Mathematics - 2022-07-16

Calvin Khor

02:29

I asked a question which got upvotes but math.se isn't giving me my green dopamine drip

>:(

Calvin Khor

02:39

ok nm it only took 10 minutes

user4539917

drip drip drip...

💧

👍🏼

Beautifully irrational

08:46

2

Q: In what sense do we take the equality in when dealing with generating functions?

I will illustrate my question with an example, consider the proof that $1 + x + x^2 + x^3 \; + \; ... = \frac{1}{1-x}$. From the finite case of a geometric series, we can achieve the equality through a limit: $$\sum_{k=0}^\infty x^k = \lim_{n\to\infty} \frac{1-x^n}{1-x} \color{red}{= \frac{1}{1-...

generating-functions

brain expanding question

Astyx

09:08

@ShaVuklia You have that $\phi_W(x) f(v) = \phi_V(x) v$ for every $v\in V$

Hold on

You don't even need that

For $\sigma$ to be an $L$-homomorphism, it needs to satisfy $\sigma(xv) = x\sigma(v)$ for all $x\in L$ and $v\in V$

(where I am not writing the representations)

So you get $C_W\circ\sigma(v) = \sum_i x_iy_i\sigma(v) = \sum_i \sigma(x_iy_iv) = \sigma(\sum_i x_iy_iv) = \sigma(C_Vv)$

Sha Vuklia

11:00

@Astyx We can't say this, because we don't work with the same basis $y_i$ dual to the trace forms necessarily

right?

Given a basis $x_i$ we have to work with a basis $y_i$ that is dual to the trace form of $\phi_V$ and a basis $z_i$ (not necessarily equal to $y_i$) that is dual to the trace form of $\phi_W$

opio

16:27

Is it possible to prove the existence of an ODE over the time interval $[0,\infty)$ by using the convergence of its implicit Euler scheme. Or can you only use it to get solutions over $[0,T]$.

Mad Spaces

16:58

Hello experianced people.
So i am a working student, this means, i work and study, my work is usually with the university, taking some jobs here and there in my department to do some work and get not much payment but enough to help me out.
me and my girlfriend want to move to an appartment to ourselves, but this is just crazy price in europe right now, and i barely get paid enough to live with hobo life style (IE no much expending)

do you have good suggestions on how i can earn money in this stage, without further impacting my studies

I did not mind the uni jobs, because i used to learn new things or make better contacts. so i am frowning upon finding job outside uni at my stage because that would be just work and no fun and would not help my studies ..

user4539917

Have you looked into applying for any financial aid through the university?

Mad Spaces

Yea i am not getting those. i tried out, sadly, since i am a foriegn student in germany, not much options are allowed to me.

Aid is usually provided for migrants, or german students.

user4539917

How about becoming a migrant?

Mad Spaces

I am not doing that.

Morally not correct, i am not being followed politically or anything.

I was thinking of trying out poker...

leslie townes

this is a hard but common problem

user4539917

17:06

Ok, just a suggestion :-)

leslie townes

several of my friends played poker in grad school

my wife got a credential and worked as a personal trainer at a gym during her phd, primarily for the flexibility of hours

Mad Spaces

You guys got a cheat sheet :)?

leslie townes

i realize this isn't inside a university

it's tough if there are legal restrictions on what you can do and where you can do it. tutoring is very easy to do 'under the table'

i did that most of my undergrad years

Mad Spaces

I do that too. But i dont get many students, i have some ads on ebay.

One student every two months or so..

Asinomás

will people think I'm weird if I leave my food an hour and a half in the shut-off oven so that I don't need to clean extra dishes?

user4539917

17:09

yes and no

Mad Spaces

I was thinking of offering some service i dont know, like making a lecture about some courses i had and selling it i make good notes and lectures.. even if my questions here are dumb

i am just really out of ideas lol

leslie townes

if you put more effort into marketing yourself as a tutor you might get more traction. i hated doing that but i did it

it's tough if it's only one person sporadically, but once i had like 4-5 students the word of mouth kept things stable

Mad Spaces

Where are good places to market my self as a tutor? i have only being using ebay

Asinomás

can I hire u?

do u take litecoin?

Mad Spaces

I am not familiar with online currencies, i much rather euros

Sure if you got something you need done

i dont mind )

leslie townes

17:10

mad i guess it depends on where you are. i worked locally with people face to face. there were physical boards on campus where people would post ads for stuff

Asinomás

I don't think I can afford u if ur local coin is euro :(

Mad Spaces

ah man amma going to go for a smoke, this is just too annoying

Asinomás

what is

user4539917

I'd suggest quite smoking to save money :-)

Asinomás

how is quiet smoking cheaper than loud ?

leslie townes

17:13

i actually met my wife while tutoring. it's a little embarrassing (we weren't dating while i tutored her) so we tend not to bring it up

user4539917

lol

shintuku

@leslietownes find a nice, dark and damp underbridge and sell kids the stuff.

