Mathematics - 2012-08-12 (page 3 of 3)

Peter Tamaroff

18:00

@BrianM.Scott Now the author says this:

Jonas Teuwen

@OldJohn Ah! Rudolf?

Peter Tamaroff

If $\mathfrak I''$ was some other topology such that $f$ was continuous, we would still have $\mathfrak I''\subset f^*(\Bbb I)$

Old John

@JonasTeuwen Nah - I prefer not to give personal names to food :)

Peter Tamaroff

"So the topology $\Bbb I$ carried to $Y$ by $f^*$ is the weakest or smalles topology such that $f$ is continuous"

Brian M. Scott

@PeterTamaroff Where did $\Bbb I$ come from ?

Peter Tamaroff

18:01

@BrianM.Scott It is the identification topology on $X/\sim_f$ induced by $\pi_f$

David Wheeler

@BrianM.Scott apparently his author uses "identification topology" instead of "quotient topology"

Brian M. Scott

Ah. I normally understand $\Bbb I$ to be the closed unit interval, and I didn’t notice that you’d defined it.

Peter Tamaroff

@BrianM.Scott =P

@BrianM.Scott Now I assume he is talking about a topology $\mathfrak I''$ such that our original "boring" $f:X\to Y$ were continuous.

Brian M. Scott

@PeterTamaroff Yes, a topology on $Y$.

Peter Tamaroff

@BrianM.Scott He says we would have $\mathfrak I''\subset f^*(\mathbb I)$ for any topology $\mathfrak I''$ on $Y$ such that $f$ is continuous. Should it be evident why this is so? Maybe I should re read?

Brian M. Scott

18:09

Have you investigated what happens if $U$ is open in $Y$ but is not the image under $f^*$ of any member of $\Bbb I$?

David Wheeler

take a typical open set U in the given topology of Y. look at its pre-image under f*.

Peter Tamaroff

@DavidWheeler OK, let me see.

David Wheeler

well, you already know that f* is continuous

Peter Tamaroff

@DavidWheeler Yes, so $f^*{}^{-1}(\mathfrak I)\subset \mathbb I$

David Wheeler

so you get an open set in the identification topology.

Peter Tamaroff

18:12

Where $\mathbb I$ is the indentification.

@DavidWheeler Yes.

David Wheeler

but what are those open sets?

Peter Tamaroff

@DavidWheeler Equivalence classes?

I mean, "composed of" equivalence classes.

David Wheeler

isn't it true that ${f^*}^{-1}(U)$ is open iff $\pi_f^{-1}({f^*}^{-1}(U))$ is open?

Peter Tamaroff

@DavidWheeler Use f^*{}^{-1}

@DavidWheeler Yes.

David Wheeler

and is it true that the last set is just $f^{-1}(U)$?

Peter Tamaroff

18:17

@DavidWheeler Yes.

David Wheeler

so if f is to be continuous.....

Peter Tamaroff

@DavidWheeler So if $f$ is to be continuous, we must have that any topology on $Y$ is contained in $f^*(\mathbb I)$

OK.

David Wheeler

it helps to deal with a "concrete" example

Peter Tamaroff

Nonsense. =P

@DavidWheeler OK. The author gives the example of "covering the circle with the real line"

And two more

David Wheeler

take X = I, the unit interval on the real line, and take Y to be the real plane

Peter Tamaroff

18:21

@DavidWheeler OK.

David Wheeler

define f(t) = (cos(2πt),sin(2πt))...this is continuous with the "usual topologies"

Peter Tamaroff

@DavidWheeler What is the domain?

$\Bbb R$?

David Wheeler

the unit interval

we could take all of R, but it doesn't make it any clearer.

if you want to use R, we can

Peter Tamaroff

@DavidWheeler No, its OK.

David Wheeler

ok, now explicitly say what $\pi_f$ is.

Peter Tamaroff

18:25

@DavidWheeler It is the set of numbers such that $t-t'=n$ an integer, right?

David Wheeler

no that is [t].

given a real number t, what is $\pi_f(t)$?

