/sci/ - Science & Math » Thread #11484354

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Anonymous Thu 19 Mar 22:17:51 2020 No.11484354 View Reply Original Report

Quoted By: >>11484398 >>11484418 >>11484526 >>11484620 >>11484706 >>11484735 >>11484908 >>11484915 >>11484938 >>11485190 >>11485199 >>11487759 >>11487780 >>11487815 >>11488017

0^0 = 1 makes no mathematical sense at all, how do you even prove this

Anonymous

Anonymous Thu 19 Mar 2020 22:30:48 No.11484398 Report

Quoted By:

>>11484354
It can be explained combinatorially, set theoretically or by insisting that certain $\mathbb{R}\rightarrow \mathbb{R}$ functions ought to be continuous. But the real reason is just that it makes some theorems easier to state.

Anonymous

Anonymous Thu 19 Mar 2020 22:33:48 No.11484405 Report

Quoted By: >>11484410 >>11484421 >>11484567

$x5 = 1 \times x \times x \times x \times x \times x$
$x3 = 1 \times x \times x \times x$
$x1 = 1 \times x$
$x0 = 1$

Anonymous

Anonymous Thu 19 Mar 2020 22:35:14 No.11484410 Report

Quoted By: >>11484421 >>11484425 >>11484567

>>11484405
fuck
$x^5 = 1 \times x \times x \times x \times x \times x$
$x^3 = 1 \times x \times x \times x$
$x^1 = 1 \times x$
$x^0 = 1$

I'm a mess with this shit

Anonymous

Anonymous Thu 19 Mar 2020 22:39:25 No.11484418 Report

Quoted By: >>11484938

>>11484354
Take a look at this, even I could understand it and I'm retarded

https://www.youtube.com/watch?v=r0_mi8ngNnM

Anonymous

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Anonymous Thu 19 Mar 2020 22:40:41 No.11484421 Report

Quoted By: >>11484439 >>11484445 >>11484507 >>11484567

>>11484405
>>11484410
$0^5 = 0$

$0^4 = 0$

$0^3 = 0$

$0^2 = 0$

$0^1 = 0$

$0^0 = 0$

Anonymous

Anonymous Thu 19 Mar 2020 22:41:28 No.11484425 Report

Quoted By: >>11484688

>>11484410
But you could still prove this without the 1, right?
i mean, you could still say
x^5=x?x?x?x?x
right?

Anonymous

Anonymous Thu 19 Mar 2020 22:42:35 No.11484429 Report

Quoted By:

test

$x^5=x*x*x*x*x$

Anonymous

Anonymous Thu 19 Mar 2020 22:45:52 No.11484439 Report

Quoted By: >>11484567

>>11484421
$5^0=1$
$4^0=1$
$3^0=1$
$2^0=1$
$1^0=1$
$0^0=1?$

Anonymous

Anonymous Thu 19 Mar 2020 22:46:41 No.11484445 Report

Quoted By:

>>11484421
$\emptyset^\emptyset \neq \emptyset$

Anonymous

Anonymous Thu 19 Mar 2020 23:15:28 No.11484507 Report

Quoted By: >>11484573

>>11484421
$0^x=1$ when x=0, but zero for all other powers, in other words, the Krönigger delta

Anonymous

Anonymous Thu 19 Mar 2020 23:20:42 No.11484526 Report

Quoted By: >>11484551 >>11484567

>>11484354
x^0 = x/x
x^-1 = (x/x)/x
x^-2 = ((x/x)/x)/x
0/0 = 1 so 0^0 = 1

Anonymous

Anonymous Thu 19 Mar 2020 23:32:30 No.11484551 Report

Quoted By:

>>11484526
>0/0=1
Try again retard

Anonymous

Anonymous Thu 19 Mar 2020 23:35:07 No.11484567 Report

Quoted By: >>11484581 >>11484938

>>11484405
not a proof.

>>11484410
not a proof.

>>11484421
not a proof.

>>11484439
not a proof.

>>11484526
not a proof.

Anonymous

Anonymous Thu 19 Mar 2020 23:37:54 No.11484573 Report

Quoted By:

>>11484507
>Krönigger delta
lol. But the delta takes two parameters and $x^y$ is not equal to the delta in the general case.

Anonymous

Anonymous Thu 19 Mar 2020 23:40:44 No.11484578 Report

Quoted By: >>11484725

If $0^0=1$ , then $0^{-1}=1$ = $\infty$

Anonymous

Anonymous Thu 19 Mar 2020 23:41:54 No.11484581 Report

Quoted By: >>11484594

>>11484567
$cring$

Anonymous

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Anonymous Thu 19 Mar 2020 23:42:10 No.11484583 Report

Quoted By:

$0_{penis}$

Anonymous

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Anonymous Thu 19 Mar 2020 23:45:27 No.11484594 Report

Quoted By:

>>11484581

Anonymous

Anonymous Thu 19 Mar 2020 23:53:41 No.11484620 Report

Quoted By: >>11484995

>>11484354
You don't.
$\lim\limits_{x \to 0} x^x(\to 0^0)=1$ but $\lim\limits_{x \to 0} x^{\frac{1}{\ln x}}(\to 0^0)=e$

