# Solution to recurrence relation with two variables: one variable known in advance from data [closed]

I want to solve the following recurrence relation:

$$a_{n+1} = \alpha a_n + c_{n+1}$$

with initial conditions: $$a_0 = 0$$ $$c_1 = \beta$$

I actually know $c_{n+1}$ from the data in advance, however I am trying to find a solution to $a_{n+1}$ in terms of $c_{n+1}$. Is this possible?

## closed as off-topic by B. Mehta, Namaste, Xander Henderson, cansomeonehelpmeout, José Carlos SantosMay 11 '18 at 21:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – B. Mehta, Namaste, Xander Henderson, cansomeonehelpmeout, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.

• Just to clarify, $\alpha$ and $c_n$ are known? Also, what have you tried yourself to solve this problem? – B. Mehta May 11 '18 at 14:29
• @B.Mehta Yes, but I wrote $c_n$ with a subscript because its values are changing. Also, $\beta$ is known, just to be completely clear. – kolonel May 11 '18 at 14:30

You can just keep using the formula and get a pattern like this:
$a_{n+1} = \alpha a_n + c_{n+1} = \alpha(\alpha a_{n-1} + c_{n})+ c_{n+1} = \alpha(\alpha(\alpha a_{n-2} + c_{n-1})+c_n)+ c_{n+1} = \cdots = \alpha^{n+1}a_0 + \alpha^nc_1 + \alpha^{n-1}c_2+ \cdots \alpha c_n + c_{n+1}$

If you want in terms of cumulative sum

$a_{n+1} = \alpha a_n + c_{n+1}$
$a_{n} = \alpha a_{n-1} + c_{n}$
$a_{n-1} = \alpha a_{n-2} + c_{n-1}$
$\cdot \ \ \ \ \ \ \ \ \ \ \ \ \ \cdot \ \ \ \ \ \ \ \ \ \ \ \ \ \cdot$
$\cdot \ \ \ \ \ \ \ \ \ \ \ \ \ \cdot \ \ \ \ \ \ \ \ \ \ \ \ \ \cdot$
$\cdot \ \ \ \ \ \ \ \ \ \ \ \ \ \cdot \ \ \ \ \ \ \ \ \ \ \ \ \ \cdot$
$a_2 = \alpha a_1 + c_2$
$a_1 = \alpha a_0 + c_1$
$a_{n+1} = (\alpha -1)\sum_{k=0}^n a_k + \sum_{k=1}^n c_k$
• Thanks. Can you express it terms of a cummulative sum of $c_n$? – kolonel May 11 '18 at 14:43
• To be clearer, by cummulative sum, I mean the following: http://mathworld.wolfram.com/CumulativeSum.html, and we denote the cummulative sum up to index $n$ by $\sigma c_n$ – kolonel May 11 '18 at 14:49
• I think it should just $\alpha$ in the last expression not $\alpha - 1$. – kolonel May 11 '18 at 15:41
• by adding we get, $$\sum_{k=1}^{n+1} a_k = \alpha \sum_{k=0}^n a_k + \sum_{k=1}^n c_k \implies a_{n+1}+\sum_{k=1}^{n} a_k = \alpha \sum_{k=0}^n a_k + \sum_{k=1}^n c_k$$ – kayush May 11 '18 at 15:44