Peter Tamaroff

@DavidWheeler Give me a second. I'll jot down something

David Wheeler

i suppose the simplest way to get a concrete expression for $\pi_f(t)$ is to use the "floor function"

Peter Tamaroff

@DavidWheeler $\pi_f$ maps each $t$ to it's equivalence class.

In this case we have that $t\sim_f t'$

whenever

David Wheeler

yes, but what are those equivalence classes?

Peter Tamaroff

18:30

$\sin(2\pi t)=\sin(2\pi t')$ and $\cos(2\pi t)=\cos(2\pi t')$

@DavidWheeler I think the function is one one.

@DavidWheeler You said the domain was $[0,1]$ right?

Oh, OK.

David Wheeler

well, specifically, $\pi_f(t) = [t - \lfloor t \rfloor]$

Peter Tamaroff

$\sin(2\pi t)=\sin(2\pi t')\iff t-t'=n$, am I crazy?

David Wheeler

no, you're not crazy

but what you wrote isn't true

Peter Tamaroff

@DavidWheeler Where?

David Wheeler

two sines can be equal without their arguments being a difference of 2pi

for example sin(pi) = sin(0)

however, if sin(x) = sin(x') AND cos(x) = cos(x'), then x - x' = $2\pi k$

where k is an integer.

Peter Tamaroff

18:39

@DavidWheeler Yes, I meant that.

My fault.

David Wheeler

so yes, [t] = {t+k: k in Z}

Peter Tamaroff

@DavidWheeler In this case $2\pi t-2\pi t'=2\pi n$ so $t-t'=n$

@DavidWheeler OK.

David Wheeler

ok, f maps 0 to (1,0), yes?

and f is 1-1 until we get to 1, and then we've "identified" 0 and 1.

Peter Tamaroff

@DavidWheeler Right.

David Wheeler

if we keep going, we just "wrap another unit around the circle"

so we're not really gaining anything from considering R instead of [0,1]

Peter Tamaroff

18:42

@DavidWheeler OK.

David Wheeler

we're just doing the same thing "more times"

[1,2] gets mapped to the same points as [0,1]

in fact f(1+t) = f(t)

Peter Tamaroff

@DavidWheeler Yes.

David Wheeler

it's "easier to think about" just the unit interval.

ok, what is an open set of the circle, under the relative topology in $\Bbb R^2$?

Peter Tamaroff

@DavidWheeler An arc.

David Wheeler

an open arc, yes...which is just an intersection of an open disk and the circle.

actually "unions of open arcs" but its ok just to consider basic open sets.

Peter Tamaroff

18:45

@DavidWheeler Yeah.

David Wheeler

now, what is the pre-image of an open arc?

Peter Tamaroff

@DavidWheeler Well, it depends.

David Wheeler

listening....

Peter Tamaroff

@DavidWheeler It the arc doesn't unclude $(0,0)$ then it is an open interval. However, if it includes $(0,0)$ we get something like $[0,a)\cup(b,1]$?

David Wheeler

yup.

Peter Tamaroff

18:48

@DavidWheeler Oh but that is open in $[0,1]$.

Se we're good.

David Wheeler

right...but the point is, f isn't an open map.

Peter Tamaroff

@DavidWheeler It doesn't map open sets of $[0,1]$ into open sets of $S^1$?

@DavidWheeler How do you define an open map?

David Wheeler

what does f map the open set (b,1] to?

Peter Tamaroff

@DavidWheeler To an arc from the fourth quadrant up to $(1,0)$

David Wheeler

yes, and including (1,0)

and in $S^1$, f((b,1]) has a boundary point....

so the image of an open set isn't necessarily open

Peter Tamaroff

18:54

@DavidWheeler In the boundary of what?

David Wheeler

how does $f^*$ fix this?

Peter Tamaroff

@DavidWheeler Give me a few.

David Wheeler

f maps the open (in I = [0,1]) set (b,1] to the half-open arc from $(\cos(2\pi b), \sin(2\pi b))$ to (1,0)

Peter Tamaroff

@DavidWheeler Yes. Iwas writing something :P

@DavidWheeler We have $f^*:[0,1]/\sim_f \to S^1$. Now there should be one peculiar open set in $[0,1]/\sim_f$,

David Wheeler

the identification topology "fixes" this like so: since 0~1, any neighborhood of 1, must also include a neighborhood of 0, so we can't just take (b,1], we have to add some [0,a) as well.

