Mathematics - 2021-04-14

leslie townes

12:00 AM

up he goes

HereToRelax

what is updog?

copper.hat

the reverse of godpu.

dc3rd

Ok, I'm going to change $l$ to $k_{j}$. And I will say $k_{j} > K$. Now since $x_{k}$ is Cauchy that should work.

HereToRelax

godpu = god poo?

I finally got cleared for a date for my undergrad thesis exam

Ted Shifrin

No, @dc3rd. At least I don't see what you're doing. Write it all out carefully and tell me the radius of the ball. I'm going for a walk, so you have 20 minutes or more.

dc3rd

12:12 AM

ok.

take deep breaths and enjoy the fresh air

Lucas Henrique

hi chat

HereToRelax

hello Lucas

Are you Lucas Podolski the great striker of the German football team?

Oh nvm that's Lukas

Ted Shifrin

12:33 AM

We have a German Lukas here from time to time, but a different one.

leslie townes

or so he would have us believe

Ted Shifrin

Well, true, you earn your living not believing (in) anyone.

dc3rd

12:49 AM

I had forgotten what the * meant in your text and I caved to temptation....I have failed you Jedi Master ...😞....but the idea of every term in the sequence is a realization and not just something abstract has been made more clear.

Ted Shifrin

Unless you got the corrected printing, there is a stupid typo in that solution. Did you catch it?

dc3rd

Was it using $j$ instead of $k$ to generalize?

copper.hat

better than a smart typo.

Ted Shifrin

No. The formula for the radius of the ball has what?

@copper Mostly I leave the smart typos to you.

dc3rd

$1 + \|X_{K + 1}\|$

leslie townes

12:53 AM

i was just wondering, i wonder if there's ever been an argument improved, or even a unintentional theorem proved, by a typo.

Ted Shifrin

That was most definitely NOT what I have (corrected or not).

leslie townes

typos only take. they never give.

Ted Shifrin

I certainly wrestled with my share of errors in proofs (not necessarily typos), @leslie.

dc3rd

oh you meant $min(\|x_{1}\|, \dots , 1 + \|x_{K+1}\|)$

Ted Shifrin

Right. What is the typo in that?

Remember the picture of what's going on here.

Actually, I had a colleague (who made many millions off well-written textbooks) who tried to make sure there was an error on each page of the answers at the back of his calculus and ODE books. I found that amusing.

This particular error should be caught immediately but is still instructive.

Clarinetist

1:00 AM

Ugh, I'm going to have to get the Benadryl out. We got SNOW today! SNOW!

Ted Shifrin

I miss snow!

HereToRelax

we never get snow where I live

what's good about it

Ted Shifrin

I lived in it for 10 years. Now at my decrepit age, I am happy just to visit it.

But fluffy beautiful white stuff is amazing. Not so pretty in a dirty big city.

HereToRelax

First time I went to snow as a kid I just jumped into it and ended up freezing like a moron

leslie townes

it will surprise no one that in my first experience with snow i couldn't stop throwing snowballs at members of my family, even when they told me to stop.

dc3rd

1:11 AM

it seems to me that the error is that it should be $\min(\|x_{1}\| + 1, \|x_{2}\| + 1 \dots , 1 + \|x_{K+1}\|)$

copper.hat

i was given a 'clean up the proofs & do the numerics' paper during my graduate work, and discovered a type like $a \le b$ where the real result was $b \le a$ which invalidated the entire result.

HereToRelax

@copper.hat how did you give them the sad news?

copper.hat

that was not the issue. the issue was i had to find a whole new proof. it was pure grind whereas their original was short.

HereToRelax

oh

Ted Shifrin

@dc3rd where did the +1 come from in the first place? You are not understanding the argument.

If leslie weren't here, I'd tell you to draw a picture.

@copper.hat good for you!

dc3rd

1:25 AM

I've drawn a picture. may not be instructive though.

