>>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?
>>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?
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there are two cardinal sins from which all others spring: Impatience and Laziness. - franz kafka
i know very (i stress very) little calc, but i always understood 0^0 to be undefined.
Logically it doesn't matter how many times you multiply zero by itself, it's still zero.
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Also logically you can't divide a number by zero. Follow the patternLogically it doesn't matter how many times you multiply zero by itself, it's still zero.
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
I concur with thantos. I though about it all day at work but you can't divide by zero
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.
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.
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
Last edited by Speedy5; 02-22-2004 at 12:09 AM.
heh, I was about to post about "indeterminate form" when I clicked on the link (which apparently has disappeared)
0! = 1
doesn't it?
Yes, but that's merely a definition. 0^0, just as 0/0, are by themselves undefined.
All the buzzt!
CornedBee
"There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
- Flon's Law
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.
Which is why it is undefined and cannot be anuthing else.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.
Not necessarily. 0 raised to itself can be either indeterminant or 1 There are a few discussions of this at PhysicsForums
And I'm obviously right, because you all know that I'm a inherent genius.
XD
Precisely.Originally posted by RoD
Which is why it is undefined and cannot be anuthing else.
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.
