Mathematics - 2019-10-17 (page 2 of 3)

Aladdin

5:01 PM

@TedShifrin Okay.z varies from(0,$x^2$+$y^2$)?

Lucas Henrique

@TedShifrin technically. I just can’t find the equivalent words in English, but it’d be “iniciação científica”

Ted Shifrin

@Aladdin: Yes, and in cylindrical?

@Lucas: Seems strange to me. It's a standard part of an abstract algebra curriculum, but you need a lot of group theory and field theory to make sense of it.

Lucas Henrique

Basically I’m paid to study a field I’m interested (so is the institution that pays me) as long as I show results of my study and keep good grades. God bless FAPESP.

Aladdin

($x^2$+$y^2$),infinity)

Simone

@TedShifrin I just bought your book :-)

Ted Shifrin

5:04 PM

@Aladdin? I'm confused.

Simone

the paperback version, the hardcover is quite too pricey for me

Ted Shifrin

@Simone: I think the paperback is smuggled from India.

Totally removed from me. But maybe I don't know.

Regardless, I hope you find it helpful and challenging.

Lucas Henrique

I’ve looked up some translations and it’d be “scientific initiation/undergraduate research”.

Simone

From India? Oh dear... what did I just buy XD

Ted Shifrin

Regardless, Lucas, I think students/universities plunge too fast into "undergraduate research" these days. But I'm old-fashioned.

LOL, @Simone. I'd be curious to know what's printed inside the cover ...

Aladdin

5:06 PM

@TedShifrin Sorry.it's (0,$r^2$)

Ted Shifrin

OK, @Aladdin. Now I'm happy.

Simone

oh God It is an indian company!

Ted Shifrin

ROFL.

Yup, and they totally stole it. I get no moneys from them.

I'm pretty sure.

Simone

I bought it on Amazon, It can't be that scammy

@TedShifrin sorry :-(

Ted Shifrin

I'm curious now. I'm going to investigate a little.

Galois in the Field

5:09 PM

@TedShifrin Okay, I guess you're right :D. The zeros in the other chart are at $1/a$ and $1/b$

Ted Shifrin

@Galois: You mean poles.

Those are, of course, the same poles we already had.

Simone

The company is: Books by Vedams

Ted Shifrin

Anyhow, @Simone, I hope you learn a lot. Unless they now have the corrected printing, you will find some mistakes, and there's a list of errata on my webpage (linked in my profile).

Lucas Henrique

@TedShifrin exactly. I’ve been studying group theory, ring theory in general, linear algebra (it’s part of my current classes) and now module theory. After module theory I’ll start field theory; my professor told me that after the introduction to field theory I’ll need a book on graduate algebra, specifically the sections about group theory and field theory.

Simone

@TedShifrin thanks for the kind words

Galois in the Field

5:12 PM

Yes poles sorry

Ted Shifrin

@Lucas: Make sure you check out Michael Artin's book Algebra.

Aladdin

@TedShifrin I got the answer now.Thnaks a lot

Ted Shifrin

Sure thing.

Galois in the Field

@TedShifrin So I've found that $D= (\frac{dz}{(z-a)(z-b)})$ and $\Omega(D)=\Bbb span_{\Bbb C}(\frac{dz}{(z-a)(z-b)})$, for the case that $D=-a-b$, and I would proceed similarly for an arbitrary $D=\sum_{i=1}^n m_i\cdot p_i$ with $\sum_{i=1}^n m_i=-2$

Lucas Henrique

Okay! I’ve been using Serge Lang’s “Undergraduate Algebra” and after I finish I’ll use his (how surprising!) “Algebra”

ÍgjøgnumMeg

5:14 PM

uh oh

Ted Shifrin

I would not recommend Lang's Algebra.

ÍgjøgnumMeg

^

Ted Shifrin

Start with something a little more friendly.

Simone

Herstein's?

Ted Shifrin

Artin's book is higher level than Lang's undergrad book. So is Dummit/Foote. So is Jacobson.

ÍgjøgnumMeg

5:15 PM

Reppin' D&F

Ted Shifrin

I'm not a fan of Herstein.

