Cubed once again

Weird fact of the day: Suppose x_1, ..., x_k \geq 0, then:

Lim_{n\rightarrow\infty}(\sum_{j=1}^{k} x_{j}^{n} )^{1/n}=max \lbrace x_1, ..., x_k \rbrace

What does it really mean? It means: if you take some positive numbers x_1, ..., x_k, raise each to the power of n, add them all together, take the n-th root of the sum, then if you start to increase n the result will start to approach the greatest of the numbers you started with.

Why is that weird? Because the procedure is symmetric with respect to all the numbers x_1, ..., x_k, and the result depends only on one of them, the greatest one (Let’s call it x_\ast). All the other k-1 numbers are pretty much completely irrelevant. You can set them all to zero and it will not change the outcome, because all the work is done by x_\ast, and by x_\ast alone. How does the procedure “determine” which number is the greatest?

How do we prove this amazing fact? Like this:

We already defined x_\ast:=max \lbrace x_1, ..., x_k \rbrace. Now:

(x_\ast^{n} )^{1/n} \leq (\sum_{j=1}^{k} x_{j}^{n} )^{1/n} \leq (k \cdot x_\ast^{n} )^{1/n} and

Lim_{n\rightarrow\infty}(x_\ast^{n})^{1/n}=Lim_{n\rightarrow\infty}(k\cdot x_\ast^{n} )^{1/n}=x_\ast, therefore Lim_{n\rightarrow\infty}(\sum_{j=1}^{k} x_{j}^{n} )^{1/n}=x_\ast, the end.

Does it work with other families of functions \varphi_n(x) or only with powers and roots? The proof only uses some monotonicity and the fact that Lim_{n\rightarrow\infty} \varphi_n^{-1}(k\cdot \varphi_n(x))=x, so it should also work, for example, when \varphi_n(x)=n^x, meaning that:

Lim_{n\rightarrow\infty}(log_n\sum_{j=1}^{k} n^x_{j} )=max \lbrace x_1, ..., x_k \rbrace

I can’t think of a good way to completely characterize the class of functions, for which the proof holds. Can you?

What does this mean geometrically? It means that as n increases, level surfaces of a function f_n(x_1, ..., x_n)=(\sum_{j=1}^{k} |x|_{j}^{n} )^{1/n} defined in a k-dimensional space, will  start to resemble a k-dimensional cube, because a cube is a level surface of a function f_\infty(x_1, ..., x_n)=max \lbrace |x_1|, ..., |x_k| \rbrace. You might remember that we already spoke about this in my previous post titled “On Norms”.

Have you made a cool-looking but completely pointless animation to demonstrate this phenomenon? I thought you’d never ask:

(you may or may not need to click on it)

Does this fact have any useful applications? Not that I know of, which proves that this is some high quality math, and not same applied rubbish. (Just kidding, applied math can be beautiful too. Once in a thousand years.)

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6 Responses to Cubed once again

  1. christopherolah says:

    > I can’t think of a good way to completely characterize the class of functions, for which the proof holds. Can you?

    Besides something like “locally monotone”, we want (\forall \epsilon > 0)(\exists n)(\phi_n^{-1}(n\phi_n(x)) -x < \epsilon).

    Obviously, a \phi that is, in some sense yet to be determined, changing fast would result in a smaller change in \phi^{-1}(n\phi(x)) (think IFT).

    But using the normal idea of differentiation entangles us in discussions about the size of \phi(x) and such. It would be much nicer to use the geometric derivative.

    In fact, it is fairly clear that we want \lim _{n\to\infty} D^*\phi_n(x) = \infty… something needs to be said about the rate of convergence, but I'm too tired to work out.

    Or my jet lag-addled mind has gone crazy. There's some evidence for that in the "it won't let me sleep despite being tired" thing…

    • George G. says:

      Let me formalize what I think you’re thinking. Let us assume that \varphi_n is monotonously growing, and so is \varphi_n'. We want to prove that:
      \varphi^{-1}_n(k \varphi_n(x))\rightarrow x, when n \rightarrow \infty.
      Let us rewrite that as:
      \varphi^{-1}_n(k \varphi_n(x))\rightarrow \varphi^{-1}_n(\varphi_n(x))
      which is equivalent to:
      \varphi^{-1}_n(k \varphi_n(x)) -\varphi^{-1}_n(\varphi_n(x))\rightarrow 0
      We assumed that \varphi_n' was growing monotonously, which gives us the bound:
      \varphi^{-1}_n(k \varphi_n(x)) -\varphi^{-1}_n(\varphi_n(x)) \leq (k-1)\varphi_n(x)\cdot \frac{d\varphi^{-1}_n(x)}{dx}\vert_{x=\varphi_n(x)}
      but:
      (k-1)\varphi_n(x)\cdot \frac{d\varphi^{-1}_n(x)}{dx}\vert_{x=\varphi_n(x)} = \frac{\varphi_n(x)}{\varphi_n'(x)}(k-1)
      This implies that \varphi^{-1}_n(k \varphi_n(x))\rightarrow x, if \forall x\geq 0: \frac{\varphi_n(x)}{\varphi_n'(x)} \rightarrow 0, so we have a sufficient condition.

