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 $


$x^{2}-2x+2=(x-1)^{2}+1$, 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).

| cite | improve this answer | |
  • $\begingroup$ 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? $\endgroup$ – joe Jun 7 '15 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 Jun 7 '15 at 1:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.