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I have come across a few different questions relating to my issue (namely this one, but the answers are not working for me. Here are my inputs;

{Slider[Dynamic[c], {0, 2}], Dynamic[c]}
Dynamic[RecurrenceTable[{a[n + 1] == (c + a[n]^2)/2, a[1] == c/2},a, {n, 1, 20}]]
Dynamic[ListPlot[RecurrenceTable[{a[n + 1] == (c + a[n]^2)/2, a[1] == c/2},a, {n, 1, 20}]]]

These work so far and are just to help me visualize the sequence and its values. Then I try

Dynamic[RSolveValue[{a[n + 1] == (c + a[n]^2)/2, a[1] == c/2}, a[Infinity], n]]
(* RSolveValue[{a[1+n]==1/2 (0.715 +a[n]^2),a[1]==0.3575},a[\[Infinity]],n] *)

And I can't seem to get it to output the limit (when one exists, namely when c is less than or equal to 1). I've tried a couple of the different solutions from that thread but I can't get it to work. Most of the time it just reprints my input like the above. I'm trying to avoid defining my function because then I have to shift the index (n+1 to n), which I don't want to make a habit of.

I'm new to Mathematica so my apologies for any mistakes. Is there a way to get this thing to give me the limit or let me know if it doesn't exist?

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  • $\begingroup$ As with tghe prior post, you can get a finite list of candidates by solving a quadratic (in this case) like so. In[919]:= Solve[ainf == (c + ainf^2)/2, ainf] Out[919]= {{ainf -> 1 - Sqrt[1 - c]}, {ainf -> 1 + Sqrt[1 - c]}}. It is not difficult to reason by induction that the first is the correct limit: c<=1 by assumption, so a[1]<=1/2 and for all n, a[n+1]<=1/2+a[n]^2/2<1/2+1/2. $\endgroup$
    – Daniel Lichtblau
    Commented Dec 12, 2023 at 17:03
  • $\begingroup$ I should add that my reasoning about convergence only applies for -1<=c<=1. For values where c<-1 some experimenting indicates we have limit cycles. $\endgroup$
    – Daniel Lichtblau
    Commented Dec 12, 2023 at 21:05

3 Answers 3

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Your problem is that Infinity is not a number. But you may choose a large number and calculate a[n] for e.g. 1000 and 10000 and check if they are the same.

E.g. you may write:

Dynamic[RSolveValue[{a[n + 1] == (c + a[n]^2)/2, a[1] == cc/2},  a[1000], n]]
Dynamic[RSolveValue[{a[n + 1] == (c + a[n]^2)/2, a[1] == cc/2}, a[10000], n]]

Then you get:

enter image description here

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  • $\begingroup$ I see! I'm guessing then that RSolveValue can only take in finite values? To that end, may I ask; would there be a way to use Limit[] and have n -> Infinity? Choosing a large number manually might leave some room for human (my) error. Additionally, it'd be more comforting to have the function tell me explicitly "There is a limit; this" or "There is no limit". $\endgroup$
    – ant
    Commented Dec 12, 2023 at 11:09
  • $\begingroup$ I haver no proof, but I think that "Limit" tries to get a symbolic solution and fails for this non linear recursion equation. $\endgroup$
    – Daniel Huber
    Commented Dec 12, 2023 at 13:13
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Another approach is to look for stable points of the iteration. For example:

Solve[astar == (c + astar^2)/2, astar]

{{astar -> 1 - Sqrt[1 - c]}, {astar -> 1 + Sqrt[1 - c]}}

So these two values 1+/-Sqrt[c] are the only candidates for stable points. If you linearize, you find

D[(c + astar^2)/2, astar]
astar

So only the first of the these (with the minus sign) is a stable equilibrium (assuming 1>c>0).

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The OP question is more appropriate for math.stackexchange.com but the limit seems to be 1 - Sqrt[1 - c] for -3<=c<=1.

a1 can be 0 instead of c/2 since it only shifts the sequence by one position.

$$\left\{-3\leq c\leq 1,a_1=0,a_{n+1}=\frac{a_n^2+c}{2},\underset{n\to \infty }{\text{lim}}a_n=1-\sqrt{1-c}\right\}$$

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