What does 0^0 equal?

Quote: 24Bingo

With a "googolplexian" of nines, no.

With infinite nines, yes.

If that is the case - I will stop entering 9's on my computer ...

I wasn't sure if I would be able to enter a googolplexian of them, but I know I can't enter an infinite number of 9's ... I plan on living to an old age, but not forever.

Quote: Doc

You just can't prove it by looking at limits that are approached, unless you can prove that at least one of these is a continuous function.

I hate to argue against myself, but can't we say with certainty that x^x is not continuous at x=0? This is because the limit as x approaches 0 of x^x from the negative side equals i. Right? So the function f(x)=x^x suddenly switches from 1 to i at 0. But where is it in that transition point right at 0? Saying it is "undefined" is so unsatisfying. While I still like 1 the best, it would be easier to support something like "La La Land" or the Wizard number than "undefined," "Error 0" or "#NUM."

The function equal to 1 at the origin, and x^x everywhere else, is continuous, but it's not the same function as x^x.

And no, as it approaches from the negative side, the limit is still 1, and every complex approach. There are ways to spiral in along the Riemann surface that will give you values other than one, but with a branch cut (i.e., treating x^x as a well-defined function by limiting the phase angle), any approach will get you arbitrarily close to 1.

Quote: 24Bingo

And no, as it approaches from the negative side, the limit is still 1.

Let's look at -0.5^-0.5.

-0.5^-0.5 =
1/-0.5^0.5 =
1/(-1 * 0.5)^0.5 =
1/(i * 0.5^0.5) =
sqrt(2)/i. =
-sqrt(2)*i

Correct me if I'm wrong. I don't think I've used an imaginary number in practice since 9th grade algebra. Anyone who knows what a Riemann surface is I have great respect for. So, don't take this as a challenge, but more as an opportunity to teach me where I'm in error.

Quote: Wizard

I hate to argue against myself, but can't we say with certainty that x^x is not continuous at x=0? This is because the limit as x approaches 0 of x^x from the negative side equals i. Right? So the function f(x)=x^x suddenly switches from 1 to i at 0. But where is it in that transition point right at 0? Saying it is "undefined" is so unsatisfying. While I still like 1 the best, it would be easier to support something like "La La Land" or the Wizard number than "undefined," "Error 0" or "#NUM."

The negative side of f(x) = x^x is not i.

f(-1) = -1
f(-2) = 1/4
f(-3) = -1/27

It's not a continuous function on the negative side as f(-1/2), f(-1/4), f(-3/2), ... are all imaginary values but f(-1/3), f(-1/5), ... are real. The values jump all over the place but for the values of x for which the function is defined, the absolute value of the function approaches 0 as x approaches negative infinity.

I confess that I had not even considered the functions for negative values of x and the fact that x^x would have complex number values. Also, 0^x would be discontinuous at the origin, jumping from a value of zero for all positive x to infinity for all negative x. It appears that of the three functions that I mentioned, only x^0 is nicely behaved with real and finite values everywhere except at x=0.

Quote: 24Bingo

The function equal to 1 at the origin, and x^x everywhere else, is continuous, but it's not the same function as x^x.

In my first post of this thread, I mentioned some weakness in both my math skills and my memory. In particular, I'm not clear on this point, which probably derives from a long absence from thinking about complex numbers.

On another topic that has been raised, I'm not really sure about the states of "undefined" vs. "really needs to be defined" vs. "has a generally-accepted definition not derivable on its own". I don't think I have any serious problem with defining 0^0=1, if it doesn't conflict with any other part of important mathematics and makes x^0 a continuous function throughout. And as I have stated, I thought that definition was the one mathematicians accepted.

Quote: Wizard

My HP-15C (the greatest calculator ever) says it is "Error 0."

HP has changed it's tune: the 48G says the answer is 1.

I think it is 1 regardless of my calculator.

If 0^0 = x, then 0ln(0) = ln(x). ln(0) is undefined, but who cares, because it is multiplied by 0. So, this simplifies to 0 = ln(x). Therefore, x=e^0 = 1.

Just wondering, are me and babs the only ones who have not the slightest clue what is being debated in this thread ?

Curious, that's all. Ignorance is bliss, but I would like to know what I am ignorant about !

