Bernoulli polynomials

Jordan Bell
March 9, 2016

1 Bernoulli polynomials

For k0, the Bernoulli polynomial Bk(x) is defined by

zexzez-1=k=0Bk(x)zkk!,|z|<2π. (1)

The Bernoulli numbers are Bk=Bk(0), the constant terms of the Bernoulli polynomials. For any x, using L’Hospital’s rule the left-hand side of (1) tends to 1 as z0, and the right-hand side tends to B0(x), hence B0(x)=1. Differentiating (1) with respect to x,


so B0(x)=0 and for k1 we have Bk(x)k!=Bk-1(x)(k-1)!, i.e. Bk(x)=kBk-1(x). Furthermore, for k1, integrating (1) with respect to x on [0,1] produces


hence 01B0(x)𝑑x=1 and for k1,


The first few Bernoulli polynomials are


The Bernoulli polynomials satisfy the following:

k=0Bk(x+1)zkk! =ze(x+1)zez-1

hence for k1 it holds that Bk(x+1)=kxk-1+Bk(x). In particular, for k2, Bk(1)=Bk(0).

Using (1),

k=0Bk(1-x)zkk! =ze(1-x)zez-1

hence for k0,


Finally, it is a fact that for k2,

sup0x1|Bk(x)|2ζ(k)k!(2π)k. (2)

2 Periodic Bernoulli functions

For x, let [x] be the greatest integer x, and let R(x)=x-[x], called the fractional part of x. Write 𝕋=/ and define the periodic Bernoulli functions Pk:𝕋 by


For k2, because Bk(1)=Bk(0), the function Pk is continuous. For f:𝕋 define its Fourier transform f^: by


For k1, one calculates P^k(0)=0 and using integration by parts,


for n0. Thus for k1, the Fourier series of Pk is11 1 cf.


For k2, n|P^k(n)|<, from which it follows that |n|NP^k(n)e2πint converges to Pk(t) uniformly for t𝕋. Furthermore, for t,22 2 Hugh L. Montgomery and Robert C. Vaughan, Multiplicative Number Theory I: Classical Theory, p. 499, Theorem B.2.


For f,gL1(𝕋) and n,

f*g^(n) =𝕋(𝕋f(x-y)g(y)𝑑y)e-2πinx𝑑x

For k,l1 and for n0,

Pk*Pl^(n) =Pk^(n)Pl^(n)

and Pk*Pl^(0)=0=-Pk+l^(0), so Pk*Pl=-Pk+l.

3 Euler-Maclaurin summation formula

The Euler-Maclaurin summation formula is the following.33 3 Hugh L. Montgomery and Robert C. Vaughan, Multiplicative Number Theory I: Classical Theory, p. 500, Theorem B.5. If a<b are real numbers, K is a positive integer, and f is a CK function on an open set that contains [a,b], then

a<mbf(m) =abf(x)𝑑x+k=1K(-1)kk!(Pk(b)f(k-1)(b)-Pk(a)f(k-1)(a))

Applying the Euler-Maclaurin summation formula with a=1,b=n,K=2,f(x)=logx yields44 4 Hugh L. Montgomery and Robert C. Vaughan, Multiplicative Number Theory I: Classical Theory, p. 503, Eq. B.25.


Since e1+O(n-1)=1+O(n-1),


Stirling’s approximation.

Write an=-logn+1mn1m. Because log(1-x) is concave,


which means that the sequence an is nonincreasing. For f(x)=1x, because f is positive and nonincreasing,


hence an>0. Because an is positive and nonincreasing, there exists some nonnegative limit, γ, called Euler’s constant. Using the Euler-Maclaurin summation formula with a=1,b=n,K=1,f(x)=1x, as P1(x)=[x]-12,


which is


as 0R(x)x-2x-2, the function xR(x)x-2 is integrable on [1,). Since 0nR(x)x-2𝑑xnx-2𝑑x=n-1,


for C=1-1R(x)x-2. But -logn+1mn1mγ as n, from which it follows that C=γ, and thus