      But \forall x\geq 0: \frac{\varphi_n(x)}{\varphi_n'(x)} \rightarrow 0 is equivalent to \forall x\geq 0: \varphi_n^\ast(x)\rightarrow \infty, because \varphi_n^\ast(x)=exp(\frac{\varphi_n'(x)}{\varphi_n(x)}), where \varphi_n^\ast is geometric derivative.

      Oh, and I think you made a typo here: (\forall \epsilon > 0)(\exists n)(\phi_n^{-1}(n\phi_n(x)) -x  0)(\exists n)(\phi_n^{-1}(k\phi_n(x)) -x < \epsilon), where k is some constant, otherwise you get a completely different problem.

      • christopherolah says:

        Yeah, the n was a mistake. I blame us using the same variable for the number of dimensions and the term in the sequence. 😛 I thought the issues I had with making the rate exceed n was odd.

        In any case, your proof is fairly straight-forward, but I can’t help but feel like it defeats the point of using the geometric derivative…

        Again, we have \varphi^{-1}_n(k \varphi_n(x))\rightarrow \varphi^{-1}_n(\varphi_n(x)) = x. Consider that, by local approximation with geometric derivatives, \varphi^{-1}_n(k \varphi_n(x)) is approximately x+\log_{f*(x)}(k). Let the geometric derivative be monotone and this is an upper bound (if I wasn’t lazy, I’d use some sort of weaker condition — I really just need to rule out the case of a negative second geometric derivative that keeps increasing and pushing it farther and farther back) .

        So we have \varphi^{-1}_n(k \varphi_n(x))\rightarrow x

        Now, as f^*(x) \to \infty, \log_{f^*(x)}(k) \to 0 so one ends up with with
        \varphi^{-1}_n(k \varphi_n(x))\rightarrow x, the desired result.

      • George G. says:

        I took the liberty of tweaking and merging two of your comments for the sake of neatness. I hope you don’t mind.

        “In any case, your proof is fairly straight-forward, but I can’t help but feel like it defeats the point of using the geometric derivative…”

        Oh, it absolutely does. I am a geometric derivatives skeptic. I think that everything that can be proven using them can be proven without them without complicating the proof much. But I may be mistaken about it.

        Anyway. It seems that you are trying to construct the same proof I came up with, based on your remarks, but using geometric derivatives. Alright. But it seems that your proof requires f^\ast
        to be growing monotonously, and I only wanted f' to be growing monotonously. Your condition is stronger (consider f(x)=x^2, f'(x)=2x, f^\ast (x)=exp(2/x)) .
        Also, I don’t quite understand why \varphi^{-1}_n(k \varphi_n(x)) is approximately x+\log_{f*(x)}(k). It may or may not be correct: I don’t claim to understand geometric calculus perfectly, but it just doesn’t look right to me. Please, clarify. f is \varphi_n, I assume.

  2. christopherolah says:

    Thanks for merging my comments!

    >I am a geometric derivatives skeptic.

    I know. I got quite a bit of amusement out of resolving this using them, just because of that.

    > I think that everything that can be proven using them can be proven without them without complicating the proof much.

    It’s a matter of elegance. Perhaps it will be more clear in my solution to

    > Also, I don’t quite understand why \varphi^{-1}_n(k \varphi_n(x)) is approximately x+\log_{f*(x)}(k).

    Actually, I’m going to step back in the proof a bit because I can make things a bit neater.

    To prove \varphi^{-1}_n(k \varphi_n(x))\rightarrow \varphi^{-1}_n(\varphi_n(x)) = x, let’s solve for \epsilon in \varphi^{-1}_n(k \varphi_n(x)) = x + \epsilon.

    Well, consider k\varphi_n(x) = \varphi_n(x + \epsilon). We expand to our first term in the geometric approximation: k\varphi_n(x) = \varphi_n(x)*\varphi^*_n(x)^\epsilon. Some simple algebraic manipulation results in \epsilon = \log_{\varphi^*_n(x)}(k).

    All this is a formalization of the fact that a high geometric derivative of a function f means that the difference between f(x) and kf(x) is in some sense small. So if it becomes bigger and bigger, they go to the same point (in terms of the value of x that would be needed to achieve the same result…).

    The reason I like this a lot more than your approach is that it is conceptually smaller. It’s a result of the basic idea of the geometric derivative,

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