Subject: Calculus

# Bernoulli Numbers

Suppose we are given with a power series of the first n natural number for positive integer exponents r which is expressed as

\sigma_r(n) = 1^r + 2^r + \cdot \cdot \cdot + n^r

For small values of r expressions for \sigma_r (n) , the values are well known. For example if we have the first three values of r, we�ll have

\sigma_1 (n) = \frac{n \left ( n+1 \right )}{2}
\sigma_2 (n) = \frac{n \left ( n+1\right ) \left ( 2n+1\right )}{6}
\sigma_3 (n) = \left [ \frac{n\left ( n+1\right )}{2}\right ]^2

Let us sum up the r^{th} powers up to n � 1. We then have

Taking the values of S_r (n) for different values of r, we have

S_0 (n) = 1^0 + 2^0 + \cdot \cdot \cdot n^0 = n
S_1 (n) = 1^1 + 2^1 + \cdot \cdot \cdot + n^1 = \frac{n \left ( n+1 \right )}{2} = \frac{n^2}{2} - \frac{n}{2}
S_2 (n) = 1^2 + 2^2 + \cdot \cdot \cdot + n^2 = \frac{n \left ( n+1\right ) \left ( 2n+1\right )}{6} = \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}
S_3 (n) = 1^3 + 2^3 + \cdot \cdot \cdot + n^3 = \frac{n^2 \left ( n-1\right )^2}{4} = \frac{n^4}{4} - \frac{n^3}{2} - \frac{n^2}{4}
S_4 (n) = 1^4 + 2^4 + \cdot \cdot \cdot + n^4 = \frac{n \left ( n-1\right )\left (2n-1 \right ) \left ( 3n^2 -3n -1 \right )}{30} = \frac{n^5}{5} - \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30}

and so on....

From this small list, we can observe some interesting properties of the polynomial S_r (n).

1. The leading term of S_r (n) is \frac{n^{r+1}}{r+1}. The next term is -\frac{n^r}{2}.
2. S_r (0) = S_r (1) = 0.
3. S_r (-n) = (-1)^{r+1} S_r (n+1) .
4. S_r (n) has factors n, (n-1). When r is even, (2n-1) is also a factor.

From the discussion of Taylor series, from the form

f(x+c) = f(x) +cf'(x) + \frac{c^2}{2!}f''(x) + \cdot \cdot \cdot

We can rewrite this into

f(x+c) = e^{c\mathbf{D}}f(x)

where \mathbf{D} is the differential operator. If we subtract S_r (n+1) and S_r (n), we get n^r. We can now write this as

(e^{\mathbf{D}}-1) S_r (n) = n^r

where \mathbf{D} here means differentiation with respect to n. Thus,

S_r (n) = \frac{1}{e^{\mathbf{D}}-1}n^r = \left ( \frac{1}{e^{\mathbf{D}}-1}\right )(\mathbf{D}\mathbf{D}^{-1}) = \frac{\mathbf{D}}{e^{\mathbf{D}}-1}D^{-1}n^r = \frac{\mathbf{D}}{e^{\mathbf{D}}-1}\left ( \frac{n^{r+1}}{r+1} + C\right )

where C is the constant of integration to be determined from S_r (0)= 0. Next we expand the term \frac{\mathbf{D}}{e^{\mathbf{D}}-1} into its power series expansion.

### Definition

\frac{z}{e^z-1} = \sum_{k=0}^{\infty }B_k \frac{z^k}{k!}

Here, B_k, k=0, 1 , 2,� are known as Bernoulli�s numbers. We know that \frac{z}{e^z-1} converges when z=\pm 2\pi ni, n=1,2,3,� or when |z|=2\pi. Next, let us examine the properties of Bernoulli�s numbers.

If we let z approaches zero, from the definition, we see that

B_0 = 1

and

\frac{z}{2} + \frac{z}{e^z-1} = \frac{z}{2} \cdot \frac{e^z+1}{e^z-1} = \frac{z}{2} \coth{\frac{z}{2}}

which is an even function of z. Thus, in its power expansion about z = 0 the odd order coefficients is also zero. So, B_1+1/2, B_3, B_5, B_7,� are zero. This means that

B_1 = \frac{1}{2}
B_{2k+1} = 0

where k = 1, 2, 3, ....

