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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

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  • $\begingroup$ Pardon once again I have been wirting in a haste. I have edited the post appropriately, i.e. $I\Delta_{-1}(I)_m=-m(I)_m$ $\endgroup$
    – Jpk
    Commented Feb 18, 2020 at 18:41
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    $\begingroup$ Can you prove a link to this paper? $\endgroup$ Commented Feb 18, 2020 at 20:01

2 Answers 2

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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.

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  • $\begingroup$ Thank you very much for this. This clearly required some practice and knowledge of linear operators with which I am not familiar. Do you happen to have a good reference on the matter? Thank a lot in advance. $\endgroup$
    – Jpk
    Commented Feb 19, 2020 at 10:33
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    $\begingroup$ @Jpk I suggest the Wikipedia article Umbral calculus as a starting point and its references. $\endgroup$
    – Somos
    Commented Feb 19, 2020 at 16:48
  • $\begingroup$ If I set $c$ to be dependent on x the same relationship does not hold as the derivative of the composed function breaks the relation you have written. Is the any other way to compute the eigen functions when $c =c(x)$? $\endgroup$
    – Jpk
    Commented Jun 14, 2020 at 23:34
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    $\begingroup$ @Jpk That is a different question which you may want to submit. I don't know the answer. $\endgroup$
    – Somos
    Commented Jun 15, 2020 at 1:28
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In the article mentionned in comment and not in the question, the author give details about the eigenvalues of $S$.

$$S_{x,y} = -y \frac{b}{a} \delta_{x,y} + y I \frac{b}{a} \delta_{x,y-1}$$

Now the author is implicitely using the fact that $S$ is triangular, therefore its eigenvalues are the diagonal components. I can provide a proof if needed but it is part of the properties you can find easily on any article about triangular matrices.

Now I did not perform the calculation of the eigenfunction, but I believe it is the general method.

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  • $\begingroup$ The computation of the eigenvalues once the change of basis is done is clear. I'm more perplexed by the calculation of the eigenfunctions; what do you mean by general method? [I'm not familar with this part of maths] $\endgroup$
    – Jpk
    Commented Feb 18, 2020 at 21:25

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