5
$\begingroup$

This is a related question to the one I posted on MO earlier:

Is there a simple proof of the following Identity for $\sum_{k=m-1}^l(-1)^{k+m}\frac{k+2}{k+1}{\binom l k}\binom{k+1}m$?

It arose in the same context: the degeneracy of umbilic points on Weingarten surfaces.

For all $l,m\in{\mathbb N}$ with $l\geq m\geq0$ the following identities appear to hold: \begin{eqnarray} &(1-(2m+1)(m+1)){\textstyle{{l+1 \choose m}}}\nonumber\\ +\sum\limits_{k=m+1}^{l+1}&(-1)^{\scriptstyle{{k+m}}}{\textstyle{{l+1 \choose k}}}\left[(1-(2k+1)(m+2)){\textstyle{\frac{2m+2}{2k+1}{k \choose m+1}}}+(1-(2k+1)(m+1)){\textstyle{\frac{2m+1}{2k+1}{k \choose m}}}\right]\nonumber\\ &= \left\{\begin{array}{ccl} 0&if& l>m\\ 2(l+1)(l+2) &if& l=m \end{array}. \right.\nonumber \end{eqnarray}

Obviously the $l=m$ case is trivial (I include it for completeness). So, any suggestions for a proof of this?

$\endgroup$
  • 2
    $\begingroup$ First step: The beauty inside the square brackets simplifies to $\left(m-4k-2km\right)\dbinom{k}{m}$. The second step could be observing that $\dbinom{l}{k}\dbinom{k}{m} = \dbinom{l}{m}\dbinom{l-m}{k-m}$, and of course the $\dbinom{l}{m}$ can be taken out of the sum. After that, I believe you end up with the $l-m$-th finite difference of a degree-$1$ polynomial in $k$. $\endgroup$ – darij grinberg Aug 6 '17 at 12:50
  • $\begingroup$ That said, I suspect there are typos in the statement. The sum goes all the way up to $l+1$, but the $\dbinom{l}{k}$ kills the $k=l+1$ addend, which doesn't smell of intent to me. $\endgroup$ – darij grinberg Aug 6 '17 at 12:55
  • $\begingroup$ @darij grinberg - correct and now corrected! $\endgroup$ – Brendan Guilfoyle Aug 6 '17 at 13:48
  • $\begingroup$ I don't think you need anything more than Darij's first step. This gives the $(l+1)$th (or is it $(l+1)$st?) difference of a polynomial in $k$ of degree $m+1$ $\endgroup$ – Ira Gessel Aug 7 '17 at 19:20
3
$\begingroup$

Following the hint @darijgrinberg stated in the comment section with respect to the beauty inside the square brackets we focus on the sum and

we obtain \begin{align*} \color{blue}{\sum_{k=m+1}^{l+1}}&\color{blue}{(-1)^{k+m}\binom{l+1}{k} \left[(1-(2k+1)(m+2))\frac{2m+2}{2k+1}\binom{k}{m+1}\right.}\\ &\qquad\qquad\qquad\qquad\quad \color{blue}{\left.+(1-(2k+1)(m+1))\frac{2m+1}{2k+1}\binom{k}{m}\right]}\\ &=\sum_{k=m+1}^{l+1}(-1)^{k+m}\binom{l+1}{k}[m-2k(m+2)]\binom{k}{m}\tag{1}\\ &=\binom{l+1}{m}\sum_{k=m+1}^{l+1}(-1)^{k+m}\binom{l+1-m}{k-m}[m-2k(m+2)]\tag{2}\\ &=\binom{l+1}{m}\sum_{k=1}^{l+1-m}(-1)^{k}\binom{l+1-m}{k}[-2k(m+2)-m(2m+3)]\tag{3}\\ &=-2(m+2)\binom{l+1}{m}\sum_{k=1}^{l+1-m}(-1)^{k}\binom{l+1-m}{k}k\\ &\qquad-m(2m+3)\binom{l+1}{m}\left([[l+1=m]]-1\right)\tag{4}\\ &=-2(m+2)\binom{l+1}{l+1-m}(l+1-m)\sum_{k=1}^{l+1-m}(-1)^{k}\binom{l-m}{k-1}\\ &\qquad-m(2m+3)\binom{l+1}{m}\left([[l+1=m]]-1\right)\tag{5}\\ &=2(m+2)(l+1)\binom{l}{m}\sum_{k=0}^{l-m}(-1)^{k}\binom{l-m}{k}\\ &\qquad-m(2m+3)\binom{l+1}{m}\left([[l+1=m]]-1\right)\tag{6}\\ &\color{blue}{=2(l+1)(l+2)[[l=m]]}\\ &\qquad\color{blue}{-(1-(2m+1)(m+1))\binom{l+1}{m}\left([[l+1=m]]-1\right)}\tag{7}\\ \end{align*} in accordance with OPs claim.

