/sci/ - Science & Math » Thread #14261977

22KiB, 518x264, x0.png

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This makes no sense

Anonymous Fri 25 Feb 01:09:15 2022 No.14261977 View Reply Original Report

Quoted By: >>14261988 >>14262092 >>14262131 >>14262217 >>14262435 >>14262444 >>14262518

It should be equal to 0

Anonymous

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Anonymous Fri 25 Feb 2022 01:12:49 No.14261984 Report

Quoted By: >>14262122 >>14262133 >>14262150

that would be a bit strange though, innit

Anonymous

Anonymous Fri 25 Feb 2022 01:15:42 No.14261988 Report

Quoted By: >>14261996

>>14261977
What does power to 0 even look like?

Anonymous

Anonymous Fri 25 Feb 2022 01:24:02 No.14261996 Report

Quoted By:

>>14261988
$x^n = x^{n-1}*x$
$\frac{x^n}{x} = x^{n-1}$
$\frac{x^1}{x} = x^{1-1}= x^0 = 1$
It's called an identity, x^0 = 1 for the same reason x *1 = x

Anonymous

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Anonymous Fri 25 Feb 2022 02:40:50 No.14262092 Report

Quoted By:

>>14261977
You have a tree that grows a root ball 1 unit in volume. You have 0 roots. What's the length of each root? 0.

Anonymous

Anonymous Fri 25 Feb 2022 03:12:24 No.14262122 Report

Quoted By: >>14262135 >>14262153

>>14261984
why did you crop it?? post the entire thing... its well defined for negatives too

Anonymous

Anonymous Fri 25 Feb 2022 03:16:27 No.14262126 Report

Quoted By:

https://en.wikipedia.org/wiki/Empty_product

Anonymous

Anonymous Fri 25 Feb 2022 03:20:17 No.14262131 Report

Quoted By:

>>14261977
(2^4)/2=2^3
(2^3)/2=2^2
(2^2)/2=2^1
(2^1)/2=2/2=1=2^0
(2^0)/2=2^-1=1/2
Makes perfect sense

Anonymous

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Anonymous Fri 25 Feb 2022 03:21:12 No.14262133 Report

Quoted By:

>>14261984
there you go nigger. terrible argument there desu. i dont see why would you use x as a base neither, you should have used a positive contant at least

Anonymous

Anonymous Fri 25 Feb 2022 03:22:49 No.14262135 Report

Quoted By: >>14262145

>>14262122
only for the integers

Anonymous

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Anonymous Fri 25 Feb 2022 03:29:55 No.14262145 Report

Quoted By: >>14262149 >>14262166

>>14262135
are you retarded? it's clearly showing very small fractions. the truth is those otehrer numbers are imaginary... but you can still calculate the modulus

Anonymous

View Same Google iqdb SauceNAO x^x.png, 5KiB, 356x252

Anonymous Fri 25 Feb 2022 03:30:56 No.14262149 Report

Quoted By:

>>14262145
**fixed

Anonymous

Anonymous Fri 25 Feb 2022 03:32:23 No.14262150 Report

Quoted By: >>14262206

>>14261984
btw 0^0 is indefined for this very same reason (it switches to imaginay numbers), a contant^0 is not

Anonymous

Anonymous Fri 25 Feb 2022 03:33:30 No.14262153 Report

Quoted By: >>14262158

>>14262122
x*2^3 = x*2*2*2
x*2^2 = x*2*2
x*2^1 = x*2
if the pattern follows then x*2^0 = x, which equals x*1 so 2^0 or any number to the zero = 1.

Anonymous

Anonymous Fri 25 Feb 2022 03:36:37 No.14262158 Report

Quoted By:

>>14262153
meant to be reply for OP

Anonymous

Anonymous Fri 25 Feb 2022 03:42:58 No.14262166 Report

Quoted By:

>>14262145
>you can still calculate the modulus
then you arent graphing x^x anymore, youre graphing a different function. retard.

Anonymous

Anonymous Fri 25 Feb 2022 04:12:07 No.14262206 Report

Quoted By: >>14262247

>>14262150
only exception: 0^0 is assumed to be 0 in the context of taylor series.

Divine differentialist

Divine differentialist Fri 25 Feb 2022 04:25:06 No.14262217 Report

Quoted By:

>>14261977
The rules and definitionf of exponentiation always result in a minimum of 1 regardless of value applied because otherwise why would one bother exponentiating a value instead of treating it as an analysis problem?

Anonymous

Anonymous Fri 25 Feb 2022 04:50:31 No.14262247 Report

Quoted By: >>14262253

>>14262206
No it's not, 0^0 is always 1 in the context of taylor series.

Anonymous

Anonymous Fri 25 Feb 2022 04:58:13 No.14262253 Report

Quoted By:

>>14262247
Wha-
>I wrote 0
Fuck me, you are of course correct.

Anonymous

Anonymous Fri 25 Feb 2022 09:01:55 No.14262435 Report

Quoted By:

>>14261977
Intuitively, you can't just nullify a definitely existing value. It at least has to collapse to the neutral element of multiplication.

Anonymous

Anonymous Fri 25 Feb 2022 09:12:04 No.14262444 Report

Quoted By:

>>14261977
It's simple to see

> X^0 = 1
> 0 * log(X) = log(1)
> log(x) = 0 / 0
> X = e^(0 / 0)

Where are you getting hung up?

Anonymous

Anonymous Fri 25 Feb 2022 10:41:17 No.14262518 Report

Quoted By:

>>14261977
Let b = x^0

a = a^1 = (a^0) * (a^1) = a^(0+1) = b * a

Multiply each side by multiplicative inverse of a, which is a^(-1):

a * a(^-1) = 1 = b * a * a^(-1) = b * 1 = b

So we get the proof that b = 1 = a^0 = x^0.

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