My understanding of the +1 is that since the sequence is Cauchy for all $\epsilon > 0$ there will be a $K$ whereby $\|x_k - x_l\|< \epsilon$. In this case we've chosen $\epsilon = 1$ since it is Cauchy this can occur and as such for $k, l > K$ we have $\|x_k - x_{K+1}\|$ where we've taken $l = K+1$

now reverse triangle inequality will give me the specific $\|x_k\| < 1 + \|x_{K+1}\|$ relationship

Ted Shifrin

For all $k>K$, yes.

So you draw a ball of radius 1 around $x_K$ and all the further terms are in there. Now what?

dc3rd

then that would mean $\|x_1 - x_K\|, \|x_2 - x_K\| < 1$ as two examples if I'm interpreting the further terms as you mean them to be.

Ted Shifrin

No. What does further mean?

I was trying to save typing math symbols on my iPad.

dc3rd

ok then further in the sequence is what you mean by further terms. i.e: $x_{k+1}, x_{k+2}$, etc

Ted Shifrin

K, but of course.

dc3rd

1:36 AM

well then each of those is less than $1$. i.e: $\|x_{k+1} - x_K\|, \|x_{k+2} - x_K\| < 1$

Ted Shifrin

So you're convincing yourself of the validity of my sentence. I asked you Now what?

dc3rd

1:52 AM

should I be comparing the distance between the further terms that are in this ball of radius 1?

Ted Shifrin

You have the damn answer, except for a typo. Figure it out. I am not wanting to be mean, but I think my book and Rudin are not realistic goals for you.

dc3rd

2:24 AM

You're entitled to have that opinion based on questions I've asked and what your years of experience have illustrated to you.

I made the choice a while ago to be all in on this endeavor regardless of the hardships I may endure....there is no Plan B. I've been face to face with death so enduring the difficulties of changing my thought patterns is soft.

I did figure out the typo and I figured out the issue with why it took me so long to see it. But I'll chill out on asking you questions for a bit until I got something that would be considered "extremely challenging" in your realm.

leslie townes

i can't even back out what the problem is, although i think i was here. i was mostly in another window.

dc3rd

there's no problem anymore. there's a typo in Ted's book that says $\min(\|x_{1}\|, \dots , 1 + \|x_{K+1}\|)$, but in reality it should say $\max(\|x_{1}\|, \dots , 1 + \|x_{K+1}\|)$

leslie townes

oh, i meant the problem that all of this was about solving. i was trying to back it out from some of the sequence stuff, formed some vague ideas, but couldn't get there.

i'm assuming this comes up in some exercise, if it was just some random inequality who knows

dc3rd

oh the question asked to show that if $x_{k}$ is a cauchy sequence, then all points lie in some ball centered at the origin.

my two mistakes were overlooking the importance of the word all and what that meant for the Cuachy sequence

I can't wrong him for getting frustrated....shrug

Zest

For a large summer party 3 large boxes of peaches have been bought. However, while decorating, a mishap happened as some of the fruit fell on the floor, making it unusable. They distributed the remaining peaches evenly on 3 tables. During the party, 4 peaches were eaten at each table. At the end of the party, there were a total of as many peaches left as had previously been on each table.

How many peaches remained intact?

It's 6, isn't it?

Andrew Micallef

2:43 AM

@TedShifrin ok, so trying to take this approach. I start with trying to find $\int{\Psi^* [-ih \frac{\partial}{\partial x}] \Psi}dx$.

Ah screw it will need to wait till i am home, this is too much tex for my phone keyboard skills to handle

Was trying to type out this

But i saw how you all feel about images this morning so sorry

Also i ommited the constant$ih$ because i can, and the limits of integration are positive and negative infinity

And i am clearly not getting what you meant

Ted Shifrin

Why are you even using integration by parts here? @Andrew

Andrew Micallef

I thought it would make life easier

Sorry after all that my lunch break just ended (i think i will post a proper Q in main when I get the chance(

Ted Shifrin

If I give you $\int ff’\,dx$ what is the better approach?

