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How to solve the following nonlinear recurrence equation using RSolve?

$$a(n+1):=\frac{1-(1-a(n)^4)^{1/4}}{1+(1-a(n)^4)^{1/4}} , \quad a(0):=\sqrt{2}-1$$

I tried:

RSolve[{a[n + 1] == (1 - (1 - a[n]^4)^(1/4))/(1 + (1 - a[n]^4)^(1/4)),
a[0] == Sqrt[2] - 1}, a[n], n]

but there is no answer.

Then I tried Wolfram Alpha, but $a(3)$ and $a(4)$ equal zero!

WolframAlpha["a[n + 1] =(1 - (1 - a[n]^4)^(1/4))/(1 + (1 - \
a[n]^4)^(1/4)), a(0)=Sqrt(2)-1", {{"Values", 1}, "Content"}]

enter image description here

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  • $\begingroup$ What makes you think there should be a closed-form solution? $\endgroup$ – masterxilo Oct 4 '16 at 21:40
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Aquick and dirty answer

 a[0] := N[Sqrt[2] - 1]
 a[n_] := a[n] = (1 - (1 - a[n - 1]^4)^(1/4))/(1 + (1 - a[n - 1]^4)^(1/4))

The repetition of a[n] is for memorization.

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    $\begingroup$ This just implements the recurrence; the OP seems to want the explicit solution. $\endgroup$ – J. M. will be back soon Oct 4 '16 at 11:28
  • $\begingroup$ $a(n)=0$ for $n\geq3$, I tried a[0] := N[Sqrt[2] - 1, 200] but nothing change $\endgroup$ – vito Oct 4 '16 at 11:33
  • $\begingroup$ In my computer with N[Sqrt[2] - 1, 200], a[1]= 0.00373489, a[2]=2.43231*10^-11,a[3]=0. Perhaps restart your kernel $\endgroup$ – cyrille.piatecki Oct 4 '16 at 11:38
  • $\begingroup$ I have tried to change the precision of the machine a[0] := SetPrecision[N[Sqrt[2] - 1, 200], 20],a[0] := SetPrecision[N[Sqrt[2] - 1, 200], 20] it change nothing for a[3] and a[4] $\endgroup$ – cyrille.piatecki Oct 4 '16 at 11:52
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    $\begingroup$ you can start with symbolic a[0] then N[a[3],20] yields ~10^-44, a[4] ~ 10^-175 is there a point to this? $\endgroup$ – george2079 Oct 4 '16 at 14:09

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