Mathematics - 2019-04-06

Tanner Swett

12:33 AM

So here's a question. How do I get a job in mathematics?

I have a bachelor's degree in math and 4 years of professional experience as a software developer.

I don't want to work as a software developer any more, unless it's something that involves a lot of math.

Ted Shifrin

With just a bachelors you end up having to do stuff that's computer-oriented and/or business-oriented with consulting.

Leaky Nun

@TedShifrin I thought the MGF $E[e^{tX}]$ was the real thing

but I was wrong

Ted Shifrin

And of course there's actuarial stuff, but I have always found that exceedingly uninteresting and less and less mathematics.

Leaky Nun

the characteristic function $E[e^{itX}]$ is where it's at

Ted Shifrin

LOL, OK, Leaky.

I know one of them showed up at the end of my probability course — the former, I believe.

Tanner Swett

12:38 AM

I've looked into doing actuarial stuff, but yeah, I don't know if it would just be boring.

Ted Shifrin

Maybe biotech, @Tanner?

Tanner Swett

Maybe!

Ted Shifrin

The mathematics in things like CT scans is fascinating ...

Of course, that's old news now.

Leaky Nun

@TedShifrin the thing is that the char. f. converges everywhere

Tanner Swett

My problem is that if a task is too boring, then I can't focus on it and I can't get it done. That's why software development hasn't proven to be a viable career for me.

Ted Shifrin

12:39 AM

There are few easy answers, @Tanner.

Tanner Swett

Usually the stuff that makes a task boring is that it's too... "easy". In other words, the fact that I can think about it for a couple of seconds and see exactly how it's going to play out.

So here's a question. Does actuarial involve a lot of tedious, routine stuff?

Ted Shifrin

I visited Grand Rapids a few summers ago, @Tanner. Saw the gardens and ate at a fabulous restaurant.

Yes, actuarial does, and it's more and more business-oriented and less math. In the old days you had to pass exam(s) on numerical analysis.

But you should talk to people who do it.

Tanner Swett

Yeah, I probably should.

Ted Shifrin

I had some students/advisees who loved it and others who hated it and quit.

Tanner Swett

I wonder... if I just go and pass a whole bunch of actuarial exams, could I then get into a lead role, where I can do the "hard" stuff myself and pass the "easy" stuff on to others?

That's... sort of a fantasy of mine. I don't know if it's actually realistic.

Ted Shifrin

12:43 AM

Honestly, I doubt it.

A number of my former charges have worked for Microsoft, Google, Yahoo, etc. Most of them have found it fulfilling, but I don't know precisely what they have done.

Tanner Swett

Maybe I should start by learning what the jobs in mathematics are.

I have no idea what actual, practical work there is out there that involves Laplace transforms, say.

Ted Shifrin

Engineering.

Hence my mentioning biotech.

Tanner Swett

Oh yeah.

Leaky Nun

breaks a wall

Tanner Swett

I'm gonna do a Google search for 'how to become an electrical engineer' and see if that gets me anything interesting.

Ted Shifrin

12:47 AM

LOL, have fun exploring.

Tanner Swett

The joke I like to make is that the perfect job for me would be "computer programmer in the 1960s".

Leaky Nun

1. study electrical engineering

2. apply for a job

3. ???

4. profit

@TedShifrin have you heard of the windmill?

Ted Shifrin

I hear it causes cancer.

Leaky Nun

exactly

I’m kinda concerned since I live right next to a windmill

Ted Shifrin

Well, cover yourself in aluminum foil.

Tanner Swett

12:55 AM

I should also go talk to my alma mater's career center to get ideas.

Ted Shifrin

Yeah, although often googling may be more helpful.

Tanner Swett

I'll do some of both. :D

Ted Shifrin

Sounds like a plan.

Really interesting mathematics in signal processing ...

Leaky Nun

@TedShifrin this is the logistic distribution right

Ted Shifrin

Don't ask me.

