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I am interested in solving recursions and while I was trying to solve this one: $a_n=a_{n-1}+4n-3$

I ran onto a problem, I don't see how to find an explicit solution using standard inhomogeneous method(finding homogeneous and particular solution). Specifically I have a problem with finding particular solution. Problem for me is how to find expression to represent $f(n)=4n-3$. I do know how to solve for $f(n)=n$ with representing $f(n)$ as polynomial $an+b$ and substituting in the recursion. Is there a way to solve this type of recursions when $f(n)$ isn't depending just on $n$. Can we solve for even more complicated expressions such as $f(n)=5n^2+2n+4$ or any polynomial of degree $n$

Raphael
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TijuanaTaxi
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1 Answers1

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Given:

$a(n) = a(n-1) + 5n^2 + 2n + 4$

We hypothesise that because the difference is quadratic, $a(n)$ is cubic. In particular:

$a(n) = a_3 n^3 + a_2 n^2 + a_1 n + a_0$

So:

$a_3 n^3 + a_2 n^2 + a_1 n + a_0 = a_3 (n-1)^3 + a_2 (n-1)^2 + a_1 (n-1) + a_0 + 5n^2 + 2n + 4$

$\Rightarrow 3 a_3 n^2 + (-3 a_3 + 2 a_2) n + (a_3 - a_2 + a_1) = 5n^2 + 2n + 4$

This must be true for all $n$. Therefore:

$3 a_3 = 5$

$-3 a_3 + 2 a_2 = 2$

$a_3 - a_2 + a_1 = 4$

Note that $a_0$ doesn't appear in these equations, so this is effectively a linear system, three equations in three unknowns. The unique solution is:

$a_3 = \frac{5}{3}$

$a_2 = \frac{7}{2}$

$a_1 = \frac{35}{6}$

And so:

$a(n) = \frac{5}{3} n^3 + \frac{7}{2} n^2 + \frac{35}{6} n + a_0$

is a solution to the recurrence equation for any $a_0$.

Pseudonym
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