Quote: wudged

The negative side of f(x) = x^x is not i.

Okay, I concede I was wrong there. Would you agree with the following statement?

Quote: 24Bingo

... as it approaches from the negative side, the limit is still 1, and every complex approach.

Quote: CrystalMath

ln(0) is undefined, but who cares, because it is multiplied by 0.

Is it possibly "undefined" because it is equal to something like negative infinity? If so, then I for one care enough not to ignore a term of zero times infinity, since it might be a non-negligible number.

Quote: Doc

Also, 0^x would be discontinuous at the origin, jumping from a value of zero for all positive x to infinity for all negative x.

You're right. Somehow I forgot that. But still, x^(-ln c / ln |csc x + cot x|) makes the point better anyway.

Actually, I've realized I made another mistake: x^x, as a function from C -> C, would NOT be continuous along the negative real line due to the branch cut (i.e., because the phase angle is going from pi to (-pi + epsilon), so there's a factor of e^(2x*pi*i)). The limit still converges at the origin, though, since if you treat it as theta = -pi it still goes to 1. As a function from R -> C, though, it would be continuous everywhere but 0, where it's undefined, and putting a 1 there would make it continuous. Anyway, on to where I didn't make a mistake:

Quote: Wizard

Let's look at -0.5^-0.5.

-0.5^-0.5 =
1/-0.5^0.5 =
1/(-1 * 0.5)^0.5 =
1/(i * 0.5^0.5) =
sqrt(2)/i. =
-sqrt*(2)*i

Correct me if I'm wrong. I don't think I've used an imaginary number in practice since 9th grade algebra. Anyone who knows what a Riemann surface is I have great respect for. So, don't take this as a challenge, but more as an opportunity to teach me where I'm in error.

The thing is, the imaginary parts get smaller and smaller as you approach 0. (-.5)^(-.5) happens to have no real part, but (-.5001)^(-.5001) and (-.4999)^(-.4999) do (-0.0004442 and .0004443, respectively). Just the same way, (-.0001)^(-.0001) has a tiny imaginary part (.0003144), and it disappears as we approach the origin.

The way you take an exponent is to express the complex number in polar form, r*e^(i*theta), where r is sqrt(a^2 + b^2) and theta is the arctangent of b/a (or if a is zero, pi/2 if b is positive and -pi/2 if negative; if b is zero and a is negative, theta = pi, as in the weirdly mythologized Euler's identity, e^(i*pi) + 1 = 0), this being equal due to Euler's formula, e^(i*theta) = cos(theta) + i*sin(theta). From this, you can give it a natural log of (ln r) + (i*theta), and then just use the formula x^a = e^(a ln x). This is why I've been talking about "branch cuts" - by adding an even multiple of pi to theta, you'll get the same value but a different natural log, so it's often useful to define the log treating the phase angle as -pi < theta <= pi.

(Incidentally, by this definition, the cube root of a negative number is nonreal, even though it has a real cube root - that's sort of the sacrifice you make to make it well-defined.)

Quote: 24Bingo

The way you take an exponent is to express the complex number in polar form, r*e^(i*theta), where r is sqrt(a^2 + b^2) and theta is the arctangent of b/a (or if a is zero, pi/2 if b is positive and -pi/2 if negative; if b is zero and a is negative, theta = pi, as in the weirdly mythologized Euler's identity, e^(i*pi) + 1 = 0), this being equal due to Euler's formula, e^(i*theta) = cos(theta) + i*sin(theta). From this, you can give it a natural log of (ln r) + (i*theta), and then just use the formula x^a = e^(a ln x). This is why I've been talking about "branch cuts" - by adding an even multiple of pi to theta, you'll get the same value but a different natural log, so it's often useful to define the log treating the phase angle as -pi < theta <= pi.

Well, I'm speechless, for now. As I admitted, my complex number arithmetic is about as rusty as the Vashon Island bicycle in a tree.


Image source: tubulocity.com

Can you just tell me what the answer is to -0.5^-0.5, if I am in error.

Also, where is MathExtremist when you need him?