Comment:

  • In (1) we use @darijgrinbergs simplified bracketed beauty.

  • In (2) we use the binomial identity $$\binom{p}{q}\binom{q}{r}=\binom{p}{r}\binom{p-r}{q-r}$$

  • In (3) we shift the index to start with $k=1$.

  • In (4) we split the sum and do some simplifications regarding $(1-1)^{l+1-m}$ using Iverson brackets.

  • In (5) and (6) we use the binomial identity $$\binom{p+1}{q+1}=\frac{p+1}{q+1}\binom{p}{q}$$ and we shift the index to start with $k=0$.

  • In (7) we do some final simplifications and adaptions to better see the relationship with OPs identity.

$\endgroup$
  • 1
    $\begingroup$ The context in which this arose is now in a paper on the arxiv: arxiv.org/abs/1709.00580 $\endgroup$ – Brendan Guilfoyle Sep 5 '17 at 17:17
7
$\begingroup$

Let $l$ and $m$ be two integers such that $l\geq m\geq0$. You want me to prove the identity \begin{align} & \left( 1-\left( 2m+1\right) \left( m+1\right) \right) \dbinom{l+1} {m}+\sum_{k=m+1}^{l+1}\left( -1\right) ^{k+m}\dbinom{l+1}{k}Q\left( k,m\right) \nonumber\\ & = \begin{cases} 0, & \text{if }l>m;\\ 2\left( l+1\right) \left( l+2\right) , & \text{if }l=m \end{cases} ,\tag{1}\label{g-pf.1} \end{align} where \begin{align*} Q\left( k,m\right) & =\left( 1-\left( 2k+1\right) \left( m+2\right) \right) \dfrac{2m+2}{2k+1}\dbinom{k}{m+1}\\ & +\left( 1-\left( 2k+1\right) \left( m+1\right) \right) \dfrac {2m+1}{2k+1}\dbinom{k}{m}. \end{align*}

Set $x=-2m-4$ and $y=-2m^{2}-3m$. Then, $y=1-\left( 2m+1\right) \left( m+1\right) $.

Every nonnegative integer $k$ satisfies $\dbinom{k}{m+1}=\dfrac{k-m} {m+1}\dbinom{k}{m}$ (by straightforward computation), and therefore the definition of $Q\left( k,m\right) $ rewrites as \begin{align} Q\left( k,m\right) & =\left( 1-\left( 2k+1\right) \left( m+2\right) \right) \dfrac{2m+2}{2k+1}\cdot\dfrac{k-m}{m+1}\dbinom{k}{m}\nonumber\\ & +\left( 1-\left( 2k+1\right) \left( m+1\right) \right) \dfrac {2m+1}{2k+1}\dbinom{k}{m}\nonumber\\ & =\left( m-4k-2km\right) \dbinom{k}{m}\tag{3}\label{g-pf.3} \end{align} (after some straightforward computation).