Peter Tamaroff

19:00

@DavidWheeler OK.

David Wheeler

naively: we "glue" or "paste" 0 to 1, so the two half-open intervals become a single open arc.

Peter Tamaroff

@DavidWheeler I see.

David Wheeler

so, the identification topology (in this instance) gets rid of two kinds of open sets: (b,1] and [0,a)

Peter Tamaroff

@DavidWheeler I see.

David Wheeler

the mapping f* becomes a homeomorphism under these conditions (its bijective, and since its open, it's inverse is continuous)

Peter Tamaroff

19:04

@DavidWheeler =) OK. I will go and eat something. I'll be back shortly.

@DavidWheeler So the unit interval glued by the ends is homeomorphic to $S^1$

I guess we con do this with any curve, right? Any path in $\Bbb R^2$

(bounded)

David Wheeler

don't jump to conclusions...the figure 8 is not homeomorphic to the circle.

anon

http://mathworld.wolfram.com/SimpleCurve.html

Peter Tamaroff

19:34

@DavidWheeler I meant an "open" curve

David Wheeler

look at anon's link

the possible "circuits" in a curve matter

Peter Tamaroff

19:51

@DavidWheeler I meant just a line from a point to another, a bounded line, just a path connecting two dots, no crossings, no weird stuff going on

Kannappan Sampath

@Matt How did your exams go?

skullpatrol

20:09

hi @FortuonPaendrag

Fortuon Paendrag

@KannappanSampath It is soo good to see you again, Kannappan!! How have you been?

Hello, skullpatrol.

anon

20:21

@Geekster: You cannot check if $1$ is in the list of natural numbers if you cannot check any equalities whatsoever. You can only compute new numbers, and pretend you don't know what they are (since it seems directly checking a number's identity is forbidden). If you can perform multiple computations on the numbers in the collection and check if and when they give the same results (which involves checking equalities), then it's possible via checking $ab=b$ in the way I described in my comment.

For any $b$ in the collection, if $ab=b$ holds true for every other $a$ then $b=0$; we can discard these from the collection and then check if there exists an $a$ such that $ab=b$ for every nonzero $b$. If there is, then it is $1$, else there is no $1$. The only exception to this should be if the collection only contains $0$'s and $1$'s.

If you want an explanation for the downvotes (beyond simply that MSE is made up of jerks), consider the following question: You have a collection of sticks, and wish to find which one is the largest, but the only thing you're allowed to do is break them or burn them (you can never directly compare two sticks). This is what it feels like reading your question.

4

skullpatrol

hi @Mike

hi @HenningMakholm

Nimza

20:48

Good night!

"chair" and "department" of faculty - are they the same thing?

synonyms?

skullpatrol

no

Peter Tamaroff

@Nimza "Chair" means "position" usually.

Like "he holds the chair of director of ....."

???

Nimza

@PeterTamaroff mmm.... here is an example of use

".. is a student at Chair of Mathematical Physics, Faculty of ..."

so it is a bad translation?

Henning Makholm

@Nimza Some universities use that nomenclature, because in principle they have a one-to-one correspondence between organizational units and named professorships, in which case "chair" becomes a quaint local name for department.

It doesn't seem to be used at the majority of universities in the western world today, though.

Nimza

@HenningMakholm so interesting, thanks

Peter Tamaroff

21:00

@HenningMakholm My answer here is not really helpfull, is it?

Henning Makholm

@PeterTamaroff Dunno. There are scary-looking formulas involving integrals that make me disinclined to even try to understand the question.

Peter Tamaroff

@HenningMakholm Heheheh OK.

I deleted it. I'll try to think about it nevertheless.

N3buchadnezzar

@PeterTamaroff Did you see that did did it?

Henning Makholm

However, perhaps what the OP needs most is something that targets his apparent understanding that there is something "indeterminate" about the limit itself. (It seems to hint that he thinks limits are something that operate on particular formulas, rather than on sequences).

