Mathematics

12:42 AM

So for 3 points (2,5) (3,2) and (4,5) if I'm asked to find the polynomial fitting the curve between them I make $p(x) = a_0 + a_1x_1 + a_2x^2$ what do I do again

oh okay so I sub the x values into the variables and equate them to the y

robjohn

@copper.hat I've had a few in the last week; puzzling, at best.

obliv: yeah, that's a general way to do it. for fitting a polynomial of degree n to n+1 points, you'll get n+1 equations in the n+1 unknowns a_0, ..., a_n, and then you can just solve that system. (in some cases but not in general, there may be more than one solution, or no solutions)

Rithaniel

1:25 AM

I encountered a generalization of Euclidean division ($p=qf+r$) for multivariable polynomials (It was "for $p(x,y)$ and $\{f_1(x,y),...,f_n(x,y)\}$ there exists $\{q_1(x,y),...,q_n(x,y)\}$ and $r(x,y)$ such that $p(x,y)=r(x,y)+\sum_{i=1}^nf_i(x,y)q_i(x,y)$," or something close to that)
I haven't been able to find more on that with my google-fu. Would someone point me in the right direction?

@leslietownes Certainly not more than one solution, and none only if you give it two different values at one point.

ted: oh, i was thinking of the "n+1 points" as a list, not as a set.

1:47 AM

I don't get it. These points are in the plane, so they're ordered pairs.

ted, {(1,0), (2,3), (1,0)} is a list of 3 points in the plane, even if it is a set of only two points in the plane

that's the sense in which i was thinking. e.g. if you were implementing this in software, if you skipped a step where you checked the input for duplicates, there could be more than one solution.

2:06 AM

Oh, good grief. OK.

2:59 AM

@robjohn i get a slow trickle, no obvious pattern.

3:20 AM

@amWhy I wouldn't say I'm a particular music fan. Well, I am but not in the traditional way. I don't own many CDs or cassettes, and frankly those I do own are probably in boxes where I haven't listened to them in a while. Of course nowadays people often purchase music online, but I've never done that. But I was a regular headbopper on the school bus when music was playing, and can go wild at weddings, etc.

@amWhy I joined the MusicFans Stack Exchange site to answer a "what's that song that's in my head that I don't even know the words to but I know the beat, or at least the beat of the chorus" question, and later had another one. I ended up answering my own questions both times (actually I think a co-worker answered one of them when it came on the radio, but I forgot the answer later figured it out myself and was like, "that's what that guy told me."

@copper.hat a common complaint as one ages. there are undergarments that can absorb this.

@amWhy I happened to be rather fortunate in that when I first joined that forum, I saw a question asking for a third name of three musicians that happened to remind me of a VH1 program I had seen, so I looked that up on YouTube and added the third name there as the answer and it turned out I was right. So I got a reputation boost pretty quickly.

@leslietownes ahh, the dribble down theory

@amWhy I really have a favorite group, and I'm not even sure I have a favorite style. Some songs I really go crazy when I hear them are September by Earth, Wind and Fire, Celebration by Kool & The Gang, and Jump by the Pointer Sisters. Upbeat, dance-type music.

@copper.hat Thank Reagan?

3:35 AM

@amWhy I do like group theory, but the reason I went on this particular chatroom is I was looking for a chatroom about lattices (Leech Lattice, etc.) and since that type of lattice is listed in Wikipedia as "Lattice (group)" I thought this might be the place.

That’s a very recondite topic, @Kevin.

@amWhy I DON'T really have a favorite group, I mean.

@TedShifrin What does recondite mean?

you don't just bump into the leech lattice in day to day life. you have to go looking for it.

It means very few people know anything about it.

@TedShifrin The 'wonder years' of Reagan/Thatcher politics.

3:37 AM

@copper.hat But that was a cute TV show.

@TedShifrin Wonder Years yes, the politics not so much

@leslietownes Ah, thanks, and thanks leslie townes for the similar answer. Nicely stated, btw

meant to credit :62868927 also

@TedShifrin meant to credit you also. I'm new at this. Can't you tell?

I had a colleague years ago who did lattice-ordered groups. I think there were literally a handful of people in the world working on this.

No need to worry about credit. We’d help if we could.

@TedShifrin Thanks. I'm glad I dropped in on this forum earlier today, as answering @amWhy 's question was fun. And it may give you a bit of an idea of what I'm all about (although I'm still figuring that out myself :) ).

also @amWhy , I'm not a parent. Don't know where you got that or if you meant something about a "parent board".

a stackexchange chat profile is associated with a 'parent' profile (an SE account). maybe that.