Leaky Nun

1:03 AM

but it's a cool program

hi chat

hi Eric.

hi eric

how goes it

tired. lol.

1:04 AM

same

Dair

nice.

Ted Shifrin

Wait 'til you're my age :P

Dair

@TedShifrin I'm literally going to be asleep 24/7 by your age.

Ted Shifrin

That sounds very boring.

Dair

it's also called being dead.

Ted Shifrin

1:12 AM

That's one alternative.

Érico Melo Silva

tbh the world wont last that long

Ted Shifrin

There is that alternative, too.

Leaky Nun

what's the $r$ in $\binom{n}{r} = C^n_r = \frac{n!}{r!(n-r)!}$ stand for?

Ted Shifrin

I write it with $k$, so I have no idea.

Leaky Nun

:o

Dair

1:13 AM

It stands for $r$

Leaky Nun

heresy

Érico Melo Silva

literally all the cool people use k

Ted Shifrin

Thank you, Eric.

Dair

Isn't $r$ some sort of british thing? :P

Ted Shifrin

or pirates?

Leaky Nun

1:15 AM

I learnt it when I was in primary school and I have always used $r$

Érico Melo Silva

honestly if i say "n choose" out loud my muscle memory immediately forces me to follow it with k

Yup, me too.

and r for me

Next topic?

yeah this well is dry

Leaky Nun

1:17 AM

what's the right definition of a sequence of distribution converging to another?

Érico Melo Silva

what do you mean by a distribution

Leaky Nun

oh, random variables

Érico Melo Silva

there are many notions of convergence and it depends what you're doing

Leaky Nun

can I say that their characteristic functions converge uniformly to the characteristic function of the limit?

or is that too strong?

Érico Melo Silva

that sounds way stronger to me than what usually comes up

Leaky Nun

1:23 AM

if I drop "uniformly" then?

looks like Lévy's continuity theorem then

but I don't like C.D.F. because it's too arbitrary

how about $L^2$ convergence of char. f.

Érico Melo Silva

what are you even trying to do

Leaky Nun

nvm the char. f. of the normal distr. isn't even $L^2$

let's say the central limit theorem

Ted Shifrin

Bye, all.

Leaky Nun

see you

Érico Melo Silva

tchau @Ted

classical CLT is typically phrased in terms of convergence in distribution of random variables

but there are like a billion versions of CLT

MrAP

5:26 AM

For $\theta_1,\theta_2,\dots,\theta_{51}\in R$, let $A(\theta_1,\theta_2,\dots,\theta_{51})$ be the average of the complex numbers $e^{i\theta_1},e^{i\theta_2},\dots,e^{i\theta_{51}}$, where $i=\sqrt{-1}$. Find the minimum and maximum value of $|A|$.

I have found the maximum value to be 1 using the triangle inequality for complex number. I cannot understand how to find the minimum value. Please help.

Leaky Nun

@MrAP can you make A=0?

user131753

Anyone here interested in a question about C Programming?

user76284

8:52 AM

@MrAP Let $\theta_i = i/51 \cdot 2\pi$. Then all the points are equally spaced around the unit circle so they all cancel out when added up.

@user170039 You can ask, though you'd probably find more help at chat.stackoverflow.com/rooms/54304/c.

NewBornMATH

11:00 AM

Hello, anyone online ?

user131753

1:15 PM

@user76284 Thanks. Already posted my question there.

user193319

2:04 PM

If $\{G_n\}_{n \in \Bbb{N}}$ is a collection of (nested?) groups, what could $\displaystyle \lim_{\rightarrow} G_n$ mean?

Secret

direct limit?

user193319

Apparently it could mean the direct limit of groups

Haha

Direct limit

In mathematics, a direct limit is a way to construct a (typically large) object from many (typically smaller) objects that are put together in a specific way. These objects may be groups, rings, vector spaces or in general objects from any category. The way they are put together is specified by a system of homomorphisms (group homomorphism, ring homomorphism, or in general morphisms in the category) between those smaller objects. The direct limit of the objects A i {\displaystyle A_{i}} , where ...

