What is the Difference Between Undefined and Indeterminate?

[Feynman]

what is the value of 0^0?

 

[arildno]

It's indeterminate, it depends on the way in which you "approach it"
For example, consider the function f(x) defined on positive numbers:
$$f(x)=0^{x}\to{f(x)}=0,x>0$$
Approaching $$0^{0}$$ by evaluating f at ever closer x's, clearly indicates that:
$$0^{0}=\lim_{x\to{0}}f(x)=0$$

Now, consider the function g(x):
$$g(x)=x^{0}\to{g}(x)=1,x>0$$

Using g in the limiting procedure, yields $$0^{0}=1$$

That is $$0^{0}$$ "by itself" is indeterminate

 

[Feynman]

so it is a complex number?

 

[Feynman]

if x < 0?///

 

[gazzo]

You're not allowed to do that? Atleast not with the set of reals?

As a consequence of one of the multiplication axioms, by definition.

x^0 = 1
iff x =/= 0

there are websites devoted to the number zero and I am sure somebody will quote one as usual :P

 

[gazzo]

gah.

http://home.ubalt.edu/ntsbarsh/zero/zero.htm

:/

 

[Feynman]

So if x=0??
What will do?
Gasso?

 

[gazzo]

it's not allowed.

you'll be banished to astrology!

:P

 

[Feynman]

So we can't calculate 0^0 like an limite
So what should we do?

 

[gazzo]

why would you want to $$0^0$$ anyway?

 

[Feynman]

What do u mean Gasso?

 

[tomkeus]

When I looked last time $$0^0$$ wasn't defined in maths so it cannot be calculated. It's somewhat similar to $$\frac{1}{0}$$. It cannot be calculated, but if some function is approaching it it may converge to some nuber, but it depends on the function.

 

[Feynman]

It can calculate but using colex nombers like 1^n

 

[tomkeus]

Then how much it is?? I'm very curious.

 

[Feynman]

What do u mean tomkeus?

 

[rayjohn01]

From my physics viewpoint zero is never zero but a small +/- dx , in this sense 0^0 also involves the non integer root of a negative number

 

[tomkeus]

I just want to say that $$0^0$$ isn't real or complex valued like $$e^{i\phi}$$ or $$2^{15}$$. It's indefinite value.

 

[Feynman]

No , i think that 0^0 has a value

 

[chroot]

No, Feynman, not in this universe. It has no specific value. Mathematicians call such objects indeterminate.

- Warren

 

[matt grime]

Why must it be defined, Feynman? Just because you can write it and think that it looks like it ought ot be a number doesn't mean it is actually such. log(x) can be defined for all real positive x, and if you're prepared to learn some complex analysis for complex non-zero x too, that doesn't mean log(0) is defined.

 

[Njorl]

So, 0^0 is indeterminate because:

lim x->0 of x^0 =1 and,
lim x->0 of 0^x=0.

assume z= y^x, y=f(x) is such that when y->0, x->0.

Can we construct a functional relationship between x and y that when they approach 0 as a limit, they produce a "z" that is some finite number between 0 and 1? Or are 0 and 1 the only allowable results.

Njorl

 

[rayjohn01]

This is exactly why I raised the physical viewpoint -- I'm not a mathematician but it appears to me that they are always in trouble with 'zero'.
A typical example is Integration -- y = int ( f(x).dx ) . dx ----> 0
IF you ignore any closed form result and start with a numerical analyisis , it forces you to choose dx because the sum of zeros IS zero. So dx=0 does not make sense but it can be a small as you like.
Nature ( isn't that what maths tries to describe) does not deal in zeros even though some objects may be VERY small ( 10^-39 ) or so . Even worse than that nature keeps objects moving in such a way as you may not even know where they are !

 

[Zurtex]

No , i think that 0^0 has a value

Think about it this way:

$$0^0 = 0^10^{-1} = \frac{0}{0}$$

 

[CrankFan]

Maybe you'll find this useful

http://groups.google.com/groups?selm=3071@mentor.cc.purdue.edu&output=gplain

 

[matt grime]

This is exactly why I raised the physical viewpoint -- I'm not a mathematician

no it would appear not

but it appears to me that they are always in trouble with 'zero'.
A typical example is Integration -- y = int ( f(x).dx ) . dx ----> 0

there are two dx's in there, is that what you really mean?

IF you ignore any closed form result and start with a numerical analyisis

closed form of what? what numerical analysis?

it forces you to choose dx because the sum of zeros IS zero.

finite or countably finite indexed sum that is, uncountable becomes moot, obviously

So dx=0 does not make sense

who said it did? dx isn't even a number

but it can be a small as you like.

i'm sorry? you're confusing delta and d, it appears: dx is not a number, though on occasion by treating it as such it may yield useful applied results.

Nature ( isn't that what maths tries to describe) does not deal in zeros

who knows, maths may be used to model naturally occurring phenomena, and it certainly does use zeroes: the cardinality of the set of elephants that are mice is zero.

even though some objects may be VERY small ( 10^-39 ) or so . Even worse than that nature keeps objects moving in such a way as you may not even know where they are !

hmm, don't think you want to introduce quantum mechanics, which is after all a mathematical model, and especially the uncertainty princple which is just a formal result of certain parts of analysis and integration, which you didn't appear to understand when you used it above.

 

[chroot]

Physics attempts to describe nature and happens to use some math along the way. Mathematics as a subject makes no attempt to describe nature. You're definitely muddling several concepts that should be disparate, rayjohn01.

- Warren

 

[loseyourname]

Here you go, Feynmann. Just as $\sqrt{-1} = i$, now $0^0 = j$. We've got ourselves a new imaginary number. If you can find a use for it, go ahead and rewrite the complex analysis books for us.

 

[Njorl]

Here you go, Feynmann. Just as $\sqrt{-1} = i$, now $0^0 = j$. We've got ourselves a new imaginary number. If you can find a use for it, go ahead and rewrite the complex analysis books for us.

This will make electrical engineers very unhappy.

Njorl

 

[gazzo]

Most electrical engineers are already very unhappy.

Then again, so are mathematicians

=(

 

[MiGUi]

$$0^0$$ can't have a value because if it had one, then arythmetics wouldn't work properly.

We can say that $$a^x = \prod_{i=1}^{x} a$$

If a = 0, then $$0^x = 0$$ because $$0^x = 0*0*0...$$

Can we mutiply a number, zero times by itself? No, we can't use the previous definition in that case. We know that a^0 = 1. Why?

Lets use the logarythm properties.

$$a^0 = 1$$
$$ln (a^0) = ln (1) = 0$$
$$0*ln(a) = 0$$

And that is true if ln(a) is defined. But ln is a function which is defined only for real positive numbers, so a can't be negative or zero to apply that property.

So, which value has $$0^0$$ ?

Like another user said, $$0^0 = \frac{0}{0} = 0*\infty$$ and that can't be solved as well. So we must accept that $$0^0$$ don't exists. It's not like "i", because "i" only makes the square root work properly, but indeterminations are big problems for mathematics :)