/sci/ - Science & Math » Thread #11674720

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Anonymous Fri 15 May 08:06:15 2020 No.11674720 View Reply Original Report

Quoted By: >>11674721 >>11674724 >>11674726 >>11674728 >>11675678 >>11675699 >>11675710 >>11675754 >>11675763 >>11675989 >>11676161 >>11676406 >>11676467 >>11676502

you should be able to solve this

Anonymous

Anonymous Fri 15 May 2020 08:06:55 No.11674721 Report

Quoted By:

>>11674720
50/50 it either happens or it doesn’t.

Anonymous

Anonymous Fri 15 May 2020 08:09:27 No.11674724 Report

Quoted By: >>11675976

>>11674720
-1/12

Anonymous

Anonymous Fri 15 May 2020 08:10:35 No.11674726 Report

Quoted By:

>>11674720
The answer is that "less than 1" isn't the same as "1".

Anonymous

Anonymous Fri 15 May 2020 08:10:53 No.11674728 Report

Quoted By: >>11675665

>>11674720
1

Anonymous

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Anonymous Fri 15 May 2020 08:17:58 No.11674746 Report

Quoted By: >>11674749 >>11675665

Anonymous

Anonymous Fri 15 May 2020 08:19:00 No.11674749 Report

Quoted By:

>>11674746
>proof by graphical inspection
Heinously based

Anonymous

Anonymous Fri 15 May 2020 08:19:49 No.11674750 Report

Quoted By: >>11675665

1/infinity is 0 and infinity ^ 0 is 1 so 1 so this is some basic high schooler math

Anonymous

Anonymous Fri 15 May 2020 15:40:27 No.11675665 Report

Quoted By: >>11675674 >>11675799 >>11675814

>>11674728
>>11674746
>>11674750
Wrong.
Let us consider the function $x^(1/y)$
Then we can try to find the limit as x and y both go to $\infty$
$\lim\limits_{(x,y)\to(\infty,\infty)} x^{1/y}$
We first consider the following limts:
$\lim\limits_{x\to\infty} \lim\limits_{y\to\infty} x^{1/y} = \lim\limits_{x\to\infty} x^0 = 1$
$\lim\limits_{y\to\infty} \lim\limits_{x\to\infty} x^{1/y} = \infty$ (since $\frac{1}{y}$ is positive)
Therefore the limit does not exist and the value of such an expression depends on how you approach it.

Anonymous

Anonymous Fri 15 May 2020 15:44:40 No.11675674 Report

Quoted By: >>11675713

>>11675665
graph dont lie

Anonymous

Anonymous Fri 15 May 2020 15:46:04 No.11675678 Report

Quoted By:

>>11674720
0.99999999999999999999999999999

Anonymous

Anonymous Fri 15 May 2020 15:53:49 No.11675699 Report

Quoted By:

>>11674720
inf^(1/inf) = inf^(0) = 1

Anonymous

Anonymous Fri 15 May 2020 15:58:14 No.11675710 Report

Quoted By:

>>11674720
put it in limit form what you mean by infinity

Anonymous

Anonymous Fri 15 May 2020 15:59:12 No.11675713 Report

Quoted By: >>11675736

>>11675674
Graph is for $x^{1/x}$ not for $x^{1/y}$ (two varibles). If you don't belive me go graph it youreslf in 3D space and see that it has the exact behavior I described retard.

Anonymous

Anonymous Fri 15 May 2020 16:04:29 No.11675736 Report

Quoted By: >>11675753

>>11675713
graph dont lie

Anonymous

Anonymous Fri 15 May 2020 16:10:28 No.11675753 Report

Quoted By: >>11675764 >>11675908

>>11675736
Go graph $x^{\frac{1}{\lnx}}$ ( $\lnx\to\infty$ when $x\to\infty$ so it is a $\infty^{\frac{1}{\infty}}$ ) and see that it is equal to e, not 1 retard.

