Mathematics - 2021-12-11 (page 2 of 2)

shintuku

17:06

@TedShifrin btw ted do past professors do research for fun?

say when there's no more pressure to publish or anything like that

Wolgwang

> Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

What's wrong in $\dfrac{4×48×47×46×45}{5!}$?

Leaky Nun

@Wolgwang why $5!$ ?

user69608

Hey! can someone give hint to this problem?codeforces.com/contest/406/problem/B

shintuku

@Wolgwang do a couple of tree branches to figure out why you don't need to divide

Wolgwang

Because we are choosing 5 cards and order doesn't matter so we subtract 5!

user69608

17:14

its not a live contest

Wolgwang

@shintuku Tree branches?

leslie townes

he's asking you to smoke marijuana before you finish your problem

2

Wolgwang

(P.S. I know $\binom{4}{1}\times \binom{48}{4}$ gives right answer)

Leaky Nun

@Wolgwang think about what $4 \times 48 \times 47 \times 46 \times 45$ represents

shintuku

@Wolgwang

Wolgwang

17:19

@LeakyNun Total number of arrangements.

shintuku

each juncture corresponds to a card choice. you start with four initial junctures. choosing a path in there, you have 48 new junctures, etc.

Leaky Nun

@Wolgwang of?

Wolgwang

@LeakyNun Of 5 cards (?)

Leaky Nun

how did you obtain the expression $4 \times 48 \times 47 \times 46 \times 45$?

what does the $4$ mean? what does $48 \times 47 \times 46 \times 45$ mean?

Wolgwang

Fundamental principle of counting... There are total 5 cards : _×_×_×_×_ In 1st card slot, 4 ace cards can be inserted and now there are 48 remaining cards which will be inserted in 2nd,3rd,4th,5th slot.

leslie townes

17:23

wolg if you order the ace first and don't have an ace anywhere else, you aren't multiply counting hands that have the ace in other positions when you compute that product

Leaky Nun

@Wolgwang so now think back about $5!$

you claimed that you have $5!$ because the order doesn't matter

is it true that the order doesn't matter?

Wolgwang

@LeakyNun Yes

Oct 3 '15 at 20:11, by Ted Shifrin

@Karim: At the end you divide by $k!$ because you don't care about the order of those $k$ things chosen.

Leaky Nun

@Wolgwang but the ace card needs to be on the first slot

Wolgwang

@LeakyNun Why? :-/

Leaky Nun

that's what the $4$ means

maybe it would help to think about "what" you're counting, every time you write an expression

so the set you're required to count is "the 5-card combinations with exactly one ace"

you have first transformed it to "5-card combinations where the first card is an ace and the rest of the cards are not"

and for that, you first count the 5-card arrangement where the first card is an ace and the rest of the cards are not

which gives you $4 \times 48 \times 47 \times 46 \times 45$

Wolgwang

17:28

@leslietownes @LeakyNun Ohk I think I have got it!

Leaky Nun

and the difference between the second set and the third set is $4!$, not $5!$

and this highlights the importance of not blindly applying formulas, but thinking about what each term means

Wolgwang

So choosing cards with an ace is a two steps process.

Ted Shifrin

Counting is tricky!

@Wolgwang Why did you quote me from 6 years ago?!

robjohn

Ted Shifrin

Kentucky is all tornadoed …

robjohn

17:33

I saw that... along with 5 other states

Prithu biswas

@robjohn huh?

shintuku

it is time we develop terraforming

Ted Shifrin

Doubtless a libtard plot.

Wolgwang

@TedShifrin Because that was relevant to the discussion and I was doing something around those lines. :)

Ted Shifrin

@Wolgwang you can also use $\binom{48}{4}$.

user69608

17:36

any hints please,I literally can't think anything. codeforces.com/contest/406/problem/B (not a live contest)

Prithu biswas

@robjohn Wait.... you mean they misspelled "meth" to "math"?

robjohn

@Prithubiswas you were commenting on the difference between "math" and "meth"

@Prithubiswas yes

Ted Shifrin

That’s an old story …

Wolgwang

@Wolgwang Ted, Yes I had tried that first.

robjohn

@TedShifrin It can still be apt

Prithu biswas

17:38

r/funny

It took me an embarrassingly long time to see the joke.

Ted Shifrin

@robjohn or inapt.

leslie townes

or inept

Ted Shifrin

Stooopid English

Yet another OP for Ted to ignore.

leslie townes

is there a site or network-specific rule for questions about primary material that is only provided in a language other than english? the other day someone posted a question in mandarin, and i thought about voting to close but didn't see applicable ground. if there is one, it might be buried in site policies and not in the summaries that are provided on a close vote.