4 Hurwitz zeta function

For 0<α1 and Res>1, define the Hurwitz zeta function by


For Res>0,


and for n0 do the change of variable t=(n+α)u,

Γ(s) =0(n+α)s-1us-1e-(n+α)u(n+α)𝑑u

For real s>1,






fN(s,u)0 and the sequence fN(s,u) is pointwise nondecreasing, and


By the monotone convergence theorem,


which means that, for real s>1,




Now, by (1), for 0<u<2π,

f(s,u) =us-1e-αu11-e-u

For k2, real s>1, and 0<u<2π, by (2),


which is summable, and thus by the dominated convergence theorem,

01f(s,u)𝑑u =01k=0(-1)kBk(α)uk+s-2k!du

Check that sk=0(-1)kBk(α)1k!1k+s-1 is meromorphic on , with poles of order 0 or 1 at s=-k+1, k0 (the order of the pole is 0 if Bk(α)=0), at which the residue is (-1)kBk(α)1k!.55 5 Kazuya Kato, Nobushige Kurokawa, and Takeshi Saito, Number Theory 1: Fermat’s Dream, p. 96. On the other hand, check that s1f(s,u)𝑑u is entire. Therefore ζ(s,α)Γ(s) is meromorphic on , with poles of order 0 or 1 at s=-k+1, k0 and the residue of ζ(s,α)Γ(s) at s=-k+1 is (-1)kBk(α)1k!. But it is a fact that Γ(s) has poles of order 1 at s=-n, n0, with residue (-1)nn!. Hence the only pole of ζ(s,α) is at s=1, at which the residue is 1.

Theorem 1.

For n1 and for 0<α1,


For n1, because ζ(s,α) does not have a pole at s=1-n and because Γ(s) has a pole of order 1 at s=1-n with residue (-1)n-1(n-1)!,

lims1-n(s-(1-n))Γ(s)ζ(s,α) =ζ(1-n,α)lims1-n(s-(1-n))Γ(s)

On the other hand, ζ(s,α)Γ(s) has a pole of order 1 at s=1-n with residue (-1)nBn(α)1n!. Therefore


i.e. for n1 and 0<α1,


5 Sobolev spaces

For real s0, we define the Sobolev space Hs(𝕋) as the set of those fL2(𝕋) such that


For f,gHs(𝕋), define


This is an inner product, with which Hs(𝕋) is a Hilbert space.66 6 See

For c, if s>r+12,


For fHs(𝕋), the partial sums |n|Nf^(n)e2πinx are a Cauchy sequence in Hs(𝕋) and by the above are a Cauchy sequence in the Banach space Cr(𝕋) and so converge to some gCr(𝕋). Then g^=f^, which implies that g=f almost everywhere.

For k1, P^k(0)=0 and P^k(n)=-(2πin)-k for n0. For k,l>s+12,

Pk,PlHs(𝕋) =n{0}-(2πin)-k-(2πin)-l¯

Thus if k>s+12 then PkHs(𝕋), and in particular PkHk-1(𝕋) for k1.

For s>r+12, if fHs(𝕋) then there is some gCr(𝕋) such that g=f almost everywhere. Thus if r+12<s<k-12, i.e. k>r+1, then there is some gCr(𝕋) such that g=Pk almost everywhere. But for k1, Pk is continuous, so in fact g=Pk. In particular, PkCk-2(𝕋) for k2.

6 Reproducing kernel Hilbert spaces

For x𝕋 and f:𝕋, define (τxf)(y)=f(y-x). We calculate

τxf^(n) =𝕋f(y-x)e-2πiny𝑑y

Let r1. For x𝕋, define Fx:𝕋 by


For n,

Fx^(n) =δ0(n)+(-1)r-1(2π)2re-2πinxP^2r(n).

Fx^(0)=1, and for n0,


For fHr(𝕋),

f,FxHr(𝕋) =f^(0)Fx^(0)¯+n{0}f^(n)Fx^(n)¯|n|2r

This shows that Hr(𝕋) is a reproducing kernel Hilbert space.

Define F:𝕋×𝕋 by

F(x,y) =Fx,FyHr(𝕋)

Thus the reproducing kernel of Hr(𝕋) is77 7 cf. Alain Berlinet and Christine Thomas-Agnan, Reproducing Kernel Hilbert Spaces in Probability and Statistics, p. 318, who use a different inner product on Hr(𝕋) and consequently have a different expression for the reproducing kernel.