Mathematics

12:09 AM

Is there a standard term for $\gcd(\{P(n) \mid n\in \Bbb Z^+$ when $P$ is a polynomial with integer coefficients?

i do not know of one. sometimes things that can be expressed in short formulas with things that do have standard terms, do not get their own standard terms.

but the fact that i don't know it doesn't mean anything. this has been my useless answer of the day.

Ted Shifrin

I don’t think about these things, but is there an example where it’s different from the gcd of the coefficients?

Oh, I see an easy example.

Sum of an even number of odd coefficient terms.

MJD

12:28 AM

For $x^2+x$ it is 2, not 1.

Ted Shifrin

12:39 AM

Yes, read what I said next.

MJD

I did.

copper.hat

1:04 AM

barely survivable heat in albany today

leslie townes

not too bad here, although i'd rather be in albany.

copper.hat

apparently gavin declared a heat emergency

there's a cool room in the french laundry i hear

leslie townes

my daughter's yelling about not being allowed to bring handfuls of dirt into the house

Ted Shifrin

As long as it’s by the truckful!

leslie townes

1:27 AM

"sorry olivia, you can't go outside. you're not old enough." explanation given to the cat for why she couldn't go with us to check the mail.

probably hears that explanation a lot at day care.

Permutator

2:13 AM

@robjohn Are you free right now?

robjohn

It's just about dinner time. Is it something quick?

OlympicComputerChairSitter

Hi chat

I mean math gurus

^_^

leslie townes

it isn't something quick. everybody be quiet and maybe the whole thing will just blow over.

EM4

yo yo yo

OlympicComputerChairSitter

2:33 AM

Hi @leslietownes @EM4

What maths are you studying lately?

EM4

I took day off rest so I read environmental anthropology haha, you?

leslie townes

i study what i see on the site and what pops into this chatterbox. i am not an active user of my own math.

OlympicComputerChairSitter

@EM4 I'm trying to work a problem in Hartshorne's AG. It's a very advanced book, but hopefully I can swim and not sink. I don't care if this problem takes me a week to get. I'm going to work on it without looking at this solution manual on github

The solution manual was written by someone 10x smarter than me at least. There is no learning if I just solution manual every problem.

leslie townes

good luck. my school had a class where the syllabus for a 15 week course was something like 'we will do chapter 1 of hartshorne and a few topics from chapter 2, time permitting.' i exaggerate but only a little.

OlympicComputerChairSitter

@leslietownes yeah, I am skipping some problems that just aren't aesthetically pleasing to me. I hope doing that doesn't prevent me from learning the subject

leslie townes

2:39 AM

my impression is that there's enough in there to keep somebody off the streets and out of trouble for a good long while.

OlympicComputerChairSitter

It's as if, if you could do all the problems in this book in 2 years. Then you already knew the material to begin with...

@leslietownes I'm starting Open University (OU) distance learning for a Bachelor's in Mathematics this upcoming September

The first course I chose is statistics, which I'm completely new to except I know a little about Lebesgue measure

After I get a bachelor's I want to persue a PhD or Master's. I think in 4 years, I will be mathematically mature enough to start up doing that

After the PhD, I want to just keep going to school for life, lol

I will get a job where I can work from home, and do distance learning when I'm not working.

I'm good in school as long as I'm not studying grassmannians

EM4

I agree with you @OlympicComputerChairSitter I am the same way.

STEM is a drug.

OlympicComputerChairSitter

what's that

Science Tech, Engineering, Math?

EM4

STEM is Science Technology Engineering Mathematics.

OlympicComputerChairSitter

Nice

Yes, it's drug-like, and so we should take ample breaks to recover and do other things

such as exercise

leslie townes

2:50 AM

or cyberbullying people on twitter

shintuku

hey congrats on obamacare yall

OlympicComputerChairSitter

What's that, what happened?

shintuku

supreme court ruled in favour of upholding

EM4

hahhahha @leslietownes

OlympicComputerChairSitter

@shintuku that's good

I heard that in Oregon, they've decriminalized posession of small amounts of any drug. Which is how it should be

leslie townes

2:52 AM

i now understand what you meant by ample breaks to recover and do other things.

