0^0?

[Beer w/Straw]

I got banned from a forum basically cause I didn't like someones response and I called the person worthless.

As far as I know mathematicians and engineers differ on the answer. Maple 18 just gave me a warning message, Mathematica 10 gives me indeterminate and some calculators give me 1.

I'm guessing the correct answer is indeterminate and this one person gave me the answer of 1 and tried to haughtily give some mathematical proof but I lack the knowledge to argue.

I would really like to see a proof, however, would it take a mathematician to answer this?

 

[Beer w/Straw]

Maybe that would come across badly and I should ask nicer.

I did express frustration.

 

[quantum_wave]

While you wait for someone who actually knows the laws of math, I'll just ask, why isn't it zero?

 

[Dr_Toad]

I hope this helps:

According to some Calculus textbooks, 0^0 is an ``indeterminate form''. When evaluating a limit of the form 0^0, then you need to know that limits of that form are called ``indeterminate forms'', and that you need to use a special technique such as L'Hopital's rule to evaluate them. Otherwise, 0^0 = 1 seems to be the most useful choice for 0^0. This convention allows us to extend definitions in different areas of mathematics that otherwise would require treating 0 as a special case. Notice that 0^0 is a discontinuity of the function x^y. More importantly, keep in mind that the value of a function and its limit need not be the same thing, and functions need not be continous, if that serves a purpose (see Dirac's delta).

 

[someguy1]

In the real number system, $$0^0$$ is undefined. That is, it simply has no value. That's because $$\lim_{x \to 0} 0^x = 0$$ but $$\lim_{x \to 0} x^0 = 1$$, so there's no consistent way to make sense of the expression in the real numbers.

However if you want to make an arbitrary convention, it's sensible to say that $$0^0 = 1$$. Here are two plausibility arguments.

* Using calculus we can show that $$\lim_{x \to 0} x^x = 1$$.

* In set theory, $$0^0$$ is the set of functions from the empty set to the empty set. There is exactly one such function, namely the empty function. So in set theory, it's correct to say that $$0^0 = 1$$.

So $$0^0$$ is undefined; but if you want to assign it a sensible value, 1 is a reasonable choice. But you can't use that assumption in a calculation involving real numbers.

 

[Beer w/Straw]

Thank you for the replies. I'll look more closely at l'hopital's rule.

1/0 has a complex number, but 0^0 has none?

 

[someguy1]

Thank you for the replies. I'll look more closely at l'hopital's rule.

I fail to see how l'Hôpital applies to the indeterminate form $$0^0$$, since l'Hôpital explicitly refers only to the indeterminate forms $$\frac{0}{0}$$ and $$\frac{\infty}{\infty}$$.

1/0 has a complex number, but 0^0 has none?

That's true only when you adjoin a hypothetical "point at infinity" to construct the one-point compactification of the complex plane, also known as the Riemann sphere. It's mathematically valid in a particular technical context, but if you use the expression $$\frac{1}{0}$$ outside of its proper context, it's still wrong.

http://en.wikipedia.org/wiki/Riemann_sphere

 

[Beer w/Straw]

I fail to see how l'Hôpital applies to the indeterminate form $$0^0$$, since l'Hôpital explicitly refers only to the indeterminate forms $$\frac{0}{0}$$ and $$\frac{\infty}{\infty}$$.

I don't feel bad for being confused. But you've got my interest. (I was thinking about Dr_Toads response.)

That's true only when you adjoin a hypothetical "point at infinity" to construct the one-point compactification of the complex plane, also known as the Riemann sphere. It's mathematically valid in a particular technical context, but if you use the expression $$\frac{1}{0}$$ outside of its proper context, it's still wrong.

http://en.wikipedia.org/wiki/Riemann_sphere

Personally, I like Mathworld better than wiki. I don't know if it is applicable.

 

[someguy1]

I don't feel bad for being confused. But you've got my interest. (I was thinking about Dr_Toads response.)

Yes, I was actually referring to Dr_Toad's response. The remark about Dirac was a little off too, since the Dirac "function" is actually a distribution and not a function. Mathematically, in any case. Physicists are a little loose with the terminology.

Personally, I like Mathworld better than wiki. I don't know if it is applicable.

My comment still applies regardless, f(z) = 1/z can be evaluated at 0 on the Riemann sphere but that does not mean that 1/0 can be used outside of a specific technical context. In complex analysis they study how some expressions that blow up at a point can be fixed and how others can't. So again there's a technical context involved, you can't just uncritically use the expression 1/0 without an understanding of the context.

Anyway I hope I shed some light on the subject, even if there are some nuances left dangling.

 

[rpenner]

I fail to see how l'Hôpital applies to the indeterminate form $$0^0$$, since l'Hôpital explicitly refers only to the indeterminate forms $$\frac{0}{0}$$ and $$\frac{\infty}{\infty}$$.

