Mathematics - 2023-11-28 (page 1 of 2)

robjohn

00:04

@Koro Does this answer help?

Obliv

01:41

How can I rewrite this using euler's formula $(e^{-i(j-1)\alpha}(e^{i\beta}-e^{-i\beta})$

Do I write out each term individually like $e^{i\theta}=\cos\theta+i\sin\theta$ for each piece

$(\cos((j-1)\alpha) + i\sin((j-1)\alpha))(e^{i\beta} -e^{-i\beta})$ or does that last part have some kind of identity

that looks like a complex conjugate actually

wait nvm the whole thing is the real part of $e^{-i(j-1)\alpha}$

Ted Shifrin

Huh?

Obliv

nvm its not

Ted Shifrin

The second term is $2i\sin\beta$.

Obliv

should I combine the exponentials first

Ted Shifrin

Stop making random guesses.

Obliv

01:53

$e^{-i(j-1)\alpha+i\beta}-e^{-i(j-1)\alpha-i\beta}$?

Ted Shifrin

Not helpful. Did you read my comment?

Obliv

Oh ok

2nd term as in $(e^{i\beta}-e^{-i\beta})$?

it's ok if it doesn't simplify, i'm just doing a derivation so I probably went wrong somewhere

Yup, was missing something.

Obliv

02:21

I wonder how to evaluate $\sum\limits_{j=1}^N e^{i(j-1)\alpha}=1+e^{2i\alpha}+e^{4i\alpha}+...+e^{i(N-1)\alpha}$

for $N$ odd

the textbook shows when $N$ is even and we set up the coordinates differently we have $\sum\limits_{j=1}^N e^{i(2j-1)\alpha}=e^{i\alpha}+e^{i3\alpha}+...+e^{i(N-1)\alpha}$

I hope it evaluates to be the same thing

oh and I have another term I forgot about earlier.

mm i'm just gonna move on lol this question is taking forever.

Ted Shifrin

Don’t you know the formula for the sum of a geometric series?

Shame on you.

robjohn

$$\sum_{k=0}^{n-1}ar^k=a\frac{1-r^n}{1-r}$$

Obliv

interesting

Ted Shifrin

@robjohn ever the helpful one while I excoriate

robjohn

I was just typing that when you asked.

sorry if I ruined things

Ted Shifrin

02:31

And my colleague was upset 25 years ago that his calculus students didn’t know the law of cosines….

robjohn

precalculus is aptly named

Ted Shifrin

This site is going downhill apace — I got told to go **** myself when someone posted a totally wrong answer and I said it was wrong.

robjohn

don't say things that might upset some sensitive person

That would cut down a lot on most conversation

Ted Shifrin

I’ll quit participating.

robjohn

I wouldn't listen to people who say things like ****

Ted Shifrin

02:37

Democracy is teetering … scary all over.

Obliv

i "cut" someone off (even though there was PLENTY of room) when taking a right on some road and the person held their horn for like 30 seconds while I was taking a left at an intersection

drove next to me and said "DIE MOTHER******" in an almost european accent

robjohn

Yeah, I'm scared of what the next year might bring.

Obliv

sometimes people are weird

I'm not violent but I really wanted to step out of the car..

Ted Shifrin

In most of the US they could have pulled out a gun and shot you … with impunity.

Xander Henderson

@TedShifrin No. I have to rederive it every time I need it. I can't seem to get the formula to stick in my mind.

But it only takes a couple of seconds.

So it's... fine.

Obliv

02:43

Lol true, time to get a cybertruck

Xander Henderson

@robjohn My money is on another Trump election. With him presiding from prison. :/

Daniel Donnelly

@mick

you come here and ping someone

with @

that's then create a room once you're both here

robjohn

@XanderHenderson that would be scary, indeed.

Ted Shifrin

@XanderHenderson I spit on your money. Nothing personal.

Xander Henderson

@robjohn indeed. I have little hope.

@TedShifrin It is not a bet I want to win...

robjohn

02:48

The devil vs the lame man

Ted Shifrin

Democracy is taking huge hits world-round.

