5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Surely the answer to this is no. $\endgroup$ – Mariano Suárez-Álvarez Feb 5 '18 at 19:06
  • 1
    $\begingroup$ @MarianoSuárez-Álvarez Is there a deep reason why not, apart from the fact that it admittedly sounds too good to be true? $\endgroup$ – Tiwa Aina Feb 5 '18 at 19:07
  • 1
    $\begingroup$ That things never work like that. But it is a claim which, by its very form, one cannot really prove. $\endgroup$ – Mariano Suárez-Álvarez Feb 5 '18 at 19:08
  • 1
    $\begingroup$ 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 $$\sum a_n b_n x^n=\frac{1}{2i\pi}\int_{|t|=r} f(t)g(\frac{x}{t})\frac{dt}{t}$$ $\endgroup$ – Kelenner Feb 5 '18 at 19:17
  • $\begingroup$ 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. $\endgroup$ – orole Feb 5 '18 at 19:20
2
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.