Simone

That's the one I have XD

Galois in the Field

Where in particular, the general form has
$$f(z)dz = \frac{\prod_{i=1}^n (z-a_1)^{e_i} dz}{\prod_{j=1}^m (z-b_i)^{f_j}},$$
with the $a_i$'s the distinct zeros with their multiplicities, and the $b_i$'s the distinct poles with their multiplicities

Lucas Henrique

@TedShifrin the monoid section uses a compact topology (?) as an example so... yeah not friendly.

Simone

I've found that Herstein is too liberal on some notation

Ted Shifrin

5:17 PM

I don't like several things about Herstein (although I sort of studied from it as an undergrad). One, he writes functions on the right. Two, he emphasizes trickiness too much. Three, and most important, he doesn't integrate linear algebra into the subject and he doesn't integrate the subject into mathematics. But it's a classic.

Simone

Trickiness?

Lucas Henrique

@TedShifrin what would you recommend as a more abstract book about linear algebra? I’ve tried Kostrkin & Manin but that’s just ridiculous...

IIRC in page 27 they prove that every vector space has a basis and in page ~50 they start Clifford algebras.

Tobias Kildetoft

@LucasHenrique I don't think Manin has written anything that can be considered easy to read

Ted Shifrin

That's harder. Jacobson's algebra (and Artin) both have some sophisticated linear algebra in there. My standard recommendation is Friedberg, Insel, Spence.

Galois in the Field

Thanks very much for your help above Ted, you're amazing

Ted Shifrin

5:23 PM

LOL, thanks, @Galois, and you're most welcome.

Simone

haha I had the same reaction when I tried reading Model theory: an introduction by David Marker... I was reading words but I didn't know what they meant XD

Ted Shifrin

@Simone: A lot of his exercises rely on sneaky tricks. To some extent one can say that about all of mathematics, but I think he's worse.

Simone

And that was after numerous textbooks on logic Godels theorems, introductions to model theoory etc

Ted Shifrin

@Simone: Make sure you check out some of the videos — they might help with different examples, intuition, etc., while you're looking at my book.

Simone

@TedShifrin got it thanks Theodore.

Sayan Chattopadhyay

5:24 PM

@AlessandroCodenotti Can I construct such a set without using or having any idea of any measure theory stuff?

Ted Shifrin

LOL @Theodore.

Simone

... isn't it your name?

Ted Shifrin

Yes, but only the official name. I really do go by Ted.

Simone

"not even my mum calls me Theodore anymore" .. right?

XD

Ted, got it

Ted Shifrin

Well, she did long ago call me by that when she was angry.

Lucas Henrique

5:29 PM

@Simone “hmm... those are certainly all words...”

Akiva Weinberger

@MartinSleziak Ah yeah I got it from that blog post

Alessandro Codenotti

@SayanChattopadhyay Yes

Simone

@LucasHenrique exactly XD: "ok but how do I actually build this time machine?!!!"

Sayan Chattopadhyay

Could you give me some hint on it?

Alessandro Codenotti

I don't really know how to give an hint without explaining the whole construction but let's see

Oh ok, that should be good, do you know the usual construction of the Cantor set?

GFauxPas

5:39 PM

How do I know how many faces $|x|+|y|+|z|=1$ has?

Akiva Weinberger

Where is it nondifferentiable

GFauxPas

Where any variable vanishes

Where a variable changes sign

Archer

6:02 PM

The dimensions of flow rate = mmm s^-1 don't make sense to me...How to exactly interpret it(flow rate)?

The book also says that F is the velocity vector field.

I meant m^3 s^-1 .

GFauxPas

What dimensions would you expect Archer?

Archer

@GFauxPas same as that of velocity m/s

It makes more sense to say fluid is flowing at the rate x m/s along the curve than saying x m^3/ s

Emolga

6:22 PM

Hi I need a quick reminder: Why is saying that a,b,c form an equilateral triangle the same as saying that the matrix with rows 1,1,1 a,b,c b,c,a singular?

That is, $\begin{vmatrix}
1 & 1 &1 \\
a &b &c \\
b &c &a
\end{vmatrix}=0$

Sayan Chattopadhyay

Yeah @AlessandroCodenotti. So I guess you would try to somehow weed rationals out of it?