"here's some link to some journal article, click through and figure out what i'm asking about" already seems wrong even if the journal article is in english, but it seems particularly wrong if the SE question is in english but i need to know russian to understand what i'm being directed to.

for my purposes, the link might as well not be there at all

Leaky Nun

@leslietownes lol

do you have a link?

Ted Shifrin

17:52

@leslie There is an English version, but it's accessible only with library privileges. The whole interaction was frustrating, but I get all the blame. It's an imposition for him to tell me the definition and clarify the main issue I asked (but apparently he didn't understand that it was a question, repeatedly).

Look at my previous link, @Leaky.

I probably once knew just enough Russian to stumble through the article, but it certainly is not worth my trouble given all this.

leslie townes

leaky: for the thing in mandarin, no. for ted's question, the link has downloadable russian (through another link) but no public english version.

Ted Shifrin

In general, MSE users are told to make their questions self-contained. It's not unreasonable to provide definitions, especially if there's ambiguity. I sort of want to think about the question, but my inner leslie says I shouldn't waste the time.

leslie townes

this strikes me as weird for the separate reason you often identify, ted. if a person isn't paying attention to definitions, why are they knee deep in this article in the first place.

what was it in the universe that brought them there.

Ted Shifrin

It's too much trouble to make the question a good question. Just help me anyway.

leslie townes

"but what is the def---" "THERE'S NO TIME!!"

Koro

17:58

🍭🍬🥜🍰

Ted Shifrin

It's related to a fundamental result in elementary differential geometry, I think, but it's not quite the standard setting. If you have a surface with $K=0$ but no umbilic points (i.e., no planar points), then it must be a ruled surface with the tangent plane constant along rulings.

@leslie Despite the attack from the other person, I do realize that my inference was stated not as a question but as an implied "confirm, please?" ... and for a non-native speaker, for example, that might be confusing. I think it was "confusing" for native speakers as well.

leslie townes

i dunno, it fairly clearly wasn't an answer (if for no other reason than, uh, it wasn't posted as one?), and it makes sense as an elaboration of the previous explicit question. i think there's an expectations barrier, as opposed to a language barrier.

Wolgwang

331

A: Do posts have to be in English on Stack Exchange?

tl;dr: Unless you're posting on a language-related site (e.g. French Language) or a site where all questions are expected to be in a different language (e.g. Stack Overflow in Spanish), yes, all posts are expected to be in English. A special case exists for the Russian language; see below. What...

leslie townes

good to know. on math.se it's not that acute of an issue. if someone states their problem simply and clearly enough, and gets the stuff in math mode right, one can often figure it out, and who really cares. but context-dispositive cites to external sources in other languages, oof.

Ted Shifrin

@Wolgwang It's a fact of life that many journal articles (and textbooks) are in different languages. There's nothing wrong with that. But good form is not to expect me to read pages of an article in order to understand or answer a question here.

The other issue is paywalls. When I say I cannot access the article, don't tell me how to do it when it evidently doesn't work. (Many, but not all, users here have access to university libraries. I no longer do.)

leslie townes

18:12

yeah. people who have institutional access can't always tell that there is a paywall for the rest of us.

Balarka Sen

18:44

I'm having trouble understanding the double centralizer. If $M$ is a left $R$-module, the "centralizer" is $\mathrm{End}_R(M) = \{T : M \to M : T(rx) = r T(x)\, \forall r \in R, x \in M\}$, basically because of $"T r = r T$. One can reformulate this as follows, I suppose: consider $m_r : M \to M$, $x \mapsto r x$, multiplication on the left. Then $T \circ m_r = m_r \circ T$.

Then they consider $\mathrm{End}_R(M)^{op}$, which is the same ring as $\mathrm{End}_R(M)$ but multiplication happening in reverse order.

$M$ is a left $\mathrm{End}_R(M)$-module by $T \cdot x = T(x)$ as $(T_1 T_2) \cdot x = T_1(T_2(x)) = T_1 \cdot (T_2 \cdot x)$. So now consider it as a right $\mathrm{End}_R(M)^{op}$-module, where $S^{op}$ is the same ring as $S$ but multiplication happening in reverse order

This makes sense: $x \cdot (T_2 T_1) = (T_1 \circ T_2)x = T_1(T_2(x)) = (x \cdot T_2) \cdot T_1$.