Obliv

What's that mean @Ted

robjohn

@TedShifrin It may be too long since people saw the alternative at work.

Ted Shifrin

Argentina on top of Italy on top of Brazil. Trompism everywhere.

Rachel Maddow is showing all the Naziism in American history of which I was unaware.

robjohn

@TedShifrin Italian sandwich on South American

Ted Shifrin

02:52

@robjohn I phrased that so perfectly for you.

And then Hamas … and right-wing Israel. I give up.

Xander Henderson

@TedShifrin not too mention AMERICAN college students chanting "From the river to the sea."

Ted Shifrin

And the three Palestian American college students just shot in Burlington.

Xander Henderson

Oh, jeebus. I hadn't heard about that. I've been away from the Internet most of the week.

Ted Shifrin

I blame most of this on Newt Gingrich and Fox,

Xander, better sanity that way.

Xander Henderson

@TedShifrin Ugh... THAT man. Indeed, I think that he and Rush are largely responsible for the current state of affairs in the US.

Anywho, I should go to bed. I have exams to write.

And my phone just turned black and white.

Obliv

03:00

And have a good night

AMDG

Man, almost like all the socialism we see today was a two-faced plot to gain power and exploit society as evidenced by the way that "corporations" exploit society and legislation to everyone's detriment while nothing gets done; and while those who espouse these unreasonable systems of government over empires and come to power end up taking turns in oppressing and exploiting. Talk about projecting amirite.

Night

Obliv

I have to do a presentation on the mesozoic era of geologic time tomorrow.. guess I better get started on that

robjohn

@Obliv Ask Ted about it before he leaves ;-)

Obliv

Oh snap

OH thank god it's actually due next week

Yay more time for procrastination

robjohn

Oh, I hadn't noticed that he'd already left.

Obliv

03:16

Yeah I think he left right before you roasted him

copper.hat

Wow, I missed twin primes and spitting.

Still reeling a bit from the news of the Dublin riot recently.

oscarmetal break

03:45

I need someone's help with a question. Given a vector space $V$, define $X$ to be the set of all nilpotent transformations on V, and G=$Aut_F(V)$. Define the conjugation on X by acting element from G. I need to show the orbit over X under this action is finite. So far all I can tell is that a nilpotent transformation after conjugation is still nilpotent but I don't know how does it help.

Ted Shifrin

@robjohn You are now in cahoots with Leslie?

@copper.hat A French cook apprentice and Brazilian the heros … ironically.

@oscarmetalbreak That means it makes sense to say you have a group action on $X$.

What is $F$?

You’re asking that there be only finitely many matrices conjugate to a fixed nilpotent matrix. I do not believe that in general.

oscarmetal break

@TedShifrin F is not specified but for convenience, I assume F is a field. Not a fixed nilpotent matrix. In this case, X is the set of all nilpotent matrices over V and G is all automorphisms over V, so that G is acting on X by conjugation.

Ted Shifrin

If $F$ is not a finite field, I do not believe it.

What are all the matrices conjugate to $\begin{bmatrix} 0&1\\0&0\end{bmatrix}$ working over $\Bbb R$?

oscarmetal break

04:05

But since G has been specified to be all automorphism over V, then I guess one can always conjugate the nilpotent matrix with another matrix that is invertible, isn't it?

So an orbit in this case on X is $\{gTg^{-1}|T \text{is nilpotent} g\in Aut_{F}(V)\}$

Ted Shifrin

Fix $T$.

oscarmetal break

Yes. T is fixed

Ted Shifrin

You wrote it wrong.

Maybe you misstated the problem. Maybe you want to show that the number of orbits is finite. Not that each orbit is finite.

oscarmetal break

Right. I should have said the number of orbits is finite

copper.hat

@TedShifrin Yep, many ironies there. Hard to determine the underlying facts. I am not sure why folks expect people from other countries to be perfectly behaved when there is plenty of antisocial local behaviour.

oscarmetal break

04:15

Is there anything like if two nilpotent transformations $T_1$, $T_2$ span the same subspace then one can always find an automorphism $g$ on V such that $gT_1g^{-1} = T_2$?

