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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.

From https://mathworld.wolfram.com/BellPolynomial.html (14),

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

From https://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html (10),

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

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.

Reference: https://en.wikipedia.org/wiki/Bell_polynomials#Applications

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

Reference: https://en.wikipedia.org/wiki/Bell_polynomials#Generating_function .

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!

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$\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.

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  • $\begingroup$ Nice! Thank you! $\endgroup$
    – BBadman
    Jun 27, 2023 at 6:13

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