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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?

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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. – 2017-01-31
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    Interesting. I expected as much...and now I'm wondering if the complex definition of $\cos$ and tweaking it may help. – 2017-01-31
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    Duplicate with [**this**](http://math.stackexchange.com/questions/2102847/interpolating-function-for-a-nested-radical-y-x-sqrtay-x-1#comment4323932_2102847) – 2017-03-24
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    @NgChungTak thanks! – 2017-03-24

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