Peter Tamaroff

@HenningMakholm Hmmm right.

Henning Makholm

21:13

@PeterTamaroff FWIW, I think it is slightly bad form to edit away the contents of the post in addition to deleting. I have trouble seeing what purpose it serves, and it only makes it more difficult for the curious to find the original text, and tends to make curiousity-inclined people actually curious.

skullpatrol

hi @JasperLoy

user19161

@skullpatrol Hi!

skullpatrol

Welcome back @FortuonPaendrag

user19161

@jonas How is her fever now? Also, I hope you are better now too.

Peter Tamaroff

21:31

@HenningMakholm I'll try and add something there ;)

Jonas Teuwen

@JasperLoy She is fine. My temperature is 34.9°C lol.

DeadMG

22:00

hey

BenjaLim

@FortuonPaendrag I asked Kannappan yesterday on skype to come back to math.se

Fortuon Paendrag

@BenjaLim It seems that it worked, if only for a short time.. :(

BenjaLim

@FortuonPaendrag Commutative Algebra

Fortuon Paendrag

@BenjaLim Ahahahaha. Definitely a nuke ;)

BenjaLim

@FortuonPaendrag Maybe only a tomahawk cruise missile?

Fortuon Paendrag

22:05

@BenjaLim Or it is a chicken that you are killing and not a mosquito.

BenjaLim

hahahahahahahhaha

@FortuonPaendrag Do you do algebra?

Fortuon Paendrag

@BenjaLim Not really. My past year was almost entirely Analytic, although I did read some algebra on my own.

user19161

@JonasTeuwen Another similarity between us. I usually have very low temperatures as well!

Jonas Teuwen

@JasperLoy Oh, you do? :-).

Fortuon Paendrag

@BenjaLim you, on the other hand, seem to be a big algebra guy

BenjaLim

22:07

@FortuonPaendrag Where are you studying now?

@FortuonPaendrag I also do fourier analysis :D

user19161

@BenjaLim Anyway, he seems so determined, so I won't ask him again.

Fortuon Paendrag

@BenjaLim At Colorado College, in the U.S. It's a liberal arts school.

BenjaLim

@FortuonPaendrag You're from dehli yes?

@FortuonPaendrag I have been to dehli.

Fortuon Paendrag

@BenjaLim India, yes. Not quite Delhi. It's a city called Hyderabad. But Delhi is cool to I suppose

user19161

@BenjaLim Do you mean Delhi?

BenjaLim

22:08

@JasperLoy yes

user19161

Dehli is not Delhi, hahahaha.

user19161

Next thing you will say is deli.

Fortuon Paendrag

Michael Krohn-Dehli

Michael Krohn-Dehli (born 6 June 1983 in Copenhagen) is a Danish footballer who currently plays as a midfielder for Danish club Brøndby IF and for the Denmark national football team. Club career He began playing football at Rosenhøj BK in Denmark, after that he played for Danish clubs Hvidovre IF and Brøndby IF. The midfielder made his debut for Ajax on 17 September 2006 in the league match Roda JC – Ajax (2–0). Krohn-Dehli played for a number of Danish clubs before joining Ajax from Brøndby IF. Krohn-Dehli excelled as a youth team player for Ajax under Danny Blind. In 2004 however, he w...

BenjaLim

@FortuonPaendrag Hydrebad.....

@FortuonPaendrag I find the way it's cooked to be very inspiring

Fortuon Paendrag

@Benjalim OF course. We make the best biryani in the world

user19161

22:10

@FortuonPaendrag That city sounds like a vegetable name.

BenjaLim

hahahahahahaha

@FortuonPaendrag I have some pakistani friends that claim that the briyani in pakistan is better than anywhere in india...

user19161

I am still waiting for them to delete 8 accounts. I am keeping 3 though.

BenjaLim

@FortuonPaendrag youtube.com/watch?v=DBY61x4lIXM

Fortuon Paendrag

@Ben Oh, they've never eaten at Paradise or at the Alpha hotel in Hyderabad.