Alessandro Codenotti

Yes. Try starting with $[e,\pi]$ and do a similar construction, but in a way that gets rid of at least one rational at each step

Emolga

I can cook up a brute-force proof, but I want to see this more geometrically

GFauxPas

6:37 PM

@Archer flow is a 3d phenomenon though. Imagine water travelling through pipes for example

I'm not sure what you mean by flowing along a curve

But fluid has 3 spatial dimensions

Shine On You Crazy Diamond

@Ultradark where's Ultra dank

P = NP

:D

Rishi

Can some one help me

Shine On You Crazy Diamond

@Rishi what's problem?

You can use mathjax here

MatheinBoulomenos

@ShineOnYouCrazyDiamond so $N=1$ or $P=0$?

Shine On You Crazy Diamond

? Complexity theory

computational

I have a proof in my head, just need to write the paper

MatheinBoulomenos

6:42 PM

Hey @Ted

Shine On You Crazy Diamond

Have been working on the smallest grammar poblem for years

Ted Shifrin

hi @Mathein

Rishi

A n side regular polygon area is given by $${na^2 \over 4} cot {π \over n}$$ . Can I use this result to find area of circle

GFauxPas

Hello Mr Shifrin

MatheinBoulomenos

I might end up TAing a grad course as an undergrad

it's not sure yet, though

ÍgjøgnumMeg

6:43 PM

One of my courses -.-

heheh

MatheinBoulomenos

:P

Shine On You Crazy Diamond

Can someone sponser my paper?

*sponsor

Ted Shifrin

@Emolga: I don't even understand what this means. $a,b,c$ are vectors in the plane, so what does it mean to write that determinant?

Emolga

@TedShifrin Complex numbers.

Rishi

Someone respond

Tobias Kildetoft

6:44 PM

@MatheinBoulomenos What course?

MatheinBoulomenos

modular forms

Tobias Kildetoft

cool

Archer

@GFauxPas but in the given example there's only 1

Hi @TedShifrin

Ted Shifrin

Aha. Should explain that, @Emolga.

So I can't do it geometrically without first doing some column operations.

Hi @Archer

Emolga

@TedShifrin fine, as long as we don't actually open the determinant

Archer

6:46 PM

44 mins ago, by Archer

43 mins ago, by Archer

The dimensions of flow rate = mmm s^-1 don't make sense to me...How to exactly interpret it(flow rate)?

@TedShifrin Could you explain this please? Why are the dimensions of flow m^3/s in 2D?

Akiva Weinberger

A 3x3 grid (9 points, 3 vertical lines and 3 horizontal lines) is essentially a reverse $K_{3,3}$ graph

The lines are the vertices and the points are the edges

Ted Shifrin

What are the units on $\mathbf F$ to start with, @Archer? I would take it to be m/sec (thinking of $\mathbf F$ as a velocity field). Then the line integral would have units of m$^2$/sec.

Rishi

Can someone please help me with sum of binomial coefficients . Where lower element is same in each term.

Archer

@TedShifrin yes the book says that F is the velocity field

@TedShifrin i has unit of meter ...So m^3/ s

Ted Shifrin

No, $i$ doesn't have units.

The coefficients have units.

Archer

6:54 PM

$\hat i$'s coefficient is m

Ted Shifrin

If you want to assign units to $i$, then those units become part of the units of $F$.

I personally do not assign units to $i$.

If you look at the vector $x\vec i$, it has units of length because of $x$.

Rishi

Some hint for $$\sum_{k=n}^{2n}{k \choose n}$$

Ted Shifrin

@Rishi: Look at the symmetry of Pascal's triangle.

Archer

@TedShifrin Yeah so let's take it like $\mathrm{F m/s. 1m \hat i. \Delta x m}$

Ted Shifrin

Huh?

Rishi

6:59 PM

@TedShifrin Can you explain a little more please

Archer

@TedShifrin take x as 1 m here, I meant.

Ted Shifrin

Do you know Pascal's triangle and what its entries are in terms of $\binom kn$, @Rishi?

So, @Archer? I don't understand what your problem is. $\vec F$ has units of m/sec, and dotting with the change in position vector gives you another unit of m.

Semiclassical

Oh hey, hockey stick identity

Archer

@TedShifrin My question is how do we physically interpret the units m^2/s as flow rate along a curve?

Ted Shifrin

I told you that's wrong.

Archer

7:03 PM

Still m^2/s doesnt make sense either.

@TedShifrin We also have a $\Delta x$ which has units of m

Rishi

@TedShifrin I haven't study Pascal triangles yet . Is there other way like break it into factorials etc

Ted Shifrin

No, @Rishi. You need to know that the entire sum from $k=0$ to $k=2n$ is the appropriate power of $2$ and use symmetry of the coefficients.

That's why m/sec * m = m^2/sec, @Archer. You're not reading what I write.

Archer

@TedShifrin m/sec * m (coming from i's coefficient 1 m) * m (coming from delta x)

Semiclassical

I’ll note that, as a physicist, I’m more familiar with the magnetostatic version of this story

Where instead of “flow rate” you have the local magnetic field

Ted Shifrin

@Archer: I already told you you're overcounting. If you count the units on $i$, then you have fewer units on the coefficient of $F$.

@Semiclassic; I hate what his book is doing. I want flow across a curve or across a surface, and circulation along the curve.

Regardless, his units are a mess, and I'm losing patience. It's lunchtime.

Semiclassical

7:07 PM

@TedShifrin agreed

Archer

@TedShifrin OK I get it...Then how to interpret m^2/s as flow rate?

Ted Shifrin

That's why you divide by the area inside and get /sec (which you can think of a rate of turning induced by the vector field — think of the wire spinning in the water and remember that radians are unitless).

Akiva Weinberger

Are you talking about the 2D equivalent of volumetric flow rate?

Ted Shifrin

Rate of fluid flow across a curve/surface, although his book is using it for circulation (which I do NOT like).

Archer

@TedShifrin But even before that book calls m^2/ s as flow rate.

Akiva Weinberger

7:12 PM

Oh OK so like if I put a stick into a flowing river

Ted Shifrin

I do not like your book.

Akiva Weinberger

and like the stick has a lot of paint on it

so that the paint leeches off into the river so that it dyes a 2D section of the river

then the amount that the "leech area" grows per second (so m^2/s) is the flow rate

Archer

@AkivaWeinberger yeah it makes sense but the book calls circulating around a curve of that paint as flow rate with units of m^2/ s

Ted Shifrin

It really works better conceptually to make $\vec F$ be density * velocity and then you can think of rate of mass flow across or along the curve.

This is how I do it (with surfaces) in my book and my video.

(Although as I recall I made left something out in the video, but caught it and corrected it later.)

OK, lunch time for me.

Abwatts

Hey guys, I am having quite a hard time understanding why it is possible to write $6(6^k)-1$ as $ (1+5)((6^k)-1)$..

sorry I meant going from here: $6(6^k)-1$ to $(1+5)(6^k)-1$ and then to $5(6^k)+((6^k)-1)$

Never mind. I figured it out.

Shine On You Crazy Diamond

7:41 PM

Can someone help me publish a paper?

Galois in the Field

8:43 PM

If $D$ is an integral divisor, how do I find a basis of $\Omega(-D)$? Riemann-Roch only gives me a bound on the degree of $D$, and I think I can take the spanning set perhaps of meromorphic differential 1-forms with poles of orders up to their order in -D

Ryan Unger

@ShineOnYouCrazyDiamond your advisor

Abwatts

Is $f^{-1}([0, 1])$ (the pre-image) of $f(x) = x^2$ is -1, 0, 1?

Galois in the Field

9:01 PM

@Abwatts No, it is not a finite set, it is the interval $[-1,1]$

Ted Shifrin

@Galois: I think you're confused. If you know $D$, you know the degree. In general, you only know how to balance the dimensions of $L(-D)$ and $L(D-K)$, but sometimes you know more (as in the example we did earlier). But this is an important topic in curve/Riemann surface theory.

@Abwatts: $[0,1]$ is the whole closed interval, is it not?

Galois in the Field

@TedShifrin Right, I don't know why I said that. Although I don't see how to determine the basis

That's the exercise I'm looking at for context (from the textbook 'A course on complex analysis and Riemann surfaces' - Schlag)

Ted Shifrin

What does integral divisor mean?

I do not know the book.

Galois in the Field

It means that D>= 0, so it consists only of zeros

Ted Shifrin

His book is crazy. The rest of that world calls it an effective divisor.

Galois in the Field

9:06 PM

Sure :)

Ted Shifrin

This is not an easy question. Even the dimension will depend on $l(-D)$, which we don't know.

