Mathematics - 2018-01-18 (page 6 of 8)

Moreover, those cultures are decidedly not isolated in one respect: as examples of human species, they’d have similar brains and similar basic requirements to survive

yep, even we and animals have much in common, which are not discovered yet! @Semiclassical

like genes

So if they happened to encounter similar circumstances, it hardly seems so bizarre that they would happen to develop similar mathematical concepts

With aliens, however, that goes out the window entirely

If there are any intelligent beings out there I would not expect them to have the same basic understanding of physics as us.

And if the physics is completely different, so would be the mathematics

5:02 PM

heheh..... never, they might be using binary, trinary or whatever even no numbers, which i don't care at all, but I've much interest, what I can do using those concepts which they are using

@BalarkaSen It is hardly obvious that they would, at any rate

@Semiclassical True

It's just not easy to draw any conclusion about such large-scale things

Have you ever read the Asimov short story Midnight?

yep

i've

midnight and school

of issac

There’s an element of that in the setting, I think, where the perpetual daylight had dramatic implications for their culture and science

And that’s just one variable

5:05 PM

@Semiclassical Have you ever Googled "project blue planet" ?

No

just give a try

I don't have the patience for this UFO thing

Agreed

As exercises in fiction and creative thinking, they’re amusing

There are about 100, UFO sightings every year over the world and most of them are more than 1km in DIAMETER as told by the viewers

no that's not fiction @Semiclassical

5:07 PM

As attempts to understand the universe, they’re crap

It's clear that this discussion is going towards a hilarious direction

I'm outta here

Though UFO stuff doesn’t bother me nearly as much as quantum mysticism

skullpatrol

cya

yea quantum is very cooooooooooooooool!

With UFOs there’s at least an element of uncertainty about what technology is possible

5:09 PM

coolest thing ever discovered , according to some theories that there are 11 dimentions

@skullpatrol I'm still in this chat, just not going to involve myself in this discussion

I don’t think it’s enough to warrant belief in ufos, but I can see the possibility

I'm not that maths and physics guy but scientist say that

Quantum mysticism, by contrast, is bullshit

how?

skullpatrol

5:10 PM

@BalarkaSen ok

@Semiclassical Quantum mysticism being the idea that consciousness participates in particle interactions?

The key word there being "mysticism"

I wanted to learn those theories, of quantum

those sound awesome!

Stuff like that, yeah

I mean, there is weirdness in QM

@Semiclassical Yeh that's kind of ridiculous

5:11 PM

I think by "quantum mysticism" you mean stuff of the form "Quantumness is weird, therefore anything is true"

mathematics guys are cool!! they are philosophical and scientific too

yep

in The h Bar, yesterday, by Secret

NB. Still find SR and GR less intuitive than quantum mechanics

But that weirdness does not translate into any notion of psychic phenomena or telekinesis

skullpatrol

Entanglement with mysticism?

yea

I'm trying to understand space-time

and timetravel

5:12 PM

Reminds me of the (joking) quote "Infinity means that anything can be true for any reason."

using speed of light!

@Semiclassical I think their point of argument is that interactions unfold in different ways depending on your observation or whatever

Right

Instead of accepting that it's a probabilistic thing they want to involve mysticism in it

also, sci-fi

5:14 PM

Well, I think that’s too sharp a cut

One important thing about quantum: Entanglement means there is influence at a distance in the form of correlations of observables, but there is no signalling, thus you cannot send a message faster than light with it

^

Another important thing about quantum: The value of the observables are not determined until being measured

@Secret Hm, that's something that always bothered me. What do you mean by there is no signaling?

