/sci/ - Science & Math » Thread #12390645

180KiB, 1280x720, math.jpg

View Same Google iqdb SauceNAO

If you were smart you would know the anser of zero to the power of zero

Anonymous Fri 27 Nov 02:28:16 2020 No.12390645 View Reply Original Report

Quoted By: >>12390723 >>12390757 >>12390829 >>12391108 >>12391110 >>12391265 >>12394126

0^0 = ?

What's the answer and tell me why.

Anonymous

Anonymous Fri 27 Nov 2020 02:42:16 No.12390678 Report

Quoted By: >>12391074

1
https://www.google.com/search?q=y%3Dx%5Ex&oq=y%3Dx%5Ex

Anonymous

Anonymous Fri 27 Nov 2020 02:49:47 No.12390688 Report

Quoted By: >>12390717

0, the square root of 0 is 0. If you staking the square roots of square roots of 0 infinitely many time you still have 0. 0^(1/?)=0^0=0

Anonymous

Anonymous Fri 27 Nov 2020 03:06:19 No.12390717 Report

Quoted By: >>12390741

>>12390688
>square root
wtf does 2 have to do with this

Anonymous

Anonymous Fri 27 Nov 2020 03:09:36 No.12390723 Report

Quoted By:

>>12390645
Is it, banana?

Anonymous

Anonymous Fri 27 Nov 2020 03:18:13 No.12390741 Report

Quoted By: >>12391239

>>12390717
Works with any root. It was just pick arbitrarily to get the point across. Take the nth root infinitely many times the result is still 0. 0^0=0^(1/?)=0

Anonymous

Anonymous Fri 27 Nov 2020 03:25:06 No.12390750 Report

Quoted By: >>12390845 >>12391219

$\begin{align*} x^n &= x^{n+1}/x \\ ... \\ x^2 &= x^3/x \\ x^1 &= x^2/x \\ x^0 &= x^1/x\end{align*}$

So $0^0 = 0/0$ which is undefined.

Anonymous

Anonymous Fri 27 Nov 2020 03:27:11 No.12390757 Report

Quoted By:

>>12390645
The answer I know is that you have to intend the powers of numbers not as the numbers multiplied by themselves n times but 1 multiplied by the number multiplied by itself n times, so when the exponent is 0 you remain with just 1

Anonymous

Anonymous Fri 27 Nov 2020 03:56:04 No.12390829 Report

Quoted By:

>>12390645
0^0=0^(1-1)=0^1*0^(-1)=0*1/0=0/0 and we can't divide stuff with 0 (but technically, u can get different answers: u can argue it's 1 with the use of limits, or it's 0 since 0/x=0, but generally we say u can't do that)

Anonymous

Anonymous Fri 27 Nov 2020 04:01:15 No.12390845 Report

Quoted By: >>12390924

>>12390750
Uhhh...
x^0=x^1/x
x^0=x/x
x^0=1
0^0=1

Anonymous

Anonymous Fri 27 Nov 2020 04:23:37 No.12390924 Report

Quoted By:

>>12390845
>x/x = 1
not true if x is 0

Anonymous

Anonymous Fri 27 Nov 2020 05:19:25 No.12391074 Report

Quoted By: >>12391106

>>12390678
/thread

Anonymous

Anonymous Fri 27 Nov 2020 05:26:52 No.12391098 Report

Quoted By: >>12391106 >>12392635 >>12392910

1

The empty product (multiplying zero numbers together) is by definition the multiplicative identity, 1.

This is just like the empty sum is by definition the additive identity, 0.

Anonymous

Anonymous Fri 27 Nov 2020 05:30:36 No.12391106 Report

Quoted By: >>12391140

>>12391074
Not thread because you can't ever select exactly 0 on that graph.

>>12391098
It all depends on the definition. In general it is undefined but in certain situations it is useful to define it as 1. Such as the binomial theorem where is makes the equation simpler.

Anonymous

Anonymous Fri 27 Nov 2020 05:31:21 No.12391108 Report

Quoted By:

>>12390645
Tell you what, OP:
You show me a PRACTICAL, REAL-WORLD APPLICATION for zero raised to the power of zero, and I'll answer your question.

Anonymous

Anonymous Fri 27 Nov 2020 05:32:17 No.12391110 Report

Quoted By: >>12391225

>>12390645
it is indeterminate. that is the answer. There is no defined value. f(x,y)=x^y has a discontinuity at (0,0). Just look at what happens when you approach 0^0 by some really rapidly approaching function to the power of a really slowly approaching function, which is completely legal.

Anonymous

Anonymous Fri 27 Nov 2020 05:42:12 No.12391140 Report

Quoted By:

>>12391106

What is generaly undefined? The empty prpduct or 0^0?

I've never seen any context where the empty product was not defined as the multiplicative identity. Can you give me an example?

As for 0^0, when n is a whole number, x^n is usually defined as x times itself n times. So the "usual" definition of 0^0 would make it 1.

I do agree that this is only a matter of which definition is being used, but as far as I know, this is the "usual" definition.

Anonymous

Anonymous Fri 27 Nov 2020 06:13:24 No.12391219 Report

Quoted By:

>>12390750

Why do you assume that x^n = x^(n+1) / x?

It's obviously not true when x = 0.
0^2 is not equal to 0^3 / 0.

Anonymous

Anonymous Fri 27 Nov 2020 06:19:34 No.12391225 Report

Quoted By:

>>12391110
I'm not quite sure I follow your argument. So what if f(x, y) = x^y is discontinuous at (0, 0)?
Doesn't that just imply that you can't use limits to find f(0, 0)?

Anonymous

Anonymous Fri 27 Nov 2020 06:28:00 No.12391239 Report

Quoted By:

>>12390741
lim_{x -> a} f(x) is only equal to f(a) if f is continuous at a.

Why do you assume that 0^x is continuous at x=0?

Anonymous

Anonymous Fri 27 Nov 2020 06:40:31 No.12391265 Report

Quoted By:

>>12390645
OP why would I want to know this?

Anonymous

Anonymous Fri 27 Nov 2020 10:07:31 No.12391635 Report

Quoted By:

It's either 0, 1, or undefined. But if you're going to say 0 or 1 you need an actual good reason, which depends on the context (which field you're working in).

Set theory clearly says 0^0 is the set of all functions from the empty set to the empty set, which turns out to be 1. This seems the most conclusive answer seeing as this is based purely on the axioms of ZF, which is consistent with most of mathematics.

Anonymous

View Same Google iqdb SauceNAO lil-bo-weep-19980102-allfamous.jpg, 59KiB, 300x300

Anonymous Fri 27 Nov 2020 17:32:59 No.12392635 Report

Quoted By:

>>12391098
>The empty product
well noticed

Anonymous

Anonymous Fri 27 Nov 2020 19:11:19 No.12392910 Report

Quoted By:

>>12391098
This is the right answer.

Anonymous

Anonymous Sat 28 Nov 2020 01:08:29 No.12394126 Report

Quoted By:

>>12390645
Indeterminate, anon.
Do you mean:
$\lim_{x\to 0} x^0$ ?
Then it's 1.
Do you mean:
$\lim_{x\to 0} 0^x$ ?
Then it's 0
Do you mean:
$\lim_{x\to 0} x^x$ ?
Then it's 1.
It depends on how you get there.

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches