Mathematics - 2017-07-14 (page 4 of 4)

SteamyRoot

19:00

o.O

First this chat goes wild about fixed points, and now about periodic points

Semiclassical

Now we just need to start talking about period 3 points and we'll be bound for chaos.

Akiva Weinberger

@SteamyRoot Whoah I think I have a counterexample to the fixed set thingy from this morning

Or at least the set-theoretic version

SteamyRoot

To what part of the fixed set thingy exactly?

Akiva Weinberger

There exists a set $X$ and a function $f:X\to X$ such that $f(\bigcap_n f^n(X))\ne\bigcap_n f^n(X)$

That is, $\bigcap_nf^n(X)$ not necessarily fixed by $f$

SteamyRoot

Okay, so what's the example?

Akiva Weinberger

19:07

Trying to think of the best way to describe it

Secret

$\{\forall a,b \in \omega_2,(-\infty,b),(a,\infty),(a,b)\}$
check convergence on $(0,0)$:

$\forall a \in \omega_2, (0,0)\subseteq(0,a)$ thus $(0,a)$ is a nbhd of $(0,0)$

Thus $(0,0)\subseteq (0,0)$ thus $(0,0)$ is a nbhd

Any countable sequence $\{(a,b)\}$, with $a,b$ both $\to 0$ converges to $(0,0)$

Uncountable nets, need to check its definition again...

SteamyRoot

o.O

Not sure what $\omega_2$ is.

Akiva Weinberger

$X=\Bbb Z_{\le0}\cup\{1'\}\cup\{1'',2''\}\cup\{1''',2''',3'''\}\cup\dotsb$

SteamyRoot

But - you're sure this space is Hausdorff and compact?

Akiva Weinberger

where $\Bbb Z_{\le0}=\{\dots,-2,-1,0\}$.

Secret

19:08

$\omega_2$ second uncountable ordinal.

I am not sure, I just try to work out its topological properties. The set is $\omega_2 \times [0,1)$ btw

Akiva Weinberger

$f$ subtracts $1$ from each element, keeping the same amount of tick marks if possibly

SteamyRoot

Oh, nevermind, for a second I thought that was Akiva's counterexample >.<

Akiva Weinberger

so $1^{\prime\prime\dots\prime}\mapsto0$ no matter how many tick marks there are, for example

Semiclassical

Ugh, I thought I'd left my days of thinking about steepest descent approximations behind.

SteamyRoot

Right... And this set has the discrete topology?

Semiclassical

19:10

But they just keep dragging me back in :P

Akiva Weinberger

@SteamyRoot I'm doing the set-theoretic version right now, so no topology

In any case, $\bigcap_nf^n(X)=\Bbb Z_{\le0}$, but that's not fixed under $f$.

Secret

nope, because it is basically the closed long ray with $\omega_1$ (first uncountable ordinal) replaced by $\omega_2$ it should have the natural topology the order topology

SteamyRoot

@AkivaWeinberger What do you mean with "set-theoretic version" ?

Akiva Weinberger

5 mins ago, by Akiva Weinberger

There exists a set $X$ and a function $f:X\to X$ such that $f(\bigcap_n f^n(X))\ne\bigcap_n f^n(X)$

SteamyRoot

Oh.

Akiva Weinberger

19:11

Dealing with sets rather than topological spaces

SteamyRoot

So no continuity conditions on $f$

Akiva Weinberger

However, it may be possible to embed this into a topological space

(Into a compact and Hausdorff one)

But does the example make sense?

For any $n$, there's a sequence of length $n$ that kind of "feeds into" $0$

but there's no infinite sequence

Secret

Hmm.... given an increasing countable sequence $\{(a,b)\}$ with $a,b \to \infty$, it does not seemed clear to me which $\omega_n$ it will converge to. I need to think about this later...

Mike Miller

@PVAL-inactive This is equivalently asking if we can start with S^n and add finitely many cells of degree n+1 or higher so that the resulting space is different but has the same homotopy groups. This seems very unlikely but also completely intractable.

SteamyRoot

Wait, why is $\cap_n f^n(X) = \mathbb{Z}_{\leq 0}$ ?

Akiva Weinberger

19:15

@SteamyRoot Well, first off, do you see why $0\in\bigcap_nf^n(X)$?

SteamyRoot

For any $k$ and any $n$, the element $f^n (k+n) = k$

Akiva Weinberger

@SteamyRoot And do you see why $1'\notin\bigcap_nf^n(X)$?

SteamyRoot

Oh, wait

Akiva Weinberger

Note that there is no element called $2'$

SteamyRoot

Yeah, those ticks.

PVAL-inactive

19:16

@MikeMiller Why do you know you only have 1 n-cell (up to homotopy type)

Leaky Nun

@Secret be specific

SteamyRoot

Right, yes, that works as a space.

Akiva Weinberger

@SteamyRoot I'm thinking maybe there should be another way to describe the set

but yeah, there are $n$ elements with $n$ ticks

PVAL-inactive

I mean you have the map $S^n \to X$ gives an isomorphism on $\pi_n$ so there's something you get.

Akiva Weinberger

so $f^n(n^{\prime\prime\dotsb\prime})=0$, yeah?

Secret

19:17

@LeakyNun well, it seemed to heavily depend on the terms in the sequence for example $\{(n,n)\}$ $n\to \Bbb{N}$, will it converge to $\omega_1$ or $\omega_2$?

Mike Miller

@PVAL We know its nth homology and you can make a space out of as few cells as its homology demands.

Akiva Weinberger

But there's nothing that'll give you $n^{\prime\prime\dotsb\prime}$

SteamyRoot

Yup.

Akiva Weinberger

(if there are $n$ ticks there)

Leaky Nun

@Secret which topology? the ultra-long line?

Akiva Weinberger

19:18

So we agree the intersection of repeated applications of $f$ is $\Bbb Z_{\le0}$ @SteamyRoot

But then $f(\Bbb Z_{\le0})=\Bbb Z_{\le-1}$

Mike Miller

Ya I know all that. There's only one map the equivalence could be. That's why you can talk about building it up from the sphere.

Secret

@LeakyNun It should be the order topology since it is basically constructed by replacing $\omega_1$ of the long line with $\omega_2$

SteamyRoot

Yeah, don't worry, I already understand the entire thing now :P

Mike Miller

I thought about it for a bit before responding I just made no progress.

SteamyRoot

It's indeed a counterexample to the "set-theoretic" version

Secret

19:19

I am trying to use it to understand higher cardinals that is not just about their sizes

Akiva Weinberger

@SteamyRoot I knew there was something bothering me this morning when some claimed to prove it using just set-theoretic techniques!

Leaky Nun

@Secret what is the long line?

Secret

$\omega_1 \times [0,1)$

Mike Miller

The picture I have is that the first level the map isn't an iso you add a cell and in a dimension above that add some cells to kill off your original pi_n+k and remake it with new relator cells.

Akiva Weinberger

@Secret Other way 'round I think

Leaky Nun

19:20

@Secret then what the hell is $(n,n)$?

SteamyRoot

Well, yeah, it would be very suspicious that the question asks it for a Hausdorff compact space and a continuous map; and you wouldn't need those at all :P

Leaky Nun

@AkivaWeinberger I don't think so

Long line (topology)

In topology, the long line (or Alexandroff line) is a topological space somewhat similar to the real line, but in a certain way "longer". It behaves locally just like the real line, but has different large-scale properties (e.g., it is neither Lindelöf nor separable). Therefore, it serves as one of the basic counterexamples of topology. Intuitively, the usual real-number line consists of a countable number of line segments [0, 1) laid end-to-end, whereas the long line is constructed from an uncountable number of such segments. == Definition == The closed long ray L is defined as the carte...