Anonymous

Anonymous Fri 20 Mar 2020 00:08:36 No.11484655 Report

Quoted By:

X^0 = X^N / X^N

Anonymous

Anonymous Fri 20 Mar 2020 00:24:09 No.11484688 Report

Quoted By:

>>11484425
But it wouldn't work for x0, as you'd have nothing in the output. If you understand exponentials as 1 multiplied n times (times as in repetition, not the multiplication operation) by a number x, that would mean x0 would be 1 multiplied by nothing

Anonymous

Anonymous Fri 20 Mar 2020 00:30:11 No.11484706 Report

Quoted By: >>11485690

>>11484354
http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJ4XngiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjEwMDB9XQ--

Anonymous

Anonymous Fri 20 Mar 2020 00:31:05 No.11484712 Report

Quoted By:

Dont be retarded. n^k is the number of maps from a k-element set to the n-element set. How many maps are there from the empty set to itself? Exactly one, the empty binary relation (which is a map). Therefore, 0^0 =1.

Anonymous

Anonymous Fri 20 Mar 2020 00:40:12 No.11484725 Report

Quoted By:

>>11484578
Big if true.

Anonymous

Anonymous Fri 20 Mar 2020 00:42:48 No.11484735 Report

Quoted By: >>11484898

>>11484354
$0^0$ is undefined.

Anonymous

Anonymous Fri 20 Mar 2020 02:11:32 No.11484898 Report

Quoted By: >>11484930

>>11484735
Any number raised to 0 is equal to 1. 0 is a number, therefore $0^0=1$

Anonymous

Anonymous Fri 20 Mar 2020 02:19:44 No.11484908 Report

Quoted By:

>>11484354
You don't, even mathematicians today don't agree on the definition

Anonymous

Anonymous Fri 20 Mar 2020 02:23:17 No.11484915 Report

Quoted By:

>>11484354
Why don't you just try learning math while pretending it's not, and then when it eventually makes a difference you'll know why.

Anonymous

Anonymous Fri 20 Mar 2020 02:31:12 No.11484930 Report

Quoted By: >>11485203

>>11484898
>Any *nonzero number raised to 0 is equal to 1.

Anonymous

Anonymous Fri 20 Mar 2020 02:33:32 No.11484938 Report

Quoted By:

>>11484567
This.
>>11484354
>makes no mathematical sense at all, how do you even prove this
You don't have to prove something that is not defined.
X^0 is well defined for any X that is not 0; X^k/X^k is both 1 and by exponent rule it is X^0 for any k.
>>11484418
>I could understand it
A limit is not the value of the function at the point. Undergrad tier calculus.
Fuck off back to high school all of you peanut brain retards

Anonymous

Anonymous Fri 20 Mar 2020 03:00:43 No.11484995 Report

Quoted By: >>11485729 >>11486808

>>11484620
x^x is continuous at x=0, so this is not an argument.

Anonymous

Anonymous Fri 20 Mar 2020 03:02:28 No.11484999 Report

Quoted By:

it's the empty product retards

Anonymous

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Anonymous Fri 20 Mar 2020 03:38:08 No.11485099 Report

Quoted By:

>0^0=0
>0^0=1
Since both are equally valid, we have 0^0=(0+1)/2=0.5. QED

Anonymous

Anonymous Fri 20 Mar 2020 04:20:57 No.11485190 Report

Quoted By: >>11485302

>>11484354
Serious question here.
After reading this post I googled a bit. Most websites define x^n as follows :

x^n = 1 * (x*x*.....*x) [n times]

Using this interpretation we can see why 0^0 is equal to 1. But how correct or true is this interpretation wrt to formal mathematics?

Anonymous

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Anonymous Fri 20 Mar 2020 04:22:51 No.11485199 Report

Quoted By:

>>11484354
Dosent that invalidate prime numbers?
there is one even prime number.
0 is only a place holder

Anonymous

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Anonymous Fri 20 Mar 2020 04:24:38 No.11485203 Report

Quoted By:

>>11484930
and a hampster born in a fishbowl is a fish.

Anonymous

Anonymous Fri 20 Mar 2020 04:58:50 No.11485302 Report

Quoted By:

>>11485190
Very. It's how exponentiation works out in groups, rings, and fields, and how the functions-from-set-to-set interpretation behaves (this is not a coincidence).

Anonymous

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Anonymous Fri 20 Mar 2020 08:52:40 No.11485690 Report

Quoted By: >>11488069

>>11484706
beautiful
basic calculus should be teached in middle school (using graphics plots) this proof in basic calculus it is not difficult at all to teach it to a child

Anonymous

Anonymous Fri 20 Mar 2020 09:27:26 No.11485729 Report

Quoted By:

>>11484995
What? I demonstrated how $0^0$ can be evaluated to be both equal to 1 and e and so it has no clear value. $x^x$ is not defined at 0 and therefore cannot be continuous there. 0 is however a limit point of its domain and we can take the limit of the function, which is equal to 1. If we consider the function $x^{\frac{1}{\ln x}}$ we can see that it is also not defined at 0 but 0 is a limit point of its domain and we can take the limit. The limit is also of the form $0^0$ and is equal to $e$

Anonymous

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Anonymous Fri 20 Mar 2020 18:37:33 No.11486768 Report

Quoted By:

How do I get started in game theory?
Is there anything I should consider watching?
Is there anything I should avoid?

