Mathematics - 2023-09-26

Lord Pooty

2:42 AM

What are some instances in Mathematics where new ideas were very slowly adopted?

Like, real world academic examples

Or, any instances which show how the adoption of an idea or piece of knowledge was very slow.

robjohn

complex numbers took a while to be accepted.

Lord Pooty

Do you have any examples?

one potato two potato

my homework solution took a while to be accepted after several claims.

Ted Shifrin

If they listen to me, they’ll keep rejecting that! 🤷‍♂️😻

robjohn

grumpy

Lord Pooty

2:59 AM

I mean academic examples

one potato two potato

my paper took a while to be accepted after several rejections.

Lord Pooty

nvm

robjohn

@LordPooty Look here.

Thomas Finley

3:29 AM

suppose f: [a,b]-> R be a differential function. let a <c<d<b. If f'(c) <0 and f'(d)> 0, then show that there exist a point p between c and d such that f'(p)=0.

I am struggling with problem

I could prove that f has an absolute maximum and an absolute minimum.

Next, I showed that c and d are not points of extremum.

What to do next?

@robjohn I need your calculus skills, please. Can you help me with it?

robjohn

@ThomasFinley with what?

Thomas Finley

@robjohn with this: chat.stackexchange.com/transcript/message/64478946#64478946 i.e the problem above...

leslie townes

4:29 AM

thomas, you give a sketch of a proof of this result in your question math.stackexchange.com/questions/4772260/… and an answerer comments on aspects of why a proposed simplification of the argument does not work. maybe start there.

take the "a" in the notation of that question to be what is now c, the "b" in that question to be what is now d, the "alpha," "beta," and "gamma" to be what are now f'(c), f'(d), and 0, respectively. the value "xi" returned by that argument will do for p here.

robjohn

4:56 AM

@ThomasFinley $f$ is decreasing at $c$ and increasing at $d$. There must be some point between $c$ and $d$ where $f$ attains its minimum on $[c,d]$ since $[c,d]$ is compact.

at that point, what is $f'$?

leslie townes

all of this and more, sketched in thomas's own question from less than a week ago

robjohn

@leslietownes are you saying that I will be answering the same question again in a week?

leslie townes

only if you're lucky

robjohn

@ThomasFinley show that the minimum point cannot be $c$ or $d$.

Ted Shifrin

5:45 AM

And I said that a week ago! But it was $a$ and $b$. In a week, letters inflate.

leslie townes

honk

Ted Shifrin

Happy goose to you.

robjohn

6:10 AM

I'm so glad I've been able to parrot Ted

Novice

Hi, I have to read a paper for school and I'm a little out of my depth. Is this snippet correct?

I ask because there are some Stack Exchange posts saying that $\alpha$-Hölder does not imply bounded variation

I may be misunderstanding something; I'm unfamiliar with these definitions

one potato two potato

there's a notion of 'depth n foliation'. How interesting.

leslie townes

novice: it might be that the author is using 'bounded variation' to mean something other than what one would understand from e.g. an analysis textbook. it might also be that they are unwittingly making an extra hypothesis that they are unaware of (indeed, holder does not imply 'bounded variation' in the analysis sense), or that they are not actually using the hypothesis of bounded variation and only using the holder condition.

if this is outside of pure math, maybe all of these are happening at once.

when [as here] an author gives a remark in passing that might be wrong, see if the hypothesis [here, of bounded variation] is ever used anywhere, or if they are maybe using something else that is the actual focus of their attention.

Novice

6:26 AM

this is happening in statistics. Thanks for confirming that the authors are probably wrong at least in this instance. I will look at the broader context and see if I can figure out what they mean and are assuming.

Ted Shifrin

It looks like they’re just wanting a “local” version of BV.

Uniform in the family given $x,y$.

Novice

I'll put up another snippet that mentions BV and Hölder, from earlier in the paper

Ted Shifrin

Can you verify their definition of bounded variation?

Novice

I doubt it's defined, since the first mention of "bounded variation" (according to pdf search) is here:

last couple lines

leslie townes

depending on the area of stat, they may have never seen the definition of 'bounded variation' that people use in analysis and may be assuming it means something simpler than what it means.

if i had to read this paper, i'd just ignore the issue and see if it ever comes up. e.g. if they are just plugging certain assumptions into theorems that they use like a black box, what are the actual assumptions of those theorems.

Ted Shifrin

6:31 AM

Does our notion of BV even make sense on a general metric space?

I only know it on an interval in $\Bbb R$.

leslie townes

some thoughts on that problem here: mathoverflow.net/questions/234251/…

Novice

So without further knowledge, it's possible that they say BV but really just mean Hölder, and there is no serious issue, but it's possible that it could also be more serious?

leslie townes

novice, withhold judgment until later. it doesn't sound as though it's going to be serious, because the author is saying something potentially confusing only as a "for example."

this suggests there will be a general thing that they paid more attention to than the example, or that maybe some stuff will come later in a worked out example where the meaning will be clearer than it is in an offhand remark.

