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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.
complex numbers took a while to be accepted.
Do you have any examples?
my homework solution took a while to be accepted after several claims.
If they listen to me, they’ll keep rejecting that! 🤷‍♂️😻
2:59 AM
I mean academic examples
my paper took a while to be accepted after several rejections.
@LordPooty Look here.
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?
@ThomasFinley with what?
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.
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 $f'$?
all of this and more, sketched in thomas's own question from less than a week ago
@leslietownes are you saying that I will be answering the same question again in a week?
only if you're lucky
@ThomasFinley show that the minimum point cannot be $c$ or $d$.
5:45 AM
And I said that a week ago! But it was $a$ and $b$. In a week, letters inflate.
Happy goose to you.
6:10 AM
I'm so glad I've been able to parrot Ted
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
there's a notion of 'depth n foliation'. How interesting.
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.
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.
It looks like they’re just wanting a “local” version of BV.
Uniform in the family given $x,y$.
I'll put up another snippet that mentions BV and Hölder, from earlier in the paper
Can you verify their definition of bounded variation?
I doubt it's defined, since the first mention of "bounded variation" (according to pdf search) is here:
last couple lines
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.
6:31 AM
Does our notion of BV even make sense on a general metric space?
I only know it on an interval in $\Bbb R$.
some thoughts on that problem here: mathoverflow.net/questions/234251/…
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?
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.
Sounds like I’m not the only one with that question, leslie. And it’s not clearly resolved.
What about a rectangle in $\Bbb R^n$?
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.
6:37 AM
I think their $V(x,y)$ is their definition.
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.
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 $\mathbb R$
highly likely that we've already spent more time thinking about this use of terminology than the author did.
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
I’m worrying about domain, not range, Novice.
6:41 AM
Oh, sorry
No need to be sorry. I’m just clarifying.
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.
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
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.
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)
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.
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.
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.
Fair enough.
Thanks for the advice.
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.
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.
1 hour later…
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 $g(x) \cdot h(y)$?
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.
$M= N\oplus M/N$
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 $\alpha$.
Consider $M=R=K[x, y]$ and $N=(x, y) $
Is $K[x, y]/(x, y) $ free?
1 hour later…
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
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
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
2 hours later…
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.
12:49 PM
@sunny $\pm(1-\alpha)t+C = |(1-\alpha)(\pm t)+C| = (1-\alpha)|t\pm C|$
isn't it?
@Jakobian hmm, my brain goes $$\begin{align}\pm(1-\alpha)t+C&=\pm(1-\alpha)\left(t+\frac{D}{1-\alpha}\right) \\ &=(1-\alpha)\pm\left(t+\frac{D}{1-\alpha}\right) \\ &=(1-\alpha)\left|t+\frac{D}{1-\alpha}\right|.\end{align}$$
@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
It turns out you get the same solution if you go with $|x|^{1-\alpha}=(1-\alpha)|t-D|$, it just confused me
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.
Eilenberg was Polish?
1:40 PM
2:30 PM
@AlessandroCodenotti Cantor scheme need not give raise to a Cantor set?
I'm asking why we define it that way then
A: Show that every uncountable and closed subset of the complete separable metric space contains a homeomorphic subset with the Cantor set.

Alessandro CodenottiFirst 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
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
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))
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.
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.
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?
@robjohn Outlook is the only MS product I have to consistently use at work. :/
@XanderHenderson halp
@XanderHenderson Ah, it's good that that is the only one, but bad that you have to deal with it consistently.
oh wait nvm
yeah that set would be empty
Q: Books for Teenager