It’s sorta like this

Suppose you and I (on opposite sides of the world) had a pair of coins which we flip repeatedly

5:17 PM

I heard it explained with, you have three boxes in one place labeled A, B, and C, and three boxes in another place labeled A', B', and C'

and they each contain either a black or white ball

If you open A and A', they'll have the same contents, similarly for B and B' or C and C'

I follow you both

And let me assume that, if we only look at our individual results , we just get fifty-fifty results

If you open A and B' (or A' and B), or B and C' (or C' and B), there's an 85% chance that they have the same contents

So from our perspective the coins look unbiased

Okay @Semiclassical @Akiva

5:19 PM

(Like, we can repeat this experiment infinitely many times, resetting the boxes so you don't know what's in them at the start of each experiment, and you notice that 85% of the time those pairs agree)

But if you open A and C', or A' and C, then they only agree 50% of the time.

True

Ah I see where this is going

Each time you perform this experiment you're only allowed to open one box from each place

But suppose we then compared our outcomes and found that, on average, we saw that our coins were correlated

@Semiclassical Mhm

The thing is, if probability and logic and whatnot were to make sense, A and C' should agree 85% of 85% of the time, that is, 72% of the time

But they don't, which means, uh, some sort of quantum weirdness must be going on

I dunno

5:21 PM

I don't see why the situation re A and B' and re A and C' should be different

If you try to encode bits and then tries to take advantage of entanglement, then even though there is a correlation when the outcome of the measurement from both subsystems were obtained and compared, to any subsystem, all they see is just random instance of 0 and 1 with no pattern whatsoever.

That is, the outcomes only look correlated when viewed with both systems taken account of, in any individual system, the result is a statistical mixture. Since you cannot control the outcome of the measurement of the individual subsystems, you cannot encode any bits on it and hence cannot send a message

I’m actually working a bit on this stuff with a prof

afterall, (A,A') (B,B') (C,C') are symmetric with respect to permutation

@LeakyNun No they're not

what are those boxes?

5:22 PM

You have them labelled

The people setting up the experiment put stuff into the boxes without telling you

oh ok...

The main trickiness of the two-party version of this is that you need statistics

And this is where my understanding is foggy, 'cause I'm not quite sure how one sets up the boxes so that the probabilities work out like that, but apparently you can use quantum weirdness to do it or something

I dunno

For a non-statistical version of it you need to have three parties

5:24 PM

You can produce correlated pairs, somehow, right?

I can talk a bit more precisely about akivas setup but I need to get on my laptop

I need to scurry off now though

You should write something up so I can read it later

It’s actuslly related to the cubic surface I was talking about a few weeks ago

So the gist is that e.g. you can measure your entangled pair and get them anticorrelated 100% (e.g. if A is 1, then B is 0 and vise versa), but to the individual experimenters recording the outcomes of A and B respectively, all they saw is a randon sequence of 0s and 1s, hence there is no message

@Secret Ah I see

The appearance of the 0's and 1's is not determined

I SEE

Well that was easy ;)

5:26 PM

Right

In the three-party case, you can have something like: If you know the measurements of two of the parties, you know the measurements of the third.

So that's clearly a form of correlation.

Mmm I see

But in order to know the measurements of two of the parties, you need to be in causal contact with both of them

so the first party only learns about the third by communicating with the second, and that's presumably done via signalling

so you're not able to learn about the other parties without some channel of signalling. hence the set-up is non-signalling

Got it. Makes sense now.

And this is why. in order to perform a quantum teleportation, a classical (slower than or equal to speed of light) channel is needed to convey the information of the measurement outcomes, which is needed for the other party to recover the quantum state at the other side

there are important details in this, of course. you might wonder if you could set it up so that the first two parties can communicate before information would have time to propagate to the third party (e.g. put observers 1,2 on opposite sides of the earth and observer 3 on mars.)

apparently, though, that set-up isn't possible.

(which is plausible to me insofar as that set-up would require a distinct asymmetry between the three parties.)

5:31 PM

How can I calculate the limit of

$\lim_{n \to \infty}\left(\frac{3n}{3n+1}\right)^{6n}$?

I am unable to form it into something like:
$\lim_{n \to \infty}\left(1 + \frac{1}{3n}\right)^{6n}$

How do limits work with reciprocals?