Maybe also need to assume $V$ is finite-dimensional

Ted Shifrin

Yes, definitely need finite dim.

Do dim 2 and dim 3. Give a list of the “standard” matrices in each orbit.

Other than the zero matrix, there’s 1 for $n=2$ and 2 for $n=3$.

oscarmetal break

I think I am close. I am now a bit uncertain. For a fixed k, there are $\binom{n}{k}$ ways of choosing the bases to form a subspace. Is it possible that I can always find $\binom{n}{k}$ nilpotent transformations of index k which can produce one of the subspaces? If it is true. I guess it will be suffice to give the proof.

$n = dim(V)$

Ted Shifrin

04:38

Index? Give me the answers for 2 and 3. Start by proving those.

oscarmetal break

what do you mean by standard matrix? I am a bit obsolete to linear algebra now.

Ted Shifrin

Don’t worry what I mean. Give me a simple nonzero matrix in the $2\times 2$ case to which every nonzero nilpotent matrix is similar.

oscarmetal break

like $\begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}$ ?

Ted Shifrin

Right. Good. What about $n=3$?

oscarmetal break

04:53

Like $\begin{pmatrix} 0&1&0 \\ 0&0&1\\ 0&0&0 \end{pmatrix}$?

I don't really know how to compute another?

copper.hat

sure you do

start with the 2x2 example.

oscarmetal break

Like $\begin{pmatrix} 0&0&1\\ 0&0&0\\ 0&0&0\end{\pmatrix}$

The first one has index 3 the second has index 2.

So they are not similar

How can you conclude this number?

Ted Shifrin

Why did you move the 1 over?

Think about eigenvectors ….

oscarmetal break

05:16

so it is like$\begin{pmatrix} 0&1&0\\ 0&0&0\\ 0&0&0\end{\pmatrix}$?

Ted Shifrin

Yes.

copper.hat

edit and remove the slash ahead of the last pmatrix or use Matlab notation

oscarmetal break

How to edit a comment?

I read that every nilpotent transformation can be written as an upper diagonal matrix. Is it how you conclude the number of them in dim = 2 and 3? ( I must have been learned this in linear algebra)

copper.hat

you can only do it for some small time period after posting, click on the little downward pointing triangle between your avatar and the comment.

oscarmetal break

and then? I saw the option of permalink.

copper.hat

05:26

the option only lasts for a short while.

it will show edit | delete | flag for moderator at the bottom of the popup

oscarmetal break

I didn't see the edit option there though

copper.hat

maybe you are not quick enough? maybe its a rep thing?

oscarmetal break

I guess it is because I am not quick enough, the comment was 14 mins ago

copper.hat

05:42

next time you comment try editing immediately :-)

oscarmetal break

I see. I will check that next time

Anyway. Thank you @TedShifrin. I believe I have got you mean. Now I can consult the Jordan form with the problem now.

robjohn

08:09

@TedShifrin Nah, I just thought that you could help with the Jurassic and I could help with the Cretaceous ;-)

one potato two potato

08:37

I think nobody will answer but is there some kind of inclusion relations between convex core and compact core of a geometrically finite (hyperbolic) 3-manifold?

hard to ask this kind of question on the main site because it's just a one line question usually regarded as a bad question.

Mad

09:33

@Thorgott i dont understand how does this relate to the step being made, where is the differential coming from?

Mad

10:15

oh apparently then the diagramm commute

i didnt know that!

Jakobian

11:53

My perfectly fine answer didn't get any attention :(

actually there's a minor mistake

mick

12:15

0

Q: Minimal distance between zero's $d = D(f(z)) = \inf_{i \neq j} |(z_i - z_j)| s.t. f(z_n) = 0 $?

Let $f(z)$ be a transcendental entire function. Hence $f$ is not a polynomial. Assume $f(z)$ has infinitely many zero's $z_n$ $$f(z_n) = 0$$ Lets say that $f(z)$ is given by a taylor series. Im interested in the distance between those zero's $$d = D(f(z)) = \inf_{i \neq j} |(z_i - z_j)| s.t. f(z_...

broad question. any ideas or references ?

Lucky Chouhan

@Jakobian is an attention seeker.

