Mathematics - 2019-04-02 (page 1 of 3)

12:06 AM

Um, no ... I've never done more than 4 or 5 ...

Although I had macros for typesetting augmented and non-augmented matrices of various sizes, but the matrix entries went in the usual {.&.&.\\.&.&.} format.

Math geek

0

Q: Show that $\{Z(p):p\in \mathscr P\}$ is a basis for the closed sets of some topology (Called the Zariski topology) on $\mathbb R^n$.

Let $n\in \mathbb N$, and let $\mathscr P$ denote the collection of all polynomials in $n$ variables. For $p\in \mathscr P$, let $Z(p)=\{(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0\}$. Show that $\{Z(p):p\in \mathscr P\}$ is a basis for the closed sets of some topology (Called the Zariski topology) on...

Rithaniel

Yeah, I'm trying to get 64 6x6 tables into a latex document in the form of tables. I've decided to type up a program to convert things from one text document to another to make it easier.

Math geek

Alternatively, Can I prove by just considering the $\{Z(p)^c:p\in \mathscr P\}$ forms a basis for open set?

Daminark

Somehow I keep clicking on the nerd chat on accident

How's it going everybody?

Math geek

Will the proof valid?

Rithaniel

12:13 AM

It is going well. Got home early today.

Érico Melo Silva

@Daminark did u manage to get into that formal languages class

Ted Shifrin

Hi Demonark.

Daminark

When I saw it was full I was just like eh I'll do probability instead

Ted Shifrin

Maybe a subbasis, @Mathgeek.

hi @Eric

Érico Melo Silva

lol word

hi Ted

Daminark

12:15 AM

Also Charlie's complex analysis class this quarter is gonna be practically disjoint from Marianna's so I think I'm gonna try to audit

Érico Melo Silva

what is he doing

Daminark

First 2 weeks is an undergrad complex analysis speedrun, then he's gonna prove PNT third week (I think not the proof we did), then do Riemann mapping theorem, then harmonic measures, Riemann surfaces/uniformization, and something called "X evolution" where I forgot what X was

Érico Melo Silva

oh maybe i’ll hop in for uniformization

Daminark

Yeah should be a fun time

Ted Shifrin

That isn't a short hop.

Érico Melo Silva

12:20 AM

the length of the hop is a function of the lecture speed i guess

Ted Shifrin

Well, of course ... not to mention clarity, preparation, etc.

Érico Melo Silva

hes a p solid teacher i think, i had him for ugrad complex and grad functional and it was aight

Daminark

The guy tends to be quick and pretty good aside from the very noble attempt to teach us Sobolev spaces

Math geek

@TedShifrin I verified all the properties of basis. I have a doubt that will the proof shows that $\{Z(p):p\in \mathscr P\}$ a basis for closed subsets?

Ted Shifrin

Which Charlie is this?

Daminark

12:22 AM

Charles Smart

Ted Shifrin

Oh, I don't know that Charlie.

I'm not smart enough.

I am not reading any of the post, @Mathgeek.

Daminark

He's into very different things than you. It's hard to put a finger on him exactly but kinda probability, kinda PDE, kiiinda dynamics?

Ted Shifrin

You certainly got enough responses.

Érico Melo Silva

smart is smart

Ted Shifrin

Fingers can be difficult for multi-focused peoples.

I'm sure @Eric

Érico Melo Silva

12:24 AM

lol

Daminark

On his website his recent projections include stochastic homogenization, the abelian sandpile, mixing flows, and Anderson localization

Whatever those things mean lmao

Érico Melo Silva

aNaLySiS

Math geek

@TedShifrin I verified not here, in book. I got $\{Z(p)^c:p\in \mathscr P\}$ basis for some topology on $X$. will that implies $\{Z(p):p\in \mathscr P\}$ form basis for closed subsets of $X$?

Ted Shifrin

I actually have never heard of a basis for a topology based on closed sets.

Daminark

But yeah that aside, had number theory and probability today. Tomorrow graph theory and civ are my official classes, and I'm thinking of auditing diffgeo

Math geek

12:27 AM

2

Q: Is my definition correct? How do I prove the finite union of elements of $\mathscr U$ lie in $\mathscr U$?

Definition of basis for a topology on $X$ given in the Foundation of Topology By Patty given by Let $(X,\mathscr T)$ be a topological space. A basis for $\mathscr T$ is a subcollection $\mathscr B$ of $\mathscr T$ with the property that if $U\in \mathscr T$ then $U=\emptyset$ or there is ...

general-topology

Ted Shifrin

What sort of diff geo, Demonark?

