## converting a new recursive polynomial set into matrices

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• This procedure connects matrix representations to combinatorial trangular sequences in a direct manner! %I A000001 %S A000001 1, -1, 1, 0, -2, 1, 2, 2, -4, 1,
Message 1 of 1 , Oct 23, 2006
This procedure connects matrix representations to combinatorial
trangular sequences in a direct manner!

%I A000001
%S A000001 1, -1, 1, 0, -2, 1, 2, 2, -4, 1, -6, 2, 11, -7, 1, 22, -24,
-37, 42, -12, 1,
-146, 204, 217, -338, 133, -20, 1, 1766, -2654, -2395, 4359, -1995, 387,
-33,
1, -36064, 56072, 46122, -91568, 45323, -9982, 1073, -54, 1, 1212048,
-1921280, -1492916, 3124562, -1616590, 381615, -46200, 2901, -88, 1,
-66338064, 106377800, 79774002, -172519714, 91629105, -22515577, 2912958,
-205269, 7732, -143, 1
%N A000001 triangular sequence from the recursive polynomial:
p(k, x) = (-Fibonacci[n] + x)*p(k - 1, x) - (n - 1)*p(k - 2, x)
%C A000001 Here are the matrices my Mathematica produces:
2by2
{{0, 1},
{1, -1}}
3by3
{{0, 1, 0},
{0, 0, 1},
{0, 2, -1}}
4by4
{{0, 1, 0, 0},
{0, 0, 1, 0},
{0, 0, 0, 1},
{-2, -2, 4, -1}}
5by5
{{0, 1, 0, 0, 0},
{0, 0, 1, 0, 0},
{0, 0, 0, 1, 0},
{0, 0, 0, 0, 1},
{6, -2, -11, 7, -1}}
6by6
{{0, 1, 0, 0, 0, 0},
{0, 0, 1, 0, 0, 0},
{0, 0, 0, 1, 0, 0},
{0, 0, 0, 0, 1, 0},
{0, 0, 0, 0, 0, 1},
{-22, 24, 37, -42, 12, -1}}
How I get matrices from recursive polynomials: connections between
sequence types:
Starting with the first easy and linear type in Fibonacci/ Bonacci
sequences:
1) Recursive transforms like
f[n]=f[n-1]+f[n-2]
are connected to vector Matrix Markovs:
2) M={{0,1}
,{1,1}}
v[1]={0,1}
v[n]=M^n.v[1]
3) characteristic polynomials of these both
give coefficient expansions: ( toral inversion forms)
f[x]=x2-x-2
g[x]=x2*f[1/x]
coefficient expansion of[ x/g[x]]
4) Solutions of the recursive transforms gives the Binet F[n] types
of sequences.
5) digraph substitution algebras also produce these sequences.
That is five different ways of doing the same thing.
Well,
I've found a similar set of relationships in triangular sequence like
those characterized by the binomial.
1) triangular sequence as in Pascal's triangle of Combinations
2) polynomial recursive formulations whose coefficients are the
triangular sequences
3) Matrices whose characteristic polynomials are the recursive polynomials
4) row sum sequences of the triangulars
5) Matrix Markovs generated by the matrices
6) coefficient expansions of the recursive polynomials

What Gary and I've been investigating lately is the relationship of
"Field" type rules:
Steinbach suggested that the sequences associated with diagonals of
polynomials
were involved in a field like structure.
What we have been looking at is sums and products of this basic matrix
type and how they relate to the base
matrix.
What I discovered was that if you have the polynomials ,
you can get matrices.
If you have polynomials, you can get a triangular sequence.
If you have a triangular sequence , you can generate matrices.
I'm sure that n baesd functions like the Binets also follw this pattern
as well.

Since the matrices are closely related to different graph and
substitution algebras as well
geomtery, differential equations and symmetry, we have one of the most
extensinve systems of interlocking
mathematics here that is known.

%F A000001 p(k, x) = (-Fibonacci[n] + x)*p(k - 1, x) - (n - 1)*p(k - 2, x)
%e A000001 {{1},
{-1, 1},
{0, -2, 1},
{2, 2, -4, 1},
{-6, 2, 11, -7, 1},
{22, -24, -37, 42, -12, 1},
{-146, 204, 217, -338, 133, -20,1},
{1766, -2654, -2395, 4359, -1995, 387, -33, 1},
{-36064, 56072, 46122, -91568, 45323, -9982, 1073, -54, 1},
{1212048, -1921280, -1492916, 3124562, -1616590, 381615, -46200, 2901,
-88, 1},
%t A000001 Clear[p, a, b, An]
p[-1, x] = 0; p[0, x] = 1;
p[k_, x_] := p[k, x] = (-Fibonacci[n] + x)*p[k - 1, x] - (n - 1)*p[k - 2, x]
Table[Expand[p[n, x]], {n, 0, 10}]
Table[Sum[CoefficientList[p[n, x], x][[m]], {m,
1, Length[CoefficientList[p[n, x], x]]}], {n, 0, 15}]
w = Table[CoefficientList[p[n, x], x], {n, 0, 10}]
Flatten[w]
An[d_] := Table[If[n == d, -w[[n]][[m]], If[m == n + 1, 1, 0]], {n, 1,
d}, {m, 1, d}]
Table[An[d], {d, 1, 10}]
Table[CharacteristicPolynomial[An[d], x], {d, 1, 10}]
Join[{{1}}, Table[CoefficientList[CharacteristicPolynomial[An[d], x],
x], {d,
1, 10}]]
Flatten[%]
Table[NSolve[CharacteristicPolynomial[An[d], x] == 0, x], {d, 2, 10}]
%O A000001 1
%K A000001 ,nonn,
%A A000001 Roger Bagula (rlbagula@...), Oct 23 2006
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