Question on solving linear recurrence relations

  • Thread starter japplepie
  • Start date
  • #1
93
0

Main Question or Discussion Point

Why does the characteristic equation of a linear recurrence relation always look like

an = series of constants multiplied by a number raised to n
 

Answers and Replies

  • #2
Stephen Tashi
Science Advisor
7,027
1,250
Why does the characteristic equation of a linear recurrence relation always look like

an = series of constants multiplied by a number raised to n
Unless a specific statement is made about the relation between the characteristic equation and the recurrence, we don't have any theorem to prove. The definition of the characteristic equation gives the form of the equation. Definitions are stipulations, not things to be proven or disproven. We can explain the motivation for the definition by doing some algebra.

You didn't establish much notation, so let's use the notation of the Wikipedia article http://en.wikipedia.org/wiki/Recurrence_relation.

In that article, the recurrence relation is:
[itex] a_n = c_1a_{n-1} + c_2a_{n-2}+\cdots+c_da_{n-d} [/itex] .

Assume [itex] a_n = t^n [/itex] for some unknown [itex] t [/itex]. Substitute a [itex]t^k [/itex] for each [itex] a^k [/itex] in the recurrence relation.

This gives:
[itex] t^n = c_1t^{n-1} + c_2t^{n-2}+\cdots+c_dt^{n-d} [/itex]

Assuming [itex] t\ne 0 [/itex] multiply both sides of the above equation by [itex] t^{d-n} [/itex] :

[itex] t^d = c_1 t^{d-1} + c_2t^{d-2}+\cdots+ c_d^0 [/itex]

Which is equivalent to:
[itex] 0 = t^d - c_1 t^{d-1} - c_2t^{d-2}-\cdots- c_d^0 [/itex]

---------
The above is the same as the characteristic equation:
[itex] 0 = t^d - c_1t^{d-1} - c_2t^{d-2}-\cdots-c_{d} [/itex]

Which comes from setting zero equal to the characteristic polynomial, which is:
[itex] p(t)= t^d - c_1t^{d-1} - c_2t^{d-2}-\cdots-c_{d} [/itex]

.
 

Related Threads on Question on solving linear recurrence relations

Replies
3
Views
5K
  • Last Post
Replies
22
Views
8K
Replies
3
Views
583
Replies
1
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
11
Views
6K
  • Last Post
Replies
12
Views
775
  • Last Post
Replies
5
Views
3K
Top