Umbral calculus - eigenfunctions of operator

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

$$$$\mathcal{S} = a^{-1}(bI\Delta_{-1} + c\Delta_{1}), \quad I \in \mathbb{Z^+}$$$$

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:

$$$$\psi_n(I) = \sum_{m=0}^{n} {n \choose m} \left(-\frac{c}{b}\right)^m(I)_{n-m}$$$$ and the eigenvalues

$$$$\lambda_n=-n\frac{b}{a}$$$$

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?

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

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.

• 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. – Jpk Feb 19 '20 at 10:33
• @Jpk I suggest the Wikipedia article Umbral calculus as a starting point and its references. – Somos Feb 19 '20 at 16:48
• 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 $c =c(x)$? – Jpk Jun 14 '20 at 23:34
• @Jpk That is a different question which you may want to submit. I don't know the answer. – Somos Jun 15 '20 at 1:28

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.

• 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] – Jpk Feb 18 '20 at 21:25