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#73
May2805, 07:00 AM

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BoTemp was right in critizing me, shmoe; I hadn't made the proper restriction on x.



#74
May2805, 07:41 AM

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How you stated it is fine to me, you were talking about a right hand limit. There's no reason for anyone to assume (or even care) if your function was defined at 0.



#75
May2805, 12:09 PM

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i agree that 0^0= 1
a^x= 1/(a^(x)) if a=x=0 then there is no way that 0^0= 0 because we'll have 0=infinite(1/0) 


#76
May2805, 12:52 PM

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If you like heuristic reasoning from identities, what about 0^x = 0? 


#77
May2805, 01:17 PM

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I don't think of 0 as being in the same number system as anything else really. It's more of a concept like infinity.
So you really can't do all of the same math with 0 as you can with other numbers. [tex]0^0[/tex] makes no sense. Nor would [tex]log(0)[/tex]. LOL... that's so funny... where I listed ['tex'] 0^0 [/'tex'] it put "infinity". EDIT: And now it's back to 0^0. Hmmm.... 


#78
May2805, 03:11 PM

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Incidentally, though, for each operation which is undefined at zero (such as 1/x), there's a corresponding operation which is undefined at one. (such as 1/(x1)) So, in a very real sense, you can do exactly as much with zero as you can do with any other real number. 


#79
May2805, 04:41 PM

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0^x = 0 defined: for all x > 0 undefined: for all x < 0, x = 0 


#80
May2805, 05:14 PM

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Wait! Did someone already do this?
[tex]0 \approx (1/\infty)[/tex] Not quite; but approximately. [tex]1/(1/\infty) \longrightarrow (1/1)/(1/\infty) \longrightarrow (\infty/1) * (1/1) \longrightarrow \infty*1 = \infty[/tex] Though [tex]0[/tex] is a little less than [tex]1/\infty[/tex]. So what value is it really? 


#81
May2805, 05:35 PM

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As far as math teachers are concerned, you may safely assume [tex]0^0 = 1[/tex] (/End deliberate handwaving mode). As a precaution, most documents or proofs that require its use (most that I've seen, anyway) will still explicitly state it as a useful interpretation before applying it.



#82
May2905, 05:01 AM

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#83
May2905, 05:25 AM

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hurkyl 0^x=0 if x is different from 0
0^m*0^m=0^0=1 or as i said before a^x=1\(a^(x)) so if a=x=0 this means 0^0= 1\(0^(0)) in this case 0^0 should b equal to one x^0= 1 even for x=0 


#84
May2905, 09:06 AM

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If you look very carefully you've just divided by zero: one of m or m is negative (i'm assuming integral exponent) 


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