Ted Shifrin

Sounds like I’m not the only one with that question, leslie. And it’s not clearly resolved.

What about a rectangle in $\Bbb R^n$?

leslie townes

hence my feeling that maybe the author is conjuring up things that they didn't intend to conjure up with a 'for example' they weren't thinking too carefully about.

my wife went to school with some stat people who wouldn't know a metric space from serge lang's proverbial hole in the ground, although that didn't stop them from spewing jargon as if they did.

Ted Shifrin

6:37 AM

I think their $V(x,y)$ is their definition.

leslie townes

that would make sense. so maybe they don't know that "bounded variation" already has a pretty common use that their use is potentially colliding with in readers' heads.

Novice

I think the Wikipedia article on BV is suggesting that the concept of BV can be extended to functions into normed vector spaces. I think the functions in this paper are into $\mathbb R$

leslie townes

highly likely that we've already spent more time thinking about this use of terminology than the author did.

Novice

Well, I want to learn more math, and I think I'm being graded (to some extent) based on getting into the technical details, so I guess this is good for me. But I agree I should try to read the paper more holistically and figure out whether there is any real problem, or just sloppiness

Ted Shifrin

I’m worrying about domain, not range, Novice.

Novice

6:41 AM

Oh, sorry

Ted Shifrin

No need to be sorry. I’m just clarifying.

leslie townes

it's just the classic question of what to do when you read a paper and there's something in it that causes you to stop and go "uhh, what." here, the remark-ish context suggests, maybe keep reading, see what if anything ends up being used, try to work back to the intent of the remark from that. almost certainly they did not intend to draw on some generalization of BV that people on mathoverflow were hypothesizing about.

Ted Shifrin

Indeed!

Novice

Thanks. I have to spend a few weeks with this paper so I will hopefully get into the details. I may come back if I have more questions (of a mathematical variety). Thanks again

leslie townes

i wonder if search engines will ultimately lead to less "on the fly" invention of terminology in papers. X years ago i don't think you'd have to worry about someone googling a particular substring of a definition or remark, wondering if it was an informal choice of phrasing or reference to an external concept. now this is one of the first things i'd worry more about if i were an author.

user400188

6:45 AM

Is there a wikipedia like page the mathematical properties of best of N games? E.g., best of three (whoever wins 2 out of the 3 wins the match).

Or, since I doubt anyone knows where one is off the top of their head, where can I look for one? (Wikipedia doesn't have one)

leslie townes

i.e. not just the old concern of "is my way of chatting about this more confusing than helpful" but "is my way of chatting about this going to lead someone to some other set of definitions that i'm definitely not intending to use here"

user: if you mean modeling things like win probabilities, when there is some theorized probability of a win on an individual game, i would search MSE or stats SE for "best of n" for particular values of n, or maybe just the parameter.

user400188

I've searched a bit on MSE already, but I'm looking for something a bit more comprehensive than single questions, single answers.

Maybe I'm asking too much.

leslie townes

i think in complete generality it's too much. how many players. what is the probability distribution on the set of sequences of games, etc.

and once you start filling in answers to those things you begin to get textbook exercises as opposed to what someone would write a textbook about.

user400188

Fair enough.

Thanks for the advice.

leslie townes

i guess my more particular guess is that the general formulas for those things are intractable in generality, such that you don't get anything to analyze until you make specializing assumptions. i'd love to be wrong, though.

user400188

6:54 AM

I've been working on a weekend project to see if I can model best of three. But I'm sort of starved for references atm. I would be very surprised if they didn't exist, but so far I haven't found many.

soupless

8:22 AM

Is it true that we can just factor out an expression involving $x$ if $x$ is common in all parameters? Like, $f(x, xy)$ becomes $g(x) \cdot h(y)$?

Sourav Ghosh

Give an example of a free module $M$ over an integral domain $R$ such that $\exits N<M$ (submodule) which is not free but the factor module M/N is free.

If $R$ is a PID such $N<M$ doesn't exist.

$0\to N\to M\to M/N \to 0$ is a split exact sequence.

$M= N\oplus M/N$

sunny

Let $0<\alpha<1$ and $x=x(t)$. In simplifying a solution to a certain ODE, someone went from $$|x|^{1-\alpha}=\pm (1-\alpha)t+C.\tag1$$ to $$|x|^{1-\alpha}=(1-\alpha)|t-D|.\tag2$$ This confused me. What is $D$ in terms of $C$? It seems like $D$ is not just some arbitrary constant, but depends on $\alpha$.