The "double centralizer" is $\mathrm{End}_{\mathrm{End}_R(M)^{op}}(M)$

Ted Shifrin

chokes

Balarka Sen

Yeah

It's nuts

So let's see, if $f : M \to M$ is in the double centralizer what should it satisfy? $f(m \cdot T) = f(m) \cdot T$ for any $T \in \mathrm{End}_R(M)^{op} = \mathrm{End}_R(M)$. Which is to say $f(T(m)) = T(f(m))$.

Why is this not the same as $\mathrm{End}_{\mathrm{End}_R(M)}(M)$?

Ted Shifrin

Shouldn't you get the property on both sides of $T$?

Balarka Sen

Which property?

Ted Shifrin

$f(T\cdot m)$ as well?

Balarka Sen

18:54

Why?

Ted Shifrin

Isn't the point of op to turn it into a right-module?

Balarka Sen

Yes

Ted Shifrin

Where is @Thor when we need him?

Balarka Sen

I don't know why we took the op. That is, what's the difference between $\mathrm{End}_{\mathrm{End}_R(M)}(M)$ and $\mathrm{End}_{\mathrm{End}_R(M)^{op}}(M)$?

Possibly, the point may be that if $R$ is treated as a left $R$-module, then $\mathrm{End}_R(R) \cong R^{op}$.

Thorgott

@BalarkaSen none

since inversion is an isomorphism between these two rings identifying those two module structures

sorry, the above is mistaken, you don't need the inversion

it's just literally the same thing

$R$-linearity and $R^{op}$-linearity, I mean

Balarka Sen

19:11

Yeah

They go on to write the ring homomorphism $R \to \mathrm{End}_{\mathrm{End}_R(M)^{op}}(M)$

$r \mapsto m_r$ where $m_r(x) = rx$

$m_r : M \to M$ is a $\mathrm{End}_R(M)^{op}$-module homomorphism... is this where op is needed or what?

$m_r(x \cdot T) = m_r(T(x)) = r T(x) = T(rx) = m_r(x) \cdot T$

Seems to work fine without the op??

Thorgott

well, op or not, the set is the same, and the $R$-linear structure on that set is the same as well

I guess the difference in whether you want that double centralizer to be an End_R(M) or an End_R(M)^op module itself?

i mean, idk this stuff either, to be clear

Balarka Sen

Yeah I am totally bewlidered by why they use op

Ted Shifrin

To be op-pressive.

@leslie Here is a good one for you. Any guesses?

Balarka Sen

One second, let's try $M = R$, left-module over $R$ itself. Then $\mathrm{End}_R(R)$ acts on $M$ by $T \cdot m = T(m)$.

Identify $R^{op} \cong \mathrm{End}_R(R)$, $x^{op} \mapsto (m \mapsto xm)$.

This is totally confusing

Elements of $\mathrm{End}_R(R)$ are of the form $\mathrm{mult}_x$. $\mathrm{mult}_{xy} = \mathrm{mult}_x \circ \mathrm{mult}_y$. What's the big deal?

$\mathrm{End}_R(R)$ is just $R$

No ops

Is my algebra professor seriously using a different convention for the ring structure on End_R(M)?

No one in their right mind would switch the order of multiplication

Maybe he denotes $ST$ to be $S$ followed by $T$, i.e., $T \circ S$

That would be insane

Thorgott

19:31

@BalarkaSen isn't this the wrong way around

Balarka Sen

Yeah

He says in the notes that $\mathrm{End}_R(R) \cong R^{op}$

I think he's using the wrong convention

@BalarkaSen If you use this then $R^{op}$ is correct

Thorgott

does he also write $(x)f$ for the value of a function

Balarka Sen

LOL

Thorgott

some algebraists in the 50s or so literally did this

Balarka Sen

Wait, but even the wiki page does this

Double centralizer theorem

In the branch of abstract algebra called ring theory, the double centralizer theorem can refer to any one of several similar results. These results concern the centralizer of a subring S of a ring R, denoted CR(S) in this article. It is always the case that CR(CR(S)) contains S, and a double centralizer theorem gives conditions on R and S that guarantee that CR(CR(S)) is equal to S. == Statements of the theorem == === Motivation === The centralizer of a subring S of R given by C R (...

See "motivation" where they keep fidgeting with left and right

Let M be a right R module and give M the natural left E-module structure, where E is End(M), the ring of endomorphisms of the abelian group M. Every map mr given by mr(x) = xr creates an additive endomorphism of M, that is, an element of E.

It's a right E-module, why on earth should it be a left E-module?

Thorgott

19:34

you apply functions from the left

Ted Shifrin

Even past that, @Thor. Herstein's eponymous book is full of that. I think it was the 60s.

