Apparently this needs its own thread because many of you retards need to be educated.
0^0=1.
>inb4 some retard calls an expression involving no limits an "indeterminate form"
>inb4 another retard claims a function cannot be defined at a point where it's discontinuous
0^0 != 1
fuck you piggot
0^0 whats this?
>0^0 is a function
lel
0w0 what's this??
>any number to the 0 is one
>0 factorial is one
>this must be two
Pack it up boys, math is over, I solved the rest of it.
it's just a matter of definition/convention
it's not very useful in a general sense
it's akin to a programmer defining 0/0=0 or 0/0=1, you can use whatever works for the particular application
0^0- (0 to the power of 0) = 1 (which is a little different and needs a further more explanation. But more examples can include this.
3^3=18 (this means 3x3x3) different than 3x3. It means 3x3=6 and 6x3=18. Get it now?
3x3 !=6 : 3x3=9
Is there a zero-th root for values other than 1?
8 = \
x^0 is smooth
sqrt(0) is undefined?, 0 ^ 0.5 = 0
0^x
0^1=0
0^.5=0
0^.000000000000000000000000000001=0
0^0=1
what are you even using it for other than for mental masturbation?
consider the following.
[math]0^a=0 [/math] for all real a, not including 0.
[math]b^0=1 [/math] for all real b, not including 0.
if a or b were 0, we would get a contradiction.
therefore, [math]0^0 [/math] is undefined, but often defined as 1 for convenient working in real analysis.
...
i don't get it but there's some dodgy infinite series manipulation and maybe division by zero somewhere
QED
QED
To conclude: 0^0= WHATEVER YOU NEED IT TO BE
0^0 is indefinite because it's the same as saying 0/0. Anything else to the power of 0, like 3^0, will be 1 since that's the same as saying dividing a number with itself once. Stop being brainlets over simple statements.
what does piggot mean?
>0^0 != 1
>0^0! = 1
>0^1 = 1
>0 = 1
Waow
>inb4 another retard claims a function cannot be defined at a point where it's discontinuous
>I don't know what "indefinite" means: the post
>I don't understand exponents: the post
Back to middle school for you!
For negative integers a, b, a^b is the number of b-tuples from a set with a elements.
Speak English nerd.
>0^0 is indefinite because it's the same as saying 0/0.
Brainlet
Ever heard of a multiplicative identity? Guess what, x^1 is not the base case, x^0 is
0! = 1
0^1 = 0
10^3 = 10 x 10 x 10 = 1000
10^2 = 10 x 10 = 100
10^1 = 10
10^0 = 10 x [1 / (10)] = 1
10^-1 = 10 x [1 / (10 x 10)] = 1/10
10^-2 = 10 x [1 / (10 x 10 x 10)] = 1/100
Stop being a computer science brainlet. You're overcomplicating what exponents are, which is multiplying a base by itself n amount of times. To use your terminology, x^1 is the base case because you don't have to use recursion to determine the product. x^0 is 1 for the trivial fact that a number divided itself is 1.
So then is 0^2 also undefined?
Because it's like saying 0^3/0. And you can't divide by 0 so...
>0^3 = 0 x 0 x 0
>attempt: 0^3/0 = 0 x 0 x 0 / 0 = 0 x 0 = 0^2
>believing that 0/0 is equivalent to 1, allowing for that abortion of an "equivalent" statement
Genius! Try this one for size: 0^2 = 0 x 0. Dumbass. I seriously hope none of you are this fucking retarded, memeing about Rudin when a simple exponent evades your comprehension.
meant to reply not to myself, but to
I'm just using the logic you used to say 0^0 is undefined.
Are you saying that your reasoning is invalid? Gasp!
If you can't divide by 0, then your counterargument doesn't make any sense. 0^3/0 is not equal to 0^2 for the reason that you can't divide by 0. So no, 0^2 is still 0 but 0^3/0 is indeterminate.
