Equation involving double sum and product

Question

I am a physicist who needs your help. I am currently working on a problem in relativity and to verify a result that I derived I should prove this terrible looking equation:

$$\binom{l}{2\alpha}\prod_{j=1}^\alpha\left(-\frac{l-2(\alpha-j)}{l-2(\alpha-j)-1}\frac{2(\alpha-j)+1}{2(\alpha-j)+2}\right)=2^{l/2}\sum_{k=0}^{l/2}\sum_{i=0}^\alpha(-1)^i\binom{k}{i}\binom{l/2-k}{\alpha-i}\binom{l/2}{k}\binom{1/2(l/2+k-1)}{l/2}$$

where $l$ is an even integer and $\alpha\leq l/2$ is an integer. The last binomial coefficient in the equation is the generalized form of the binomial coefficient: https://en.wikipedia.org/wiki/Binomial_coefficient#Generalization_and_connection_to_the_binomial_series. I've tried induction but even the base step was not trivial due to this generalized form.

I have checked this equation in Mathematica up until $l=100$. I know no advanced math is required for this and it is just a matter of tedious algebra. So my question: do any of you guys know tricks on how to solve such equations?

Any help would be more than welcome!

Answer 1

Following the work by @epi163sqrt we seek to show that

$$2^m \sum_{k=0}^m \sum_{j=0}^p (-1)^j {k\choose j} {m-k\choose p-j} {m\choose k} {1/2(m+k-1)\choose m} = (-1)^p {m\choose p}^2.$$

We will suppose $m\ge p.$ Re-ordering we find

$$2^m \sum_{k=0}^m {m\choose k} {1/2(m+k-1)\choose m} \sum_{j=0}^p (-1)^j {k\choose j} {m-k\choose p-j}.$$

We get for the inner sum

$$[z^p] (1+z)^{m-k} \sum_{j\ge 0} (-1)^j {k\choose j} z^j.$$

Here we have extended to infinity because of the coeffient extractor. Continuing,

$$[z^p] (1+z)^{m-k} (1-z)^k.$$

With the outer sum we have

$$2^m [z^p] (1+z)^m \sum_{k=0}^m {m\choose k} {1/2(m+k-1)\choose m} (1+z)^{-k} (1-z)^k \\ = 2^m [z^p] (1+z)^m [w^m] (1+w)^{1/2(m-1)} \sum_{k=0}^m {m\choose k} (1+w)^{1/2 k} (1+z)^{-k} (1-z)^k \\ = 2^m [z^p] (1+z)^m [w^m] (1+w)^{1/2(m-1)} \left[1+\frac{\sqrt{1+w} (1-z)}{1+z}\right]^m \\ = 2^m [z^p] [w^m] (1+w)^{1/2(m-1)} [1+z + \sqrt{1+w} (1-z)]^m \\ = 2^m [z^p] [w^m] (1+w)^{1/2(m-1)} [1+\sqrt{1+w} + z (1 - \sqrt{1+w})]^m \\ = 2^m {m\choose p} [w^m] (1+w)^{1/2(m-1)} (1-\sqrt{1+w})^p (1+\sqrt{1+w})^{m-p}.$$

The contribution from $w$ is

$$\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{m+1}} (1+w)^{1/2(m-1)} (1-\sqrt{1+w})^p (1+\sqrt{1+w})^{m-p}.$$

Now we put $1-\sqrt{1+w} = v$ so that $w = v(v-2)$ and $dw = 2(v-1) \; dv$ (map takes zero to zero) to get

$$\;\underset{v}{\mathrm{res}}\; \frac{1}{v^{m+1} (v-2)^{m+1}} (1-v)^{m-1} v^p (2-v)^{m-p} 2(v-1) \\ = 2 (-1)^{m-p+1} \;\underset{v}{\mathrm{res}}\; \frac{1}{v^{m-p+1} (v-2)^{p+1}} (1-v)^{m} \\ = 2^{-p} (-1)^m \;\underset{v}{\mathrm{res}}\; \frac{1}{v^{m-p+1} (1-v/2)^{p+1}} (1-v)^{m} \\ = 2^{-p} (-1)^m \sum_{q=0}^{m-p} {m\choose q} (-1)^q {m-q\choose p} 2^{-(m-p-q)} \\ = 2^{-m} (-1)^m \sum_{q=0}^{m-p} {m\choose q} (-1)^q {m-q\choose p} 2^q.$$

