I am confused about 0^0

[The Jugglernaut No.5811300]

I am a student in a math class and this problem brainfucks the hell out of me. Can anyone help me?

[Anonymous No.5811305]

>>5811300
0^0=1
It's the convention.

Some people don't write it like that but call it "indeterminate form"

[Anonymous No.5811308]

><span class="math">\sqrt[4]{4}=4[/spoiler]
what?

[Anonymous No.5811310]

>0^(1/0)
wat

[Anonymous No.5811316]

I have a semi-related question. Is it pure convention that x^0=0, or is there some underlying mathematics which make it true? If the latter, can someone explain to me?

[Anonymous No.5811318]

http://www.wolframalpha.com/input/?i=0%5E0

[Anonymous No.5811323]

>>5811316
x^0=!0
x^0=1

[Anonymous No.5811327]

>>5811323
...Huh?

[Anonymous No.5811330]

>>5811327
x^0=1, not 0
where did you even get the zero from, it's wrong

[Anonymous No.5811332]

>>5811316
For natural numbers, m^n equals the number of ways of making lists of length n out of m objects (repetition allowed).

And in general, you want to be able to make x^0 = 1 in order to write power series nicely.

[oilerscup20014 !S0ooHAHDTU No.5811337]

lim[f(x)->0, g(x)->0]f(x)^g(x) is indeterminate. It can equal ANY value depending on the function.

[Anonymous No.5811338]

testing: <span class="math">0^0 = 1[/spoiler]

[Anonymous No.5811347]

x^0 = 1
1 = 0.999999...
Therefore, x^0 = 0.99999999...

[Anonymous No.5811353]

There is a more conceptual explanation than simple considerations of continuity.
If x,y are nonnegative numbers, then x^y is the cardinality of the set of functions from a set with y elements to a set with x elements.
0^0 is the number of functions from the empty set to itself. One such function (its graph is the empty set) exists.

[Anonymous No.5811355]

I see what you're doing, but the first one isn't really pattern matching. The base doesn't matter in those examples, and then you fail to realize that the number of the root is in the denominator of the exponent, making it a simple divide by zero error.

Trying to prove this by induction isn't gonna help you much, because 0 is exceptional as a argument and as a number of arguments. Any number of basic functions will freak out at 0, and your pattern-matching attempts would lead to the wrong answer every time.

0^x is undefined for negative x, so making it equal to one does not mess with the continuity of the function. However, x^0 is defined for all real x and=1, so making 0^0 preserves the continuity of the function. So we really don't miss out on much by declaring 0^0 to be 1 by convention.

[Anonymous No.5811384]

>>5811300
<span class="math">0^{0} = \frac{0}{0}[/spoiler] which is indeterminate

just as

<span class="math">1^{0} = \frac{0}{0} = 1 [/spoiler]

[Anonymous No.5811388]

Yeah, nah.

[Anonymous No.5811418]

>>5811388

I don't think anyone here is dumb enough to fall for this one.