MathGroup Archive 2009

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: help in constructing a binomial consequence

  • To: mathgroup at
  • Subject: [mg98277] Re: help in constructing a binomial consequence
  • From: "Sjoerd C. de Vries" <sjoerd.c.devries at>
  • Date: Sun, 5 Apr 2009 06:33:46 -0400 (EDT)
  • References: <gr4jm9$ac5$>

Hi Galina,

I'm not sure why your topic title uses the word binomial. It suggests
the coefficients of your sequence should be binomial but your example
doesn't show any such coefficients. Furthermore, your series is a bit
irregular. The first few start with x having the highest power, then
decreasing, but F[7] - F[9] do not adhere to that pattern.

Assuming you're mistaken here and F[7] should be x^3, F[8] x^2 y, F[9]
x y^2 and F[10] y^3, F can be given by

F[n_] :=
 Module[{t, m},
  t = \[LeftFloor]1/2 (-1 + Sqrt[1 + 8 (-1 + n)])\[RightFloor];
  m = n - 1 - 1/2 t (1 + t);
  x ^(t - m) y^m

I leave it as an exercise to you to determine why the variables t and
m have these specific values.

A brute-force approach, but easier to understand, would be

F[i_] := Flatten[Table[x ^(n - r) y^r, {n, 0, i}, {r, 0, n}]][[i]].

It does more calculations than necessary, but I assume i will never
get very high.

Cheers -- Sjoerd

On Apr 3, 11:08 am, Galina <Galina.Pil... at> wrote:
> Hello all,
> I need  to find an eigenvalues of the matrix M [N*N] where elements
> are of the type F[i]*F[j]. I need help to construct the elements F[i]
> which must be in the following order: F[1]=1, F[2]=x, F[3]=y, F[4]
> =x^2, F[5]=x*y, F[6]=y^2, F[7]=y*x^2, F[8]=x*y^2, F[9]=x^3 an=
d etc....
> Could someone help, please?
> Thanks a lot,
> Galina

  • Prev by Date: Re: Interpolation -> segfault?
  • Next by Date: Re: Joust in Mathematica
  • Previous by thread: Re: help in constructing a binomial consequence
  • Next by thread: Find out if LinearProgramming has found a solution