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10254963 No.10254963 [Reply] [Original] [archived.moe]

[math] 0^0 = 1 [/math] and here is my quick proof:

[eqn] \displaystyle \lim_{x \rightarrow 0} x^x = \lim_{x \rightarrow 0} e^{x\ln{(x)}} = e^{\lim_{x \rightarrow 0} x\ln{(x)}} = \Bigg( \lim_{x \rightarrow 0} x \Bigg)\Bigg(\lim_{x \rightarrow 0} x\ln{(x)} \Bigg) [/eqn]

so by squeeze theorem:

[eqn] \displaystyle \Bigg( \lim_{x \rightarrow 0} x \Bigg)\Bigg(\lim_{x \rightarrow 0} x\ln{(x)} \Bigg) = 0 \implies \lim_{x \rightarrow 0} x^x = e^0 = 1 [/eqn]

Q.E.D :)

>> No.10254974

Your proof shows that [eqn]\lim_{x \rightarrow 0} x^x = 1[/eqn]. That part is not not disputed, I think. But that does not mean that 0^0 = 1.

It is, mind you. But it doesn't follow from this limit.

>> No.10254986
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10254986

Math is the shittiest meme to ever come out of humanity

>> No.10255016

>>10254963
Okay, but where's your proof that exponentiation is continuous.

>> No.10255024

>>10254974
how come it doesnt??
>>10255016
why do I need to proof that??

>> No.10255048

>>10255024
>how come it doesnt??
Why would it?

>> No.10255051

>>10255024
x^x approaches 1 on the real axis, but not anywhere on the complex plane

>> No.10255052

>>10254974
>He doesn't understand limits

>> No.10255055

>>10255052
Do tell.

>> No.10255061

>>10255055
im not this anon>>10255052
[math] \displaystyle \lim_{x \to c} f(x) = a [/math] doesnt imply that [math] f(c) = a [/math]

thats why even though sin(x)/x approaches 1 as x approaches 0, sin(0)/0 isnt equal to one, its undefined.

>> No.10255063

>>10255061
Indeed.

>> No.10255095
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10255095

>>10255061
>partial sum definition of infinite series ignores this
and that's why its the worst way to define infinite series

>> No.10255098

>>10254963
Isn't a^n just 1×a×...×a n times?
So a^0 = 1 by defn
Also 0^0 is then 1 without ambiguity.

Just like 0! = 1 by def

>> No.10255101 [DELETED] 
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10255101

>>10254963
Wolframed

>> No.10255103

>>10255098
Not quite.
[math]a^0=1[/math] and [math]a^n=aa^{n-1}[/math] define exponentiation.

>> No.10255407

>>10255103
[math]\frac{a^n}{a^n}=a^{n-n}=a^0=1[/math]

>> No.10255417

>>10255407
Yeah, but if you just define [math]a^ba^c=a^{b+c}[/math] you still have to define [math]a^1=a[/math], which is really short but still gives longer proofs.

>> No.10255437

>>10254963

here is a definition of the exponent:

a(n,0)=1
a(n,k+1)=n*a(n,k)

> every natural number is 0 or of the form k+1 for some k
> this always halts for any input because the second argument decreases
> the proof that a(0,0)=1 is obvious

This is why sometimes 0^0 = 1. Attempting to prove this with calculus is beyond retarded.

>> No.10255600

>>10255437
>its 1 by definition
faggot

>> No.10255613

>>10255024
are you retarded? by definition the function doesn't have to be defined in the exact point.

>> No.10255632

>>10254974
based and redpilled answer.

Here is a real proof that 0^0=1.
def: for cardinal numbers A, B, we define A^B as the number of functions from a set of size A to a set of size B.
Since there is precisely one function from the empty set to itself, 0^0=1

>> No.10255634

>>10254963
whoa dude! You're a genius!

Doing a similar thing you can prove that:
[math]\frac{1}{0}=infinity[/math]

>> No.10255637

>>10255632
ah yes, its obvious that 0^0 is 1 if you only work with the nonnegative integers

>>
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