Anonymous

Anonymous Fri 15 May 2020 16:12:02 No.11675754 Report

Quoted By:

>>11674720
Im a retarded 19 yo undergraduate w/ 0 papers to my name and lots of opinions

Anonymous

Anonymous Fri 15 May 2020 16:17:11 No.11675763 Report

Quoted By:

>>11674720
Lim(x^(-1/x)) = lim exp(ln(x^(-1/x))) = lim exp(-1/x)^ln(x) = 1

Anonymous

Anonymous Fri 15 May 2020 16:17:10 No.11675764 Report

Quoted By: >>11675828

>>11675753
>not 1 retard
Correct, you're retarded enough for two people.

Anonymous

Anonymous Fri 15 May 2020 16:30:01 No.11675799 Report

Quoted By: >>11675828

>>11675665
Retard

Anonymous

Anonymous Fri 15 May 2020 16:33:34 No.11675814 Report

Quoted By:

>>11675665
What is 1 divided by infinity? 0. What is infinity to the 0 power? 1 easy pz. I’m not sure what sort of schizo math this is.

Anonymous

Anonymous Fri 15 May 2020 16:36:58 No.11675828 Report

Quoted By: >>11675840

>>11675764
>>11675799
He's right is what he is. What is it about the OP that makes you interpret this as the limit of x^(1/x)? There is nothing in the OP to indicate that?

Anonymous

Anonymous Fri 15 May 2020 16:39:02 No.11675840 Report

Quoted By: >>11675955

>>11675828
Why do you interpret as x ^ (1/x)
It’s infinity ^ (1/infinity)
We can both accept 1/infinity is 0
Infinity ^ 0 is 1 I’m not sure what sort of schizo math they are trying to pulll.

Anonymous

Anonymous Fri 15 May 2020 16:55:13 No.11675908 Report

Quoted By: >>11675955 >>11676019

>>11675753
Don't be stupid... Evaluate $\frac{x}{x}$ and $\frac{x}{x^2}$ as $x \rightarrow \infty$ . If everything goes to infinite it is the same thing, right?

Anonymous

Anonymous Fri 15 May 2020 17:12:43 No.11675955 Report

Quoted By: >>11676019 >>11676037

>>11675840
Well x^(1/lnx) is also infinity^(1/infinity) and it's limit is e

>>11675908
The first is 1, the second is 0. They both evaluate $\frac{\infty}{\infty}$ which depends on how you approach it too.

Anonymous

Anonymous Fri 15 May 2020 17:16:55 No.11675976 Report

Quoted By:

>>11674724
Pop math retard a show your proof

Anonymous

Anonymous Fri 15 May 2020 17:19:16 No.11675989 Report

Quoted By: >>11676009

>>11674720
undefined
https://www.wolframalpha.com/input/?i=inf%5E%281%2Finf%29

Anonymous

Anonymous Fri 15 May 2020 17:19:41 No.11675991 Report

Quoted By:

graph don't lie

Anonymous

Anonymous Fri 15 May 2020 17:24:57 No.11676009 Report

Quoted By: >>11676104

>>11675989
wolfram lies

Anonymous

Anonymous Fri 15 May 2020 17:27:06 No.11676019 Report

Quoted By: >>11676099

>>11675955
imagine getting gtfo by >>11675908
but still squirming around like an oily pig

Anonymous

Anonymous Fri 15 May 2020 17:31:16 No.11676037 Report

Quoted By: >>11676099

>>11675955
it doesnt matter how you approach it if you graph x/x^2 it will go to 0 not some other crazy schizo shit you are trying to say it will go to if you start saying "well infinity equivalent to infinity"

Anonymous

Anonymous Fri 15 May 2020 17:52:22 No.11676099 Report

Quoted By:

>>11676019
>>11676037
Reatrds it matters how you approach $\frac{\infty}{\infty}$
If you approach it with x/x you get a different result from when you approach it with x/x^2. That was my original point. The same is true for inf^(1/inf). Imagine being this reatrded.