"hey kids, want a bit of tutoring?"

Asinomás

I left my toothbrush in the hotel

leslie townes

first lesson is free

Asinomás

I need to buy a new toothbrush

I'll go get one in the chinese bazaar

hopefully it doesn't cause my death in 30 years

robjohn

17:18

@Asinomás when you're 63, buying a toothbrush that won't cause one to die for 30 years sounds like a good toothbrush.

Asinomás

I think that's also true for me now that you mention it

I don't think it does that though :'(

user4539917

Talking about getting old, TIL Edward Witten's father is 101 years old.

robjohn

@user4539917 I'm surprised Ed Witten is still alive

I guess he didn't look that old when I used to see him at Princeton

user4539917

Yup, I think he's 70ish

leslie townes

vietoris had a long life

Asinomás

17:25

Leopold Vietoris

Leopold Vietoris (; German: [viːˈtoːʀɪs]; 4 June 1891 – 9 April 2002) was an Austrian mathematician, World War I veteran and supercentenarian. He was born in Radkersburg and died in Innsbruck. He was known for his contributions to topology—notably the Mayer–Vietoris sequence—and other fields of mathematics, his interest in mathematical history and for being a keen alpinist. == Biography == Vietoris studied mathematics and geometry at the Vienna University of Technology. He was drafted in 1914 in World War I and was wounded in September that same year. On 4 November 1918, one week before t...

leslie townes

i remember a prof saying "this guy died, like, six months ago"

haha asinomas

Asinomás

I was going to post it :)

He lived and did lots of things that are unusual nowadays it seems O.o

leslie townes

lennart carleson is still alive, which blows my mind

polite proofs

Good morning professor

Ted Shifrin

Howdy, robjohn, Munchkin’s pet, polite.

leslie townes

17:32

hi ted. daughter slept without a pacifier last night! progress.

user4539917

don't let her use her thumb as a substitute

leslie townes

yeah, we won't. she did great

user4539917

coolio

copper.hat

morning

user4539917

hello

Mad Spaces

17:42

give me ideas that make me rich and i will return 10% investment as a gift. GO!

Leaky Nun

be lucky and invest just the right stocks at just the right moment

leslie townes

a lot of people have used lesliecoin to realize their dreams

user4539917

rob a bank

Balarka Sen

leslie townes

be born to a rich person

Mad Spaces

17:46

Okay i wil make this problem even harder

Balarka Sen

@AkivaWeinberger Kirby diagram for $T^4$ (if I am not mistaken)

leslie townes

touch balarka's jellyfish thing and it will give you three wishes

Mad Spaces

How can i get enough money using Mathematics or physics.
GO!

I already tried tackling riemann and i failed

geocalc33

what amount $X$ do you define as making you rich

shintuku

quantitative finance?

Mad Spaces

17:46

1k per month am happy

i am around 600 per month rn

need more

shintuku

engineering degree

Mad Spaces

Do you guys wanna laugh?

i sent my prof a riemann proof idea using mirror images of physics

sadly, it was bullcrap

so much for the million buckeros

@shintuku sorry sir what are you doing in this chat room?

Engineering? begone!

leslie townes

have you considered providing legal services

shintuku

i those are your only two options for money with math

Mad Spaces

illegal you mean surely?

shintuku

17:48

quantitative finance and engineering

user4539917

chemical engineers make the big bucks

Mad Spaces

If i make a lecture, with latex, stuff like that, for some topic, do you think anyone will be interested in buying that?

user4539917

> the average income of chemical engineers is $114,820, more than double the national average for all occupations, $56,310.

leslie townes

you can also 'break bad' and sell meth for more money

Mad Spaces

Leslie i am a good boy :( i dont do that.

leslie townes

17:57

you can alternatively sell meth even without a background in chemical engineering

user4539917

11 mins ago, by user4539917

rob a bank

leslie townes

if you need money right now, robbery is the only way

user4539917

It's the quickest

Koro

Suppose a countably infinite set of real numbers has a finite non zero limit point, then the set can be arranged in a sequence $(a_n)$ such that $a_n^{1/n}$ converges.

But I don't know how to prove this. :(

But is the statement really true? If I take the set as $\mathbb Q$, then how can I arrange elements of Q in a sequence $(a_n)$ such that $(a_n)^{\frac 1n}$ converges?

