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## Is x^x injective?

lets take the derivative. we get x^x= e^[xln(x)] so the deriv is [x^x][ln(x) +1].

now x^x is always positive for positive x, and ln(x) + 1 is positive whenever ln(x) > -1, i.e. for x > 1/e.

thusm x^x is strictly monotone increasing for x in the interval [ 1/e, infinity).

moreover, on (0,1/e) for the same reason it is strictly decreasing.

so it is not injective on R+, but is injective on each of the two intervals where it is monotone.

it is of course not even defined on many negative numbers like -1/2, since the square root of -1/2 is undefined.

 Quote by Mystic998 Just something bugging me...f(0) doesn't equal f(1). f isn't defined at 0.
For a function to be continuous at a point we need it not to be defined, but only that the limit of the function as x goes to a is a real number. This because continuity is assured if we define the value of the function at that point as the limit in question - Data's argument holds.

 Quote by Izzhov While we're on the subject of this function, I have another question: exactly what kind of function is $$x^x$$, anyway? It's transcendental, but would it be considered exponential? When I looked up exponential function on Wikipedia, it said an exponential function is one that raises a constant to a variable power, which is not the case with $$x^x$$. So would it just be classified as a miscellaneous transcendental function?
i think this function is exponentialo-power, i do not know if my naming is right, but it is how to say it an exponential and a power function meanwhile.

it is like this the general form:

[u(x)]^v(x)

 Quote by AlphaNumeric It's injective on the positive integers, you don't need the function to be continuous to define injectivity/surjectivity, only consider the images/preimages of the mapping (it's a well defined notion on discrete sets). $$n^{n}$$ is clearly increasing as integer n does, $$n^{n} < (n+1)^{n}$$ for n>0. Therefore $$(n^{n})^{-1} = n^{-n}$$ is strictly decreasing as n increases, so no equalities for $$m\not= n$$. Then considering $$(-n)^{-n} = (-1)^{-n}n^{-n}$$ so you just get an oscillating pattern. Since $$|(-n)^{-n}| \not= |(-m)^{-m}|$$ by the strictly decreasing nature of $$n^{-n}$$, then the alternating sign of the sequence doesn't give equality either. Therefore $$(-n)^{-n} \not= (-m)^{-m} \quad \forall n \not= m$$.
 On the negative integers $x^x$ is certainly injective, since for n,m natural $$\frac{(-1)^n}{n^n} = \frac{(-1)^m}{m^m}$$ iff m=n. (in fact injectiveness on the positive integers implies injectiveness on the negative integers in this case, since if $f(x) = x^x$ then for integers n, $f(-n) = (-1)^n/f(n)$)

 Quote by Izzhov I was asking about negative integers, not positive ones.
That's easy.
Extend your considerations to complex function $f(z)=z^z;z\in\mathbb{C}$ and have a party.

 Quote by Izzhov I was asking about negative integers, not positive ones.
I know, hence why a bunch of minus signs are in my post. I was showing that because the function is injective for the positive integers by obviousness, you can see that it's also injective for the negative integers by putting in minus signs to make it about the negative integers.

 Quote by Data Clearly $1^1 = 1$. You can easily see that $x^x$ is differentiable for x>0 and that $$\lim_{x \rightarrow 0^+} x^x = 1.$$ So, the MVT (or, more appropriately, Rolle's theorem, to see that there's a max or min) tells you that it's impossible for $x^x$ to be injective on the positive reals (and thus on the reals).
I am sorry, but I can not prove lim(x^x) =1 when x approaches +0. Can you help?

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 Quote by Werg22 For a function to be continuous at a point we need it not to be defined, but only that the limit of the function as x goes to a is a real number. This because continuity is assured if we define the value of the function at that point as the limit in question - Data's argument holds.
In order that a function be continuous at a point we do need it to be defined: that's part of the definition of "continuous". What you mean to say is that if the limit at that point exists, then it has a removable discontinuity- and we can make it into a continuous function by defining it to be equal to the limit at that point. However, when you have done that you have a new function that is continuous at the point.

