# X^x. this is a weird one

1. Sep 10, 2007

### tony873004

I wrote my own graphing software, and I thought it had a bug in it, because when I graphed x^x, the curve starts out at (1,1), dips down a little, then rises again to pass through (1,1) as expected, before travelling to the wild blue yonder.

It seemed like this function has no range in the negative numbers. But then I was thinking (-1)^(-1) is just 1/-1=-1. Lo and behold, if I stare real close, it did plot a pixel there. And another one at (-2, 0.25). And others that seemed to asymptotically approach 0. I guess a graphing calculator would have been more obvious since the pixels are much bigger. But why no pixels inbetween?

Playing with the calculator, I realized that only (-integer)^(-integer) has a valid solution. Is there a plain English explanation as to why non-integers fail?

It also got me to thinking.... 0^0. Why is this 1? It seems like it should be indeterminate, like 0/0. Zero divided by anything is 0, and anything divided by 0 is infinity, setting up a paradox for 0/0, which is what I thought defined an indeterminate. 0^anything = 0. anything^0=1, so 0^0 should be another paradox, hence indeterminate. But the calculator sees things differently.

2. Sep 10, 2007

### Hurkyl

Staff Emeritus
Did you try plugging in, say, x = -1/2, and computing by hand?

Also, your calculator probably hasn't plotted a pixel at (0, 1). What you are seeing is that the limit of your function as x goes to zero (from the positive side) is positive 1, but no point actually at (0, 1).

3. Sep 10, 2007

### tony873004

That's a good way to put it. (-1/2)^(-1/2) = 1/(-1/2)^(1/2)=1/sqrt(-1/2), and I can't take the square root of a negative number because nothing times itself will equal a negative number since (-)*(-)=+. Thanks for the plain English explanation.

But you can take the cube root of a negative number. (-1/3)^(-1/3) = 1/(-1/3)^(1/3)^(1/3) = 1/crt(-1/3). What times itself, times itself again = -1/3? Well, the cube root of (+1/3) = 0.693361274350635. If I make this negative and cube it, I get (-0.693361274350635)^3 = -0.333333333333334, which is -1/3, except for that 4 at the end, which should have rounded to a 3, I'm guessing. So it seems there is a number that when cubed can equal -1/3. So why does the calculator consider 1/crt(-1/3) invalid?

re: 0^0, my software, which I wrote myself, is at the mercy of how the programming language does the math. When it arrives at x=0, it performs the operation 0^0, which it considers to be 1 (same result as Google Calculator). So, arriving at an answer it considers to be vaild, its next instruction is to plot (0,1), so that pixel does exist. I would have just expected some sort of overflow error, not a nice clean answer, because it makes more sense the way you phrased it, that it's limit from the positive side is 1, but the actual point does not exist.

4. Sep 10, 2007

### Gib Z

You know, sometimes you have to take more confidence in yourself :P Don't ask "why is this 1?", ask "why does that calculator think its 1". Calculators are prone to mistakes. After all, Calculators are humans too :P

5. Sep 10, 2007

### JonF

One good reason is because 0 = -0, and we want x^-a = 1/x^a

6. Sep 10, 2007

### Hurkyl

Staff Emeritus
So now, you should be able to guess what's going on: you know that if you have a number that is exactly a rational number with an odd denominator, you can compute x^x... how often do you think your calculator tried such a value of x?

There's another issue, by the way... if your calculator decides that it should use the principal value of exponentation, then numbers like (-1/3)^(-1/3) are actually complex. This expression is trivalued: any of the following three numbers work:
$$-\sqrt[3]{3},$$
$$\sqrt[3]{3} \left( \frac{1 + i \sqrt{3}}{2} \right),$$
$$\sqrt[3]{3} \left( \frac{1 - i \sqrt{3}}{2} \right).$$

I belive the usual standard for picking which value to return says to use
$$\sqrt[3]{3} \left( \frac{1 - i \sqrt{3}}{2} \right).$$

7. Sep 10, 2007

### hotvette

My favorite equation! I think 0^0 is a candidate for L'Hopital's rule. I seem to recall working this out a while back and convincing myself of the result.

