Re: Help generalizing Liouville's Polynomial Identity
- To: mathgroup at smc.vnet.net
- Subject: [mg103618] Re: [mg103596] Help generalizing Liouville's Polynomial Identity
- From: Tito Piezas <tpiezas at gmail.com>
- Date: Wed, 30 Sep 2009 05:01:39 -0400 (EDT)
- References: <200909291140.HAA25752@smc.vnet.net> <4AC23ED6.5000000@wolfram.com>
Hello Dan, Thanks again for the help. I've updated my site http://sites.google.com/site/tpiezas/001 with the identities and added your name (and a few other guys who responded). By the way, someone observed that the numerical factor is generated by the formula 2^(m-1) Binomial[3(m-1), m-1], where m is the number of objects taken at a time per term in the RHS. Once again, thanks for your time. :-) Sincerely, Tito On Tue, Sep 29, 2009 at 11:07 AM, Daniel Lichtblau <danl at wolfram.com> wrote: > TPiezas wrote: > >> Hello all, >> >> "Liouville's polynomial identity" is given by >> http://mathworld.wolfram.com/LiouvillePolynomialIdentity.html and can >> be concisely encoded as, >> >> 6(x1^2 + x2^2 + x3^2 + x4^2)^2 = Sum(x_i +/- x_j)^4 >> >> To determine the number of terms of the summation, since we are to >> choose 2 objects from 4, then this is Binomial[4,2] = 6. But as there >> are 2 sign changes, then total is 2 x 6 = 12 terms, given explicitly >> in the link above. Going higher, and choosing 3 objects out of n >> variables, I found that, >> >> 60(x1^2 + x2^2 + ... + x7^2)^2 = Sum(x_i +/- x_j +/- x_k)^4 >> >> is true. The RHS contains 4 x 35 = 140 terms. (If the x_i are non- >> zero, one cannot express the square of the sum of less than 7 squares >> in a similar manner.) >> >> Question: If we take the next step, >> >> a(x1^2 + x2^2 + ... + x_n^2)^2 = Sum(x_i +/- x_j +/- x_k +/- x_m)^4 >> >> for some positive integer "a", then what is the least n, if we are to >> choose 4 objects at a time out of the x_n? >> >> I believe the RHS is a simple matter for Mathematica to calculate, and >> one can incrementally test n = 5,6,7,...etc until a neat identity is >> found. >> >> - Titus >> > > Some code: > > vars[n_,x_] := Array[x,n]; > > iterators[n_,j_] := Module[ > {iters=Table[{j[k],{-1,1}}, {k,n}]}, > iters[[1,2,1]] = 1; iters] > > quarticGenerator[n_,vars_] := Module[{j}, > Sum[(vars.Array[j,n])^4, > Apply[Sequence,iterators[n,j]]]] > > liouvilleQuartic[n_,k_,x_] := Total[ > Map[quarticGenerator[k,vars[k,x]/. > Thread[Range[k]->#]]&, > Subsets[Range[n],{k}]]]/2 > > Examples: > > Table[Factor[liouvilleQuartic[n,3,x]], {n,3,8}] > > This verifies your n=7 remark. > > Going one higher: > > InputForm[Table[Factor[liouvilleQuartic[n,4,x]], {n,4,12}]] > > This shows that n=10 works. We get for that case > > 672*(x[1]^2 + x[2]^2 + x[3]^2 + x[4]^2 + x[5]^2 + x[6]^2 + > x[7]^2 + x[8]^2 + x[9]^2 + x[10]^2)^2 > > The conjecture to prove is that this always works for n = 3*k-2. A bit of > further validation: > > In[10]:= InputForm[Factor[liouvilleQuartic[13,5,x]]] > > Out[10]//InputForm= > 7920*(x[1]^2 + x[2]^2 + x[3]^2 + x[4]^2 + x[5]^2 + x[6]^2 + x[7]^2 + x[8]^2 > + > x[9]^2 + x[10]^2 + x[11]^2 + x[12]^2 + x[13]^2)^2 > > It does start to get slow, though: > > In[11]:= InputForm[Factor[liouvilleQuartic[16,6,x]]] // Timing > Out[11]= {245.845, 96096*(x[1]^2 + x[2]^2 + x[3]^2 + x[4]^2 + x[5]^2 + > x[6]^2 + x[7]^2 + x[8]^2 + x[9]^2 + x[10]^2 + x[11]^2 + x[12]^2 + > x[13]^2 + x[14]^2 + x[15]^2 + x[16]^2)^2} > > The constant factors do not seem to be the initial part of any sequence in > the OEIS. > > Daniel Lichtblau > Wolfram Research >
- References:
- Help generalizing Liouville's Polynomial Identity
- From: TPiezas <tpiezas@gmail.com>
- Help generalizing Liouville's Polynomial Identity