Took the below answer from askamathematician.com ,I don't understand any of the math, but basically it's saying that in the proofs mathematicians use to show that 0^0= 1, the x is being defined as a number >0 (which is what mathematicians use), if you define x as being 0 the answer would be zero (I would, of course, define x as being 0 since the equation is 0^0, however mathematicians don't use this definition, they define x as having a value greater than 0.). And, really, I wouldn't use any proof to try to define 0^0, I would take it at face value that nothing multiplied by nothing nothing times equals nothing. So, the real question is not whether 0^0 is actually zero, but whether x is being defined by the mathematician as being 0, or is being defined as a number greater than 0. That's the math trick here.

Taken from askamathematician.com :

Mathematician: Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true.

Let’s consider the problem of defining the function f(x,y) = y^x for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:

y^x := 1 \times y \times y \cdots \times y

where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get

y^{x} = 1 \times y.

However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving

y^{0} = 1

which holds for any y. Hence, when y is zero, we have

0^0 = 1.

Look, we’ve just proved that 0^0 = 1! But this is only for one possible definition of y^x. What if we used another definition? For example, suppose that we decide to define y^x as

y^x := \lim_{z \to x^{+}} y^{z}.

In words, that means that the value of y^x is whatever y^z approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.

[Clarification: a reader asked how it is possible that we can use y^z in our definition of y^x, which seems to be recursive. The reason it is okay is because we are working here only with z>0, and everyone agrees about what y^z equals in this case. Essentially, we are using the known cases to construct a function that has a value for the more difficult x=0 and y=0 case.]

Interestingly, using this definition, we would have

0^0 = \lim_{x \to 0^{+}} 0^{x} = \lim_{x \to 0^{+}} 0 = 0

Hence, we would find that 0^0 = 0 rather than 0^0 = 1. Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what y^x means for all positive real numbers x and y, and it does preserve continuity of the function as we approach x=0 and y=0 along a certain line.

So which of these two definitions (if either of them) is right? What is 0^0 really? Well, for x>0 and y>0 we know what we mean by y^x. But when x=0 and y=0, the formula doesn’t have an obvious meaning. The value of y^x is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what y^x means for positive values is not enough to conclude what it means for zero values.

But if this is the case, then how can mathematicians claim that 0^0=1? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use 0^0=0 or if we say that 0^0 is undefined. For example, consider the binomial theorem, which says that:

(a+b)^x = \sum_{k=0}^{\infty} \binom{x}{k} a^k b^{x-k}



where \binom{x}{k} means the binomial coefficients.

Now, setting a=0 on both sides and assuming b \ne 0 we get

b^x

= (0+b)^x = \sum_{k=0}^{\infty} \binom{x}{k} 0^k b^{x-k}

= \binom{x}{0} 0^0 b^{x} + \binom{x}{1} 0^1 b^{x-1} + \binom{x}{2} 0^2 b^{x-2} + \hdots

= \binom{x}{0} 0^0 b^{x}

= 0^0 b^{x}

where, I’ve used that 0^k = 0 for k>0, and that \binom{x}{0} = 1. Now, it so happens that the right hand side has the magical factor 0^0. Hence, if we do not use 0^0 = 1 then the binomial theorem (as written) does not hold when a=0 because then b^x does not equal 0^0 b^{x}.

If mathematicians were to use 0^0 = 0, or to say that 0^0 is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to k=0. We gain elegance and simplicity by using 0^0 = 1.

There are some further reasons why using 0^0 = 1 is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians. The choice is not “right”, it is merely nice.

Quote: Wizard

Also, where is MathExtremist when you need him?

I don't have much to add that hasn't been said elsewhere, specifically here:
http://mathforum.org/dr.math/faq/faq.0.to.0.power.html
Bottom line, 0^0 is usually defined as 1 because it's more important that it's 1 than 0 or indeterminate. That's certainly true for algebra and combinatorics, wherein if 0^0 != 1, then a bunch of stuff doesn't work. I like Knuth's explanation best:

Quote: Concrete Mathematics (Graham, Knuth, Patashnik)

Some textbooks leave the quantity 0^0 undefined, because the functions 0^x and x^0 have different limiting values when x decreases to 0. But this is a mistake. We must define x^0=1 for all x , if the binomial theorem is to be valid when x=0 , y=0 , and/or x=-y . The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant.

Seriously, try making the binomial theorem work if 0^0 != 1.

