## Assistance with an analytic solution to radical

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ShaiHulud
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Joined: Wed Apr 06, 2011 2:45 am UTC

### Assistance with an analytic solution to radical

Hehe, I thought I would check out the forums and noticed there's a mathematics section. Nice

I've been working on this problem on an off for about 2 years with no luck

$r = \sqrt(1+\sqrt(2+\sqrt(4+\sqrt(8 + ...))))$.

As you can guess, it's the powers of 2 as terms in an infinitely nested radical. I know it converges since it's greater than zero, and $\sqrt(2^{2^0} + \sqrt(2^{2^1} + ... )$ converges.

I've basically only tried denesting, pulling out radicals, and recursive definitions to no avail. Any suggestions would be greatly appreciated. Thank you.

Qaanol
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### Re: Assistance with an analytic solution to radical

ShaiHulud wrote:Hehe, I thought I would check out the forums and noticed there's a mathematics section. Nice

I've been working on this problem on an off for about 2 years with no luck

$r = \sqrt(1+\sqrt(2+\sqrt(4+\sqrt(8 + ...))))$.

As you can guess, it's the powers of 2 as terms in an infinitely nested radical. I know it converges since it's greater than zero, and $\sqrt(2^{2^0} + \sqrt(2^{2^1} + ... )$ converges.

I've basically only tried denesting, pulling out radicals, and recursive definitions to no avail. Any suggestions would be greatly appreciated. Thank you.

Clever problem. I don’t have an explicit solution or proof, but the sequence of partial results appears to go to a limit of about 1.783165809264099.
wee free kings

phlip
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### Re: Assistance with an analytic solution to radical

I got the same number. I plugged it into Wolfram|Alpha in the hope it'd give some kind of simple closed form, which would give a clue as to how to tackle it... but it didn't give anything useful. Which means that if it does have a simple closed form, it's going to be something that Alpha can't auto-detect.

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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Qaanol
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### Re: Assistance with an analytic solution to radical

phlip wrote:I got the same number. I plugged it into Wolfram|Alpha in the hope it'd give some kind of simple closed form, which would give a clue as to how to tackle it... but it didn't give anything useful. Which means that if it does have a simple closed form, it's going to be something that Alpha can't auto-detect.

I’d like to know the syntax for entering an infinitely nested radical on Wolfram Alpha
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phlip
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### Re: Assistance with an analytic solution to radical

No, I mean I typed in 1.78etc into alpha, and it came up with a bunch of really unhelpful suggestions (look in the "Possible Closed Form" section).

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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jp26
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### Re: Assistance with an analytic solution to radical

I tried generalising the equation to:
$r_c = \sqrt{1+\sqrt{c+\sqrt{c^2+\sqrt{c^3 + ...}}}}$
and can prove that for [imath]c=1[/imath] we have [imath]r_1[/imath] as the golden ratio [imath]\frac{1+\sqrt{5}}{2}[/imath] and for [imath]c=4[/imath] we have [imath]r_4 = 2[/imath] by using
$\sqrt{4^k+\sqrt{4^{k+1}+\sqrt{4^{k+2} + ...}}} = 2^k + 1$

Cranica
Posts: 42
Joined: Thu Sep 18, 2008 1:58 am UTC

### Re: Assistance with an analytic solution to radical

I don't know too much about convergence, but how does the second statement in the OP imply that the radical does in fact converge?

iamacow
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Joined: Sun Feb 03, 2008 2:34 am UTC

### Re: Assistance with an analytic solution to radical

Cranica wrote:I don't know too much about convergence, but how does the second statement in the OP imply that the radical does in fact converge?

if you consider the original r as the limit of a sequence, like r0 = sqrt(1), r1=sqrt(1+sqrt(2)), and so on, this is positive and increasing. Considering the other as a limit of a sequence call it sn, we know that rn < sn for any n, so that rn is increasing and bounded above by the limit of sn, so it must converge.

I'm pretty sure.

phlip
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### Re: Assistance with an analytic solution to radical

Right. Since sqrt(2), sqrt(2+sqrt(4)), sqrt(2+sqrt(4+sqrt(16))), sqrt(2+sqrt(4+sqrt(16+sqrt(256)))), ... converges (specifically, to sqrt(2)*phi), and is strictly greater than the sequence in question, we have that the sequence is increasing and bounded from above, so it must converge to something...

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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Cranica
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### Re: Assistance with an analytic solution to radical

Okay, that makes sense. I just didn't see an obvious setup for the convergence of the second radical.

I showed this to a professor earlier and I'm fairly sure it derailed him for the rest of the day - he at least said he'd seen it before, but didn't remember how it was done.

phlip
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### Re: Assistance with an analytic solution to radical

Cranica wrote:I just didn't see an obvious setup for the convergence of the second radical.

The 2^2^n one?
Spoiler:
It's the expansion of sqrt(2)*sqrt(1+sqrt(1+sqrt(1+...)))

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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Cranica
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### Re: Assistance with an analytic solution to radical

Yeah, I figured that out in the intervening hours I wonder if maybe multiplying it by, say, 2 might help. Then we replace each 2^n term in the radical with 2^(2^n + n), and double the value of the whole radical. So we have $2r = 2*\sqrt(1+\sqrt(2+\sqrt(4+\sqrt(8 + ...)))) = \sqrt(8+\sqrt(32+\sqrt(1024 + ...)))$

I have no idea if that's useful or not.

Rgeminas
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### Re: Assistance with an analytic solution to radical

Can't we just say, naívely, that

$r=\sqrt{1+\sqrt{2+...}}=\sqrt{1+\sqrt{2}r}$?

Or, more rigorously, that r is the limit of the sequence:
$\{r_n\} | r_1=1, r_{n+1}=\sqrt{1+r_n\sqrt{2}}$,

It is not hard to prove that the sequence is bounded above by [imath]\frac{1+\sqrt{3}}{\sqrt{2}}[/imath], and that it is strictly increasing.

Therefore, we can say that [imath]r^2=1+r\sqrt{2}[/imath], where [imath]r[/imath] is the limit of [imath]r_n[/imath], by using the definition of the sequence and properties of limits. Then, we solve the equation and find the limit.

phlip
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### Re: Assistance with an analytic solution to radical

Rgeminas wrote:Can't we just say, naívely, that

$r=\sqrt{1+\sqrt{2+...}}=\sqrt{1+\sqrt{2}r}$?

If you expand that out, though, you get$r = \sqrt{1+\sqrt{2+\sqrt{8+\sqrt{128+\dots}}}} = \sqrt{2^{2^0-1} + \sqrt{2^{2^1-1} + \sqrt{2^{2^2-1} + \dots}}}$which is closely related to the$\sqrt{2^{2^0} + \sqrt{2^{2^1} + \sqrt{2^{2^2} + \dots}}}$that the OP mentioned they'd already solved, and which reduces to\begin{align} r &= \sqrt{2}r' \\ r' &= \sqrt{1+\sqrt{1+\sqrt{1+\dots}}} \\ &= \sqrt{1+r'} \\ &= \phi \\ r &= \sqrt{2} \phi \end{align}

Also, if you're going to be fancy with the diacritics, it's "naïve", not "naíve".

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enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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Cranica
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### Re: Assistance with an analytic solution to radical

There might very well be no analytic solution known for this - Wolfram claims there's none known for the one below, anyway.
$r = \sqrt(1+\sqrt(2+\sqrt(3+\sqrt(4 + ...))))$