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Consider the following sequence: $$a_{n+1}=ka_n+\frac{1}{a_n}$$ where $k<1$ and $a_0=100$.

Is there a closed form for $a_n$? I've seen a similar sequence that converges to $\sqrt{2}$ here but is it possible without using hyperbolic geometry as much as possible?

A. P.
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wyvern
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  • The case $k=\frac12$ is very special : it is known as the "babylonian method" for extracting square roots. 2) You must mean 'hyperbolic trigonometry" not geometry. 3) Is $k$ > 0 ?
    • – Jean Marie Feb 01 '23 at 07:40
  • For such sequences like the sequence $x_n=\sqrt{2}\coth\left(2^{n-1}\mathrm{arccoth} (x_1)\right)$, you can notice that it can be written under the form $x_n=a f\left(2^{n-1}f^{-1}(x_1)\right)$ (like in linear algebra you have "conjugation" transformation $M'=FMF^{-1}$). This will exactly be the same for your sequence (with a general value of $k$) for a certain function $f$ and a constant $a$. Is there a simple enough function $f$ for your case ? – Jean Marie Feb 01 '23 at 08:35
  • @chankelvin Do you really need the closed form, or you just want to compute the limit? Your sequence converges to $\sqrt{\frac{1}{1-k}}$. – PierreCarre Feb 01 '23 at 09:53