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I'm a little confused on how convolution of formal series works mostly the indexing. This is the problem I'm working on.

Let $n$ and $k$ be fixed. Calculate $$\sum_{i=0}^{n}(-1)^{i}\binom{n}{i}i^{k}$$

My idea was to use generating functions I got

$$A(x)=\sum_{n\geq 0}\left(\sum_{i=0}^{n}(-1)^{i}\binom{n}{i}i^{k}\right)x^{n}$$ This looks like the convolution of two series, $(-1)^{i}i^{k}$ and $\binom{n}{i}$.

But I'm confused on the indexing. Would

$A(x)=B(x)C(x)$ where $B(x)=\sum_{n\geq0}\binom{n}{n-i}x^{n}$ and $C(x)=\sum_{n\geq0}(-1)^{n}n^{k}$? I think $B(x)$ is incorrect but what should be the index for $B(x)$?

Thanks.

TheNumber23
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1 Answers1

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Permit me to re-write this as $$(-1)^n \sum_{k=0}^n {n\choose k} (-1)^{n-k} k^m$$ because I use $i$ in complex variable calculations. Then we see immediately that $$A(z) = \sum_{n\ge 0} (-1)^n \frac{z^n}{n!} = \exp(-z).$$ Furthermore $$B(z) = \sum_{n\ge 0} n^m \frac{z^n}{n!} = z \exp(z) \sum_{q=1}^m {m \brace q} z^{q-1},$$ which may be verified by induction. Here we have used the Stirling numbers of the second kind.

It follows that $$A(z) B(z) = z \sum_{q=1}^m {m \brace q} z^{q-1}.$$ Therefore the sum is $$(-1)^n n! [z^n] A(z) B(z) = (-1)^n n! [z^{n-1}] \sum_{q=1}^m {m \brace q} z^{q-1} = (-1)^n n! {m \brace n}.$$

Addendum. The induction proof of the formula for $B(z)$ goes like this. Suppose $$B_m(z) = \exp(z) \sum_{q=1}^m {m \brace q} z^q.$$ Then $$B_{m+1}(z) = z \frac{d}{dz} B_m(z),$$ which gives $$B_{m+1}(z) = \exp(z) \sum_{q=1}^m {m \brace q} z^{q+1} + \exp(z) \sum_{q=1}^m {m \brace q} q z^q\\ = \exp(z)\left({m \brace m} z^{m+1} + \sum_{q=1}^m \left(q {m \brace q} + {m \brace q-1}\right)z^q \right) \\= \exp(z) \sum_{q=1}^{m+1} {m+1 \brace q} z^q$$ by the defining recurrence of the Stirling numbers of the second kind.

There is another calculation of this type at this MSE link -- I and at this MSE link -- II.

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Marko Riedel
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