# Bernoulli Numbers

• Jan 4th 2009, 09:31 AM
HTale
Bernoulli Numbers
I'm unsure where this should go, but because it is related to a formula for $\displaystyle \zeta(2k)$, I've put it in here. The question is regarding this recursive formula for Bernoulli numbers

$\displaystyle \displaystyle \sum^{n-1}_{k=0} {n \choose k} B_k = 0$

Essentially, we therefore have the following;

$\displaystyle \displaystyle {0 \choose k} B_0 + {1 \choose k} B_1 + {2 \choose k} B_2 + \ldots + {n-1 \choose k} B_{n-1} = 0$

Given we also have

$\displaystyle \displaystyle {n-j \choose k} = {n \choose k} \quad (\forall \,\, 1 \leq j \leq n-1)$

Then we can pair each $\displaystyle k$ for $\displaystyle 1 \leq k \leq n-1$, to obtain

$\displaystyle \displaystyle {0 \choose k} B_0 + {1 \choose k}(B_{n-1}+ B_1 ) + \ldots + {j \choose k} (B_{n-j} + B_j)= 0$

Where each $\displaystyle (n-j) + j = n$. I then wanted to consider $\displaystyle n$ even and $\displaystyle n$ odd, but I'm unsure if this is even the right way to go about. If so, how on earth do you validate that recursive formula?

HTale.

• Jan 4th 2009, 11:14 AM
galactus
Are you trying to derive the formula for $\displaystyle B_{n}$, the nth Bernoulli number?.

For the Bernoulli polynomials, we have:

$\displaystyle \int P_{n}(x)dx=0 \;\ n=1,2,3,.....$

The resulting polynomials are our Bernoulli polynomials.

And are denoted by $\displaystyle B_{n}$

$\displaystyle B_{n}$ is a poly with degree n with rational coefficients. The first few are:

$\displaystyle B_{0}(x)=1$..............[1]

$\displaystyle B_{1}(x)=x-\frac{1}{2}$

$\displaystyle B_{2}(x)=\frac{1}{2}x^{2}-\frac{1}{2}x+\frac{1}{12}$

$\displaystyle B_{3}(x)=\frac{1}{6}x^{3}-\frac{1}{4}x^{2}+\frac{1}{12}x$

$\displaystyle B_{4}(x)=\frac{1}{24}x^{4}-\frac{1}{12}x^{3}+\frac{1}{24}x^{2}-\frac{1}{720}$

To display the Bernoulli polys in a form in which their coefficients become nice and elegant, we can write the constant terms as:

$\displaystyle B_{n}(0)=\frac{B_{n}}{n!}, \;\ n=0,1,2,3,.....$

Then we get:

$\displaystyle B_{0}(x)=B_{0}$

$\displaystyle B_{1}(x)=\frac{x}{1!}+\frac{B_{1}}{1!}$

$\displaystyle B_{2}(x)=\frac{x^{2}}{2!}+\frac{B_{1}}{1!}\cdot\fr ac{x}{1!}+\frac{B_{2}}{2!}$

$\displaystyle B_{3}(x)=\frac{x^{3}}{3!}+\frac{B_{1}}{1!}\cdot\fr ac{x^{2}}{2!}+\frac{B_{2}}{2!}\cdot\frac{x}{1!}+\f rac{B_{3}}{3!}$

$\displaystyle B_{4}(x)=\frac{x^{4}}{4!}+\frac{B_{1}}{1!}\cdot\fr ac{x^{3}}{3!}+\frac{B_{2}}{2!}\cdot\frac{x^{2}}{2! }+\frac{B_{3}}{3!}\cdot\frac{x}{1!}+\frac{B_{4}}{4 !}$

and so on.

in general:

$\displaystyle B_{n}(x)=\frac{1}{n!}\sum_{k=0}^{n}\binom n k\cdot B_{k}x^{n-k}$............[2]

This ca be verified by induction.

But, as we know, $\displaystyle B_{0}, \;\ B_{1}, \;\ B_{2}, \;\ ....$ are the Bernoulli numbers.

So, from [1], we see we get:

$\displaystyle B_{0}+1, \;\ B_{1}=\frac{-1}{2}, \;\ B_{2}=\frac{1}{6}, \;\ B_{3}=0, \;\ B_{4}=\frac{-1}{30}$

Their computation can be found by the observation that, for n=1,2,3,...

$\displaystyle 0=\int_{0}^{1}B_{n}(x)dx=\int_{0}^{1}B'_{n+1}(x)dx =B_{n+1}(1)-B_{n+1}(0)$

and therefore

$\displaystyle B_{n+1}(0)=B_{n+1}(1)$

Thus, bu [2]:

$\displaystyle B_{n+1}=\sum_{k=0}^{n+1}\begin{pmatrix}n+1\\k\end{ pmatrix} \cdot B_{k}$

$\displaystyle 0=\sum_{k=0}^{n}\begin{pmatrix} n+1\\k\end{pmatrix}\cdot B_{k}$

This gives us a recursive procedure for finding [tex]B_{n}[tex] once $\displaystyle B_{0}, \;\ B_{1}, \;\ ...., \;\ B_{n-1}$ are known.

