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It's a power series that I found during the computation for my research.

\begin{equation*} \sum_{k=0}^n \binom{n}{k}\frac{n!}{(n-k)!}x^{n-k}(-1)^k. \end{equation*}

Without the annoying term of $\frac{n!}{(n-k)!}$, it is clearly simplified to $(x-1)^n$ due to binomial theorem.

Isn't there any name for this series? Can I simplify it as a closed-form polynomial?

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2 Answers 2

up vote 3 down vote accepted

You can write it this way: $$\sum_{k=0}^n \binom{n}{k}\frac{n!}{(n-k)!}x^{n-k}(-1)^k = \sum_{k=0}^n \binom{n}{k}(-1)^k\frac{d^k}{dx^k} (x^n) = \left(I-\frac d{dx}\right)^n x^n $$

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Interesting. I found out that it's an associated Laguerre polynomial. By the way, would you recommend a textbook or reference on how to handle differentiation as a linear map? I don't have much background on this field and I need to learn myself on this topic. –  Federico Magallanez Mar 31 at 15:50
    
I don't think there is much to know about on this topic. –  mookid Mar 31 at 17:06
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It can be rewritten as $\displaystyle\sum_{k=0}^n{n\choose k}^2k!x^{n-k}(-1)^k$, and, unless you're willing to accept hypergeometric functions as closed form, the answer is no. My advice would be for you to try and approximate it asymptotically.

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To be honest,I dont understand the other answer. –  Sabyasachi Mar 31 at 5:37
    
Differentiation is a linear map; the identity map is a linear map; linear maps form a ring, so $(I-d/dx)^n$ is a well defined linear map. The series given above is what $(I-d/dx)^n$ maps $x^n$ to. –  Mike Miller Mar 31 at 5:40
    
@Sabyasachi ? why don't you comment on my answer? –  mookid Mar 31 at 6:41
    
@mookid i upvoted anyway. it looked legit. –  Sabyasachi Mar 31 at 7:27
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