 Quote by pixel01 I am sorry, but I can not prove lim(x^x) =1 when x approaches +0. Can you help?
Well, $x^x = e^{x\ln x}$, so it suffices to look at

$$\lim_{x \rightarrow 0^+}{x \ln x} = \lim_{x \rightarrow 0^+}\frac{\ln x}{\frac{1}{x}};$$

Apply l'Hopital.

 Quote by Data Well, $x^x = e^{x\ln x}$, so it suffices to look at $$\lim_{x \rightarrow 0^+}{x \ln x} = \lim_{x \rightarrow 0^+}\frac{\ln x}{\frac{1}{x}};$$ Apply l'Hopital.
Thanks Data. It's so simple but great.

 Quote by Izzhov While we're on the subject of this function, I have another question: exactly what kind of function is $$x^x$$, anyway? It's transcendental, but would it be considered exponential? When I looked up exponential function on Wikipedia, it said an exponential function is one that raises a constant to a variable power, which is not the case with $$x^x$$. So would it just be classified as a miscellaneous transcendental function?
No,xx is not really exponential function.
ex is exponential and transcendental function,but is classified as elementar function.
OTOH,xx is trancendental,but nonelementar one.
"Miscellanous transcendental" function ?What does that mean?
Anyway,I think it's a beautiful function with many interesting features..

Important property is (that nobody mentioned) it is a convex function over the whole interval of its' definition ( means that it's graph curve may have only 1 or 0 extremes).Of course,in this case it's one point of minimum at x0=1/e.
But consider it's inverted syster :function (x-1)x-1 .It's also convex ,and defined over the same interval.But ,it has no stationary points.Prove that.
There are more interesting things like inequality $x^x\geq x$ holds for x>0 .Equality occurs only for x=1.
Therefore the graph curve of the function f(x)=xx ,in Kartesian plane,is located "above" line y=x.
f'(1)=1 means that in point (1,1) tangent slope on the graph is exactly 45°.That's also interesting ,isn't it?
 Recognitions: Homework Help Science Advisor an elementary function is usually defined to be one obtained from the field of rational functions by adjoining algebraic elements, and also exponentials and logarithms of functions already obtained. In this sense, x and log(x) are both elementary, and hence so is the exponential of their product, i.e. in that sense, e^(xln(x)) = x^x is an elementary function. see e.g. the article in the math monthly from 1972, on integration in finite terms, by maxwell rosenlicht.
 mathwonk,this is strange to me ,but if that's really definition you're right (thank you!). composition of standard elementary functions x,e^x,and ln(x) creates xx.Indeed,I thought elementary functions were those obtainable from standard elementary functions only by means of finite number of arithmetical operations (+,-,*,:) among them.

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 Quote by tehno mathwonk,this is strange to me ,but if that's really definition you're right (thank you!). composition of standard elementary functions x,e^x,and ln(x) creates xx.Indeed,I thought elementary functions were those obtainable from standard elementary functions only by means of finite number of arithmetical operations (+,-,*,:) among them.

throw in composition and you are there.

i.e. ln(ln(x)) is also elementary.

I guess the definition was made this way to try to encompass functions we usually try to antidifferentiate, like x^2 e^(x^3).

by the way in omitting trig functions, i was tacitly assuming the functions are complex valued, so that trig functions and their inverses are a special case of exponentials and logs.

e.g. arctan is the same as log except it twirls around i and -i instead of 0 and infinity. so if you compose with an automorphism like (1-i)/(1+i) they become almost the same.

to see this, just think of log as path integral of 1/z along paths that avoid 0 and infinity, and arctan as path integral of 1/(1+z^2) along paths that avoid i and -i.
 You pro-mathematicians certainly know *all rules * of your game.And We computer scientists must obey it without objections . Maybe,I'll be hang for telling you this but most of us consider only polynomials over Q-field truly elementary functions

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