A Nice Substitution « Foxmaths! 2.0

July 21, 2009

A Nice Substitution

Filed under: Maths — Fox @ 9:54 pm

I wasn’t always a math person. I used to be much more into reading books. I would read just about anything. Fast, too. But then … I stopped. I remember, the first book I bought but distinctly didn’t read was P.D. James’ Death of an Expert Witness. I picked it up again yesterday, years after the fact, and so far … it’s not bad. Maybe I will get into this ‘reading’ thing again.

But, a small calculation is in order, one that has been bothering me.

Consider the function,

$f(x) = x^{x^{x^{x^{...}}}}$

An infinite power tower of x’s. I’ve done some things with this function before, but not this, I believe.

It is certainly questionable, what such a power tower actually means. For the purpose of this calculation, we imagine taking a number x, then raising x to that power, then raising x to the resulting power, then raising x again, etc. This creates a sequence of finite power towers, and we define f(x) to be the limit of that sequence, as the height of the tower goes to infinity.

For some values of x, clearly this sequence diverges to infinity. 2 to the 2nd, etc, etc, etc, rapidly becomes unmanagable. 2^2^2^2^2 has 19729 digits. f(x) for values such as these can be taken as infinite.

But! Over some x, f(x) is actually well defined, finite, seemingly smooth, and many other nice things as well. Consider f(1), which equals 1. f(0) is another matter all together…

That being so, what is the maximal x for which f(x) is finite, and what is f(x) at that point?

My thinking went something like this: I have no idea. But it might be kind of cool to calculate the derivative of f(x).

Looking at the definition of f(x), there’s one substitution that’s just itching to be made.

$f(x) = x^{f(x)}$

If we have an infinite power tower of x’s, and drop one off (the bottom most), we still have an infinite power tower of x’s, and the substitution can be made. The rest is standard implicit differentiation.

$ln( f(x) ) = f(x)*ln(x)$

$f'(x) = \frac{ f(x)^2/x }{1 - f(x)*ln(x)}$

While the above isn’t really a nice looking function, it’s certainly much more reasonable that having infinitely tall towers flopping about.

And it does suggest something interesting. Notice, being a rational function, if the denominator goes to zero, f’(x) is going to explode to infinity. If that happens, f(x) is going to, effectively, explode as well.

At the very least, if the denominator goes to zero … it would be bad. This all suggests it might have some bearing on our problem. Letting the denominator go to zero, we can solve for f(x) at that point as

$f(x) = 1/ln(x)$

This, along with our initial definition of f(x), gives us something to work with.

$ln( f(x) ) = f(x)*ln(x)$

$ln( 1/ln(x) ) = ln(x)/ln(x)$

$ln( ln(x) ) = -1$

$ln(x) = 1/e$

$x = e^{1/e}$

So, at the above x value, the derivative blows up, and the function goes to infinity. Interestingly, our condition on the denominator of f’(x) also gives us an easy way of calculating f(x) at that point.

$f(x) = 1/ln(x)$

$f(e^{1/e}) = 1/(1/e)$

$f(e^{1/e}) = e$

And, fiddling with Mathematica seems to confirm all this.

But … what’s bothering me is that I feel like I did too much work. The first time I saw this problem, this quiet little Asian kid whipped out the e^(1/e) like it was a trivial and self evident result. Was he seeing something that I’m not seeing? I don’t know…

I think it’s worth thinking about, in any case. Not too long, perhaps, but worth thinking about. And now, back to murder and mayhem. After all, isn’t that what summer is all about?

Comments (6)

6 Comments »

I believe I just lefted a comment few minutes ago, but they are not showing up, maybe its caught in the spam filter
…Or I didn’t actually pressed submit before I closed the tab. xD
in the previous comment, I said I know that x^(1/x) is reaches maximum at e. That’s just a those fact I learned when I was doing this problem.
“partition 100 into positive integers, so the product of those integers are maximum”. I’m sure this is a popular problem because I see it everywhere…
With that equipped, this problem becomes obvious.

still, I don’t see how x^(1/x) could fit in, so I looked over your post again and noticed that:
f(x) = x^f(x) => x = f(x)^(1/f(x))
, which I overlooked many times because I was trying to figure out maximum for f(x) instead x and trying to keep f(x) by itself in the lhs.
so now…
x is maximum when f(x) = e
x = e^(1/e)

Comment by Mgccl — July 22, 2009 @ 1:17 am

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The sun is hiding! It looks like a well-manicured fingernail. Do you think I’ll get any cool superpowers from the eclipse?

Comment by Kitten — July 22, 2009 @ 1:19 am

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Eclipse! That is too cool. Lucky you. If you do get powers, don’t get annoying like Heroes…

Comment by Fox — July 22, 2009 @ 1:53 am

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Maximizing for x … Very clever! I will have to remember that.

Comment by Fox — July 22, 2009 @ 1:55 am

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Oh and wonderful maths you got there. I remember some kid in my calculus class wanted to solve for the derivative of that function (this was a very long time ago) and I had no clue how to even begin. Your solution is simply fantastique

Comment by Kitten — July 22, 2009 @ 4:55 am

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you could use Lambert’s W function
http://en.wikipedia.org/wiki/Lambert%27s_W_Function
(sorry for the poor formatting
y=x^y
yx^(-y)=1
ye^(-y ln x)=1
(-y lnx) e^(-y lnx)=-lnx
-y ln x= W(-lnx)
y=W(-lnx)/(-lnx)

W(x) is only defined for x≥-1/e, so W(-lnx) is only defined for x≤e^(1/e)

Comment by anonymous — July 24, 2009 @ 2:34 pm

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