@robjohn have you ever met @leslietownes ?

Jakobian

@LuckyChouhan that's not what it means

I want appreciation for what I'm writing

attention seeking is a behaving with the goal of acquiring attention

shintuku

12:32

destigmatize attention seeking, respond with proper psychological care and affirmation

Jakobian

I have nothing against people that are attention seeking on its own

but I don't think like someone that's seeking attention, I don't need attention, I need people to read my answers

so I know that I'm not just wasting my time with this

its completely different

I'm not really satisfied with just having someone attention, my goal is to seek something further - this is the difference, the end goal you're doing it with

yes it is attention, and yes I am seeking it, but I'm not seeking attention only

I am seeking a specific type of attention

its not like, oh yeah someone paid attention to me, now I'm happy. No

lets learn what subtlety is

Alessandro Codenotti

14:54

@Jakobian do you see why this continuum, obtained by attaching together two buckethandle continua at their endpoints, is circle-like?

Jakobian

15:12

Not really. Because wouldn't the point of attachment be problematic?

unless I'm misunderstanding what it means to attach those

Alessandro Codenotti

15:56

It's not clear to me either. But Illanes and Nadler claim this space is both arc-like and circle-like. While I see the former, I don't understand why the latter is true

Jakobian

16:19

@TedShifrin math.stackexchange.com/a/4816103/476484

I think this would be my answer to the question I've asked yesterday. Just simple application of mean value theorem

Honestly, I'm not even sure if the assumptions imply that $F$ is continuous

continuity of those partial derivatives does heavy lifting here

Ted Shifrin

Andrew made the same comment I did (about local extensions), but then you did the straightforward proof I suggested reducing it just to the single-variable case. Glad it worked out.

Demonic @Alessandro I know I'll regret asking this, but ... what does circle-like mean?

Xander Henderson

@TedShifrin I means "like a circle". Duh.

Ted Shifrin

What if I dislike circles?

Jakobian

It means any cover has a circle-like refinement

Xander Henderson

Then you aren't invited.

Alessandro Codenotti

16:26

For every $\varepsilon>0$, there is an open cover $U_1,\ldots,U_n$ of the space made of pieces with diameter $<\varepsilon$ and such that $U_i\cap U_j\neq\varnothing$ iff $|i-j|\leq 1$ or one of them is $U_1$ and the other is $U_n$

Jakobian

meaning its a chain of open sets $U_1, ..., U_n$ such that $U_i\cap U_{i+1} \neq \emptyset$ and $U_1\cap U_n\neq\emptyset$ but all other pairs are disjoint

so it resembles a circle in some way

Ted Shifrin

Oh, so a wedge product of finitely many circles is circle-like.

Alessandro Codenotti

I don't think so

Ted Shifrin

This definition makes it sound like the Cech first homology should be $\Bbb Z$, but your picture suggests not.

Jakobian

wedge product of two circles should break because of the point they're attached to, I think

Ted Shifrin

16:28

Oh, no, what I said is wrong. But then why is that creature circle-like?

Alessandro Codenotti

(turns out that being circle-like, for a compact connected metric space $X$, is equivalent to saying that $X$ can be written as the inverse limit of a system of circles with surjective maps)

Ted Shifrin

Yes @Jakobian

Alessandro Codenotti

@TedShifrin I have no clue. I can kind of see that it is arc-like (same definition but $U_1$ doesn't touch $U_n$) and I'd like to understand why it is circle-like, because I believed that a space cannot be both but apparently I was wrong

Ted Shifrin

Why don't you have the trouble at the junction point, just as Jakobian said about my comment?

Jakobian

@AlessandroCodenotti isn't pseudo-arc both?

Alessandro Codenotti

16:32

Is it? Why is it circle-like?

Jakobian

I think I remember my topology professor mentioning it, yes

oh maybe it was circle of pseudo-arcs

not pseudo-arc

okay no it was the pseudo-arc

leslie townes

think of all the misspent time that led to all of the statements in that paragraph. at least it presumably kept those people off the street for a while.