Sourav Ghosh

Consider $M=R=K[x, y]$ and $N=(x, y) $

Is $K[x, y]/(x, y) $ free?

Jakobian

9:41 AM

@soupless no. For example, if $f(0, 0)\neq 0$ then $h(y)$ must be constant so $f(x, xy)$ must be independent of $y$. This is in general false

leslie townes

soupless as jakobian notes this is not true in general, although if you had context for why you might expect or want it to be true in some specific setting, there might be more to say

it strikes me as an odd thing to want, suggesting that maybe there is some extra context in which it would make more sense, or maybe the thing you would want to achieve by factoring is achievable in some other way, or both

Jakobian

10:44 AM

@sunny We have $|x|^{1-\alpha} = (1-\alpha)|t\pm C|$, so $D = \pm C$

it not only depends on $C$, but also on the sign that's given

sunny

12:38 PM

@Jakobian hmm, how did you obtain $|x|^{1-\alpha} = (1-\alpha)|t\pm C|$? I must be missing something...

To give you some more context, I'm solving $x'=|x|^\alpha, x(0)=0$. One solution is $x(t)\equiv 0$. Suppose $x\neq 0$, then the ODE reads $\frac{x'}{|x|^\alpha}=1$. Now consider $$\frac{\mathrm{d}}{\mathrm{d}t}\left(|x|^{1-\alpha}\right)=\mathrm{sgn}(x)(1-\alpha)|x|^{-\alpha}x'=\pm (1-\alpha).$$ Then, integrating, we obtain $$|x|^{1-\alpha}=\pm (1-\alpha)t+C.\tag1$$ And here is where a TA claimed that $(1)$ is equivalent to $$|x|^{1-\alpha}=(1-\alpha)|t-D|.$$ This I do not understand.

Jakobian

12:49 PM

@sunny $\pm(1-\alpha)t+C = |(1-\alpha)(\pm t)+C| = (1-\alpha)|t\pm C|$

isn't it?

sunny

@Jakobian hmm, my brain goes $$\begin{align}\pm(1-\alpha)t+C&=\pm(1-\alpha)\left(t+\frac{D}{1-\alpha}\right) \\ &=(1-\alpha)\pm\left(t+\frac{D}{1-\alpha}\right) \\ &=(1-\alpha)\left|t+\frac{D}{1-\alpha}\right|.\end{align}$$

Jakobian

@sunny oh right

$\pm(1-\alpha)t+C = |(1-\alpha)(\pm t)+C| = (1-\alpha)|t\pm \frac{C}{1-\alpha}|$

but yeah, there is still a $\pm$ here

sunny

It turns out you get the same solution if you go with $|x|^{1-\alpha}=(1-\alpha)|t-D|$, it just confused me

Thorgott

1:05 PM

@SouravGhosh There's this very fun trick called the Eilenberg swindle. Say $P\oplus Q=F$ with $F$ being a free module. Then, $\bigoplus_{n\in\mathbb{N}}F=(P\oplus Q)\oplus(P\oplus Q)\oplus\dotsc=P\oplus(Q\oplus P)\oplus(Q\oplus P)\dotsc=P\oplus\bigoplus_{n\in\mathbb{N}}F$ and $\bigoplus_{n\in\mathbb{N}}F$ is clearly a free module. The direct summands of free modules are called projective modules and this trick shows that any non-free projective module gives rise to an example as you want it.

Jakobian

Eilenberg was Polish?

Thorgott

1:40 PM

yes

Jakobian

2:30 PM

@AlessandroCodenotti Cantor scheme need not give raise to a Cantor set?

why?

I'm asking why we define it that way then

3

A: Show that every uncountable and closed subset of the complete separable metric space contains a homeomorphic subset with the Cantor set.

First of all note that if $X$ is a Polish space (completely metrizable and separable space), and $F\subseteq X$ is closed, then $F$ is also Polish (usually with a different metric). The result you want can be shown to hold with a combination of the following two classical results: Theorem 1: Let...

reading this answer of yours for reference about what Cantor scheme is

Alessandro Codenotti

2:53 PM

Uhm not sure, usually you have extra conditions from which you do get a Cantor set

But I guess in principle a Cantor scheme is just an infinite binary tree of subspaces

Thomas Finley

3:51 PM

I am so out of ideas trying to solve part (h) in the image above

I have no idea how to solve this.

@TedShifrin , @EE18 , @SoumikMukherjee Will you please take a look at this?

I could solve part (g) (i.e the problem preceeding (h))

Xander Henderson

ARG! Why is Outlook such a pain in the ass!

I have turned off auto-correction over and over and over again, and it keeps getting turned back on.

WHY!?!

robjohn

Because someone at Microsoft thinks you can't spell?