Thorgott

but I don't get why they start with a right $R$-module

Balarka Sen

In wiki's convention, (m)(T o S) = T(S(m)) = S(m) o T = (m o S) o T

I just switched my hands, I'll play their game. Even then their convention is garbage

They think TS is T followed by S

Idiots

Ted Shifrin

It is with the right action of functions convention.

Thorgott

at some point we should start distinguishing between conventions and mistakes

Balarka Sen

19:38

The double centralizer of a left $R$-module $M$ are just all self-maps of $M$ which commute with all left $R$-module endomorphisms of $M$.

Ted Shifrin

@Thor @Astyx: Are you familiar with the "word" càdlàg? It just appeared on main.

Apparently it's a French acronym, but I've never in my life seen that before.

Thorgott

that's some probability thing, isn't it?

some semi-continuous whatever function

Balarka Sen

Yeah, cdf's are cadlags

Ted Shifrin

I'm just stunned.

Balarka Sen

OK, let $R$ be a simple ring and $I \subset R$ be a nonzero left-ideal. The claim is $R \to D(I)$ is an isomorphism, where $D(I)$ is my short hand for double centralizer of $I$ considered as a left $R$-module.

It's injective by simplicity. Need to prove surjective.

$D(I)$ consists of all maps $\phi : I \to I$ such that $\phi$ commutes with all left $R$-module endo's $T : I \to I$

Thorgott

19:42

ok, I feel stupid right now

but I figured out part of the issue

leslie townes

the academie francaise should dispatch a team of armed people by helicopter who rappel down and stop people from writing cadlag.

Thorgott

@BalarkaSen this is actually correct

leslie townes

i would have thought it was an untranslatable element of one of copper's jokes, although i guess the accents go the wrong way for that

Thorgott

you need to send an element to right multiplication by it, not left multiplication

left multiplication is not linear if your module structure is left too

in the non-commutative case

Balarka Sen

Oh ok

Thorgott

19:46

so if $R$ is a left $R$-module, you need $R\rightarrow\mathrm{End}_R(R),r\mapsto(x\mapsto xr)$ to get a well-defined map, and this reverses multiplication

this shit is always so awful

the sign errors of category theory

Balarka Sen

But for any left $R$-module $M$, $M$ is a left $\mathrm{End}_R(M)$-module, or no?

$T \cdot m = T(m)$.

Thorgott

yes, that works tautologically

Balarka Sen

Take $M = R$. Explain me how $R$ is a left $R^{op}$-module

I am very confused

Thorgott

a left $R$-module structure on an abelian group $M$ is the same thing as a ring hom $R\rightarrow\mathrm{End}_{\mathbb{Z}}(M)$. in this case, that's just the inclusion $\mathrm{End}_R(M)\subseteq\mathrm{End}_{\mathbb{Z}}(M)$

@BalarkaSen ok, let's unwind the definition

take $x\in R$ (the module) and $r\in R^{op}$ (the scalar). then $r$ corresponds to the element $(x\mapsto xr)$ in $\mathrm{End}_R(M)$, which acts on $x$ by mapping it to $xr$

the left $R^{op}$-module structure is just right multiplication

Balarka Sen

Oh ok

Thorgott

19:50

it corresponds to the tautological right $R$-module structure of $R$

but I'm still not quite sure what this tells us about the general case

Ted Shifrin

@Thorgott This is what I was suggesting earlier to @Balarka. But this stuff intimidates me.

@leslie Is munchkin going to the duck pond today?

leslie townes

we just got back!

Ted Shifrin

Ah, I missed it :(

Thorgott

yeah, it's intimidating in the same way signs are

no mathematician can ever escape the terror these things exude

Fargle

@TedShifrin One hit around 30 miles from me. Heck of a night.

robjohn

19:57

@leslietownes It has been a long time since we've been to our duck pond. I should see if I can convince the park group to go there this weekend.

Balarka Sen

@BalarkaSen Let me see if I can do this without paying attention to right vs left. So $D(I)$ is basically self-maps of $I$ which commute with all left $R$-module endo's $I \to I$. A particular class of left $R$-module endo's are $\mathcal{R}_a : I \to I, x \mapsto x a$, right multiplication by elements of $a \in I$. $\phi \circ \mathcal{R}_a = \mathcal{R}_a \circ \phi$ implies $\phi(xa) = \phi(x)a$ for all $x \in R, a \in I$.

Ted Shifrin

I'm glad you're OK, @Fargle.

leslie townes

there was almost nobody at the park today. all of the folks that play soccer and other outdoor sports appear to have packed it in for the season.