BenjaLim

@FortuonPaendrag Once some friends from india made briyani the traditional way

with copper pots sealed with dough

user19161

22:11

@ben Can you tell me if Munkres Algebraic Topology is any good?

BenjaLim

@JasperLoy It's a lot more rigorous than hatcher :D

user19161

@BenjaLim Ah good. It seems there are a number of people who think Hatcher is handwaving too much.

Fortuon Paendrag

@JasperLoy Hatcher has lots of pictures, that should make you cautious.

Way too many pictures to be a respectable math textbook.

user19161

@FortuonPaendrag Yes, though some people like it. Rudin's 3 books have no pictures!

Fortuon Paendrag

@BenjaLim Wow. That must have taken a lot of effort. Biryani is not easy to make.

BenjaLim

22:13

IIRC it took about 3 - 4 hours

user19161

@BenjaLim But it is very seldom used in courses these days if you look around.

BenjaLim

That is because he does not treat $\pi_1$

user19161

@BenjaLim Which is now covered in his first book.

BenjaLim

@FortuonPaendrag Why do you think I'm big on algebra?

Fortuon Paendrag

@JasperLoy IMHO, Rudin's real and complex analysis is the perfect example of a Math Textbook. Starts from scratch and will exhaust you. In a good way.

user19161

22:15

@FortuonPaendrag Though it leaves many important things to the exercises sometimes.

user19161

@BenjaLim Since you are always talking about it?

Fortuon Paendrag

@BenjaLim Algebraic Topology and Commutative Algebra and Matrix Lie Groups are all I've heard you talk about. Hint much? ;)

user19161

I also hate those books which omit too many proofs.

user19161

@ben tb says though that Bredon is full of errors. But if you can fill in the gaps I think it is superb.

Fortuon Paendrag

@JasperLoy I guess the first part of spivak's calculus on manifolds would have been a better example

user19161

22:19

@FortuonPaendrag Spivak is a weird guy. He writes a calculus book that is damn long and a calculus on manifolds book that is damn short.

Fortuon Paendrag

@JasperLoy Come on. You must admit that the contents are very good.

user19161

@FortuonPaendrag He also wrote 5 volumes of differential geometry haha!

user19161

But I wonder how many people bother reading through all of them.

Fortuon Paendrag

@JasperLoy Five? Wow..

user19161

@FortuonPaendrag Yes, I thought they are so famous everyone knows.

Fortuon Paendrag

22:22

I must be very ignorant then. :(

BenjaLim

@JasperLoy thanks for the heads up on bredon

@FortuonPaendrag :D yes that is true...

@FortuonPaendrag Just wondering, I think that $\Bbb{Q} \otimes_\Bbb{Z} \Bbb{Q}$ is not finitely generated as a $\Bbb{Z}$ - module yes?

@JasperLoy I find AM more readable than Hatcher.

Old John

Good evening folk(s)

DeadMG

how can you solve a boolean formula which is so impossibly large, you can't ever hope to read it?

BenjaLim

I'm off now guys

bye!!

anon

@BenjaLim $\Bbb Q\otimes_{\Bbb Z}\Bbb Q\cong\Bbb Q$ I believe, which is not finitely generated

BenjaLim

22:35

@anon really?

anon

let me write it out

BenjaLim

Yes I know $\Bbb{Q}$ is not finitely generated as a $\Bbb{Z}$ - module.

@anon Yes I think it is true that as abelian groups $\Bbb{Q} \otimes_\Bbb{Z} \Bbb{Q} \cong \Bbb{Q}$.

anon

$$\frac{a}{b}\otimes\frac{c}{d}=d(\frac{a}{db})\otimes \frac{c}{d}=\frac{a}{bd}\otimes c=\frac{ac}{bd}\otimes1$$

whoah what the

Jonas Teuwen

Scary man...

Get away from me, freak!

BenjaLim

@anon Also I think you can show that you have a unique linear map from $\Bbb{Q}$ to any other module $N$

if you have a bilinear map $B : Q \times Q \to N$

anon

22:38

--

Old John

@JonasTeuwen who?

