# I Recursive square root inside square root problem

#### gvitalie

Good morning, Dear Friend Gionata, even if it is evening.

Best Regards,
one of Yours many friends,
Vitalie.

I have been debating this issue for days:

I can't find a recursive function of this equation:
$\large{\sqrt{2+\pi \sqrt{3+\pi\sqrt{4+\pi\sqrt{5+\dotsb}}}}}$

Starting value 2 always added with pi

has been trying to find a solution this for days now, is what I have achieved so far:
This sucession result not correct:
$f\left(n\right)=\sqrt{n+1+\pi f\left(n+1\right)}$

http://www.wolframalpha.com/input/?i=f\left(n\right)=\sqrt{n+1+\pi+f\left(n+1\right)}
Unfortunately, I do not know how to move forward, thanks a lot!

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#### Quasimodo

After working on the problem for a day I have found:

Starting with the more general expression, $\large{\sqrt{1+\pi \sqrt{2+\pi\sqrt{3+\pi\sqrt{4+\dotsb}}}}}$ , we can get the backward recursive relation: $$f\left(n\right)=\sqrt{-1+\pi \sqrt{n+1+f\left(n+1\right)^2}}$$
Putting $f\left(n+1\right) = i$, we work our way down until we reach $f\left(1\right)$.

For example, let $n=3$.

$f\left(4\right) = i$, $f\left(3\right) = \sqrt{-1+\pi \sqrt{4+i^2}} = \sqrt{-1+\pi \sqrt{3}}$, until we reach finally: $$f\left(1\right) = \large{\sqrt{-1+\pi \sqrt{1+\pi \sqrt{2+\pi\sqrt{3}}}}}$$.

Squaring, adding one and dividing by $\pi$, we are getting $\sqrt{1+\pi \sqrt{2+\pi\sqrt{3}}}$ as required.

For your particular sequence getting as far down as $f\left(2\right)$ only, will suffice...

As easy as that!

#### Quasimodo

After working on the problem for a day I have found:

Starting with the more general expression, $\large{\sqrt{1+\pi \sqrt{2+\pi\sqrt{3+\pi\sqrt{4+\dotsb}}}}}$ , we can get the backward recursive relation: $$f\left(n\right)=\sqrt{-1+\pi \sqrt{n+1+f\left(n+1\right)^2}}$$
Putting $f\left(n+1\right) = i$, we work our way down until we reach $f\left(1\right)$.
And here is the Mathematica code for $\sqrt{2+\pi \sqrt{3+\pi\sqrt{4+\pi\sqrt{5+\dotsb}}}}$, for n=100:

z = I; For[n = 100, n > 1, n = n - 1,
z = N[Sqrt[-1 + \[Pi] Sqrt[n + 1 + z^2]], 100]]; (z^2 + 1)/\[Pi]

"Recursive square root inside square root problem"

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