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Topic: x^x=5 (Read 5569 times) |
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Obob
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Re: x^x=5
« Reply #25 on: Jan 27th, 2008, 10:44pm » |
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I respect your opinion Icarus, but you weren't around much when srn347 turned this forum into a wasteland of complete nonsense. I agree I may have been a bit hasty in my denouncing "temporary"'s posts, but the amount of abuse that I and other regulars on the forum endured from srn has made me very impatient to what is no doubt his reincarnation.
Sure there is the Lambert W function, but that is nearly a cop out as far as this "riddle" is concerned: its practically like saying, if you want to solve the equation f(x) = c, then the correct answer is "the inverse function of f, evaluated at c." The actual content to the solution is debatable at best.
And my initial reaction in the second post of the thread was to his hint, which is completely false and meaningless. In fact, so long as n is at least 4+epsilon there isn't even a chance that it could be true, since the tower of exponents on sqrt(n) will be bigger than 2, rendering the sqrt(n)^(tower of exponents) bigger than n.
I do admit that this thread has for the most part turned into a decent discussion, but putnam is most emphatically not the place for riddles based on mathematical misunderstandings to be discussed.
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Obob
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Re: x^x=5
« Reply #26 on: Jan 27th, 2008, 11:30pm » |
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Also, just to clarify, my statement to "take your nonsense elsewhere" was in particular directed at his insistence that xx=0 has a solution, exactly the kind of thing that srn used to wildly profess, despite any arguments to the contrary.
Srn had this way of receiving replies to his threads or opinions and dismiss them, even when all he did was spout nonsense. I am convinced he didn't learn a single thing from all his time here; previously I thought maybe he learned that people don't have to put up with him if they didn't want to, and that they have ways of dealing with him without disrupting the entire community. It appears that he doesn't even respect that.
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temporary
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Re: x^x=5
« Reply #27 on: Mar 15th, 2008, 10:02pm » |
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on Jan 27th, 2008, 8:53pm, Icarus wrote:There is NO solution in  . 0^0 = 1, and if x  0, then xy 0 either, for any y, and any acceptable definition of exponentiation.
I'm guessing that you are thinking something like (- )- . But infinities don't work that way. (-)- is an undefined form, and for very good reason. |
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(-)- is correct, which I should have noticed you saying sooner. It is not undefined, although which zero it will become(negative zero, positive zero, regular zero) is undefined, it is also irrelevant.
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My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
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Obob
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Re: x^x=5
« Reply #28 on: Mar 16th, 2008, 5:59am » |
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on Mar 15th, 2008, 10:02pm, temporary wrote:
(-)- is correct, which I should have noticed you saying sooner. It is not undefined, although which zero it will become(negative zero, positive zero, regular zero) is undefined, it is also irrelevant. |
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Wow.
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Icarus
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Re: x^x=5
« Reply #29 on: Mar 16th, 2008, 11:50am » |
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Alright, Obob. You've got me on this one! 
temporary:
First of all, +0 = -0 = 0. They are the same number.
Second, it is easy to get non-zero numbers out of (-)- forms, so to say that it is 0 in any way is wrong. There is no accepted convention that defines it to be 0.
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| « Last Edit: Mar 16th, 2008, 7:02pm by Icarus » |
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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temporary
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Re: x^x=5
« Reply #30 on: Mar 25th, 2008, 6:24pm » |
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(-)- can be rewritten as 1/(-infinity)^infinity. Since (-infinity)^infinity is +-infinity, it's reciprocal must be zero.
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My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
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