So, $\displaystyle B_{n}=0$ when n is odd and greater than 1. And $\displaystyle B_{n}$ alternates in sign whenever n is even.

Now, we can derive the formula for the sum of consecutive powers.

For every real number x,

$\displaystyle B_{n+1}(x+1)-B_{n+1}(x)=\frac{x^{n}}{n!}, \;\ n=0,1,2,3,....$

When n=0, the right side is equal to 1.

Proceeding with induction, we assume its tre for n and show that:

$\displaystyle B_{n+1}(x+1)-B_{n+1}(x)=\frac{x^{n+1}}{(n+1)!}$

Skipping ahead, we can derive Euler's beautiful formula for the sum of the reciprocals of all even powers:

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{2k}}=\frac{(-1)^{k-1}\cdot B^{2k}\cdot (2\pi)^{2k}}{2(2k)!}$

For k=1, $\displaystyle \sum_{1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2}}{ 6}$

k=2, $\displaystyle \sum_{1}^{\infty}\frac{1}{n^{4}}=\frac{{\pi}^{4}}{ 90}$

k=3, $\displaystyle \sum_{1}^{\infty}\frac{1}{n^{6}}=\frac{{\pi}^{6}}{ 945}$

There is much more, but I am tired of typing. I may come back with more. I done research on these in undergrad some time back. I hope this was a nice tutorial.
• Jan 4th 2009, 11:39 AM
HTale
Thank you ever so much. So, these Bernoulli polynomials, how do I actually know they all have rational coefficients? I'm guessing if you can prove that the Bernoulli polynomials have rational coefficients, then it follows that each Bernoulli number is also rational. How would you go about showing that?
• Jan 4th 2009, 12:18 PM
galactus
That is evident from $\displaystyle \int_{0}^{1}P_{n}(x)dx$

The polynomials $\displaystyle 1, \;\ x, \;\ \frac{x^{2}}{2!}, \;\ \frac{x^{3}}{3!}, \;\ \frac{x^{4}}{4!}, ......$

which are very prominent in Taylor series, have the property that each is the antiderivative of the preceeding one.

$\displaystyle D(x)=1$

$\displaystyle D(\frac{x^{2}}{2!})=x$

$\displaystyle D(\frac{x^{3}}{3!})=\frac{x^{2}}{2!}$

$\displaystyle D(\frac{x^{4}}{4!})=\frac{x^{3}}{3!}$

and so on. Where D is the differentiation operator.

In general, If $\displaystyle P_{n}$ is the nth polynomial in the sequence, then

$\displaystyle P_{0}(x)=1, \;\ P'_{n}(x)=P_{n-1}(x)$

The resulting polynomials are our Bernoulli polys, as stated before.

Now, it is evident that $\displaystyle B_{n}$ is a polynomial of degree n with rational coefficients.

But, what you want is a proof of the recursive formula, am I right?.

Here goes:

We have already shown that $\displaystyle B_{n}(0)=B_{n}(1)$ whenever

$\displaystyle n\geq 2$, and thus both sides agree when x=0. To see they agree everywhere if just suffices to show that they have the same derivative.

Since by the induction hypothesis,

$\displaystyle D\left[B_{n+1}(x+1)-B_{n+1}(x)\right]=B_{n+1}(x+1)-B_{n+1}(x)=D\left[\frac{x^{n+1}}{(n+1)!}\right]$

Corollary:

The sum of the nth powers of the first N-1 positive integers is given by:

$\displaystyle 1^{n}+2^{n}+3^{n}+....+(N-1)^{n}=n!\left[B_{n+1}(N)-B_{n+1}(0)\right]=n!\int_{0}^{N}B_{n}(x)dx$

Proof

The first formula is an almost immediate consequence of

$\displaystyle B_{n+1}(x+1)-B_{n+1}(x)=\frac{x^{n}}{n!}$

Simply take $\displaystyle x=0,1,2,....,N-1$ and then add the resulting equations. We then get:

$\displaystyle 1^{n}+2^{n}+....+(N-1)^{n}=n!\sum_{x=0}^{N-1}\left[B_{n+1}(x+1)-B_{n+1}(x)\right]=n!\left[B_{n+1}(N)-B_{n+1}(0)\right]$

The second formula follows from the first(by the fundamental theorem of calculus)since $\displaystyle B'_{n+1}=B_{n}$

Bernoulli's formula can also be written symbolically as:

$\displaystyle 1^{n}+2^{n}+...+(N-1)^{n}=\frac{1}{n+1}\left[(N+B)^{n+1}-B^{n+1}\right]$...........[3]

Here, the term $\displaystyle (N+B)^{n+1}$ is to be expanded by means of the binomial theorem, and each power of $\displaystyle B_{i}$ is to be replaced by the corresponding Bernoulli number $\displaystyle B_{i}$.

