## Thursday, October 26, 2006

### Bernoulli Polynomials

In today's blog, I introduce Bernoulli polynomials and show how they can be used to present an even shorter version of the summation formula we found for Bernoulli numbers.

The content in today's blog is taken from Concrete Mathematics (Graham, Knuth, and Patashnik, 1989).

Definition 1: Bernoulli Polynomial

where n ≥ 0 and bk is the Bernoulli number (see Definition 1, here)

I will now show how Bernoulli Polynomials can be used to create a concise summation formula.

Lemma 1: ∑ (k=0,n) ekz = [enz - 1]/[ez - 1]

Proof:

(1) Let Sn = ∑ (k=0,n-1) ekz = ∑ (k=0, n) (ez)k = e0 + ez + e2z + ... + e(n-1)z

(2) Sn + (ez)n = (ez)0 + ∑ (k=0,n-1) (ez)k+1

(3) (ez)Sn = (ez)∑ (k=0,n-1) (ez)k = ∑ (k=0,n) (ez)k+1

(4) Putting #2 and #3 together gives us:

Sn + (ez)n = (ez)Sn + 1

(5) So, solving for Sn gives us:

Sn - (ez)Sn = 1 - (ez)n = Sn(1 - ez)

(6) Dividing both sides by (1 - ez) gives us:

Sn = [1 - (ez)n]/[1 - ez] *(-1)/(-1) = [enz - 1]/[ez - 1]

QED

NOTE: If you are not familiar with the idea of generating functions, start here.

Corollary 1.1: Generating Function for Bernoulli Polynomials

∑ (m ≥ 0) Bm(n)zm/m! = (zexz)/(ez - 1)

Proof:

(1) Let T(z,n) be the generating function for Bernoulli polynomials Bm(n) so that we have:

T(z,n) = B0(n)z0/0! + B1(n)z1/1! + .... = ∑ (m=0,∞) Bm(n)zm/m!

(2) Inserting the definition for Bernoulli polynomials (see Definition1 above), we have:

T(z,n) = ∑ (m=0,∞) ∑(k=0; m) [(m!/[k!][m-k]!)*Bkxn-k] zm/m!

(3) Using a previous result on binomial convolution (see Lemma 1, here), we can reverse this to get:
∑(m=0; ∞) ∑(k=0;m) [(m!/[k!][m-k]!)*Bkxn-k] zm/m! = [∑(m=0; ∞) Bmzm/m!]*[∑(m=0, ∞) xmzm/m!]

(4) From a previous result (see Theorem, here), we have:

∑ (m ≥ 0) (Bm)zm/m! = z/[ez - 1]

(5) Further, using the power series for e (see Lemma 2, here), we have:

exz = ∑ (m ≥ 0) xmzm/m!

(6) Inserting these results gives us:

T(z,n) = ∑(m=0; ∞) ∑(k=0;m) [(m!/[k!][m-k]!)*Bkxn-k] zm/m! = [∑(m=0; ∞) Bmzm/m!]*[∑(m=0, ∞) xmzm/m!] =

= (z/[ez - 1])*(exz) = (zexz)/(ez - 1)

QED

Theorem: Sm-1(n) = (1/m)(Bm(n) - Bn(0))

Proof:

(1) Sm(n) = 0m + 1m + ... + (n-1)m = ∑ (k=0,n) km [See Definition2, here for definition of Sm(n)]

(2) Let S(z) = ∑ (m ≥ 0) Sm(n) zm = S0(n) + S1(n)z + S2(n)z2 + ... = ∑ (m ≥ 0) ∑ (k=0,n) kmzm

(3) Using the power series for ez (see Lemma 2, here), we know that:

ez = ∑ (n ≥ 0) zn/n! = z0/0! + z1/1! + ... + zn/n!

(4) We can see that:

ekz = ∑ (m ≥ 0) (kz)m/m! = ∑(m ≥ 0)kmzm/m!

(5) Let G(z,n) be the generating function for S(z) so that we have:

G(z,n) = S0(n)x0/0! + S1(n)z1/1! + S2(n)/z2/2! + ... = ∑ (m ≥ 0) Sm(n)zm/m!

(6) Using step #2, then we have:

G(z,n) = ∑ (m ≥ 0) ∑ (k=0, n) kmzm/m! = ∑ (k=0,n) ekz

(7) Using the Lemma 1 above, we have:

G(z,n) = [enz - 1]/[ez - 1]

(8) Let T(z,n) be the generating function for the Bernoulli Polynomial

(9) Using Corollary 1.1 above, we have:

T(z,n) = ∑ (m ≥ 0) Bm(n)zm/m! = (zexz)/(ez - 1)

(10) This then gives us that:

T(z,n) - T(z,0) = (zenz)/(ez - 1) - (ze0z)/(ez - 1) = z[enz - 1]/[ez-1]

(11) This shows that z*G(z,n) = [T(z,n) - T(z,0)]

(12) So that we have:

S0(n)z1/0! + S1(n)z2/1! + S2(n)/z3/2! + ... = [(B0(n)z0/0! + B1(n)z1/1! + ....) - (B0(0)z0/0! + B1(0)z1/1! + ....)]

(13) Lining up, equal powers of zi gives us:

Sm-1(n)/(m-1)! = Bm(n)/m!] - Bm(0) /m!

(14) Multiplying both sides by (m-1)! gives us:

Sm-1(n) = (1/m)[Bm(n) - Bm(0)]

QED

References