I'm very new to umbral caluclus and I have come across a paper that makes use of some results in this area, which I do not quite understand.
The problem I have is the following.
Consider the following operator
\begin{equation}
\mathcal{S} = a^{-1}(bI\Delta_{-1} + c\Delta_{1}), \quad I \in \mathbb{Z^+}
\end{equation}
where $\Delta_h[f(I)] =f(I+h)-f(I), h \in \mathbb{Z}$
It is claimed by the paper that given that $\Delta_{1}(I)_m=m(I)_{m-1}$ and $I\Delta_{-1}(I)_m = -m(I)_m$ [where with $(I)_m$ we denote the falling factorial] the eigenfunctions $\psi(I)$ of the operator are computable as:
\begin{equation}
\psi_n(I) = \sum_{m=0}^{n} {n \choose m} \left(-\frac{c}{b}\right)^m(I)_{n-m}
\end{equation}
and the eigenvalues
\begin{equation}
\lambda_n=-n\frac{b}{a}
\end{equation}
Note that $\mathcal{S}(I)_m = -m\frac{b}{a}[(I)_m – \frac{c}{b}(I)_{m-1}]$
How is this caluculation performed? I have looked at Rota's book but I see no reference to the computation of the eigenfunctions. Can anybody point me out in the right direction?
Thank you all in advance
Best Answer
Begin with the study of polynomials in $X$ and some linear operators. Define the shift linear operator $$ E_h[f(X)] \!:=\! f(X\!+\!h), \tag{1} $$ and the difference linear operator $$ \Delta_h[f(X)] := f(X\!+\!h)\!-\!f(X), \tag{2} $$ and the derivative linear operator $$ D[f(X)] := \frac{d}{dX} f(X), \tag{3} $$ where $\,f()\,$ is any polynomial.
Define the falling factorial linear operator for monomials $$ L[X^n] := X(X-1)\dots(X-n+1). \tag{4} $$ Without loss of generality, define the $\,S\,$ linear operator by $$ S[f(X)] := (X\Delta_{-1}+c\,\Delta_{1})[f(X)]. \tag{5}$$ Applying this to falling factorials gives $$ S[(X)_n] = -n ( (X)_n - c\, (X)_{n-1}). \tag{6} $$ Rewriting this using the $\,L\,$ operator gives $$ S[L[X^n]] = -n L[ X^{n-1} (X-c)]. \tag{7} $$
Rewrite this using the derivative operator gives $$ S[L[f(X)]] = -L[ D[f(X)](X-c)]. \tag{8} $$ Apply this to the case $\,f(X)=E_h[X^n]\,$ to get $$ S[L[E_h[X^n]]] = -L[D[E_h[X^n]](X-c)]. \tag{9} $$ Apply this to the case $\,h=-c\,$ to get $$ S[L[(X-c)^n]] = -L[D[(X-c)^n](X-c)]. \tag{10} $$ But we know the derivative of $\,(X-c)^n\,$ and so get $$ S[L[(X-c)^n]] = -n\,L[(X-c)^n]. \tag{11} $$ Define the eigenvector function $$ \psi_n(X) \!:=\! L[(X\!-\!c)^n] \!=\! \sum_{m=0}^n {n \choose m} (-c)^m(X)_{n-m}. \tag{12} $$ with eigenvalue $\,-n\,$ and is expanded into a finite sum using the binomial theorem.