No. 118 (essentialsaltes) wrote,
No. 118
essentialsaltes

Freaky math thing du jour

                  ...
               (√2)
            (√2)
Consider (√2)


That is, √2 raised to the power of √2 raised to the power of √2 ad infinitum.

At first blush (and maybe second and third blush) it looks freaking infinite. I mean if you go 2^2^2^2... step by step you get 2, 4, 8, 16, 32 ... and it blows up pretty quick.

Ok, root-2 isn't as big as 2, but it's still bigger than 1. And if you raise it to a power bigger than 1 then it's gonna get bigger. And it'll continue to get bigger and bigger each time, over and over again. Bingo, infinite.

But let's just play with it. Since the exponentiation goes on for ever, then if we call that thingy x, then we can say that

x = (√2)^x

O... kay, that equation looks pretty hopeless too.

[a subtle but extremely important point, and a key to understanding the problem correctly (or at least, not wrongly): you can't just break off the 'last' one and say that x = x^(√2), because there is no last one if our beastie goes to infinity.]

Anyway, if we take the log-base-2 of both sides, we get:

log2(x) = x log2 (√2) = x * (1/2)

2 log2(x) = x

That's pretty shitty, too. Let's exponentiate it

x^2 = 2^x

Hey, that's pretty, at least. I still can't solve that fucker without cheating. Watch me closely as I misdirect you. Okay, the right side has to be a power of 2, so on the left side, x must be a power of 2.

If x is 1, then we have 1 = 2. Nope
If x is 2, then we have 4 = 4. Yep
If x is 4, then we have 16 = 16. Yep
If x is 8, then we have 64 = 1,024. Nope
If x is 16, then we have 256 = 65,536. Nope

It is trivial to show (* i.e. I am too lazy to verify, but I have a hunch that) that we're not going to get any more solutions, since the exponential is blowing up way too fast.

Maybe we should have just been satisfied with our original idea that there were zero finite solutions, because now we have two finite answers. That's pretty fucked up. It's probably infinite, but maybe it's 2. Unless it happens to be 4. It certainly can't be both 2 and 4, or the universe would immediately collapse.

So now what?

When analytical tools fail... draw pretty pictures.
If we just go ahead and brute force iterate the successive exponentiations, starting with √2 and then plugging in the new value back in as our new x, we find that indeed, the answer keeps getting bigger. But contrary to my Bingo, infinite conclusion above, we see that there's some diminishing returns, and the value approaches 2 as the limit.

The answer of 4 (although a solution of the derived equation) is basically irrelevant to the original problem, since starting from √2, you get to 2. However, if you examined that same graph near 4, the lines cross the other way round, so that 4 actually drives the pretty pictures away, unlike 2, which acts as an 'attractor'.

So the answer is 2. Weird.
Tags: math
Subscribe
  • Post a new comment

    Error

    Anonymous comments are disabled in this journal

    default userpic

    Your reply will be screened

    Your IP address will be recorded 

  • 11 comments