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#1 2006-03-12 05:23:00

fgarb
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A Powerful Puzzle

I like this one

That's x raised to the power of x raised to the power of x, going on forever equals 2. Solve for x.

Last edited by fgarb (2006-03-12 05:23:41)

#2 2006-03-12 06:36:22

ashwil
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Re: A Powerful Puzzle

This one is hurting my brain!

#3 2006-03-12 07:04:27

fgarb
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Re: A Powerful Puzzle

It still hurts my brain, and I know the answer!  Incidentally, if anyone can solve this, then I have a followup. Don't try them in the reverse order though, if you do it has the potential to be seriously confusing!

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#5 2006-03-12 12:44:08

ashwil
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Re: A Powerful Puzzle

If I am right about the first one, I reckon that solving for 10 would only give a very slightly higher answer, but I don't have the time right now!

#6 2006-03-12 13:15:29

mathsyperson
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Re: A Powerful Puzzle

Well, not entirely, but it is right if you've roundeded it. But the real question is... what is 1.414 more commonly expressed as?

Oh, and the solution for 10 is actually lower. Get your head around that one!

Why did the vector cross the road?
It wanted to be normal.

#7 2006-03-12 15:58:59

ganesh
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Re: A Powerful Puzzle

Let

Therefore,

***someone continue from where I have left...***

Character is who you are when no one is looking.

#8 2006-03-12 16:37:12

fgarb
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Re: A Powerful Puzzle

That is definitely correct, and it should be straightforward from where Ganesh left it. Now, it should be pointed out, as mathsyperson said, that the solution you get for y=10 in this way is lower than for y=2.

But if you think about it, for x and y > 1, x < y implies that x^x^x^... is always less than y^y^y^... , so this implies that 10 < 2, which is nonsense. I'm still thinking about this, but it should mean that there is a cutoff value for y above  which there is no x solution.

So my final puzzle related to this is: find the largest value of y such that

has a solution. Unfortunately, I think you'll need to know calculus to be able to figure this out, but the answer makes me wonder if there's something really deep going on here that I don't understand. I find this really interesting!

Last edited by fgarb (2006-03-12 16:38:16)

#9 2006-03-12 17:07:20

ganesh
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Re: A Powerful Puzzle

I think I can declare that there is no solution for the above equation. The highest value of y for

is y=e or 2.7182818284 approximately and the highest value of x for finite y is x=1.444667861 approximately. I am sure it can be proved that

has no solution.

Character is who you are when no one is looking.

#10 2006-03-12 17:21:19

fgarb
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Re: A Powerful Puzzle

That is what I get as well. If anyone has any idea why e ends up a solution to this problem, I'd love to hear it! That annoying constant seems to have a way of popping up everywhere.

#11 2006-03-12 21:06:05

ashwil
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Re: A Powerful Puzzle

mathsyperson wrote:

Well, not entirely, but it is right if you've roundeded it. But the real question is... what is 1.414 more commonly expressed as?

Mathsy, yes, I did round it and I do I know what it is expressed as - just wanted to leave something in the puzzle for someone else!

#12 2006-03-12 22:47:21

mathsyperson
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Re: A Powerful Puzzle

Ah. Fair enough. Sorry.

Wouldn't the cut-off point be y=e?
It makes sense that the cut-off point is e, because that's when the gradient of x starts becoming negative. But maybe I've missed something.

Edit: I've done some research in Excel and you're right.

The 10th root of 10 is implied to be the solution of x for y=10, but using that value makes y converge to 1.371288574.

Why did the vector cross the road?
It wanted to be normal.

#13 2006-03-30 05:52:55

sabujakash
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Re: A Powerful Puzzle

Well to phrase the problem in some other way....

Consider the sequence {x, x^x, x^x^x,...} for which values of x the series is converging???

Last edited by sabujakash (2006-03-30 05:56:08)

#14 2006-03-30 07:56:38

MathsIsFun

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Re: A Powerful Puzzle

Yes sabujakash, above the magic number 1.44466786... the series diverges.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

rm
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