Jakobian

oh I think my professor was talking about pseudo-circles

Ted Shifrin

Well, this gives you a perfect opportunity to rant about the bad person that was RL Moore.

Jakobian

actually

Ted Shifrin

16:39

I think all this arcane point set topology is pseudo-math. There — I said it.

Jakobian

I think it might be pretty applicable

there is lakes of Wada for example and its also kinda bizarre

or Mandelbrot set

Ted Shifrin

Applicable to other arcane point-set topology, you mean?

leslie townes

the dustbin of history called, and it wants its math (and beloved lead hero figure) back.

Jakobian

No, to actual life, maybe

Ted Shifrin

Much as I want to complain about how arcane scheme theory is in algebraic geometry, sometimes it just shows up very naturally.

LOL @actual life, ... maybe.

Xander Henderson

16:43

@Jakobian Neither is particularly interesting from a topological point of view.

Jakobian

Continua arise when talking about dynamical systems, no? And those are all things relating to dynamical systems

that sounds applicable in some way

Xander Henderson

@Jakobian Sure. But it is not the topology of these spaces which are particularly interseting.

leslie townes

i want to found an art school where the gimmick is that the students are blank slates who never look at the work of other artists, except periodically i give them pictures of garfield and ask them to make more pictures of garfield.

i guarantee that the results will be as interesting as point set topology.

Jakobian

@XanderHenderson this answer suggests otherwise math.stackexchange.com/a/2622423/476484

leslie townes

or maybe i just tell them the attributes of garfield.

Xander Henderson

16:47

@Jakobian That doesn't really convince me. It is either path connected or it isn't. The fact that it is hard to answer the question does not, in my mind, make the question itself terribly interesting.

Jakobian

The answer is saying that if local connectedness holds then some other interesting (for someone studying Mandelbrot set) things hold

Xander Henderson

From the answer:

> Note that experts are not so much interested in this question by itself (local connectedness of the Mandelbrot set) but rather are interested in a conjecture called genericity of hyperbolicity, which is known to be true IF the Mandelbrot set is locally connected.

The topology itself is not particularly interesting.

Jakobian

The thing you cited implies it is interesting

Ted Shifrin

runs to kitchen to pop popcorn

Xander Henderson

@Jakobian I have told you that I don't find it interesting. You are telling me that I am wrong to not find it interesting. Thank you for invalidating my opinion. This matters a lot to me.

Jakobian

16:52

We are talking about if this is interesting to someone studying the particular object in the particular community and the answer clearly is yes, because it has potential to answer questions that experts would like to know

Xander Henderson

@Jakobian That seems to be what you are talking about. I was very clear that I was expressing an opinion.

Koro

Let's talk history.

Jakobian

I hope its about math at least

Koro

Did Russia (the Russian empire) ever have colonies?

Xander Henderson

@Koro Yes, depending on what you mean by "colony".

Russia colonized Siberia, and had settlements in Alaska and North America.

Pretty sure they also had settlements in the Pacific (I've been to one of their Hawai'ian settlements, for example).

Koro

16:56

I mean taking over another government or monarchy or whatever it was at that time.

@XanderHenderson But I read that Alaska was Russia's and that Russia sold it to US at 10 million dollars. related

Xander Henderson

@Koro That seems like a fairly restrictive notion of "colony".

@Koro Yes. history.state.gov/milestones/1866-1898/alaska-purchase

Jakobian

Poland had a state between Russia and China

Koro

you mean the country Mangolia?

Jakobian

me?

Koro

yes

Jakobian

16:59

huh?

Koro

@Jakobian I didn't know that and I don't think it's true.

Jakobian

its true

Jaxa (state)

Jaxa (Chinese: 雅克薩; Polish: Jaxa, Jaksa) was a 17th-century microstate in North Asia with its capital in Albazino existing between 1665 and 1674. It was located on the border of the Tsardom of Russia and Qing China, by the Amur river. Its population was made up of Polish and Ukrainian refugees from the Tsardom of Russia, and the indigenous Evenks and Daurs. It was established from the territory of the Tsardom of Russia in 1665 by Nikifor Chernigovsky and his men, who fled Russia, and existed until 1674 when it was incorporated back to that country. == Name == The name of Jaxa originates from the...

