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It is a combinatorics proof. Anyone has any idea on how to prove

$$\sum \limits_{i=0}^{l} \sum\limits_{j=0}^i (-1)^j {m-i\choose m-l} {n \choose j}{m-n \choose i-j} = 2^l {m-n \choose l}\;$$

We need to prove this equation holds for all $l$.

I know that $\sum {n \choose j}{m-n \choose i-j}$ equals to ${m \choose i}$ but has no idea if there is a $(-1)^j$, it seems like a PIE but actually not....

Could anyone help me move forward in the process?

Xiaonan
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2 Answers2

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It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. We can write this way \begin{align*} [z^m](1+z)^n=\binom{n}{m}\tag{1} \end{align*}

We obtain \begin{align*} \sum_{i=0}^l&\sum_{j=0}^i(-1)^j\binom{m-i}{m-l}\binom{n}{j}\binom{m-n}{i-j}\\ &=\sum_{i=0}^l\binom{m-i}{m-l}\sum_{j=0}^\infty(-1)^j[z^j](1+z)^n[u^{i-j}](1+u)^{m-n}\tag{2}\\ &=\sum_{i=0}^l\binom{m-i}{m-l}[u^i](1+u)^{m-n}\sum_{j=0}^{\infty}(-1)^ju^j[z^j](1+z)^n\tag{3}\\ &=\sum_{i=0}^l\binom{m-i}{m-l}[u^i](1+u)^{m-n}(1-u)^n\tag{4}\\ &=\sum_{i=0}^{\infty}[z^{m-l}](1+z)^{m-i}[u^i](1+u)^{m-n}(1-u)^n\tag{5}\\ &=[z^{m-l}](1+z)^m\sum_{i=0}^\infty(1+z)^{-i}[u^i](1+u)^{m-n}(1-u)^n\tag{6}\\ &=[z^{m-l}](1+z)^m\left(1+\frac{1}{1+z}\right)^{m-n}\left(1-\frac{1}{1+z}\right)^n\tag{7}\\ &=[z^{m-l}](1+z)^m\cdot\frac{(2+z)^{m-n}}{(1+z)^{m-n}}\cdot\frac{z^n}{(1+z)^n}\\ &=[z^{m-n-l}](2+z)^{m-n}\\ &=[z^{m-n-l}]\sum_{j=0}^{m-n}\binom{m-n}{j}2^jz^{m-n-j}\tag{8}\\ &=2^l\binom{m-n}{l} \end{align*}

Comment:

  • In (2) we apply (1) to the binomial coefficients of the inner series and set the upper limit of the index $j$ to $\infty$ without changing anything since we add only zeros.

  • In (3) we use the linearity of the coefficient of operator, do some rearrangements to prepare for the next step and use the rule \begin{align*} [u^{p-q}]P(u)=[u^p]u^qP(u) \end{align*}

  • In (4) we apply the substitution rule \begin{align*} A(u)=\sum_{j=0}^\infty a_ju^j=\sum_{j=0}^\infty u^j[z^j]A(z)\\ \end{align*}

  • In (5) we apply (1) to the binomial coefficient of the first series

  • In (6) we do again some rearrangement similar to (3)

