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mathelord
Oct23-05, 01:19 PM
a friend [no longer a user of the forum]showed me this and i felt i should as well show this to you all
let x^x=x
x^[1/x]=x
x^[1/x]-x=0
x[x^(1/x-1)-1]=x
so x is 0 and 1.
which evntually gives 0^0 as 0
in sloving with anither method,he also got -1 as x.
how true is the topic 0^0=0.
he explained that if you have nothing and you raise it to nothing you eventually get nothing.

hypermorphism
Oct23-05, 01:29 PM
...
x^[1/x]-x=0
x[x^(1/x-1)-1]=x
...
How did you get from the first step to the second ?

Karlsen
Oct23-05, 02:09 PM
If you have x = 0, you get 1/0 which is invalid.
I also fail to see how you got from the first to the second equation.

roger
Oct23-05, 02:14 PM
I don't actually see how the first equation is valid , never mind the second one.

Could somebody please explain ?

Karlsen
Oct23-05, 02:16 PM
It's valid because he wants to solve x^x = x, for x.

benjamincarson
Oct23-05, 02:38 PM
a friend [no longer a user of the forum]showed me this and i felt i should as well show this to you all
let x^x=x
x^[1/x]=x
x^[1/x]-x=0
x[x^(1/x-1)-1]=x
so x is 0 and 1.
which evntually gives 0^0 as 0
in sloving with anither method,he also got -1 as x.
how true is the topic 0^0=0.
he explained that if you have nothing and you raise it to nothing you eventually get nothing.

mathelord, this is all pretty trivial. What is it that's confusing you? If you "have nothing and you raise it to nothing", then you don't have anything to exponentiate! You don't "eventually" get nothing, it was always nothing.

hypermorphism
Oct23-05, 02:50 PM
mathelord, this is all pretty trivial....
Factoring x from 0 and getting x is some pretty nontrivial algebra. :rofl:

benjamincarson
Oct23-05, 03:02 PM
Factoring x from 0 and getting x is some pretty nontrivial algebra. :rofl:
I don't see how he was "factoring x from 0 ". Anyway, It wouldn't be hard to construct a proof that shows that the equation x^{x}=x is only valid for 0 and 1. So......
0^{0}=0
1^{1}=1

Fiddle-dee-do

Zurtex
Oct23-05, 03:05 PM
Well:

\lim_{x \rightarrow 0} x^x = 1

\lim_{x \rightarrow 0} 0^x = 0

Some times 00 = 1 is defined as for usefulness, but in general its not determined.

hypermorphism
Oct23-05, 03:13 PM
I don't see how he was "factoring x from 0 "....
Look at his fourth step.

It wouldn't be hard to construct a proof that shows that the equation LaTeX graphic is being generated. Reload this page in a moment. is only valid for 0 and 1....

It would be impossible. 00 is an indeterminate form. See here (http://mathworld.wolfram.com/Indeterminate.html) and here (http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/).

Robokapp
Oct23-05, 04:46 PM
x^0=1 any number raised to zeor power equals 1. x*x*x*x...*x zeor ammount of times equals 1 because 1 is the null value in multiplication. a number times himself zero times is equal to 1.
0^x=0 zero raised to any power equals zero. 0*0*0*0*0*...0=0 because of multiplication property. no matter how many zeors you have...you still have zero.

0^0=? Well...is it 1 becasue it is raised to zeor parts? or is it 0 becasue the 0 is raised to a power? Can't be both, but it can be none. answer: undefined.