Is x^x injective?

mathwonk

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.

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.

sutupidmath

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)

sorry data allready explained this, i did not read through all threads.

Izzhov

I was asking about negative integers, not positive ones.

Data

On the negative integers [itex]x^x[/itex] is certainly injective, since for n,m natural

[tex]\frac{(-1)^n}{n^n} = \frac{(-1)^m}{m^m}[/tex]


iff m=n.

(in fact injectiveness on the positive integers implies injectiveness on the negative integers in this case, since if [itex]f(x) = x^x[/itex] then for integers n, [itex]f(-n) = (-1)^n/f(n)[/itex])

tehno

That's easy.
Extend your considerations to complex function [itex]f(z)=z^z;z\in\mathbb{C}[/itex] and have a party.

AlphaNumeric

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.

Please actually read my post.

pixel01

I am sorry, but I can not prove lim(x^x) =1 when x approaches +0. Can you help?

HallsofIvy

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.

Data

Well, [itex]x^x = e^{x\ln x}[/itex], so it suffices to look at

[tex]\lim_{x \rightarrow 0^+}{x \ln x} = \lim_{x \rightarrow 0^+}\frac{\ln x}{\frac{1}{x}};[/tex]

Apply l'Hopital.

pixel01

Thanks Data. It's so simple but great.

tehno

No,x^x is not really exponential function.
e^x is exponential and transcendental function,but is classified as elementar function.
OTOH,x^x 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 [itex]x^x\geq x[/itex] holds for x>0 .Equality occurs only for x=1.
Therefore the graph curve of the function f(x)=x^x ,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?

mathwonk

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.

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 x^x.Indeed,I thought elementary functions were those obtainable from standard elementary functions only by means of finite number of arithmetical operations (+,-,*,:) among them.

mathwonk

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.

tehno

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