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 b_k 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) e^kz = [e^nz - 1]/[e^z - 1]

Proof:

(1) Let S_n = ∑ (k=0,n-1) e^kz = ∑ (k=0, n) (e^z)^k = e⁰ + e^z + e^2z + ... + e^(n-1)z

(2) S_n + (e^z)ⁿ = (e^z)⁰ + ∑ (k=0,n-1) (e^z)^k+1

(3) (e^z)S_n = (e^z)∑ (k=0,n-1) (e^z)^k = ∑ (k=0,n) (e^z)^k+1

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

S_n + (e^z)ⁿ = (e^z)S_n + 1

(5) So, solving for S_n gives us:

S_n - (e^z)S_n = 1 - (e^z)ⁿ = S_n(1 - e^z)

(6) Dividing both sides by (1 - e^z) gives us:

S_n = [1 - (e^z)ⁿ]/[1 - e^z] *(-1)/(-1) = [e^nz - 1]/[e^z - 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) B_m(n)z^m/m! = (ze^xz)/(e^z - 1)

Proof:

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

T(z,n) = B₀(n)z⁰/0! + B₁(n)z¹/1! + .... = ∑ (m=0,∞) B_m(n)z^m/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]!)*B_kx^n-k] z^m/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]!)*B_kx^n-k] z^m/m! = [∑(m=0; ∞) B_mz^m/m!]*[∑(m=0, ∞) x^mz^m/m!]

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

∑ (m ≥ 0) (B_m)z^m/m! = z/[e^z - 1]

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

e^xz = ∑ (m ≥ 0) x^mzm/m!

(6) Inserting these results gives us:

T(z,n) = ∑(m=0; ∞) ∑(k=0;m) [(m!/[k!][m-k]!)*B_kx^n-k] z^m/m! = [∑(m=0; ∞) B_mz^m/m!]*[∑(m=0, ∞) x^mz^m/m!] =

= (z/[e^z - 1])*(e^xz) = (ze^xz)/(e^z - 1)

QED

Theorem: S_m-1(n) = (1/m)(B_m(n) - B_n(0))

Proof:

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

(2) Let S(z) = ∑ (m ≥ 0) S_m(n) z^m = S₀(n) + S₁(n)z + S₂(n)z² + ... = ∑ (m ≥ 0) ∑ (k=0,n) k^mz^m

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

e^z = ∑ (n ≥ 0) zⁿ/n! = z⁰/0! + z¹/1! + ... + zⁿ/n!

(4) We can see that:

e^kz = ∑ (m ≥ 0) (kz)^m/m! = ∑(m ≥ 0)k^mz^m/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) S_m(n)z^m/m!

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

G(z,n) = ∑ (m ≥ 0) ∑ (k=0, n) k^mz^m/m! = ∑ (k=0,n) e^kz

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

G(z,n) = [e^nz - 1]/[e^z - 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) B_m(n)z^m/m! = (ze^xz)/(e^z - 1)

(10) This then gives us that:

T(z,n) - T(z,0) = (ze^nz)/(e^z - 1) - (ze^0z)/(e^z - 1) = z[e^nz - 1]/[e^z-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:

S_m-1(n)/(m-1)! = B_m(n)/m!] - B_m(0) /m!

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

S_m-1(n) = (1/m)[B_m(n) - B_m(0)]

<sub>QED

References</sub>

• "Bernoulli Polynomials", Wikipedia.com
• Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, Addison-Wesley, 1989