I think this is the right interpretation.

Alessandro Codenotti

2:24 PM

Just to be clear if your groups are nested then the morphisms for the direct system are assumed to be inclusions

Perturbative

2:39 PM

So I read this proposition in my diff geom notes. "Let $f : M^n \to \mathbb{R}$ be a smooth function. If $df = 0$ then the surface $S_C = f^{-1}(c)$ for some $c \in \operatorname{Im}(f)$ is an $(n-1)$-dimensional manifold. Isn't this just a conseqeunce however of the fact that if $df = 0$, then $f$ is a constant function, so the map $f : M \to \mathbb{R}$ has constant rank $1$,

and so the level set $f^{-1}(c)$ is a properly embedded submanifold of codimension $1$ in $M^n$, yielding $S_C$ to be a $(n-1)$-dimensional manifold.

Leaky Nun

rip latex

Secret

3:01 PM

statisticsbyjim.com/regression/…

A small $s_{y/x}$ is often good news because it means even if your model is way off, it is way off in a consistent way, allowing you correct the deviation as a systematic error. Meanwhile a large $s_{y/x}$ means the model is such a poor fit that data points can be anywhere away from the trend in question, meaning that it is dominated by noise and not a trend, making correction of deviation harder

James Groon

3:49 PM

If $\sum_{n=1}^{\infty} a_{n} + b_{n}$ converges, must $\sum_{n=1}^{\infty} a_{n}$ and $\sum_{n=1}^{\infty} b_{n}$ converge as well ?

I'm guessing not , but can't think of anything useful

Sam

4:04 PM

How do you expand $5(x+\Delta x)^2$ ?

James Groon

4:22 PM

oh an=n , bn=-n

silly me

Soham

4:53 PM

Has anyone read atiyah's proof of convexity of the moment map ?

vzn

5:28 PM

@Semiclassical hi whats new, have a proposition for you, plz drop by here for more info :) chat.stackexchange.com/rooms/9446/theory-salon

Ted Shifrin

@Perturbative This is nonsense. If $f$ is constant, it has rank $0$. The correct hypothesis is that $df_x\ne 0$ for all $x\in f^{-1}(c)$.

@Sam Do you know the formula for $(a+b)^2$? If not, expand it out ...

vzn

@Secret SE has this dynamic where some rare rooms reach critical mass & its hard to reach, would like to see chgs in the software/ design/ mgt/ culture etc to counteract that.

from long chat participation it seems to me theres too much fixation on particular chat rooms. people want to be free to talk about whatever in particular established rooms. it seems very strange to me that people rarely exercise their freedom to create new rooms and yet SE allows anyone to do that. think maybe part of the solution is something that would help increase user mobility between related or popular chat rooms. eg "related rooms" in the sidebar or something like that... some of whats going on here is the Physics Chat room is one of the most popular on SE... — vzn Feb 6 '17 at 23:03

Daminark

Hey Ted!

Rithaniel

5:52 PM

So, if $I\subsetneq R$ with $R$ a commutative ring with identity, then $I$ is saturated. I can easily show that $I$ is multiplicatively closed, but I don't see how to get from there to $I$ being saturated.

Ted Shifrin

6:21 PM

Hi Demonark :)

Perturbative

@TedShifrin Yep that was nonsense, was just trying to sketch something of a proof

Ultradark

7:23 PM

Say you have a number of cyclic groups, the first with 1 element, second with 2 elements and so on

generally set up like the picture

Akiva Weinberger

@JamesGroon Well, if $a_n,b_n>0$…

Ultradark

with the nucleus being the group with 1 element

Rithaniel

Those are not exactly cyclic groups. If you have a full shell and you add one electron, you don't go back to 1. It's more like a monoid, if I understand monoids correctly.

