# recurrence relation

• Oct 17th 2010, 11:10 PM
dunsta
recurrence relation
Hi
Am I close with this or way off?

First I have to decide if this is a linear constant coefficient recurrence relation
$a_{k+1} + a^2 _{k} + k^2$

I think a. The equation is not linear because the base $a^2_{k}$ this $a$ is squared.

b. The equation is not constant coefficient recurrence relation because one of the coefficients $k^2$ is not constant.

but, I also feel the equation could be linear because (excuse my simplicity, but I'm a beginner here), because the equation moves down in steps a_(k+1) --> a_(k)

thanks
• Oct 17th 2010, 11:52 PM
CaptainBlack
Quote:

Originally Posted by dunsta
Hi
Am I close with this or way off?

First I have to decide if this is a linear constant coefficient recurrence relation
$a_{k+1} + a^2 _{k} + k^2$

I think a. The equation is not linear because the base $a^2_{k}$ this $a$ is squared.

b. The equation is not constant coefficient recurrence relation because one of the coefficients $k^2$ is not constant.

but, I also feel the equation could be linear because (excuse my simplicity, but I'm a beginner here), because the equation moves down in steps a_(k+1) --> a_(k)

thanks

It is non-linear as you say.

The $k^2$ term is not a coefficient

CB
• Oct 18th 2010, 12:13 AM
dunsta
Thanks Capn,
I want to find $a_2$and $a_4$
when $a_0 = 1$
Am I on the right path

when $a_0 = 1$
$a_{k+1} + a^2 _{k} + k^2$ for $a_1$

we know $a_0 = 1$ so
$a_{0+1} + a^2 _{0} +0^2 = 1$
$a_{1} + 1^2 + 0 = 1$
$a_{1} + 1 + 0 = 1$
therefore $a_{1} = 0$

so now we know $a_{1} = 0$
$a_{1+1} + 0^2 + 1^2 = 0$
$a_2 = -1$

so now we know $a_{2} = -1$
$a_{2+1} + -1^2 + 2^2 = -1$
$a_3 + 1 +4 = -1$
$a_3 = -6$

so now we know $a_{3} = -6$
$a_{3+1} + -6^2 +3^2 =-6$
$a_{4} + 36 +9 =-6$
$a_{4} =-51$

have I worked through this correctly?

thanks for helping.
• Oct 18th 2010, 01:40 AM
emakarov
Is it $a_{k+1} + a^2 _{k} + k^2$ or $a_{k+1} = a^2 _{k} + k^2$? The former is not an equation. I strongly suspect that the first + is a typo.
• Oct 18th 2010, 02:00 AM
dunsta
sorry, it's
$a_{k+1} + a^2 _{k} + k^2 = 0$
• Oct 18th 2010, 03:15 AM
emakarov
In this case, $a_{k+1}=-a_k^2-k^2$. Therefore, a_1 = -1 - 0 = -1; a_2 = -1 - 1 = -2; a_3 = -4 - 4 = -8; a_4 = -64 - 9 = -73.