## A generalized Taylor series operator: Umbral shift, binomial transform, and interpolation

A generalized Taylor series operator represented as two different infinite sums of differential operators related by a binomial transform can provide some intuition on umbral substitution and interpolation of umbral coefficients. (These notes reprise those in other earlier posts.)

Consider the generalized shift operator acting on the basic integral powers of $x$ as the following differential operator with $D = d/dx$ and $(a.)^n = a_n$:

$T_{x,a.} x^n = e^{a. D}x^n = \sum_{k \ge 0} a_k \frac{D^k}{k!} x^n = \sum_{k=0}^{n} \binom{n}{k} a_k x^{n-k}=(a.+x)^n.$

Then also

$T_{x=0,a.}x^n = e^{a.D} |_{x=0} x^n = (a.)^n = a_n = e^{-(1-a.)D}|_{x=1} x^n = T_{x=1,a.-1}x^n,$

and we can identify the two reps, related through a binomial transform, when acting on polynomials; i.e.,

$T_{x=0,a.}=e^{a.D}|_{x=0}=e^{-(1-a.)D}|_{x=1}=T_{x=1,a.-1}.$

When $a_n = z^n$, the generalized shift operator acting on a function analytic at $x=0$ and $x=1$, gives the Taylor series of the function about those points:

$f(z)= e^{zD}|_{x=0}f(x) = e^{(z-1)D}|_{x=1}f(x),$

which will evaluate the same at a point as long as the two circles of convergence overlap at that point, i.e., when both series are convergent at that point.

A particularly enlightening example is $f(x)=x^s$ for $s$ non-integral. Then $T_{x=0,a.}= e^{a.D}|_{x=0}$ gives divergent derivatives at some order, but

$T_{x=1,a.-1}x^s = e^{-(1-a.)D}|_{x=1}x^s = \sum_{k \ge 0} (-1)^k \binom{s}{k} \sum_{j=0}^{k} (-1)^k \binom{k}{j} a_j$

$= (1-(1-a.))^s = (a.)^s=a_s,$

a finite difference interpolation that can be taken as the value of $a_s$ when convergent. Note that $s= n$, a positive integer, gives results consistent with the prior discussion. As a sanity check, try $s=1/2$ and $a_n = (1/2)^n.$

This result is also consistent with formal interpolation via the Mellin Transform:

$\int_0^\infty \frac{x^{s-1}}{(s-1)!}e^{-a.x}dx = a.^{-s}= a_{-s}= \int_0^\infty \frac{x^{s-1}}{(s-1)!}e^{-x}e^{(1-a.)x}dx$

$= \sum_{n \ge 0} (1-a.)^n \int_0^\infty \binom{s+n-1}{n} \frac{x^{s+n-1}}{(s+n-1)!} e^{-x}dx= \sum_{n \ge 0}(-1)^n \binom{-s}{n} (1-a.)^n$

$= (1-(1-a.))^{-s}.$

A similar duality of differential ops involving the binomial transform exists for umbral scaling op reps

$S_{a.}x^n = e^{a.\widehat{xD}_{x=0}} x^n = (a.x)^n=a_nx^n=e^{-(1-a.)\widehat{xD}_x}x^n=(1-(1-a.))^nx^n,$

where $\widehat{xD}^n = x^nD^n.$

Acting on the exponential, we obtain the binomial transform for an exponential generating function (e.g.f.):

$e^{a.\widehat{xD}_{x=0}} e^x = e^{a.x} = e^{-(1-a.)\widehat{xD}_x}e^x =e^x e^{-(1-a.)x}.$

Taking the Borel-Laplace transform of the two equivalent e.g.f.s generates the analogous binomial transform for the corresponding ordinary generating functions (o.g.f.s):

$\int_0^\infty \frac{1}{p} e^{-\frac{x}{p}} e^{a.x}dx = \sum_{n \ge 0} a_n p^n=\frac{1}{1-a.p}$

$= \int_0^\infty \frac{1}{p} e^{-\frac{x}{p}} e^{-x} e^{(1+a.)x}dx = \frac{1}{p+1}\sum_{n \ge 0} (1+a.)^n (\frac{p}{p+1})^n=\frac{1}{p+1}\frac{1}{1-(1+a.)\frac{p}{p+1}}= \frac{1}{1-a.p}$

In contrast to the binomial transformation of the e.g.f.s, for which both expressions are convergent or divergent together, one of the expressions for the o.g.f. may be divergent while another is convergent.