Akiva Weinberger

I think he's just using the picture as inspiration

Secret

What is your full question, it isn't clear what is this overall structure is constructing

Akiva Weinberger

7:31 PM

(Dunno what for yet though)

Rithaniel

That's fair enough.

Secret

Are you talking about a group $G$ whoose subgroups are $\Bbb{Z}_{18},\Bbb{Z}_{8},\Bbb{Z}_{2},\{1\}$ in that order?

Rithaniel

I'm still stuck on this whole "showing an ideal is saturated" problem.

Akiva Weinberger

What's a saturated ideal?

Secret

It means if $xy \in I$, then $x,y \in I$ for all $x,y$

Rithaniel

7:34 PM

A subset $S$ of a ring $R$ is saturated if $xy\in S\implies x,y\in S$

I'm also given that this particular ring is commutative with identity.

Secret

i.e. the divisors of $xy$ are within the ideal

Akiva Weinberger

Ah, interesting.

So for example if I do the set of even numbers within the ring of integers

that's not saturated because $2\cdot3=6$ but $3$ isn't even

Yeah?

Rithaniel

Like, I'm trying to assume that $x,y\notin S$ and arrive at a contradiction, but I don't see the thread. Also, yes.

Akiva Weinberger

But the set of even numbers ($\langle2\rangle$ or $2\Bbb Z$ or however you wanna write it) is an ideal, and $\Bbb Z$ is a commutative ring with identity

so why doesn't this have the hypotheses?

Rithaniel

But wait, the question asks to show that any proper ideal of commutative ring with identity is saturated.

Now my head hurts.

Akiva Weinberger

7:37 PM

Either that's a counterexample or I'm misunderstanding something

Ultradark

I'm a little late in answering, but, the picture is more for inspiration. The first group would be the nucleus, and the second group(1st shell) would have two elements, the third group would have three elements (2nd shell). So every shell would have 1 more than the previous shell

Rithaniel

Ah, there is a line about $0\notin S$, but it's awkwardly placed. I'll copy/paste.

Let $R$ be a commutative ring with identity. We say that a nonempty subset $S\subseteq R$ (with $0\notin S$) is multiplicatively closed if $s,t\in S$ implies that $st\in S$. We say that the set $S$ is saturated if for all $x,y\in R$, if $xy\in S$ then $x,y\in S$.
(5 pts) If $I\subsetneq R$ is a proper ideal, show that $I$ is saturated.

Akiva Weinberger

Isn't $0$ always an element of an ideal?

Rithaniel

Ah, yeah, it would have to be.

Akiva Weinberger

I'm confused

Also it sounds like $1$ has to be an element of any saturated set

Secret

7:40 PM

@Ultradark but how are all the groups related to each other. What I am seeing so far is you have a sequence of groups each with one more element than the previous one. Are they subgroups of a group, or something else?

Akiva Weinberger

and the only ideal containing $1$ is $R$ itself

I think the question is wrong

Rithaniel

You don't necessarily have multiplicative inverses.

Secret

@AkivaWeinberger That should be saturated? Any product of even numbers is even, and any even numbers can be split into a product of even numbers?

Akiva Weinberger

@Secret $2\cdot3\in S$ but $3\notin S$

Secret

ah right...

Akiva Weinberger

7:47 PM

Wait. Do you want both products to be in $S$, or just at least one?

Ultradark

@Secret Can the sequence of groups be itself a larger group?

Akiva Weinberger

(As an example of something that seems like it is saturated, take the set of powers of 2)

@Ultradark Well there's an idea of a direct limit of groups, where you have a sequence of nested groups and kinda piece them together

Rithaniel

It might have been a typo. Maybe he wanted us to show that an ideal is multiplicatively closed, which is infinitely easier.

Akiva Weinberger

Or is that the right word for it?

