Geoff wrote:Why is X raised to the 0 power = 1 and not 0 ?
Notice that 0^0 is a discontinuity of the function f(x,y) = x^y, because no matter what number you assign to 0^0, you can't make x^y continuous at (0,0), since the limit along the line x=0 is 0, and the limit along the line y=0 is 1.
The discussion of 0^0 is very old. Euler argues for 0^0 = 1 since a^0 = 1 for a not equal to 0 . The controversy raged throughout the nineteenth century, but was mainly conducted in the pages of the lesser journals: Grunert's Archiv and Schlomilch's Zeitshrift.
Sounds like a calculus limit problem with indeterminate forms. You use logarithms to solve the 0^0 ones.I have a function and I have to know if it equals zero or not and it contains 0^0, and if that is 1 then the function in some cases makes more sense in others in makes not sense at all if it does equal 1.
The controversy raged throughout the nineteenth century, but was mainly conducted in the pages of the lesser journals: Grunert's Archiv and Schlomilch's Zeitshrift.
ThomasGR wrote:Very well so far, 0^0=1, but what happened with 0^x?
I mean 0^1=0, 0^2=0, etc.
and when we get 0^0, than suddenly we have a value 1?
ThomasGR wrote:Does it go also for negative numbers?
Episcopus wrote: I suppose I can't fathom how this would be of any use in real life. You're better off doing weights.
Emma_85 wrote:I think his proof is quite a good one actually, Episcopus. And now I know more about 0^0 than my maths teacher .
classicalclarinet wrote:just like Latin
Eureka wrote:I'm not sure what you mean by this, but in these two cases it doesn't matter whether 0 is approached from above or bellow. (I presume that's what you're talking about.)
Teachers, *pech* my primary school teachers used to tell me that a/0=0. I tried arguing with them, but that never works. Teachers know everything. :Rolling Eyes: :Smile:
ThomasGR wrote:I think there is also a problem with negative numbers.
4*0^(-4)
=
4/(0^4)
4/(0)
=
0?
ThomasGR wrote:Maybe your teacher was correct, after all?
ThomasGR wrote:I think there is also a problem with negative numbers.
Eureka wrote:hmmmmmm… 0^(-4):
x^(-4) as x-->0 = infinity
0^x as x-->-4 = 0
Eureka wrote:…Still, for most purposes, a/0=infinity.
Democritus wrote:Maybe what you meant to say was, as b approaches zero, the expression a/b increases to infinity.
Democritus wrote:Eureka wrote:hmmmmmm… 0^(-4):
x^(-4) as x-->0 = infinity
0^x as x-->-4 = 0
I think the problem is that the expression "0^x as x-->-4" is itself invalid, because when x is near -4 it's negative, and that means we have zero in the denominator. In other words, 0^x is undefined, for all x<0.
Users browsing this forum: No registered users and 5 guests