Akiva Weinberger

Well, the continuous and compactness stuff was used to make sure the intersection is nonempty

@SteamyRoot

Mike Miller

But this so chaotically and drastically changes higher homotopy groups you have no reason to guess some number of cells will magically end up changing all the higher ones back.

Secret

Since long line (and also the "ultralong line") has the order topology, its points are of the form $(a,b)$

PVAL-inactive

19:21

@Mike I think you can do that, and there should be a proof that something like that never terminates analogous to the infinite dimensional ness of Eilenberg MacLane spaces.

Secret

so $\{(n,n)\},n\in \Bbb{N}$ is the sequence $\{(0,0),(1,1),(2,2,),...\}$

Leaky Nun

@Secret no, (1,1) isn't in the topology. $(a,b)$ where $a \in \omega_2$ and $b \in [0,1)$.

Mike Miller

You can do that but I don't expect a proof it can't terminate

Akiva Weinberger

@Secret What's the ultralong line, $\omega_2\times[0,1)$? I thought that wasn't locally homeomorphic to $\Bbb R$

You get problems at like $\langle\omega_1,0\rangle$

PVAL-inactive

I suspect most sequences of homotopy groups can't be realized by a f.d. complex.

Secret

19:23

I am not currently interested with $\Bbb{R}$, I want to understand the bahaviour of higher cardinal sets and I am pretty sure it is not just about their size

Akiva Weinberger

Oh OK sure

Paul Plummer

He has the lexographical order topology on it

Akiva Weinberger

@SteamyRoot So the question is… can we embed that into a Hausdorff compact space

SteamyRoot

No.

Because the theorem holds for a Hausdorff compact space :P

Secret

uh wait a sec.. I am confused temporarily...

Akiva Weinberger

19:24

Oh, wait, I think the reason is that we can consider the sequence $(1',2'',3''',\dots)$

Axoren

Why can't all curves be straight lines?

Akiva Weinberger

It's a compact space so it must have a convergent subsequences and that messes it up somehow

@Axoren What?

Leaky Nun

@AkivaWeinberger it boggles me that $\omega_1 \times [0,1)$ must have cardinality $\mathfrak c$...

Secret

@LeakyNun Order topology is generated by the base $\{(-\infty,b),(a,b),(a,\infty)\}$, right?

Akiva Weinberger

@LeakyNun Is that true??

Danu

19:25

But what's the historical reason behind "arithmetic" @MikeMiller?

Akiva Weinberger

Oh, in ZFC probably

Not necessarily in ZF

I think

Leaky Nun

sure

Axoren

@AkivaWeinberger I'm just kind of frustrated that some formulas I'm working on are complicated by the fact that I'm not guaranteed straight lines.

Akiva Weinberger

What subject?

PVAL-inactive

@Mike for instance if the map wasn't surjective for exactly one of the $\pi_k$ you could kill off the original sphere and then get a f.d. Eilenberg MacLane space.

Leaky Nun

19:26

@Secret yes

but beware that $b$ is also a tuple

Secret

ah ok

Axoren

@AkivaWeinberger I guess Differential Geometry, but it's more like most subjects of math all at once. I'm trying to get a simple formulation for gradient descent of a specific cost function.

SteamyRoot

@AkivaWeinberger That's sequentially compact, though :P

Axoren

And the integral curve doesn't end up being a geodesic, so I need to calculate it's curve rather than just using the exponential map.

Secret

The cool thing about higher cardinals is that you can no longer count or rely on $\Bbb{R}$ thus working with them force you to think with infinities

(paraphasing thinking with portals)

PVAL-inactive

19:28

actually nevermind this probably messes with higher homotopy groups.

Paul Plummer

When do you think about $\mathbb{R}$ when thinking about cardinals?

Are you talking about $\omega$ and everything else is higher

Secret

$\aleph_1$ is the cardinality of $\Bbb{R}$ if continuum hypothesis holds

Paul Plummer

Really?