Anonymous

Anonymous Fri 20 Mar 2020 18:49:06 No.11486808 Report

Quoted By:

>>11484995
Prove it.
>continuous = limit exists, value is equal to the limit
Literally assuming the answer

Anonymous

Anonymous Sat 21 Mar 2020 00:05:14 No.11487563 Report

Quoted By:

we approach $0^0$ by defining $x^x$ as a function which will be approach at 0. we call its limit L
$\lim\limits_{x \to 0} x^x = L$
$\lim\limits_{x \to 0} \ln{ x^x } = \ln L$
$\lim\limits_{x \to 0} x \times \ln x = \ln L$
$x \times \ln x = \frac{\ln x}{ \frac{1}{x} }$
Using L'Hospital, as it approaches $\frac{0}{0}$ :
$\frac{\ln x}{ \frac{1}{x} } \to \frac{\frac{1}{x}}{\frac{1}{x^2}} = -x$
thus
$\lim\limits_{x \to 0} x \times \ln x = \lim\limits_{x \to 0} -x = \ln L$
$\lim\limits_{x \to 0} e^{-x} = L$
$e^0 = L = 1$

Anonymous

Anonymous Sat 21 Mar 2020 01:19:49 No.11487759 Report

Quoted By:

>>11484354
>what is indeterminate?

Anonymous

Anonymous Sat 21 Mar 2020 01:23:59 No.11487780 Report

Quoted By:

>>11484354
It's convenient in a number of cases.

Anonymous

Anonymous Sat 21 Mar 2020 01:34:40 No.11487815 Report

Quoted By: >>11487883 >>11488069 >>11488087

>>11484354
you can't
but check this out

0.9^0.9 = 0.9095...
0.8^0.8 = 0.8365...
0.7^0.7 = 0.7790...
0.6^0.6 = 0.7360...
0.5^0.5 = 0.7071...
0.4^0.4 = 0.6931...
0.3^0.3 = 0.6968...
0.2^0.2 = 0.7247...
0.1^0.1 = 0.7943...

0.09^0.09 = 0.8051...
0.08^0.08 = 0.8170...
0.07^0.07 = 0.8301...
0.06^0.06 = 0.8446...
0.05^0.05 = 0.8608...
0.04^0.04 = 0.8791...
0.03^0.03 = 0.9001...
0.02^0.02 = 0.9247...
0.01^0.01 = 0.9549...

from this
0^0 approaches...
?

Anonymous

Anonymous Sat 21 Mar 2020 01:56:57 No.11487883 Report

Quoted By: >>11488063

>>11487815
Interesting, What's that dip b/t 0.3^0.3 and 0.4^0.4?

Anonymous

Anonymous Sat 21 Mar 2020 03:01:06 No.11488017 Report

Quoted By:

>>11484354
It's an arbitrary convention which often makes the math simpler but is not really fundamental or anything.
Don't overthink it.

Anonymous

Anonymous Sat 21 Mar 2020 03:19:56 No.11488063 Report

Quoted By: >>11489182

>>11487883
If you're curious the turning point is around
(0.3679...)^(0.3679...) = 0.6922...

Or, if we wanted to be more elegant, we could write it as
(1/e)^(1/e) = 0.6922...

Anonymous

Anonymous Sat 21 Mar 2020 03:23:26 No.11488069 Report

Quoted By:

>>11487815
>>11485690
Thank you two

Anonymous

Anonymous Sat 21 Mar 2020 03:28:49 No.11488087 Report

Quoted By:

>>11487815
that's quite appealing

Anonymous

Anonymous Sat 21 Mar 2020 13:55:38 No.11489182 Report

Quoted By:

>>11488063
Right after I posted I also realized it was very close to ln(2).
Very interesting.

Anonymous

Anonymous Sat 21 Mar 2020 14:12:20 No.11489218 Report

Quoted By:

It can not be proven because it is not generally true. Nobody ever needs tobbn interpret the expression 0^0 unless there's already a context for which it is defined. For example it is defined that way for Taylor series. Not because it's mathematically true that 0^0=1 but because it's understood that 0^0 only shows up in the Taylor series because x^0 shows up, and the Taylor series would only ever give the correct value of an otherwise well defined function when x=0 if 0^0 is assumed to be 1. That doesnt mean that it's 1, but that it's not worth wondering "but what if" in that particular context.

Anonymous

Anonymous Sat 21 Mar 2020 14:18:09 No.11489231 Report

Quoted By:

>zero times itself zero times is 1

how does math do it? What will math think of next?
0 is a number meant to describe nothing, but it would seem that 0 has a value greater than the true value of nothing.

Since 00 means there are zero (no) instances of 0, what 00 really means is nothing. Empty space. "00" becomes " ".
and somehow true nothing = 1, rather than 0.
True nothing is greater than 0.

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