We once bordered China

Alessandro Codenotti

Fun fact: in 1989 Poland bordered three countries, in 1993 it bordered seven, none of which is among the three from 1989

Regarding dynamical systems and continua I spend a lot of time thinking about actions of $\mathrm{Homeo}(X)$ on things, where $X$ is a continuum, and the topology of $X$ is often useful in extracting dynamical properties. Lately I've been thinking about Peano continua in general, before that I was focussing on dendrites

Jakobian

@AlessandroCodenotti with what topology?

Koro

I thought Russia being the country with the largest area had the most no. (total 11) of timezones but it's not true.

Alessandro Codenotti

17:04

compact-open

Jakobian

Oh I see

Koro

France has the most no. of timezones !!

Alessandro Codenotti

And the longest land border of France is with Brazil

France borders the Netherlands too, but only in the carribeans

Ted Shifrin

17:25

Now we've merged into arcane geography. I nominate arcane foodstuffs from different cultures.

user 85795

17:58

How was the popcorn

Ted Shifrin

Stale.

user 85795

Freshly popped?

Alessandro Codenotti

@Jakobian What is this from?

Jakobian

18:21

@AlessandroCodenotti I didn't download that one, but PSEUDO-CIRCLES AND UNIVERSAL CIRCULARLY CHAINABLE CONTINUA by James Ted Rogers, Jr, discusses some similar things

Obliv

Ello' I require assistance with a pre-calc question

Xander Henderson

"Ello'"?

Ted Shifrin

Apostrophe dislocated. Summon the paramedics.

Xander Henderson

@TedShifrin Glad I'm not the only one thrown off by that. :D

Obliv

I have $\frac{x}{1-x-x^2}$ I reckon it's $\frac{-x}{x^2+x-1}$ but i forget how to do partial fractions

i realized my mistake it shoulda been 'ello?

CHEERIO

Xander Henderson

18:26

@Obliv What does partial fractions have to do with it? You multiplied upstairs and downstairs by $-1$. Isn't all you've done?

I'm confused.

Obliv

does this link work for you

my prof posted a generating function for fibonnaci numbers but I didn't understand the partial fraction part (expectedly)

Ted Shifrin

Factor the quadratic, of course..

Obliv

in case it doesn't work

Yea I could do $\frac{-x}{(x+1)(x-1)+x}$ or $\frac{-x}{x(x+1)-1}$

Xander Henderson

@Obliv There is a general principle at play: you have some expression, and you want to write it in a different way. So write out the alternative, using variables to represent any unknown parameters, and then compare that to the original and work out the parameters.

@Obliv What? Why?

Factor the polynomial.

Obliv

Huh?

Xander Henderson

18:30

$x^2 - x - 1 = (x-r_1)(x-r_2)$ for some (possibly complex) numbers $r_1$ and $r_2$.

If you are hoping for a partial fractions decomposition, the first step is to factor the denominator, no?

Obliv

Hmm..

I didn't know you could just factor numbers like that

Xander Henderson

@Obliv That's essentially the fundamental theorem of algebra...

Ted Shifrin

Factor numbers?

Obliv

Fact or fiction?

I was just staring at paul's online math notes section for partial fractions and didn't immediately see that situation

but thank you for that @XanderHenderson !

Ted Shifrin

Remember the quadratic formula?

Obliv

18:34

u know,... i didn't until my diff eq class in the spring needed it

so now i do!

Xander Henderson

@Obliv If you haven't factored the denominator yet, you are not yet to the point where anything about partial fractions is likely to be helpful...

Obliv

$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

surely that's correct

Xander Henderson

Once you factor that denominator, the goal is to write $$\frac{-x}{x^2-x-1} = \frac{\text{something}}{x-r_1} + \frac{\text{something else}}{x-r_2}.$$

Obliv

for linear terms we have constants for $\text{something}$ and $\text{something else}$ right

Xander Henderson

By some relatively straigh-forward degree arguments, you can argue that both "something" and "something else" must be degree $0$ polynomials, i.e. constants.

So, suppose that they are constants, and try to work out what they are.