@Zest What is $6$?

leslie townes

3:06 AM

i feel like the guy in the theater who needs to alert the projectionist and shouts FOCUS!

Ted Shifrin

LOL ... to me?

P.S. I've done that before.

leslie townes

no to the blurry photos. if we must have photos i want them well composed.

Ted Shifrin

Oh, you're almost as picky as @robjohn.

leslie townes

did projectors just get more reliable? i guess there's a lot of digital projection now? i don't remember hearing or yelling focus for a long time. it's probably been 15 years.

Zest

@TedShifrin @TedShifrin the amount of peaches that remained intact.

Ted Shifrin

3:10 AM

The peaches that fell on the ground are gone. Aren't they asking how many are left on the three tables?

Zest

Yes, that's as i understood it as well. But i'm confused because others saight the answer would be 18 peaches.

leslie townes

that first line sounds like the beginning of a sad poem.

where are the peaches on the ground? ay, where are they?

Ted Shifrin

And it is correct,

Zest

Which answer is correct @TedShifrin? 6 or 18 peaches?

Ted Shifrin

Nothing like irrelevancy in word problems!

18, of course.

Back to 5th grade math!

Zest

3:14 AM

Thank you very much @TedShifrin. I got it wrong since i solved the equation

$3(x-4) = x$ which returned 6

Ted Shifrin

What is $x$?

Zest

the amount of peaches that has been put on each table initially.

Ted Shifrin

Always write a sentence defining $x$.

OK. So why is $x$ the answer ?

Zest

I looked at it this way: Let $x$ be the amount of peaches put on each table. From each table $4$ peaches have been eaten, so on each table there remain $x-4$ peaches. It's said that "at the end of the party, there were a total of as many peaches left as had previously been on each table."

Therefore i computed

$(x-4) + (x-4) + (x-4) = x$

Ted Shifrin

Maybe @leslie will draw a picture :)

I understand your equation. It's fine. You ignored my question, and that makes me impatient.

leslie townes

3:19 AM

i'm not a wild person, but this sounds like a dull party. offer your guests something more than peaches.

Ted Shifrin

Martinis would be a good start.

Especially today.

leslie townes

yeah. you have to build to the peaches.

Zest

Oh i'm sorry, i'm not sure if i understood you correctly.

Ted Shifrin

At least peach melba!

What does “why is $x$ the answer?” possibly mean?

@copper said I was very patient. I’m thinking those days are disappearing.

Zest

Well we have eaten 4 peaches of each table but in the end there are as many intact peaches left as we initially put on each table. That's why i solved the equation for $x$ and obtained $x=6$ as the total amount of intact peaches.

Ted Shifrin

3:23 AM

Total amount? Think, man.

Reread what you said $x$ was.

Zest

Ah, of course. Now i see it.

Ted Shifrin

When you define $x$ you should also say what the answer to the question is in terms of $x$.

The answer is not automatically $x$.

Zest

You're absolutely right. I somehow confused my answer with the definition of $x$.

Thank you very much @TedShifrin.

Ted Shifrin

@leslie I need a sabbatical. Maybe legal secretary?

Just remember what you’ve learned from this, @Zest.

Zest

Yes, indeed. I brushed over it and got confused where i'm wrong. Should have taken a step back.

leslie townes

3:29 AM

motivational speaker. that's what i'm going to do.

Ted Shifrin

It's good to write a sentence saying explicitly what the question wants you to find, once you have defined your variable.

I'm not feeling motivational.

copper.hat

@TedShifrin compared to me you are very patient.

Zest

That's indeed a very good advise @TedShifrin. Highly appreciating your help.

Ted Shifrin

You're welcome.

leslie townes

this is going to sound slightly stereotypic, but everybody knows that irish blood boils at very low temperatures

Ted Shifrin

3:31 AM

Not much longer, copper. I'm coming over to your side.

What about Russian/Polish blood?

leslie townes

i usually just fill in the blank with the ancestry but i already used it once on the irish. i got nothing.

also you really don't want to rile some of those folks up. the irish sometimes have a sense of humor about it.

