## 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

## Monday, October 23, 2006

### A Generating Function for the Bernoulli Numbers

In my previous blog, I defined the Bernoulli numbers using a recurrence relation:

In today's blog, I will show how the Bernoulli numbers can be used with a generating function. A generating function is a power series, that is, a compact expression that defines an infinite sum. For background on generating functions, I recommend the wikipedia article (see reference) or Graham et al's Concrete Mathematics (see reference).

The content in today's blog is taken from Concrete Mathematics. The term "binomial convolution" is taken from this work. A "convolution" is the multiplication of two generating functions. This lemma is called a "binomial convolution" by Graham et al because the result involves binomial coefficients (see binomial theorem, here, for more details).

Lemma 1: Binomial Convolution

Let:

h(z) = f(z)*g(z)

Then there exists dn such that:

with:

Proof:

(1) Multiplying f(z) with g(z) gives us:

(2) Grouping the result by zi gives us:

(3) If we set ci to each coefficient. For example, if we set c0 to (a0b0)/(0!0!), then we have the following formula:

(4) Now, let's define a value di such that:

(5) This then gives us that:
di/(i!) = ci

(6) Combining step #5 with step #2 and step #3 gives us:

QED

Thereom: Generating Function for the Bernoulli Numbers

z/(ez - 1) is the generating function for the Bernoulli numbers

Proof:

(1) Let G(z) = ∑ (n ≥ 0) Bnzn/n! where Bi is the Bernoulli number for i.

(2) We know from a previous result (see Lemma 2, here) that:

ez = ∑ (n ≥ 0) zn/n!

(3) So we can use Lemma 1 above to show that:

(4) Now, the definition for Bernoulli numbers (see Definition 1, here) is:

for all cases except for B0 where B0 = 1.

(5) If we let n=m+1 and add Bn to both sides, then we have:

or Bn + 1 in the case where n=m+1=1.

(6) This enables us to simplify the result in step #3 to get:

Note: the z at the bottom comes from the fact that at n=1, our result is (B1+1)z1/1! = B1z1/1! + z.

(7) If we subtract G(z) from both sides, we get:

(ez)G(z) - G(z) = z =

G(z)(ez - 1) = z

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

G(z) = z/(ez - 1)

QED

Corollary: B2i+1=0 if i ≥ 1.

Proof:

(1) z/(ez - 1) = ∑ (n ≥ 0) Bnzn/n! [This follows from the theorem above]

(2) Now, B1 = -(1/2) [See Definition 1, here]

(3) So,

z/(ez - 1) + (z/2) = ∑ (n ≥ 0 and n ≠ 1) Bnzn/n!

We subtract (-z/2) instead of -1/2 because we are removing this item from a generating function where each value in the sum z/(ez-1) pairs up with a unique term of the sum.

In this case, B1 matches up with the term B1(z1)/1! which equals -z/2.

(4) z/(ez - 1) + (z/2) = (2z + z[ez - 1])/(2[ez-1]) = (2z -z + z[ez])/(2[ez-1]) =

= (z/2)(e
z+1)/(ez - 1)

(5) (z/2)(ez + 1)/(ez-1) = [e-(z/2)/e-(z/2)]* (z/2)(ez + 1)/(ez-1) =

= (z/2)[(ez/2 + e-z/2)/(ez/2 - e-z/2)]

(6) So, we see in step #5, that if we replace z with (-z), the identity remains unchanged.

(-z/2)[(e-z/2 + e-(-z/2)]/(e-z/2 - e-(-z)/2) =

= -z/2[e-z/2 + ez/2]/(e-z/2 - ez/2) =

= z/2[ez/2 + e-z/2]/(ez/2 - e-z/2)

In the language of mathematics, a function is called even if f(-x)=f(x) and we call a function odd if f(-x) = -f(x). [See here for more details]

Example: cos(x) is an even function while sin(x) is an odd function. [See here for review of sin and cos]

(7) But this means that the same must hold true for ∑ (n ≥ 0 and n ≠ 1) Bnzn/n!

Since by Step #3, we showed that they are equal functions.

So that we have:

∑ (n ≥ 0 and n ≠ 1) Bnzn/n! = ∑ (n ≥ 0 and n ≠ 1) Bn(-z)n/n! = ∑ (n ≥ 0 and n ≠ 1) (-1)n(Bnzn/n!)

(8) Matching up each of the terms according to powers of zn, this gives us that:

Bn = (-1)nBn

(9) But this can only be true if Bn = 0 when n is odd.

QED

References