Maybe it's useful that 0^0 != 1 for some other branch of math, but not for anything I've ever submitted to gaming labs. :)

Quote: Wizard

Quote: wudged

The negative side of f(x) = x^x is not i.

Okay, I concede I was wrong there. Would you agree with the following statement?

I would say the limit is -1 as x approaches 0 from the negative side.

Quote: MathExtremist

I don't have much to add that hasn't been said elsewhere, specifically here:
http://mathforum.org/dr.math/faq/faq.0.to.0.power.html
Bottom line, 0^0 is usually defined as 1 because it's more important that it's 1 than 0 or indeterminate. That's certainly true for algebra and combinatorics, wherein if 0^0 != 1, then a bunch of stuff doesn't work. I like Knuth's explanation best:

Seriously, try making the binomial theorem work if 0^0 != 1.

Maybe it's useful that 0^0 != 1 for some other branch of math, but not for anything I've ever submitted to gaming labs. :)

ME, once again making sense of the nonsense in my head.

Even No Thing is Some Thing. It's the only way to make Things work.

Quote: wudged

I would say the limit is -1 as x approaches 0 from the negative side.

How do you figure?

I was just thinking that x ln x = x (ln |x| + pi*i) still approaches zero, so e^(x ln x) approaches 1. Where do you get the -1?

The problem I have with Knuth is that then a lot of functions go from having singularities with simply no value to having singularities with a value floating off somewhere, simply because it makes more common formulae cleaner to use his convention. It sounds like a physicist's argument, and for physical things, makes a lot of sense, but from a mathematical standpoint...?

(Oh, and by the way, yes, (-.5)^(-.5) is -i/sqrt(2).)

Quote: 24Bingo

How do you figure?

I was just thinking that x ln x = x (ln |x| + pi*i) still approaches zero, so e^(x ln x) approaches 1. Where do you get the -1?

For values of -1 < x < 0, as x approaches 0 the function is going to be taking the nth root of x. Since x is negative, the function must evaluate to either a negative number or a complex number.

I didn't read through all the posts, but 0^0 is called an indeterminate form. In which case you must use l'Hôpital's rule

The rule is named after the 17th-century French mathematician Guillaume de l'Hôpital, who published the rule in his book Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes (1696), the first textbook on differential calculus.
However, it is believed that the rule was discovered by the Swiss mathematician Johann Bernoulli.

Short version of rule
lim x->c f(x)/g(x) = lim x->c f'(x)/g'(x)

EXAMPLE #1)Consider the function F=x^x as x approaches zero. ln(F) = x*ln(x) = ln(x)/x^-1 which is of the form f/g where function f and function g are both approaching infinity as x -> 0. In this case the limit will be the same as f'/g' or (1/x)/(-x^-2) = -x . So the function approaches zero, so ln(F)=0 or F->1

In this particular case 0^0 approaches 1

EXAMPLE #2)Consider the function F=sin(x)^sin(x) as x approaches zero. ln(F) = ln(sin(x))*sin(x) = ln(sin(x))/sin(x)^-1 which is of the form f/g where function f and function g are both approaching infinity as x -> 0. In this case the limit will be the same as f'/g' or (cos(x)/sin(x)))/(-sin(x)^-2*cos(x)) = -c/s /(c/s^2)=s= sin(x). So the function approaches zero, so ln(F)=0 or F->1.

In this particular case 0^0 also approaches 1


I can't think of an example off hand that comes out to something other than 1, but anything is possible. It is late, if I think of one I will post.

I taught this stuff for years.

(If I post "x^(-ln c/ln |csc x - cot x|)" again, someone's going to make it their signature, aren't they? Whoops.)

You've sort of answered it yourself - it's L'Hopital's rule. If f(x)^g(x) approaches 0^0, what you have is, essentially, e^(ln(f(x))/(1/g(x))), so L'Hopital's rule gives you e^(-f'(x)*g²(x)/(f(x)*g'(x))). Apparently, someone's proven that goes to e^0 or e^inf for all f(x) and g(x) analytic at 0 (...it looks like there should be an analytic extension of the exponent I came up with?), but I don't know the proof.

I'm so verbal with math that I can't look at the question and not ask, doesn't this mean to take nothing and do nothing with it?