Here's one way, using l'Hôpital's rule:
$$f(x) = x^x
\\ g(x) = \ln f(x) = x \ln x
\\ \lim_{x\to 0^+} g(x) = \lim_{y \to +\infty } g(\frac{1}{y}) = \lim_{y \to +\infty } \frac{1}{y} \ln \frac{1}{y} = - \lim_{y \to +\infty } \frac{\ln y }{y} =^? - \frac{\infty}{\infty}
\\ - \lim_{y \to +\infty } \frac{\ln y }{y} = - \lim_{y \to +\infty } \frac{\frac{d \quad}{dy} \ln y }{\frac{d \quad}{dy} y} = - \lim_{y \to +\infty } \frac{\frac{1}{y} }{1} = - \lim_{y \to +\infty }\frac{1}{y} = 0
\\ \lim_{x\to 0^+} x^x = \lim_{x\to 0^+} e^{g(x)} = e^{\lim_{x\to 0^+} g(x)} = e^0 = 1$$

Likewise because $$\lim_{x\to 0^+} f(-x) = \left( \lim_{x\to 0^+} (-1)^{-x} \right) \times \frac{1}{\lim_{x\to 0^+} f(x) } = \lim_{x\to 0^+} \cos{ \pi x } - i \sin { \pi x } = 1$$ we can say $$\lim_{x\to 0} x^x = 1$$.

Here's another way without using l'Hôpital's rule: $$f(x) = x^0 \\ \lim_{x\to 0^+} f(x) = 1 \\ \lim_{x \to 0^-} f(x) = 1 $$

Of course $$\lim_{x\to 0^+} 0^x = 0$$ and $$\lim_{x\to 0^-} 0^x $$ is a non-starter.

No one is saying $$0^0 = \lim_{x\to 0} x^x $$ as fact, but for many purposes in analysis, this a natural definition. Likewise 0! = 1 is also a natural definition which cannot be derived from 1! = 1 and (n+1)! = (n+1) × n!

 

[Beer w/Straw]

I'll look into that.

 

[Dr_Toad]

Yes, I was actually referring to Dr_Toad's response. The remark about Dirac was a little off too, since the Dirac "function" is actually a distribution and not a function.

Apologies. My quote was simply the intro to the page I tried to link. Rpenner has laid the beast to rest in his usual concise way, I think.

 

[Beer w/Straw]

No apologies. You quoted my university.

 

[danshawen]

Great question, BWS!

Various answers to the question from various levels of mathematical elegance can be found here:

http://www.askamathematician.com/20...ematicians-and-high-school-teachers-disagree/

The binomial theorem seems to prefer 0^0 =1, and I think this trumps L'Hopital's rule for sheer utility.

Another idiosyncratic math practice that drives me nuts (because someone asked me on another thread) is naming every nuance of every theorem for any famous mathematician who ever touched it. L'Hopital's is a good example of a pedestrian bit of calculus more easily remembered for what it does than for the name of the individual who claims discovering it.

Leibniz did a lot of excellent work, and part of the reason for that was because he wasn't bogged down remembering and crediting the names of every mathematician who ever did similar work before him.

 

[Beer w/Straw]

And why does Mathematica give me Indeterminate?

 

[Jason.Marshall]

And why does Mathematica give me Indeterminate?

Can you give me an opinion on my thread Xi function??

 

[Jason.Marshall]

Great question, BWS!

Various answers to the question from various levels of mathematical elegance can be found here:

http://www.askamathematician.com/20...ematicians-and-high-school-teachers-disagree/

The binomial theorem seems to prefer 0^0 =1, and I think this trumps L'Hopital's rule for sheer utility.

Another idiosyncratic math practice that drives me nuts (because someone asked me on another thread) is naming every nuance of every theorem for any famous mathematician who ever touched it. L'Hopital's is a good example of a pedestrian bit of calculus more easily remembered for what it does than for the name of the individual who claims discovering it.

Leibniz did a lot of excellent work, and part of the reason for that was because he wasn't bogged down remembering and crediting the names of every mathematician who ever did similar work before him.

Can you as well give an opinion sorry for the intrusion am just really curious about this?

 

[Beer w/Straw]

What?

I can't edit the quote.

 

[rpenner]

And why does Mathematica give me Indeterminate?

Because not all of mathematics is combinatorics and in analysis $$x^y$$ does not unambiguously reach the limiting value 1 as (x,y) approach (0,0) from all directions.

The exponentiation operation in analysis is distinct from the operation in set theory's cardinal arithmetic.

Thus, the authors of Mathematica made a choice to code Indeterminate rather than give an answer that often works but sometimes does not.

 

[danshawen]

Can you as well give an opinion sorry for the intrusion am just really curious about this?

I'm with rpenner (and also Wolfram). Its value largely depends on the context of the expression it is being used in. And I will grudgingly admit, L'Hopital's rule is a useful arbitrator.

Other than "A New Kind of Science", Wolfram can be relied upon. Like mathematics, cellular automata is a useful tool, but a "new kind of science" that completely eliminates the need ever to do science the old way, it is not. Too many symbolic logic halting problems like this one, for starters. Wolfram's 20 year tome falls flat on motivation among other things. Principia it is not.

rpenner, however, is completely reliable.

Gödel's incompleteness theorem wins again. Another math problem without a definitive answer, completely consistent and definitely a number and not an infinity (because 1 or 0 are both numbers). Think we're all done? Think again: square root (- 0^0 / 0^0). Don't forget to change the -1 in the expression to e^(i*pi).