5.54.7 Finding linear recurrences : reverse_rsolve
reverse_rsolve takes as argument a vector
v=[v0...v2n−1] made of the first 2n terms of a sequence (vn)
which is supposed to verify a linear recurrence relation of
degree smaller than n
where the xj are n+1 unknowns.
reverse_rsolve returns the list x=[xn,...,x0]
of the xj coefficients (if xn≠ 0 it is reduced to 1).
In other words reverse_rsolve solves the linear system of
n equations :
xn*vn+...+x0*v0 | = | 0 |
... |
xn*vn+k+...+x0*vk | = | 0 |
... |
xn*v2*n−1+...+x0*vn−1 | = | 0
|
|
The matrix A of the system has n rows and n+1 columns :
A=[[v0,v1...vn],[v1,v2,...vn−1],...,[vn−1,vn...v2n−1]] |
reverse_rsolve returns the list x=[xn,...x1,x0] with xn=1
and x is the solution of the system A*revlist(x).
Examples
-
Find a sequence satisfying a linear recurrence of degree at
most 2 whose first elements 1, -1, 3, 3.
Input :
reverse_rsolve([1,-1,3,3])
Output :
[1,-3,-6]
Hence x0=−6, x1=−3, x2=1 and the recurrence relation is
Without reverse_rsolve, we would write the matrix of the system :
[[1,-1,3],[-1,3,3]] and use the rref command :
rref([[1,-1,3],[-1,3,3]])
Output is [[1,0,6],[0,1,3]] hence x0=−6 and x1=−3
(because x2=1). - Find a sequence satisfying a linear recurrence of degree at
most 3 whose first elements are 1, -1, 3, 3,-1, 1.
Input :
reverse_rsolve([1,-1,3,3,-1,1])
Output :
[1,(-1)/2,1/2,-1]
Hence so, x0=−1, x1=1/2, x2=−1/2, x3=1, the recurrence
relation is
vk+3 − | | vk+2 + | | vk+1 −vk =0 |
Without reverse_rsolve, we would write the matrix of the system :
[[1,-1,3,3],[-1,3,3,-1],[3,3,-1,1]].
Using rref command, we would input :
rref([[1,-1,3,3],[-1,3,3,-1],[3,3,-1,1]])
Output is [1,0,0,1],[0,1,0,1/-2],[0,0,1,1/2]]
hence x0=−1, x1=1/2 and x2=−1/2 because x3=1),