Example:

$\displaystyle 1^{3}+2^{3}+....+(N-1)^{3}=\frac{1}{4}\left(N^{4}+4N^{3}B_{1}+6N^{2}B_ {2}+4NB_{3}\right)$

$\displaystyle =\frac{1}{4}\left(N^{4}-2N^{3}+N^{2}\right)$

$\displaystyle =\frac{1}{4}N^{2}\left(N-1\right)^{2}$

Notice also taking N=1 in formula [3], we get:

$\displaystyle \frac{1}{n+1}\left[(1+B)^{n+1}-B^{n+1}\right)=0$

or

$\displaystyle (1+B)^{n+1}=B^{n+1}, \;\ \forall \;\ n=1,2,....$

This is precisely the recursive formula:

$\displaystyle B_{n+1}=\sum_{k=0}^{n+1}\begin{pmatrix}n+1\\k\end{ pmatrix}\cdot B_{k}$

BUT, tucked away in the second half of the corollary is the statement:

$\displaystyle 1^{n}+2^{n}+....+(N-1)^{n}=n!\int_{0}^{N}B_{n}(x)dx$

This is imporant, the representation of a sum in the form of an integral.

This is what led the illustrious Euler to derive his aforementioned formula for finding the sum of the reciproicals of the even powers.
• Jan 4th 2009, 12:38 PM
HTale
You are an absolute legend! Thank you so much!
• Jan 4th 2009, 12:43 PM
ThePerfectHacker
Let $\displaystyle B_0=1$ and define, $\displaystyle B_n = - \frac{1}{n+1}\sum_{k=0}^{n-1} {{n+1}\choose k}B_k$.
The sequence $\displaystyle B_0,B_1,B_2,...$ is the Bernoulli sequence.

Lemma 1: The generating funtion for $\displaystyle \frac{z}{e^z - 1}$ is $\displaystyle \sum_{n=0}^{\infty} B_n \frac{z^n}{n!}$.

Proof: We will show $\displaystyle z = (e^z - 1)\sum_{n=0}^{\infty} B_n \frac{z^n}{n!}$.
Write as power series, $\displaystyle z = z\left( \sum_{j=0}^{\infty} \frac{z^j}{(j+1)!} \right) \left( \sum_{n=0}^{\infty} B_n \frac{z^n}{n!} \right) = z\sum_{k=0}^{\infty}a_k \tfrac{z^k}{(k+1)!}$ where $\displaystyle a_k = \sum_{i=0}^k \frac{B_i}{i!} \cdot \frac{(k+1)!}{(k-i+1)!}$
Thus, $\displaystyle a_k = \sum_{i=0}^k {{k+1}\choose i}B_i = 0$ for $\displaystyle k\geq 1$ and $\displaystyle a_0=1$ by comparing coefficients.
This is the definition of the Bernoulli sequence. Thus the identity is proven.

Lemma 2: We have that $\displaystyle \cot z = \frac{1}{z} - 2\sum_{n=1}^{\infty} \frac{z}{\pi^2 n^2 - z^2}$

Proof: We know that $\displaystyle \frac{\sin z}{z} = \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{\pi^2 n^2} \right)$
Now take the logarithm of both sides followed by the derivative of both sides.
---

We can now derive formula for $\displaystyle \zeta (2k)$.
First we have,
$\displaystyle z\cot z = 1 - 2\sum_{n=1}^{\infty} \frac{z^2}{\pi^2 n^2 - z^2}$
By geometric series,
$\displaystyle \frac{z^2}{\pi^2 n^2 - z^2} = \sum_{m=1}^{\infty} \left( \frac{z}{\pi n} \right)^m$
Therefore,
$\displaystyle z\cot z = 1 - 2 \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{z^{2m}}{\pi ^{2m} n^{2m}} = 1 - 2 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{z^{2m}}{\pi ^{2m} n^{2m}}$$\displaystyle =1 -2\sum_{m=1}^{\infty} \frac{\zeta(2m)}{\pi^{2m}}z^{2m}$

Now, $\displaystyle \cot z = \frac{\cos z}{\sin z}$ where $\displaystyle \cos z = \cosh(iz) = \tfrac{e^{iz} + e^{-iz}}{2}$ and $\displaystyle i\sin z = \sinh (iz) \implies \sin z = \tfrac{e^{iz} - e^{-iz}}{2i}$

Therefore, $\displaystyle z\cot z = iz + \frac{2iz}{e^{2iz} - 1} = iz + \sum_{m=0}^{\infty} B_m \frac{(2iz)^m}{m!} = 1 + \sum_{m=2}^{\infty}B_m \frac{(2iz)^m}{m!}$ by above Lemma.

Thus, $\displaystyle \sum_{m=2}^{\infty}B_m \frac{(2iz)^m}{m!} = -2\sum_{m=1}^{\infty} \frac{\zeta(2m)}{\pi^{2m}}z^{2m}$

Compare coefficients to get, $\displaystyle B_{2m} \frac{(-1)^m 2^{2m}}{(2m)!} = - \frac{2\zeta(2m)}{\pi^{2m}} \implies \zeta (2m) = (-1)^{m+1} \frac{2^{2m-1}\pi^{2m}}{(2m)!}B_{2m}$
• Jan 4th 2009, 04:07 PM
HTale
Thank you, ThePerfectHacker. That's another interesting approach angle, I will look at both.