Fargle

Me too.

robjohn

@Fargle was the weather scary near you?

leslie townes

19:59

most of the herons were on a little island in the pond instead of their usual perches closer up, which was disappointing.

robjohn

it's the island in the middle of our pond that is the good spot for seeing the ducks. The geese wander around the grass away from the pond.

Balarka Sen

So if $\Phi : R \to D(I)$ is the map $\Phi(x)(i) = x i$, then, $\phi \circ \Phi(x) = \Phi(\phi(x))$.

Fargle

@robjohn Somewhat; it could have been worse. I still remember the tornado that hit town when I was around 3 years old; it was still some few miles from us, but we could hear it from our basement. This time it was just thunder and lightning and some big gusts.

robjohn

@Fargle I have heard dust devils in the high desert near us, and it sounds like a freight train passing by. Then the awning from the house down the road lands in your yard.

Balarka Sen

Here $x \in I$, which means $\phi \Phi(I) \subseteq \Phi(I)$, for all $\phi \in D(I)$. So $D(I) \Phi(I) \subseteq \Phi(I)$

$D(I) = D(I) \Phi(R) = D(I) \Phi(I) \Phi(R) \subseteq \Phi(I) \Phi(R) = \Phi(R)$

Using $IR = R$ repeatedly, which is true since $IR$ is 2-sided and $R$ is simple

Astyx

20:04

@TedShifrin it has a wikipedia article, but first time I hear of it

continue à droite, limite à gauche

Ted Shifrin

Yeah, someone else found the wiki. Crazy!! How did your pizza turn out?

Astyx

I should have put the dough in the refrigerator, it had fallen back down the next day

Balarka Sen

What's an example of a semisimple ring which is not Artinian by the way? Sounds like a myth.

Buddhini Angelika

Hi.

Ted Shifrin

Yeah, if you don't use it within two hours, you should refrigerate/freeze.

Buddhini Angelika

20:06

Hope all of you are keeping fine and the covid situations are getting better in your regions.

Astyx

still edible, but I'm a bit disappointed, especially since it looked like it swelled (?) a lot

I was hoping the cold temperature of incoming winter would be enough

Ted Shifrin

rose (past of rise)

robjohn

@Astyx had risen

Astyx

oh right

Balarka Sen

@BalarkaSen There's no such thing

Ted Shifrin

20:07

@BuddhiniAngelika Thanks, and same to you. Looks like it's all going to get worse again.

Astyx

I google translated "la pâte gonfle" because all I could think about was "inflate"

Ted Shifrin

Comme le pneu? :)

robjohn

@BalarkaSen it's nice that you can carry on a conversation with yourself when no one else can ;-)

Ted Shifrin

Thor conversed quite conversantly, @robjohn.

Balarka Sen

Just trying to keep the math alive in this chat

With a personal agenda, namely, I should learn this stuff for exams

Buddhini Angelika

20:10

When we are asked to find the modulus and argument of a complex number, say $z=3+4i$ for finding argument is it correct to write as a) $arg(z)= tan^{-1}(4/3)$ OR b)$arg(z) = tan^{-1}(4/3)+2n \pi$ OR c)$Arg(z)=tan^{-1}(4/3)$?

Eventhough Arg(z) is the principle value of the argument in some examples I've seen that they take just that as the answer

leslie townes

buddhini, depends on who is asking (might be worth going back a few pages if it's a textbook). to me, 'the' argument suggests one answer, and if a choice of 'prinicipal' argument has been made somewhere, i would use that.

Buddhini Angelika

And in some cases they just say arg(z)=tan^{-1}(4/3) without the 2n \pi

As in answer a)

@TedShifrin Thanks @TedShifrin Yes in our country too, its the same

leslie townes

it really depends on context. if they've distinguished notationally between arg as one chosen value and arg as a set (e.g. via "arg" and "Arg") i would follow the notational cue. if they aren't consistent about their own notation, maybe a question for whoever prepared the notes.

i wouldn't think it would matter at all on human-graded work, but if this were in a multiple-choice setting, you'd basically have to ask somebody to know what was expected.

Buddhini Angelika

@leslietownes Hmm. I'm not particularly using a text book, but I want to make a not so I've referred many web sites and notes.

Astyx

If it's for a specific test, ask the one who's making you take the test. If it's for your own understanding, there is no standard rule, and all that matters is that you understand what's going on, how you write it is up to you. I've almost never seen these functions used in practice

leslie townes

20:17

what astyx says. the fact that you've noticed the concept is enough for your own understanding.

