# Using RSolve for recurrence equation

I'd like to use Mathematica to verify the solution to a recurrence equation. I have the following equation:

$$Q_{k+1} = Q_k + \alpha(r_{k+1} - Q_k)$$.

I also have a derivation showing how to obtain a solution for any $$k$$:

$$Q_k = Q_{k-1} + \alpha(r_k - Q_{-1})$$

$$\ \ \ \ = \alpha r_k + (1 - \alpha)Q_{k-1}$$

$$\ \ \ \ = \alpha r_k + (1 - \alpha)\alpha r_{k-1} + (1 - \alpha)^2Q_{k-2}$$

$$\ \ \ \ = (1 - \alpha)^kQ_0 + \sum_{i=1}^k\alpha (1 - \alpha)^{k-i}r_i$$,

where $$Q_0$$ is some arbitrary constant. However, when I use RSolve, I get a different answer.

RSolve[Q[k] == Q[k - 1] + \[Alpha] (Subscript[r, k] - Q[k - 1]), Q[k], k]

gives me the solution:

$$(1 - \alpha)^{k-1}\mathbb{c}_1+(1-\alpha)^{-1+k}\sum_{K[1]=0}^{-1+k}(1-\alpha)^{-K[1]}\alpha r_{1+K[1]}.$$

This is close but not exactly what I want. So what am I missing here?

You are missing the initial condition Q[0] == Q0.

Q[k] /. RSolve[{Q[k] == Q[k - 1] + α (Subscript[r, k] - Q[k - 1]), Q[0] == Q0},
Q[k], k][[1]] // FullSimplify

This is equivalent to

$$(1-\alpha )^k Q0+\sum _{K[1]=1}^k \alpha\ (1-\alpha )^{k-K[1]}\ r_{K[1]}$$

• Great that works. I thought the $\mathbb{c}_1$ constant was the value for the initial condition, but I guess not. Do you know where it comes from if it's not the value for Q[0]? May 14, 2020 at 20:54
• @whaaswijk I do not know that. May 14, 2020 at 21:15
Clear["Global`*"]

Format[Q[k_]] := Subscript[Q, k];
Format[Q0] = Subscript[Q, 0];

Include the initial condition in RSolve

sol = (RSolve[
{Q[k] == Q[k - 1] + \[Alpha] (Subscript[r, k] - Q[k - 1]), Q[0] == Q0},
Q[k], k][[1]] // Simplify) /. K[1] -> i

Translate the index of summation

sol2 = ((sol /. Sum -> Inactive[Sum]) /.
Inactive[Sum][expr_, {i, imin_, imax_}] :>
Inactive[Sum][(expr /. i -> i - 1), {i, imin + 1, imax + 1}]) //
Collect[#, Q0] &

Simplify the summation term

(sol3 = sol2 /. (expr1_ * Inactive[Sum][expr2_, {i, imin_, imax_}]) :>
Inactive[Sum][expr1*expr2, {i, imin, imax}]) // Activate

• This is very nice, but it's using some advanced Mathematica magic that I'm not familiar with. I had no idea that you could do things like translating the index of summation. I'll have to look at the documentation to see exactly how this all works. May 14, 2020 at 20:58
• That was some deft use of Inactive and Collect. May 14, 2020 at 21:09