For what it's worth, the calculator at http://web2.0calc.com/ has the answer to any number to the zeroth power as 1, except zero to the zeroth = undefined.

I did not cheat, but solved it as a word-concept.

I have nothing, and raised it to the nothing power... I have nothing.
Sounds like poker to me.

If I start with a lemon and try to change it into an orange, unless it becomes exactly an orange, then it is still partially a lemon.

***I had to edit here***

Since e^(ax ln x) got involved...

solve for x... e^x = 0.

If the OP answer zero is incorrect... undefined is correct. Kudos to Don S. for making me rethink the theorm.

Quote: ThatDonGuy

I say "undefined" because:

(a) 0^x = 0 for all x <> 0, so the limit as x approaches 0 of 0^x = 0.

(b) for all real numbers a, x^ax = e^{(ax ln x)}.
As x approaches 0, 1/x approaches positive infinity, as does -ln x.
Using L'Hopital's rule, the limit of x * (-ln x) = the limit of (-ln x) / (1/x) = the limit of (-1/x) / (-1/x²) = the limit of x.
Therefore, the limit of ax ln x as x approaches 0 = a * the limit of (x ln x) as x approaches 0 = a * the limit of x as x approaches 0 = 0.
The limit of e^{ax ln x} as x approaches 0 = e⁰ = 1.

NOW FACTOR OUT THE CONSTANTS...e and ln(e). OOOOPS!

It's been one heck of a long time since I had any reason to apply L'Hopital's rule to any problem. As I recall, it is useful in cases where you are searching for the limit of a function f(x) as x approaches some value but the calculations for f(x) approach an indeterminate value, such as approaching 0/0. L'Hopital's rule gives an additional opportunity to find a determinate value for that limit of f(x).

I don't think that is the difficulty in the original problem of this thread. I think the function(s) of interest approach limits that we can reasonably well determine (other than those that, for negative x, approach through a complex number space that many of us don't understand very well). However, we cannot establish that the function is continuous and has a value at x that is equal to the limit as x approaches that point, i.e.,

lim_x-->af(x) may or may not equal f(a).

That is, knowing the limit approached by a function as the form of the function approaches 0^0 does not tell us the value of 0^0.

Am I repeating myself too many times on that point?

Doc,

That horse, he is still twitchin'; best give him another kick for good measure.

IOW, no. You're making perfect sense within the parameters of your argument. To someone. Somewhere else on this board. I'm sure of it. Me, (glug).

Wikipedia is always right.

Quote: s2dbaker

Wikipedia is always right.

Well, I can't agree with that statement, but the link does point to a pretty good treatment of this issue.

Quote: Buzzard

I came up with -4.28 And I think I can prove it.

hahahahahahahah. You should take a Real Analysis course. Not enough humor in those.

There are three ways to look at this:

(1) 0^x always equals 0 EXCEPT when x = 0. 0^x always = 0. There would be an incontinuous function if you jump from 0 to 1 if 0^0 = 1. This would be a continuous function if 0^0 = 0.
(2) x^0 always equals 1 EXCEPT when x = 0. This is a continuous function is 0^0 = 1.
(3) The function x^x approaches 1 as x -> 0. But this is a function of math, meaning that the exponent rules the equation. This might be the right answer.

Really, what is the meaining of the exponential function? And does it take precedence over the natural number itself?

Therefore, I can't say definitely that 0^0 = 0 or 1. The wizard's argument is compelling as always, and if a gun was held to my head and had to make a choice, I'd be forced to say "1".

Quote: s2dbaker

Half dimensions don't happen in real life.

Tell that to atomic suborbitals!

This is my favorite thread. Two important things: out of 48 votes no one voted "I don't care." And by convention 0!=1, so my life will stay intact for now.

I did mistype o^0 into Wikipedia and it said, "o^0 does not exist. You can ask for it to be created." (the first o is the letter in this case). So I will be working on o^0 from now on.

deleted

Quote: drussell0208

This is my favorite thread. Two important things: out of 48 votes no one voted "I don't care." And by convention 0!=1, so my life will stay intact for now.

I did mistype o^0 into Wikipedia and it said, "o^0 does not exist. You can ask for it to be created." (the first o is the letter in this case). So I will be working on o^0 from now on.