Unless I'm still screwing up his backwards notation.

Galois in the Field

I think the answer is almost given in the text, but I can't yet decipher it from this passage:

My problem is 2-fold potentially. I can't see why 2\leq s, rather than 1\leq s is all that is needed in general, and I can't see how to restrict this spanning set to a basis, since I don't know why I would have control over the zeros for example

Ted Shifrin

Oh, I see. It does come immediately out of RR because of his backwards notation.

Galois in the Field

Wait it does :O?

Ted Shifrin

So $\Omega(-D)$ is going to be mero forms with poles on $D$. RR matches that up with $l(D) = 0$ (unless $D=0$).

Galois in the Field

9:11 PM

Why would we know $l(-D)$? Wouldn't that require knowing $dim(\Omega(-D))$

Xirema

Hey, would anyone here be able to assist me with this question I posted a few days ago? Because—and I mean absolutely no disrespect to the user who posted an answer, whom I assume was trying to assist in a way that probably would have been extremely helpful to a user more advanced than me—but the response is pretty useless for me, a basic user, unfortunately.

Ted Shifrin

I swear his notation will kill me.

@Xirema: It's way too complicated for me to even look at it.

Xirema

And I've been hacking away at that problem for the better part of several weeks without much luck. Most of my efforts have just ended up springing OOM errors from creating Rational Numbers that ballooned too big for my computer to store them.

Ted Shifrin

Generating functions are a basic tool.

So I would suggest you learn about them.

Galois in the Field

@TedShifrin $\Omega(-D) = \{\omega \in M\Omega\mid (\omega)\geq -D,\omega =0\}$ in which case the meromorphic forms need not actually have all the poles of $-D$ right?

Ted Shifrin

9:14 PM

Right, those are the worst poles allowed.

Xirema

@TedShifrin If I learn how Generating Functions work, is that going to allow me to get the kind of information I want, i.e. the Mode, Median, individual outcome probabilities?

Ted Shifrin

@Xirema: I don't know. I didn't read the question (it would take me hours). I'm merely noticing that the answer was given in terms of generating functions. I am no expert on probability stuff.

The point, @Galois, is that RR balances $l(-D)$ and $l(D-K) = \dim\Omega(-D)$. Since $D$ has all positive coefficients ...

Or maybe it's $\Omega(D)$. I give up.

Galois in the Field

Haha the indexing is painful, here for inputting $-D$ I get $l(D)=deg(-D) - g + 1 +l(-D-K)$ and I'll compare what you say now

So if $D>0$ strictly, then $l(D)=0=deg(-D) - g +1 + l(-D-K)$ so $$dim(\Omega(-D))=l(-D-K)=g-deg(-D)-1$$ and if $D=0$ then $$g = l(-D-K)=dim(\Omega(-D))$$

Ted Shifrin

Yeah, that looks right. Drives me nuts.

Galois in the Field

brb 2 min, need to get water

Ted Shifrin

9:23 PM

I'm disappearing, anyhow.

Galois in the Field

Thanks for your help :). I'll try to work out this dang basis

Xirema

Alright, well if anyone can assist me, either ping me here or here, which is where I usually do most of this number crunching busywork. You can dig through the transcript of that room back several weeks (it's a low activity room) if you want to see what we've done so far to try to hack away at it.

Jacksoja

10:31 PM

@ÍgjøgnumMeg Hi

ÍgjøgnumMeg

Hullo @Jacksoja

Jacksoja

I have a small question about the gaussian integers

ÍgjøgnumMeg

Go ahead, I may or may not be able to help

Jacksoja

okay , given a PID for example I = (2+i)

If we take the quotient by that

what is the strategy to understand the quotient?

ÍgjøgnumMeg

(you mean principal ideal)

Jacksoja

10:33 PM

yes sorry

should this be something obvious or should one do some work

with the arithmatic on Z[i]

Ted Shifrin

You should at least start by drawing a picture of what the quotient ring would look like.

How many elements will it have?

Ultimately, a proof will require something like the fundamental homomorphism theorem.

Jacksoja

Hi Ted, a picture?

MatheinBoulomenos

Z[i]=Z[x]/(x^2+1) can help

ÍgjøgnumMeg

^

sniped

Ted Shifrin

Well, I had in mind drawing the grid, drawing the ideal in the grid, and figure out what all the equivalence classes are.

