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In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form. The indeterminate forms include 0^0, 0/0, 1^∞, ∞−∞,∞/∞, 0 × ∞, and ∞/0.........so it's also a indeterminate form you have given....
0^0 = 0^(m-m) = 0^(m) . 0^(-m) ....
now as, 0^(-m) = 1 / 0 which is undefined...
so 0^0 = undefined .....well first I tried to prove it like this but then I found that it wasn't a good proof as it had contradictions . :(
easy to proof.
imagine that the exponent is absolutely like an element of carbon molecule.
exponent is like a box that contain something, copy the box and paste then imagine that they will combine and has more energy, it will be two thing that are same like x*x in mathematic.
how about x^0?
so easy.
it means that there are no element, like an empty box.
because the box still present in universe, it mean there must be a symbol to represent it, that is the 1 or one.
@Aditya Parson
–
0^0 is neither 1 nor indeterminate, it is undefined...
there is a difference between indeterminate and undefined.
e.g.
app.0/app.0 is indeterminate where as 0/0 is undefined
similarly, 0^0 is undefined as 0^0 = e^(0log0) where log 0 is undefined, but (app. 0)^app.0 is indeterminate whose value is 1...if both functions which are approaching 0 used in (app0)^app0 are the same.
if both functions are x.
lim (x app0) x^x can be found by equating it to e^xlogx where if xapp. 0 xlogx approaches 0[this can be found by using LH rule] and hence x^x approaches 1(as e^0 = 1)
Easy Math Editor
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exact zero raise to the power exact zero is NOT DEFINED.
Read this funny yet informative article 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/
Watch this video
In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form. The indeterminate forms include 0^0, 0/0, 1^∞, ∞−∞,∞/∞, 0 × ∞, and ∞/0.........so it's also a indeterminate form you have given....
http://www.wolframalpha.com/input/?i=lim+x+is+approaching+to+0+x%5Ex
it is indeterminate form in chapter limits....
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but mathematicians prove through limits that it it comes to 1?
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ok......can u xplain me how?
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read the article posted by Saurabh.
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ok thanks....i'll do that...
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limits tell us that n/0 = inf but we know that n/0 is undefined
0^0 = 0^(m-m) = 0^(m) . 0^(-m) .... now as, 0^(-m) = 1 / 0 which is undefined... so 0^0 = undefined .....well first I tried to prove it like this but then I found that it wasn't a good proof as it had contradictions . :(
really nice question....... i think so it is undefined.
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I agree with that. You can easily prove that it is an indeterminate form in mathematics,
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in laws of exponent they never mention that base a is not equal to 0
If 0^0=1 what's the proof and if 0^0= undefined what's the proof ???
look at this.
4^2 = 16
4^1 = 4
4^0 = 1
why 4^0 = 1?
easy to proof. imagine that the exponent is absolutely like an element of carbon molecule. exponent is like a box that contain something, copy the box and paste then imagine that they will combine and has more energy, it will be two thing that are same like x*x in mathematic.
how about x^0?
so easy. it means that there are no element, like an empty box. because the box still present in universe, it mean there must be a symbol to represent it, that is the 1 or one.
0^0=0/0-------------not defined
yes we can prove it by log theorem? take log and calculate.
well someone can give me the clear proof of any number raised to 0 is 1.
it should be zero..
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But many take it to be 1.
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LIke google! https://www.google.com/search?q=0^0
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Yes.
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0^0 is neither 1 nor indeterminate, it is undefined... there is a difference between indeterminate and undefined. e.g. app.0/app.0 is indeterminate where as 0/0 is undefined similarly, 0^0 is undefined as 0^0 = e^(0log0) where log 0 is undefined, but (app. 0)^app.0 is indeterminate whose value is 1...if both functions which are approaching 0 used in (app0)^app0 are the same. if both functions are x. lim (x app0) x^x can be found by equating it to e^xlogx where if xapp. 0 xlogx approaches 0[this can be found by using LH rule] and hence x^x approaches 1(as e^0 = 1)
only google calculator is showing 1! but other scientific calculators finds math error.
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I know that.
Just a consideration
The way I was taught, even though 0^0 is indeterminate, mathematicians will consider it to be 1.
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considering is different thing...sometimes in physics also we consider that 1/0 = infinity...but that is wrong.. 0^0 is undefined
undefined. bt nt equal 2 1.