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Is x^x injective?
#1
I have recently taken great interest in studying the properties of the function f(x) = x^x , and I was wondering: is there any way to prove whether f(x) = x^x is an injective (i.e. one-to-one) function? I realize that if I can prove that if the inverse of f(x) = x^x is also a function, then f(x) = x^x is injective. The problem is: f^{-1}(x) is non-algebraic, so I can't figure out whether it's a function or not. Does anyone know another way to prove whether or not f(x) = x^x is injective?

NOTE: The domain of this function is real numbers.
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#2
(11/16/2009, 04:53 AM)Amherstclane Wrote: I have recently taken great interest in studying the properties of the function f(x) = x^x , and I was wondering: is there any way to prove whether f(x) = x^x is an injective (i.e. one-to-one) function? I realize that if I can prove that if the inverse of f(x) = x^x is also a function, then f(x) = x^x is injective. The problem is: f^{-1}(x) is non-algebraic, so I can't figure out whether it's a function or not. Does anyone know another way to prove whether or not f(x) = x^x is injective?

NOTE: The domain of this function is real numbers.

Did you ever take a look at the graph of ?
First we have do termine the range of definition.
Surely should be greater 0, because non-integer powers are not defined for , e.g. is not defined or is not defined. At 0 we have sometimes the difficulty to compute which is not defined in Analysis, so we also exclude this number. So our range of definition is the interval .

Now you take some math software and plot , say on (because we can not show an infinitely long x-Axis):
   

There you see that the function is not injective on (0,1).
It is an interesting exercise to determine the minimum of the function. By solving . This is also the way you can prove that the function is injective on , i.e. by showing that the derivative is always greater 0.

PS: Can you please adapt your math skill level in your profile.
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#3
(11/16/2009, 04:53 AM)Amherstclane Wrote: NOTE: The domain of this function is real numbers.

If the domain is real numbers, then its codomain would have to be complex, because for negative reals, x^x is not real (except for integers).
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#4
It is not injective. You can prove this by counterexample: f(1/4)=f(1/2).
Not only that, but I would go so far as to say that all values between 1/e^1/e and 1 are not injective, even though I haven't really tried to prove it by seeing if the function x^x is made of "monotonic pieces" or not.
I am a spam bot
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#5
Dear forum-members,

I just want to point your attention to the fact that amherstclane and nettson are actually spam bots. This seems unbelievable as they sound quite reasonable.
But as you can see their posts are just copied from this forum:
http://www.physicsforums.com/archive/ind...64901.html

I will leave nettson some time in the system, just changing his signature to reveal that it is a spam bot.
So you can follow his actions and how far AI went already.
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#6
(12/22/2009, 03:31 PM)bo198214 Wrote: So you can follow his actions and how far AI went already.

Henryk - don't now whether this discussion is useful or a contact with the administrators of that forum could be informative. I found some comments about spam-bots in another forum -maybe you already know this (they are of april 2009).

Gottfried

a couple of related msgs in http://www.tricky.org
Gottfried Helms, Kassel
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