OK, anything Brian Scott says about point set topology has to be super-correct.

But I'm not going to think about this.

Daminark

Grad with Calegari. Here's math.uchicago.edu/~dannyc/courses/riem_geo_2019/…

This is the last time he taught it a few years ago, with psets: math.uchicago.edu/~dannyc/courses/riem_geo_2013/…

Math geek

@TedShifrin I got $\{Z(p)^c:p\in \mathscr P\}$ basis for some topology on $X$. will that implies $\{Z(p):p\in \mathscr P\}$ form basis for closed subsets of $X$?

Ted Shifrin

That seems like a second course, Demonark. I wonder if everyone knows the first course.

Daminark

Not everyone did the bootcamp (and not everyone who did it learned/retains much), though the prereq is the first two quarters of the sequence, doing AT/difftop

Ted Shifrin

12:34 AM

Can you just prove that in general, @Mathhgeek?

There's a lot of stuff in a first manifolds course ...

Math geek

You mean that result true?@TedShifrin

Ted Shifrin

I don't know.

I've already said I'm not going to think about it.

Math geek

Sir, I am humbly requesting you. Please help me. I don't have a teacher. I am studying self @TedShifrin

Daminark

Oh I thought you meant first course meaning first diffgeo (curves/surfaces). Grad difftop this year did some of the technical stuff about smooth manifolds like Whitney embedding and all but not much I feel

Ted Shifrin

I'm sorry. Chat is not here for demands.

Demonark, there's a lot more of basics ... Lie derivatives, flows, forms and Stokes, vector bundles, etc.

Math geek

12:41 AM

@TedShifrin okay

Ted Shifrin

OK, I'm leaving for kitchen duty.

Daminark

Yeah not much of that stuff happened really, it was just, okay here's a manifold, smooth map, tangent space, immersion/submersion, Whitney embedding, tubular neighborhood, orientable manifold, and then it was topology. And alright, have fun! :)

Érico Melo Silva

@TedShifrin chicago difftop varies, my year did those things, but sometimes it happens that there are no courses that would teach u the important basic notions

Secret

1:22 AM

Last night dream, a research paper becomes a question for a high school maths exam paper

Basically, in that dream's history, there is a longstanding debate on the average mark students get from colouring questions C and the difference in average mark students get from fill in the numbers vs multiple choices questions N-M. Then in 2013 in that dream history, 3 researchers published a paper which they designed an integrated exam question such that C and N-M can be measured simultaneously. The authors hoped that will finally settled the debate

However, the result is so many students got stumped at that question because the question wording itself is too vague (asking students to draw an object in the first row, then the same object in the second row after taking a rest, and then the same object again but more detailed in the 3rd row)

(while the correct answer is, draw 3 objects inside the box provided, color them in 3 different colours, and then repeat the same pattern down the column, and then use different sets of 3 colours for the 2nd and 3rd column, resulting in something like a 3 color puzzle and for each box, the 3 objects looks like they tiled the boxes properly)

user10478

1:38 AM

Hello, is it possible for a vector field in $R^2$ to have nonzero curl at a point, or along a line, in a finite rectangle, etc., yet zero curl everywhere else?

Secret

along a line is easy, you stack the vector field (y , -x) (which should form a rotating like vector field) along the z axis, then the line (x,y,z)=(0,0,z) is nonzero curl

user10478

Hmm, that wouldn't be a vector field in $R^2$ anymore though.

Secret

o right, I misread

$\frac{\partial x f(x,y)}{\partial x}-\frac{\partial -y f(x,y)}{\partial y} = f(x,y)+ x\frac{\partial f(x,y)}{\partial x}-f(x,y)-y\frac{\partial f(x,y)}{\partial y}=x\frac{\partial f(x,y)}{\partial x}-y\frac{\partial f(x,y)}{\partial y}$

$=\binom{x}{-y} \cdot \nabla f(x,y)$

Want
$\binom{a}{-b} \cdot \nabla f(a,b) > 0$

and

$\binom{x}{-y} \cdot \nabla f(x,y) = 0, (x,y)\neq (a,b)$

solve for such $f$ to get your required vector field

I am suspecting something like $(\frac{y}{x^2+y^2}, \frac{-x}{x^2+y^2})$ but I don't recall the example except I have seen it before, where the basic idea is vectors get longer and longer the closer they are to the origin to compensate for less number of vectors encircled by any infinitesimal region, so the curl cancels out everywhere except at the origin

user10478

2:00 AM

Ahh, you're talking about $\begin{bmatrix}-\frac{y}{x^2+y^2}\\\frac{x}{x^2+y^2}\end{bmatrix}$ I think, the one with zero curl, but nonzero circulation?