On the other hand, it is known that any three integers $a$, $b$ and $c$ satisfying $b\geq c$ satisfy \begin{equation} \dbinom{a}{b}\dbinom{b}{c}=\dbinom{a}{c}\dbinom{a-c}{b-c}\tag{5}\label{g-pf.5} \end{equation} (this is the so-called trinomial revision formula, in Knuth's terminology). Now, \begin{align} & \sum_{k=m+1}^{l+1}\left( -1\right) ^{k+m}\dbinom{l+1}{k} \underbrace{Q\left( k,m\right) }_{\substack{=\left( m-4k-2km\right) \dbinom{k}{m}\\\text{(by \eqref{g-pf.3})}}}\nonumber\\ & =\sum_{k=m+1}^{l+1}\left( -1\right) ^{k+m}\dbinom{l+1}{k}\left( m-4k-2km\right) \dbinom{k}{m}\nonumber\\ & =\sum_{k=m+1}^{l+1}\left( -1\right) ^{k+m}\left( m-4k-2km\right) \underbrace{\dbinom{l+1}{k}\dbinom{k}{m}}_{\substack{=\dbinom{l+1}{m} \dbinom{l+1-m}{k-m}\\\text{(by \eqref{g-pf.5}, applied to }a=l+1\text{, }b=k\text{ and }c=m\text{)}}}\nonumber\\ & =\sum_{k=m+1}^{l+1}\left( -1\right) ^{k+m}\left( m-4k-2km\right) \dbinom{l+1}{m}\dbinom{l+1-m}{k-m}\nonumber\\ & =\sum_{k=1}^{l+1-m}\underbrace{\left( -1\right) ^{k+m+m}}_{=\left( -1\right) ^{k}}\underbrace{\left( m-4\left( k+m\right) -2\left( k+m\right) m\right) }_{\substack{=xk+y\\\text{(by straightforward computation)}}}\dbinom{l+1}{m}\dbinom{l+1-m}{k}\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we have substituted }k+m\text{ for }k\text{ in the sum}\right) \nonumber\\ & =\sum_{k=1}^{l+1-m}\left( -1\right) ^{k}\left( xk+y\right) \dbinom {l+1}{m}\dbinom{l+1-m}{k}.\tag{7}\label{g-pf.7} \end{align} But \begin{align} & \dbinom{l+1}{m}\sum_{k=0}^{l+1-m}\left( -1\right) ^{k}\left( xk+y\right) \dbinom{l+1-m}{k}\nonumber\\ & =\sum_{k=0}^{l+1-m}\left( -1\right) ^{k}\left( xk+y\right) \dbinom {l+1}{m}\dbinom{l+1-m}{k}\nonumber\\ & =\underbrace{\left( -1\right) ^{0}}_{=1}\underbrace{\left( x\cdot 0+y\right) }_{\substack{=y\\=1-\left( 2m+1\right) \left( m+1\right) }}\dbinom{l+1}{m}\underbrace{\dbinom{l+1-m}{0}}_{=1}\nonumber\\ & \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum_{k=1}^{l+1-m}\left( -1\right) ^{k}\left( xk+y\right) \dbinom{l+1}{m}\dbinom{l+1-m}{k}}_{\substack{=\sum _{k=m+1}^{l+1}\left( -1\right) ^{k+m}\dbinom{l+1}{k}Q\left( k,m\right) \\\text{(by \eqref{g-pf.7})}}}\nonumber\\ & =\left( 1-\left( 2m+1\right) \left( m+1\right) \right) \dbinom{l+1} {m}+\sum_{k=m+1}^{l+1}\left( -1\right) ^{k+m}\dbinom{l+1}{k}Q\left( k,m\right) .\tag{11}\label{g-pf.11} \end{align} Thus, the left-hand side of the equality \eqref{g-pf.1} is the left-hand side of \eqref{g-pf.11}.