3:51 AM

Why are so few people working on that @TedShifrin

aren't lattices widely applicable

although the only thing I can think of are snowflakes in physics lol

@leslietownes so people will come here on the chat who don't participate in the main mathematics.se forum?

as long as you have I think 50 rep, you can talk in any chat room on stackexchange

I will give a shameless plug for my most recent question here: math.stackexchange.com/questions/4623074/…

if you want to render latex check out the sidebar on the right to get chatjax

I wonder if there will come a day when reading that question comes easy or makes sense :P @KevinM.Lamoreau

I think I made it through dummit & foote's abstract algebra up to rings but it was a painful initiation

and I don't recall anything about it anymore lol

@Obliv yeah, I can see it can be confusing. Basically there's a class of lattices the "the densest that is built from the densest that is built from the densest, etc", and sometimes there a ties, but the winning "offspring" of tied lattices only share the title if they themselves are tied. I was wondering what additional lattices might be in that category if the densest lattice including any given tied lattice in one dimension lower was included. Think "share of inheritance."

Thanks to whomever upvoted my question

"aren't lattices widely applicable" @Obliv there are two very different mathematical concepts called lattices. I don't know if you were thinking about the same one I was or the "Lattice (order)". See en.wikipedia.org/wiki/Lattice#Mathematics .

4:09 AM

Yeah I don't even know what a lattice is, so I definitely can't be of help :P But good luck

thanks @Obliv

Oh, so lattices in physics aren't the same as the mathematical ones?

I think the lattices in physics use one of the two concepts of lattices from mathematics. Repeating arrangements of points.

5:43 AM

@Jakobian @AlessandroCodenotti: I learnt another way which does not require the fact that Y is Hausdorff, I have posted that here as an answer to my own post math.stackexchange.com/a/4628027/266435

I request your review of this please. I think that it is correct but I want your opinion and also how you would have solved it.

PM 2Ring

7:06 AM

@TedShifrin I assume that you know that Danica McKellar from The Wonder Years is a mathematician, and a successful author of maths books. Her original target audience was middle-school and high-school girls, but her latest books have been aimed at younger kids. danicamckellar.com/math-books FWIW, Danica has an Erdős number of 4.

I suspect Danica knows a thing or two about lattices... iopscience.iop.org/article/10.1088/0305-4470/31/45/005

@Obliv Almost. There are actually 3 distinct chat networks: one for Stack Overflow, one for Meta Stack Exchange, and one for everywhere else.

8:09 AM

It is quite surprising that for normed space X, $X\simeq X''$ may not imply reflexivity of X.

is it? i mean, while i think any examples of such X are pretty weird as banach spaces get, i don't know why i would expect there not to be any.

R.C. James showed that.

The example is known as James space.

Reflexive space

In the area of mathematics known as functional analysis, a reflexive space is a locally convex topological vector space (TVS) for which the canonical evaluation map from X {\displaystyle X} into its bidual (which is the strong dual of the strong dual of X {\displaystyle X} ) is an isomorphism of TVSs. Since a normable TVS is reflexive if and only if it is semi-reflexive, every normed space (and so in particular, every Banach space) X {\displaystyle X} is reflexive if...

The first para mentions it.

@leslietownes I didn't know this. I thought isometry should imply reflexivity.

Until I saw the following exercise: If X is reflexive, Y is linearly isomorphic isometric with X, then Y is also reflexive.

and the explicit mention of this fact in Kreyszig's functional analysis book.

8:28 AM

if you step away from the specifics a little, "A is reflexive" means that a particular map from A into A'' is an isomorphism, while "A and A'' are isomorphic" means only that there is some map from A into A'' that's an isomorphism. if you phrase it this way, it shouldn't be at all clear why the second property should imply the first.

if you just think of sitting down and trying to prove it, it's not at all clear how you would even make use of the hypothesis.

not to imply that james's example isn't cool, and more generally there are tons of "counterexample" spaces in banach space theory with really delicate constructions, where it's not at all simple to see, OK, there's some path that's going to lead to that example.

but i guess, as a default expectation, if the proof doesn't just fall out of the basic axioms and fit in the first chapters of a book on normed spaces, a safe default assumption is that a counterexample is more likely to exist than not, even if it might be very complicated or an open problem to find one.

right, I don't find this surprising now as at first I had misunderstood 'reflexivity' as 'X is isomorphic with X'' '. But reflexivity is a particular isomorphism.

with this, as you mention there is no reason to believe why a counterexample like James space should not exist.

i never got too deep into banach space theory, but my rough impression is, a lot of the cleverness of various examples has yet to be translated into anything resembling generally applicable strategies or ideas. this may be because in complete generality, banach spaces just don't have enough structure for that to be possible.

kinda like arbitrary topological spaces, where you run into really ugly "counterexamples" pretty quickly unless you impose more structure.

although maybe i just never paid much attention. i went to a talk by someone who was knee deep in banach spaces and he was describing stuff that sounded really complicated to me as if it was a normal, everyday thing that you could have intuition about.

