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Take the function:

f(x) = sqrt( ... sqrt(sqrt(sqrt(sqrt(sqrt(x)-1)+1)-1)+1)-1 ...) +1

I want to find the limit of the function for a given value of x as the number of iterations tends towards infinity.

How? I am using Wolfram Alpha, could not get the "Nest" functions to work (and unsure if they'd help anyway).

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    $\begingroup$ Can you use mathematica? Bacause questions about WA syntax are off topic. $\endgroup$
    – Kuba ♦
    Jul 7, 2017 at 6:07

2 Answers 2

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You can use Nest as follows:

Nest[Sqrt[Sqrt[#] - 1] + 1&, 2., 100]

1.4534

Or, you can use the usual trick for these kinds of things:

NSolve[r == Sqrt[Sqrt[r] - 1] + 1, r]

{{r -> 1.4534}, {r -> 1.}}

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  • $\begingroup$ Thanks. Nest function sadly doesn't work on WA but I thought it was a general Wolfram stack exchange. Your answer is correct for Mathematica so I'll check it. (I ended up just doing it manually on my TI 89 lol). $\endgroup$
    – CommaToast
    Jul 7, 2017 at 7:12
  • $\begingroup$ @CommaToast The following WA input works for me. $\endgroup$
    – Carl Woll
    Jul 7, 2017 at 18:17
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There are 2 fixed points. You can solve and illustrate,e.g.:

Solve[f[x] == x, x]
f[x_] := Sqrt[x - 1] + 1
nf[x_, n_] := Nest[f, x, n]
Manipulate[
 DiscretePlot[nf[x0, j], {j, 1, 10, 1}, GridLines -> {None, {2}}, 
  PlotRange -> {0, 5}], {x0, 1.1, 20, Appearance -> "Labeled"}]

enter image description here

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