Secret

yeah

user10478

It's related to the reason I asked my question, but that one isn't defined at the origin.

So there isn't really a point with nonzero curl.

Secret

I have no idea, somehow my gut feeling told me you cannot have a continuous vector field that behave that manner, but I don't know enough cohomology to work out why

You should ask that question on main, I think there will be interesting answers

user10478

Okie, I asked a similar question that got no interest as of yet (math.stackexchange.com/questions/3159687/…), but this question is slightly different, so I'll try it too.

usukidoll

3:16 AM

is mathematical modeling systems analysis?

Secret

System analysis is larger, as it can involve purely reason based things that does not use mathematic models:

Policy analysis

Policy analysis is a technique used in public administration to enable civil servants, activists, and others to examine and evaluate the available options to implement the goals of laws and elected officials. The process is also used in the administration of large organizations with complex policies. It has been defined as the process of "determining which of various policies will achieve a given set of goals in light of the relations between the policies and the goals." Policy analysis can be divided into two major fields: Analysis of existing policy, which is analytical and descriptive – ...

usukidoll

so what is systems analysis and how does it relate to math?

Secret

You can use mathematical modelling to analyse components of a system and how it interacts with each other, but not all systems need maths to analyse

For example, legal systems and policymaking mostly is about referring to case studies and then make deductions that way

(as well more dodgy things like how to smear your opponents or other campaigns to attract votes)

Thus maths is just one of the languages that can be used to reason about systems

usukidoll

it's for a cover letter
I've used Matlab and R so I'm trying to find a link between them and systems analysis

Secret

Ah in that case, the systems dealt with will be mostly technical in nature, hence maths will be essential to understand them. Try tailor how your math lab and R experience is related to the system your company use as a subject of investigation in question

usukidoll

3:30 AM

I'm still a student and the company is a health insurance firm... data analyst at the entry level

Secret

Matlab and R are very good at analysing huge volumes of data, which is something the health industry routinely need to deal with. I think you might not need to talk about system analysis, just relate how your knowledge of these two software can help the company to extract information from their client health record data or something like that

I think that will be a better way to tailor the letter since it is more obvious what the contributions will be

usukidoll

ahhh. ... I'm getting feedback on the cover letter tomorrow in the afternoon with a career counselor

Secret

anymore detail you might have check the specific position on what you are applying for to see how data analysis can help the company to determine how much the premiums to their clients should be based on their health records. But yeah, check with your career counsellor, they have better experience in the industry than us chat users am

usukidoll

I filled up a page really fast so at least there are some things to say. :3

BAYMAX

is there an online platform where i can caluclate the error of the Taylor series

for $f(x) = ln(1 + 2x)$, a = 4, n = 3, 3.7 ≤ x ≤ 4.3

usukidoll

3:38 AM

wolframalpha?

oh waitttt didn't see the xtraz

Secret

3:56 AM

@BAYMAX I don't think there is an online platform that does the computation directly, but you can probably get Wolfram alpha to compute the error term (the n+1 th order term) for you and use that to estimate the error

Buddhini Angelika

4:08 AM

A union of two edge disjoint 1-factors of a graph always form a 2-regular subgraph, right?

BAYMAX

0

Q: $R_{3}(x)$ in the Taylor series expansion of $f(x) = \ln(1 + 2x)$, centered at $a = 4, n = 3, 3.7 ≤ x ≤ 4.3$

I was trying to compute the error or remainder of the third order $|R_{3}(x)|$ here. For $f(x) =\ln (1 + 2x), a = 4, n = 3, 3.7 ≤ x ≤ 4.3$ $|R_{3}(x)| = \dfrac{M}{(4!)} (x-a)^{3+1}$ I got $M = \dfrac{48}{(1+2x)^4}$ in which I substitute $x = 3.7$ and It gave me $R_{3}$ as $0.00000325385$, is t...

calculus analysis logarithms taylor-expansion

@Secret

Secret

4:53 AM

math.pitt.edu/~sparling/23014/23014convergence/node7.html

Your error term is always larger than the difference between f(x) and T_n(x)

thus compute $T_4(4)|4.3-4|^{4}$ and $T_4(4)|3.7-4|^{4}$ then take the larger value as your upper bound to your error

Sir Cumference

5:09 AM

Love how this is a real thing

Abstract nonsense

In mathematics, abstract nonsense, general abstract nonsense, generalized abstract nonsense, and general nonsense are terms used by mathematicians to describe abstract methods related to category theory and homological algebra. More generally, “abstract nonsense” may refer to a proof that relies on category-theoretic methods, or even to the study of category theory itself. == Background == Roughly speaking, category theory is the study of the general form, that is, categories of mathematical theories, without regard to their content. As a result, mathematical proofs that rely on category-theoretic...

