How to solve recurrence relation when it is equal to a non-zero constant?

Question

My Initial Question :

I understand how to solve a recurrence relation like such:

$t(n) = t(n-1) + t(n-2)$.

I would turn it into:

$t(n) - t(n-1) - t(n-2) = 0$, then $r^2 - r - 1 = 0$, then you solve for roots and such.

But what do I do if the recurrence relation looks like this?

$t(n) = t(n-1) + t(n-2) + 4$.

Can I follow the same pattern and do:

$t(n) - t(n-1) - t(n-2) = 4$, then $r^2 - r - 1 = 4$, then $r^2 - r - 5 = 0$ and continue on?

I was able to solve my problem using the following method:

I decremented n by 1 turning:

$t(n)=t(n−1)+t(n−2)+4$

into

$t(n-1)=t(n−2)+t(n−3)+4$

I then subtracted the second equation from the first to get:

$t(n)= 2t(n−1)-t(n-3)$

After which I was able to carry on fine.

You could try $t(n) - t(n-1) - t(n-2) = t(n-1) - t(n-2) - t(n-3)$ which is $t(n)-2t(n-1)+t(n-3)=0$ and then $r^3-2r^2+1=0$ which is $(r-1)(r^2-r-1)=0$ — Henry, Mar 15 '17 at 00:43
See this previous Question, Non-homogeneous recurrence relation — hardmath, Mar 15 '17 at 00:58
I would highly encourage you to read generatingfunctionology — Santana Afton, Mar 15 '17 at 01:15
These constant coefficient (but non-homogeneous) relations are easily solved by making a change of unknowns. Split the constant term among the "right-hand side" terms, e.g. $t_n - t_{n-1} - t_{n-2} = 4$ becomes $s_n - s_{n-1} - s_{n-2} = 0$ when $s_n = t_n + 4$. — hardmath, Mar 15 '17 at 01:16
Thank you everyone for all the answers and readings. Most of what I read seems to be stuff I haven't covered in class yet, so I hope it will be helpful for me soon. After reading through the links I came back and re-read Henry's comment, this time understanding it fully, and used that to solve my assignment. Thanks. — Gharrot, Mar 15 '17 at 03:02

score 2 · Answer 1 · answered Mar 15 '17 at 01:42

The short answer is no.

A slightly less short answer is that you will have the solution to the recurrence is going to be the homogeneous solution plus the particular solution. The homogeneous solution will be found as normal for the related recurrence where there was no "extra" stuff added at the end, in this case $t(n)=t(n-1)+t(n-2)$. The particular solution will be of the same "form" as the nonhomogeneous part (with some slight complications to avoid overlapping with the homogeneous solution). In this case since it was $t(n)=t(n-1)+t(n-2)\color{red}{+4}$ (and since $1$ is not a root of the characteristic polynomial) the particular solution will be a constant.

So, $t(n)=t(n-1)+t(n-2)+4$ will have a closed form looking like $t(n)=c_1\lambda_1^n+ c_2\lambda_2^n + d$ where $\lambda_1,\lambda_2$ are the roots of the polynomial $\chi(x)=x^2-x-1$, $c_1$ and $c_2$ are constants which are determined by the initial conditions, and $d$ is a constant which can be solved for. (in this case it will be $-4$)

A longer answer can be found in any textbook under the heading "Solving a nonhomogeneous linear recurrence." See for example wikipedia or generatingfunctionology.

How to solve recurrence relation when it is equal to a non-zero constant?

My Initial Question :

I was able to solve my problem using the following method:

1 Answers1