Obliv

18:38

I recall but forget there was a fast way to do these, using some "cover up" method by an engineer

Xander Henderson

@Obliv You are focusing on specific techniques (e.g. Heaviside's "cover up"). Think about what you are trying to do. Don't obsess over the specific methods which you don't remember.

Obliv

the way I did it was $\frac{-x}{x^2-x-1}=\frac{A}{x-r_1}+\frac{B}{x-r_2}\to -x = A(x-r_2)+B(x-r_1)$?

then put in the value for $x$ to cancel one of them

so $-r_2 = B(r_2-r_1)$

and $-r_1=A(r_1-r_2)$

Xander Henderson

Sure. That works.

Obliv

but isn't that 4 unknowns and 2 equations

er wait i'll keep reading the file

Xander Henderson

No, $r_1$ and $r_2$ are the things you were supposed to have already determined, by factoring the denominator.

Obliv

18:43

well it's somehow the case $r_1 = \frac{1+\sqrt{5}}{2}$

oh

Xander Henderson

14 mins ago, by Xander Henderson

$x^2 - x - 1 = (x-r_1)(x-r_2)$ for some (possibly complex) numbers $r_1$ and $r_2$.

Obliv

so thats why ted asked me if I knew the quadratic formula..

Lol

$\frac{1\pm\sqrt{1+4}}{2}=\frac{1\pm\sqrt{5}}{2}$

got it, thank you @XanderHenderson

Thorgott

@TedShifrin I told him the same thing!

Obliv

nvm

Thorgott

I guess the issue is, though, that while Cech cohomology is compatible with inverse limits, the inverse limit of a sequence of maps between $\mathbb{Z}$s isn't necessarily just that

(those have actually been completely classified by Katsuya Eda iirc)

Obliv

18:55

why does it become negative from step 1 to 2

actually what even is happening in that first step

Alessandro Codenotti

@Jakobian hmm I was looking for a reference about the circle-likeness of the pseudoarc

Xander Henderson

@Obliv Look at the actual numbers you are working with? What are $\varphi$ and $\rho$?

What can you say about their reciprocals?

Do some computations!

Jakobian

@AlessandroCodenotti it might be there too

Alessandro Codenotti

I didn't see it, but I only skimmed

Obliv

oh wait yeah they're related by $p=-\frac{1}{\phi}$

is the infinite sum part an identity

Jakobian

19:00

@AlessandroCodenotti jstor.org/stable/2045086

@leslietownes

this one is for all my fans

Of course, integrals here are gauge integrals

Alessandro Codenotti

@Jakobian Thanks!

Jakobian

where I used chain rule on general rectangles, a theorem I proved here on math.stackexchange (the proof is easy)

If $T$ is a closed or half-closed interval, the theorem implies differentiability at the endpoints and all functions should also be differentiable at the endpoints whenever stated

Assumptions of differentiability theorem are that $x\mapsto f(x, t)$ is measurable for all $t$ (follows from continuity), integrable for some $t$ (follows from continuity), $\partial_t f$ exists and is bounded by two integrable functions

So this is satisfied by first three points so indeed we can use it here

The properties of continuity of $f$ and $\partial_t f$ are used to prove the partial derivatives of $F$ are continuous, so this is important

Probably can be weakened though, but then it'd be harder to prove this theorem and differentiability at the endpoints (personally I don't want to jump into that rabbit hole)

leslie townes

19:17

boo

👎

user 85795

hiss 🐍

Ted Shifrin

Are we redoing Halloween?

Thorgott

@Thorgott ah no, the Eda result is much harder and classifies inverse limits of free groups of at most countable rank

inverse limits of free groups of rank 1 can be classified by hand

Ted Shifrin

19:33

Thor is now mumbling to himself ….

user 85795

Isn't that called thinking outloud.

Aka, self-dialogue.

Thorgott

you can consider yourself glad that I don't use this place to do my mumbling about category theory and homological algebra :P

user 85795

Where did you do that :^)

(please say on paper :)

Xander Henderson

Paper!? Real mathematicians draw figures in the sand with a stick!

3

user 85795

Please don't disturb my circles.

