Why do non-integer powers fail in graphing x^x?

[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 traveling 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.

 

[Hurkyl]

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).

 

[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.

 

[Gib Z]

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.

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

 

[JonF]

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.

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

 

[Hurkyl]

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?

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

 

[hotvette]

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.

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 = x^x, what is y'?

 

[murshid_islam]

matt grime said in this thread that 0^0 = 1.

 

[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 = x^x

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 = e^ln(00) = 0⁰

 

[JonF]

matt grime said in this thread that 0^0 = 1.

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

 

[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?

 

[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:

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

 

[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.

 

[HallsofIvy]

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.

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.

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

Pretty standard Calculus I problem.

 

[murshid_islam]

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

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.

 

[murshid_islam]

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

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

 

[cristo]

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

Indeed it is.

 

[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!