Jonas Teuwen

@OldJohn Nobody. Just wanted to say this 8-).

BenjaLim

You have $p : Q \times Q \to Q$ that sends $(a,b)$ to $ab$

@anon no just want $\Bbb{Q}$ to satisfy the universal prop.

Old John

@JonasTeuwen Great :)

David Wheeler

yes, rational multiplication works.

Old John

22:39

@JonasTeuwen I sometimes shout something similar when I am out with my wife - just to annoy her :)

BenjaLim

@DavidWheeler and then you define the map $L : Q \to N$ as $L(q) = B(1,q)$

Jonas Teuwen

@OldJohn :P. Haha.

BenjaLim

the map on "elementary rationals" @DavidWheeler

Old John

"Who ARE you?" ... "Why are you following me???"

BenjaLim

@JonasTeuwen What anon did above was just algebraic manipulation

and you have to be careful when doing it because the tensor product was just a $\Bbb{Z}$ - module

and not over $\Bbb{Q}$.

Jonas Teuwen

22:41

I know. Just... whatever.

@OldJohn Mine might be a bit stranger to say.

anon

I only used $\Bbb Z$-linearity of $\otimes$

David Wheeler

@BenjaLim you mean "only pull out integers in front"?

BenjaLim

@anon Here's a cooler way:

no I can't do it my way.

I have to extend scalars first.....

@DavidWheeler You can't put $q(a \otimes b)$ where $q$ is rational

@FortuonPaendrag I have work on fourier theory now.

bye guys!

David Wheeler

no, you have to do what anon did

fortunately, Q is divisible.

anon

We're all clear on what's allowed and what isn't, Benja's just saying one needs to be careful when the scalars are restricted as a sort of general feature of using algebraic manipulations to make arguments about tensor products.

I think

David Wheeler

22:46

i was thinking of the map $\frac{a}{b} \otimes \frac{c}{d} \mapsto \frac{ac}{bd}$

that works, doesn't it?

Jack Schmidt

23:03

@DavidWheeler Yes. The calculation should look suspiciously similar to chat.stackexchange.com/transcript/message/5746931#5746931

In general if $f:R \to S$ is an epimorphism of rings and $A_S, {}_S B$ are $S$-modules thought of as $R$ modules, then $A \otimes_S B \cong A \otimes_R B$. Or more simply $S \otimes_R S \cong S$. Two standard cases are $S=R/I$ where it is easy, and $S = T^{-1} R$ is a localization of $R$ (by inverting some elements $T$). The latter case is proved exactly as in the previous calculation.

If you see notation like $k(\mathfrak{p})$ lying around, this little fact can sometime greatly simplify your understanding of what is going on: they are just saying a bunch of tensor products don't care what ring you tensor over.

Jack Schmidt

23:27

@KannappanSampath Core exercises of chapter 6 (Isaacs CTFG) are #1,2,3,7,12,17,18 ; I've not found M-groups helpful in my work (4,5,6,9,10,11) ; I've found 8,13,14,16,19,20,21 helpful in my work.

David Wheeler

@JackSchmidt i still struggle with what a tensor product is. not "a construction" of something that satisfies the universal property...but something a bit more hard to explain

BejaLim has sort of gone "tensor-crazy" he sees them 'everywhere" now

anon

It's a loading dock for bilinear maps to pass down to. :-)

Fortuon Paendrag

@BenjaLim: Sorry I left you hanging. You are correct. You are correct, the tensor product is not finitely generated, since it is equal to $\mathbb Q$ no?.

Jack Schmidt

@DavidWheeler I just think of it as letting you multiply modules together. This is particularly sane if one of the modules is a ring. "Bilinear" is code for "distributive multiplication". Good examples are $k[x] \otimes_k k[y] = k[x,y]$ and $\mathbb{Q}[i] \otimes_\mathbb{Z} \mathbb{Q}[\sqrt{2}] = \mathbb{Q}[i,\sqrt{2}]$. Also $\mathbb{Z}/n\mathbb{Z} \otimes A = A/nA$ says make sure $n$ acts as 0.

