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        riddles >> easy >> sqrt(1/2)*sqrt(sqrt(1/2))*sqrt(sqrt(sqrt(1/2)))*..
        
(Message started by: towr on Oct 21st, 2003, 1:48am)
Title: sqrt(1/2)*sqrt(sqrt(1/2))*sqrt(sqrt(sqrt(1/2)))*..
Post by towr on Oct 21st, 2003, 1:48am
Probably too simple, but what is
[sqrt](1/2) * [sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt][sqrt][sqrt](1/2) * ...
Title: Re: sqrt(1/2)*sqrt(sqrt(1/2))*sqrt(sqrt(sqrt(1/2))
Post by BNC on Oct 21st, 2003, 2:04am

on 10/21/03 at 01:48:50, towr wrote:
Probably too simple...


Probably...
::[hide]
Define A = [sqrt](1/2) * [sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt][sqrt][sqrt](1/2) * ...

But:
[sqrt](1/2) * [sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt][sqrt][sqrt](1/2) * ... = [sqrt][(1/2) * [sqrt](1/2) * [sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt][sqrt](1/2) * ...] = [sqrt][(1/2)*A]

So A=[sqrt][(1/2)A]
A2=0.5A.
A > 0
Hence A = 1/2
[/hide]::
Title: Re: sqrt(1/2)*sqrt(sqrt(1/2))*sqrt(sqrt(sqrt(1/2))
Post by towr on Oct 21st, 2003, 2:57am
There is another way to solve it, which is probably more usefull in general..

try this one:
http://www.ai.rug.nl/~towr/PHP/FORMULA/formula.php?md5=6c24cd181c1c949e08905b302d3dd9bc
[e]euhm, let's limit that to n factors ;) [/e]
Title: Re: sqrt(1/2)*sqrt(sqrt(1/2))*sqrt(sqrt(sqrt(1/2))
Post by william wu on Oct 21st, 2003, 4:55am
BNC, that proof is too complicated  :D

[hide]
[sqrt](1/2) * [sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt][sqrt](1/2) * [sqrt][sqrt][sqrt][sqrt][sqrt](1/2) * ...
= (1/2)1/2*(1/2)1/4*(1/2)1/8*...
= (1/2)[sum]2^-k | k = 1 to inf
= (1/2)1
= 1/2
[/hide]
Title: Re: sqrt(1/2)*sqrt(sqrt(1/2))*sqrt(sqrt(sqrt(1/2))
Post by Icarus on Oct 21st, 2003, 5:47pm
You can come up with a nice finite limit for the infinite sequence in towr's second challenge if 0 [le] A < 1.

An advantage William's approach has over BNC's is that BNC simply assumes the series converges. But William's approach can be made to show that it does.

This is more critical than you might think. I have seen many divergent sequences such that if you assume that there is a limit, you will be able to come up with a nice answer for what that limit is from the sequence's definition, much as BNC did. Unfortunately, that answer is meaningless unless you are a particle physicist!
Title: Re: sqrt(1/2)*sqrt(sqrt(1/2))*sqrt(sqrt(sqrt(1/2))
Post by towr on Oct 22nd, 2003, 12:38am

on 10/21/03 at 17:47:09, Icarus wrote:
You can come up with a nice finite limit for the infinite sequence in towr's second challenge if 0 [le] A < 1.
Yes, but unfortunately that send the wrong message, because I wanted to convey that the approach of "every factor is smaller than 1 thus it converges to 0" doesn't allways work.. (of course no one here fell for it)
Thanks btw for having taught me how to properly deal with converging infinite products :P (long ago in another thread)

A = e is also a nice case (without n going to [infty] of course)..
Title: Re: sqrt(1/2)*sqrt(sqrt(1/2))*sqrt(sqrt(sqrt(1/2))
Post by Icarus on Oct 22nd, 2003, 3:32pm

on 10/22/03 at 00:38:38, towr wrote:
Yes, but unfortunately that send the wrong message, because I wanted to convey that the approach of "every factor is smaller than 1 thus it converges to 0" doesn't allways work.. (of course no one here fell for it)


I'm not sure how that would apply here, since it does converge to zero when A < 1, though not simply because every term is less than 1!


Quote:
Thanks btw for having taught me how to properly deal with converging infinite products :P (long ago in another thread)


You're welcome! It's nice to know that something I said made sense at some point! ;)
Title: Re: sqrt(1/2)*sqrt(sqrt(1/2))*sqrt(sqrt(sqrt(1/2))
Post by SWF on Oct 22nd, 2003, 7:21pm
This reminds me of the Nested Radicals (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1054060550) riddle.  That one also had people driving Icarus up the walls by ignoring convergence yet still coming up with the correct answer.
Title: Re: sqrt(1/2)*sqrt(sqrt(1/2))*sqrt(sqrt(sqrt(1/2))
Post by Icarus on Oct 22nd, 2003, 7:46pm
It was not coming up with the right answer that was a problem. If an answer exists, the techniques employed will definitely find it. Their only downfall is that if a solution does not exist, they will still give an answer without any indication that it is purely garbage.

These functional equation solutions identify what could be called "equilibrium points" of an iterated process. But they will not tell you if the equilibrium point is a stable or unstable one. If its stable, the sequence will converge to it nicely. If it is unstable, and no stable point exists, the sequence will diverge. Hence, a complete solution to any such problem has to include some sort of proof of convergence.

I have to admit that I don't see what is particularly interesting about ee1/2e1/3e1/4...

Of course, Hn = [sum]k=1n 1/k approximates ln(n+1), so you would assume some connection, but looking at the numbers produced shows nothing to me that makes it more interesting than any other value of A.
Title: Re: sqrt(1/2)*sqrt(sqrt(1/2))*sqrt(sqrt(sqrt(1/2))
Post by towr on Oct 22nd, 2003, 11:10pm

on 10/22/03 at 19:46:56, Icarus wrote:
Of course, Hn = [sum]k=1n 1/k approximates ln(n+1), so you would assume some connection, but looking at the numbers produced shows nothing to me that makes it more interesting than any other value of A.
meh.. perhaps you're right.. it wouldn't be the first time ;)
Dividing by eeuler_gamma helps to make it look nicer though.. And if you floor it you just get n back..


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