Secret

Nonsense as a word is orders of magnitude more tame than "absurd"

usukidoll

5:45 AM

cake

Secret

grabs an amorphous cake

onlinelibrary.wiley.com/doi/pdf/10.1002/jps.24284

0xbadf00d

Does anybody know a book where the statement "a function $f:\mathbb R\to\mathbb R$ is of bounded variation if and only if it is the difference of bounded nondecreasing functions" is proved? I know how to proof it, but I need a reference which I can use in a thesis.

Secret

7:53 AM

in The h Bar, 1 min ago, by PM 2Ring

@Secret Yes, it's closely related to the halting problem. You can't just test stuff & see what happens. You have to use a clever logical proof, like how we prove that a plain Turing machine cannot be a halting oracle for plain Turing machines.

hmm... sometimes I wonder how does a deductive system with semantic meaning on each symbols, can accomplish something in finite steps that will otherwise take infinite steps to finish

Somehow quantifiers $\forall, \exists$ have this power to tackle infinite objects in just one step

But it does show one thing: Infinity is also bounded by logic

Secret

8:10 AM

in The h Bar, 41 secs ago, by Secret

@PM2Ring Sometimes these proofs always amazes me on how they can bypass the explicit need to compute infinite steps to get the answer. Like the proof of the halting problem relies on the halting program to produce a logical contradiction very similar to the way the liar sentence work. It's as if despite these objects are infinite in practice, they have properties that can be handled in a finite number of steps, hence bypassing the computation

Secret

8:20 AM

in The h Bar, 1 min ago, by PM 2Ring

In Life there's a fairly small pattern that calculates prime numbers. Let the cycle period = T. Then on generation nT it emits a glider iff n is prime. The pattern is essentially a prime sieve that grows over time. A modified version of this pattern emits a glider iff n-2 & n are twin primes. So a proof that this pattern never stops emitting gliders is a proof of the twin primes conjecture.

Leaky Nun

8:51 AM

Let Bn be the n-th occurrence of an event following a Poisson process with parameter β. Given B1<=1<B2, then B1 ~ U(0,1)?

Tobias Kildetoft

@LeakyNun I am not sure that is an entirely meaningful question

Given B1 and B2, it no longer makes sense to talk of the distribution of B1

Leaky Nun

this is conditional distribution

Tobias Kildetoft

How is that defined?

Leaky Nun

you can define the c.d.f. first via P(B1<=t | B1<=1<B2) := P(B1<=t and B1<=1<B2) / P(B1<=1<B2)

Tobias Kildetoft

hmm

this is too far outside my comfort zone

ask me again 12 years ago

Leaky Nun

9:04 AM

ok

user123

9:21 AM

any chance someone can help with that question?

Tobias Kildetoft

@user123 Well, it has two natural actions, so which one is meant?

user123

amm , i guess multipication from left

Tobias Kildetoft

Ok, so let's look at the orbits of some elements of a natural form. What does the orbit of a diagonal matrix look like?

user123

hmm.

Tobias Kildetoft

Even more explicit: What if it is diagonal and has a $1$ in the upper left?

user123

9:31 AM

im trying to think for example about $Orb(diag(p,1)) $ but i dont see how to get a nice representation of it

Tobias Kildetoft

Well, what does multiplication from the right by that matrix do?

user123

we get that the first column is $ap,cp$ and the second is the $b,d$ if we multiply by matrix with a,b,c,d entries

so the second column is the same

so it multiply by p the first column

and does nothing to the second

Tobias Kildetoft

Ok, now, think a bit about why we cannot get all matrices with a given determinant this way

(or maybe: what would we need to change to make this give us all matrices)

user123

there are elements that their first column is not divisible by p

so we can't get all matrices with determinant p..