You can do it purely algebraically, as @Mathein is suggesting.

ÍgjøgnumMeg

10:36 PM

Also $N(2+i) = ?$

Jacksoja

Hi mathein

ÍgjøgnumMeg

where $N$ is the norm

MatheinBoulomenos

hi @Jacksoja @ÍgjøgnumMeg @Ted

ÍgjøgnumMeg

Hi @Mathein

Ted Shifrin

hi Mathein

Jacksoja

10:36 PM

my question is, is there theory already written about this or should one work each case out?

Z[x] / (x^2+1) is very clever

Ted Shifrin

My advice is to work examples out until you discover the secrets.

Jacksoja

we use the isomorphism theorem

Ted Shifrin

You shouldn't always ask for the answer in a slick way.

Jacksoja

okay thank you Ted

no I was asking just not to waste effort on something that might be very hard to do without more theor

theory

Ted Shifrin

You should know some theory, though, that might shed some light. Do you know about prime and maximal ideals?

Jacksoja

10:38 PM

Yes

the example being worked in my notes

Ted Shifrin

So that's worth thinking about a little bit, but I still say you should develop intuition with the picture as I suggested. Just like you can draw a picture of $\Bbb Z/(5)$, for example.

ÍgjøgnumMeg

what ArE piCtHure ?:ß?

Jacksoja

does not really explain how they arrive at the conclusion.

Ted Shifrin

But the 2D picture is more interesting.

Jacksoja

am not sure what you mean Ted, the points are lattice points in R^2

of Z[i]

Ted Shifrin

10:40 PM

OK.

And now draw the ideal in there.

And then what are the equivalence classes in the quotient given by?

Hint: Squares.

Jacksoja

I will check this with couple of points but it will be very surprising for me if this works haha, a good surprise I would say

@MatheinBoulomenos @TedShifrin @ÍgjøgnumMeg How did they came up with norms for eucledian rings?

MatheinBoulomenos

do you mean the concept of an euclidean ring?

Jacksoja

I understand it all comes from the division with rest

MatheinBoulomenos

I mean, you have integers and polynomials over a field, both are natural examples and they have similar properties, so it seems natural to look for a generalization that captures both examples

I don't know the actual history

Jacksoja

I mean this | s-n| < 1/2

for Z[i]

we take nearest integer or something of that sort

let me check my notes to be correct

MatheinBoulomenos

10:48 PM

@Jacksoja yes. The idea is that if you want to do division with rest, you need to be able to say that the rest is smaller than the element you divided by. Otherwise you could always say that $a=0 \cdot q + a$, so we have divided $a$ by $q$ with rest $a$. This is not very interesting

that's why you need a function to measure "how large" your ring elements are in some sense

Jacksoja

@MatheinBoulomenos I see, but even in the proof in the book, they show us that it works, but not how they found it

it raises many questions like, are norms unique in such rings?

MatheinBoulomenos

no

Leaky Nun

@MatheinBoulomenos if $A \otimes_\Bbb Z k \cong B \otimes_\Bbb Z k$ for all field $k$ then do we have $A \cong B$?

Jacksoja

or does this norm work for other rings etc

Leaky Nun

no

Jacksoja

10:50 PM

Okay am convinced ! haha

I will explore more thanks @MatheinBoulomenos @TedShifrin @LeakyNun

MatheinBoulomenos

@LeakyNun take $0$ and $\Bbb Q/\Bbb Z$

Leaky Nun

@MatheinBoulomenos nice

@MatheinBoulomenos is the category of left k[GxH]-modules equivalent to the product of the category of left k[G]-modules and the category of left k[H]-modules?

MatheinBoulomenos

@LeakyNun no I don't think so

Leaky Nun

ok

Jacksoja

@MatheinBoulomenos in this Z[x] / (x^2+1) = Z[i] , did you come up with that from this Z[i] / (0) = Z[i] ?

MatheinBoulomenos

10:58 PM

take G=H=1. k-Mod is not equivalent to k-Mod x k-Mod

@Jacksoja err, not exactly. R/(0)=R for any ring R. That's not really interesting

Jacksoja

I mean when you mod out by x^2+1

you are saying that x^2 +1 = 0

Transcript for

$Mathematics$ Mathematics