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I am trying to turn the recurrence relation into a linear equation in order to sum some probabilities. I've done this with some recursive stuff similar to the Fibonacci sequence, but I was told by a friend that this is much harder because I have an alternating variable.

$x_n=(1-x_{n-1})*(\frac{1}{36+12(-1)^{n}}),$

$x_1=\frac{1}{24}, x_2=\frac{23}{1152}, x_3$ should equal $\frac{1129}{27648}$

The recursive sequence works as intended, and I could definitely program it to approach the eventual number I want, but I'd rather get the proper linear one. I tried to apply the method some people on here explained to me for fibonacci-esque sequences, but I couldn't get it to work. Honestly I don't think I completely get what exactly is happening when I translate from one to the other, so that might be even more important than this specific equation. If I did not explain anything right please ask me, thank you.

gamelizard
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  • Are you sure it is linear? I may be misunderstanding the question but the points $(1, \frac{1}{24})$, $(2, \frac{23}{1152})$, and $(3, \frac{1129}{27648})$ do not lie on the same line. $y-\frac{1}{24}=\frac{-25}{1152}(x-1)$ is the equation of the line that goes through the first two points, but not the third. I don't completely understand the situation, but maybe have an equation for odd values of $n$, and another for even values of $n$. – Lucien Jaccon Feb 25 '24 at 03:27
  • It is an inhomogeneous first-order linear recurrence relation with non-constant coefficients. You may solve it explicitly by following this answer of mine : https://math.stackexchange.com/questions/4610074/recurrence-relation-in-reduction-formula-for-integral-sinnx/4610252#4610252 (from the line "$c_n = \ldots$"). – Abezhiko Feb 25 '24 at 08:19
  • Yes this one I don't think is linear, I am trying to translate it from recursive to linear. So really another way I wrote it down is that the first part of the equation is $(1-x_1) * (1/24) or (1/48)$ and it alternates. The way I made it alternate is to have it be (36+-12). I see what you mean in terms of 2 separate equations, and I will definitely consider that. – gamelizard Feb 25 '24 at 09:37

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Alternate terms of your sequence satisfy linear recurrences with constant coefficients: \begin{align} x_{2n+1}&=\frac{1}{24}\big(1-x_{2n}\big)\\ &=\frac{1}{24}-\frac{1}{24\cdot48}\big(1-x_{2n-1}\big)\\ &=\frac{47}{1152}+\frac{x_{2(n-1)+1}}{1152}\ . \end{align} The solution of this recurrence is \begin{align} x_{2n+1}&=\frac{x_1}{1152^n}+47\left(\frac{1152^n-1}{1152^{n+1}-1152^n}\right)\\ &=\frac{1}{24\cdot1152^n}+\frac{47\big(1152^n-1\big)}{1151\cdot1152^n}\\ &=\frac{1128\cdot1152^n+23}{27624\cdot1152^n}\ , \end{align} and the evenly indexed terms of the sequence are then given by \begin{align} x_{2n}&=1-24x_{2n+1}\\ &=\frac{23\big(1152^n-1\big)}{1151\cdot1152^n}\ . \end{align}

lonza leggiera
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