Tobias Kildetoft

Right, if we had done this over the rationals instead, we would just get a single orbit

so in the orbit of this given matrix, do we get all matrices whose first column is divisible by $p$ and which have determinant $p$?

user123

9:38 AM

yes

Tobias Kildetoft

ok, great. So this might give us an idea of what sort of condition could determine the orbits

Nick

Can someone help me interpret what accuracy means in this paper http://www0.cs.ucl.ac.uk/staff/ucacpjb/UJBEC4.pdf

They somehow apply stochastic optimisation to the subjective task of recommendation (ie, no globally correct solution) and get a final result as "active users" vs "prediction accuracy" bar chart, how can I believe these results when I can't see the implementation?

user123

So the orbits are - first column divisible by p , second column divisible by p and none divisible by p ? @TobiasKildetoft

im not sure about the third one :P

Tobias Kildetoft

@user123 the last one does not sound right to me. Why would that be a single orbit?

(also, here it is probably good to think briefly about why the given conditions really are mutually exclusive)

Nick

10:00 AM

@AkivaWeinberger can you help me?

Akash. B

anyone?

a help pls

Nick

@Akash.B ask away.

Akash. B

okay

first let me ask a simple question

is it possible to draw a straight line on a sphere?

no,right?

Nick

@Akash.B a straight line and a sphere could meet at a point or set of points. Any line on the surface of a sphere would be a curve.

Akash. B

what? infinitely many lines or straight lines?

Nick

10:05 AM

@Akash.B I was thinking internally, like chords. From point to point. That's possible, and infinitely many.

Akash. B

@Nick yeah so no straight lines,right?

Nick

well, there will always exist a shorter path than a surface path joining any two points on a curved surface. That's true.

Anush

Let $z = f(x,y) = x^2 + 2xy$

how is $\frac{dz}{dx}$ different from $\frac{\partial z}{\partial x}$ ?

Akash. B

@Nick do you understand what I really meant?

Nick

@Anush when you take partial derivative, you assume all other variables as constant. In your case, 2y will be treated as a constant. You don't need to apply chain rule for y.

Anush

10:09 AM

@Nick That makes sense. So what would $\frac{dz}{dx}$ be in this case?

I am not sure what the derivative with respect to x means if we don't treat y as a constant

Nick

@Anush 2x + {...} <-- apply chain rule: derivative of f(g(x)) is f'(g(x))⋅g'(x)

Akash. B

@Nick

Nick

so, how would you differentiate x*y? (ie, d(xy)/dx )

@Akash.B do you need to prove shortest paths are on rectilinear surfaces? You might also want to look into Mobius strips. (no, i don't get your question)

Akash. B

@Nick no

@Nick when I was solving a problem on trigonometry

Nick

Spherical geometry

Spherical geometry is the geometry of the two-dimensional surface of a sphere. It is an example of a geometry that is not Euclidean. Two practical applications of the principles of spherical geometry are navigation and astronomy. In plane (Euclidean) geometry, the basic concepts are points and (straight) lines. On a sphere, points are defined in the usual sense. The equivalents of lines are not defined in the usual sense of "straight line" in Euclidean geometry, but in the sense of "the shortest paths between points", which are called geodesics. On a sphere, the geodesics are the great circles...

Anush

10:16 AM

@Nick I don't know. I meant it looks like g(x) = x and f(x) = xy in your example

but that doesn't help

Akash. B

okay okay you saw that picture?

@nick

we can only obtain a curved line on drawing a line on a sphere

right?

@Nick can you pls respond?

Nick

@Anush lol, I'd just consider y as a function of x and be done with it. check this: wolframalpha.com/input/?i=differentiate+z+%3D+x%5E2+%2B+2xy

@Akash.B you're comprehending this correctly, you don't have to be in shock about it.

Alessandro Codenotti

@Nick That's not true on every surface, take the plane without the origin and two points symmetric with respect to the origin, what's the shortest path between them? The Hopf-Rinow theorem gives conditions to guarantee a minimal geodesic between any pair of points on a Riemannian manifold

Anush

@Nick d(x^2 + 2 x y) = 2 (x + y) dx + 2 x dy

hmmm :)

Akash. B

@Nick okay.so according to this is trigonometry possible?

we would not get a proper base for triangle

Secret

10:27 AM

Spherical trigonometry

Spherical trigonometry is the branch of spherical geometry that deals with the relationships between trigonometric functions of the sides and angles of the spherical polygons (especially spherical triangles) defined by a number of intersecting great circles on the sphere. Spherical trigonometry is of great importance for calculations in astronomy, geodesy and navigation. The origins of spherical trigonometry in Greek mathematics and the major developments in Islamic mathematics are discussed fully in History of trigonometry and Mathematics in medieval Islam. The subject came to fruition in Early...