Ted Shifrin

Do some research.

leslie townes

i will moonlight as an ethnographer.

Ted Shifrin

Copper has a sense of humo(u)r?

Rover

3:48 AM

Hey Hi Guys Can anyone please tell what really is Maths? What significance it has of it's own and is it just about numbers and no visualisation of what that equation is really signifying ?

@TedShifrin what do you say ?

leslie townes

it means different things for different people. for some it is all numbers. for some the numbers do not play a big role. people do try to figure out what things, equations included, are signifying, although ahem not always through visualization.

copper.hat

4:29 AM

i know there is an irish stereotype regarding tempers, but would say that, drinking situations excluded, irish are more likely to be socially constrained in pressure situations.

of cource, the other stereotype means that the measure of the exclusion is pretty large.

you are definitely more patient in general @TedShifrin. if i think someone is making a genuine effort i will go the extra mile, but otherwise i will switch off quickly. you persist and will continue to engage.

leslie townes

not a lot of tempers on that side of my family. just a bunch of people who mostly never talked to each other

copper.hat

mmm, well, if we are talking family then that is a different story :-).

robjohn

4:58 AM

@TedShifrin $$ \begin{align} \int vv'\,\mathrm{d}x &=vv-\int vv'\,\mathrm{d}x\\ &=\frac12vv+C \end{align} $$

roundabout, but it works.

oops, I misread the fuzzy image

Not that I'm picky ;-p

leslie townes

FOCUS

copper.hat

a lot of entendres in here

leslie townes

5:16 AM

i don't mind fuzzy images as fuzzy images i just don't like the sensation that i'm losing my eyesight more than i actually am

Ted Shifrin

@robjohn Yes, I realize that works, but I think @Andrew would profit from seeing the immediate substitution as the more natural approach. :P

robjohn

@TedShifrin I know that. I was just being picky!

leslie townes

5:34 AM

:(

copper.hat

5:55 AM

those optical illusions that had multiple focus solutions used to give me a headache.

leslie townes

if you stare long enough into that FOCUS image, you see a giraffe in 3d

copper.hat

a klein giraffe with its neck... i'll leave that to your imagination.

leslie townes

hahaha

copper.hat

good Eine kleine Nacht

my mom loved mozart's music

robjohn

7:16 AM

@copper.hat I've toyed with the idea of generating some of those to do some 3-D illustrations.

Rithaniel

Re: "What is Math"
It's abstraction. Instead of talking about "3 apples" or "3 meters per second" or "3 volts," you just consider the abstract notion of "3" itself. The advantage of abstraction is that it simplifies things, and the abstract result can then be applied to any of hundreds of real world situations.
One of the drawbacks of abstraction is that suddenly you don't have anything to experiment with, which means you have to fall back (hard) on logic and reasoning, so you could also say that math is the rigorous application of logic.

3

CowperKettle

8:26 AM

> Why was 2019 afraid of 2020?
Because they had a fight and 2021.

Snapdragon-X

9:43 AM

What is math if not 3 objects persevering... cnet.com/a/img/qVQMRM1NE2KelrmjcUC99uXgXI0=/1200x630/2021/03/05/…

Sha Vuklia

9:57 AM

@LeakyNun u here?

Leaky Nun

@ShaVuklia ja

Sha Vuklia

cool, kan ik iets vragen ~

Leaky Nun

hoe kan ik jou helpen

Sha Vuklia

xD

so, my algebra is a bit rusty ;x

but I think the argument should be simple

I don't understand why $K(C)$ is an algebraic extension of $K(t)$

and they mentioned three things

eh, but I don't see why what they say yields the conclusion?

oh, maybe I should give some definitions

$C$ is a projective curve

$K(C)$ is its function field

(ehh, actually I don't know if there is some ambiguity in the definitions?)

