Infinity to the 0 power, or Zero to the power infinity

[JEDI]

What do you get?

i say 1 and 0, respectively.

anything to the 0 power is 1.

0 x anything = 0.

Right/wrong?

 

[Lorn]

NO!

 

[deftron]

0 times anything is 0
Anything times 0 is 0
Anything to the 0 power is 1




Go back to kindygarden

 

[EyeMWing]

Originally posted by: deftron
0 times anything is 0
Anything times 0 is 0


Go back to kindygarden

0^inf, or 0*0*0*0*0... is indeed 0.
inf^0 has nothing to do with multiplying anything times 0. It's 1 by definition.

Go back to the 6th grade.

 

[chuckywang]

Learn limits....they are good for you.

 

[Bv3]

Infinity is not a number so "Zero to the power infinity" is a nonsensical statement and has no answer. It's like asking "What do you get if you draw a four-sided triangle?" or "What happens when an unstoppable force meets an immovable object?"

 

[deftron]

Originally posted by: EyeMWing

Originally posted by: deftron
0 times anything is 0
Anything times 0 is 0


Go back to kindygarden

0^inf, or 0*0*0*0*0... is indeed 0.
inf^0 has nothing to do with multiplying anything times 0. It's 1 by definition.

Go back to the 6th grade.

Why'd you delete a line out?

 

[pray4mojo]

If I remembered my calculus I'd tell you, but I don't.

 

[stan394]

infinity is not a number

 

[IAteYourMother]

0* infinity=infinity

 

[chuckywang]

Originally posted by: EyeMWing

Originally posted by: deftron
0 times anything is 0
Anything times 0 is 0


Go back to kindygarden

0^inf, or 0*0*0*0*0... is indeed 0.
inf^0 has nothing to do with multiplying anything times 0. It's 1 by definition.

Go back to the 6th grade.

You're right about 0^inf always being 0, but inf^0 is not always 1.

Consider (2^n)^(1/n) as n->infinity.

It converges to 2.

 

[chuckywang]

Originally posted by: IAteYourMother
0* infinity=infinity

uh...no.

 

[EyeMWing]

Originally posted by: chuckywang

Originally posted by: EyeMWing

Originally posted by: deftron
0 times anything is 0
Anything times 0 is 0


Go back to kindygarden

0^inf, or 0*0*0*0*0... is indeed 0.
inf^0 has nothing to do with multiplying anything times 0. It's 1 by definition.

Go back to the 6th grade.

You're right about 0^inf always being 0, but inf^0 is not always 1.

Consider (2^n)^(1/n) as n->infinity.

It converges to 2.

Oh, now I remember. We actually had a 3-day argument about this in my calc class. It was a BEAUTIFUL time killer. I believe we never actually reconciled the differences between the definition and the limits.

 

[chuckywang]

Originally posted by: EyeMWing

Originally posted by: chuckywang

Originally posted by: EyeMWing

Originally posted by: deftron
0 times anything is 0
Anything times 0 is 0


Go back to kindygarden

0^inf, or 0*0*0*0*0... is indeed 0.
inf^0 has nothing to do with multiplying anything times 0. It's 1 by definition.

Go back to the 6th grade.

You're right about 0^inf always being 0, but inf^0 is not always 1.

Consider (2^n)^(1/n) as n->infinity.

It converges to 2.

Oh, now I remember. We actually had a 3-day argument about this in my calc class. It was a BEAUTIFUL time killer. I believe we never actually reconciled the differences between the definition and the limits.

There's really no argument. inf^0 is one of the seven indeterminate forms.

 

[eLiu]

Originally posted by: EyeMWing

Originally posted by: deftron
0 times anything is 0
Anything times 0 is 0


Go back to kindygarden

0^inf, or 0*0*0*0*0... is indeed 0.
inf^0 has nothing to do with multiplying anything times 0. It's 1 by definition.

Go back to the 6th grade.

infinity is not a number...it's a concept. In analysis it exists in limit form only (or more rigorously, as the supremum of the reals). Wander in to complex #s (and beyond) and infinity becomes an even more fuzzy idea.

Anyway...inf^0 in the limit can be ANYTHING.

Here:
lim x->inf of exp(A*x)^(1/x).
= lim x->inf of exp(1/x * ln(exp(A*x))
= lim x->inf of exp(A).
= exp(A).

Now...A can be any real number, so we can achieve any positive real number in this way (stick a negative sign out front to hit any negative real #). As A->-infinity, the above limit goes to 0.

edit: the above argument is in no way rigorous but it illustrates the point

also, 0^inf is indeed 0. (at least in the real# system)