\centerline{\bf Bernoulli numbers\strut}
\centerline{Miguel Carri\'on \'Alvarez\strut}

\bigskip

Let me start by pointing out that the spectrum of the algebra of
entire functions is~${\bf C}$, while the spectrum of the algebra of formal
power series is just~$\{0\}$, because a formal power series is
invertible if, and only if, its constant term is nonzero, so the only
character of~${\bf C}[[z]]$ is the ``evaluation at $z=0$''.

\proclaim 1. Taylor's theorem.

$$
f(z+t)=\sum_{k\ge 0}{t^kf^{(k)}(z)\over k!}=\sum_{k\ge 0}{\bigl[(ta)^kf\bigr](z)\over k!}=\Bigl[\sum_{k\ge 0}{(ta)^k\over k!}\Bigr]f(z)=(e^{ta}f)(z).
$$

\proclaim 2. First difference operator.

$$
(\Delta f)(z)=f(z+1)-f(z)=(e^af)(z)-f(z)=(e^a-1)f(z).
$$

\proclaim 3. Fundamental theorem of difference calculus.

$$
F(n)-F(0)=\sum_{i=0}^{n-1}\bigl[F(i+1)-F(i)\bigr]=\sum_{i=0}^{n-1}(\Delta
F)(i)=\sum_{i=0}^{n-1}f(i).
$$

\proclaim 4. Integration is a one-sided inverse of differentiation.

It is well-known that the derivative of an integral is the integrand,
but the integral of a derivative may differ from the function to be
differentiated by any constant. That is, $aa^{-1}=1$ but $a^{-1}a\neq
1$.

\proclaim 5, 7, 8, 9. Identities involving Bernoulli numbers.

Clearly, 
$$
x=(e^x-1){x\over e^x-1}=\sum_{k\ge 1}{x^k\over k!}\sum_{n\ge
0}B_n{x^n\over n!}.
$$
This means that
$$
x=\sum_{m\ge 1}x^m\sum_{0\le n<m}{B_n\over n!(m-n)!},
$$
so
$$
B_0=1
\qquad\hbox{and}\quad
0=\sum_{0\le n<m}{B_n\over n!(m-n)!}={1\over m!}\sum_{0\le n<m}B_n{m\choose n},
$$
and
$$
\pmatrix{1 & 0 & 0 & 0 & 0 \cr
         1 & 2 & 0 & 0 & 0 \cr
         1 & 3 & 3 & 0 & 0 \cr
         1 & 4 & 6 & 4 & 0 \cr
         1 & 5 & 10 & 10 & 5 \cr}\pmatrix{B_0\cr B_1\cr B_2\cr B_3\cr
         B_4\cr}=\pmatrix{1\cr 0\cr 0\cr 0\cr 0\cr}.
$$
so
$$
B_0=1;\qquad
B_1={-1\over 2};\qquad
B_2={1\over 6};\qquad
B_3=0;\qquad
B_4={-1\over 30}.
$$

Since the identity 
$$
x=\sum_{k\ge 1}{x^k\over k!}\sum_{n\ge 0}B_n{x^n\over n!}.
$$
involves cancellations of finite sums, so 
$$
a=\sum_{k\ge 1}{a^k\over k!}\sum_{n\ge 0}B_n{a^n\over n!}.
$$
for any operator~$a$.

\proclaim 6. The inverse of~$\Delta$.

$$
\Delta\Delta^{-1}=\Delta{a\over e^a-1}a^{-1}=aa^{-1}=1.
$$
Obviously, since~$\Delta f=0$ whenever~$f$ is constant,~$\Delta$
cannot have a left-side inverse, so~$\Delta^{-1}\Delta\neq 1$.

\proclaim 10. 

Let~$f(z)=z^4$. Then, 
$$
\sum_{i=0}^{n}i^4=\sum_{i=0}^{n}f(i)=\sum_{i=0}^{n}(\Delta\Delta^{-1}f)(i)=(\Delta^{-1}f)(n+1)-(\Delta^{-1}f)(0). 
$$
But~$(a^{-1}f)(z)={1\over 5}z^5$, and
$$
(\Delta^{-1}f)(z)={1\over 5}\sum_{n\ge 0}{B_n\over n!}{d^nz^5\over
         dz^n}={1\over 5}\sum_{n\ge 0}B_n{5\choose n}z^{5-n},
$$
So
$$
\sum_{i=0}^{n}i^4={1\over 5}\sum_{k\ge 1}B_k{5\choose
k}(n+1)^{5-k}={1\over 5}(n+1)^5 -{1\over 2} (n+1)^4 +
{1\over 3} (n+1)^3 -{1\over 30} (n+1),
$$

\proclaim 11. A trick.

$$
\eqalign{
1^p+\cdots+n^p=&{1\over p+1}\sum_{k\ge 1}{p+1\choose
k}B_k(n+1)^{p+1-k}={1\over p+1}\Bigl(\sum_{k\ge 0}{p+1\choose
k}B_k(n+1)^{p+1-k}-(n+1)^{p+1}\Bigr)=\cr
=&{(B+n+1)^{p+1}-(n+1)^{p+1}\over p+1}}
$$
if we interpret~$B_k$ to mean $B^k$.  If we take $n = 0$ this
reduces to
$$
0 = (B+1)^{p+1} - B^{p+1}
$$
which is the recursive formula for the Bernoulli numbers from
part 8.
\bye