Jakobian

18:13

Positive reals?

Koro

ohh yes.

please read 'real numbers' as positive real numbers above.

Then I take $\mathbb Q^+$ and the rest of my question stays.

Jakobian

I think $a^{1/n}$ should converge uniformly to $1$ for $a\in [1/r, r]$, $r>1$

So this is about how fast do we want to place the numbers which are close to 0/infinity

Or even like, if we take $a_n$ to have the property that $1/n \leq a_n\leq n$

for large n

Leaky Nun

@Koro try to see if you can prove that this is a counter-example XD

Jakobian

Take a sequence $q_n\in \mathbb{Q}$ such that $q_n\to q\neq 0$ and then wait till $q_n\in [1/n, n]$ for all $n\geq N$

Koro

@LeakyNun the book has also given a solution for this (I don't understand the solution). But I think that I don't understand the statement itself due to Q+

Jakobian

18:22

Let the rest of those be enumerated as $p_n$

You want to wait till you can have $1/n \leq p_m\leq n$ and then place $p_m$ in your sequence, and otherwise just keep on adding elements from $q_n$

if that makes sense

This will be a sequence such that $1/n \leq a_n\leq n$ for all large $n$, so $a_n^{1/n}\to 1$

But you want to make sure that you'll exhaust all of the $q_n$ as well

So maybe do this process for odd $n$, and for even $n$ always place member of $(q_n)$ next

Ted Shifrin

18:56

If the countable set is $e^n, e^{-n}$, all $n\in \Bbb N$, how can it be true?

You need indices to grow very fast compared to $n$. Can you reuse elements of the set?

leslie townes

i was thinking about this. this has a subsequence vs. subnet flavor to it.

graffe

19:12

If you start a symmetric random walk on the number line at the origin and have the rule that if you are at 0 and try to move left you stay at 0, how can you compute the expected time to get to x?

robjohn

@graffe Does the random walk always have integer steps? Is it always $1$?

graffe

@robjohn yes it is always 1

x is a positive integer

Daniel Adams

@graffe cant you set up one of those difference equations?

graffe

@DanielAdams it's a good idea but I don't know

My guess is it is x(x+1)

graffe

19:31

@robjohn do you know how to argue it is x(x+1)?

Which I now believe

(though numerical experiments)

robjohn

@graffe where does it start?

Mathematical Emergency

19:44

@robjohn at the beginning.

graffe

19:56

@robjohn at 0

@robjohn math.stackexchange.com/questions/4494262/…

Balarka Sen

20:14

@graffe Say $\tau_{x, y}$ is the expected time this modified random walk reaches $y > 0$, starting at $x > 0$. Then $\tau_{x, y} = 1/2 \cdot \tau_{x-1, y} + 1/2 \cdot \tau_{x+1, y} + 1$. Moreover, $\tau_{0, y} = 1/2 \cdot \tau_{0, y} + 1/2 \cdot \tau_{1, y} + 1$.

And $\tau_{y, y} = 0$. These give you a recurrence you can solve

graffe

@BalarkaSen could you post this as an answer?

the question is now at math.stackexchange.com/questions/4494262/…

Balarka Sen

Feel too lazy, but maybe someone else can give something more detailed.

graffe

thanks anyway

Balarka Sen

I do think $\tau_{0, y} = y(y+1)$ solves the above.

So your guess is right.

graffe

that's cool!

hooray for numerical simulations

if anyone likes games, I just asked math.stackexchange.com/questions/4494257/which-button-to-press

Balarka Sen

20:28

hi @KarlKroningfeld

Karl Kroningfeld

hey how's it going?

Ted Shifrin

Hi, a @Balarka, @Karl

Balarka Sen

hi Ted

good on this side, what about you

Karl Kroningfeld

doing fine thanks

hey Ted

Balarka Sen

just saw an infamous mathematician answering a random question in the main page

Ted Shifrin

20:30

More infamous than us?

Balarka Sen

very. he's been accused of multiple harassments

had no idea he is active in MSE

Ted Shifrin

OK, who?

Or what question?

I Know nothing.

user4539917

I know even less.

polite proofs

Why can I sometimes write long messages and something I can't?

