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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?

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

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  • $\begingroup$ Just to clarify, $\alpha$ and $c_n$ are known? Also, what have you tried yourself to solve this problem? $\endgroup$ – B. Mehta May 11 '18 at 14:29
  • $\begingroup$ @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. $\endgroup$ – kolonel May 11 '18 at 14:30
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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 $
$a_2 = \alpha a_1 + c_2$
$a_1 = \alpha a_0 + c_1$
Adding


$a_{n+1} = (\alpha -1)\sum_{k=0}^n a_k + \sum_{k=1}^n c_k$

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  • $\begingroup$ Thanks. Can you express it terms of a cummulative sum of $c_n$? $\endgroup$ – kolonel May 11 '18 at 14:43
  • $\begingroup$ 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 $\sigma c_n$ $\endgroup$ – kolonel May 11 '18 at 14:49
  • $\begingroup$ @kolonel check if the edited answer is in the correct form $\endgroup$ – kayush May 11 '18 at 14:58
  • $\begingroup$ I think it should just $\alpha$ in the last expression not $\alpha - 1$. $\endgroup$ – kolonel May 11 '18 at 15:41
  • $\begingroup$ by adding we get, $$\sum_{k=1}^{n+1} a_k = \alpha \sum_{k=0}^n a_k + \sum_{k=1}^n c_k \implies a_{n+1}+\sum_{k=1}^{n} a_k = \alpha \sum_{k=0}^n a_k + \sum_{k=1}^n c_k$$ $\endgroup$ – kayush May 11 '18 at 15:44

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