  • In (7) we apply the substitution rule again

  • In (8) we select the index $j=l$ to obtain the term with power $m-n-l$

Markus Scheuer
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  • (+1). It is good to see that you are getting better at these, attacking more involved sums and making considerable progress. I checked it once and it seems sound (will check it several more times). I would say we have the same generating functions but you are not using the theory of complex variables which can be a plus sometimes. Congratulations! We now have a quality page where readability should profit from variety. – Marko Riedel May 06 '16 at 20:41
  • @MarkoRiedel: Thanks for your nice comment. No doubt, the essence of our answers is the same. It's just a matter of different representations. I admire Egorychevs approach (of course :-)). I found the notation which I've used here the first time in Mathematics for the Analysis of Algorithms by D.H. Greene and D.E. Knuth. They state there that I. Gessel attributed this method of coefficients to Egorychev and so we've come full circle. :-) (+1 btw for your more original notation :-)) – Markus Scheuer May 06 '16 at 21:01
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    As for original notation of course one must be careful. One favorite of mine is to refer to your substitution rule as an annihilated coefficient extractor which I think accurately represents what is happening. I always use the notation by D. E. Knuth for Stirling numbers of both kinds and for falling and rising factorials, simply because it is intuitive and works well. – Marko Riedel May 06 '16 at 21:07
  • @MarkoRiedel: I agree. I like this paper from him regarding Stirling numbers. It also gives me a glimpse of what he may have thought when he was creating TeX :-) – Markus Scheuer May 06 '16 at 21:16
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    The advent of TeX and LaTeX is comparable to Gutenberg but that is stating the obvious. – Marko Riedel May 06 '16 at 21:19
  • Hi, I am a little confused on $(2) \Rightarrow (3)$, I agree that $[u^{p-q}]P(u) = [u^p]u^q P(u)$, but how you can take the $u^q$ out? I am confused for this part... – Xiaonan May 06 '16 at 22:05
  • @Xiaonan: Let's say it in words. The identity states: The coefficient of $u^{p-q}$ of $P(u)$ is equal to the coefficient of $u^p$ of $P(u)$ multiplied with $u^q$. You might want to check it with small examples like $x^2^4=x^2=6$ and $[x^3]x(1+x)^4=x^3=6$. Do you see the relationship? – Markus Scheuer May 06 '16 at 22:30
  • Yes...I can understand this example, but I am confused that then you just got $u^j$ out and change from $[u^i] u^j (1+u)^{m-n}$ to $u^j [z^j] \cdots$...would you please explain this briefly? – Xiaonan May 06 '16 at 23:57
  • From my opinion, $ x [x^3] (1+x)^4 = 4x?$ – Xiaonan May 06 '16 at 23:58
  • @Xiaonan: In (3) we have $z^j^n$. The factor $u^j$ belongs to $[u^i]$ left from it. – Markus Scheuer May 07 '16 at 00:46
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    @MarkusScheuer Sweet!!! Thank you! – Xiaonan May 07 '16 at 01:25
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Suppose we seek to verify that

$$\sum_{p=0}^l\sum_{q=0}^p (-1)^q {m-p\choose m-l} {n\choose q} {m-n\choose p-q} = 2^l {m-n\choose l}$$

where $m\ge n$ and $m-n\ge l.$

This is

$$\sum_{p=0}^l {m-p\choose m-l} \sum_{q=0}^p (-1)^q {n\choose q} {m-n\choose p-q}.$$

Now introduce the integral

$${m-n\choose p-q} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p-q+1}} (1+z)^{m-n} \; dz.$$

Note that this vanishes when $q\gt p$ so we may extend the range of $q$ to infinity, getting for the sum

$$\sum_{p=0}^l {m-p\choose m-l} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} (1+z)^{m-n} \sum_{q\ge 0} (-1)^q {n\choose q} z^q \; dz \\ = \sum_{p=0}^l {m-p\choose l-p} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} (1+z)^{m-n} (1-z)^n \; dz.$$

Introduce furthermore

$${m-p\choose l-p} = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{l-p+1}} (1+w)^{m-p} \; dw.$$

This too vanishes when $p\gt l$ so we may extend $p$ to infinity, getting

$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{l+1}} (1+w)^{m} \\ \times \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z} (1+z)^{m-n} (1-z)^n \sum_{p\ge 0} \frac{w^p}{z^p} \frac{1}{(1+w)^p} \; dz \; dw.$$

The geometric series converges when $|w/z/(1+w)|\lt 1.$ We get

$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{l+1}} (1+w)^{m} \\ \times \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z} (1+z)^{m-n} (1-z)^n \frac{1}{1-w/z/(1+w)} \; dz \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{l+1}} (1+w)^{m} \\ \times \frac{1}{2\pi i} \int_{|z|=\epsilon} (1+z)^{m-n} (1-z)^n \frac{1}{z-w/(1+w)} \; dz \; dw.$$

Now from the convergence we have $|w/(1+w)|<|z|$ which means the pole at $z=w/(1+w)$ is inside the contour $|z|=\epsilon.$ Extracting the residue yields (the pole at zero has disappeared)

$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{l+1}} (1+w)^{m} \left(1+\frac{w}{1+w}\right)^{m-n} \left(1-\frac{w}{1+w}\right)^n \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{l+1}} (1+2w)^{m-n} \; dw \\ = 2^l {m-n\choose l}.$$

Marko Riedel
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