I think I can only write long replies

Ted Shifrin

I always get cut off after 5 lines or so.

polite proofs

20:42

Okay, well, here is a proof of FTOC part 1:

It is given that $f$ is continuous at $c$, so it follows that for every $\varepsilon > 0$, there is some $\delta > 0$ such that $-\varepsilon + f(c) < f(x) < \varepsilon + f(c)$ whenever $0 < |x - c| < \delta$. Now by our lemma, we have that $$h(-\varepsilon + f(c)) < \int_c^{c+h} f < h(\varepsilon + f(c)).$$ Since $h > 0$, we may divide by $h$ to get $$-\varepsilon + f(c) < \frac{F(c+h) - F(c)}{h} < \varepsilon + f(c).$$

Therefore, we clearly have that $$\left|\frac{F(c+h) - F(c)}{h} - f(c)\right| < \varepsilon$$ whenever $0 \le h < \delta$. This shows that $\lim_{h \to 0^+} \frac{F(c+h) - F(c)}{h} = f(c)$. If $-\delta < h \le 0$, then note that $\int_{c+h}^c f = -\int_c^{c+h} f$, and so we have $$(-h)(-\varepsilon + f(c)) < -\int_c^{c+h} f < (-h)(\varepsilon + f(c)).$$ Therefore, dividing by the positive number $-h$, we get the same inequality chain as before:

$$-\varepsilon + f(c) < \frac{F(c+h) - F(c)}{h} < \varepsilon + f(c),$$ and so $$\left|\frac{F(c+h) - F(c)}{h} - f(c)\right| < \varepsilon$$ whenever $-\delta < h \le 0$. This shows that $\lim_{h \to 0^-} \frac{F(c+h) - F(c)}{h} = f(c)$, and so therefore we may conclude that $F'(c) = f(c)$.

I think that works, but it's drastically different from Spivak's proof.

I literally can't figure out how to rigorously prove that $\lim_{h\to0} m_h = \lim_{h\to0} M_h = f(c)$ in Spivak's proof, where $m_h = \inf_{[c,c+h]} f$ for $h>0$ and $m_h = \inf_{[c+h,c]} f$ for $h<0$ and $M_h$ is similarly defined for $\sup$

There must be some trick that I am missing

Ted Shifrin

Not drastically different at all. BTW, this is the proof I had you do last chapter. I’ve forgotten in what context.

polite proofs

It is quite different I think. We don't define functions $m(h)$ and $M(h)$ at all. And showing those limits exist seems very tricky. I don't know why Spivak writes it off as 'because $f$ is continuous at $c$' when that doesn't really help.

@TedShifrin Last chapter?

Ted Shifrin

Of course it does.

polite proofs

We want to show that $\forall \epsilon > 0, \exists \delta > 0, 0 \le h < \delta$ implies $|f(c) - m_h| < \epsilon$. At this point, you (or at least me) get completely stuck

(Similarly for $M_h$)

Ted Shifrin

It’s immediate from your argument. But his approach is important conceptually.

Officially, you get $\le \epsilon$, but it’s easily repaired.

polite proofs

21:01

Okay, we get $-\epsilon + f(c) < \inf f < \epsilon + f(c)$, then I suppose since $-\epsilon + f(c) \ge -\epsilon + \inf f$ and $\epsilon + f(c) \ge \epsilon + \inf f$, so I suppose $-\epsilon + \inf f \le -\epsilon + f(c) \le \epsilon + \inf f \le \epsilon + f(c)$

Ted Shifrin

The first one should have $\le$, no?

polite proofs

Well, we are trying to prove that $0 \le h < \delta \Rightarrow |\inf f - f(c)| < \epsilon$

Ted Shifrin

I said that’s not right.

polite proofs

Why not?

Ted Shifrin

Think about the meaning of inf,

polite proofs

21:07

No but I mean that's what $\lim_{h\to 0^+} m_h = f(c)$ means though, or am I wrong?

Ted Shifrin

You’re wrong.

polite proofs

On which part?

Ted Shifrin

It doesn’t mean what you said.

polite proofs

If I am wrong then I've been trying to prove the wrong thing all along, so maybe this will make me feel better in some way

Here, I am using $\inf f$ to mean $\inf \{ f(x) : c \le x \le c + h \}$ for brevity

If that's still wrong, then could you please write the correct definition?

Or accentuate the exact error

Ted Shifrin

If $x<c$ for all $x\in A$, is $\sup A <c$?

polite proofs

21:14

In that case, $c$ is an upper bound of $A$, but it may not be the smallest such upper bound, which means $\sup A \le c$

However I don't know why that applies to the definition of a limit

Ted Shifrin

It doesn’t. It applies to an error in your reasoning. For the defn of limit, you should prove that having $\le\epsilon$ is equivalent to having $<\epsilon$.