OlympicComputerChairSitter

I don't do a lot of drugs, mostly coffee and nicotine gum

*gum

Andrew Micallef

Cyberbulling is a great way to recover from math infatuation

shintuku

i think coffee becomes drugs if you take enough

OlympicComputerChairSitter

It is a drug

shintuku

yes but not drugs

OlympicComputerChairSitter

2:54 AM

80% of the world is addicted to caffeine

leslie townes

i'm not addicted, i just like the way my tea tastes. eight or nine cups a day.

EM4

I am water guy LOL.

OlympicComputerChairSitter

Water can be addictive, hypernatremia

leslie townes

i'll die if i don't get my water.

someone had to say it

shintuku

i do a 2 day detox of coffee once a month

OlympicComputerChairSitter

2:57 AM

Also, the aire we breath is addictive

shintuku

it is painful but it seems like drinking coffee adds up after a while

OlympicComputerChairSitter

Therefore, you should hold your breath for 1 minute every hour to detox

Also, sleep can be addictive. You should stay awake 1 / week to detox

leslie townes

i think it's pretty dehydrating if it's all you drink but otherwise mostly OK. i vary in a lot of noncaffeinated herbal 'teas' so i'm not buzzing all day

OlympicComputerChairSitter

@leslietownes that's good. Sometimes we just want a cup of hot water with herbs

shintuku

why is tea in single quotes haha

sounds like it is something much worse

Andrew Micallef

2:59 AM

@OlympicComputerChairSitter ... i dont think you are using that word properly

leslie townes

right now i'm drinking burdock root we'll see what happens.

OlympicComputerChairSitter

I know, hyponatremia, or something

I heard that in order to fully understand Grassmann's work you need to be high on a certain drug...

And that drug is caffeine

EM4

what work of Grassmann?

OlympicComputerChairSitter

Hermann Grassmann

Hermann Günther Grassmann (German: Graßmann, pronounced [ˈhɛʁman ˈɡʏntɐ ˈɡʁasman]; 15 April 1809 – 26 September 1877) was a German polymath, known in his day as a linguist and now also as a mathematician. He was also a physicist, general scholar, and publisher. His mathematical work was little noted until he was in his sixties. == Biography == Grassmann was the third of 12 children of Justus Günter Grassmann, an ordained minister who taught mathematics and physics at the Stettin Gymnasium, where Hermann was educated. Grassmann was an undistinguished student until he obtained a high mark on the...

EM4

what did he do?

OlympicComputerChairSitter

3:05 AM

Grassmannian

In mathematics, the Grassmannian Gr(k, V) is a space that parameterizes all k-dimensional linear subspaces of the n-dimensional vector space V. For example, the Grassmannian Gr(1, V) is the space of lines through the origin in V, so it is the same as the projective space of one dimension lower than V. When V is a real or complex vector space, Grassmannians are compact smooth manifolds. In general they have the structure of a smooth algebraic variety, of dimension k ( n − k ) . {\displaystyle k(n-k).} The...

It's a coincidance that the grassmannian is like a bundle of long blades of grass

leslie townes

i know a grass man who lives in oregon. i wonder if they're related.

OlympicComputerChairSitter

Yes, they are related, since they are both of the same species

Manjul Bhargava

Manjul Bhargava (born 8 August 1974) is a Canadian-American mathematician. He is the Brandon Fradd, Class of 1983, Professor of Mathematics at Princeton University, the Stieltjes Professor of Number Theory at Leiden University, and also holds Adjunct Professorships at the Tata Institute of Fundamental Research, the Indian Institute of Technology Bombay, and the University of Hyderabad. He is known primarily for his contributions to number theory. Bhargava was awarded the Fields Medal in 2014. According to the International Mathematical Union citation, he was awarded the prize "for developi...

He holds a professorship at many different colleges. Einstein of our time

maa.org/sites/default/files/pdf/upload_library/22/Hasse/…

That's a paper of his that seems accessible to an undergrad

leslie townes

interesting. i used to love reading AMM for that reason, you'd see really good exposition in there. and sometimes from the mathematical equivalent of celebrities.