Akash. B

@Secret so is my scale really straight?

Nick

@Akash.B @Secret SohCahToa still applies, so do the corresponding identities.

Secret

yeah I am aware of that, where each rule has a slightly twist due to the curvature

Straight lines get generalised into something called geodesics in curved space. They are what is "straight" in such spaces

Nick

@Secret This can't be seen as euclidean, can it? Because those would be the shortest absolute distances.

Secret

Spherical geometry is non Euclidean space, one can realise that when neighbouring geodesics intersect each other (i.e. parallel lines will intersect not at infinity, thus violating one of Euclid's postulates)

Akash. B

10:31 AM

@Secret why it doesn't appear to me as curved?

Secret

because spherical geometry is a manifold, which means locally it looks like Euclidean space

Akash. B

@Secret okay

@Secret so if we draw a line in space will it be a straight line?

Nick

trigonometry on wobbly-wobbly lines and spaces next plz x)

Secret

what space? Euclidean, Outer space or Spherical space?

Akash. B

outer space

Nick

10:35 AM

if I 3d draw a 3d line in 3d space i would have 3d problems

but ideally, it would be a straight line, expressing the shortest distance between some (x1, y1, z1) and (x2, y2, z2)

Akash. B

@Nick what?

Secret

Our universe is something like a 4 dimensional version of a dodecahedron, which means it is curved. Straight lines drawn locally may not be a geodesic if it is allowed to extend at cosmic scales

so it depends on how you drew it

Akash. B

@Secret who found this?

Tobias Kildetoft

@Secret Why would you single out a dodecahedron there?

Akash. B

on what basis?

Nick

10:38 AM

@Akash.B I just meant to say it would slant to all angles. Expressible by c1x + c2y + c3z = 0 relation in space. Unlike Secret, I assume my universe a 3d box, with dimension 400x400x400, units unknown, probably galactic pixels.

Secret

@TobiasKildetoft Our universe is most likely a Poincaré dodecahedral space

Akash. B

@Nick btw what's that third coordinate ? is it time?

Secret

which is basically a lorentzian version of a 120 cell

Nick

@Akash.B no, depth. You have three spatial dimensions.

Akash. B

okay

yess, hoooray

I will complain my authorities to replace my maths text

Nick

10:41 AM

...uh, a lot of work went into that ncert text. Like several decades, trust me, it has enough a standard.

Akash. B

@Nick ours is not ncert

its scert

fed up of learning mistakes

I should get marks for my answers

Secret

If there is one thing I want, I want to turn all India exams into Dedekind finite sets so that I can slice them out into oblivion

Nick

@Akash.B Oh, yes yes. Then, do complain... or simply make a humble request regarding your marks to change, if need be, in the form of a written formal letter to those concerned.

Akash. B

@Nick okay

Secret

Almost all questions from India exams lead to boring questions in this chat

Nick

10:44 AM

@Secret did you take MSc. Maths? How did you end up like this?

Secret

I am not from India, I took a different kind of public exam to get to uni

my Hong Kong public exam is still not a good exam though because it encourage memorisation

It's something called the HKCEE, and it is now replaced by HKDSE

I would say I really learn most of my maths proper in undergrad level

Nick

@Secret do you have an option to write in both english and mandarin?

i learnt most of my maths lounging here listening to mathematicians talk.

Secret

They use English in all their exams except for the Chinese language and Chinese history subjects

Akash. B

@Secret so you speak Chinese

Nick

@Secret It's interesting to me we learn so much in our youth and get to apply so little of that knowledge in practical life.

Secret

10:49 AM

Yes, like Leaky and Loch, we born in hong kong

thus we all speak cantonese

@Nick a lot of daily life maths really don't go beyond arithmetic and possibly a little bit of statistics and database manipulations

Akash. B

@Nick why we need to study so much then?

Secret

It is only when you enter technical fields that the higher level maths matter

The education system is still running in the manner that it tries to pump people into technical fields or something

and thus stuff us with a lot of facts, but not much thinking

Akash. B

@Secret everyone just look how good our memory is

not potentials

bored of it

user193319

12:14 PM

Is $\sum_{n=1}^\infty \int_{E} |f_n| < \infty$ stronger than $\sum_{n=1}^\infty |f_n|$ being integrable?

I was thinking the Beppo Levi theorem would say so, but now I am not sure.