There's a power $n$, it is not that simple I think...

we are dealing with something of the form $f(n)^n$

i.e. if $\lim_{n\to \infty} f(n)=F$ then what can you conclude about $\lim_{n\to\infty} \frac{1}{f(n)}$?

$1/F(n)$?

Liad

5:33 PM

@jublikon That'd mean that $F(n)$ was n-dependent, which doesn't make a lot of sense for a limit as n->infinty

Liad

why does $\mu(N) = 0$ ?

@Semi so I have to calculate the integral of my function ?

@Semiclassical

What?

I thought that big F means antiderivative of f

no. it can, but in this context I am just using it to denote $\lim_{n\to\infty}f(n)$

5:36 PM

ah

If you like, I can do $f(\infty)$ instead. (that's a bit abusive but it captures the point)

yes, I see now what you mean

I must be honest that I still do not understand how to go on

So if $\lim_{n\to \infty}f(n)$ exists and equals $f(\infty)$, what can you say about the limit of $1/f(n)$?

Let me further assume that this limit is nonzero (do you see why?)

the limit of $1/f(n)$ is the reciprocal/inverse of $f(\infty)$

I think

reciprocal, yes

the only issue is if $f(n)\to 0$ as $n\to\infty$

since then you've got 1/0 problems

5:39 PM

There's also something more subtle about entanglement in that it violates an inequality called Bell Theorem. But other than it basically means there are no local hidden variables (that is, a model where the outcome is actually preexisted, instead of being fundementally indetermined), I still don't fully grasp what it means

I am bad at handling inequalities

(for instance, the sequence 1,-1/2,1/3,-1/4,1/5,-1/6... converges to zero but the sequence of reciprocals 1,-2,3,-4,5,-6... grows without bound in both directions)

we are talking about the limit of $\lim_{n \to \infty} 1/f(n)$. Do we know the limit of $f$?

So if $\lim_{n\to\infty} (1+\frac{1}{3n})^{6n}$ exists and is nonzero, what can we conclude about $$\lim_{n\to\infty} \frac{1}{(1+\frac{1}{3n})^{6n}}?$$

@Semiclassical It can be $0$ some constant or $\infty$?

is that correct?

Semi: Ok that's clever, I now see what you are doing there to solve this problem

5:43 PM

Does that match what we agreed about $\lim_{n\to\infty} f(n)$ and $\lim_{n\to\infty}\frac{1}{f(n)}$ ?

(Do you see which $f(n)$ I'm suggesting you use?)

I think yes. Because if $\lim_{n \to \infty}f(n)=\infty$ then $\lim_{n \to \infty}\frac{1}{f(n)}=0$, if $\lim_{n \to \infty}f(n)=\text{some constant}$ then $\lim_{n \to \infty}\frac{1}{f(n)}=\text{some constant}$, if $\lim_{n \to \infty}f(n)=0$ then $\lim_{n \to \infty}\frac{1}{f(n)}=\text{undefinded}$

right, with the constants being related as reciprocals.

but I do not see which$f(n)$ you suggest to use

Parts of this are strangely funny

Well, I'm asking you about $\lim_{n\to\infty}(1+\frac{1}{3n})^{6n}$

not a lot of options for $f(n)$

5:46 PM

DeepMind Control Suite

why $\lim_{n \to \infty}(1+\frac{1}{3n})^{6n}$? The original task was $\lim_{n \to \infty}(\frac{3n}{3n+1})^{6n}$

always with power $6n$ of course

Well, what’s the reciprocal of that?

https://twitter.com/OriolVinyalsML/status/948675218259800066

...shit. I think the ac adapter for my laptop may have finally bit the dust

ok

$\frac{3n+1}{3n}^{6n}$?

