2
$\begingroup$

Recently, i was trying to solve this recurrence relation $$ a_{n+4} = \frac{-\alpha(x)}{(n+4)} \cdot a_{n+3} +\frac{-\beta(x)}{(n+3)\cdot (n+4)} \cdot a_{n+2} $$

But i can't solve for $a_n$

I've tried to solve using Wolfram|Alpha and i got this result.
Actually i'm really good at recurrence relations, but i can't understand how do i get this result and how to solve it.
Edit: n = 0,1,2,3....

$\endgroup$
6
$\begingroup$

Hint Multiplying by $(n+4)!$ you get $$(n+4)!a_{n+4} = -\alpha(x) *(n+3)! a_{n+3} -\beta(x) *(n+2)!a_{n+2}$$

Let $b_n=n! a_n$ then, your recurrence is $$b_{n+4}= -\alpha(x) \cdot b_{n+3} -\beta(x) \cdot b_{n+2}$$ which is a standard second order recurrence.

Solve it, and then $$a_n=\frac{b_n}{n!}$$

$\endgroup$
  • $\begingroup$ when i solve this with wolfram|alpha for b result is this. This result is very different from the solution above is there something else i'm missing $\endgroup$ – Murat Güven Feb 16 at 22:05
  • $\begingroup$ @MuratGüven WA solves the second order recurrence, the solution is given by the roots of the characteristic equation (which is what WA has in bracket, under the extra assumption that they are different). $\endgroup$ – N. S. Feb 16 at 22:33
  • $\begingroup$ why beta is in the denominator here? $\endgroup$ – Murat Güven Feb 16 at 23:15
  • 2
    $\begingroup$ @MuratGüven That is irrelevant... $\frac{\sqrt{\alpha^2-\beta}}{\beta}c_1$ is just a contant, so writing that or $C_1$ is the same thing...That particular form probably appears because of the way in which WA solves the recurrence, but most of those constants can be put inside $C_1,C_2$. $\endgroup$ – N. S. Feb 16 at 23:24

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.