You're not using any logic. 0/0 isn't equal to 1 so you can't cancel numbers like the way you chose to "demonstrate" that 0^3/0. It's not. 0/0 is still easily defined. Go back to middle school and get a refresher on exponents.
Well then 0^0 isn't 0/0
that's not what he's saying at all
He's calculating x^(n-1) as x^n/x.
It doesn't work when x is 0. Because that's not how you do it.
you're an idiot
Fuck man I'm too tired for this crap. Extremely sloppy explanations with typos. Let me try one more time.
Your argument is premised on the idea that 0/0 is equal to 1, allowing you to cancel out iterations of 0. For instance, you tried to make the statement that 0^3/0 is equal to 0^2 in order to demonstrate that x^1 cannot be the base case and that 0^0 is not equal to 0/0. The crux is of your reasoning is that such a definition would make any iteration of 0^n greater than 2, such as 0^2 or 0^10, indeterminate when it is not. Correct?
Rewrite 0^3 as 0 x 0 x 0. Group [0 x 0] together and [0/0] together. [0 x 0] is still 0, so you can pull that out of the expression. [0/0], however, is still indeterminate, so it's not equivalent to 1. Thus, 0^2 is determinate, but 0^3/0 is indeterminate, because 0^3/0 is not equivalent to 0^2.
Given 0^n, any n greater than 0 is determinate. Everything else is indeterminate. 0^0 is simply 0 divided by itself once, making it indeterminate. That is simply how exponents work by multiplying a base by itself n amount of times. You can still multiply fractions together if you consider that anything to a 0 or negative power is a base times itself, which is how exponents work anyway (10^-2 = 1/100). 0^1 is the base case because there are no other recursive iterations.
>He's calculating x^(n-1) as x^n/x.
No, YOU'RE calculating x^(n-1) as x^n/x. Like I said before, 0^3/0 is not equal to 0^2. Stop trying to disprove that 0^0 is = 0/0 by holding 0/0 to simultaneously both 1 and indeterminate. Your argument is self-defeating and only proves me further right. Maybe that's the simplest way I can beat your retardation out of your head.
0^0 is objectively undefined under normal axioms i.e. a space where you would do analysis and linear algebra
0^3 = 0 x 0 x 0 = 0
0^2 = 0 x 0 = 0
0^1 = 0
0^0 = 0/0 = indeterminate
0^-1 = 0 x [1/(0 x 0)] = 0 x [1 / 0] = 1 x [0/0] =
0^-2 = 0 x [1/(0 x 0 x 0)] = 0 x [1 / 0] = 1 x [0/0] = 0/0 = indeterminate
All of this shit still works because 0^3/0 isn't equivalent to 0^2, 0, etc. The system is logically consistent.
>Your argument is premised on the idea that 0/0 is equal to 1
Lol no. You really are too tired for this
Define "indeterminate". This should be good for a laugh
So for positive powers you just "factor" out a zero to get the next lower power of 0. OK I suppose but why do you suddenly divide by zero to get 0^0? That's not what you did for the other powers of 0. Seems like a convenient way for you to keep claiming it's undefined
0/0 = indeterminate
any number of 0s fits into 0, so it could be 1
You're the retard who thinks that 0^3/0 = 0^2. That requires canceling out 0/0 which is impossible. Go back to middle school.
Indeterminate... undefined... same thing... *sheepish laugh*
>So for positive powers you just "factor" out a zero to get the next lower power of 0.
No, you simply have one iteration less. There is no factoring. This is an algorithmic approach to exponential functions.
>OK I suppose but why do you suddenly divide by zero to get 0^0?
Because n^1 is the base case, meaning that it requires no action/no iteration/no recursion to get the product. n > 1 means multiplying n iterations of a base. n < 1 means dividing n iterations of a base. How do you think negative exponents work in a continuous, consistent fashion if you don't believe that n^0 is n/n?
>That's not what you did for the other powers of 0.
