What is Zero Raised to Itself?

[Shinobiku]

I was asked in another forum what zero raised to itself was. Any other number raised to zero equals one, but zero raised to any number equals 0. But what about zero itself. I already tried explaining it through thinking of zero as a limit of getting really small, but now this question has got me thinking. Any suggestions on how to think through this?

 

[chroot]

It is indeterminate.

- Warren

 

[Organic]

Hi Shinobiku,


I think that to answer to your question, we have to examine the logic system itself.

The most known logic system is the Boolean logic, which is based on 0 XOR 1 connective.

Fuzzy logic is based on fading between 0 XOR 1.

A non-Boolean logic is based on 0 AND 1.


In Boolean or Fuzzy logic 0^0 is unknown because any base value is some cardinal X >= 0.

Any cardinal X>0 is based on some non-empty set {x}, where cardinal X=0 is based on the empty set {}.

So, what we have here is |{x}|^0 and |{}|^0.

By Boolean and Fuzzy logic |{x}|^0=1, but |{}|^0 is not well-defined because it is based on no set's content.

So, I think the answer to your question is beyond the limitations of Boolean and Fuzzy logic.


What do you think?


Organic

 

[HallsofIvy]

Organic: Since you asked "what do you think", I think you chattered and chattered and finally said YOU were unable to answer the question.


The question is very simply answered: 0 to the 0 power is "undetermined" (that's an adjective: noun form is "indeterminant").

Notice that is NOT the same as saying "undefined". 1/0 is undefined because "1/0= x" is the same as "1= 0*x" which is untrue for all possible x. 0/0 is "undetermined" because "0/0= x" is the same as "0= 0*x" which is TRUE for all possible x. Another way of saying that is if we have f(x)->1, g(x)->0, then lim f(x)/g(x) does not exist while there are functions f(x)->0, g(x)->0 such that lim f(x)/g(x) exists and can be any given number. Similarly, given any number, we can find functions f(x)->0, g(x)->0 such that lim f(x)^g(x) exist and is equal to that number.

 

[Organic]

Hi HallsofIvy,


Thank you for your corrections about "1=0*x"(which is UNTRUE for all possible x) and "0=0*x"(which is TRUE for all possible x).

What i mean is this: any nonempty set's content {x} which is some singleton, can be used as some base for x^0=1 where 1 means base value (some singleton) exists.

In the case of the empty-set our content is NOTHING and by Boolean logic NOTHING^0 is gibberish.

By non-Boolean logic we can define another set's content, which is not {} ,not {.) (singleton) and not a collection of singletons {...}.

This object has the ability to simultaneously be in two different states like
> AND = ,0 AND 1, closed AND opened, and so on.

Therefore it exists between any distinguished singletons as a non-localized object {.___.}.

There are exactly 0 singletons in this object therefore its base value = 0, but because it exists ( unlike the empty set content, then __^0=1 or |{__}|^0=1, exactly as x^0=1 and |{x}|^0=1 in Boolean or Fuzzy logic.


Organic

 

[Shinobiku]

Alright, here's what I've come up w/ before i read anyones replies, which all had something to say, but left me w/ everyone saying it's indeterminate.

Let y = f(x)^g(x), take the ln of both sides

ln(y) = ln(f(x)^g(x))

ln(y) = g(x)*ln(f(x)), now exponentiate both sides

e^ln(y) = e^g(x) * e^ln(f(x))

y = f(x)*e^g(x), then take the limit as f(x)->0 and g(x)->0

y = 0*1 = 0

Can anyone see any flaws? Although I don't see anything wrong, it doesn't feel right to me. I believe this can only work if both f(x) and g(x) are finite polynomials.

Any thoughts?

EDIT: I just caught my own mistake (30 min. later)
go back to the third line:

ln(y) = g(x)*ln(f(x)), now take the limit as both f(x) and g(x)->0

I believe that ln(f(x)) approaches -infinity faster than g(x) approaches 0. If so, then

ln(y) = -inf.

but since this is a limit, then as

ln(y) -> -inf., means that

y -> 0

But this depends on the rate at which g(x)->0 and ln(f(x))-> -inf.

Any thoughts?

