# recurrence relation

• Oct 17th 2010, 10:10 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
\$\displaystyle a_{k+1} + a^2 _{k} + k^2\$

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

b. The equation is not constant coefficient recurrence relation because one of the coefficients \$\displaystyle 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, 10:52 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
\$\displaystyle a_{k+1} + a^2 _{k} + k^2\$

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

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

CB
• Oct 17th 2010, 11:13 PM
dunsta
Thanks Capn,
I want to find \$\displaystyle a_2 \$and \$\displaystyle a_4\$
when \$\displaystyle a_0 = 1\$
Am I on the right path

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

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

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

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

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

have I worked through this correctly?

thanks for helping.
• Oct 18th 2010, 12:40 AM
emakarov
Is it \$\displaystyle a_{k+1} + a^2 _{k} + k^2\$ or \$\displaystyle 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, 01:00 AM
dunsta
sorry, it's
\$\displaystyle a_{k+1} + a^2 _{k} + k^2 = 0\$
• Oct 18th 2010, 02:15 AM
emakarov
In this case, \$\displaystyle 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.