That's easy, o^0 is very clearly lowercase "L":

o^0 = l

Quote: MathExtremist

That's easy, o^0 is very clearly lowercase "L":

o^0 = l

Post of the month !

deleted

Quote: MathExtremist

That's easy, o^0 is very clearly lowercase "L":

o^0 = l

*applause* It was so obvious, how did I not see it?

OK, so we have the "limit" gang, who are actually not answering the question (cf. Doc's argument), then the "undefined" crowd, and the "convention" throng, most of whom say it is equal to 1, who are just giving the "undefined" answer another coloration.

The only way to cut the cake, in my opinion, is to ask the OP (Wizard) what is the aim of the question. In what context do you need the answer? Is there a context where an exact value for 0^0 is needed, instead of a limit?

Clearly, the Knuth answer is working in this line: combinatorics is one field where such a value is needed because, working with integers, they cannot resort to limits. And the requisite of consistency implies that, for those uses, the consistent answer is 1.

Now maybe someone sees another field (probably in number theory) wher 0^0 is needed? Wizard, what did you have in mind?

Quote: kubikulann

The only way to cut the cake, in my opinion, is to ask the OP (Wizard) what is the aim of the question. In what context do you need the answer? Is there a context where an exact value for 0^0 is needed, instead of a limit?

The answer shouldn't be different according to the reason the question is being asked. I'm just looking for a general answer that will result in correct answers for applications where one might encounter 0^0.

Quote: Wizard

The answer shouldn't be different according to the reason the question is being asked. I'm just looking for a general answer that will result in correct answers for applications where one might encounter 0^0.

This is why, as pacomartin pointed out, it is typically called an indeterminate form. The answer always relies upon the context in which it is needed.

Quote: Wizard

Let's example what x^x approaches as x gets close to 0. We'll start with x=8 and keep dividing by 2.

x	x^x
8.00000000	16777216.00000000
4.00000000	256.00000000
2.00000000	4.00000000
1.00000000	1.00000000
0.50000000	0.70710678
0.25000000	0.70710678
0.12500000	0.77110541
0.06250000	0.84089642
0.03125000	0.89735454
0.01562500	0.93708382
0.00781250	0.96280297
0.00390625	0.97857206
0.00195313	0.98788970
0.00097656	0.99325384
0.00048828	0.99628396
0.00024414	0.99797136
0.00012207	0.99890064
0.00006104	0.99940789
0.00003052	0.99968275
0.00001526	0.99983079
0.00000763	0.99991010
0.00000381	0.99995241
0.00000191	0.99997488
0.00000095	0.99998678
0.00000048	0.99999306
0.00000024	0.99999636
0.00000012	0.99999810
0.00000006	0.99999901
0.00000003	0.99999948
0.00000001	0.99999973

I submit for the consideration of the forum that as x approaches 0, x^x approaches 1.

Ok, for those who don't know me, I am no mathematician. So I am approaching this from a logical standpoint, and there is what appears to me to be a discontinuity in rejecting the argument made above. Please clarify.

Here's why I accept this as proof;

If 1/3 = .333333333333333333....(infinity) and
if 2/3 = .666666666666666666....(infinity), then
1/3 + 2/3 = 1, but they also equal .999999999999....(infinity); never actually reaching 1. Yet, these are mathematically accepted values as decimal representations of those fractions, with the .6 rounded up at whatever point is pertinent to the calculation.

The Wiz has shown, with his table above, that the logical progression of x^x where x approaches 0, must also end in an infinite sequence of .999's. Why, then, is one acceptable as an equivalency of 1, and the other not? Thanks in advance.

Quote: beachbumbabs

Quote: Wizard

Let's example what x^x approaches as x gets close to 0. We'll start with x=8 and keep dividing by 2.

x	x^x
8.00000000	16777216.00000000
4.00000000	256.00000000
2.00000000	4.00000000
1.00000000	1.00000000
0.50000000	0.70710678
0.25000000	0.70710678
0.12500000	0.77110541
0.06250000	0.84089642
0.03125000	0.89735454
0.01562500	0.93708382
0.00781250	0.96280297
0.00390625	0.97857206
0.00195313	0.98788970
0.00097656	0.99325384
0.00048828	0.99628396
0.00024414	0.99797136
0.00012207	0.99890064
0.00006104	0.99940789
0.00003052	0.99968275
0.00001526	0.99983079
0.00000763	0.99991010
0.00000381	0.99995241
0.00000191	0.99997488
0.00000095	0.99998678
0.00000048	0.99999306
0.00000024	0.99999636
0.00000012	0.99999810
0.00000006	0.99999901
0.00000003	0.99999948
0.00000001	0.99999973

I submit for the consideration of the forum that as x approaches 0, x^x approaches 1.