 

[HallsofIvy]

ln(y) = g(x)*ln(f(x)), now exponentiate both sides

e^ln(y) = e^g(x) * e^ln(f(x))

No, that's wrong. If ln(y)= g(x)*ln(f(x)) then
e^ln(y)= e^g(x)*ln(f(x)
= e^{ln(f(x)^g(x))}

or y= f(x)^g(x) which is exactly what you started with.

 

[Shinobiku]

Plz notice the edit

 

[chroot]

This thread was pointless after the second post.

- Warren

 

[Shinobiku]

I don't really like the stuff i came up w/, I've asked some ppl at my college, as well as tested it in mathematica using different methods and this is what I've come up w/:
I'm just going to post here what i just posted in my other forum.

I've talked w/ a friend of mine who is a math major (senior), and he likes to think of it as any number to the 0 power is the same as that number divided by itself, so

a^1/a^1 = a^(1-1) = a^0

but in the case of zero, 0^1/0^1 is already indeterminate.

That's one way to look at it. And then i went ahead and plugged a few things into mathematica:

In[2]:=[/color] 0^0

Power::indet : Indeterminate expression 0.^0. encountered.[/color]

Out[2]=[/color] Indeterminate

In[3]:=[/color] Limit[x^x, x->0]

Out[3]=[/color] 1

In[4]:=[/color] Table[{n,n^n},{n,.1,0,-.01}]//TableForm

Out[4]//TableForm=[/color]

0.10 0.794328
0.09 0.805159
0.08 0.817047
0.07 0.830151
0.06 0.844674
0.05 0.860892
0.04 0.879189
0.03 0.900147
0.02 0.924742
0.01 0.954993
0.00 Indeterminate

So, from those 3 methods, it seems that it is apporoaching 1 as a solution

 

[chroot]

Originally posted by Shinobiku
I don't really like the stuff i came up w/, I've asked some ppl at my college, as well as tested it in mathematica using different methods and this is what I've come up w/:

0^0 IS INDETERMINATE. PERIOD. THERE IS NO QUESTION, NOR ANY ROOM FOR DISCUSSION.

- Warren

 

[HallsofIvy]

Well, you're no fun at all!

 

[StephenPrivitera]

Originally posted by Shinobiku
So, from those 3 methods, it seems that it is apporoaching 1 as a solution

Although I'm quite afraid that chroot will kill me for continuing this thread, I will say that the fact that lim_x-->af(x) exists does not imply that f(a) exists.

 

[chroot]

Originally posted by StephenPrivitera
Although I'm quite afraid that chroot will kill me for continuing this thread

*BANG*

You're dead.

- Warren

 

[Shinobiku]

Indeterminate does not mean there is no answer, it just implies that the current method is not suitable. Just because it is indeterminate does not imply there can be no discussion. If you don't like the topic, then there is no need for you to read it.

Stephen: From what i know, when something doesn't exist at a particular value, then for many purposes in mathematics, it is then appropriate to look at it's behavior in the limits that it approaches the value at which it does not exist. This said, the answer lies in what context the question is brought up, and what situation the solution would be used.

 

[chroot]

Originally posted by Shinobiku
Indeterminate does not mean there is no answer

Right, it means there are an infinite number of answers. Since you can't choose just one (they're all perfectly valid), you can only say that the form does not have a definite meaning -- it is indeterminate.

- Warren

 

[Hurkyl]

I was trying to refrain from posting on this, but I couldn't resist.

0^0 is fairly ambiguous; there are lots of mathematican entities represented by '0', and '^' means different things in different contexts.


The others have focused on the context of the form of limits in calculus. If you want to know:

f(x)^g(x) as x → 0
where
f(x) → 0 as x → 0
and
g(x) → 0 as x → 0

then we have the indeterminate form 0^0, which means you have to do some more work to determine the answer.


If we're talking about exponentiation over the real numbers, then 0^0 is undefined, not indeterminate.


If we're talking about cardinal numbers, then 0^0 = 1, as can be verifed by direct computation. In fact, 0^x=1 for all cardinal numbers x.


When working with polynomials (or power series), one often permits x^0 as a valid polynomial (or power series) which is defined to be a synonym for 1. This is occasionally written as "use the convention that 0^0 = 1" when evaluating polynomials (or power series).