Besides, why pick out Outlook as being a Pain In The Ass? I'd ask about Microsoft.

Obliv

4:16 PM

does $(\mathbb{R}\times\mathbb{N}) \setminus (\mathbb{Z}\times \mathbb{N})$ mean (a,b) where a is in R but not Z, b is in N but not N?

Xander Henderson

@robjohn Outlook is the only MS product I have to consistently use at work. :/

Obliv

@XanderHenderson halp

robjohn

@XanderHenderson Ah, it's good that that is the only one, but bad that you have to deal with it consistently.

Obliv

oh wait nvm

yeah that set would be empty

Koro

1

Q: Books for Teenager

My daughter is 14 and really into maths right now. She's already thinking about studying mathematics. I'd love that, and I'd love to help her, so I thought this forum might be the right one to ask for interesting mathematics books for kids. I don't think that a book with a lot of exercises might ...

I want to scream: Algebraic topology by Hatcher

soupless

4:23 PM

Honestly, the starting parts of calculus feel alluring and abstract algebra feels intimidating, but from how I see it, wait until you get to the next parts.

Koro

@XanderHenderson yeah. Besides, its in-search also sucks. Why don't they replace outlook by something else in the firms?

Soumik Mukherjee

4:33 PM

@ThomasFinley What is $e$? and the construction of infinite basis was discussed a few days earlier, right?

@Koro All the Mathematics You Missed: But Need to Know for Graduate School by Thomas A. Garrity

soupless

4:55 PM

Ok, so I'm trying to solve for the last two digits of $$2^{3^{4^{\cdots^{2020}}}}$$

And I'm using CRT to do this. $\bmod 4$ is easy, $\bmod 25$ is not so obvious to me yet.

Am I correct to say $3^{4^{\cdots^{2020}}} \equiv 3^{4^{5^{\cdots^{2020}} \bmod }\varphi(25)}\bmod 25$?

robjohn

5:19 PM

Just thinking... $4^{5^k}\equiv0\pmod{4}$ and $4^{5^k}\equiv-1\pmod{5}$ so $4^{5^k}\equiv4\pmod{20}$ for $k\ge1$.

Ted Shifrin

@Obliv No. Note that $\Bbb R\times\Bbb N\setminus\Bbb Z\times\Bbb N = (\Bbb R\backslash\Bbb Z)\times\Bbb N$.

soupless

Wait. I got $1 \bmod 20$

robjohn

5:42 PM

@soupless How can an odd power of $4$ be $1\pmod{20}$?

soupless

Oh, no. I got it wrong

I just realized, I started with 3 and used mod 20 there

Soumik Mukherjee

Given an $n$-dimensional smooth manifold $M$, $N(M)$ is defined as the minimal cardinality of an open cover $\{U_i\}$
of $M$ such that each $U_i$ admits a
smooth embedding into $\Bbb{R}^n$. What will be $N(\Bbb{R}P^3)$?

I figured out that $N(\Bbb{R}P^2)$ will be $3$, but I don't have any idea for the above

user 726941

@XanderHenderson perhaps they hard wired it into the program thinking people need to spell "correctly"

Jakobian

@Koro hope you're joking

Auden Young

i think i may be making life overly complicated for myself - i'm trying to find examples of unique metric spaces, and one i found is the metric you can define on a graph, where the metric is just the length of the shortest path between two vertices. i'm currently considering a simple, connected, finite graph, and i've proved for any such graph all subsets are clopen, and thus that any function from the graph to itself is continuous.
however, i'm having trouble exploring what sequences in this space could look like. since it's a finite graph, my first instinct was that all sequences converge…

Jakobian

5:55 PM

@AudenYoung a finite metric space is discrete

if you see graph as points with common edge connected by an interval, then this is very much not discrete (unless has no edges), and also not every subset of this is clopen

people actually study distances on graphs, this is pretty much explored iirc

user 726941

Distance = | a - b |

Jakobian

?

user 726941

That's an elementary fact.

Jakobian

you're not making any sense

user 726941

I'm just starting from the basics.

Jakobian

6:00 PM

starting what?

user 726941

a chat?

Jakobian

starting a chat from basics?

?

¿

Ted Shifrin

OK, enough, children.

Jakobian

Anyway, there is a book called, Geometry of Cuts and Metrics
Considering distances on graphs leads to some cool things

Thomas Finley

6:33 PM

Can anyone give me some ideas on how to prove, that " All the linearly independent sets of a finite dimensional vector space of dimension say, n have, atmost n elements" ?

Ted Shifrin

How about the definition of dimension?

Thomas Finley

@TedShifrin I used it.

Ted Shifrin

Apparently not.

