For the recurrence relation:

$f_n = 2a_{n-1} - 2a_{n-2}$ I got the characteristic equation that had complex roots:

$x^2 - 2x + 2 = 0$ that gave roots $i, -i$ and I wasn't sure how to continue the solving the recurrence relation with the complex roots.

Another problem I found was with the recurrence relation:

$a_n = 4a_{n-1} - 4a_{n-2}$

I got the characteristic equation that had only 1 root:

$x^2 - 4x + 4 = 0$

$r_1$ = 2

And I wasn't sure what form to use for the general solution as it's typically of the form $a_n = c_1 * (r_1)^n + c_2 * (r_2)^n $


1 Answer 1


$x^{2}-2x+2=(x-1)^{2}+1$, and that roots $1-i$ and $1+i$. Now in general, if your characteristic eqution with real coeffiecients has two imaginary roots, they are conjugate, and can be written as $re^{i \theta}$ and $re^{-i \theta}$,$r>0$ and $\theta \in ]0,2\pi[$ then $a_{n}=k_{1}r^{n} \cos n\theta+k_{2}r^{n}\sin n\theta$. Now, if you have a real double root $r_{0}$, $a_{n}=k_{1}r_{0}^{n}+k_{2}nr_{0}^{n}$, in the case of a degree 2 real equation! (If you've seen ODEs, you will notice a lot of similarities).

  • $\begingroup$ For the case in my question where I have only 1 root could I just leave it as: $a_n = c_1 r_0 ^n$ since I don't have the second root? $\endgroup$
    – jn025
    Commented Jun 7, 2015 at 1:51
  • $\begingroup$ It's $a_{n}=c_{1}r_{0}^{n}+c_{2}nr_{0}^{n}=2^{n}(c_{1}+c_{2}n)$ $\endgroup$
    – mich95
    Commented Jun 7, 2015 at 1:53

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