Question

i need a verification on 0^0

Answer 1

okay

Answer 2

nice profile pic. :)

Answer 3

lol

Answer 4

same to you

Answer 5

its undefinedddddddddd

Answer 6

thanks. :)

Answer 7

correct un defined

Answer 8

i like the new pic better lol

Answer 9

same here

Answer 10

i need to get to next class the verification for correcting the teachers error

Answer 11

No its defined as 1

Answer 12

a^0=1 for all real a not equal to 0
a^0 is undefined when a=0
@alekos

Answer 13

No, that's incorrect. It is definitely mathematically defined as 1

Answer 14

http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

Answer 15

@alekos

Answer 16

Yes, that's wonderful and I know all of this, but in the end it's been defined mathematically and the modern convention is to define 0^0=1, for good reason.
Why? Because it lets us manipulate exponentials without adding special cases.
Donald Knuth set things straight in 1992, Donald Knuth is a professor Emeritus at Stanford University.

Answer 17

https://www.math.hmc.edu/funfacts/ffiles/10005.3-5.shtml
@alekos

Answer 18

I think it is more of a convenience to define \(0^0\) to be 1. The limit of \(x^y\) does not exist as (x,y) to (0,0). Different direction of approach will yield different results.

Answer 19

there's a difference between pure maths and what's regarded as a definition.
this is one of those cases.
kind of similar to 0!=1

Answer 20

You could argue 0!=1 as (n-1)!=n!/n so 0!=1!/1=1. For 0^0 you can't really make such argument and it is for convenience only IMO.

Answer 21

Yes, I don't disagree with you.
But that is the way it has been defined, and it has to be treated in that regard.