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Good day everybody,

I am looking to prove this:

$$\delta_{a,b}=\left(\frac{1}{2}\right)^{2\left(a-b\right)}\sum_{\gamma=0}^{a-b}\left(-1\right)^{a-b-\gamma}\frac{b+\gamma}{b}\left(\begin{array}{c} 2a\\ a-b-\gamma \end{array}\right)\left(\begin{array}{c} 2b-1+\gamma\\ 2b-1 \end{array}\right)$$

for $a \ge 0 $ and $b \ge 1$.

The case $b=0$ is an exception to the rule but is also of interest for me. Therefore I would actually most appreciate a proof of matrices $D$ and $C$ to be inverse:

$$D_{mn}=\begin{cases} \left(-1\right)^{\frac{m-n}{2}}\left(\begin{array}{c} m\\ \frac{m-n}{2} \end{array}\right)\left(\frac{1}{2}\right)^{m} & \text{ if }n\leq m\text{ and }n-m\text{ is even}\\ 0 & \text{else} \end{cases} $$ and

$$C_{ij}=\begin{cases} 1 & \text{if }i=j=0\\ 2^{j}\left[\left(\begin{array}{c} \frac{i+j}{2}-1\\ j-1 \end{array}\right)+2\left(\begin{array}{c} \frac{i+j}{2}-1\\ j \end{array}\right)\right] & \text{else if }i\geq j\text{ and }i-j\text{ is even}\\ 0 & \text{else} \end{cases}$$

where indexing starts from zero (and thus contains also $b=0$ case).

Thank you.

F. Jatpil
  • 538

1 Answers1

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We write $n$ instead of $\gamma$, focus on the essentials and skip the constant $(-1)^{a-b}2^{-2(a-b)}$.

We obtain for integers $a\geq b>0$ \begin{align*} \color{blue}{\sum_{n=0}^{a-b}}&\color{blue}{(-1)^n\frac{b+n}{b}\binom{2a}{a-b-n}\binom{2b-1+n}{n}}\tag{1}\\ &=\sum_{n=0}^{a-b}\frac{b+n}{b}\binom{2a}{a-b-n}\binom{-2b}{n}\tag{2}\\ &=\sum_{n=0}^{a-b}\binom{2a}{a-b-n}\binom{-2b}{n} -2\sum_{n=1}^{a-b}\binom{2a}{a-b-n}\binom{-2b-1}{n-1}\tag{3}\\ &=\binom{2a-2b}{a-b}-2\sum_{n=0}^{a-b-1}\binom{2a}{a-b-n-1}\binom{-2b-1}{n}\tag{4}\\ &=\binom{2a-2b}{a-b}-2\binom{2a-2b-1}{a-b-1}[[a>b]]\tag{5}\\ &\color{blue}{=[[a=b]]} \end{align*}

Comment:

  • In (1) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

  • In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

  • In (3) we split the sum and apply $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$ to the right-hand sum.

  • In (4) we apply Chu-Vandermonde's identity to the left sum and shift the index of the right sum by one to start with $n=0$.

  • In (5) we use Iverson brackets in the right expression, since the corresponding sum in the line before is zero if $a=b$. We also use the binomial identity $\binom{2p}{p}=2\binom{2p-1}{p-1}$.

Markus Scheuer
  • 108,315
  • (+1). I suppose you are using what Wikipedia calls Chu-Vandermonde (Vandermonde for generic complex arguments). – Marko Riedel Sep 11 '17 at 21:08
  • @MarkoRiedel: Thanks for the review, Marko. I've corrected the naming in order to be precise. – Markus Scheuer Sep 11 '17 at 21:13
  • For readers: from (4) to (5) the summation is also done using "Chu-Vandermonde's identity". Still: why Iverson bracket? The summation from (4) to (5) does not hold for $a=b$? Presumably the negative number in the lower part of the binomial coefficients is an issue, yet I do not see clearly why. You create "by hand" $\delta_{a,b}$ by pointing the special behaviour of $a=b$ case and summing other cases. Why is it special? – F. Jatpil Sep 12 '17 at 08:03
  • @F.Jatpil: The upper bound of the sum in (4) is $-1$ if $a=b$, which means the sum is $0$ in this case. This is not reflected in the right-hand expression in (5) which is equal to $1$ if $a=b$. So, the right-hand side in (5) is $-2\binom{2a-2b-1}{a-b-1}$ if $a>b$ and $0$ if $a=b$, which is the same as $-2\binom{2a-2b-1}{a-b-1}[[a>b]]$. We conclude only in the case $a>b$ we $\binom{2a-2b}{a-b}-2\binom{2a-2b-1}{a-b-1}=0$ while in case $a=b$ we have $\binom{2a-2b}{a-b}=1$ which is the same as $[[a=b]]$ or $\delta_{a,b}$. – Markus Scheuer Sep 12 '17 at 08:40
  • You are fully right, all I am saying is that for a negative $m$ terms which look like ${n}\choose{m}$ cannot be summed using Chu-Vandermonde. In this case the formula is invalid and this creates the delta. Why the summation is invalid for negative numbers must be hidden in the details of the proof of the Chu-Vandermonde summation. The upper bound $-1$ means there are two terms $n=0$ and $n=-1$ which can be summed, but not with Chu-Vandermonde formula - a special treatment (Iverson bracket) is needed. – F. Jatpil Sep 12 '17 at 08:41
  • @F.Jatpil: I fully agree with your first part. A sum $\sum_{k=p}^q a_k$ is per definition equal to zero if $p>q$. We do not sum backwards. You might find the reference in this answer helpful which also provides valuable information about using the sigma symbol. – Markus Scheuer Sep 12 '17 at 08:44
  • Well, this is clearly an unclear issue, see: https://math.stackexchange.com/questions/35080/upper-limit-of-summation-index-lower-than-lower-limit I think the context requires you to make summation of the two terms ($n=0,n=-1$). But if you do the summation you indeed find that the summation is zero... and your arguments apply. – F. Jatpil Sep 12 '17 at 08:50
  • @F.Jatpil: I see your point, but the top voted answer also agrees in that point. I consider the statements of D.E. Knuth as authoritative. I strongly recommend to read chapter 2: Sums of the referred book. It will help you to develop your own view about this topic. – Markus Scheuer Sep 12 '17 at 08:55