Here's another challenge. If y = xx, what is y'?

8. Sep 14, 2007

### murshid_islam

9. Sep 14, 2007

### hotvette

I agree. Like I said in my previous post, I think it can be shown using L'Hopital's rule (unless I applied it incorrectly, which is entirely possible!)

Here goes:

y = xx

ln(y) = x*ln(x) = ln(x) / x-1 = f(x)/g(x)

Limit of both f(x) and g(x) is infinity as x -> 0. Therefore, candidate for L'Hopital's rule.

f'(x) = x-1
g'(x) = x-2

f'(x)/g'(x) = x

lim f'(x)/g'(x) = 0 as x->0

Therefore the lim as x->0: ln(y) = 0 = ln(00)

e0 = 1 = eln(00) = 00

10. Sep 14, 2007

### JonF

And he was right, it’s a matter of definition, there isn’t really anything to argue about.

11. Sep 14, 2007

### Gib Z

matt grime did NOT say that 0^0 = 1. He said that that definition is convenient for many definitions, although he knows perfectly well that the proper limit does not exist.

hotvette, when you are using your limit, you use x^x, you assume the exponent approaches zero at the same rate as the base. The more general limit is $$\lim_{x\to 0, y \to 0} x^y$$. This way, when zero is approaches from different directions, no limit exists.

The reason 0^0 = 1 works nicely for the reasons matt grime stated, such as power series, is because they are only in the 1 variable, x, so approaching 0 from only 1 direction.

Btw, nice to have you back murshid, where have you been?

12. Sep 14, 2007

### hotvette

Huh? I didn't know there was any stipulation about the rate at which the variable in various pieces tend to zero. Let me ask this. Following is from my old college calculus textbook:

$$\lim_{x\to 0} \frac{ln(1 + x)}{x} = \lim_{x\to 0} \frac{1}{(1+x)} = 1$$

Therefore:

$$\lim_{x\to 0} e^\frac{ln(1 + x)}{x} = e^1 = e = \lim_{x\to 0}(1 + x)^{1/x}$$

How is this any different than the following?

$$\lim_{x\to 0} \frac{ln(x)}{x^{-1}} = \lim_{x\to 0} \frac{x^2}{x} = x = 0$$

Therefore:

$$\lim_{x\to 0} e^\frac{ln(x)}{x^{-1}} = e^0 = 1 = \lim_{x\to 0}x^x$$

13. Sep 15, 2007

### Gib Z

I never disputed your computation of the limit $$\lim_{x\to 0} x^x$$. What you did IS correct, except the reason you define your limit in terms of only 1 variable, x, is because you assume the exponent and base are approaching zero, and are always equal. The actual limit that must be solved to rigorously define 0^0 for everything is the limit i gave above, where x and y are both approaching zero, but not necessarily equal.

14. Sep 15, 2007

### HallsofIvy

Do you mean a limit that evaluates as 00? There a many different limits that evaluate like that and the result, using L'Hopital's rule, depends upon the precise function used. That's one reason why 00 is undefined.

Pretty standard Calculus I problem.

15. Sep 15, 2007

### murshid_islam

i am a business student now. i am doing my MBA right now. so i guess i don't get much time for maths. but i will try to be regular from now on.

16. Sep 15, 2007

### murshid_islam

is it $$x^x(1 + \ln x)$$

17. Sep 15, 2007

### cristo

Staff Emeritus
Indeed it is.

18. Sep 15, 2007

### HallsofIvy

Here's a cute point: In differentiating $$f(x)^{g(x)}$$ there are two kinds of mistakes we could make.

We could ignore the fact that g(x) is variable and use the power rule: take the derivative to be $$g(x)f(x)^{g(x)- 1}$$.

We could ignore the fact that f(x) is variable and use the exponential rule: take the derivative to be $$ln(f(x))f(x)^{g(x)}$$

Of course, neither of those is correct. The correct derivative is their sum!