Thomas Finley

Ok my attempt is:

I assumed the basis B=v_1,v_2,...,v_n

I considered a set L to be linearly independent

For the sake of contradiction, I assumed $l_1,l_2,...,l_m\in L$ such that $m>n$

I took some scalars $c_1,c_2,...,c_m$ such that $\sum_{i=1}^mc_il_i=0$

Ted Shifrin

Huh?

Thomas Finley

6:41 PM

Then, as, $l_i\in V$ so, I considered nos. $A_{i1},A_{i2},...,A_{in}$ such that $A_{i1}v_1+A_{i2}v_2+..+A_{in}v_n=l_i$

Ted Shifrin

You have already assumed they're linearly independent. So what are you doing?

Thomas Finley

Finally, I got a system of equations $A_{i1}v_1+A_{i2}v_2+..+A_{in}v_n=l_i$ for i=1,2,...,m which I substituted in $\sum_{i=1}^mc_il_i=0$ @TedShifrin

Ted Shifrin

Are you trying to prove instead that $m>n$ vectors in an $n$-dimensional vector space must be linearly dependent?

Thomas Finley

@TedShifrin yeah

Is there any other way rather than showing this?

Ted Shifrin

Yes.

Thomas Finley

6:44 PM

@TedShifrin what's that?

Ted Shifrin

Any linearly independent set, if it is not already a spanning set, can be expanded to become a basis.

Thomas Finley

@TedShifrin I found a proof of it, but in almost all cases they assume that the linearly independent set is finite

Ted Shifrin

The proof you were doing is part of the usual proof that dimension is well-defined.

So what? Your hypothesis was that you're in a finite-dimensional space.

You don't need to look for proofs all the time. You can use your brain.

Thomas Finley

@TedShifrin Will then all the linearly independent sets be finite?

Ted Shifrin

We went through this literally weeks ago.

Thomas Finley

6:47 PM

@TedShifrin Wait, now I am confused. I don't quite remember it!

Ted Shifrin

Literally weeks ago I told you that if you have an infinite set of linearly independent vectors, then it contains finite subsets of arbitrarily large cardinality.

Thomas Finley

Sorry for my weak memory :?)

@TedShifrin ok, but how does that contradict the fact that the vector space has a finite dimension

Ted Shifrin

Go back to the beginning. If you have $k$ linearly independent vectors, either they span or they don't. If they span, the dimension is $k$. If they do not, the dimension must be $>k$.

Thomas Finley

What I am really interested in, is that : Is a linearly independent subset of a finite dimensional vector space always finite?

@TedShifrin ok

Then?

@TedShifrin Everything is fine with this statement.

Ted Shifrin

Then you're done.

If the dimension is not $>k$, then you have $k$ or fewer linearly independent vectors.

Thomas Finley

6:53 PM

@TedShifrin ok...

@TedShifrin in that case, we're done, but what if dimension is >k ?

Ted Shifrin

Indeed, you need to understand completely the following: If $\dim V = n$, then $n$ linearly independent vectors must span and $n$ spanning vectors must be linearly independent.

You're being silly now. Set $k=\dim V$.

Thomas Finley

@TedShifrin I proved it already...

@TedShifrin ok

Ted Shifrin

You have done proofs apparently without understanding the concepts.

Sha Vuklia

@Thorgott are you ok with a numb thy question?

Thomas Finley

@TedShifrin i am sorry but I dont get this at all...

Ted Shifrin

6:55 PM

You keep falling in the same holes. That suggests a lack of complete comprehension.

Thomas Finley

How to show that any linearly independent subset of a finite dimensional vector space is finite? If this claim is at all true

Till now, I know that if dim of the subset of a linearly independent subset of V, is finite say, m then m is lesser than or equal to the dimension of V

(V is the vector space)

Xander Henderson

@ThomasFinley What is your definition of "dimension"?

Thomas Finley

Also, if the linearly independent set L has, $dim(V)$ no. of vectors then, $L$ is the basis of $V$

@XanderHenderson the cardinality of the basis of a vector space

Xander Henderson

@ThomasFinley What is your definition of "basis"?

Thomas Finley

@XanderHenderson The linearly independent subset L of the vector space such that span(L)=the vector space

Xander Henderson

7:00 PM

@ThomasFinley So a basis is a set which (a) spans the space and (b) is linearly independent.

What does it mean to span the space?

Thomas Finley

@XanderHenderson yes.

@XanderHenderson the set of all linear combinations of the vectors in that set is equal to the vector space itself

Xander Henderson

In other words, any other vector in the space is a linear combination of your basis set.

Thomas Finley

@XanderHenderson yes.

Xander Henderson

And what is your definition of linearly independent?