Buddhini Angelika

Okay, thanks a lot @leslietownes and @Astyx so if the idea of the principle argument is also mentioned then I'd better write with the 2n \pi as in the above answer b), I guess

It's not for a test, but I want to give my note to someone to refer :)

leslie townes

i wouldn't assume that "principal value of ___" means the same thing to all people, for example. it's worth spelling out if you're using it in front of an unfamiliar audience. there is a whole lot of consistency among different sources for a lot of this stuff, but a commonly made arbitrary choice is still an arbitrary choice.

Buddhini Angelika

But when I was learning we were given as in answer a)

But I felt doubtful about it

Ted Shifrin

Unless they say principal value of the argument, I would say that all the possible answers should be given. (Non-principal) argument is definitely multi-valued.

Buddhini Angelika

And I'm unable to ask from my teacher because it's some time ago and now I'm not in contact..

Astyx

20:23

The correct pedantic way to write it would be $\tan^{-1}(4/3) + 2\pi\mathbb N$ because the argument really lives in the quotient $\mathbb R/\mathbb 2\pi \mathbb N$

Buddhini Angelika

@TedShifrin Okay thanks @TedShifrin

@Astyx Thank you very much @Astyx :)

user10478

Astyx

but no one does that, people usually just say "let $\theta\in \mathbb R$ be such that $z=|z|e^{i\theta}$

user10478

What is the equation for this graph? It seems like it should be $r = (1 + f)^.6 * (1 - f)^.4$ based on the parameters provided and the formula provided further down on the Kelly Criterion Wikipedia page, but it is certainly not.

Astyx

and it's not presented as such usually because people who see the argument for the first time are probably not accustomed to quotients, and people who know quotients probably already know about the argument enough to not have to consider it as a function anyway

robjohn

20:29

@user10478 it looks like that function minus $1$

user10478

Interesting, it does, doesn't it.

I guess that $-1$ just gets omitted from some calculations ultimately aimed at finding the maximum because the derivative would eat it.

leslie townes

a lot of stuff on kelly betting has this kind of stuff in it. the page's r is a 'geometric' growth rate, where r = 1 would correspond to no growth. they're plotting how this compares against that baseline. maybe this isn't consistent with how they do it elsewhere on the page.

the various pieces of the page and that picture could conceivably have been written by different people at different times.

user10478

Yeah, like the more you think about it, the more it gets in the way and complicates things.

Leaky Nun

@BalarkaSen what's the motivation of this?

Buddhini Angelika

Then, there is another question that: We can prove rules like $arg(z_1 z_2) = arg(z_1) + arg(z_2)$ for any two complex numbers $z_1, z_2$. Then, if $z_1 =r_a (cos(\theta_1)+i sin(\theta_1)$, then we take arg(z_1) = \theta_1$. If I represent arg(z) by showing all the values as in answer b) by adding 2n \pi, then should I write $Arg$ when proving these rules?

Or is it better to say that when I prove for one value of arg(z_1) (and arg(z_2)) then it applies the same for all values of arg(z_1) in the same way?

Ted Shifrin

20:43

It’s false if you use just one value.

Buddhini Angelika

Generally, for these type of properties of arguments of complex numbers, they always write arg. then how should I justify that?

Ted Shifrin

Do you see an example with principal value only that is false?

Buddhini Angelika

@TedShifrin Hmm no @TedShifrin

There, then eventhough we write as arg(z_1) = theta_1, (showing only one value, without the 2n \pi), we should understand that it represents all the values, as the (non-principle) argument is supposed to represent?

leslie townes

buddhini, whatever you do, don't adopt someone else's confusing notation because you think it is a convention. this doesn't sound right.

it's reasonably common to define "arg" so that it is a "single-valued" function, but then you definitely don't have that identity for all z_1 and z_2; you have it for a set of z_1 and z_2 defined by conditions relating to the choice you made to define "arg"

Balarka Sen

@LeakyNun You can prove Artin-Wedderburn with this double centralizer machinery. Suppose $R$ has a minimal left ideal $I$, then double centralizer says $R \cong D(I)$. Now $D(I) = \mathrm{End}_{\mathrm{End}_R(I)^{op}}(I)$; note $\Bbb D := \mathrm{End}_R(I)$ is a division ring as $I$ is a simple left $I$-module and moreover $I$ is a finite-dimensional $\Bbb D$-module.

This shows $R \cong M_n(\Bbb D)$.

Buddhini Angelika

20:54

@leslietownes Yes, thanks @leslietownes I'm bit confused. So I'm trying to rewrite everything using according to the definitions. Then arg should represent multiple values. But I don't understand how to write when proving the properties of arguments, like the one mentioned above.

leslie townes

OK, this is very much about what the definitions are. what is "arg(z)." is it a set, or a single number. if it's a set, the + in "arg(z_1) + arg(z_2)" is a slightly different kind of + then the + in z_1 + z_2, with z_1 and z_2 complex numbers.