But it is well-known (and follows, e.g., from the binomial formula) that \begin{equation} \sum_{k=0}^{N}\left( -1\right) ^{k}\dbinom{N}{k}= \begin{cases} 1, & \text{if }N=0;\\ 0, & \text{if }N>0 \end{cases} \tag{12}\label{g-pf.12} \end{equation} for every nonnegative integer $N$. Hence, for every positive integer $N$, we have \begin{equation} \sum_{k=0}^{N}\left( -1\right) ^{k}\dbinom{N}{k}= \begin{cases} 1, & \text{if }N=0;\\ 0, & \text{if }N>0 \end{cases} =0\tag{13}\label{g-pf.13} \end{equation} (since $N>0$). Now, for every positive integer $N$, we have \begin{align} \sum_{k=0}^{N}\left( -1\right) ^{k}k\dbinom{N}{k} & =\underbrace{\left( -1\right) ^{0}0\dbinom{N}{0}}_{=0}+\sum_{k=1}^{N}\left( -1\right) ^{k}\underbrace{k\dbinom{N}{k}}_{=N\dbinom{N-1}{k-1}}\nonumber\\ & =\sum_{k=1}^{N}\left( -1\right) ^{k}N\dbinom{N-1}{k-1}=N\sum_{k=1} ^{N}\left( -1\right) ^{k}\dbinom{N-1}{k-1}\nonumber\\ & =N\sum_{k=0}^{N-1}\underbrace{\left( -1\right) ^{k+1}}_{=-\left( -1\right) ^{k}}\dbinom{N-1}{k}\nonumber\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we have substituted }k+1\text{ for }k\text{ in the sum}\right) \nonumber\\ & =-N\underbrace{\sum_{k=0}^{N-1}\left( -1\right) ^{k}\dbinom{N-1}{k} }_{\substack{= \begin{cases} 1, & \text{if }N-1=0;\\ 0, & \text{if }N-1>0 \end{cases} \\\text{(by \eqref{g-pf.12}, applied to }N-1\text{ instead of }N\text{)}}}=-N \begin{cases} 1, & \text{if }N-1=0;\\ 0, & \text{if }N-1>0 \end{cases} \nonumber\\ & = \begin{cases} -N, & \text{if }N-1=0;\\ 0, & \text{if }N-1>0 \end{cases} = \begin{cases} -N, & \text{if }N=1;\\ 0, & \text{if }N>1 \end{cases} \nonumber\\ & = \begin{cases} -1, & \text{if }N=1;\\ 0, & \text{if }N>1 \end{cases} \tag{15}\label{g-pf.15} \end{align} (since $-N=-1$ in the case when $N=1$). Hence, for every positive integer $N$, we have \begin{align*} \sum_{k=0}^{N+1}\left( -1\right) ^{k}\left( xk+y\right) \dbinom{N}{k} & =x\underbrace{\sum_{k=0}^{N+1}\left( -1\right) ^{k}k\dbinom{N}{k} }_{\substack{= \begin{cases} -1, & \text{if }N=1;\\ 0, & \text{if }N>1 \end{cases} \\\text{(by \eqref{g-pf.15})}}}+y\underbrace{\sum_{k=0}^{N+1}\left( -1\right) ^{k}\dbinom{N}{k}}_{\substack{=0\\\text{(by \eqref{g-pf.13})}}}\\ & =x \begin{cases} -1, & \text{if }N=1;\\ 0, & \text{if }N>1 \end{cases} +y0= \begin{cases} -x, & \text{if }N=1;\\ 0, & \text{if }N>1 \end{cases} . \end{align*} Applying this to $N=l+1-m$ (which is a positive integer since $l+1>l\geq m$), we obtain \begin{align*} \sum_{k=0}^{l+1-m}\left( -1\right) ^{k}\left( xk+y\right) \dbinom {l+1-m}{k} & = \begin{cases} -x, & \text{if }l+1-m=1;\\ 0, & \text{if }l+1-m>1 \end{cases} \\ & = \begin{cases} -x, & \text{if }l=m;\\ 0, & \text{if }l>m \end{cases} . \end{align*} Now, \eqref{g-pf.11} yields \begin{align*} & \left( 1-\left( 2m+1\right) \left( m+1\right) \right) \dbinom{l+1} {m}+\sum_{k=m+1}^{l+1}\left( -1\right) ^{k+m}\dbinom{l+1}{k}Q\left( k,m\right) \\ & =\dbinom{l+1}{m}\underbrace{\sum_{k=0}^{l+1-m}\left( -1\right) ^{k}\left( xk+y\right) \dbinom{l+1-m}{k}}_{= \begin{cases} -x, & \text{if }l=m;\\ 0, & \text{if }l>m \end{cases} }\\ & =\dbinom{l+1}{m} \begin{cases} -x, & \text{if }l=m;\\ 0, & \text{if }l>m \end{cases} = \begin{cases} -\dbinom{l+1}{m}x, & \text{if }l=m;\\ 0, & \text{if }l>m \end{cases} \\ & = \begin{cases} 0, & \text{if }l>m;\\ -\dbinom{l+1}{m}x, & \text{if }l=m \end{cases} = \begin{cases} 0, & \text{if }l>m;\\ 2\left( l+1\right) \left( l+2\right) , & \text{if }l=m \end{cases} \end{align*} (because $-\dbinom{l+1}{m}x=2\left( l+1\right) \left( l+2\right) $ in the case when $l=m$ (this follows by trivial computations)). This proves \eqref{g-pf.1}.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.