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I found the following formula in a book without any proof:

$$\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}.$$

I don't know how to prove this at all. Could you show me how to prove this? Or If you have any helpful information, please teach me. I need your help.

Update : I crossposted to MO.

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  • $\begingroup$ For simpler equations of this form, the trick would be to let $S_{2m} = \sum^{2m}_{k=0}(-1)^k\binom{2m}{k}^3$ and try to solve the equation $S_{2(m+1)} - S_{2m} = ...$. If this doesn't work, perhaps try induction? $m = 0$ is straightforward to check. However the induction bit will certainly hardcore. $\endgroup$ – user88595 Sep 18 '13 at 14:08
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    $\begingroup$ @user88595: Thanks, but I suspect that your idea doesn't work. $\endgroup$ – mathlove Sep 18 '13 at 14:49
  • $\begingroup$ @user88595, concerning the answer below; I think the choice of the polynomials are wrong. However if you look at the coefficient of ${xy}^{2m}$ in $$ (1-x)^{2m}(1+xy)^{2m}(1+y)^{2m} $$ then you get the left hand term. To get the right hand side I was working for a while but no clue for the moment. $\endgroup$ – Arash Sep 18 '13 at 16:22
  • $\begingroup$ Related: Combinatorial Proof of Dixon's Identity $\endgroup$ – Grigory M Dec 9 '13 at 16:44
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// I thought it would be nice to have an actual solution on this page (especially if that is not one of solutions posted @ MO).

OK, so we want to compute the coefficient $$ [z^{2m}w^{2m}]\left\{(1-z)^{2m}(1-w)^{2m}(z-w)^{2m}\right\} $$ i.e. the residue $$ \operatorname*{res}_{z,w}\frac{(1-z)^{2m}(1-w)^{2m}(z-w)^{2m}}{z^{2m}w^{2m}}\frac{dz}z\frac{dw}w; $$ after the substitution $Z=z/(1-w)$, $W=w/(1-z)$ (using that $z-w=\frac{Z-W}{1-ZW}$ and $\frac{dz}z\frac{dw}w=\frac{dZ\,dW}{ZW(1-ZW)}$) we see that it's equal to $$ \operatorname*{res}_{Z,W}\frac{(Z-W)^{2m}}{(1-ZW)^{2m}Z^{2m}W^{2m}}\frac{dZ\,dW}{ZW(1-ZW)}= [(ZW)^{2m}]\left\{\frac{(Z-W)^{2m}}{(1-ZW)^{2m+1}}\right\} $$ which is equal to $(-1)^m\binom{2m}{m}\binom{3m}m$.

// The substitution looks somewhat magical. I've learned it from Gessel–Stanton paper mentioned by Todd Trimble @ MO.

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Good$^\dagger$ proof

0. Let's prove more general form of Dixon's identity: $$ \sum(-1)^k\binom{a+b}{b+k}\binom{b+c}{c+k}\binom{c+a}{a+k}=\binom{a+b+c}{a,b,c} $$ (the version from OP corresponds to $a=b=c=n$).

1. Trinomial coefficients satisfy — and are defined by — a simple recurrence. So we only need to show that LHS — let's denote it $F(a,b,c)$ — satisfies $$ F(a,b,c)=F(a-1,b,c)+F(a,b-1,c)+F(a,b,c-1). $$

2. Let's rewrite LHS in 'Dyson style': it's the constant term of $$ (-1)^{a+b+c}\frac{(x-y)^{a+b}(y-z)^{b+c}(z-x)^{c+a}}{x^{2a}y^{2b}z^{2c}}. $$ But now the recurrence follows immediately from the identity $$ \frac{y^2}{(x-y)(y-z)}+\frac{z^2}{(y-z)(z-x)}+\frac{x^2}{(z-x)(x-y)}=-1 $$ (which is due to Euler, I believe).


$^\dagger$‘Good’ as in ‘I.J.Good. Short Proof of a Conjecture by Dyson’

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This is not a (full) answer, but it won't fit to a comment (and I hope it can be finished; but I also hope that there is a much simpler solution). The LHS is the coefficient $C_{2n}$ at $1$ in $((1-z)(1-w)(1-(zw)^{-1}))^{2n}$. We have $$C_n=\frac{1}{(2\pi i)^2}\oint\oint((1-z)(1-w)(1-(zw)^{-1}))^{n}\frac{dz}{z}\frac{dw}{w}$$ (we integrate over the unit circles). Let us compute $F(t):=\sum_n C_n t^n$. We just sum a geometric series and get

$$F(t)=\frac{1}{(2\pi i)^2}\oint\oint\frac{1}{zw-t(1-z)(1-w)(zw-1)}dz\,dw.$$ Let's compute the integral wrt. $z$ using residues. In the denominator we have a quadratic polynomial in $z$ with one root in the unit disk, so the residue is $\sqrt{}$ of its discriminant, and we get (modulo calculation mistakes) $$F(t)=\frac{1}{2\pi i}\oint\frac{1}{\sqrt{(w+t(w^2-1))^2-4t^2w(1-w)^2}}dw.$$ Under the square root we have a quartic polynomial in $w$, hence $F(t)$ is a complete elliptic integral of the first kind. Now we need to bring it to the normal form, or compute its Taylor series directly; at least for now I lack the energy to do it.

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  • $\begingroup$ Should that be $(1-z)(1-w)(1-(zw)^{-1})^{2n}$? $\endgroup$ – David E Speyer Sep 27 '13 at 15:49
  • $\begingroup$ @DavidSpeyer you mean why $n$ and not $2n$? I'm just computing a bit more (i.e. $C_n$, not just $C_{2n}$), though in the end I didn't really compute anything. $\endgroup$ – user8268 Sep 27 '13 at 16:20
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I'm posting an answer just to inform that the question has received an answer by Igor Rivin on MO.

https://mathoverflow.net/questions/143334/proving-sum-k-02m-1k-binom2mk3-1m-binom2mm-binom3mm

Mark Wildon mentioned that this is Dixon's identity.

http://en.wikipedia.org/wiki/Dixon%27s_identity

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Try this: use Newton's binomial on both sides of $$(1-x)^{2n}(1+y)^{2n}(1+z)^{2n}=[(1-x)(1+y)(1+z)]^{2n}$$

then track down and equate the coefficients at terms of the form $(xyz)^k$ on both sides. This should yield the required sum.

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    $\begingroup$ The idea of using polynomials is fine however I doubt how to get the left side of the equality $\endgroup$ – Arash Sep 18 '13 at 16:26

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