Akiva Weinberger

@SteamyRoot Compact implies sequentially compact

@Secret Speedy set goes in, speedy set comes out

Leaky Nun

@PaulPlummer I just imagine a very large amount.

Paul Plummer

19:29

If it is false $\aleph_2$ could be the cardinality of $\mathbb{R}$,

Secret

yup

SteamyRoot

@AkivaWeinberger wut

Secret

So my idea of higher cardinals is $\aleph_n,n > 1$

Paul Plummer

It might be helpful for countable cardinals, since you get embedding results (but you can just think about the rationals

Akiva Weinberger

@SteamyRoot Doesn't it? It's the reverse that fails

SteamyRoot

19:30

Both fail in general.

Akiva Weinberger

Every sequence in a compact set has a convergent subsequence

SteamyRoot

That's just the definition on sequential compactness :/

Akiva Weinberger

@SteamyRoot Really??

Do we need compact and Hausdorff?

Paul Plummer

Not really seeing where $\mathbb{R}$ comes in. Do you have an example where you rely on $\mathbb{R}$ to get results about $\aleph_1$ (that isn't about reals) @Secret

Akiva Weinberger

I thought for sure this was true…

Secret

19:32

I am working under ZFC where GCH is true, so $|\Bbb{R}|=\aleph_1=2^{\aleph_0}$

SteamyRoot

Time to check Counterexamples in Topology

Secret

uh, wait a sec.. I might have misunderstood you question, in that case I am not sure except for the cardinality

SteamyRoot

$I^I$ is a compact Hausdorff space that is not sequentially compact

Paul Plummer

I am no expert, but pretty sure most people interested in cardinals find GCH makes the theory quite boring. I don't really think the reals give a good intuition about $\aleph$'s in either case anyways (you don't have an explicit bijection or way to think about such an ordering of the reals).

I could be mistaken though

Axoren

"Erik Christopher Zeeman tried for 7 years to prove that one cannot untie a knot on a 4-sphere. Then one day he decided to try to prove the opposite, and succeeded in a few hours."

9

Akiva Weinberger

19:39

@SteamyRoot What's the sequence

Axoren

"A "theorem" of Jan-Erik Roos in 1961 stated that in an [AB4*] abelian category, lim1 vanishes on Mittag-Leffler sequences. This "theorem" was used by many people since then, but it was disproved by counterexample in 2002 by Amnon Neeman."

Akiva Weinberger

@Axoren What, a regular one-dimensional knot?

Isn't that well-known?

Axoren

@Akiva let me check the specifics. I'm just looking at disproven mathematical hypotheses right now. Many of them are well-known now, such as irrational numbers.

SteamyRoot

According to Counterexamples: "$I^I$ is not sequentially compact since the sequence of functions $\alpha_n \in I^I$ defined by $\alpha_n(x) = $ the $n$th digit in the binary expansion of $x$ has no convergent subsequence. "

Akiva Weinberger

Oh god

Wow

SteamyRoot

19:43

The Stone-Cech compactification of the integers is supposedly another counterexample, but I'm not even going to try understand that one.

Mike Miller

@Danu I am wrong. Hirzebruch introduced it.

Just win baby

\o @MikeMiller

Secret

$\omega_2 \times [0,1)$

Base: $\{(a,b)\},a \in \omega_2, b \in [0,1)$

Check $(0,0)$ convergence

Since $(0,0) \subseteq (0,0)$ thus $(0,0)$ is nbhd of itself. Now $(0,0) \subseteq (0,b), b \in [0,1)$ therefore sequences and nets that converge to $(0,0)$ must be of the form $\{(a,b),a,b \to 0\}$. For example $\{(0,n)\}$ where n are the dyadic rationals $\frac{1}{2^m},m\in \Bbb{N}$ arranged in increasing powers of $\frac{1}{2}$

Check $(\omega_1,0)$ convergence

Ok it's getting late (early?) 5:44, will continue this tmr...

Axoren

@AkivaWeinberger I can't find the specifics of what type of knot, but what might be well-known now might not have been back when he proved it.