David Wheeler

i've heard several "colloquial" descriptions...a "most general bilinear thingy", a way to turn bilinear maps to morphisms

Fortuon Paendrag

23:35

Oh, anon already clarified it. @anon, thanks!

David Wheeler

right...but there seem to be several "tensor identities" (strictly speaking, isomorphism theorems) that people become familiar with..i have no "intuition" about these...

Jack Schmidt

If it wasn't such complicated notation, I would suggest always thinking of tensor product as being part of a triple: $\operatorname{Hom}(A \otimes B,C)$ is the abelian group consisting of ways to multiply elements of $A$ and $B$ and get an element of $C$.

David Wheeler

that's what i mean by "turning bilinear maps to morphisms"

you have A, and B, and instead of a bilnear map from AxB to C, you want "something" that turns that into a morphism straight-across

Jack Schmidt

yeah, it's like an “unevaluated” multiplication. It represents what the multiplication could be. A “formal product” rather than an actual one.

David Wheeler

well, sometimes it's an "actual product"

Jack Schmidt

23:40

the formal product is the actual product in polynomial rings, because polynomial rings are themselves pretty formal :-)

Like what does $x^3+2$ mean? “You take $x$ times $x$ times $x$ and add $2$” but what is $x$ times $x$? It is $x^2$ of course! It makes sense because math can handle the abstract and formal, but it also doesn't make “real” sense until you actually substitute $x=3$ to get 3 times 3 times 3 plus 2 is 29.

David Wheeler

but then you're using the evaluation homomorphism

Jack Schmidt

exactly. An element of $\operatorname{Hom}(k[x],k)$ that makes sense of the formal products.

Fortuon Paendrag

@Jack To be really lame: What does a formal polynomial say to a polynomial function? Suit up, Bro.

David Wheeler

i always think of polynomials in a single variable as "finite sequences in R (the ring)"...is this wrong?

Jack Schmidt

My wife says the polynomials are going to be legend dairy.

@DavidWheeler It is fine. I just consider that to be formal. I prefer to think of them as recipes for using ring operations.

well, part of a recipe. They are the instructions on how to combine the ingredients, but without the ingredient list! The ingredient list is the evaluation homomorphism.

Fortuon Paendrag

23:47

@David It is ok, because there is an isomorphism from the direct sum of countably many copies of $R$ to $R[x]$.

David Wheeler

well, what lead me to think of them that way, was the defintion of a vector you sometimes see as a function from {0,1,...,n} to F.

Jack Schmidt

@DavidWheeler That is a very good definition. Many things make more sense when you view them as functions. (I don't usually think of them that way, though I didn't understand transfinite sequences until I understood sequences as functions $a_i = a(i)$).

I think it's important to have both a formal and an informal understanding of an idea. The formal definition will usually generalize, and you be sure that a proof using only the formal properties will still be true in the more general context. However, without the informal idea, you cannot be sure the generalization is really the same idea.

If everything is a function, then suddenly everything is a generalization of itself, and you are lost in a land of grey ooze. So it is better if some things are functions, and some things are like functions, and some things can be thought of functions if it is useful that day.

David Wheeler

hmmm...the way i think (internally): if we want something "general" the functional point of view is "portable" (we can do little tricks with the domain and co-domain to get various "instances". but if you're "illustrating" (specifying) you need something more "down-to-earth": 'things" (objects).

JGord

Hello all, I have a quick question

If I have a set of points on the x,y plane

What is the best way of determining which points have a neighbor less than d distance away?

David Wheeler

you can connect the dots, and make a picture? erm....sorry.

JGord

23:58

It's kind of a clustering problem, but not really becuase I don't care about the clusters

David Wheeler

do you have a LOT of points?

JGord

I might

Jack Schmidt

@JGord Look for "plane sweep" algorithms. I'm no expert, but I think this problem was in several of the computational geometry books I read.

JGord

let's hear an answer for yes and no

Jack Schmidt

if just a few points: check all distances. If lots of points, that is quadratically many checks. Plane sweep should be linear.

David Wheeler

23:59

well, if the number of points is 'small" you can calculate the distance between any two.

Transcript for

$Mathematics$ Mathematics