With a trompster standing behind you.

Xander Henderson

19:43

@user85795 The orange man is BEHIND ME?!

user 85795

Oh well, the tides of time will wash it all away eventually...

oscarmetal break

21:22

Is there any way that I can show $x^2+1$ in reducible in a field $F_{4k+1}$ where $4k+1$ is prime?

Is it possible in general? I know it is true for 5

Jakobian

@oscarmetalbreak If $k$ is a square then $x^2+1 = x^2-4k = (x-2\sqrt{k})(x+2\sqrt{k})$

If there are counter-examples, they can't be squares. Like $k = 3$ maybe

For $k = 3$ its reducible as well... hmm

Ted Shifrin

Do you remember Wilson’s Theorem?

Thorgott

I don't

Jakobian

$(n-1)!\equiv -1 \pmod{n}$ iff $n$ is prime or $n = 1, 4$

I think it was

Ted Shifrin

If $x=1\cdot 2\cdot\dots\cdot (p-1)/2$, then what is $x^2 \pmod p$ when $p=1\pmod 4$?

Jakobian

21:31

I see

oscarmetal break

I don't get it

Jakobian

Another hint is that $k \equiv -(p-k)\pmod{p}$

Well maybe this will be easier

You know that $1\cdot 2\cdot ...\cdot (p-1) \equiv -1 \pmod{p}$

so now what if you group the terms $1, 2, ..., (p-1)/2$ and $(p+1)/2, ..., p-1$

oscarmetal break

Since p is a prime then the square $1,2\dots, (p-1)$ equvilent to 1 \pmod{p}$?

ZaWarudo

22:09

I was reading online that since the harmonic series diverges, for each $n \in \mathbb{N}$ there exists $n_0 \in \mathbb{N}$ such that $\sum_{k=1}^{n_0} \frac{1}{k} < n < \sum_{k=1}^{n_0+1} \frac{1}{k}$. I tried to prove this, arguing as follows: since the harmonic series diverges, for each $M \in \mathbb{R}$ there exists $n_M \in \mathbb{N}$ such that $n \ge n_M \implies \sum_{k=1}^{N_M} \frac{1}{k}>M$. So, fixing an arbitrary $n \in \mathbb{N}$, this holds in particular for $M=n$. Moreover, $\sum_{k=1}^{N_M} \frac{1}{k}<\sum_{k=1}^{N_M+1} \frac{1}{k}$ because the sum is a positive terms sum.

Is this correct? Or is there another reason why that inequality holds?

oscarmetal break

22:29

Is it possible to argue it with a well-ordering principle? Given a N, consider the set of partial sum of the series less than N and the partial sum greater than N. There exists supremum and infimum in both sets, so if there exists any other partial sum of the series between the supremum and the infimum will be a contradiction.

Summerday

can someone help me here:

0

Q: How can I show that a Poisson process with my definition below has stationary and independent increments?

We had the following definition: Let $(\Omega, \mathcal{F}, (\mathcal{F}_t)_t, \Bbb{P})$ be a filtered probability space. An $(\mathcal{F}_t)_t$ Poisson process $(N_t)_{t\geq 0}$ is a right continuous adapted process s.t. $N_0=0$ and for $0\leq s\leq t, k\in \Bbb{N}$, $\Bbb{P}(N_t-N_s=k|\mathcal...

probability probability-theory conditional-probability poisson-distribution poisson-process

Jakobian

@ZaWarudo But it isn't true

No matter which convention for N you'll take

Because for either n = 0 or n = 1 you can't take such n_0

@oscarmetalbreak Not well-ordering principle

Rather that N is well-ordered

@oscarmetalbreak no

oscarmetal break

22:47

I see. I still can't get that $x^2+1$ is reducible in a field of $F_{4k+1}$. However, I find it is possible with $x^2-1$ using the Wilson theorem. Is there any other way to tell there exists an isomorphism between $F[x]/(x^2+1)$ and $F\times F$ for a field $F_{4k+1}$. I try to construct a map from $F[x]$ to $F\times F$ but don't know how does it have a kernal of $(x^2+1)$.

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