Anonymous

Anonymous Fri 15 May 2020 17:53:30 No.11676104 Report

Quoted By:

>>11676009
ok schizo

Anonymous

Anonymous Fri 15 May 2020 18:09:45 No.11676161 Report

Quoted By: >>11676198

>>11674720
Lets look at $\lim_{x\to\infty} \text{ln}(x^{1/x})$ instead.
We then have $\text{ln}(x^{1/x})=\frac{\text{ln}(x)}{x}$ which by by L'Hopital approaches
$\lim_{x\to\infty}-\frac{1}{x^3}$ as its limit. So taking the exponential we see $\lim_{x\to\infty} x^{1/x}=1$

Anonymous

Anonymous Fri 15 May 2020 18:26:42 No.11676198 Report

Quoted By: >>11676218

>>11676161
$\displaystyle<br />\lim_{x \to \infty} 1^x = 1 \\<br />1^ {\infty} ~~ undefined <br />$

Anonymous

Anonymous Fri 15 May 2020 18:30:44 No.11676218 Report

Quoted By: >>11676224 >>11676288 >>11676354 >>11676456

>>11676198
How can that be undefined, if 1 is 1 literally in every power. I would rather agree that 1^0 is undefined, but I saw the graph.

Anonymous

Anonymous Fri 15 May 2020 18:33:21 No.11676224 Report

Quoted By:

>>11676218
https://www.wolframalpha.com/input/?i=1%5Einf

Anonymous

Anonymous Fri 15 May 2020 18:49:14 No.11676288 Report

Quoted By: >>11676295

>>11676218
You are thinking about the function 1^x which is 1 for all x and (therefore) has limit 1 as x goes to $\infty$ . But $1^\infty$ is not the limit of that function. It is the limit of any function $(f(x))^x$ where the limit of f is 1 as x goes to $\infty$ . It depends on how your function approaches 1. For exapmle (x+1)/x approaches 1 but ((x+1)/x)^x doesn't go to 1 (even though it's $\frac{1}{\infty}$ ). This is true for other values like $\frac{\infty}{\infty}$ and $0^0$ but not for all such expressions (i.e. $\frac{1}{\infty}$ is always 0). I've been trying to explain this in this thread but apparently nobody (with few exceptions) understands this stuff here. (And they are calc 1 stuff)

Anonymous

Anonymous Fri 15 May 2020 18:51:19 No.11676293 Report

Quoted By:

If you don't understand basic fucking limit issues you need to leave this board.

Anonymous

Anonymous Fri 15 May 2020 18:51:23 No.11676295 Report

Quoted By:

>>11676288
>(even though it's $\frac{1}{\infty}$ )
Meant (even though it's $1^{\infty}$ )

Anonymous

Anonymous Fri 15 May 2020 19:12:35 No.11676354 Report

Quoted By:

>>11676218
but where is that operation used?

Anonymous

Anonymous Fri 15 May 2020 19:31:50 No.11676406 Report

Quoted By:

>>11674720

In what case would I actually need to solve this though
What does an infinite root even represent

Anonymous

Anonymous Fri 15 May 2020 19:44:34 No.11676456 Report

Quoted By: >>11676553

>>11676218
>I would rather agree that 1^0 is undefined
Then you are retarded. You don't even need limits to show that.
Consider $a \neq 0$ and $n > 0$
$a^0=a^{n-n}=\frac{a^n}{a^n}=1$
Therefore $a^0=1, \, \forall a \neq 0$

Anonymous

Anonymous Fri 15 May 2020 19:47:31 No.11676467 Report

Quoted By:

>>11674720
1

Anonymous

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Anonymous Fri 15 May 2020 19:56:27 No.11676502 Report

Quoted By:

>>11674720
Nooo, you just can't get the non number root of a non numberinooo

Anonymous

Anonymous Fri 15 May 2020 20:09:57 No.11676553 Report

Quoted By:

>>11676456
I don't deny that it's 1, I only say that 1^inf = 1 is even more obvious.

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