The $n$th term of this infinitely nested radical sequence

Asked Jan 31 '17 at 21:14

Active Jan 31 '17 at 21:27

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While playing around with some friends, we found the closed form to the sequence that goes $y_1=0$ and $y_{n+1}=\sqrt{2+y_n}$ using the half angle-formula for cosine, which you may derive to be $y_n=2\cos(\pi/2^n)$. This let's us easily show that

$$\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}=\lim_{n\to\infty}2\cos(\pi/2^n)=2$$

Howeven, this does not generalize to the problem of any number beneath the radical:

$$y_{n+1}=\sqrt{a+y_n},\ y_1=0\implies y_n=?$$

Does anyone know how to derive a closed form for the general case of the $n$th term in the sequence?

edited Jan 31 '17 at 21:27

asked Jan 31 '17 at 21:14

Simply Beautiful Art

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The inverse map $x_{n+1}=x_n^2-a$ is essentially a logistic map (https://en.wikipedia.org/wiki/Logistic_map) and it leads to a closed form solution in very few cases. Of course $$\lim_{n\to +\infty}y_n = \frac{1+\sqrt{1+4a}}{2},$$ but I would not bet on the existence of a simple closed form for $y_n$, in the general case. – Jack D'Aurizio Jan 31 '17 at 22:27
Interesting. I expected as much...and now I'm wondering if the complex definition of $\cos$ and tweaking it may help. – Simply Beautiful Art Jan 31 '17 at 22:29
Duplicate with this – Ng Chung Tak Mar 24 '17 at 08:02
@NgChungTak thanks! – Simply Beautiful Art Mar 24 '17 at 10:42

The $n$th term of this infinitely nested radical sequence

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