Max MustemalMy 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
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.
@XanderHenderson yeah. Besides, its in-search also sucks. Why don't they replace outlook by something else in the firms?
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
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 $3^{4^{\cdots^{2020}}} \equiv 3^{4^{5^{\cdots^{2020}} \bmod }\varphi(25)}\bmod 25$?
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 $k\ge1$.
@Obliv No. Note that $\Bbb R\times\Bbb N\setminus\Bbb Z\times\Bbb N = (\Bbb R\backslash\Bbb Z)\times\Bbb N$.
Wait. I got $1 \bmod 20$
5:42 PM
@soupless How can an odd power of $4$ be $1\pmod{20}$?
Oh, no. I got it wrong
I just realized, I started with 3 and used mod 20 there
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
@XanderHenderson perhaps they hard wired it into the program thinking people need to spell "correctly"
@Koro hope you're joking
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
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
Distance = | a - b |
That's an elementary fact.
you're not making any sense
I'm just starting from the basics.
6:00 PM
starting what?
a chat?
starting a chat from basics?
OK, enough, children.
Anyway, there is a book called, Geometry of Cuts and Metrics
Considering distances on graphs leads to some cool things
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" ?
How about the definition of dimension?
@TedShifrin I used it.
Apparently not.
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 $\sum_{i=1}^mc_il_i=0$
6:41 PM
Then, as, $l_i\in V$ so, I considered nos. $A_{i1},A_{i2},...,A_{in}$ such that $A_{i1}v_1+A_{i2}v_2+..+A_{in}v_n=l_i$
You have already assumed they're linearly independent. So what are you doing?
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
Are you trying to prove instead that $m>n$ vectors in an $n$-dimensional vector space must be linearly dependent?
@TedShifrin yeah
Is there any other way rather than showing this?
6:44 PM
@TedShifrin what's that?
Any linearly independent set, if it is not already a spanning set, can be expanded to become a basis.
@TedShifrin I found a proof of it, but in almost all cases they assume that the linearly independent set is finite
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.
@TedShifrin Will then all the linearly independent sets be finite?
We went through this literally weeks ago.
6:47 PM
@TedShifrin Wait, now I am confused. I don't quite remember it!
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.
Sorry for my weak memory :?)
@TedShifrin ok, but how does that contradict the fact that the vector space has a finite dimension
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 $>k$.
What I am really interested in, is that : Is a linearly independent subset of a finite dimensional vector space always finite?
@TedShifrin ok
@TedShifrin Everything is fine with this statement.
Then you're done.
If the dimension is not $>k$, then you have $k$ or fewer linearly independent vectors.
6:53 PM
@TedShifrin ok...
@TedShifrin in that case, we're done, but what if dimension is >k ?
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 $k=\dim V$.
@TedShifrin I proved it already...
@TedShifrin ok
You have done proofs apparently without understanding the concepts.
@Thorgott are you ok with a numb thy question?
@TedShifrin i am sorry but I dont get this at all...
6:55 PM
You keep falling in the same holes. That suggests a lack of complete comprehension.
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)
@ThomasFinley What is your definition of "dimension"?
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
@ThomasFinley What is your definition of "basis"?
@XanderHenderson The linearly independent subset L of the vector space such that span(L)=the vector space
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?
@XanderHenderson yes.
@XanderHenderson the set of all linear combinations of the vectors in that set is equal to the vector space itself
In other words, any other vector in the space is a linear combination of your basis set.
@XanderHenderson yes.
And what is your definition of linearly independent?
@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, $c_1=c_2=...=c_n=0$
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?
@XanderHenderson yes, sure.
Okay, so what if you choose a different basis? Can the dimension of the space change?
@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.
@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" ?
@XanderHenderson yes.
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...?
@XanderHenderson the cardinality of that set is equal to the dimension of the vector space
@ThomasFinley That seems to answer your question...
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.
@XanderHenderson how, the question doesn't say, that the linearly independent set spans the vector space?

 The Periodic Table

Haikus are awesome / Chemistry's even better / So pull up a chair
7:12 PM
@ThomasFinley So what happens if it doesn't?
@user726941 thankyou so much sorry for the disruption.
@XanderHenderson What happens? Well, then the set is not a basis?
@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.
Or in other words the set has cardinality higher than the dimension of the vector space?
7:16 PM
@ThomasFinley How could that possibly happen?
@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.
You are just parroting back to me things we've already gone over.
thinks quietly ... gee, didn't we just go through all this ....
@XanderHenderson I am sorry but I have no clue what to say.
7:18 PM
@TedShifrin yes, surely
Time to stop and review @ThomasFinley
Take your time.
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?
@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
Go back to the definition of dimension.
Cardinality of a basis ir cardinality of a linearly independent set whose cardinality is same as the vector space...
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 $\gamma$?
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?
'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 .
@TedShifrin maybe you can help e further?
But I don't know or seem to get, why this is true?
Oh wait
Yes, it's true!
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
7:32 PM
Thanks a ton, @XanderHenderson
I get the thing now.
@user123234 I had to think through the question and leslie beat me to the punch. Think about what $f(\gamma)$ means :)
@leslietownes but in my case I take $K=\Bbb{F}_p$ right?
@Xander quite literally had you go through exactly what we had already been through. But sometimes it takes 5 or 10 times.
Yes, @user123234.
@TedShifrin Indeed.
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
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!
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
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.
I can say that $f(x)=\sum_{i=0}^n a_{n-i} x^{n-i}$ right?
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
But I mean if I don't take the full generality of polynomials, how can I say that if $i\geq\ord(\gamma)$ then $x^i=1$?
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
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
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
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 $\gamma^n=1$
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
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?
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]
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?
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]
8:18 PM
FYI, for the Americans among us: covid.gov/tests .
xander: thanks for the reminder.
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
And the new vaccine is now available, @Xander @leslie
@Hades easier to knock yourself over the head
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
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 $a \le x \le b$.
8:29 PM
Great book.
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 $x$?
@TedShifrin Yup. Already have an appointment to get it.
Global maximum on what domain?
@XanderHenderson Me too. The first appointment was canceled.
@TedShifrin true, forgot that
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
8:38 PM
I'm trying to figure out where I can go to see the eclipse in two weeks.
I wanna see a total eclipse before I die
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. :/
@XanderHenderson Don't worry. Hades is capable of summoning a permanent, total eclipse.
@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.
8:54 PM
@XanderHenderson You thought wrong, obviously.
9:23 PM
Whelp, boomers used to sing about the total eclipse of the heart; so, who knows when they'll "grow-up"
@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
@user726941 I hate growing up.
They don't call it "growing pains" for nothing.
@ShaVuklia sure, though I'm not necessarily confident in my ability to answer
9:38 PM
@Thor Well, confidence is over-rated. :)
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
Are you thinking about majoring in mathematics @AudenYoung
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
@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.
9:46 PM
Agreed, the boundaries are fuzzy.
@AudenYoung they'e eventually constant
@TedShifrin yup, we're always being humbled by math itself after all
anyway not sure why you're saying they're discrete
is unit interval discrete?
10:14 PM
@XanderHenderson We’re still the best generation
I mean Gen Z

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