Back to the question of 0 as a power or divisor


I just happened to remember a joking argument by Donald Knuth:

If n/0 = inf.

Then n = inf. * 0

IOW, any number is a product of infinity and 0.



HP's latest calculators have 0^0 = 1.

[result for the 30b]


HP's latest calculators have 0^0 = 1.

[result for the 30b]

Does 48sx qualify as latest ?

This 0^0=1 seems to be the accepted mathematical convention today.
How about it, Don?


Accepted only as a limit as you approach 0 from the right. In every other maths it is undefined (in my opinion).


In middle school, we usually just say it's "undefined," like division by zero. The kids accept that. Some calculators may give other answers, but I think the TI-83 family gives you an error message when you do 0^0.


In middle school, we usually just say it's "undefined," like division by zero. The kids accept that.

(Emphasis mine)

The kids may accept that, but that doesn't necessarily mean it makes sense. When I was in high school, I once had a discussion with a few classmates about the value of 0!. We'd been told that 0! = 1, and that that was "just how it was defined".

I argued that there was no need to treat 0! as a special case:

How many ways are there to sort 3 items? answer: 3! = 6. (everyone nods)

How many ways are there to sort 2 items? answer: 2! = 2. (everyone nods)

How many ways are there to sort 0 items? answer: 0! = 1. (everyone disagrees, saying there are 0 ways to sort 0 items)

It's not an issue to get worked up over; while I felt a deep sense of disconnect at that point of the discussion, this is still obviously not something that's going to seriously trip anyone up later in life. But still, I felt then, and I still feel now, that the notions of an "empty sum" and an "empty product" are useful, and they make some borderline cases like x^0 and 0! fall into place quite nicely, without introducing any new problems; an "empty sum" is a sum with no terms, equalling the additive identity, 0; an "empty product" is a product with no factors, equalling the multiplicative identity, 1. No need to treat 0^0 or 0! as special in any way; they're both empty products, hence equal to 1. The fact that there really is exactly one way to order 0 items, and the fact that 0^0 = 1 makes sense from a continuity point of view (everything else to the zeroth power equals 1; why should 0 be an exception?) is just a bonus. :-)

Edited: 18 Sept 2010, 5:21 p.m.


Yes, it makes total sense that 0!=1. x! is a continuous function and it even continuously approaches 1 as x approaches 0. There are dozens (or millions, who knows) of ways of defining the factorial function, and they all give 0!=1.

It mostly makes sense that 0^0=1.

To be specific, for any functions f and g that are analytic at 0 and for which f->0 and g->0 as x->0, (and if f is not the constant 0), then f(x)^g(x)->1 as x->0

Of course, that theorem gives you hints of how you can find f^g->0^0 that does not tend to 1.

Say, f=e^(-1/x^2), g=x. Then f^g will tend to 1 as x approaches 0 from positive numbers, but will tend to infinity as x approaches 0 from negative numbers. (Of course, f only ->0 as x->0 when x is on the real line, anyway. In the complex plane, the limit doesn't exist, which is why the function isn't analytic.)


Err, it tends to 0 as x ->0 from positive numbers. One thing that example is supposed to show, sort of, is that f^g=0^0 can equal 0, like 0^x, if f tends to 0 much faster than any analytic function.


For my money, Crawl's explanation really hits the nail on the head, so to speak. Math is full of instances where we define values of functions based on convenience for the mathematical model rather than a provable theory. For example, x^n is well defined for integers. You can further define it for rational numbers, but for irrational numbers, we simply say "uh, k^x when x is irrational is the value that makes f(x) = k^x continuous."

I think the same is true with x!. It's a continuous function, but only because we've dictated that it should be and because there is a continuous function that equals x! (gamma(x+1)).

As for 0^0, at first glance you could say it equals any number (0^0 = 0^k / 0^k = 0/0; 0/0=x; 0=0*x; x = any number). We decide to pick the one that makes the most sense, which is the one that the limit approaches as Crawl describes.


There is a theorem that states how unique the usual continuous extension to the factorial function is.

So, you could define it with an integral, or with Euler's product formula, or whatever, but however you do it, you're gonna get the same thing.


Here are maybe some "controversial" definitions:

We should only use the factorial function. Everyone should drop using the Gamma function like a bad habit. That stupid argument offset helps nothing.

B(1) (the 1st Bernoulli number) should equal +1/2, not -1/2. Note that the HP50g's IBERNOULLI function gives the "wrong" answer for this argument.


I agree, it's not an issue to get worked up over, and believe me the kids don't get worked up over it. In middle school, we teach the kids rules. One rule says that 0 to any power is 0. Another rule says that anything to the zero power is 1. Now, unless I bring up the conundrum about 0^0 seemingly falling between the cracks, few kids will even be aware of it. Kids, amazingly enough, aren't intrigued by these things! Sometimes I bring it up, to get them thinking. And I usually leave it that it's like division by zero, not allowed.

In middle school, we measure success by kids actually knowing their times tables and the basics of algebra. 0^0 isn't worth worrying about.


This 0^0=1 seems to be the accepted mathematical convention today.

Of course, this has been debated here a few times over the years, but I did not bookmark any threads...

I insist that it's incorrect to declare that "0^0=1 by definition", because:

  1. To evaluate it directly causes a DBZ or logarithm error.

  2. No unambiguous limit exists, as
    x->0+ 0x = 0

    x->0- 0x = DBZ (+inf)

    x->0 x0 = 1

In most cases, however, letting 0^0 = 1 is more useful and fits the problem. It's essentially implicit as the first term in the Taylor series for cos(x) and e^x, expanded about a = 0 (the last of the three limits is taken for the two derivations).

-- KS

Edited: 23 Sept 2010, 1:55 a.m.


There was a previous discussion of this issue in this thread.

It seems HP calculators had 0^0 return an error (undefined) in all pre-RPL models, and 0^0=1 in RPL models, seemingly in accordance with an IEEE standard for computer languages.

I am definitely out of school here, but that's the way I see it.


Well, the new math library has a flag that can either cause 0^0 to return an error (ER0pow0) or 1. :-)


Edited: 18 Sept 2010, 9:19 p.m.

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