Leaky Nun

are you happy with the fact that K(t) embeds into K(C)

Sha Vuklia

10:03 AM

sure

Leaky Nun

so t is transcendental in K(C)

Sha Vuklia

um

Leaky Nun

but since K(C) has transcendence degree 1, this means {t} is a transcendence basis of K(C), so K(C) is algebraic over K(t)

Sha Vuklia

@LeakyNun I didn't realise this, let me see

Leaky Nun

t is transcendental in K(t)

this can be seen by "uniformizer"

t is the uniformizer, in a DVR sense, of the local ring of the stalks at P

so t is transcendental

Sha Vuklia

10:07 AM

ah

because

$(t^n)\neq 0$ for each $n$?

or

ye I get it, maybe I didn't give the best reason

Leaky Nun

yeah

great

Sha Vuklia

cool, thx!

Leaky Nun

np

Sha Vuklia

hm, maybe I don't see it after all

You're also giving a slightly different argument right? Because they also use that $K(C)$ is finitely generated over $K$ (or maybe they need to mention that to conclude that $t$ is transcendental? since they only mention $t\notin K$)

Andrew Micallef

I really should apologise for my shitty fotos. I should just wait until I have time to talk properly instead of hurridly scribbling things down on paper.

@TedShifrin Thanks for your patience. If I follow your suggestion I think that takes me back to your advice from a few weeks ago, and I need to just find a profitable substitution instead of integrating by parts. I will try take that approach.

Sha Vuklia

10:22 AM

Oh wait, $K(C)$ being finitely generated over $K$ should then yield that the algebraic extension is finite

oh, I see it I think (why $t$ is transcendental)

the order wouldn't be well-defined

okok, I'm good

user863565

guys I don't know how to know if partial differential equation is linear or not?

I don't understand that function F(x_1,...x_n,u_1,....,u_n)=0 gibberish

BTW wth is this F(....) thing?

en.m.wikipedia.org/wiki/…

Sha Vuklia

10:51 AM

@LeakyNun I'm afraid I don't see it after all ;x would you mind to elaborate why $t$ is transcendental?

Is it in general true that if we have a field extension $L/K$, and $L$ had transcendence degree $k\geq 1$ over $K$, then any $x\in L\setminus K$ is transcendental over $K$?

Leaky Nun

@ShaVuklia let the valuation be normalized with v(t) = 1, suppose t^n + a1 t^(n-1) + ... + an = 0. dividing by powers of t, you may assume an != 0, so v(t^n + a1 t^(n-1) + ... + an) = v(t(something) + an) = v(an) = 0

@ShaVuklia take $K = \Bbb Q$ and $L = \Bbb Q(\sqrt2)(t)$

Sha Vuklia

Oh, oops

I was thinking more in terms of powers of the maximal ideal, and I didn't realise that I could also evaluate

thx

@LeakyNun and thx for this as well x')

Leaky Nun

yeah it's useful to use the valuation

Andrew Micallef

11:12 AM

Ohhhh, (I realise we are clearly in different timezones but I want to say thanks before heading to sleep...I'm not sure how to make the next step (what to do about the complex conjugate), but I think I should muse over it myself until the weekend. For the time being I have this on my sheet of paper:

$\int{f \cdot f' dx}$ let $u = f$, then $\frac{du}{dx} = f'$ so $dx = \frac{du}{f'}$

By substitution I get $\int{f \cdot f' dx} = \int{u \cdot f' \frac{du}{f'}}$,
The $f'$ cancels, and now I get $\int{u}du$.

Sha Vuklia

12:26 PM

@Leaky ;oo

mind if I ask another thing?

