\[{\large 0^0}=\]
Anything to the power 0 = 1
\[\huge \mathrm{\mathbf{indeterminate}}\]
?
yes indeterminate
Commonly it's taught that 0 to the zero power would be 1.
\[0^0=\frac {0^1}{0^{-1}}\]
ops typo
\[0^0=0^10^{-1}\]
0^0 is indeterminate because: 1) it cannot be determined whether it is 1 or 0 because of conflicting ideas that anything raised to 0 is 1...and 0 raised to anything is 0 2) it came from 0^a/0^a let us assume a is any number except 0... 0/0 would be indeterminate
\(\Large \color{MidnightBlue}{\Rightarrow 0^0 = {0^1 \over 0^1} }\) UNDEFINED. Maybe that gets it.
the first reason is the more valid one rather than the second
because the first one is the real reason...the second one i just made up..but does make sense
This is the conflict of two statements. 1) Anything raised to the power 0 is 1. 2) Anything divided by 0 is undefined. Well, so I guess you can't raise 0 to the zero power. So it is undefined.
conflicting ideas? aren't these ideas ment to be mathematically true, .. and provable.
on a fun note though...who can tell me why \[\huge 1^{\infty} = indeterminate\] conflicting because 0^0 satisfies both conditions
But lgba, \(\Large \color{MidnightBlue}{\Rightarrow 1^{n} = 1 }\)
0! = 1 Does that fact lead to\[0^{0}=1\]
exactly...so why is that indeterminate?
Because n is indeterminate.
good point @kropot72
According to some Calculus textbooks, 0^0 is an "indeterminate form." What mathematicians mean by "indeterminate form" is that in some cases we think about it as having one value, and in other cases we think about it as having another. Source: http://mathforum.org/dr.math/faq/faq.0.to.0.power.html
that refers to the 0^0 may be 1 or 0 so it is indeterminate
\[\frac{0\times0\times0}{0\times0}=\frac{0^3}{0^2}=\]
And anything divided by 0 is undefined. Yes. \(\Huge \color{MidnightBlue}{INDETERMINATE }\)
I like how we're sharing our ideas.
0^0 = ?; 0^1= ? 2^1 = 2; 2^0= 1 1^1 = 1; 1^0= 1
0^1=0
Check out this article, i thought it was pretty interesting. http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/
\(\Large \color{MidnightBlue}{\Rightarrow 2^1 = {2^2 \over 2} = 2 \text { and } 2^{0} = {2^1 \over 2} = 1 }\) Similarly, \(\Large \color{MidnightBlue}{\Rightarrow 0^1 = 0 \text{ and } 0^{0} = {0^1 \over 0} }\) But, that'd be undefined.
@ParthKohli, 0/0 is indeterminate. However n/0 = infinity, where n is not equal to zero
TYPE IN YOUR CALCULATOR \[\log 1 \div \log 0\] and see if the answer is 0
MATH ERROR
n/0 is not infinity. the limit as x approaches 0 of n/x is infinity. n/0 is undefined.
Anything over 0 = UNDEFINED
But guys, isn't \(\infty\) = Undefined?
that's because there's no such thing as log 0 but if 0^0 = 1 then log_0 1 must be equal to 0...but since it's not...0^0 =/= 1
infinity is a just a term. its not a number.
Neither is undefined. Undefined is also a term.
\[\infty = undefined\] BUT \[undefined \ne \infty\]
SO, if i ever measured 0^0 in a situation, it would depend on what that situation was as to the values that corresponds to the measurement 0^0
it actually makes more sense than you think
@UnkleRhaukus thmn you use calculus lol
then*
The thing is, we wouldn't ever go through this kind of terms. But, once in a while, it's good to know all about this fancy stuff.
these*
this*
well i am studying physics so numbers are real things that can be measured,
@lgbasallote I was correcting the this in the first sentence.
Perhaps this quote is pertinent: "Having to evaluate an expression of the form 0/0 is the penalty you have to pay for over-idealisation and is an indication that the treatment of the problem is at variance with nature".
\[ a^b = e^{b\ln a}\] \[ 0^0 = e^{0\ln 0} = e^{0 \times \infty}\] Indetermintae of zero times infinity ... not sure if you can really do this ... just a thought!!
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