I forget which is the direct limit and which is the inverse limit

I'm sure there's some sense in which the finite cyclic groups come together to form $\Bbb Z$

Secret

1. Well you can have e.g. $\Bbb{Z}_{288}$ which will contain the above groups you want as subgroups and they probably contain each other in a chain
2. You can also have something like $\Bbb{Z}_2 \times \Bbb{Z}_3 \times \Bbb{Z}_4 \cdots$
3. $\Bbb{Z}$ contains all $\Bbb{Z}_n$ as subgroups for integers $n$

Akiva Weinberger

7:50 PM

You have maps $\Bbb Z_{mn}\mapsto\Bbb Z_n$, yeah? So maybe by using those families of maps

@Secret 3 is wrong

We have surjections $\Bbb Z\to\Bbb Z_n$ but we don't have injections $\Bbb Z_n\to\Bbb Z$, i.e. $\Bbb Z$ doesn't have subgroups that look like $\Bbb Z_n$

$\Bbb Z$ has quotients that look like $\Bbb Z_n$

Rithaniel

Yeah, $\mathbb{Z}$ contains only one finite subgroup.

Secret

ok maybe I am thinking about $n\Bbb{Z}$ which in that case 3 is still wrong because some of these $n\Bbb{Z}$ are not subgroups

Rithaniel

emailed the prof.

Akiva Weinberger

@Secret Hm? $n\Bbb Z$ is always a subgroup of $\Bbb Z$, no?

The set of multiples of $n$

Secret

$2\Bbb{Z}$ does not have $1$ in it, no?

Akiva Weinberger

7:55 PM

The identity element of $\Bbb Z$ is zero

It's an additive group

If you wanted the multiplicative group, you'd want $\Bbb Z^\times$, which has underlying set $\Bbb Z\setminus\{0\}$

Rithaniel

If an ideal contains a multiplicative identity it is equal to the parent ring. An ideal, meanwhile, has to contain the additive identity, as it's a subgroup.

Secret

ok

Akiva Weinberger

Oh wait

$\Bbb Z^\times$ isn't a group

It doesn't have inverses

Well

Actually I think if $R$ is a ring then $R^\times$ has underlying set the set of invertible elements of $R$

which would make $\Bbb Z^\times=\{-1,1\}\simeq\Bbb Z_2$

ÍgjøgnumMeg

8:11 PM

@Akiva right

Luigi M

Hi chat! can someone help me with Gromov monotonicity lemma? it's a lemma that provides a lower bound on the area of a J-holomorphic curve in a ball of radius r in $\Bbb C^n$. Apparently it's obvious but I can't prove that the lower bound exists and it's $\pi r^2$

Poline Sandra

8:25 PM

hello, can someone see if my answer is right math.stackexchange.com/questions/3177354/…

Ted Shifrin

DogAteMy: Yes, $R^\times$ denotes the (group of) units in $R$.

@Luigi: Write it (at least locally) as a graph over the tangent space at the center of the ball. For a constant map you get the area of the disk.

user193319

9:21 PM

0

Q: Norm Topologies on an Infinite Dimensional Vector Space

I know that all norms on a finite dimensional vector space induce the same topology. Moreover, I know that there are infinite dimensional vector spaces with norms that don't induce the same topology. But I am curious about the existence of two infinite dimensional vector spaces: (1) Is there ...

Akiva Weinberger

9:41 PM

Visualizing Seven-Manifolds

►

No idea what's going on^

(Not much else on that channel)

Rithaniel

I need a lot more vocabulary to understand that video, I think.

Nice music during the animation, though.

$\xi_{8,-7}$ on the $S^7_{15}$ exotic sphere is going crazy.

Luigi M

10:29 PM

@TedShifrin Hi Ted, thanks for the answer. So you are suggesting to observe that the area of a graph (given by my curve) over a little disk at 0 of radius $\epsilon$ is at least $\pi \epsilon^2$, is that right? How do you conclude for this local observation?

Akiva Weinberger

11:11 PM

Israeli elections on Tuesday

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$Mathematics$ Mathematics