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What is the exact value of $$R=\sqrt{-3+2\sqrt{-5+3\sqrt{-7+4\sqrt{-9+\dots}}}}$$
I tried to solve it like $\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}$, i.e. I tried to find the sequence function for this expression. I got that for $x=1$ expression became $$R=x\sqrt{-2x-1+(x+1)\sqrt{-2x-3+(x+2)\sqrt{-2x-5+\dots}}}$$ Now it is obvious that $$f(x)=x\sqrt{-2x-1+f(x+1)}$$ for some function $f$ such that $R=f(1)$. Problem is how to solve this recurrence equation. After squaring both sides it become more complicated. Is there an easy way to solve this recurrence equation. If you have better solution, please explain how to find the exact value of $R$.

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    $\begingroup$ Perhaps explain what it means... Do we want the first approximation to be $\sqrt{-3}$, the second approximation to be $\sqrt{-3+2\sqrt{-5}}$ and so on? $\endgroup$ – GEdgar Jan 16 '15 at 15:06
  • $\begingroup$ @GEdgar. No, just see $f(x)$. I don't think we can find first or second or any approximation because $\sqrt{-3}\notin\mathbb{R}$. $\endgroup$ – user164524 Jan 16 '15 at 15:08
  • $\begingroup$ So, actually you don't want to evaluate the meaningless nested radical, instead you want to solve a functional equation? $\endgroup$ – GEdgar Jan 16 '15 at 15:10
  • $\begingroup$ @GEdgar. I just want to find the exact value of this expression. I computed it with Wolfram Mathematica and it converges to $1$, so there must be a way to prove it. $\endgroup$ – user164524 Jan 16 '15 at 15:13
  • $\begingroup$ So, you do want to use those complex approximations I gave, and see that they approach 1. $\endgroup$ – GEdgar Jan 16 '15 at 15:14
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Nested Radical

$$ c_0 = (n + a)^2 \quad {\small\textit{added for conciseness}}\\ x + n + a = \sqrt{c_0 + ax + x\sqrt{c_0 + a(x + n) + (x + n)\sqrt{c_0 + a(x + 2n) + (x + 2n)\sqrt{...}}}} $$

This equation was discovered by Ramanujan.

Your equation is a case of this type of nested radical with $x = 2,\ n = 1,\ a = -2$.

$$ x + n + a = 1 $$

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  • $\begingroup$ I included the acknowledgement of Ramanujan and cleaned up the equation. $\endgroup$ – Axoren Jan 16 '15 at 16:20
  • $\begingroup$ the page talks about nested radicals with positive terms, which is obviously not the case here $\endgroup$ – mercio Jan 16 '15 at 16:48
  • $\begingroup$ @mercio You only read the first paragraph and assumed it represented the entire page. Herschfeld's convergence theorem only holds true for such a nested radical, however many of the remaining solutions and identities do not have that restriction. $\endgroup$ – Axoren Jan 16 '15 at 17:21
  • $\begingroup$ The conditions for Ramanujan's formula have been completely omitted at mathworld, it seems. $\endgroup$ – GEdgar Jan 17 '15 at 15:09
  • $\begingroup$ There are no restrictions on Ramanujan's formula as it was derived algebraically without conditions on the values $x,\ n,\ \text{and } a$. $\endgroup$ – Axoren Jan 17 '15 at 17:18
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Maple agrees that this converges to $1$ (or something very close to $1$).
The plot shows $z_1=\sqrt{-3}$, $z_2=\sqrt{-3+2\sqrt{-5}}$ and so on (joined in order by line segmants):

seq

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Note that $f(x)=x^2$ solves the difference equation that you should use: $$ f(x) = x\sqrt{-2x-1+f(x+1)} $$

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  • $\begingroup$ Yes, it is obvious that $f(x)=x^2$, but I want to know how to solve difference equations like this one when solution isn't obvious. $\endgroup$ – user164524 Jan 16 '15 at 17:26
  • $\begingroup$ Usually, the solution of a difference equation is not expressible in closed form. $\endgroup$ – GEdgar Jan 16 '15 at 18:01

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