5:50 PM

Ahah, not dead yet

^^

Right. But $\frac{3n+1}{3n} = 1+\frac{1}{3n}$

@Semiclassical how Did you do that ? wolfram says $1-\frac{1}{3n+1}$

$$\Bbb Z_p[[X]] \cong \Bbb Z_p$$

Then you're probably missing a parentheses

Actually, I shoudl put it differently

5:53 PM

@jublikon Semi wants you to apply two tricks: first invert the limit and then make some changes to the fraction and you'll discover similarities to some known limit (there's probably only one limit which I could mean).

$1-\frac{1}{3n+1}=\frac{(3n+1)-1}{3n+1}=\frac{3n}{3n+1}$

@Semi I have seen it now, too ^^

but I'm talking about $\frac{3n+1}{3n}=1+\frac{1}{3n}$

@MatheinBoulomenos ^^^^^^^

@LeakyNun No

5:55 PM

altogether we have $$\frac{1}{\left(\frac{3n}{3n+1}\right)^{6n}} = \left(\frac{3n+1}{3n}\right)^{6n} = \left(1+\frac{1}{3n}\right)^{6n}$$

@MatheinBoulomenos es ist wahr

dangit mathjax y r u mad at me

@semi you're missing a minus in the exponent no?

No. I flipped the factor inside instead.

Both $\Bbb Z_p$ and $\Bbb Z_p[[X]]$ are local rings, but they can't be isomorphic, because the maximal ideal of $\Bbb Z_p$ is prinicipal, whereas the maximal ideal of $\Bbb Z_p[[X]]$, namely $(p,X)$ is not principal

5:56 PM

@MatheinBoulomenos it's a half-joke, and it is a true statement, using some interpretation

you can say $\Bbb Z_p \cong \Bbb Z[[X]]/(X-p)$

@Semiclassical that is great

So how are the limits of $\left(\frac{3n}{3n+1}\right)^{6n}$ and $\left(1+\frac{1}{3n}\right)^{6n}$ related? @jublikon

not sure what what interpretation you mean. Everything from that has a well-defined meaning

hint: there is more than one meaning to $\Bbb Z_p$

5:58 PM

But $\Bbb Z_p$ is not isomorphic to $\Bbb F_p[[X]]$

wait, what?

My version of that kind of joke: Suppose $f(x,y)=x^2+y^2$. What is $f(r,\theta)$?

$\Bbb Z_p$ has characteristic $0$

mein ganzen leben ist eine Luge

oh god, right

i'm stupid

@Semiclassical Wolfram says $\lim_{n \to \infty}(1+\frac{1}{3n})^{6n}= \infty$

6:00 PM

@Semiclassical eh, $r$?

and thus our original limit is $0$

Source?

@LeakyNun I think you mean $r^2$.

wait...

$e^2$

sorry

..

Ding

So what's your original limit?

1/e^2 ?

*!

6:01 PM

Right

:D

nice thank you! that was detailed help :)

Here's another way to see it. First, let's change the power from 6n to 3n

all that does is take the square root, and that won't change things appreciably

so now the limit is for $\left(\frac{3n}{3n+1}\right)^{3n}$

I did not know that it is allowed to take the square root just like that

In general, it's not. But the sequence here is of positive values

I would compute this limit like so

$\lim_{n \to \infty} (\frac{3n}{3n+1})^{6n} = \lim_{n \to \infty} (\frac{3n+1}{3n})^{-6n} = \lim_{n \to \infty} (1+\frac{1}{3n})^{-6n} = \lim_{k \to \infty} (1+\frac{1}{k})^{-2k} = \lim_{k \to \infty} \big( (1+\frac{1}{k})^{k} \big)^{-2} = \big( \lim_{k \to \infty} (1+\frac{1}{k})^{k} \big)^{-2} = (e)^{-2}$

where $k:=3n$ which goes to $\infty$ also as $n \to \infty$.

6:04 PM

And for positive numbers the square root function is monotonic, so taking the limit isn't a problem

@philmcole Yeah, that's what I had in mind altogether.

