## Fractional calculus and interpolation of generalized binomial coefficients

Draft

Interpolation of the generalized binomial coefficients underlie the representation of a particular class of fractional differintegro operators by convolution integrals and Cauchy-like complex contour integrals.

The basic differintegro operators can be characterized by their action on divided powers:

with $x >0$ and $FP$ denoting the Hadamard finite part,

$\displaystyle \frac{x^{\beta - \alpha}}{(\beta - \alpha)!} = D_x^{\alpha} \frac{x^{\beta}}{\beta!}= FP \int_0^x \frac{(x-t)^{-\alpha-1}}{(-\alpha-1)!} \; \frac{t^\beta}{\beta!} \; dt = FP \frac{1}{2 \pi i} \oint_{|z-x|=x} \frac{\alpha!}{(z-x)^{\alpha+1}} \; \frac{z^\beta}{\beta!} \; dz .$

An $x^{\beta - \alpha}$ can be factored out to reduce the relations to

$\displaystyle \frac{1}{(\beta - \alpha)!} = D_{x=1}^{\alpha} \frac{x^{\beta}}{\beta!}= FP \int_0^{1^{-}} \frac{(1-t)^{-\alpha-1}}{(-\alpha-1)!} \; \frac{t^\beta}{\beta!} \; dt = FP \frac{1}{2 \pi i} \oint_{|z-1|=1^{-}} \frac{\alpha!}{(z-1)^{\alpha+1}} \; \frac{z^\beta}{\beta!} \; dz .$

$\displaystyle = D_{x=1}^{-\beta-1} \frac{x^{-\alpha-1}}{(-\alpha-1)!} = FP \int_0^{1^{-}} \frac{t^{-\alpha-1}}{(-\alpha-1)!} \; \frac{(1-t)^\beta}{\beta!} \; dt = FP \frac{1}{2 \pi i} \oint_{|z-1|=1^{-}} \frac{z^{-\alpha-1}}{(-\alpha-1)!} \; \frac{(-\beta-1)!}{(z-1)^{-\beta}} \; dz,$

which can also be expressed in terms of the modified Mellin transform

$\displaystyle MT_{t \rightarrow s} [g(t)]= FP \int_0^\infty g(t) \; \frac{t^{s-1}}{(s-1)!} \; dt$

as

$\displaystyle \frac{1}{(\beta - \alpha)!} = MT_{t \rightarrow \beta+1} \; [H(1-t) \; \frac{(1-t)^{-\alpha-1}}{(-\alpha-1)!}] = MT_{t \rightarrow -\alpha} \; [H(1-t) \; \frac{(1-t)^\beta}{\beta!}] \; .$

By using the binomial expansion and Euler’s reflection formula for the gamma function, these integrals can be related to sinc function interpolation of the generalized binomial coefficient:

$\displaystyle \frac{1}{(\beta-\alpha)!} = FP \int_0^{1^{-}} \frac{t^{-\alpha-1}}{(-\alpha-1)!} \; \frac{(1-t)^\beta}{\beta!} \; dt = \frac{1}{(-\alpha-1)! \; \beta!} \sum_{n \ge 0} (-1)^n \binom{\beta}{n} FP \int_0^{1^-} t^{n-\alpha-1} \; dt$

$\displaystyle = \frac{\alpha!}{\beta!} \; \frac{-sin(\pi \; \alpha)}{\pi} \sum_{n \ge 0} (-1)^n \binom{\beta}{n} \frac{t^{n-\alpha}}{n-\alpha} \; |_{t \rightarrow 1^{-}} = \frac{\alpha!}{\beta!} \; \sum_{n \ge 0} \binom{\beta}{n} \frac{sin(\pi \; (\alpha-n))}{\pi(\alpha-n)} \; t^{n-\alpha} \; |_{t \rightarrow 1^{-}}$

$\displaystyle = \frac{\alpha!}{\beta!} \; \binom{\beta}{\alpha} \; ,$

and, with $\displaystyle z = 1 + r \; e^{i \theta} \; ,$

$\displaystyle \frac{1}{(\beta-\alpha)!} = FP \frac{1}{2 \pi i} \oint_{|z-1|=1^{-}} \frac{\alpha!}{(z-1)^{\alpha+1}} \; \frac{z^\beta}{\beta!} \; dz = \frac{1}{2 \pi } \int_{-\pi}^{\pi} \frac{\alpha!}{(r \; e^{i \theta})^{\alpha+1}} \; \frac{(1+r \; e^{i \theta})^\beta}{\beta!} \; r \; e^{i \theta} \; d\theta\; |_{r \rightarrow 1^{-}}$

$\displaystyle = \frac{\alpha!}{\beta!} \; \frac{1}{2 \pi} \int_{-\pi}^{\pi} \sum_{n \ge 0} \binom{\beta}{n} r^n \; e^{i (n-\alpha) \theta} \; d\theta \; |_{r \rightarrow 1^{-}} = \frac{\alpha!}{\beta!} \; \sum_{n \ge 0} \binom{\beta}{n} r^n \; \frac{sin(\pi \; (\alpha-n))}{\pi(\alpha-n)} \; |_{r \rightarrow 1^{-}}$

$\displaystyle = \frac{\alpha!}{\beta!} \; \binom{\beta}{\alpha}\; ,$

so underlying the integral reps is a basic sinc function interpolation of the generalized binomial coefficient:

$\displaystyle \sum_{n \ge 0} \binom{\beta}{n} r^n \; \frac{sin(\pi \; (\alpha-n))}{\pi(\alpha-n)} \; |_{r \rightarrow 1^{-}} = \binom{\beta}{\alpha} \; .$