The article wikipedia uses for reference is behind a paywall, and I can't find any other mention of it

SteamyRoot

The proof for $I^I$ is actually really doable. If $(\alpha_n)_n$ has a convergent subsequence $(\alpha_{n_k})_k$, say it converges to $\alpha$. Then for any $x \in I$, $\alpha_{n_k}(x)$ converges to $\alpha(x)$.

Now let $p \in I$ such that $\alpha_{n_k}(p) = (n_k \mod 2)$. Then the sequence $\alpha_{n_k}(p)$ is just $0, 1, 0, 1, \dots$, which does not converge.

Just win baby

19:48

@Axoren by "opposite" do they mean converse?

PVAL-inactive

@Axoren This quote stinks of ignorance on whoever was commenting.

SteamyRoot

@Axoren The way this is written, it sounds like he could've skipped all those years and proven the counterexample right away.

PVAL-inactive

I mean the other quote.

I don't think Zeeman or really anyone after Reidmeister thought you couldn't unknot a circle in 4-space.

I suspect the theorem Zeeman proved was quite a bit better of a result than that.

Daminark

Hey everyone!

Ted Shifrin

Heya Demonark

Just win baby

19:55

Hi

Ted Shifrin

And, yes, $I^I$ is not sequentially compact. I used to give this example every time I taught point set topology.

Daminark

How's everything going?

Semiclassical

same Zeeman as Zeeman effect?

Daminark

@Ted would locally compact spaces have some sort of Heine-Borel type of theorem?

Ted Shifrin

By which you mean?

Just win baby

20:08

@Semiclassical i doubt a physicist would spend 7 years on a math proof :P

Semiclassical

depends on the physicist :P

Just win baby

True dat

Daminark

What I have in mind is basically Heine-Borel, but maybe there's something similar to it but not precisely the same

Ted Shifrin

What do you mean by Heine-Borel?

Semiclassical

Ah, nope. Christopher Zeeman vs. Pieter Zeeman

Ted Shifrin

20:09

That in $\Bbb R^n$ compact is equivalent to closed and bounded?

Daminark

Yeah

Ted Shifrin

That doesn't even generalize to metric spaces ...

Daminark

Well metric spaces aren't necessarily locally compact, take a Banach space

Ted Shifrin

It doesn't generalize to $\Bbb R$ with a bounded metric.

Which is totally locally compact.

Daminark

Oh yikes, okay :/

Ted Shifrin

20:12

The right theorem invokes total boundedness.

And then it holds in any complete metric space, if I recall.

Daminark

Yeah

Ted Shifrin

G'night, @MikeM.

Mike Miller

hi

Ted Shifrin

Ain't seen much of you lately ...

I hope that's a good sign.

Alessandro Codenotti

hi chat

Ted Shifrin

20:23

Heya Alessandro.

Alessandro Codenotti

a subset of a metric space is compact iff it is totally bounded and complete @Dami

that's a generalization of Heine-Borel, I don't know how useful it is though

Daminark

I'm aware of that, I'm just wondering what's the class of spaces on which something closer to HB holds. Right now I'm wondering if vector spaces on the p-adics would do it

Which should probably reduce to proving that closed balls are compact in the p-adics

Astyx

20:39

Rehi chat

skullpatrol

Rewelcome pal

Liad

$X$ is normal. $p : X\to X \ ^ *$ is a quotient map that is closed. i want to show $X \ ^ * $ is normal.
Given $A \subset V$ where $A $ is closed and $V $ is open i want to find $U$ open nhbd of $A$ s.t $\overline U \subset V$.
So, $p \ ^ {-1}(A) \subset p \ ^ {-1}(V) $ , $X$ is normal so there is $U$
$p \ ^ {-1}(A) \subset U \subset \overline U \subset p \ ^ {-1}(V) $ .
Now
$A \subset p (U) \subset p (\overline U) \subset V$. and $p (\overline U) $ is closed because $p$ is closed. how to continue from here?

skullpatrol

No @HenningMakholm this is skullpatrol.

