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deathstryke
02-20-2004, 01:18 PM
We had this issue come up in calc while discussing exponential functions and couldn't come up with a clear answer. Any one know anything useful.

Thantos
02-20-2004, 01:22 PM
My calculator (TI-86) gave me a domain error. I have class in about 30 mins and I'll ask the profs what they think.

So for now I'm saying undefined

deathstryke
02-20-2004, 01:25 PM
I know that that is what the 83 says, but apparantly there are reasons for it having certain values, but that was all my teacher could remember from some class he took ages ago.

Thantos
02-20-2004, 01:29 PM
I think the confusion comes due to conflicting rules

0^x = 0
x^0 = 1

so 0^0 = would equal 0 and 1.

I'll ask though. The prof that has the class before mine knows a great deal and my prof just got out of college a few years ago so she might still remember :)

axon
02-20-2004, 01:41 PM
google does miracles...ever hear of it?

http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/

deathstryke
02-20-2004, 01:56 PM
That is really interesting, so apparantly they can't make up their mind about it.

UnregdRegd
02-20-2004, 02:15 PM
Off hand, I think my calculus textbook said something about the function 0^x approaching 1 as x approaches 0. However, 0^0 itself is considered to be undefined.

Thantos
02-20-2004, 03:29 PM
I asked and the answer I got is that it is generally considered to be undefined but some text do define it as 1 or 0.

Edit: I forgot they mention that when the text's do define it they do it by convention in the same manner that 0! is defined to be 1. Ok time to get rid of this headache :)

rainmanddw
02-20-2004, 04:44 PM
#include <iostream>
#include <cmath>
using namespace std;

int main()
{
cout << pow(0.0,0.0);
cin.get();
return 0;
}


out put:

1

CornedBee
02-21-2004, 06:28 AM
My university "textbook" (a collection of sheets actually) leaves it undefined. The whole thing is approached in the context of limits however. According to my textbook, the limit of something that results in an indeterminate form must be determined by looking at the functions involved. Usually the rule of l'Hospital is used.
There are some nice examples of how the limit of another indeterminate form, 0/0, depends on the functions involved.

adrianxw
02-21-2004, 06:48 AM
Windows calculator says 0^0 = 1.

CornedBee
02-21-2004, 07:39 AM
I think the Windows calculator might well use the pow function internally, which obiously uses the optimization of testing the second parameter for 0 and automaticaly returning 1 in case it is.

laserlight
02-21-2004, 09:41 AM
Using a C++ program or the Windows calculator would be unreliable in this case, since neither is an authority on such a definition, but is merely implementation dependent.

Magos
02-21-2004, 10:57 AM
It doesn't exists, since the limit differs depending on which axis you move along.

lim (x -> 0) x^0 = 1
lim (y -> 0) 0^y = 0

RoD
02-21-2004, 01:12 PM
Originally posted by Magos
It doesn't exists, since the limit differs depending on which axis you move along.

lim (x -> 0) x^0 = 1
lim (y -> 0) 0^y = 0

I was following along fine until i hit this post.....thnx magos

axon
02-21-2004, 01:17 PM
>>I was following along fine until i hit this post.....thnx magos

thats why you need to get yer ass to college buddy....calculus is great, aint it?

RoD
02-21-2004, 01:36 PM
i know very (i stress very) little calc, but i always understood 0^0 to be undefined.

RobR
02-21-2004, 06:40 PM
Logically it doesn't matter how many times you multiply zero by itself, it's still zero.

Thantos
02-21-2004, 07:00 PM
Logically it doesn't matter how many times you multiply zero by itself, it's still zero.
Also logically you can't divide a number by zero. Follow the pattern :)

x^2 = (x * x) / 1
x^1 = x / 1
x^0 = x / x
x^-1 = 1 / x
x^-2 = 1 /(x * x)

Fill in 0 for x and you find the problem ;)

linuxdude
02-21-2004, 07:28 PM
I concur with thantos. I though about it all day at work but you can't divide by zero

laserlight
02-21-2004, 11:19 PM
I suppose we get it from:

x^0 = x^(n-n) = x^n / x^n = 1, but x = 0 => 0 / 0
where n is some (real, complex?) number.