PM 2Ring

8:49 AM

Not to be confused with Banacek spaces en.wikipedia.org/wiki/Banacek

A: A Banach space is reflexive if and only if its dual is reflexive

9:22 AM

30

It can be shown directly: Let $J : X \to X^{**}$ and $J_*: X^* \to X^{***}$ be the canonical injections. Suppose by contradiction that $JX \subsetneq X^{**}$; using Hahn-Banach theorem, there exists $\zeta \in X^{***}$ such that $\zeta \neq 0$ and $\zeta \equiv 0$ on $JX$. Because $X^*$ is ref...

Here, why does there exist such $\zeta$ such that it is non zero?

Don't we need closedness of JX for such existence?

yes. but JX is closed. if you have been working with normed spaces generally, i'm not sure that it's true in that generality. completeness of X is probably used here.

if A and B are banach spaces and J: A to B is an isometry then J(A) is a closed subspace of B. to give an idea of one proof, maybe note that you get the same result if instead of J being an isometry one knows only that J is bounded and that there is some c with ||Jx|| >= c ||x|| holding for all x.

Leslie: Suppose that I take $y_0\in$ X**- JX. I define $f:span (JX\cup \{y_0\})\to F$ as follows: $f(J(x)+ \lambda y_0)= \lambda$, then I can extend it to all of X** by Hahn Banach.

F= R or C.

yes. note, eventually you should think of one of these theorems guranteeing the existence of a nonzero functional that vanishes on a subspace as "the" hahn-banach theorem or "a" hahn-banach theorem, instead of mentally going back to definitions and extensions of linear functionals every time. mentally wrap those arbitrary choices together in a theorem.

yeah :-).

@leslietownes I think we can show such c to be 1: $\|J\|=\sup_{x\ne 0} \frac{||Jx||}{\|x\|}=\sup_{x\ne 0} \frac{||x||}{\|x\|}=1$, since $\|Jx\|=\|x\|$.

J is bounded, hence continuous.

Suppose that $Jx_n, x_n\in A$ is a Cauchy sequence in J(A). Given $\epsilon>0$, there is a natural no. N, $\|Jx_n- Jx_m\|<\epsilon/2$ for all n,m>N. By isometricity, $\|x_m-x_n\|<\epsilon/2$ so (x_n) is a Cauchy sequence in A, which is complete so there is x in A such that $x_n\to x$

Now there exists N' such that for all n>N', $\|x_n-x\|<\epsilon/2$. By isometricity, $\|Jx_n-Jx\|<\epsilon/2$. Hence $Jx_n\to Jx\in J(A)$ so J(A) is closed.

continuity was not needed.

9:51 AM

isometricity implies continuity, which you are using at least implicitly when you pass from norm convergence of x_n to x, to norm convergence of Jx_n to Jx.

ohh

$\|x_n-x\|= \|J(x_n-x)\|=\|Jx_n-Jx\|$

but yes, that is the idea. again it helps to filter away some of the details. if (J x_n) is cauchy, because J is an isometry (or because of that more general condition above) (x_n) is cauchy, hence convergent to x, and then by bounedness (Jx_n) is convergent to Jx. not strictly needed to go to the epsilon/2 or the N to see the structure of the argument.

sometimes the next level of detail becomes important when you are no longer assuming that you are working with bounded operators, if you are assuming enough about the operator that enough inequalities still work.

unbounded operators are also abundant in functional analysis?

I know we can always create one on infinite dimensional normed space.

sometimes you need em, or at least want to work with em, yeah.

😮😮

I learn much more here than in my class here at college.

9:58 AM

for example the fact that (e^(kt))' = k e^(kt) shows you that you can't norm a function space that contains all of the e^(kt)'s (whether the domain is R or [0,1] or anything else interesting) so that differentiation becomes a bounded operator. it is going to be an unbounded operator.

but you might still be able to say things about it.

@leslietownes yes

PM 2Ring

10:49 AM

Interesting HNQ: Should math for elementary teachers content be taught under the direction of the math department?