Actually, on second thought I think that $\sum_{n=1}^\infty |f_n|$ being integrable is stronger than saying $\sum_{n=1}^\infty \int_{E} |f_n| < \infty$

Leaky Nun

@user193319 domain of $f_n$?

Mats Granvik

web.math.princeton.edu/generals

user193319

Reals.

Manolis Lyviakis

When we say

quotient space of A/B

we mean the elements of B are equivallent?

Tobias Kildetoft

@ManolisLyviakis We don't say quotient space of A/B

We either say "a quotient space of A" or "the quotient space A/B"

But yes, if B is a subspace, then we equate all elements of B

Manolis Lyviakis

12:29 PM

ok thanks

and that is indeed an equivalnce relationship

Tobias Kildetoft

not without more details, no

( we usually do more than identify the elements of B)

Manolis Lyviakis

So for 2 topological spaces

i cant just say

the quotient space A/B

math.uchicago.edu/~may/VIGRE/VIGRE2010/REUPapers/…

take a look here at the beginning

notations

Tobias Kildetoft

sure, once we are in a specific context, we can do that

Manolis Lyviakis

so a~b iff a,b in B

Tobias Kildetoft

But earlier, you just said that "that" was an equivalence, without specifying what the actual relation was

Manolis Lyviakis

12:31 PM

is an equivalence

Tobias Kildetoft

no

user193319

@LeakyNun I think it is true. If $\sum_{n=1}^\infty |f_n|$ is integrable, then it is an integrable function which dominates the sequence $\sum_{n=1}^N |f_n|$. Since the sequence converges pointwise everywhere to $\sum_{n=1}^\infty |f_n|$, the Lebesgue Dominated Convergence Theorem tells us that $\sum_{n=1}^\infty \int |f_n| = \int \sum |f_n| < \infty$.

Manolis Lyviakis

So what equivalnce is used to make A/B

Tobias Kildetoft

you need the reflexive closure of that

Leaky Nun

@user193319 LGTM

Manolis Lyviakis

12:32 PM

given a subspace B

a~a if a in B @TobiasKildetoft sorry i dont see why it doesnt work?

Tobias Kildetoft

@ManolisLyviakis But if a is not in B then a would not be related to a

Manolis Lyviakis

ohh

you mean

a~b iff a,b in B otherwise x~x

right?

Tobias Kildetoft

no, that would not be correct

Manolis Lyviakis

>.<

Tobias Kildetoft

You need to learn how to define these things precisely if you want to do topology

Manolis Lyviakis

12:38 PM

i want

Give a,b in A : a~b if a,b in B and a~a if a in $B^{c}$ is wrong

Tobias Kildetoft

You need a statement of the form a~b if [something]

you already have part of it, namely "a,b in B"

so now you need a second part

Manolis Lyviakis

what im trying to write is all elements on B for one equivelance class and the rest elements be the class of their self

a~b iff a=b or a,b in B and x~x if x no in B ? my best try

Tobias Kildetoft

what do you need the x~x for?

Manolis Lyviakis

ye i just noticed

Tobias Kildetoft

anyway, it is probably a good exercise to try to write that one up completely correctly, including quantifiers and everything

Manolis Lyviakis

12:48 PM

?

quantifiers?

ohh the symbols

sorry my native language is not english

Tobias Kildetoft

yes, you know "for all" and "there exists". Though in this case, probably you should just write it using set builder notation

Leaky Nun

1:17 PM

poll: computer mouses or computer mice?

Akiva Weinberger

Mouses

Rithaniel

1:37 PM

Mousen

3

user76284

I go with mice.

Rithaniel

Question: Is there a term for a general magma $Z$ with an element $f\in Z$ such that $fz=f$ for all $z\in Z$?

user76284

Magma with a left-annihilator, or left-absorber?

Rithaniel

Well, as I know annihilators, they are defined by the existence of a zero element. I'm more asking for a term for a "magma with a zero element"

Ah, yeah, it's left and right, should fix that.

Tobias Kildetoft

I think absorber is more common in this generality

Rithaniel

1:40 PM

Too late to edit, darn.

Yeah, $fz=zf$ for the kind of $f$ I'm curious about.

I will look up "absorber," as I am unfamiliar with the term.

user76284

Absorbing element

In mathematics, an absorbing element (or, annihilating element) is a special type of element of a set with respect to a binary operation on that set. The result of combining an absorbing element with any element of the set is the absorbing element itself. In semigroup theory, the absorbing element is called a zero element because there is no risk of confusion with other notions of zero. In this article the two notions are synonymous. == Definition == Formally, let (S, •) be a set S with a closed binary operation • on it (known as a magma). A zero element is an element z such that for all s in...