Koro

3:30 AM

Hi Leslie

What is the difference between Herstein's Topics in Algebra and Herstein's Algebra

?

leslie townes

3:44 AM

i think one is more advanced or comprehensive and the other isn't, but i never got too familiar with either one.

my officemate did own something that said herstein algebra.

Koro

I already had Herstein's algebra book and I have been reading it. Today in one of the proofs it said : for more proofs refer our book "Topics in Algebra"

leslie townes

or maybe it's a pure cash grab, an attempt to confuse consumers into buying the same thing twice. like after 2pac died and his label kept putting out 'new' songs with old verses on them.

Koro

It seems that in Herstein's algebra many theorems have been shifted to exercises

For example: the famous relation: $|HK|= \frac{|H||K|}{|H\cap K|}$ etc.

barista

I already know that (xdy-ydx)/(x^2+y^2) is closed but not exact on $\Bbb R^2-0$ But what if closed 1-form $w$ on $\Bbb R^2-0$ is continuous on some disk centered at the origin?

Ted Shifrin

No,Topics in Algebra was the original ballbuster book. The next was watered down for merely mortal students.

barista

3:51 AM

I want to show in that case, it's exact

Ted Shifrin

@barista if it’s smooth and closed on a simply connected region, then it's exact.

Koro

Ted, I didn't know that before, I thought both were same. I've almost finished the group part in Herstein's algebra and I use Gallian's book also along with Herstein's.

Ted Shifrin

That said, Herstein is one of my least favorite texts ever.

Koro

I still feel like I struggle with isomorphisms...

barista

@TedShifrin Yes but mine is smooth only on $\Bbb R^2-0$ not $\Bbb R^2$.

Ted Shifrin

3:52 AM

Gallian is rings first, isn’t it?

Koro

@TedShifrin You have expressed that few days earlier also while suggesting Artin. I remember :-)

@TedShifrin Rings first? Rings are introduced after Groups in Gallian

Ted Shifrin

Then we don’t have Green’s Theorem necessarily.

Oh, ok, I misremembered.

barista

To do that I first tried to prove the integral of $w$ over the boundary of that disk is zero

Maybe I can use stokes' theorem but well I wonder if it's a smooth mfd

Ted Shifrin

Oh, apply Green on an annulus. It works.

Green = Stokes here

barista

@TedShifrin You mean like a Key hole thing?

Ted Shifrin

3:56 AM

Show that the inner line integral goes to $0$ As you shrink the radius.

No, just plain annulus .

barista

4:09 AM

Ok I let $\gamma(t)=(rcost, rsint)$ where $r$ is radius of inner circle $C_r$. Then $\int_{C_r}w = \int_0^{2\pi}\gamma^*w$. Writing $w = w_1dx+w_2dy$, $=\int_0^{2\pi}-w_1(\gamma(t))sint dt+ cost w_2(\gamma(t))dt$

Koro

I find it very difficult to prove two groups isomorphic (To be more specific: in writing the explicit isomorphism). I feel like I am missing something in isomorphisms despite having understood the concept. I am not confident when I see "isomorphism". For example: $\frac {Z_4 \times Z_{12}}{<(2,2)>}$ is isomorphic to which of these: 1) $Z_8$ 2) $Z_2\times Z_2 \times Z_2$ 3) $Z_4 \times Z_2$.

barista

@TedShifrin I think continuity of $\gamma$ should work here

Koro

I have rejected 1) and 2). For, there is no 8 order element in the aforementioned quotient group and there is one order 4 element in the quotient group so 2) is also not correct. 3) seems to be but I don't know how to write the explicit isomorphism :)

Please guide. Thanks.

barista

Oh I missed $r$

leslie townes

for small groups the orders of elements can decide the issue, particularly if they are abelian groups. i wouldn't discourage you from thinking in terms of maps but a lot of finite abelian group theory can be approached one level back from that. at the cost of some investigation and maybe some machinery

barista

4:16 AM

Ok I missed $r$. I can write $\int_{C_r}w=r\int_0^{2\pi}(cost w_2(\gamma(t))-sint w_1(\gamma(t)))dt$. Since $w$ is continuous on $\Bbb R^2$, the integrand is integrable, so as $r\to 0$ the whole integral converges to $0$. What do you think @TedShifrin ?

leslie townes

disclaimer, i am not an algebraist. just a guy who took a few algebra classes once.

yoour book may have interesting isomorphism theorems involving direct sums/products that bear on quotienting.