Thomas Finley

@XanderHenderson A subset S of a vector space V is linearly independent iff for a finite number of vectors $v_1,v_2,...,v_n\in S$ and scalars $c_1,c_2,...,c_n$ the equation, $c_1v_1+...+c_nv_n=0$ implies, $c_1=c_2=...=c_n=0$

Xander Henderson

7:06 PM

Okay, great. So, look back---your definition of "dimension" says that it is the cardinality of "the" basis. Can a space have more than one basis?

Thomas Finley

@XanderHenderson yes, sure.

Xander Henderson

Okay, so what if you choose a different basis? Can the dimension of the space change?

Thomas Finley

@XanderHenderson As till now, I only learnt about finite dimensional vector spaces so, I will answer according to that. If $V$ is finite dimensional all the basis for V have the same cardinality.

Xander Henderson

@ThomasFinley Since we are only talking about finite dimensional spaces, we only need to consider finite dimensional space, no?

34 mins ago, by Thomas Finley

Can anyone give me some ideas on how to prove, that " All the linearly independent sets of a finite dimensional vector space of dimension say, n have, atmost n elements" ?

Thomas Finley

@XanderHenderson yes.

Xander Henderson

7:09 PM

@ThomasFinley So, decompose this statement: if any set of vectors is (a) linearly independent and (b) spans the space, then the cardinality of that set is...?

Thomas Finley

@XanderHenderson the cardinality of that set is equal to the dimension of the vector space

Xander Henderson

@ThomasFinley That seems to answer your question...

sanya

Not relevant here but I wanted to know if theres a chat room for chemistry if so please share the link as soon as possible. Thankyou.

Thomas Finley

@XanderHenderson how, the question doesn't say, that the linearly independent set spans the vector space?

user 726941

The Periodic Table

Haikus are awesome / Chemistry's even better / So pull up a chair

Xander Henderson

7:12 PM

@ThomasFinley So what happens if it doesn't?

sanya

@user726941 thankyou so much sorry for the disruption.

user 726941

np

Thomas Finley

@XanderHenderson What happens? Well, then the set is not a basis?

Xander Henderson

@ThomasFinley Yes, and? Think about the question you are trying to answer. If it is a linearly independent set and it doesn't span the space, what can you say about its cardinality?

Use your brain. Think about what could possibly be going on.

Thomas Finley

Or in other words the set has cardinality higher than the dimension of the vector space?

Xander Henderson

7:16 PM

@ThomasFinley How could that possibly happen?

Thomas Finley

@XanderHenderson If the cardinality of that set was equal to the dimension of V then as the set is linearly independent so, the set will be a basis for V i.e it will span V, but here we assumed that it doesn't span V.

Xander Henderson

You are just parroting back to me things we've already gone over.

Ted Shifrin

thinks quietly ... gee, didn't we just go through all this ....

Thomas Finley

@XanderHenderson I am sorry but I have no clue what to say.

Xander Henderson

Think.

Thomas Finley

7:18 PM

@TedShifrin yes, surely

user 726941

Time to stop and review @ThomasFinley

Take your time.

🤔

Xander Henderson

You might try to think about the possible cases. Suppose that $V$ is an $n$-dimensional vector space, and that $L$ is a linearly independent subset. What happens if the cardinality of $L$ is greater than $n$? equal to $n$? less than $n$? Are any of these cases impossible?

Thomas Finley

@XanderHenderson if cardinality of L is <= n then we are done.

The thing is, if it's greater than n.

I think this is impossible.

But I dont know why

Xander Henderson

Go back to the definition of dimension.

Thomas Finley

Cardinality of a basis ir cardinality of a linearly independent set whose cardinality is same as the vector space...

user123234

7:26 PM

I want to consider the ring homomorphism $ev_\gamma:\Bbb{F}_p[x]\rightarrow \Bbb{F}$ that fixes $\Bbb{F}_p$ given by $f(x)\mapsto f(\gamma)$. I want to find $\gamma$ s.t. $ev_\gamma$ is surjective. They wrote as a hint that if I have a finite field $K$ then $K^\times$ is cyclic. Can someone help me how to get this $\gamma$?

Xander Henderson

Suppose that $L$ is a set of $k > n$ linearly independent vectors in a space of dimension $n$. Is it not true that $L$ contains a linearly independent subset of cardinality $n$? Then what?

Thomas Finley

'Is it not true that $L$ contains a linearly independent subset of cardinality $n$? ' - I think this is the part where the problem lies. Why is it true?

If it's true then the linearly independent set L has the basis and so, it generates the vector space and hence, k=n, a contradiction.

So, the cardinality of L cannot be greater than n, i.e it must be finite.

But all of this is valid, if, "Is it not true that $L$ contains a linearly independent subset of cardinality $n$? " is true @XanderHenderson .

user123234

@TedShifrin maybe you can help e further?

Thomas Finley

But I don't know or seem to get, why this is true?