This sort of thing shows up zillions of times in analysis.

polite proofs

Okay, well, $\inf f \le f(c)$, so $\inf f + \epsilon \ge f(c)$ for some $\epsilon > 0$. Since if we assume $\inf f + \epsilon < f(c)$, then $\inf f + \epsilon$ is a lower bound of the image of $f$, which implies that $\inf f + \epsilon < \inf f$, which is a contradiction?

And so... because $\inf f + \varepsilon \ge f(c)$, we have $\varepsilon \ge f(c) - \inf f$

@TedShifrin Not in Spivak's book so far?!

Ted Shifrin

21:45

polite proofs

Ah, okay. So we get $|\inf f - f(c)| \le \epsilon$ for free

polite proofs

22:05

@TedShifrin Here's a write-up of what I have: We have $-\varepsilon + f(c) \le f(x) \le \varepsilon + f(c)$ whenever $0 \le h < \delta$ by continuity of $f$ at $c$. By definition, $\inf f \le f(x)$, so $\inf f \le \varepsilon + f(c)$. Next, we shall prove that $-\varepsilon + f(c) \le \inf f$. Assume, to the contrary, that $-\varepsilon + f(c) > \inf f$.

Then we must have $f(c) > \inf f + \varepsilon$. That implies that $\inf f + \varepsilon$ is a lower bound of $f$, and so $\inf f + \varepsilon \le \inf f$, which is a contradiction.

I think that proves it, for $h \to 0^+$ at least.

Ted Shifrin

22:22

I don’t see why there is any contradiction needed here. Keep it simple.

polite proofs

@TedShifrin How can I avoid it?

Ted Shifrin

22:48

Just definition of inf and sup.

polite proofs

That's what I use though

Ted Shifrin

All we need is $x\le c$ for all $x\in A$ means $\sup A\le c$.

polite proofs

Oh, I can just say that since $-\varepsilon + f(c) \le f(x)$, we have $-\varepsilon + f(c) \le \inf f$

🤦‍♂️

Thank you professor

leslie townes

ted: munchkin's new thing is that if she's asked to do something that she regards as inconvenient, she does a very theatrical "aaaarrrghhh"

robjohn

@politeproofs what do you mean by long messages?

polite proofs

22:56

@robjohn Like the full proof above, I've been able to post similarly long proofs in one message.

Ted Shifrin

@leslietownes Did Olivia teach her this?

leslie townes

it does appear to be a version of the cat's howling

robjohn

@politeproofs If you paste in a comment with returns in it, then it becomes a different kind of comment and can go much longer and bring up the (see full text) message.

also if you include a shift
return

but use that sparingly as an unnecessarily tall comment eats a lot of screen real estate

an unnecessarily tall comment eats lots of screen real estate

leslie townes

bytes are free, robjohn. it's like lesliecoin. the sky is the limit.

it's free real estate

shintuku

i see no abuse of screen real estate

something something tragedy of the commons is false, etc.

robjohn

23:08

@shintuku I removed it

shintuku

i would like to rent the middle part of chat

Ted Shifrin

23:27

That’s high rent district!

polite proofs

Dr. Shifrin, do you think I should just do the # and * exercises in your book? That's what I've been doing

Ted Shifrin

You’re missing lots of stuff. Are you trying to learn multivariable computation or theory?

polite proofs

@TedShifrin I would say both. But for the hardcore theory run I'll just read through Munkres later

So maybe an idea of the theory without going through all the proofs in depth

Ted Shifrin

There’s not much in Munkres that’s not in mine. My exercises are better.

polite proofs

23:42

Really? Why did you decide against calling it an analysis book?

Ted Shifrin

Because it is not.

polite proofs

But Munkres' is?

Ted Shifrin

Munkres assumes linear slg snd familiarity with multivariable. I do not. I find his book exceedingly dry.

The only thing he does that I don't in terms of "analysis" is the development of differential forms in terms of tensor product. I don't see the point of it for "beginners." ...

polite proofs

Interesting, well, that's even better then

Ted Shifrin

At any rate, here are the links for my homework assignments the last time I taught the course. All the computation was on WeBWork, so I don't list enough computations ... but you can see the proof problems I made my students do (some not in the book). This is what DCIII has been doing for the last year.

polite proofs

23:46

And how far is DCIII in the book?

Ted Shifrin

alpha.math.uga.edu/~shifrin/MATH3500 and alpha.math.uga.edu/~shifrin/MATH3510

I think he's in chapter 5 now.

polite proofs

Not working daily then

Ted Shifrin

He took lots of time off for other things.

polite proofs

Totally understand.. way too well unfortunately

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$Mathematics$ Mathematics