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Math Help - infinite power tower

  1. #16
    Super Member Random Variable's Avatar
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    Quote Originally Posted by pomp View Post
    Sorry, but the function is not defined on this range, so I don't see how it can be continuous. By defining it in the way that you have it is equivalent to the expression involving the Lambert W function that I gave in the PDF, the point of putting it in this form was to show that the function is only defined for certain values of x.

    You would need to define the function differently for it to make sense over the range x \in (e^{1/e},\infty)
    Yep. I'm an idiot.
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  2. #17
    Grand Panjandrum
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    Quote Originally Posted by Random Variable View Post
    Let  f(x) = x^{x^{x^{x^{...}}}}

    What is the maximum value of x such that f(x) is finite?
    You have been reading the current edition of Mathematical Gazette and I claim my 64,000.

    CB
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  3. #18
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    Quote Originally Posted by chisigma View Post
    a)  x>1

    In this case for all  n>0 is  f_{n} (x) >1 so that is...

    f(x)= \lim_{n \rightarrow \infty} f_{n} (x) = +\infty
    ...
    Anywhere also in this case the answer to the question proposed by R.V. is  x=1 ...

    Kind regards

    \chi \sigma
    Although this seems the intuitive answer, it is actually incorrect. You're in good company though, Eisenstein himself actually quite naively stated the same when working on the function. If you have the time I suggest having a quick skim through the PDF near the top of this thread.

    EDIT: As a quick test to reassure yourself, iterate f_n(x) for x=\sqrt{2}. Clearly x>1 but you should note some limiting behaviour..

    Spoiler:
    \lim_{n \to \infty} f_n(\sqrt{2}) = 2
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  4. #19
    Newbie Deco's Avatar
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    Logic?

    Without even looking at any numbers, I can logically see that x=1 must be true. Assuming any situation, a^b, for any b<1 results in decay. Unless of course a<1 aswell, but that would work contrary to obtaining a maximum. For b>1, a<1, a^b \rightarrow 0. For both variables being positive also results in a^b \rightarrow \infty.

    Therefore the only answer that would allow for a base being raised an infinite amount of times to the same value would be when x=1

    I know this isn't rigorous, but god-damnit it's right.
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  5. #20
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    Quote Originally Posted by Deco View Post
    I know this isn't rigorous, but god-damnit it's right.
    No it isn't, refer to other posts in this thread.
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  6. #21
    Newbie Deco's Avatar
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    Clearly x>1 but you should note some limiting behaviour..
    Are there no bounds to this? This would imply 10^{10^{10^{10^{...}}}} doesn't approach \infty? Am I the only only having a hard time to grasp this. And since x>1, doesn't that mean there is no maximum value of x?
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  7. #22
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    Quote Originally Posted by Deco View Post
    Are there no bounds to this? This would imply 10^{10^{10^{10^{...}}}} doesn't approach \infty? Am I the only only having a hard time to grasp this. And since x>1, doesn't that mean there is no maximum value of x?
    There is an upper bound as well as a lower bound. It was stated in another post and in the PDF near the begining of this thread that the domain of convergence is [e^{-e},e^{1/e}]
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  8. #23
    Super Member Random Variable's Avatar
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    I still can't believe I stated that the function is continous where it's not defined. Thanks for bumping this thread and reminding me.
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  9. #24
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    Quote Originally Posted by Random Variable View Post
    I still can't believe I stated that the function is continous where it's not defined. Thanks for bumping this thread and reminding me.
    haha sorry! On the bright side, at least you managed to get your head around the counter intuitive part! Seems to be throwing some people off.
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