

%%%%%plain TeX file of iossif.tex a paper by  D. Zeilberger%%%
 
%begin macros
\nopagenumbers
\def\C{{\cal C}}
\def\L{{\cal L}}
\def\M{{\cal M}}
\baselineskip=14pt
\parskip=10pt
\def\Tilde{\char126\relax}
\def\halmos{\hbox{\vrule height0.15cm width0.01cm\vbox{\hrule height
 0.01cm width0.2cm \vskip0.15cm \hrule height 0.01cm width0.2cm}\vrule
 height0.15cm width 0.01cm}}
\font\eightrm=cmr8 \font\sixrm=cmr6 
\font\eighttt=cmtt8
\magnification=\magstephalf
\def\lheading#1{\bigbreak \noindent {\bf#1} 
}
\parindent=0pt
\overfullrule=0in
%end macros
\bf
\centerline{
Proof of an Identity
Conjectured by Iossif Polterovitch that}
\centerline
{Came Up in the Agmon-Kannai Asymptotic Theory of the Heat Kernel}
\medskip
\it
\centerline{Doron ZEILBERGER\footnote{$^1$}
{\baselineskip=9pt
\eightrm  \raggedright
Department of Mathematics, Temple University,
Philadelphia, PA 19122.
{\baselineskip=9pt
\eighttt zeilberg@math.temple.edu ; 
\break http://www.math.temple.edu/\Tilde 
zeilberg/ }
Supported in part by the NSF. Aug. 10, 2000.
Exclusive to the "Personal Journal of Ekhad and Zeilberger".
I thank Amitai Regev for his invitation to stay at the
Weizmann Institute between July 24 and Aug. 2, 2000.
}}
\medskip
\it
\qquad\qquad Dedicated to 
my Teacher, Yakar Kannai, who taught me PDEs
and much more
\rm
 
Yakar Kannai's brilliant student, Iossif Polterovitch,
who did beautiful work in extending and
elaborating the classical Agmon-Kannai
theory of the heat kernel, proved, via  that theory,
the $a=3n$ case of the following identity:
$$
G(a,n;x):=
\sum_{j=0}^{a} {{1} \over {(a-j)!(j+n)!(2j+1)}}
\sum_{k=-j}^{j} {{(-1)^k (x+k)^{2j+2n}} \over {(j-k)!(j+k)!}}=0 \quad,
\quad (a \geq 2n) \quad ,
$$
that he discovered empirically,
using {\tt Mathematica}. He
also conjectured that $G(2n-1,n,x)=-1/(2 \cdot 3^n n! )$. In this
note I will prove his conjecture.
 
Introducing the shift operator $Ef(x)=f(x+1)$, we have
that $G(a,n;x)$ equals
$$
\sum_{j=0}^{a} {{(-1)^j} \over {(a-j)!(j+n)!(2j+1)}}
\sum_{k=-j}^{j} {{(-1)^{j+k} E^k} \over {(j-k)!(j+k)!}} x^{2j+2n}
=\sum_{j=0}^{a} {{(-1)^j} \over {(a-j)!(j+n)!(2j+1)!}}
(E^{1/2}-E^{-1/2})^{2j} x^{2j+2n} \, \,.
$$
Using $E=e^D$ (Taylor's theorem), 
and writing  $f(D)=((2\sinh D/2)/D)^2$ (note that
$f(D)$ is an even function ($f(D)=1+D^2/12+O(D^4) $)), we have that
this equals
$$
\sum_{j=0}^{a} {{(-1)^j} \over {(a-j)!(j+n)!(2j+1)!}}
f(D)^{2j} D^{2j} x^{2j+2n}
=\sum_{j=0}^{a} {{(-1)^j} \over {(a-j)!(j+n)!(2j+1)!}}
f(D)^{j} {{(2j+2n)!} \over {(2n)!}} \,\, x^{2n} \quad
$$
$$
={{1} \over {a! n!}}{}_2 F_1 (n+1/2,-a;3/2;f(D)) x^{2n} 
={{1} \over {a! n!}}
{}_2 F_1 (1-n,3/2+a;3/2;f(D)) 
(1-f(D))^{a-n+1} \,\,x^{2n} \quad ,
$$
by Euler's transformation:
${}_2F_1(a,b;c;z)=(1-z)^{c-a-b}{}_2F_1(c-a,c-b;c;z)$
([B], p. 2).
If $a \geq 2n$, then, since $1-f(D)=O(D^2)$, this equals $0$.
If $a=2n-1$, then, since $f(D)-1=D^2/12+O(D^4)$ this equals
$$
{{(-1)^n} \over {(2n-1)! n!}}
{}_2 F_1 (1-n,1/2+2n;3/2;1) {{(2n)!} \over {12^n}}
\quad ,
$$
which turns out to be $-1/(2 \cdot 3^n n!)$, by Gauss's
evaluation ([B], p. 3). \halmos
 
{\bf REFERENCE}
 
[B] W.N. Bailey, ``Generalized Hypergeometric Series'', Cambridge
University Press, 1935.
 
\bye
 
 
 
 
