# Is x^x injective?

1. Apr 9, 2007

### Izzhov

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.

Last edited: Apr 9, 2007
2. Apr 9, 2007

### Werg22

I suppose you are focusing on the interval x > 0, or else the function is distincontinuous. This said, the interval that poses a difficulty is 0<x<1, as it is very simple to show that if x is greater or equal to one, x^x is monotincally increasing. A quick computation with Excel suggests that the function has a minima on the interval on the interval [0, 1]. This value would obviously be 1/e since it's the only solution to the equation

$${x^{x}}^{'} = {e^{x ln x}}^{'} = x^{x} (1 + ln x) = 0$$

Now we have simplified the question; is x monotonically decreasing on the interval [0, 1/e] and is it montonically increasing on the interval [1/e, 1]? I haven't got much time now so I won't be able to investigate this as I would like, but you can work in that direction.

Last edited: Apr 9, 2007
3. Apr 9, 2007

### Manchot

It is not injective. You can prove this by counterexample: f(1/4)=f(1/2).

4. Apr 9, 2007

### Werg22

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.

Last edited: Apr 9, 2007
5. Apr 9, 2007

### 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).

Last edited: Apr 9, 2007
6. Apr 9, 2007

### Werg22

But this involves that the function is monotonic on the intervals [0,1/e] and [1/e, 1], something that has to be proven. Also this can't be shown with ways of the derivative because the derivative involves the function itself...

Last edited: Apr 9, 2007
7. Apr 9, 2007

### rbj

what do you need to prove it? that there is a unique minimum and it is at 1/e?

8. Apr 9, 2007

### 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?

9. Apr 9, 2007

### Werg22

No, that's not it. I'm wondering if we have the condition $$0<x<x_{2}<1/e$$, then

$$x^x > x_{2} ^ {x_{2}}$$

As of now, I do not see this as a certainty. The function is continuous on the interval, yes, but I am not sure if it's monotonic (the fact that there is only 1 minimum does not exclude the possibility of non-monotonicity). Maybe I'm not thinking right, so I am open to correction.

Last edited: Apr 9, 2007
10. Apr 9, 2007

### rbj

it's because $f(x) = x^x$ and all derivatives are continuous for all x>0 and, with only one place where the derivative is zero (at 1/e) and that for 0<x<1/e, the derivative is negative. if the function is continuous in that region and if the derivative is defined and negative in that region, then it must be strictly decreasing in that region. that should prove it sufficiently.

11. Apr 9, 2007

### Data

Monotonicity on any interval is not necessary: Any real function, continuous on [a,b] (with a<b), is injective on domain [a,b] iff it is either strictly increasing or strictly decreasing on (a,b).

In this case, $f(x) = x^x$ has f(0) = f(1), is continuous on [0,1] and is differentiable on (0,1). Rolle's theorem says there's a point with derivative 0 in between, and it's easy to see that this implies f is either constant or increasing at some points and decreasing at others on (0,1). In any case f isn't strictly decreasing or strictly increasing on [0,1], so f can't be injective.

It's not any standard form, so "misc. transcendental function" is good enough for me! Some people will refer to it as a "power tower of order 2," or a "tetration." You could also describe it as the composition of an exponential and $x\log x$.

Last edited: Apr 9, 2007
12. Apr 9, 2007

### Werg22

That's what was bothering me exactly! I wasn't sure the function was piecewise in in terms of monotonicity (I wasn't sure if there is an "infinite" number of oscillations on the interval or not) as I wasn't realizing that this condition is assured by the fact that $$x(1 + log x)$$ is strictly negative for x inferior to 1/e and positive for x superior to 1/e (thanks rbj)

Last edited: Apr 9, 2007
13. Apr 9, 2007

### Data

Yeah, but you don't need to worry about the explicit derivative at all. All you need to know is that it is either always zero or sometimes negative and sometimes positive (we don't care where).

Furthermore, the simple fact that f(0) = f(1) is enough to show non-injectiveness on the reals. To show non-injectiveness on the positive reals requires a little bit of argument (as posted above).

Last edited: Apr 9, 2007
14. Apr 10, 2007

### Izzhov

Well, I understand now that it is non-injective on positive real numbers. I believe, however, that $$x^x$$ is injective on negative integers. I'm not sure how you could go about proving it, though, seeing as it's not really a continuous function in that domain...

15. Apr 10, 2007

### 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 $$(-n)^{-n} \not= (-m)^{-m} \quad \forall n \not= m$$.

16. Apr 10, 2007

### Mystic998

Just something bugging me...f(0) doesn't equal f(1). f isn't defined at 0.

17. Apr 10, 2007

### Data

I am using the (often ill-advised, in the context of analysis) definition $0^0 = 1$. In general there's no definition for $0^0$ (many many combinatorial and series identities necessitate the definition $0^0 = 1$ - but in analysis it usually makes more sense to leave it undefined and remove point discontinuities piecewisely).

It's inconsequential in this case, as $\lim_{x \rightarrow 0^+} x^x = 1$ (which is what I used originally, and avoids the confusion) and the argument still works (this is of course why I later went ahead and used the convention $0^0 = 1$ heedless of any possible difficulties. $0^0=1$ makes this particular function continuous on the closed interval [0,1]).

Last edited: Apr 10, 2007
18. Apr 10, 2007

### 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.

19. Apr 10, 2007

### 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.

20. Apr 10, 2007

### 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.

Last edited: Apr 10, 2007
21. Apr 10, 2007

### Izzhov

I was asking about negative integers, not positive ones.

22. Apr 10, 2007

### Data

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, $f(-n) = (-1)^n/f(n)$)

Last edited: Apr 10, 2007
23. Apr 11, 2007

### tehno

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

24. Apr 11, 2007

### 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.