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I want to find the closed form of $$ G_n(k) = \sum_{k=0}^n k! \bigg\lbrace {n \atop k}\bigg\rbrace x^k $$

Suppose the closed form of $$ E_n(k) = \sum_{k=0}^n k!\, x^k $$ is known; call it $ A_n(k)$. And suppose also that the closed form of $$ F_n(k) = \sum_{k=0}^n \bigg\lbrace {n \atop k}\bigg\rbrace x^k $$ is known; call it $B_n(k) $.

Using these generating functions and their closed forms, can I acquire a closed form for $G_n(k)$ ? I know I can't just multiply the two functions because of the powers of $x$, but I don't think a typical convolution will give a congenial result.

actinidia
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    Surely the answer to this is no. – Mariano Suárez-Álvarez Feb 05 '18 at 19:06
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    @MarianoSuárez-Álvarez Is there a deep reason why not, apart from the fact that it admittedly sounds too good to be true? – actinidia Feb 05 '18 at 19:07
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    That things never work like that. But it is a claim which, by its very form, one cannot really prove. – Mariano Suárez-Álvarez Feb 05 '18 at 19:08
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    Only a remark. If you have two power series $f(x)=\sum a_n x^n$ and $g(x)=\sum b_n x^n$, with non zero radius of convergence, then, for $0<r$ small, you have $$\sum a_n b_n x^n=\frac{1}{2i\pi}\int_{|t|=r} f(t)g(\frac{x}{t})\frac{dt}{t}$$ – Kelenner Feb 05 '18 at 19:17
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    To make the problem well posed you need to fix the set of closed forms. Once you fix it, you can certainly prove it. If the set of closed forms is closed by term-by-term multiplication, then certainly it will follow that that product has a closed form. If, for example, the set of closed forms are the smaller class of the elementary functions, then one can give examples like $\sum_{n\geq1}\frac{x^n}{n}$ being elementary but $\sum\frac{x^n}{n^2}$ is not. – orole Feb 05 '18 at 19:20
  • What do you mean by "closed form"? – Dr Potato Jan 29 '20 at 05:54

1 Answers1

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Your series above can be re-written as (infinite) power series

\begin{align*} e_{n}(x) & = \sum_{k=0}^\infty k! \mathbf{1}_{k \leq n} x^k \\ f_n(x) & = \sum_{k=0}^\infty \bigg\lbrace {n \atop k}\bigg\rbrace \mathbf{1}_{k \leq n} x^k \\ g_n(x) & = \sum_{k=0}^\infty k! \bigg\lbrace {n \atop k}\bigg\rbrace \mathbf{1}_{k \leq n} x^k \\ & = \sum_{k=0}^\infty\bigg(k!\, \mathbf{1}_{k \leq n}\bigg)\bigg( \bigg\lbrace {n \atop k}\bigg\rbrace \mathbf{1}_{k \leq n} \bigg)x^k \end{align*}

where we recognise the last line as the Hadamard product of the power series $e_n, \, f_n$ (i.e. the term-wise product).

According to the Wikipedia page above, if closed forms are known for $e_n, f_n$ (equivalently $E_n, \, F_n$) then

$$ G_n(x) = g_n(x) =\frac1{2\pi} \int_{0}^{2\pi} e_n\big(\sqrt{z} e^{zt}\big) \, f_n\big(\sqrt{z} e^{-zt}\big) dt$$

owen88
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