 

[chroot]

You're too smart for your own good, Hurkyl.

- Warren

 

[Integral]

Originally posted by Shinobiku
I don't really like the stuff i came up w/, I've asked some ppl at my college, as well as tested it in mathematica using different methods and this is what I've come up w/:
I'm just going to post here what i just posted in my other forum.

I've talked w/ a friend of mine who is a math major (senior), and he likes to think of it as any number to the 0 power is the same as that number divided by itself, so

a^1/a^1 = a^(1-1) = a^0

but in the case of zero, 0^1/0^1 is already indeterminate.

That's one way to look at it. And then i went ahead and plugged a few things into mathematica:

In[2]:=[/color] 0^0

Power::indet : Indeterminate expression 0.^0. encountered.[/color]

Out[2]=[/color] Indeterminate

In[3]:=[/color] Limit[x^x, x->0]

Out[3]=[/color] 1

In[4]:=[/color] Table[{n,n^n},{n,.1,0,-.01}]//TableForm

Out[4]//TableForm=[/color]

0.10 0.794328
0.09 0.805159
0.08 0.817047
0.07 0.830151
0.06 0.844674
0.05 0.860892
0.04 0.879189
0.03 0.900147
0.02 0.924742
0.01 0.954993
0.00 Indeterminate

So, from those 3 methods, it seems that it is apporoaching 1 as a solution

Proof via a computer can not reveal anything of mathematics, it can only reveal the mind of the programer.

 

[Organic]

Any proof is under the limitations of some consistent axiomatic system.

As godel showed, no consistant system can be a complete system, because it depends on information beyond its own system, therefore we always have to open our minds when we deal with Math language.


Organic

 

[Njorl]

Originally posted by chroot
0^0 IS INDETERMINATE. PERIOD. THERE IS NO QUESTION, NOR ANY ROOM FOR DISCUSSION.

- Warren

What is the basis for this statement. I usually agree with you, so when I don't, I'd like to know why. I have always operated under the belief that 0^0=1, and never run into any difficulty.

My reasoning:

Take a number, 5 say, and multiply it by 2 six times. This is 5x2^6 and it equals 320. Now, take 5 and muliply it by zero, zero times. If you are doing it zero times, it is just the same as not multiplying it by anything, which is the same as multiplying by the multiplicative identity, 1. So, 0^0 =1.

Is there some complication that I am missing?

Thanks,
Njorl

 

[S = k log w]

On the other hand, it could be argued that 0^0 = 1

If the set of 0 (that is), [0] is raised to it's own set [O], is there not one set of a set? O.K., it is one set. Is one set = 1?

 

[chroot]

Originally posted by Njorl
Is there some complication that I am missing?

Try taking the log of both sides of the equation 0⁰ = 1.

0 log 0 = 0
0 * -infinity = 0

Obviously something's wrong there. ;) The left hand side is an indeterminate form, and does not equal the right hand side.

Here's a page in your favor, that 0⁰ is 1:

http://www.mathforum.org/dr.math/faq/faq.0.to.0.power.html

Here's another, from the same site, which argues against it:

http://www.mathforum.org/library/drmath/view/57322.html

The math gurus here are probably more able to effectively argue this than I.

- Warren

 

[Njorl]

Thanks. Those links do offer other view-points. I think they delve a bit into what Hurkyl was getting into. I will have to think about it a bit.

Njorl

 

[jcsd]

xⁿ/x = x^n-1, x¹ = x => x⁰ = x/x => 0⁰ = 0/0

Though it isn't necessarily a proof of equality, but it shows that defing 0⁰ as a single value requires arbitarily breaking of some of the normal algerbraic rules.

 

[S = k log w]

Originally posted by chroot
Try taking the log of both sides of the equation 0⁰ = 1.

0 log 0 = 0
0 * -infinity = 0

Obviously something's wrong there. ;) The left hand side is an indeterminate form, and does not equal the right hand side.

Here's a page in your favor, that 0⁰ is 1:

http://www.mathforum.org/dr.math/faq/faq.0.to.0.power.html

Here's another, from the same site, which argues against it:

http://www.mathforum.org/library/drmath/view/57322.html

The math gurus here are probably more able to effectively argue this than I.

