6.13.2  Values of a recurrence relation or a system: seqsolve

(See also Section 6.13.3.)

The seqsolve command finds the terms of a recurrence relation.

• seqsolve takes three arguments:
  • exprs, an expression or list of expressions that define the recurrence relation.
  • vars, a list of the variables used.
  • a, the starting value.
• seqsolve(exprs,vars,a) returns a formula for the nth term of the sequence.

For example, if a recurrence relation is defined by u_n+1 = f(u_n,n) with u₀ = a, the arguments to seqsolve will be f(x,n), [x,n] and a. If the recurrence relation is defined by u_n+2 = g(u_n,u_n+1,n) with u₀ = a and u₁ = b, the arguments to seqsolve will be g(x,y,n), [x,y,n] and [a,b].

The recurrence relation must have a homogeneous linear part, the nonhomogeneous part must be a linear combination of a polynomials in n times geometric terms in n.

Examples.

• Find u_n, given that u_n+1 = 2u_n + n and u₀=3.
  Input:
  seqsolve(2x+n,[x,n],3)
  Output:

  −n−1+4· 2n

• Find u_n, given that u_n+1 = 2u_n + n 3ⁿ and u₀=3.
  Input:
  seqsolve(2x+n*3^n,[x,n],3)
  Output:

  ⎛ ⎝

  n−3

  ⎞ ⎠

  · 3n+6· 2n

• Find u_n, given that u_n+1 = u_n + u_n−1, u₀ = 0 and u₁=1.
  Input:
  seqsolve(x+y,[x,y,n],[0,1])
  Output:

  5  ⎛ ⎜ ⎜ ⎝ − √ 5 +1 2 ⎞ ⎟ ⎟ ⎠ n   −4  ⎛ ⎜ ⎜ ⎝ − √ 5 +1 2 ⎞ ⎟ ⎟ ⎠ n     √ 5 −5  ⎛ ⎜ ⎜ ⎝ − √ 5 +1 2 ⎞ ⎟ ⎟ ⎠ n   +5  ⎛ ⎜ ⎜ ⎝ √ 5 +1 2 ⎞ ⎟ ⎟ ⎠ n   +4  ⎛ ⎜ ⎜ ⎝ √ 5 +1 2 ⎞ ⎟ ⎟ ⎠ n     √ 5 −5  ⎛ ⎜ ⎜ ⎝ √ 5 +1 2 ⎞ ⎟ ⎟ ⎠ n   20

• Find u_n and v_n, given that u_n+1=u_n + 2v_n, v_n+1 = u_n + n + 1 with u₀ = 1, v₀ = 1.
  Input:
  seqsolve([x+2*y,n+1+x],[x,y,n],[0,1])
  Output:

  ⎡ ⎢ ⎢ ⎣

  −2 n− ⎛ ⎝ −1 ⎞ ⎠ n+4· 2n−3 2

  ,

  ⎛ ⎝ −1 ⎞ ⎠ n+2· 2n−1 2

  ⎤ ⎥ ⎥ ⎦