Leaky Nun

@ShaVuklia hoe kan ik jou helpen

Sha Vuklia

I also posted it on stack:
https://math.stackexchange.com/questions/4101868/question-about-proof-2-4-b-in-silverman-injection-of-function-fields-of-curves
But let me recap it here

I don't see why their $\phi$ maps into $C_2$ in the first place

(and I also don't see why not all the $g_i$ can be constant, but I worry less about that, and more about the first question)

Sha Vuklia

12:47 PM

Oh, I also think there is a typo: it should say $C_2\subset\mathbb P^N$

Leaky Nun

@ShaVuklia if $C_2$ is $y^2 = x^3 + 1$ (i.e. $Y^2 Z = X^3 + Z^3$) then $\iota(y)^2 = \iota(x)^3 + 1$

something like this should work, sorry i'm busy

Sha Vuklia

Right, that helps, thanks!

Leonhard Euler

2:16 PM

does calc 4 exist

love_sodam

2:38 PM

If $f:D^n\to D^n$ is continuous map such that $f|_{\partial D^n}$ is a homemomorphism, then $f$ is a surjection?

Leonhard Euler

nvm my calc 4 question it is silly

but i have heard that some universities refer vector calculus to as calc 4

Thorgott

@love_sodam you mean homeomorphism onto its image?

love_sodam

Ah, $f|_{\partial D^n}:\partial D^n\to \partial D^n$

sorry

Thorgott

aha, that's a useful assumption

convince yourself you can assume $f\vert_{\partial D^n}=\operatorname{id}_{\partial D^n}$ WLOG

love_sodam

2:55 PM

hmm....

Thorgott

ok, you don't actually need this tbf

love_sodam

why can I assume like that?

Thorgott

the above step reduces this to the fact that a compact manifold can't retract onto its boundary, but you can also just directly do a similar proof here by looking at the top homology

the hint for that WLOG assumption is radial extension

love_sodam

Ok I get it thanks

Ted Shifrin

3:36 PM

@Andrew My humble apologies. I had forgotten you were working with complex functions and the complex inner product. So of course you are right that you need to differentiate $|f|^2 = ff^*$ by the product rule. Undoing the product rule is, of course, integration by parts.

@Thor: How was your lecture?

Thorgott

4:22 PM

it went fine

sadly my time management was as terrible as always, so I didn't manage to get to the examples

they're in my writeup, though, so it's not like they were worked out in vain

Karim Mansour

Hi @TedShifrin

Ted Shifrin

@Thor: Examples should always be a high priority. I often do them before proofs. :)

Hi, Karim.

Thorgott

I don't think this was a scenario in which discussing examples first is the better order, but I do regret not being able to present them

in other news, the Hodge index theorem is really cool

Ted Shifrin

4:47 PM

Remind me which theorem?

Thorgott

signature of an even-dimensional compact Kähler manifold is computed by $\sum_{p,q}(-1)^ph^{p,q}$

Ted Shifrin

Oh, I forgot that.

Sha Vuklia

Is there an easy way to see why $\operatorname{deg}(\phi)=5?$ By definition, the degree of $\phi$ is the degree of the field extension $\overline K(\mathbb P^1)$ over $\phi^*(\overline K(\mathbb P^1))$

Wait, maybe I can figure this out

Ok, so I basically have to show that $[K(X):K(X^3(X-1)^2)]=5$ (where K is some (algebraically closed) field)

Thorgott

5:22 PM

you can guess the minimal polynomial

Sha Vuklia

5:48 PM

@Thorgott Ye, that should be $T^5+2T^4-T^3-X^3(X-1)^2$, but I'm a bit out of it to see why it's this one, and not one with a smaller degree

But I guess I can try to write it all out

love_sodam

If A,B are R-algebra (R is commutative ring with 1) R-algebra homomorphism f is just a R-linear map with f(aa')=f(a)f(a') for a,a' in A right? wiki described K-algebra homomorphism using term 'K-linear' but not on R-algebra homomorphism

Sha Vuklia

ye ok, I think I see it

Thorgott

@love_sodam and f(1)=1

an R-algebra is something that's both a ring and an R-module in a compatible way and a map between those is a map that respects both structures, i.e. somethings that's both a ring homomorphism and R-linear

love_sodam

6:03 PM

If A is finitely generated R-module, then A is isomorphic to a quotient of polynomial ring R[x_1,...,x_n]. But we can also write $A = R[a_1,...,a_n]$ where a_i in A yes?