Ted Shifrin

heya @Astyx. Tu vis toujours?

Alessandro Codenotti

hi @Astyx

Astyx

20:50

Pour le moment on dirait @Ted

Peut être moins au moment des resultats

Ted Shifrin

OK ... on verra.

Alessandro Codenotti

how was your exam? @astyx

Ted Shifrin

Tu n'es jamais content :P

Astyx

Je serai content si je suis admis :p

Liad

nvm about my question , solved it.

Astyx

20:52

It's over now @AlessandroCodenotti, that's what matters :p

I'll see when I get the results

I don't want to make suppositions

Ted Shifrin

good, @Liad. :)

Alessandro Codenotti

wise decision :P @astyx

Astyx

It's probably the best attitude to have anyway :P

Alessandro Codenotti

I got the results of my exam back in the meantime, it went better than expected

I'll have to take the oral part of the exam on Monday

Ted Shifrin

Oh, that's cool, @Alessandro.

Astyx

20:57

Oh, so soon ?

I'm glad for you :)

skullpatrol

Oral is generally considered easier than written.

Astyx

Depends really

I think I'm better at written exams

skullpatrol

Not to mention lab exams :P

Astyx

Which are the worst of them all imho :)

skullpatrol

Yeah, they are a mix.

Alessandro Codenotti

21:04

I prefer oral exams over written ones, however the oral part of the probability exam is mandatory only if you get a good grade in the written part so I was kinda hoping I could avoid doing it

Ted Shifrin

So you wanted to keep failing the written?

Alessandro Codenotti

there are a lot of options between failing and a good grade :P

Ted Shifrin

So they reward you for a mediocre pass by letting you skip the oral? I guess the oral gives you a chance at distinction?

Alessandro Codenotti

I don't understand it either, it's the first exam I have organized like that

Ted Shifrin

What I conjectured is the only thing that makes sense to me, but I was sure you'd know what was going on :)

skullpatrol

21:09

I guess they wanna see who are the "used car salespeople" out there.

Alessandro Codenotti

this random variable was used only once per week by an old lady to decide how many groceries to buy, it's as good as new

Ted Shifrin

interesting that skull should mention car ... I still have nightmares that Alessandro is running me over ...

Alessandro Codenotti

lol

maths exams look easier than the driving one at the moment

Ted Shifrin

well, keep that attitude :P

Astyx

All my driving exams are both harder and more difficult than my math exams

Daminark

21:17

What's the distinction between being hard and being difficult?

Alessandro Codenotti

is there one?

Daminark

I didn't think so. Otherwise we can just say that Astyx is from the department of redundancy department

SteamyRoot

Time for an inappropriate joke

@Daminark I'm fine with a woman being difficult, but hard... :/

5

Alessandro Codenotti

@Daminark if there's one thing I learned from the cryptography part of the abstract algebra course is that redundancy is great

Astyx

I've never been redundant nor have I ever said the same thing twice in a row

6

skullpatrol

21:20

Lier, lier; neither have I.

Ted Shifrin

Well, Astyx is definitely an outlier.

Daminark

Yeah you're being a dishonest lier

Astyx

(by the way I meant to say "easier and more difficult")

Daminark

@Alessandro explain?

Alessandro Codenotti

@TedShifrin I was thinking about an outlier joke as well, I had too much probability

Ted Shifrin

21:21

well, until skull corrects his spelling, it stands ...

Astyx

I'm not sure wether me being an outlier is a good or a bad thing

Alessandro Codenotti

@Dami that's the basic idea behind error correcting codes, we add bits of redundancy to a message that are used only to detect (and sometimes correct) errors in transmitted message, rather than being used to send more informations

a classical example is the (7,4) Hamming code, every $7$ bits $4$ are used to encode the message we want to send and $3$ are parity bits that allow to find and correct a single error in any one of the $7$ bits

Daminark

Ah

skullpatrol

Have you watched Hamming's lectures on YouTube?