Ok, for those who don't know me, I am no mathematician. So I am approaching this from a logical standpoint, and there is what appears to me to be a discontinuity in rejecting the argument made above. Please clarify.

Here's why I accept this as proof;

If 1/3 = .333333333333333333....(infinity) and
if 2/3 = .666666666666666666....(infinity), then
1/3 + 2/3 = 1, but they also equal .999999999999....(infinity); never actually reaching 1. Yet, these are mathematically accepted values as decimal representations of those fractions, with the .6 rounded up at whatever point is pertinent to the calculation.

The Wiz has shown, with his table above, that the logical progression of x^x where x approaches 0, must also end in an infinite sequence of .999's. Why, then, is one acceptable as an equivalency of 1, and the other not? Thanks in advance.

In order for the 0.999...(infinity) to be reached, x would need to be 0.0 (infinite zeroes) followed by a 1, which isn't a possible number, because you can always add more zeroes before the 1.

In other words, to achieve the 0.999...(infinity) to make that argument valid, you would need to have an impossible number as x.

Edited to add:

Or does 0.0(infinite 0s)1 = 0 ?

Quote: wudged

Or does 0.0(infinite 0s)1 = 0 ?

Yes, and you'll never (in finite time) find that 1.

Quote: wudged

Or does 0.0(infinite 0s)1 = 0 ?

Your summarizing what French mathematition Lemaire stated in his 2003 "symposium on small and large imaginary numbers"

0.000000 ... 01 which he states as di01 (decimal infini zero un) is not equal to 0.

He also stated that 1/3 is not equal to di33 and 2/3 is not equal to di66, thus 1/3 + 2/3 IS = 1, but di33 + di66 is NOT = 1

Quote: wudged

Or does 0.0(infinite 0s)1 = 0 ?

It does.

Quote: NewToCraps

Your summarizing what French mathematition Lemaire stated in his 2003 "symposium on small and large imaginary numbers"

0.000000 ... 01 which he states as di01 (decimal infini zero un) is not equal to 0.

He also stated that 1/3 is not equal to di33 and 2/3 is not equal to di66, thus 1/3 + 2/3 IS = 1, but di33 + di66 is NOT = 1

I like this guy, Lemaire. I'll have to remember the name. A mathematician with a flare for the obvious.

What I can say with certainty is that the limit of ten raised to an infinitely negative quantity is zero. Zero multiplied by any number is zero. Zero plus any number is that number. I don't know who this "Lemaire" is, but he's either a crank or dealing with a set of numbers not the same as the real and complex numbers we're familiar with. They may be interesting, or even in some way useful, but treating conclusions drawn from them as if they apply to real numbers is absurd.

Quote: kubikulann

OK, so we have the "limit" gang, who are actually not answering the question (cf. Doc's argument), then the "undefined" crowd, and the "convention" throng, most of whom say it is equal to 1, who are just giving the "undefined" answer another coloration.

The only way to cut the cake, in my opinion, is to ask the OP (Wizard) what is the aim of the question. In what context do you need the answer? Is there a context where an exact value for 0^0 is needed, instead of a limit?

Clearly, the Knuth answer is working in this line: combinatorics is one field where such a value is needed because, working with integers, they cannot resort to limits. And the requisite of consistency implies that, for those uses, the consistent answer is 1.

Now maybe someone sees another field (probably in number theory) wher 0^0 is needed? Wizard, what did you have in mind?

The manager asks, "So, what is the number?"

The Industrial Engineer replies, "What do you want the number to be...?"


ADDED: FWIW, I agree with (ulp!) Wikipedia. The term 0^0 is defined as 1. The quantity 0^0 is undefined. Contradictory statements can be so meaningful!