Oh wait

Yes, it's true!

leslie townes

user: "K^x is cyclic" means there's at least one element of K_x that generates it as a group. try evaluation at one of those

Thomas Finley

7:32 PM

Thanks a ton, @XanderHenderson

I get the thing now.

Ted Shifrin

@user123234 I had to think through the question and leslie beat me to the punch. Think about what $f(\gamma)$ means :)

user123234

@leslietownes but in my case I take $K=\Bbb{F}_p$ right?

Ted Shifrin

@Xander quite literally had you go through exactly what we had already been through. But sometimes it takes 5 or 10 times.

Yes, @user123234.

Xander Henderson

@TedShifrin Indeed.

leslie townes

user: well, you could take that as an example, but there are finite fields other than F_p. you don't need to know anything about the structure of the field other than (1) it's finite and (2) has a generator, so no harm in doing the general case

Thomas Finley

7:33 PM

@XanderHenderson Thanks for putting up with me. I was indeed acting like a foolish guy

@TedShifrin But somehow it didn't click me...

It took time

But @TedShifrin thanks to you as well!

leslie townes

but also no harm in just doing the F_p case, i guess. the work is in translating what it means for something to generate K^times as a finite group

Ted Shifrin

The proof that $K^\times$ is cyclic without module theory is sneaky. We were discussing pieces of it yesterday, but I don't think the proof was quite there.

user123234

I can say that $f(x)=\sum_{i=0}^n a_{n-i} x^{n-i}$ right?

leslie townes

user: if you want, although if you like having a_ and x^ take the same index i wonder why you wouldn't do sum a_k x^k

you won't need the full generality of polynomials in F_p[x] to show that the map is surjective

user123234

But I mean if I don't take the full generality of polynomials, how can I say that if $i\geq\ord(\gamma)$ then $x^i=1$?

leslie townes

7:46 PM

i should rephrase my remark because "the full generality of polynomials in F_p[x]" is vague and being misinterpreted. to say that a function f from F_p[x] to some set S is surjective is to say that {f(q): q in F_p[x]} = S. all i meant by the above remark is that there is a smaller and simpler subset E of polynomials in F_p[x], not equal to F_p[x] for which {f(q): q in E} is also S

i.e. although you are indexing arbitrary polynomials now, you may find that a lot of the notation you're setting up to discuss arbitrary polynomials is not used later

i'm not sure of the significance of "if i geq org(gamma) then x^i = 1", but if i'm interpreting this correctly it isn't true. for example {1, -1, i, -i} as a subset of C is a finite field, every element has order dividing 4, but (for example) 5 is greater than 4 and i^5 is not 1

user123234

Sorry I don't get what you want to tell me. So I mean I know that there is a generator $\gamma$ s.t. $\langle \gamma \rangle =\Bbb{F}_p^{x}$. Now I want to show that $ev_\gamma$ is surjective. But I need to understand $f(\gamma)$ for all $f\in \Bbb{F}_p[x]$

but to compute $f(\gamma)$ I need this representation of $f$ in terms of finite sums no?

Because I know that if $n\geq ord(\gamma)$ then $\gamma^n=1$ since $\gamma$ is a cyclic element

leslie townes

you really don't need to understand f(gamma) for all f, i guess that's my point. you just need to know that {gamma^n: n an integer} is F_p

i.e. you only need to consider f of the form f(x) = x^n, n a nonnegative integer

and to repeat a remark above, "if n >= ord(gamma) then gamma^n = 1" is not true. for example {1, -1} is a finite field generated by gamma = -1 which has order 2, and 3 is larger than 2, and (-1)^3 is not 1

user123234

But I mean my map goes from $\Bbb{F}_p[x]$ to $\Bbb{F}$ not to $\Bbb{F}_p$ right?

@leslietownes ah sorry if $n=\Bbb{N}\ord(\gamma)$ then $\gamma^n=1$

leslie townes

i didn't look closely enough at the problem statement to know where your map goes. it's basically the same problem whether F is an arbitrary finite field of characteristic p, or whether it's F_p, because the only thing you're really using about it in either case is that it's a finite field

start with F_p. i think there are enough moving parts in this problem as it is, without generalizing

user123234

ahh, so I know that $ev_\gamma: \Bbb{F}_p^x[x]\rightarrow \langle \gamma \rangle$ but now if I take $x\in \langle \gamma \rangle$ i.e. $x=g^n$ for some $n, then I can find $f(x)=x^n\in \Bbb{F}_p^x[x]$ s.t. $f(\gamma)=x$ thus $ev_\gamma$ is surjective?

leslie townes

8:02 PM

user: there are some missing words in the middle of that and the mathjax is a little wonky, but yes, more or less :). the key fact you are using here, which implicitly uses finiteness of the field, is that if q is a nonzero element of your codomain, then you can represent it as [generator]^n for a nonnegative integer n (all that group theory would give you without some finiteness assumption, is an integer n that might not be nonnegative)

you need this for the "x^n" relied on in that argument to be in F_p[x]

user123234

perfect thanks a lot!

