# Recurrence Relation Problem...

• Oct 10th 2010, 01:30 PM
aamiri
Recurrence Relation Problem...

Solve the recurrence relation :
an
= 2an−1 2an−2
subject to the initial conditions

a0 = 1, a1 = 3.
• Oct 10th 2010, 01:44 PM
pickslides
First solve the characteristic equation $s^2-2s+2= 0$

What do you get?
• Oct 10th 2010, 01:55 PM
aamiri
Quote:

Originally Posted by pickslides
First solve the characteristic equation $s^2-2s+2= 0$

What do you get?

Imaginary roots,

1+1i , 1-1i
• Oct 10th 2010, 02:02 PM
pickslides
That sounds right.

The next step is to put these solutions into polar form. Once you have done this, the general solution to the recurrance relation will be of the form

$\displaystyle a_n = r^n(\alpha \cos n\theta +\beta \sin n\theta)$

using $a_0=1,a_1=3$ to solve for $\alpha$ and $\beta$
• Oct 10th 2010, 02:12 PM
aamiri
Quote:

Originally Posted by pickslides
That sounds right.

The next step is to put these solutions into polar form. Once you have done this, the general solution to the recurrance relation will be of the form

$\displaystyle a_n = r^n(\alpha \cos n\theta +\beta \sin n\theta)$

using $a_0=1,a_1=3$ to solve for $\alpha$ and $\beta$

Thanks a lot.

But, after converting the 1+1i and 1-1i into polar forms, how do I get the achieve that form of equation. Also , in the end , am I likely to end up with a simultaneous equation while solving for the two variables.
• Oct 10th 2010, 02:41 PM
pickslides
Quote:

Originally Posted by aamiri
Thanks a lot.

But, after converting the 1+1i and 1-1i into polar forms, how do I get the achieve that form of equation.

That equation is the general form of the solution. Just substitute the values you find for r and $\theta$ into it.

Quote:

Originally Posted by aamiri
am I likely to end up with a simultaneous equation while solving for the two variables.

Yes (Wink)