Twoering inferno
What is the limit value of:
sqrt(2)^(sqrt(2)^(sqrt(2)^(sqrt(2)^(... ?
That's an infinite power tower of sqrt(2)s.
sqrt(2)^(sqrt(2)^(sqrt(2)^(sqrt(2)^(... ?
That's an infinite power tower of sqrt(2)s.
Labels: mathschallenge
posted by Chris at
6:21 AM
16 Comments:
2
im not sure abt my answer but here is how i got it
let x = the whole thing
so x = (rt 2)^x
this is oly satisfied by x=2.
i remember the 2 because in another problem i had solved i had got x^x^2 = 2
so find x, the answer here was root 2. But i twisted it to get the answer for this one
A, you did it in what I consider to be the very best way possible. You made the observation directly. And it was first post. Well done.
I was half expecting numerical method approaches, to be followed by the dawn of realisation.
I might milk this power tower stuff a little more yet ;)
There are problems with publishing comments. Now it's just a check.
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OK, my solution.
Define a(n) is sqrt 2 in the power of n other powers of sqrt 2. For example a(1)=(2^0.5)^(2^0.5). Thus
a(n+1)=2^(a(n)/2). Note, that if
a(n)<2, then a(n+1)<2. But a(1)<2,
therefore a(n)<2 for each n. But a(n)<2, hence it's 2.
I've noticing that to. So I've also kicked some through by publishing another, then deleting it. Leave yours up so this one makes sense, it might help others.
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quantense, I don't understand your recursion. I'd expect a recursion:
(use rt2 for convenience, on this blog)
a(n+1) = rt2^a(n).
I'll give you a chance to explain the "2" limit before ripping into that ;)
I posted the problem, because at first you might expect that the series diverges. I stumbled across it when researching the xth root of x. I'm not sure if there is another "nice" power tower, or not.
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I surrender! Mail me if you want to know my solution.
LOL. That's Ok, I won't give you a bad time.
I just got myself a copy of Mathematica (being dying for that for years, never had a good excuse for it before). My excuse (which I announce occassionally) is that I'm stuck at home with a broken leg. I used Mathematica to plot graphs so I could numerically find the solution, that's how I knew for sure it was 2. But I also knew that from researching the xth roor of x - but I wasn't convinced. Still don't want to let the cat right out of the bag (re inspiration) as I might come up with a couple of related problems to post. Greetz.
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As a minor "curiosity" I notice the following:
As before let x = rt2^x then x^(1/x) = rt2. Both equations are satisfied by x = 2 or 4. Go figure:)
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