With umbral substitutions and Euler transformations, this can be related to a Newton series interpolation of the generalized binomial coefficients. The Bell / Touchard polynomials, $\displaystyle \phi_n(x)$, and the falling factorials $(x)_n$, are umbral compositional inverses, so formally umbrally substituting the Bell polynomials for $\beta$ in the sinc interpolation formula gives

$\displaystyle \sum_{n \ge 0} \binom{\phi.(x)}{n} r^n \; \frac{sin(\pi \; (\alpha-n))}{\pi(\alpha-n)} \; |_{r \rightarrow 1^{-}} = \sum_{n \ge 0} \frac{x^n}{n!} \; \frac{sin(\pi \; (\alpha-n))}{\pi(\alpha-n)} = \binom{\phi_.(x)}{\alpha}$

$\displaystyle = e^{-x} \sum_{n \ge 0} \binom{n}{\alpha} \frac{x^n}{n!} \; .$

The last equality follows from the generalized Dobinski formula. Using the Euler transformation on this last expression gives

$\displaystyle \binom{\phi_.(x)}{\alpha} = \sum_{n \ge 0} \frac{x^n}{n!} \; \frac{sin(\pi \; (\alpha-n))}{\pi(\alpha-n)} = e^{-x} \sum_{n \ge 0} \binom{n}{\alpha} \frac{x^n}{n!} = \sum_{n \ge 0} (-1)^n [\sum_{k =0}^{n} (-1)^k \binom{n}{k} \binom{k}{\alpha}] \; \frac{x^n}{n!} \;$

$\displaystyle = \sum_{n \ge 0} (-1)^n [\triangle^n_k \; \binom{k}{\alpha}] \; \frac{x^n}{n!} \; ,$

where $\displaystyle \triangle^s_k \; c_k = \sum^\infty_{k =0} (-1)^k \; \binom{s}{k} \; c_k \; ,$

and we can identify

$\displaystyle \frac{sin(\pi \; (\alpha-n))}{\pi(\alpha-n)} =(-1)^n \sum_{k =0}^{n} (-1)^k \binom{n}{k} \binom{k}{\alpha} \; = (-1)^n \triangle^n_k \; \binom{k}{\alpha}.$

Now umbrally substituting the falling factorials $\displaystyle (\beta)_n$ for $\displaystyle x^n$ gives the Newton series interpolation

$\displaystyle \binom{\beta}{\alpha} = \sum_{n \ge 0} \binom{\beta}{n} \; \frac{sin(\pi \; (\alpha-n))}{\pi(\alpha-n)} = \sum_{n \ge 0} (-1)^n \binom{\beta}{n} \; \sum_{k =0}^{n} (-1)^k \binom{n}{k} \binom{k}{\alpha}$

$\displaystyle = \triangle^\beta_n \; \triangle^n_k \; \binom{k}{\alpha} .$

These maneuvers can be intimately related to the Mellin transform rep. using an operator rep for the Bell polynomials and the definition $\displaystyle :xD:^n = x^nD^n$. For $x < 0$ and $0 < \sigma < 1$ ,

$\displaystyle \binom{\phi.(x)}{\alpha} = e^{-x} \; \binom{\phi.(:xD:)}{\alpha} \; e^x = e^{-x} \; \binom{xD}{\alpha} \; e^x$ $\displaystyle = e^{-x} \sum_{n \ge 0} \binom{n}{\alpha} \; \frac{x^n}{n!}= \sum_{n \ge 0} (-1)^n \triangle^{n}_{k} \binom{k}{\alpha}\; \frac{x^n}{n!}$

$\displaystyle = e^{-x} \; \binom{xD}{\alpha} \; \frac{1}{2 \pi i} \; \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{sin(\pi \; s)} \; \frac{(-x)^{-s}}{(-s)!} \;ds = e^{-x} \; \frac{1}{2 \pi i} \; \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{sin(\pi \; s)} \; \binom{-s}{\alpha}\frac{(-x)^{-s}}{(-s)!} \;ds \; .$

Then we have, for $x > 0$ , the inverse Mellin transform

$\displaystyle \frac{1}{2 \pi i} \; \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{sin(\pi \; s)} \; \binom{-s}{\alpha}\frac{x^{-s}}{(-s)!} \;ds = e^{-x} \; \binom{\phi.(-x)}{\alpha} = \sum_{n \ge 0} \binom{n}{\alpha} \; \frac{(-x)^n}{n!}$

$\displaystyle = e^{-x} \; \sum_{n \ge 0} \triangle^{n}_{k} \binom{k}{\alpha} \; \frac{x^n}{n!} \; ,$

which implies the Mellin transform and Newton series, for $0 < Real(s) < 1$ ,

$\displaystyle \binom{-s}{\alpha} = \int_0^\infty \; e^{-x} \; \sum_{n \ge 0} \triangle^{n}_{k} \binom{k}{\alpha} \; \frac{x^n}{n!} \; \frac{x^{s-1}}{(s-1)!} \; dx = \sum_{n \ge 0} \triangle^{n}_{k} \binom{k}{\alpha} \; \int_0^\infty \; e^{-x} \;\frac{x^n}{n!} \; \frac{x^{s-1}}{(s-1)!} \; dx$

$\displaystyle = \sum_{n \ge 0} \triangle^{n}_{k} \binom{k}{\alpha} \binom{s-1+n}{n} = \sum_{n \ge 0} \triangle^{n}_{k} \binom{k}{\alpha} (-1)^n \; \binom{-s}{n} = \triangle^{-s}_n \; \triangle^n_k \; \binom{k}{\alpha} .$

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