# Closed Form for Geometric-like Finite sum of Bell Polynomials

I'm trying to see if there's a nice closed form expression for the following sum:

$$\sum_{k=0}^{M} \cos(\pi k t) B_k(x)$$

where $$M \in \mathbb{N}$$, $$t \in (0,1)$$, and $$x \in \mathbb{R}^+$$.

Notation: Here $$B_{n}(x)$$ denotes the $$n$$th Bell polynomial, $$n \in \mathbb{N}$$, of $$x$$, $$x \in \mathbb{R}^+$$, $$S(n,k)$$ denotes the Stirling number of the second kind.

$$B_{n}(x) = \displaystyle{\sum_{k=0}^{n}} S(n,k) x^k$$.

$$S(n,k) = \frac{1}{k!} \displaystyle{\sum_{j=0}^{k} (-1)^j {k \choose j} (k-j)^n}$$.

Truthfully my knowledge of these polynomials is limited to some informal skimming of the wikipedia and other pages/papers I could find, so I feel I am probably not taking the most strategic approaches in trying to solve this.

Some things I've attempted:

I've tried making use of Faà di Bruno's formula, to see if I could manipulate certain functions to produce the desired finite sum, but with no luck unfortunately.

Same idea was tried with the generating functions of the Bell polynomials, but I can't seem to get it down.

I've also titled it a geometric like sum as it can also be viewed as $$\Re[\sum_{k=0}^{M} (e^{i \pi t})^k B_k(x)]$$.

I'd be curious to learn via any examples possibly provided of what would maybe be a better way to approach these types of problems in general for future reference! Would greatly appreciate any help. Thank you!

$$\def\Re{\operatorname{Re}} \def\B{\operatorname B}$$

Using $$\B_k(x)=e^{-x} \sum\limits_{j=0}^\infty\frac{j^k x^j}{j!}$$, we can interchange sums as one is a finite one, to get:

$$\Re\left(\sum_{j=0}^\infty\frac{e^{(M+1)\pi i t-x}j^{M+1}x^j}{(e^{\pi it}j-1)j!}-\sum_{j=0}^\infty\frac{e^{-x}x^j}{(e^{\pi i t}j-1)j!}\right)$$ the second sum is a lower incomplete gamma function expression:

$$\sum_{j=0}^\infty\frac{e^{-x}x^j}{(e^{\pi i t}j-1)j!}=e^{-\pi i t-x}(-x)^{e^{-\pi i t}}\gamma(-e^{-\pi i t},-x)$$

Also, the first sum uses a hypergeometric function, defined by a Pochhammer symbol $$(m)_n$$ sum. We get:

$$\sum_{j=0}^\infty\frac{e^{(M+1)\pi i t-x}j^{M+1}x^j}{(e^{\pi it}j-1)j!}= \frac{e^{(M+1)\pi i t-x}}{e^{\pi i t}-1}\sum_{j=0}^\infty\frac{(2)_j^M}{(1)_j^M}\frac{(1-e^{-\pi i t})_j x^{j+1}}{(2-e^{-\pi i t})_jj!}=\frac{xe^{(M+1)\pi i t-x}}{e^{\pi i t}-1}\,_{M+1}\text F_{M+1}(2,\dots,2,1-e^{-\pi i t};1,\dots,1,2-e^{-\pi i t};x)$$

Although the sum already is in a closed form, as a finite one, we get:

$$\sum_{k=0}^M\cos(\pi kt)\B_k(x)=\Re\left(\frac{xe^{(M+1)\pi i t-x}}{e^{\pi i t}-1}\,_{M+1}\text F_{M+1}(2,\dots,2,1-e^{-\pi i t};1,\dots,1,2-e^{-\pi i t};x)-e^{-\pi i t-x}(-x)^{e^{-\pi i t}}\gamma(-e^{-\pi i t},-x)\right)$$

There are $$M$$ $$1$$’s and $$M$$ $$2$$’s in the hypergeometric function.

• Nice! Thank you! Jun 27, 2023 at 6:13