Next observe that

$${m\choose q} {m-q\choose p} = \frac{m!}{q!\times p! \times (m-p-q)!} = {m\choose p} {m-p\choose q}.$$

Collecting everything we find

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

This is the claim. (Also goes through for $m=p.$)

Answer 2

Here is a starter. We can considerably simplify the product at the left-hand side of OPs assumed identity. \begin{align*} \binom{2m}{2\alpha}&\prod_{j=1}^{\alpha} \left(-\frac{2m-2(\alpha-j)}{2m-2(\alpha-j)-1}\,\frac{2(\alpha-j)+1}{2(\alpha-j)+2}\right)\\ &=2^{m}\sum_{k=0}^{m}\sum_{j=0}^\alpha(-1)^j\binom{k}{j}\binom{m-k}{\alpha-j}\binom{m}{k}\binom{\frac{1}{2}(m+k-1)}{m}\tag{1} \end{align*} where we conveniently set $l=2m$. We show the following is valid: \begin{align*} \color{blue}{\binom{2m}{2\alpha}}&\color{blue}{\prod_{j=1}^{\alpha} \left(-\frac{2m-2(\alpha-j)}{2m-2(\alpha-j)-1}\,\frac{2(\alpha-j)+1}{2(\alpha-j)+2}\right) =(-1)^{\alpha}\binom{m}{\alpha}^2}\tag{2} \end{align*}

We obtain \begin{align*} \color{blue}{\binom{2m}{2\alpha}}& \color{blue}{\prod_{j=1}^{\alpha} \left(-\frac{2m-2(\alpha-j)}{2m-2(\alpha-j)-1}\,\frac{2(\alpha-j)+1}{2(\alpha-j)+2}\right)}\\ &=(-1)^{\alpha}\binom{2m}{2\alpha} \prod_{j=0}^{\alpha-1}\left(\frac{2m-2j}{2m-2j-1}\,\frac{2j+1}{2j+2}\right)\tag{3.1}\\ &=(-1)^{\alpha}\binom{2m}{2\alpha} \frac{(2m)!!}{(2m-2\alpha)!!}\,\frac{(2m-2\alpha-1)!!}{(2m-1)!!}\,\frac{(2\alpha-1)!!}{(2\alpha)!!}\tag{3.2}\\ &=(-1)^{\alpha}\binom{2m}{2\alpha} \frac{2^mm!}{2^{m-\alpha}(m-\alpha)!}\,\frac{(2m-2\alpha)!}{2^{m-\alpha}(m-\alpha)!}\, \frac{2^mm!}{(2m)!}\,\frac{(2\alpha)!}{2^{\alpha}\alpha!2^{\alpha}\alpha!}\tag{3.3}\\ &=(-1)^{\alpha}\frac{(2m)!}{(2\alpha)!(2m-2\alpha)!}\,\frac{m!}{(m-\alpha)!}\,\frac{(2m-2\alpha)!}{(m-\alpha)!}\,\frac{m!}{(2m)!}\,\frac{(2\alpha)!}{\alpha!\alpha!}\tag{3.4}\\ &=(-1)^{\alpha}\frac{m!}{\alpha!(m-\alpha)!}\,\frac{m!}{\alpha!(m-\alpha)!}\tag{3.5}\\ &\,\,\color{blue}{=(-1)^{\alpha}\binom{m}{\alpha}^2}\tag{3.6} \end{align*} and the claim (2) follows.

Comment:

• In (3.1) we factor out the minus sign. We also shift the index $j$ to start with $j=0$ and we change the order of the terms $j\to \alpha-1-j$.

• In (3.2) we write the factors using double factorials .

• In (3.3) we use the identities \begin{align*} (2m)!!=2^mm!\qquad\textrm{and}\qquad (2m)!=(2m)!!(2m-1)!! \end{align*}

• In (3.4) we cancel powers of $2$ and we also expand the binomial coefficient.

• In (3.5) we simplify and do some more cancellation.

• In (3.6) we write the factors using a binomial coefficient again.

The following is left to show: \begin{align*} \color{blue}{2^{m}\sum_{k=0}^{m}\sum_{j=0}^\alpha (-1)^j\binom{k}{j}\binom{m-k}{\alpha-j}\binom{m}{k}\binom{\frac{1}{2}(m+k-1)}{m} =(-1)^{\alpha}\binom{m}{\alpha}^2} \end{align*}