Finding linear recurrences : reverse_rsolve

### 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

xn*vn+k+...+x0*vk=0

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

  vk+2 −3vk+1 −6 vk =0

  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 − |
  1
  2
  | vk+2 + |
  1
  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),