Balarka Sen

If $R$ is semisimple then as a left $R$-module it is a sum of finitely many minimal left-ideals $I_1, \cdots, I_n$.

Then A-W follows

Buddhini Angelika

@leslietownes I'm not very clear about "you have it for a set of z_1 and z_2 defined by conditions " @leslietownes

@leslietownes Oh, Okay I think I get it now :)

leslie townes

if arg( ) is a set, then arg(z_1 z_2) = arg(z_1) + arg(z_2) is a statement about sets of complex numbers (and the + on the right is not the addition of complex numbers). set equalities are proved differently from equalities of numbers. a standard format of proving A = B with A, B sets is to prove that A is a subset of B, and then to prove that B is a subset of A.

Balarka Sen

@LeakyNun Actually you need simplicity of $R$ to argue $I$ is an f.d. $\Bbb D$-module above.

$\Bbb D^{op}$-module, whatever

Buddhini Angelika

20:59

Hmm, yeah, so the proof I meant earlier was for a single valued arg

Balarka Sen

But you need it even before that for $R \cong D(I)$, of course.

Buddhini Angelika

Thank you very much @leslietownes :)

leslie townes

this is why some books try to expressly avoid or suppress multi-valued or set-valued anything for as long as possible. leaving them in the odd position of doing things like having to explain why the argument principle is called the argument principle.

Buddhini Angelika

Hmm, Okay :) :)

Xander Henderson

I hate the phrase "mutli-valued function". :(

Buddhini Angelika

21:05

Many many thanks again!

Ted Shifrin

One of my least favorite things about Stein/Stakarchi was the fact that the discussion of multivalued log was about halfway through the book.

leslie townes

i hate the phrase "single-valued" function for the same reason.

Buddhini Angelika

Many thanks to every one!! :)

Xander Henderson

@leslietownes Indeed. A function is defined by the property that if $(x,y_1), (x,y_2) \in f$, then $y_1 = y_2$.

leslie townes

ahlfors at least apologizes for it and refers to it as a "pleonastic term." triple word score, lars.

Xander Henderson

21:06

"set-valued function" is fine, though.

Ted Shifrin

We are allowed to progress past precalculus level mathematics.

Buddhini Angelika

:)

Ted Shifrin

OK, we shall henceforth refer to log as an analytic/holomorphic relation, no longer using the word function.

Xander Henderson

@TedShifrin Heh.

Thorgott

multi-valued function is just a bad way of saying "function defined on a branched covering" or something along those lines

Ted Shifrin

21:08

Branched coverings are generally a different animal.

But, yes, on a Riemann surface ... will do.

Xander Henderson

@TedShifrin Yes, I was just typing something like this.

Balarka Sen

Let $f : \Bbb C \to \Bbb C$ be a meromorphic multifunction.

leslie townes

ahlfors's introduction of arg leaves a little to be desired. in the first chapter he implicitly treats arg as a function to a quotient space, by saying that arg(z_1 z_2) = arg(z_1) + arg(z_2) is "an equation of angles and not numbers." but if you weren't raised in his generation, good luck figuring that out on the first exposure.

Balarka Sen

It is up to the reader to judge, on the basis of careful consideration, what the domain and codomain actually are.

leslie townes

it doesn't confuse me, but i wouldn't wanna teach from that.

copper.hat

21:15

i like a nice green tea made from arg leaves

leslie townes

i'm having some green tea now, it's great.

copper.hat

nice healthy guys breakfast of bacon, fried tomatoes, fried eggs & hash browns

Ted Shifrin

calls paramedics for copper

leslie townes

i was gonna say, follow that by a trip to the hospital and you might be back in time for lunch.

copper.hat

@TedShifrin its strange, since my hip thing my diet has disimproved significantly.

leslie townes

21:17

i'll have mine with a side of defibrillator, please.

copper.hat

surprisingly we have essentially no heart disease in the family.

my side, i mean.

Balarka Sen

snacks project

copper.hat

onto a linear variety?

leslie townes

that's an idea. someone who dispenses snacks instead of smacks.

Balarka Sen

leslie townes 1000 page long manuscript dedicated to ted shifrin "pursuing smacks"

leslie townes

21:20

a la recherche du snacks perdu

copper.hat

snacks for smacks

Ted Shifrin

des snacks perdus

leslie townes

whoops

copper.hat

hores douvre

to go lower than any man before...