Russian Bot Who Knows Your IP

Open source

Koro

@leslietownes There are 1st homomorphism theorem etc. but not yet introduced..

leslie townes

one thought without any theorems is to figure out or recall that Z_4 x Z_12 is Z_4 x Z_4 x Z_3, see what (2,2) looks like under that isomorphism, and then see if that helps the quotienting operation look simpler to you.

these thoughts could be worthless.

Ted Shifrin

@barista And where is Stokes?

barista

4:32 AM

@TedShifrin If I let $D_r$ the be the annulus obtained by removing radius $r$ disk from the given disk of radius $R$, then by stokes, $\int_{D_r} dw = \int_{\partial D_r} w = \int_{C} w+\int_{C_r} w$ where $C$ is circle of radius $R$. $dw =0$ by assumption, letting $r\to 0$, $0=\int_{C}w$

Ted Shifrin

OK, except watch orientation.

Koro

@leslietownes I wrote all the elements of the quotient group (tedious task) and then using Cayley table got some intuition... I hope this will help write the isomorphism

barista

@TedShifrin You mean orientation of inner circle runs clockwise right?

Ted Shifrin

Right.

leslie townes

it's generally pretty annoying to define maps from finite groups to one another without a tiny toolkit of theorems. you have to think very much in terms of elements or basically end up proving aspects of those theorems.

barista

4:38 AM

@TedShifrin From this result, I'd like to use the following fact to conclude $w$ is exact : If $w$ is a smooth covector field in an open set $U\subset Bbb R^n$ then $w$ is exact if and only if the line integral of $w$ is path-independent.

My $U =\Bbb R^2-0$ is obviouly open

Koro

It seems so. @Leslie

Ted Shifrin

So you have that theorem or do you need to prove it?

barista

I have that theorem

Ted Shifrin

OK.

barista

I first thought if the region enclosed by to paths contains the origin then we can find some small ball centered at 0 that does not intersects two paths. Then some argument but I wonder if that works

I mean I only know the integral over the boundary of the given disk is zero

Ted Shifrin

4:45 AM

It is an annoying proof to get rigorous. You have to worry about how many times a closed path goes around the origin .

barista

@TedShifrin Well then would be easier to show $w$ is conservative?

Ted Shifrin

That doesn’t make sense, but it is no different.

barista

Right..

uhoh

6:17 AM

3

Q: How does the fractional Fourier transform apply to an out-of-focus imaging system? Do we use the fractional distance to the focal plane?

In Fourier optics it is sometimes convenient to think of lenses as "Fourier transformers". For an imaging system between two planes with a pupil in the center, the amplitude in the pupil is the FT of the image. A simple way to think about this is that a lens swaps angles and positions (for a plan...

asked in Physics SE, perhaps I should have posted in Mathematics SE instead

Larry

7:12 AM

$f:X\rightarrow Y$ and is not surjective so some points in Y are not "hit". Then what is the preimage of these points that aren't hit? Would it be the empty set? Or is it invalid question?

robjohn

8:22 AM

@Larry it would be the empty set.

$E=\{x\in X:f(x)\in\{y\in Y:(\lnot\exists x\in X:f(x)=y)\}\}$

$E=\{x\in X:f(x)\in\{y\in Y:(\forall x\in X,f(x)\ne y)\}\}$

suppose that $e\in E\subset X$, then $f(e)\in\{y\in Y:(\forall x\in X,f(x)\ne y)\}$

that is, $f(e)\in Y$ and $\forall x\in X,f(x)\ne f(e)$

but $e\in X$, so we have a contradiction, and thus, $E=\{\}$

Monty

10:01 AM

Fix $x\in H$ where $H$ is a Hilbert space. Let $\phi$ be a convex functional on $H$, why is it obvious that the above should hold?

barista

11:19 AM

@TedShifrin My mistake. I suspect you already know that the proof I stated before shows $\int_{C_r}w =0$ where $C_r$ is a circle of radius $r$ centered at $0$. But the fact that $w$ is exact is still in mystery.