but isn't there a problem that I can only represent $F^x$ as $\langle \gamma \rangle$ and not $F$ in the codomain?

leslie townes

0 is in the image of every ring homomorphism. (here it's the image, for example, of the constant polynomial 0)

you can't really change what happens to constant polynomials under this homomorphism (i.e. it will be the same no matter what you're evaluating at.) f(1) is required to be the unit of the codomain by the ring homomorphism property alone, and this determines f(k) for all constant polynomials k in F_p[x]

Xander Henderson

8:18 PM

FYI, for the Americans among us: covid.gov/tests .

leslie townes

xander: thanks for the reminder.

冥王 Hades

I have another vaccine due in about a week. I hate it, not the vaccine, but the needle. I've always hated needles

Gonna need to find someone to hold my hand for a bit while I get it

Ted Shifrin

And the new vaccine is now available, @Xander @leslie

@Hades easier to knock yourself over the head

冥王 Hades

I'm too strong for my own good, unfortunately. As my profile says, I can shatter universes with my fists. I'll be risking the planet

sunny

In Differential Equations with Applications and Historical Notes by Simmons, there is a theorem stating:

> Let $f(x,y)$ be a continuous function that satisfies a Lipschitz condition $$|f(x,y_1)-f(x,y_2)| \le K|y_1-y_2|.\tag1$$ on a strip defined by $a \le x \le b$ and $-\infty < y < \infty$. If $(x_0,y_0)$ is any point of the strip, then the initial value problem $$y'=f(x,y),\qquad y(x_0)=y_0$$ has one and only on solution on the interval $a \le x \le b$.

Ted Shifrin

8:29 PM

Great book.

sunny

Cool :)

My question: suppose $\frac{\partial f(x,y)}{\partial y}$ has a global maximum. Then by the MVT, $(1)$ is always satisfied and doesn't depend on the interval $a\le x\le b$. According to the theorem, does this mean the solution is defined for all $x$?

Xander Henderson

@TedShifrin Yup. Already have an appointment to get it.

Ted Shifrin

Global maximum on what domain?

@XanderHenderson Me too. The first appointment was canceled.

sunny

@TedShifrin true, forgot that

冥王 Hades

Back when I was in the states and got the vaccine I could just hold my mother's hand, since she works at the hospital. This time I don't have that luxury

Xander Henderson

8:38 PM

I'm trying to figure out where I can go to see the eclipse in two weeks.

冥王 Hades

I wanna see a total eclipse before I die

Xander Henderson

There are some really great places nearby, but they are all on the Navajo Nation, and the Navajo are a bit tetchy about eclipses. I'm pretty sure that they don't want some biligaana mucking about on their land.

Next best option is three to four hours north of here, or three to four hours east of here. Though Gallup (only 90 minutes away) might be okay.

@冥王Hades This is an annular eclipse, not a total eclipse. :/

冥王 Hades

@XanderHenderson Don't worry. Hades is capable of summoning a permanent, total eclipse.

Xander Henderson

@冥王Hades You know, every once in a while, I think "Hey, maybe I can treat Hades like an adult." And then you say things like that.

冥王 Hades

8:54 PM

@XanderHenderson You thought wrong, obviously.

user 726941

9:23 PM

Whelp, boomers used to sing about the total eclipse of the heart; so, who knows when they'll "grow-up"

Auden Young

@Jakobian right, i was treating edges of graphs as of unit length - if i treat them as an interval (which seemed more natural when considering, e.g., graphs with weighted edges), i'm aware it gets more complicated

and i know i've basically obtained the discrete topology, which is complete, but i don't know what that implies for what sequences look like in the space

冥王 Hades

@user726941 I hate growing up.

user 726941

They don't call it "growing pains" for nothing.

Thorgott

@ShaVuklia sure, though I'm not necessarily confident in my ability to answer

Ted Shifrin

9:38 PM

@Thor Well, confidence is over-rated. :)

Auden Young

though it does seem clear that no sequence can be Cauchy, since the minimum distance between two different vertices is zero, so the space is automatically complete

so the question is whether there's an interesting infinite series that converges in such a space

user 726941

Are you thinking about majoring in mathematics @AudenYoung

Auden Young

i'm an applied physics major that happens to be taking a lot of math classes

i might accidentally get a math minor, at this rate

user 726941

LoL

Xander Henderson

@user726941 That's not a boomer song, really.

Though it is kind of borderline between the boomers and gen x.

Still, it is the kind of song which was aimed at teenagers. In the early 80s. Which would consist of the last of the boomers, and gen x.