

f(x)ex = k=0B kxk k! k=0xk k! = k=0 i=0kB ixi i! xki (k i)! = k=0 i=0k Bi i!(k i)! xk = k=0 i=0kk i Bi xk k! .

Let

ck := i=0kk i Bi = i=0k1k i Bi + Bk = k + Bk.

Then

f(x)ex = k=0(k + B k)xk k! = k=1 xk (k 1)! + f(x) = xex + f(x).

Thus, the f we seek satisfies

f(x) = xex ex 1

and is called the generating function. Since ex > 0 for all x , we may rewrite f(x) as

f(x) = x 1 ex. (3)

The entire exercise of seeking f can be generalised to complex numbers and

f(z) = z 1 ezz  with 0 (z) < 2π.

A word of caution that idenitities (2) and (3) are different from the standard formulae because we have derived them for second Bernoulli numbers, viz., with B1 = 12. The standard convention is to work with first Bernoulli numbers, viz., with B1 = 12. The first Bernoulli numbers can be obtained by following the approach of summing the k-th powers of first n 1 natural numbers, for any given n.

The Bernoulli numbers with appeared while computing Sk(n) is appears in many crucial places.

(a)
In the expansion of tan z.
tan z = k=1(1)k122k(22k 1)B 2k 2n! z2k1.

for all |z| < π2.

(b)
In computing the sum of Riemann-zeta function
ζ(s) = n=1 1 ns

for positive even integers s. The case s = 2 is the famous Basel problem computed by Euler to be π26.

Theorem 1 (Euler). For all k ,

ζ(2k) = (1)k+1(2π)2k 2(2k)!B2k.

Further, the relation ζ(2k) = B2k+1 2k+1 gives the trivial zeroes of the Riemann zeta function.

(c)
The Bernoulli numbers were also used as an attempt to prove Fermat’s last theorem (already discussed in a previous article/blog).

Definition 1. An odd prime number p is called regular if p does not divide the numerator of Bk, for all even k p 3. Any odd prime which is not regular is called irregular.

The odd primes 3, 5, 7,, 31 are all regular primes. The first irregular prime is 37. It is an open question: are there infinitely many regular primes? However, it is known that there are infinitely many irregular primes.

Theorem 2 (Kummer, 1850). If p is a regular odd prime then the equation

ap + bp = cp

has no solution in .