Ted Shifrin

whores devoirs?

copper.hat

21:32

:-). the connection between spelling & pronunciation in French is always a little tenuous for me.

Ted Shifrin

well, hors d'oeuvre means literally outside the work (or main opus)

leslie townes

like when i bring a big bag of flamin' hot cheetos in as my appetizer before a main course at an expensive french restaurant

Ted Shifrin

yeah, like that

Balarka Sen

Hm. If $A$ is central simple and $B$ is simple, then $A \otimes_k B$ is simple. How do I prove this?

Ted Shifrin

"Simple Simon met a pieman // Going to the fair ..."

Balarka Sen

21:39

Try the 2-sided ideal generated by $a \otimes b$. Then $AaA = A$, and $BbB = B$

So stuff like $1 \otimes b$ and $a \otimes 1$ for all $a \in A, b \in B$ belongs, becomes the full ring

copper.hat

always found that rhyme depressing

Ted Shifrin

I've forgotten all the rest of it.

Houshou Rattengod

Hey all, I'm stumped with my homework and I was hoping someone could clarify what I am supposed to use.

I'm taking a Statistics Class, and I'm working with the Central Limit Theorem.

My current issue, is that I have both the Population Standard Deviation, and my Sample Deviation. And I'm unsure which one I am supposed to use.

Ted Shifrin

This sounds like a question for your classmates or instructor.

Derivative

21:59

does anyone want to help me prove that if $f:\mathbb R\to\mathbb R$ is a function so that $f^{-1}(A)$ is open for all open sets $A$, then $f$ is continuous?

leslie townes

derivative: what definition of 'continuous' are you using? epsilon-delta? sequential convergence of f(x_n) for convergent sequences (x_n)? my gut feeling is that either should be on math.se somewhere.

Derivative

epsilon-delta

I searched for it but I only found the opposite direction, which was easy to prove

Ted Shifrin

Didn't we talk about this a week ago?

Derivative

maybe you did but not with me

Ted Shifrin

It was in this chat. Taking $A$ to be a very specific ball of very specific radius.

Derivative

22:02

okay I tried a bunch of balls with a bunch of radii but none of them worked

copper.hat

:-)

leslie townes

math.stackexchange.com/questions/2675843/… has the relevant ideas in some form. it does use the notion of 'neighborhood' which might be a slight twist on open set. but the idea is there.

Ted Shifrin

Write down the precise statement you're trying to prove.

leslie townes

math.stackexchange.com/questions/1752871/… is another one without neighborhoods.

Derivative

you mean that as a rhetorical device or do you actually want me to do that in the chat?

leslie townes

22:03

i'll stop dumping links.

Ted Shifrin

Since you asked us for help, I figured you should tell us the precise statement you are aiming to prove.

Derivative

okay. If $f:\mathbb R\to\mathbb R$ is such that if $A$ is open then $f^{-1}(A)$ is open then $\lim_{x\to t}f(x)=f(t)$ for all $t\in\mathbb R$

Ted Shifrin

Write the precise $\delta$-$\epsilon$ statement.

Derivative

for all $\epsilon>0$ there exists $\delta>0$ so that $|x-t|<\delta$ implies $|f(x)-f(t)|<\epsilon$

Ted Shifrin

So now let's discuss what open set $A$ is sitting in that sentence so that $f^{-1}(A)$ might be relevant.

It might be that your choice of letters makes it less obvious what is fixed and what is varying.

Derivative

22:09

well the obvious choice was $A=B_\delta(x)$ or so I thought, because I couldn't get it to work

Ted Shifrin

No. If you're going to take the inverse image under $f$, where should this open set live? In the domain or in the range?

And my worry that you don't know who is constant and who is variable is a valid worry. So you'd better verbalize that. In your definition sentence, who is fixed and who is varying?

Derivative

$t$ is fixed, $x$ is varying

also $\epsilon$ is fixed and we need to choose $\delta$ as a function of it

Ted Shifrin

OK, so any ball you choose needs to be centered at a fixed point, not a variable point. Now go back to my first question there. If you're going to take inverse image under $f$, where should you be?

Derivative

in the range? Then maybe $A=B_\epsilon(f(t))$?

Ted Shifrin

Perfect.

Derivative

22:15

okay, I'll see if I finish from here thanks

Ted Shifrin

Good. Just remember to think through things like this, though. This should be a method for you.

Wojciech Kulma

23:54

Hi! Does the maximizer of this function converge?
$f_n(x) = n x \exp(-n x ^2)$

looks like this: https://i.imgur.com/L61XZlt.png

i would say yes it does, "from right to left"

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$Mathematics$ Mathematics