# annihilated coefficient extractor (ACE)

Please, someone can explain me how works the annihilated coefficient extractor technique (ACE). I have seen some examples but i can't understand. Is there some text to read about this specific argument?

• Where did you hear about it? Could you reference the source? – Yuriy S Feb 4 '18 at 14:47

The ACE also called substitution rule is demonstrated in the example below. The key is the representation mentioned in (3).

We obtain \begin{align*} \color{blue}{\sum_{j=0}^k}&\color{blue}{\binom{j+r-1}{j}\binom{k-j+s-1}{k-j}}\\ &=\sum_{=0}^k\binom{-r}{j}(-1)^j\binom{-s}{k-j}(-1)^{k-j}\\ &=(-1)^k\sum_{j=0}^\infty[z^j](1+z)^{-r}[u^{k-j}](1+u)^{-s}\tag{1}\\ &=(-1)^k[u^k](1+u)^{-s}\sum_{j=0}^\infty u^j[z^j](1+z)^{-r}\tag{2}\\ &=(-1)^k[u^k](1+u)^{-s}(1+u)^{-r}\tag{3}\\ &=(-1)^k[u^k](1+u)^{-r-s}\\ &=(-1)^k\binom{-r-s}{k}\\ &\color{blue}{=\binom{k+r+s-1}{k}} \end{align*} and the Chu-Vandermonde identity follows.

Comment:

• In (1) we apply the coefficient of operator twice and set the upper limit of the series to $\infty$ without changing anything since we are adding zeros only.

• In (2) we use the linearity of the coefficient of operator and apply the rule $[z^{p}]z^qA(z)=[z^{p-q}]A(z)$.

• In (3) we apply the substitution rule of the coefficient of operator with $u=z$
\begin{align*} A(u)=\sum_{j=0}^\infty a_j u^j=\sum_{j=0}^\infty u^j [z^j]A(z) \end{align*}

Here are some more examples using the substitution rule:

• Ex. 1: Is there an explicit expression for $\sum_{j= k+1}^{2n}{j\choose k}{n\choose j-n}$?.

• Ex. 2: A strange combinatorial identity: $\sum\limits_{j=1}^k(-1)^{k-j}j^k\binom{k}{j}=k!$.

• Ex. 3: Sum Involving Bernoulli Numbers : $\sum_{r=1}^n \binom{2n}{2r-1}\frac{B_{2r}}{r}=\frac{2n-1}{2n+1}$ by Marko Riedel

Note: The substitution rule stated as formula 1.6 and a lot of other powerful techniques were introduced by G.P. Egorychev in Integral Representation and the Computation of Combinatorial Sums.