- Warren

What is the Fibbonici of 0?

 

[ahrkron]

What do you mean "the fibbonacci of" a number?
What would be the fibbonacci of 1? of 2?

 

[jcsd]

F₀ (i.e. the zeroth Fibonacci number) is usually defined as 0.

 

[S = k log w]

The formula for finding a fibonacci is:

f(n)=f(n-1)+f(n-2) for any real value n. So there would be no
fiboncci for 0.
For complex, and non interger numbers

See this interesting web site:

http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibFormula.html#Binetneg



O/1 is infinity, or indeterm., or undefined, or whatever. For the sake of argument, I will call this number "(01/01)"

(0/1) x (0/1) = 1

Therefor 0^0 = 1

but if X x X = 1 and X = 0 then X = 0

so 0 x 0 is either 0 or it is one, thus (indetermin)

unless

(indetermin) x (indetermin) = 1

 

[jcsd]

Originally posted by S = k log w
The formula for finding a fibonacci is:

f(n)=f(n-1)+f(n-2) for any real value n. So there would be no
fiboncci for 0.
For complex, and non interger numbers

See this interesting web site:

http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibFormula.html#Binetneg



O/1 is infinity, or indeterm., or undefined, or whatever. For the sake of argument, I will call this number "(01/01)"

(0/1) x (0/1) = 1

Therefor 0^0 = 1

but if X x X = 1 and X = 0 then X = 0

so 0 x 0 is either 0 or it is one, thus (indetermin)

unless

(indetermin) x (indetermin) = 1

No, you can only really use the formula to find a Fibonacci number when n is a natural number. But as F₂ = 1 and F₁ = 1, you can define F₀ as 0 from the recurssive formula and this is how it is conventially defined.

0/1 is undefined and it's pretty easy to show that it cannot be a real or a complex number and thus you cannot perform algebraic operations on it.

 

[S = k log w]

Originally posted by jcsd
No, you can only really use the formula to find a Fibonacci number when n is a natural number. But as F₂ = 1 and F₁ = 1, you can define F₀ as 0 from the recurssive formula and this is how it is conventially defined.

0/1 is undefined and it's pretty easy to show that it cannot be a real or a complex number and thus you cannot perform algebraic operations on it.

A good point. Using Fibinocci it can not be proven.

Not using Fibinocci, rather using set theory,
is there a quantity of 1 of the set [undefined]
x a quantity of 1 of the set [undefined]

might that = 1?

If it were so, then 0 x 0 = 1
0 x 1 = 0

Would there be such a thing as -0 ?
If so, what would it be?

 

[HallsofIvy]

S = k log w wrote:
O/1 is infinity, or indeterm., or undefined, or whatever.

and then

jcsd wrote:
0/1 is undefined and it's pretty easy to show that it cannot be a real or a complex number and thus you cannot perform algebraic operations on it.

Am I missing something? I have been suffering under the delusion that 0/1 was equal to 0 for some years! Or is it possible that you meant either 1/0 or 0/0??

 

[jcsd]

Oops back to school for a good dose of reading comphrehension for me, I thought it was 1/0 (though why I wrote 0/1 I don't know).

 

[S = k log w]

This is really interesting.

Look at this:

SQRT AND SQUARE


((+2)+(+2)) = +2^2
((-2)+(-2)) = -2^2
((+2)+(-2)+(+2)+(-2)) = 0^2

0^0 = ?

 

[selfAdjoint]

Originally posted by S = k log w
This is really interesting.

Look at this:

SQRT AND SQUARE


((+2)+(+2)) = +2^2
((-2)+(-2)) = -2^2
((+2)+(-2)+(+2)+(-2)) = 0^2

0^0 = ?

Umm, ((-2)+(-2)) = -2^2? Want to rethink?

 

[chroot]

Originally posted by selfAdjoint
Umm, ((-2)+(-2)) = -2^2? Want to rethink?

I think (hope) he meant -(2^2).

Don't forget PEMDAS...

- Warren

 

[S = k log w]

Originally posted by chroot
I think (hope) he meant -(2^2).

Don't forget PEMDAS...

- Warren

Thank you.