I did that for negative powers as well. Which are also undefined.
What?
>1^0 = 2^0
>1 = 2
what did mathfags mean by this?
>1*0 = 2*0
>1 = 2
>1+2 = 2+1
>1 = 2
I think that he's simply doing arithmetic operations with infinity which isn't properly defined.
The first part fails if you introduce the strict order > : every term of the sum alpha is > 0
therefore alpha > 0 and 0>0, a contradiction
The sequence 0^b where b->0 can't converge to 1, hence your hypothesis is false, newmathfag
...
>unlabeled axes
...
askamathematician.com
>read this whole page
>get to the "Matematician" part
>mfw
it's a product of a family indexed by the empty set, hence its value is the neutral element for the multiplicative law, 1
why?
Sorry
haha fukn rekt
ARE YOU EVEN TRYING!>>>?
He is the whitest looking black dude I've ever seen.
Anyway, can you amerifags explain this to me. Why the hell does this shit keep happening, when it seems like almost everyone is against it?
coalburners who are rebelling against daddy
At the end of the day, only the numerics matters.
0^0=1
OK so I define it as 0
>for negative integers
...
are you sure you're in the right thread
matlab confirmed for being inferior to mathematica
...so what is [math]\displaystyle\frac{0}{0^0}[/math]?
Mathematica is for math autists. MATLAB is for scientists i.e people that can get things done.
Good luck with for example Taylor series if you don't define 0^0 as one
Nope. Since a^b is defined as e^(b*log(a)), here what we have is by definition e^(0*log(0)), and log(0) is undefined. So the whole thing has to be undefined.
Now, if we take the LIMIT of this, ie Lim(x->0)(x^x), we find that it's 1, but don't confuse that with equality.
(Sorry, typing this on my phone so no TeX)
Lim(x->0)(0^x)
where is your god now
I'm still pretty sure that's undefined, for the same reason: you still have Lim(x->0)(e^(x*log(0))), and log(0) is still undefined.
weak b8 lad
>>>>>MATLAB
AHAHAHAhAHAHAHAHAHAHAAH
Also Python gives 0**0=1.
Can someone explain to me why 0!=1 ?
I never understood why, only that it makes series more convenient to calculate.
Top kek
>2014b
user, you're 3 years behind
What did he mean by this?
>it's akin to a programmer defining 0/0=0 or 0/0=1
it's nothing akin to that, Cletus
Let this Taylor series make it clear. Clearly 0^0 has to be 1 if you put x=0. Also 0! must be 1 becouse the first term is one.
Also an empty set can be arranged 0! ways. But empty set can arranged only 1 way. So 0!=1
Haven't got time to update. Maybe soon.
>Python
trash
then explain why mathematica which is for "math autists" consider 0^0 to be undefined whereas programming/engineering shit like python and matlab define 0^0 to be 1
alpha is divergent you dum nigher
Sry pal no one defines a^b that way
compsci brainlet detected
>a^b is defined as e^(b*log(a))
nice circular definition
This nigga gets it.
I can see what you mean with the empty set. But I don't see how you could prove that.
Don't give the autist trying to force the word he made up attention, please.
it doesn't mean anything, i think it's supposed to be an insult like brainlet or something, but it's only like one guy who uses it try to force it as a Veeky Forums meme
e^(b*log(a)) is defined as c^((b*log(a))*log(e))
Check chapter 19 of Spivak for a full explanation, but basically he defines [math]log(x) = \int_{1}^{x}\frac{1}{t}dt[/math], which in turn defines [math]e^{x}[/math] as its inverse function. Then all other exponentials are defined in terms of that.
you what. 3^3 is 27 and 3x3 is 9, are you preschooler or what
Multiplying nothing by a factor of nothing equals nothing. Nothing is 1. Logic.
It can be expressed as such.
n^x=(n^(x+1))/n
So if n=x=0 then
0^0=(0^(0+1))/0=0/0
EZ - PZ