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Re: trouble with a Binet in a generalized Pell recursion

  • To: mathgroup at smc.vnet.net
  • Subject: [mg106632] Re: trouble with a Binet in a generalized Pell recursion
  • From: Roger Bagula <roger.bagula at gmail.com>
  • Date: Tue, 19 Jan 2010 05:14:20 -0500 (EST)
  • References: <201001141046.FAA19694@smc.vnet.net> <hip8bk$sto$1@smc.vnet.net>

On Jan 15, 12:17 am, Daniel Lichtblau <d... at wolfram.com> wrote:
> RogerBagulawrote:
> > The Pell equations come from the recursion:
> > a(n)=2*a(n-1)+a(n-2)
> > with two different starting points:{0,1}, and {1,1}.
> > I generalized that to:
> > a(n)=a0*a(n-1)+a(n-2)
>
> > Three ( almost) different ways to do Pell recursions.
> > 1) simple recursion
> > 2) Binet root forms
> > 3) Matrix Markov
> > The Binet form for the second function type doesn't work.
>
> > The modulo two patterns are simple patterns  and not the fractals I
> > was hoping for
> > when I thought of this last night.
>
> > Mathematica:
> > Clear[f, g, a, b, v1, v2, n, a0, b0, f1, g1]
> > f[0, a_] := 0; f[1, a_] := 1;
> > f[n_, a_] := f[n, a] = a*f[n - 1, a] + f[n - 2, a]
> > g[0, a_] := 1; g[1, a_] := 1;
> > g[n_, a_] := g[n, a] = a*g[n - 1, a] + g[n - 2, a]
> > Table[f[n, a], {n, 0, 10}, {a, 1, 11}]
> > Table[g[n, a], {n, 0, 10}, {a, 1, 11}]
> > b0 = x /. Solve[x^2 - a*x - 1 == 0, x][[1]]
> > a0 = x /. Solve[x^2 - a*x - 1 == 0, x][[2]]
> > FullSimplify[a0 - b0]
> > FullSimplify[a0 + b0]
> > f1[n_, a_] := (a0^n - b0^n)/Sqrt[4 + a^2]
> > g1[n_, a_] := (a0^n + b0^n)/a
> > Table[FullSimplify[ExpandAll[f1[n, a]]], {n, 0, 10}, {a, 1, 11}]
> > Table[FullSimplify[ExpandAll[g1[n, a]]], {n, 0, 10}, {a, 1, 11}]
> > v1[n_, a_] = MatrixPower[{{0, 1}, {1, a}}, n].{0, 1}
> > v2[n_, a_] = MatrixPower[{{0, 1}, {1, a}}, n].{1, 1}
> > Table[FullSimplify[ExpandAll[v1[n, a][[1]]]], {n, 0, 10}, {a, 1, 11}]
> > Table[FullSimplify[ExpandAll[v2[n, a][[1]]]], {n, 0, 10}, {a, 1, 11}]
> > ListDensityPlot[Table[Mod[f[n, a], 2], {n, 0, 32}, {a, 1, 33}],
> >     Mesh -> False]
> > ListDensityPlot[Table[Mod[g[n, a], 2], {n, 0, 32}, {a, 1, 33}], Mesh -
> >> False]
>
> > Respectfully, Roger L.Bagula
> > 11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
> >http://www.google.com/profiles/Roger.Bagula
> > alternative email: roger.bag... at gmail.com
>
> Not sure if a question is being raised. This response is in regard to
> the remark "The Binet form for the second function type doesn't work."
> Your definition of g1 does not agree with the result of RSolve for g.
>
> In[136]:= Together[
>   Expand[g[n, a] /.
>      RSolve[{g[n, a] == a*g[n - 1, a] + g[n - 2, a], g[0, a] =
== 1,
>        g[1, a] == 1}, g[n, a], n]][[1]]]
>
> Out[136]= (1/Sqrt[4 + a^2])2^(-1 -
>    n) (-2 (a - Sqrt[4 + a^2])^n + a (a - Sqrt[4 + a^2])^n +
>     Sqrt[4 + a^2] (a - Sqrt[4 + a^2])^n + 2 (a + Sqrt[4 + a^2])^n -
>     a (a + Sqrt[4 + a^2])^n + Sqrt[4 + a^2] (a + Sqrt[4 + a^2])^n)
>
>  From this I form what seems to be a corrected version.
>
> g2[n_, a_] :=
>   1/(2*Sqrt[4 + a^2])*((Sqrt[4 + a^2] + a - 2)*
>       b0^n + (Sqrt[4 + a^2] - a + 2)*a0^n)
>
> This appears to agree with your recursive definition for g[n,a].
>
> Daniel Lichtblau
> Wolfram Research

Daniel Lichtblau
Thanks for your help.
The recursions:
f(n)=f(n-1)+a*f(n-2)
with characteristic polynomials
x^2-x-a
can be treated the same way.
Matrix:
{{0,1},{a,1}}
for the Markov vector form.
Roger Bagula


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