To solve for the Bernoulli numbers, we use a recurrence formula as follows

\frac{z}{e^z-1}e^z = z + \frac{z}{e^z-1}
\left ( \sum_{k=0}^{\infty }B_k \frac{z^k}{k!} \right ) \left ( \sum_{m=0}^{z^m}\frac{z^m}{m!} \right )= z + \sum_{n=0}^{\infty }B_n \frac{z^n}{n!}

On the left hand side there is the product of two power series. The first is from the definition of the Bernoulli numbers, and the second is the power series for the exponential function. We recall the rule for multiplying two power series. If

\left ( \sum_{k=0}^{\infty }a_k z^k \right ) \left ( \sum_{m=0}^{z^m}b_m z^m \right )= \sum_{n=0}^{\infty }c_n z^n

then

c_n = \sum_{k=0}^{n} a_kb_{n-k}

Applying this rule to

\left ( \sum_{k=0}^{\infty }B_k \frac{z^k}{k!} \right ) \left ( \sum_{m=0}^{z^m}\frac{z^m}{m!} \right )= z + \sum_{n=0}^{\infty }B_n \frac{z^n}{n!},

We'll have

\sum_{n=0}^{\infty } \left ( \sum_{k=0}^{n}\frac{B_k}{k!\left ( n-k \right )!} \right )z^n = z + \sum_{n=0}^{\infty }B_n \frac{z^n}{n!}

We then compare the coefficients of z^n on both sides for n>1 to obtain

\frac{B_n}{n!} = \sum_{k=0}{n}\frac{B_k}{k!\left ( n-k \right )!}

Since B_0 and B_1 are already known, we only care about n>1. Thus

B_n= \sum_{k=0}{n}\frac{n!}{k!\left ( n-k \right )!}B_k = \sum_{k=0}^{n}\begin{pmatrix}n \\ k \end{pmatrix}B_k

This relation can be symbolically written as

B_n = (1+B)_n

where (1+B)_n is to be expanded just like a binomial expansion except that instead of taking superscripts to get powers such as B_k we take subscripts to get the various Bernoulli numbers B_k. Actually since the B_n term cancels from both sides, we get a relation involving Bernoulli numbers until B_{n-1}. So

B_0 + nB_1 + \frac{n\left (n-1 \right )}{2} + \cdot \cdot \cdot + \begin{pmatrix}n \\ n-2 \end{pmatrix}B_{n-2} + nB_{n-1} + B_n = B_n

actually gives

B_{n-1} = -\frac{1}{n}\left ( B_0 + nB_1 + \frac{n\left (n-1 \right )}{2}B_2 + \cdot \cdot \cdot \begin{pmatrix}n \\ n-2 \end{pmatrix}B_{n-2} \right ) = -\frac{1}{n}\sum_{k=0}^{n-2}\begin{pmatrix}n \\ k \end{pmatrix}B_k

In this formula nearly half the terms on the right hand side do not contribute anything since B_3 = B_5 = B_7 = � = 0. Using the above equation we can compute B_2 and B_4 as follows:

B_2 = -\frac{1}{3}(B_0 +3B_1) = -\frac{1}{3}(1+3(-\frac{1}{2})) = \frac{1}{6}\$

B_4 = -\frac{1}{5} (B_0 + 5B_1 + 10B_2 + 10B_3) = -\frac{1}{5}\left [1 + 5(-\frac{1}{2}) + 10(\frac{1}{6})+ 0 \right ] = -\frac{1}{30}

Other values are shown in the table below

nBernoulli's Number
21/6
4-1/30
61/42
8-1/30
105/66
12-691/2730
147/6
16-3617/510
1843867/798
20-174611/330
22854513/138
24-236364091/2730
268553103/6
28-23749461029/870
308615841276005/14322
32-7709321041217/510
342577687858367/6
36 -26315271553053477373/1919190
3838 2929993913841559/6
40-261082718496449122051/13530

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