From: Stephen Montgomery-Smith <stephen@math.missouri.edu>
Subject: Re: The most beautiful integral
Date: Tue, 17 Jul 2001 18:01:49 GMT
Newsgroups: sci.math
Summary: Abel's identity via operators on polynomial rings

Bill Taylor wrote:
> 
> Oh well.    Here's another fave series not really related to these, but will
> evoke equally awe-inspired insussuration amongst those with some math knowledge,
> but who haven't seen it before...
> 
>            n
>          -----
>          \       / n \
>           >     (     )  (k-1)      (n-k)       n
>          /       \ k /  k      (n-k)        =  n
>          -----
>          k = 1
> 
> HOWBOUT THAT!   A kind of "pseudo-binomial theorem".
> 

Is this Abel's identity?  Not so easy to prove.  Here is a more general
statement.  Define a (non-commutative) operator x*n = x(x+n)^{n-1}. 
Then

(x+y)*n = sum_{k=0}^n (n choose k) x*(n-k) y*k

Here is a slick way to prove this.  Define Df(x) = f'(x-1).  Note that
D(f(x+k)) = (Df)(x+k) for any k, that D(x*n) = n x*(n-1), and that the
kernel of D are the constant functions, so D^{-1}(x*n) = x*(n+1)/(n+1) +
constant.

Prove the formula by induction.  Easy if n=0.  If it is true for n,
apply D^{-1} to both sides, which gives the formula for n+1 up to an
additive constant.  Find the constant by setting x=0.

-- 
Stephen Montgomery-Smith
stephen@math.missouri.edu
http://www.math.missouri.edu/~stephen
==============================================================================

From: ayatollah potassium <sanguis@bibimus.edu>
Subject: Re: The most beautiful integral
Date: Thu, 19 Jul 2001 15:31:52 -0400
Newsgroups: sci.math

Stephen Montgomery-Smith wrote:

> > How prove
> >             n
> >           -----
> >           \       / n \
> >            >     (     )  (k-1)      (n-k)       n
> >           /       \ k /  k      (n-k)        =  n
> >           -----
> >           k = 1
> > from
> > (x+y)*n = sum_{k=0}^n (n choose k) x*(n-k) y*k

There is a linear operator T on polynomials in y,
taking y*k to y#k,  where y#k = (y+k)^k  and
the Abel polynomial y*k = y(y+k)^(k-1).

The details of what T is, are unimportant here.
It exists because both families of polynomials
behavior the same under the differential operator
controlling the problem,  f(y) -> f'(y-1).

To the binomial formula for Abel polynomials, apply operator T:

T  [sum (n choose k)  x*k y*(n-k) = (x+y)*n]

sum (n choose k)  T ( x*k y*(n-k)) = T (x+y)*n
sum (n choose k)  x*k T y*(n-k) = (x+y)#n
sum (n choose k) x*k y#(n-k) = (x+y)#n

Inserting the definitions,

sum [ (n choose k) x (x+k)^(k-1) (z+n-k)^(n-k) ]   =  (x+z+n)^n

This looks like it should be more general than the stated identity.
Indeed, thinking of both sides as polynomials in x and looking
at the coefficient of x, produces:

sum {k=1 to n} [(n choose k) k^(k-1) (z+n-k)^(n-k)]  = n (z+n)^(n-1)

which when z = 0 specializes to what we want.