I am studying the behaviour of the recursive function defined by

$x_{n+1}=x_n-\frac{1}{x_n}$, $x_0=2$

To get a better understanding of it, I would like to be able to see the "transition" between plots for different values of $x_0$, while I can move $x_0$ freely between some values. My initial idea on how to achieve this, is to use the following code:

{a[n + 1] == a[n] - 1/a[n], a[0] == t}, a, {n, 1000}]], {t, -3, 3}]

However, this code just makes Mathematica freeze a bit and then display $Aborted in the manipulate window, not doing what I want it to do. Why does this not work? How could I approach this problem using different code?

I am using Mathematica 11.0.


2 Answers 2


Try this. The problem is that you initial setting takes toooo long to evaluate. There is a 5 seconds limit build in.

But may be you should also limit n to something lower than 1000 like 25 or so? (Updated it to use N@t in code below, instead of just t to make it load faster initially)

 {t, -3, 3},
 SynchronousInitialization -> False,
 TrackedSymbols :> {t},
 Initialization :> (
   f[t_] := 
    RecurrenceTable[{a[n + 1] == a[n] - 1/a[n], a[0] == N@t}, 
     a, {n, 50}]

enter image description here

Mathematica graphics

So before you had the default True and that is why it aborted

Mathematica graphics

  • $\begingroup$ Hey thanks for your answer. You make a good point about the evaluation time, when using the integer 2, the eval time increases significantly for larger n. Somewhere between 25 and 30 it becomes uncomfortably large. That's why I put a 2. instead of a 2 in my original plot, which evaluated much, much quicker. Can something similar be done here to decrease the loading time? $\endgroup$
    – Tyron
    Oct 28, 2017 at 15:41
  • $\begingroup$ @Tyron yes you can. Just changed the t to be numerical. Updated my answer. So you have to decide if you want t to by real or integer. With integer it is much slower to load initially and so you have to use SynchronousInitialization->False $\endgroup$
    – Nasser
    Oct 28, 2017 at 15:51
  • $\begingroup$ @Nasser: In your answer above, you have a single difference equation: RecurrenceTable[{a[n + 1] == a[n] - 1/a[n], a[0] == N@t}, a, {n, 50}]. Can you tell me how your code needs to be changed if you have, say, 10 recurrence equations. How would you replicate your code for this extended problem? Thanks. ` $\endgroup$ Oct 29, 2018 at 3:24

Somehow the recursion poses a problem, but the iterative counterpart is rather straightforward. You can do

f[u_] := u - 1/u;
 a = {N@t};
 Do[AppendTo[a, f[Last[a]]], 1000];
 ListPlot[a, PlotRange -> {-50, 50}],
 {t, -3, 3}]

enter image description here


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