Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2006
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2006

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

Search the Archive

Re: new procedure for converting a new recursive polynomial set into matrices

  • To: mathgroup at smc.vnet.net
  • Subject: [mg70906] Re: new procedure for converting a new recursive polynomial set into matrices
  • From: Roger Bagula <rlbagula at sbcglobal.net>
  • Date: Wed, 1 Nov 2006 03:55:47 -0500 (EST)
  • References: <ehkcmn$jn9$1@smc.vnet.net>

I found a good reference for going from a tridiagonal matric to a 
recursive polynomial:
http://projecteuclid.org/Dienst/UI/1.0/Summarize/euclid.pjm/1102708710
Joanne Dombrowski
Tridiagonal matrix representations of cyclic selfadjoint operators.
Source: Pacific J. Math. 114, no. 2 (1984), 325?334

It appears to work well in Mathematica as well!

%I A000001
%S A000001 1, 1, -1, 0, -2, 1, -1, -3, 4, -1, -3, -6, 14, -7, 1, -14,
-24, 72, -48, 12,
-1, -109, -172, 586, -449, 143, -20, 1, -1403, -2103, 7718, -6375,
2296,
-402, 33, -1, -29354, -42588, 163595, -141144, 54448, -10718, 1094,
-54, 1,
-996633, -1416535, 5597100, -4956116, 1990080, -418458, 47881, -2929,
88, -1,
-54785461, -76870204, 309093440, -278042336, 114356068, -24994552,
3050819,
-208922, 7768, -143, 1
%N A000001 Fibonacci central triadiagonal matrices as a triangular
sequence from a recursive polynomial definition
%C A000001 Matrices:
{{1}},

{{1, -1},
{-1, 1}},

{{1, -1, 0},
{-1, 1, -1},
{0, -1, 2}},

{{1, -1, 0, 0},
{-1, 1, -1, 0},
{0, -1, 2, -1},
{0, 0, -1, 3}},

{{1, -1, 0, 0, 0},
{-1, 1, -1, 0, 0},
{0, -1, 2, -1, 0},
{0, 0, -1, 3, -1},
{0, 0, 0, -1, 5}},

{{1, -1, 0, 0, 0, 0},
{-1, 1, -1, 0, 0, 0},
{0, -1, 2, -1, 0, 0},
{0, 0, -1, 3, -1, 0},
{0, 0, 0, -1, 5, -1},
{0, 0, 0, 0, -1, 8}}
The Dombrowski paper defines a recursive polynomial form from the
tridiagonal matrices:
p[1,x]=1,p[2,x]=(x-b[1])/a[1]
p[n,x]=((x-b[n-1])*p[n-1,x]-a[n-2]*p[n-2,x])/a[n-1]
As long as
b[n-1]/a[n-1]
and
a[n-2]/a[n-1]
behave well ( rationally or like Integers)
this definition is a good recursive polynomial on a tridiagonal
matrix.
Here I use:
a[n]=-1
and
b[n]=Fibonacci[n]
%D A000001 Joanne Dombrowski,
Tridiagonal matrix representations of cyclic selfadjoint operators,
Pacific J. Math. 114, no. 2 (1984), 325?334
%F A000001 M(n,m)=If[ n == m, Fibonacci[n], If[n == m - 1 || n == m +
1, -1, 0]]
%e A000001 Triangular sequence:
{1},
{1, -1},
{0, -2, 1},
{-1, -3, 4, -1},
{-3, -6, 14, -7, 1},
{-14, -24, 72, -48, 12, -1},
{-109, -172, 586, -449, 143, -20, 1},
{-1403, -2103, 7718, -6375,2296, -402, 33, -1},
{-29354, -42588, 163595, -141144, 54448, -10718, 1094, -54, 1}
%t A000001 T[n_, m_] := If[ n == m, Fibonacci[n], If[n == m - 1 || n ==
m + 1, -1, 0]];
M[d_] := Table[T[n, m], {n, 1, d}, {m, 1, d}];
Table[M[d], {d, 1, 10}]
Table[Det[M[d]], {d, 1, 10}]
Table[Det[M[d] - x*IdentityMatrix[d]], {d, 1, 10}]
a = Join[M[1], Table[CoefficientList[Det[M[d] - x*IdentityMatrix[d]],
x], {d,
1, 10}]];
Flatten[a]
%O A000001 1
%K A000001 ,nonn,
%A A000001 Roger Bagula and Gary Adamson (rlbagula at sbcglobal.net), Oct
30 2006
RH
RA 192.20.225.32
RU
RI

>
>  
>


  • Prev by Date: Graphics--Exporting Graphics, drawing multiple objects at once and what not
  • Next by Date: Re: new procedure for converting a new recursive polynomial set into matrices
  • Previous by thread: Graphics--Exporting Graphics, drawing multiple objects at once and what not
  • Next by thread: Re: new procedure for converting a new recursive polynomial set into matrices