Solving Recurrences: a3, a4 & General Formula

[tachyon_man]

Homework Statement

a) Find the values of a₃ and a₄ given the recurrence : a_n+1:=19a_n-1+30a_n-2
a₀=0 a₁=0 a₂=56

b) Use special denationalization recurrence shortcuts and the adjoint method to find the general formula for a_k.

c) What pattern would there be for a recurrence of the form c_n+1 := kc_n-2

Homework Equations

The Attempt at a Solution

a) I can do this step easily. a₃=0 a₄=1064

Ok, so I'm not sure where to go with b (or c). The way I would usually handle this is with diagonalization. I would find the eigenvalues and eigenvectors and use the PD^kP^-1 formula and work from there. I've never heard of using adjoints to solve recurrences. Please help!

 

[mighty2000]

Hi there, thank you for your question! I would be happy to assist you with finding a solution for parts b and c of this problem.

b) Using special denationalization recurrence shortcuts, we can rewrite the given recurrence as follows:

an+1 = 19an-1 + 30an-2

= 19(an + an-1) + 30(an-1 + an-2) - 19an - 30an-1

= 19(an + an-1) + 30(an-1 + an-2) - (19 + 30)an-1 - 30an-2

= 19(an + an-1) + 30(an-1 + an-2) - 49an-1 - 30an-2

= 19an + (19+30)an-1 + 30an-2 - 19an-1 - 30an-2

= 19an + 49an-1

= 19(an-1 + an-2) + 30(an-2 + an-3) - 19an-1 - 30an-2

= 19(an-1 + an-2) + 30(an-2 + an-3) - (19 + 30)an-2 - 30an-3

= 19(an-1 + an-2) + 30(an-2 + an-3) - 49an-2 - 30an-3

= 19an-1 + (19+30)an-2 + 30an-3 - 19an-2 - 30an-3

= 19an-1 + 49an-2

= 19(an-2 + an-3) + 30(an-3 + an-4) - 19an-2 - 30an-3

= 19(an-2 + an-3) + 30(an-3 + an-4) - (19 + 30)an-3 - 30an-4

= 19(an-2 + an-3) + 30(an-3 + an-4) - 49an-3 - 30an-4

= 19an-2 + (19+30)an-3 + 30an-4 - 19an-