Assistance with an analytic solution to radical

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ShaiHulud
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Assistance with an analytic solution to radical

Postby ShaiHulud » Wed Apr 06, 2011 2:58 am UTC

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

[math]r = \sqrt(1+\sqrt(2+\sqrt(4+\sqrt(8 + ...))))[/math].

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 [math]\sqrt(2^{2^0} + \sqrt(2^{2^1} + ... )[/math] converges.

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

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Qaanol
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Re: Assistance with an analytic solution to radical

Postby Qaanol » Wed Apr 06, 2011 6:46 am UTC

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

[math]r = \sqrt(1+\sqrt(2+\sqrt(4+\sqrt(8 + ...))))[/math].

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 [math]\sqrt(2^{2^0} + \sqrt(2^{2^1} + ... )[/math] 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.
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phlip
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Re: Assistance with an analytic solution to radical

Postby phlip » Wed Apr 06, 2011 8:12 am UTC

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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Qaanol
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Re: Assistance with an analytic solution to radical

Postby Qaanol » Wed Apr 06, 2011 8:36 am UTC

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

Postby phlip » Wed Apr 06, 2011 10:09 am UTC

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╰};
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jp26
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Re: Assistance with an analytic solution to radical

Postby jp26 » Fri Apr 08, 2011 9:55 pm UTC

I tried generalising the equation to:
[math]r_c = \sqrt{1+\sqrt{c+\sqrt{c^2+\sqrt{c^3 + ...}}}}[/math]
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
[math]\sqrt{4^k+\sqrt{4^{k+1}+\sqrt{4^{k+2} + ...}}} = 2^k + 1[/math]

Cranica
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Re: Assistance with an analytic solution to radical

Postby Cranica » Mon Apr 11, 2011 4:01 pm UTC

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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Re: Assistance with an analytic solution to radical

Postby iamacow » Mon Apr 11, 2011 4:51 pm UTC

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.

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Re: Assistance with an analytic solution to radical

Postby phlip » Tue Apr 12, 2011 12:40 am UTC

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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Cranica
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Re: Assistance with an analytic solution to radical

Postby Cranica » Tue Apr 12, 2011 7:31 am UTC

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.

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phlip
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Re: Assistance with an analytic solution to radical

Postby phlip » Tue Apr 12, 2011 9:15 am UTC

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╰};
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Cranica
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Re: Assistance with an analytic solution to radical

Postby Cranica » Tue Apr 12, 2011 9:33 am UTC

Yeah, I figured that out in the intervening hours :P

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 [math]2r = 2*\sqrt(1+\sqrt(2+\sqrt(4+\sqrt(8 + ...)))) = \sqrt(8+\sqrt(32+\sqrt(1024 + ...)))[/math]

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

Rgeminas
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Re: Assistance with an analytic solution to radical

Postby Rgeminas » Fri Apr 15, 2011 7:16 pm UTC

Can't we just say, naívely, that

[math]r=\sqrt{1+\sqrt{2+...}}=\sqrt{1+\sqrt{2}r}[/math]?

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

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.

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Re: Assistance with an analytic solution to radical

Postby phlip » Sat Apr 16, 2011 3:03 am UTC

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

[math]r=\sqrt{1+\sqrt{2+...}}=\sqrt{1+\sqrt{2}r}[/math]?

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

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╰};
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Cranica
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Re: Assistance with an analytic solution to radical

Postby Cranica » Sun Apr 17, 2011 2:56 am UTC

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

http://mathworld.wolfram.com/NestedRadicalConstant.html


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