Though, note that $\lim_{n\to\infty} f(n)=L\implies \lim_{n\to\infty} f(3n)=L$ but not vice versa

Yeah that's important to note

So it's a bit better to start from $\lim_{k\to\infty}(1+1/k)^k=e$ and pass to the relevant subsequence

You could also look at this as $$\left(\dfrac{3n}{3n+1}\right)^{6n}=\left(1-\dfrac{1}{3n+1}\right)^{6n} = \left[ \left(1-\dfrac{1}{3n+1}\right)^{3n+1}\right]^2 \left(1-\dfrac{1}{3n+1}\right)^{-3}$$

Then you can get the first factor from $\lim_{k\to\infty}(1-1/k)^k=e^{-1}$ and the second factor just goes to 1.

So it's again $e^{-2}$

But I don't like that as much...

Vrouvrou

Hello, someone knows Luzin theorem ?

or Lusin theorem?

Someone have an idea about that: math.stackexchange.com/questions/2610350/…

Q: Integral of a Gaussian multiplied with a Confluent Hypergeometric Function?

6:28 PM

This looks disgusting

0

$$\int_0^{\infty}\!\!\mathrm{d}x~x^2\,e^{-\alpha x^2+i\beta x}\,_1F_1(a,2,icx)$$ Here, $\alpha,c>0$, $\beta\in\mathbb{R}$, $a\in\mathbb{C}$ and $_1F_1(\dotsi)$ is the Confluent hypergeometric function of the first kind [see, http://mathworld.wolfram.com/ConfluentHypergeometricFunctionoftheFirstKi...

@Clarinetist do not want

Though tbf you can write out $_1F_1$ as a power series and in principal integrate term-by-term

Sad part is, I don't even know what Hypergeometric Functions are. All I know is that I should be scared if they're involved in anything

lol

There's a few equivalent ways to define them

One is as solutions to an ODE belonging to a certain family of such ODEs, with the parameters in front telling you what the coefficients of the ODE is

that's kind of painful, though

Another definition is deceptively simple. First, a bit of notation: Let $(a)_n=a(a+1)\cdots(a+n-1)$

so basically like the factorial except you're going up starting at $a$ and you have exactly $n$ factors

Interesting...

Gauss's hypergeometric function then has the series definition $_2F_1(a,b;c|x)=\sum_{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n}\frac{x^n}{n!}$

Which is bad looking, but it's basically a power series whose coefficients are a certain kind of rational function in $n$

6:35 PM

o_o and that apparently converges?

with some conditions imposed on $a$, $b$, and $c$ I imagine

A lot of the time, yes.

Right

Note that $(a)_{n+1}/(a)_n = a+n$

Ah, good ol' ratio test

I'll do that computation some other time :P

so the ratio test tells you it has radius of convergence $R=\lim_{n\to\infty} \frac{(n+1)(n+c)}{(n+a)(n+b)}=1$, I think?

Something like that

the series terminates if $a$ or $b$ is a finite negative integer, mind, since then $(a)_n$ eventually becomes zero identically

6:39 PM

BTW random remark... moving back up to MN has been a wake-up call. It really snows up here. I was spoiled having lived in Madison and in Iowa from 2014-2017

lol

welcome to semi-siberia

And just be glad you don't actually live in siberia

citation: theweathernetwork.com/news/articles/… @Clarinetist

(non-paywall version)

Oh mannnnn

Terrrrrible

jesus

yeeeah

Fun observation: If someone tells you that the temperature is colder than -40, is that ambiguous?

looool

well it probably means i'm already dead

so no confusion possible

6:44 PM

-40...apples?