Alessandro Codenotti

I didn't know they were recorded, interesting

skullpatrol

21:31

Good stuff.

Alessandro Codenotti

I'll look them up after I'm done with my exams

I'm going to sleep now though, bye everyone

skullpatrol

Cya

Daminark

See you!

Astyx

Bye

What's this story about net neutrality ?

Ted Shifrin

In the US they want to spy on all our net activity and email and ...

SteamyRoot

21:35

They want to legalize the spying they kind of already do :^)

skullpatrol

...trump mania

Astyx

So this is specific to the US ?

Ted Shifrin

we presume so, Astyx.

SteamyRoot

Unless you somehow use an American ISP in France, yes :P

Astyx

(btw I still don't know wether "outlier" has a pejorative connotation, or any connotation ?)

SteamyRoot

21:36

If you feel left out, we get spied on too, don't worry.

Astyx

Yeah, I guess so :p

Ted Shifrin

I was mostly just punning on skull's misspelling.

outlier = non-conformist

Astyx

Yeah, I know, I'm just curious

Is it pejorative ?

skullpatrol

Nah

Ted Shifrin

I know you like to think I'm always insulting you, Astyx, but I'm really not :D

skullpatrol

21:38

Like a rebel is an outlier in a sense.

Astyx

Right, thanks :p

Eric Silva

hi chat

net neutrality is more about allowing ISPs to charge you more money for less services while screwing over competition than it is abt spying

skullpatrol

Money is always the bottom line.

hi btw

SteamyRoot

My university's ISP is better than any publicly available one... so if something similar gets legalised here, I'm moving to my office - permanently.

Astyx

Aren't they already allowed to do that ?

Eric Silva

21:52

@Astyx i mean slow down internet depending on what it's used for basically

which isn't legal

bc internet is treated like a utility, like water

Astyx

Oh, yeah, that would be terrible

SteamyRoot

As soon as you listen to the Soviet anthem, your internet slows down so hard you can't get past 5 seconds :^)

Astyx

I wouldn't be able to waste my time on the internet as efficiently as I am now

Eric Silva

yeah it would let companies like Spectrum which have services that compete with netflix slow down your internet if you try to access netflix

SteamyRoot

As much of a joke as that may be, you're basically allowing censorship through that way :P

Eric Silva

21:53

which is suuuuuper scummy

skullpatrol

The market is getting more nasty.

Astyx

The market always has been nasty

Hi @Dami

Eric Silva

yeah i think it's less abt the market in this case than corporate interests trying to shut down the market and control it

skullpatrol

Isn't that what happened to Microsoft?

Simply Beautiful Art

22:50

Say, if $x$ is algebraic, is it necessarily the case that $|x|$ is also algebraic?

(assume $x\in\mathbb C$)

SteamyRoot

23:20

Oh, wait, nevermind.

It's too early in the morning to think properly.

@SimplyBeautifulArt I'd guess so. If $z$ is algebraic, probably its complex conjugate is too. Then the product of those two, the modulus squared, is algebraic too. And finally, the root of the modulus squared should be algebraic too...

Fargle

23:39

Hello chat.

Las Vegas Raiders

Hi pal.

Simply Beautiful Art

@SteamyRoot Ah yes, that was simple xD

xD You know, in every instance when I used fractional calculus, it was always for a fractional calculus problem, until I just solved this problem

Basically, the following series is not extremely immediate to solve:
$$\sum_{n=0}^\infty\frac{x^n}{\Gamma(n+\alpha)}$$

For any $\alpha\in\mathbb N$, this trivial reduces to exponential functions, and this can be seen by differentiating the Taylor expansion for $e^x$.

In the same spirit, one can take fractional derivatives of $e^x$ to solve the above problem, with an end result of incomplete gamma functions

Transcript for

$Mathematics$ Mathematics