Re: Help generalizing Liouville's Polynomial Identity
- To: mathgroup at smc.vnet.net
- Subject: [mg103623] Re: Help generalizing Liouville's Polynomial Identity
- From: pfalloon <pfalloon at gmail.com>
- Date: Wed, 30 Sep 2009 05:02:34 -0400 (EDT)
- References: <h9srs3$pc1$1@smc.vnet.net>
On Sep 29, 9:42 pm, TPiezas <tpie... at gmail.com> wrote: > Hello all, > > "Liouville's polynomial identity" is given byhttp://mathworld.wolfram.com= /LiouvillePolynomialIdentity.htmland 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 Hi, The following Mathematica implementation may be useful: First, define functions to generate the lhs (without multiplicative constant) and rhs of the putative identity, for n terms and subsets of length k (n=4,k=2 in the original identity): (* left side with n terms *) lhs[n_] := Sum[x[i]^2, {i,n}]^2 (* right side with n terms, subsets of length k *) rhs[n_,k_] := Module[{sets, terms}, (* get all distinct subsets with k elements *) sets = Subsets[Array[x,n], {k}]; (* get all sums involving +/- over each subset (with first term always positive) *) terms = Table[Tuples[Join[{{First@set}}, Table[{-t,t}, {t,Rest@set}]]], {set, sets}] // Flatten[#,1] &; (* take sum^4 *) Total[(Plus@@@terms)^4] ] Search for values of n,k where the ratio of rhs/lhs is a numeric quantity: res = Do[expr = FullSimplify[rhs[n,k]/lhs[n]]; If[NumericQ[expr], Sow[{n,k,expr}]], {n,4,10}, {k,2,n} ] // Reap // Last // Flatten[#,1]& {{{4, 2, 6}, {7, 3, 60}, {10, 4, 672}}} I believe this is *all* the values for which such an identity holds for n<=10, 2<=k<=n. Note there are the two solutions you mentioned together with another one for n=10, k=4. Assuming the identity exists, the constant needed on the lhs can actually be found just by looking at the coefficient of, say, x[1]^4. On the lhs it is the constant, while on the rhs it is given by: multiplier[n_,k_] := 2^(k-1) Binomial[n-1,k-1] This is because there are Binomial[n-1,k-1] subsets of length k which contain x[1], and for each subset there are 2^(k-1) possible combinations of +/- signs. Check that this holds: multiplier@@@res[[All,;;-2]] == res[[All,-1]] True Hope this helps! Cheers, Peter.