(I understand the joke)

(That's a common way my high school teacher respond whenever soething that requires units is not mentioned)

Tuki

helo

Daminark

@BalarkaSen people can survive -40 if they bundle up well enough

7:00 PM

i cant

New York generally ranges from 0F to 100F

and New England as well

I do not want to be in any place that goes to -40

Incidentally, that's the reason Fahrenheit is better than Celsius. (a) New York City is the center of the world, and (b) weather here ranges roughly from 0F to 100F, making it better for describing weather

Daminark

Yeah Celsius is where the metric system is def inferior. But yeah Chicago in previous years has apparently hit -40 (not sure if this is with or without windchill)

Actually yeah it was with windchill

7:15 PM

Hm, yeah, I forgot about windchill

A: Derivative of the sine function when the argument is measured in degrees

Decided to put my opinions up on MSE:

0

This annoyed me when I revisited this material years later and had to teach this material to someone else, so I'm posting an answer here. (The other answer is perfectly fine, but I wanted to give a more complete exposition.) Now here's the thing: you're told to find the derivative of $\sin(\thet...

I'm trying to find a record of windchill temperatures in New York and I can't find one

Roadmap of the Post-Apocalyptic U.S. Part 2: Missouri to Oregon (ASMR)

►

Sounds legit!

@Clarinetist Incidentally, there's a cool identity involving trigonometric functions and pi that doesn't depend on using radians or degrees

Say you are measuring angles in some units, and $r$ is the smallest positive root of $\cos(x)$. ($90^\circ$ if you're doing degrees)

Then $\cos(x)\ge\dfrac\pi4\left(1-\dfrac{x^2}{r^2}\right)$.

Furthermore, $\frac\pi4$ is the only constant for which $\cos(x)\ge C\left(1-\dfrac{x^2}{r^2}\right)$ holds.

In degrees, that's $\cos(x)\ge\dfrac\pi4\left(1-\dfrac{x^2}{90^2}\right)$.

That's really interesting

7:26 PM

desmos.com/calculator/4nfpczew1f

^Demonstration. You can move C around on the slider

or click on the colored button to the left of an equation to make it invisible

7:49 PM

@Clarinetist my impulse would be to say $\sin(x^\circ)$, with $x$ a pure number, should really be understood as $\sin(\pi x/180)$

you differentiate with respect to $x$ to get $180/\pi \cos(\pi x/180)=(180/\pi)\cos(x^\circ)$

so one 'should' have $\frac{d}{dx}\sin (x^\circ)=\frac{180}{\pi}\cos(x^\circ)$

...which is pretty much what you do. so I guess we have similar mindsets :P

Essentially $^\circ$ is a constant

with ${^\circ}=\frac\pi{180}$

@Semiclassical Er, upside-down

But yeah, it ends up being ${^\circ}\cdot\cos(x^\circ)$

I don't like using degrees

It's unnatural

What temperature is it where you are?

18C

No degrees

About $\pi/10$ radians Celsius then

7:54 PM

$2\pi$ noscoping

Maybe $4\tau$ noscoping... :P

Just $\tau$

That's the whole point of $\tau$

I'll admit, 90 degrees psychologically 'feels' right to me

isn't $2\tau = \pi$ or am i misremembering

that's not a scientific statement, mind

Ba-dum tish? @Semiclassical

7:56 PM

$\tau=2\pi$

lol, I hadn't even intended that

weird, because $\tau \! \tau = \pi$

@BalarkaSen $\tau$ is the circle

You get that the $n$th root of unity is $e^{\tau i/n}$ and stuff

meh

I learned about angle measurements in degrees, so radians always feel foreign in a way that degrees don't

and yet, I certainly agree that radian measure makes more sense

Yeah, I always have to do a conversion in my head when I see, like, $\tan\left(\dfrac\pi6\right)$

7:58 PM

right

yeah i mean, i also use degrees

which is $1/\sqrt3$ incidentally

but that's not a good case against it's naturality

or I have to think "okay, pi is a semicircle, so pi/3 is a third of a circle"

Or $\sqrt3/3$

7:58 PM

@Semiclassical 2pi/3 is a third of a circle

whereas if I see "60 degrees" I know immediately what angle that is