# I Recursive square root inside square root problem

#### gvitalie

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

Let me please to share this example, maybe it could be of help for You.
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]

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