Quote: beachbumbabs

Ok, for those who don't know me, I am no mathematician. So I am approaching this from a logical standpoint, and there is what appears to me to be a discontinuity in rejecting the argument made above. Please clarify.

Here's why I accept this as proof;

If 1/3 = .333333333333333333....(infinity) and
if 2/3 = .666666666666666666....(infinity), then
1/3 + 2/3 = 1, but they also equal .999999999999....(infinity); never actually reaching 1. Yet, these are mathematically accepted values as decimal representations of those fractions, with the .6 rounded up at whatever point is pertinent to the calculation.

The Wiz has shown, with his table above, that the logical progression of x^x where x approaches 0, must also end in an infinite sequence of .999's. Why, then, is one acceptable as an equivalency of 1, and the other not? Thanks in advance.

Well, actually, babs, I don't think there is much of a difference. I believe the thing is that mathematicians take a lot of privilege with math. From the answer I took from askamathematician.com the 0^0 = 1 is used as mathematical "code", or a "cheat", or "short cut" in order to make complex theorems and calculations easier to do. They could have chosen 0 or undefined or 1, all have proofs for them, but they chose 1 because it's easier to work with. Understand that they say they could have used 0 and the theorems would still hold true, but it would be much more complex and most math guys wouldn't be able to do the math because they only know how to work the equations with the "cheats" and "shortcuts" they were given in school. In other words, 0^0 = 1 is just a "cheat" mathematicians used to save them a bunch of steps in a calculation. If you look at the case for 0^0 being undefined you will see that it is undefined because you divide by 0 in the proofs. Get that in the actual multiplying of nothing nothing times there's no division being done at all, but in some mathematicians proof he divided by 0, so he now claims 0^0 is undefined. But, you can't explain to people that their proofs are no good, they have to use another proof, or they could just multiply nothing nothing times.

Got a little side tracked there, but, the point is 0^0 =1 and .999...=1 are just "cheats" and "shortcuts" mathematicians use. They don't actually believe that 0^0 is anything other than 0 anymore than they believe .999... is anything other than .999... , but they are common "codes" or "cheats" or "short cuts" that are used by every mathematician. The difference is some mathematicians understand that they are just using a "cheat" or "short cut" and others just believe things like 0^0=1 is a straight up fact as opposed to a mathematical "code" or "cheat".

So, babs, getting back to your question. You've got 0^0 = .999... and .999... = 1. But, you never got 0^0=1. You just used a "short cut" or "cheat". Which is fine, but 0^0 still equals 0, as much as .999... still equals .999... you never proved any different. Hint. an accurate proof would show 0^0 = 0 . Anything else would prove, your proof is wrong, if you can see that making sense. We already know 0^0 = 0 and we know .999... is .999... we just want a way to prove it. If you can't prove that then you have nothing, except a common "cheat" or "short cut" mathematicians use.

This is pretty great as a response, Jy; thanks for thinking it through on paper. More food for thought.

The Wizard said I could come here to see who's good at math. This generally would be regarded as a calculus limit problem.
0⁰ is not too meaningful in itself, but becomes tractible in the form lim x^x as x -> 0 which we'll take to mean approaching 0 as decreasing positive values of x. Here's one way to find the limit:

y = x^x; log each side:

ln y = x ln x ; express x ln x as (ln x)/(1/x)

Differentiate numerator and denominator with respect to x (L'Hospital's rule)

(1/x)/(-1/x²) = -x

so lim ln y = 0 as x -> 0

lim y = 1 as x -> 0 since the antilog of 0 is 1

This is similar to ThatDonGuy's explanation but I would say that 0⁰ = lim x^x as x -> 0 = 1 at least if the limit is approached from the right, so my answer is 1. The Wizard has given a numerical example and that is the result that calculators should be programmed to give.

Some subsequent posts have objected to my saying that 0⁰ is not meaningful. What I mean is that it is not too productive to try to interpret it semantically. Different people can come to different conclusions, and in this case all of them can be correct. A zero exponent indicates something divided by itself; in this case 0/0, which can be any number at all.

Also, my presentation is not original. This is a fairly standard derivation of this limit and I wouldn't be surprised to find this question as a problem or an example in some calculus textbook.