They also call it a zero element.

Secret

Absorbers basically are the generalisation of $0z=0$ to arbitrary magmas. Similar to identities, if it is a two sided absorber, it is unique.

Rithaniel

Alright, so it's just a "magma with a zero element" and there is no specific name.

Leaky Nun

Dec 3 '18 at 9:50, by Leaky Nun

Z_2 the 2-adic integers and C_2^N are both compact Hausdorff topological groups; the homeomorphism is not a group isomorphism so the Haar measures should be different; how exactly is it different?

Rithaniel

(I like the examples of absorbers on the wikipedia page)

anakhro

1:48 PM

hey all u hot cats

Leaky Nun

but I'm a hot dog :c

Rithaniel

and I am a cold cat.

anakhro

hey all u <insert adjective here> <insert plural noun here>

@LeakyNun what math have you been up to lately?

Leaky Nun

nothin' much, you?

anakhro

Just working on my thesis

S. Crim

2:06 PM

hey i have a question about stochastic calculus, does anyone reckon they can help me out?

anakhro

>Just ask; don't ask to ask.

S. Crim

ok im very sorry

anakhro

Don't be sorry. It's fine.

S. Crim

Given that M is a L2 bounded martingale, do we have that $\mathbb{E}[\langle M\rangle_t]=\mathbb{E}[M_t^2]$?

Im quite sure that that holds, but I can't really figure out why

and I'd like to know why it holds

anakhro

What do the langle/rangle mean in this case. I don't know anything about this.

S. Crim

2:12 PM

the quadratic variation

anakhro

2:26 PM

I don't know how expected value changes with limits.

When do they commute?

I doubt that helps though

When I look for the result, @S.Crim I see things that seem to require that $M_0 = 0$

S. Crim

I found something about that aswell, but it didn't look like a straight answer to the question

anakhro

eventuallyalmosteverywhere.wordpress.com/tag/…

ctrl + f for (**)

That seems to be why they say it needs $M_0 = 0$

Whether or not that s helpful, I am not sure

(i am only but a tourist to the subject)

isaac9A

The open interval of integers (2, 4) and the closed interval of integers [3] are both equal to the set containing the integer 3, right?

S. Crim

@anakhro thanks alot, I'll look through it

anakhro

@isaac9A $(2,4)\cap\mathbb Z = \{3\}$

If that is what you are wondering.

@S.Crim I am curious as to the solution to your problem, so if it does help, come back and tell me how you figured it out.

S. Crim

2:37 PM

Sure

isaac9A

@anakhro but does (2,4) ={3}?

anakhro

@isaac9A it depends heavily on the context.

isaac9A

if we are talking about intervals in the integers? o

anakhro

Yes, if your (2,4) is taken as a subset of Z, then sure.

However, that is not standard notation for integers.

isaac9A

ok gotcha thanks

anakhro

2:38 PM

It's notation we use almost exclusively for the real numbers.

S. Crim

@anakhro I found the solution

anakhro

@S.Crim how does it work out?

S. Crim

was rather easy actually as it was literally a theorem in my lecture notes, but I was overlooking it

isaac9A

also is there a complex number z that you can raise the complex number c = 0 + i to, such that the modulus of c^z is greater than 1?

anakhro

Oh that's convenient. :P

S. Crim

2:42 PM

it is just given that for M a continuous local martingale, that <M>_t is the unique monotone increasing process with <M>_0=0 such that (M_t^2-<M>_t) is a local martingale

so there it instantly follows

used some easier formatting than in Latex to type it quicker

hope its still readable

anakhro

@isaac9A what have you tried?

@S.Crim it's readable, thanks!

isaac9A

My intuition is that there is no real number r such that the modulus of i^r is greater than one

I tried separating i^(a + bi) into (i^a)(i^bi) but that didn't help much, then tried euler's formula

but since you are just dealing with sines and cosines which themselves never sum to greater than 1 I think you cannot raise i to any power, real or complex that will give you a modulus gerater than one

I then plugged i^z =54 into wolfram alpha and got a solution but it looks strange: I got that z = (2 (2 n π - i (log(2) + 3 log(3))))/π

anakhro

What is euler's formula for i, @isaac9A?

Semiclassical

2:59 PM

One way to take the modulus is to multiply by the complex conjugate

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