## Assistance with an analytic solution to radical

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### 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

[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.

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.

### 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

[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.

wee free kings

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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.

Code: Select all

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void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

### 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

wee free kings

- 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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void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

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

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]

[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]

### 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?

### 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 r

_{0}= sqrt(1), r

_{1}=sqrt(1+sqrt(2)), and so on, this is positive and increasing. Considering the other as a limit of a sequence call it s

_{n}, we know that r

_{n}< s

_{n}for any n, so that r

_{n}is increasing and bounded above by the limit of s

_{n}, so it must converge.

I'm pretty sure.

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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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void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

### 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.

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:**

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void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

### 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 [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.

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.

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

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.

[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.

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

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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void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

### 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.

[math]r = \sqrt(1+\sqrt(2+\sqrt(3+\sqrt(4 + ...))))[/math]

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

[math]r = \sqrt(1+\sqrt(2+\sqrt(3+\sqrt(4 + ...))))[/math]

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

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