Speedy5
02-21-2004, 11:52 PM
Well see here, conflicting theories:

x ^ 0 = x / x

Normally this would equal 1, however:

if x = 0, x ^ 0 = 0 / 0

0 / 0 = undefined

But since any number divided by itself is 1, then it could also be one. But then you can do elementary school reasoning. How many times does 0 go into 0? 0 times.

So we have three logical answers to 0 ^ 0:

undefined
1
0

-------------------------------------

0 ^ 0 = ans

log[base 0] (ans) = 0

log ans / log 0 = 0

log 0 = undefined

So, even more proof that 0 ^ 0 = undefined.

Speedy5
02-22-2004, 12:03 AM
If we say 0 ^ 0 = 1, then there is an identity that denies that.
log[base b] (b) = 1

0 ^ 0 = 1
a ^ b = c
log[base a] (c) = b
<substituting>
log[base 0] (1) = 0

It does not follow the identity so 0 ^ 0 can't be 1.

However, another identity states:
log[base b] (1) = 0

0 ^ 0 = 1
a ^ b = c
log[base a] (c) = b
<substituting>
log[base 0] (1) = 0

Since both identities clash, 0^0 must be undefined. And see the above above post. By definition, a log must not have a base of 0. Therefore you really can't have an exponent with a base of 0. Wierd.

Anyways I believe: 0 ^ 0 = undefined

laserlight
02-22-2004, 12:23 AM
heh, I was about to post about "indeterminate form" when I clicked on the link (http://www.mathforum.org/dr.math/faq/faq.0.to.0.power.html) (which apparently has disappeared) :)

rainmanddw
02-22-2004, 02:38 AM
0! = 1

doesn't it?

CornedBee
02-22-2004, 05:41 AM
Yes, but that's merely a definition. 0^0, just as 0/0, are by themselves undefined.

Zach L.
02-22-2004, 11:58 PM
0! = 1 because it is the same as gamma(1)
Similarly, 1! = gamma(2), 2! = gamma(3), etc. The gamma function itself is quite nasty though.

The problem with saying 0^0 is anything is that there are many different ways to get to 0^0, and they don't all have the same limit.

RoD
02-23-2004, 12:14 AM
Originally posted by Zach L.
The problem with saying 0^0 is anything is that there are many different ways to get to 0^0, and they don't all have the same limit.

Which is why it is undefined and cannot be anuthing else.

gcn_zelda
02-23-2004, 04:11 PM
>>Which is why it is undefined and cannot be anuthing else.

Not necessarily. 0 raised to itself can be either indeterminant or 1 There are a few discussions of this at PhysicsForums (http://physicsforums.com)

And I'm obviously right, because you all know that I'm a inherent genius.

XD

Zach L.
02-23-2004, 08:12 PM
Originally posted by RoD
Which is why it is undefined and cannot be anuthing else.

Precisely.

It can't be '0 or indeterminate'. It is not well defined (because of the limits)... Not 0... Not anything else. Certainly, there are instances when taking it to be a certain value (e.g. 1 or 0) is convenient, and there is nothing preventing you from treating it as such in those instances, provided that it is understood that the value being used is the limit of a particular function.

RoD
02-23-2004, 08:59 PM
indeterminate sounds like another way to say undefined. So my statement stands.

CornedBee
02-24-2004, 05:44 AM
"Sounds like" isn't "is".

Indeterminate means that the thing itself is undefined, but can be assigned a value in some specific context, like the limit of an expression that would result in 0/0.

golfinguy4
02-24-2004, 03:40 PM
Originally posted by RoD
indeterminate sounds like another way to say undefined. So my statement stands.

Idiot. When you learn calculus (which tells you how to solve the problem), you can talk. Until then, STFU. You are talking out of complete ignorance.

You need L'Hopital's rule or some algebra (depending on the problem).