Deadcode

This has been upvoted to 11, but hasn't received any comments or answers. Would it be frowned upon if I cross-posted it on Math.SE? What can be proven regarding the differences in power between unary ECMAScript regex functions and primitive recursive functions?

Also, if anybody just wants to comment here on the things discussed in there, I'd welcome it.

leslie townes

i don't know the applicable etiquette but i doubt it would be more likely to get answers on math.SE. there are obvious and close connections to math but there's a lot of regex knowledge required here. math.SE also tends to like one question per post.

if you could distill one of your questions into something perhaps a little simpler than that, maybe it would be more likely to attract attention on math.SE. but you do seem to have targeted the question appropriately already.

in my very humble opinion :)

Deadcode

Okay, thanks :)

leslie townes

side note, they are very interesting questions.

Deadcode

Thank you.

Euler 2

6:18 PM

Yeah i also think so

leslie townes

i would conjecture that the answer to your question 1 is 'no,' but i have no explanation for this, it's just a vibe.

Deadcode

That was the first one I'd consider asking on Math.SE.

leslie townes

conjecture is too strong a word. guess, but not wildly guess.

Euler 2

i asked it once but i am asking again: can someone recommend a good book for cryptography

neither too basic nor too complicated

leslie townes

also a very good question. all of the ones i have seen are one or the other.

Euler 2

6:31 PM

There was a chapter on cryptography in one of my nt books

I liked it and became interested

aWxvdmVjcnlwdG9ncmFwaHk=

leslie townes

gesundheit

Anselm678

7:10 PM

can someone give two examples of non metrizable topological spaces?

leslie townes

the weak star topology on the dual of a banach space

*infinite dimensional banach space

pick two non-isomorphic examples of such and i just gave you two examples

Anselm678

okay thanks

leslie townes

i was slightly kidding there. there are probably more natural examples but infinite dimensional vector spaces and their duals provide a host of them.

robjohn

@Deadcode I assume you've seen FRACTRAN

leslie townes

a lot of examples proceed by first proving that metrizability implies some more abstract property. like first countability, or other stuff about countably many balls being enough to do x or y.

then you go to your library of examples of bad spaces that don't have that property.

wow, i really like nate eldridge's answer here: mathoverflow.net/questions/52032/…

it's like he crawled inside my brain, went back in time, and wrote my answer. we might as well call it my answer.

Thorgott

7:22 PM

take sth non-hausdorff

leslie townes

there's the more natural answer.

Andrew Micallef

8:36 PM

By that, I take it this would be wrong:
$\frac{d}{dy}\int{f^* [\frac{\partial}{\partial x}] f}dx$ (where $f$ is a function of $x,y$, and $f^*$ the complex conjugate of $f$)
let $u$ = $f$
then $\frac{du}{dx} =? \frac{\partial}{\partial x}f$ (should the LHS be a partial derivative?)
so by the same logic as before: $\frac{d}{dy}\int{u^*}du = \frac{d}{dy}[\frac{{u^*}^2}{2} + C]$

which then becomes $\frac{d}{dy}[\frac{{f^*}^2}{2} + C]$

Ted Shifrin

Yup, wrong. I missed the conjugate.

Andrew Micallef

:P

okay, well at least I am getting some cognative excercise with this one

Balarka Sen

@TedShifrin Trying to learn spectral theorem for bounded self-adjoint operators

Ted Shifrin

Cool.

Andrew Micallef

I was so pleased with myself for that and evrything. I seam to be going in circles with the integration by parts: Apply the formula $\int{uv'} = uv - \int{vu'}$
So my case I let $u = f^*$, then $u' = \frac{\partial}{\partial x}f^*$, so $v' = \frac{\partial}{\partial x}f$ and $v = f$.
then I get $$f^*\cdot f - \int{f \frac{\partial}{\partial x}f^*}$$

also, side note $u$ and $v$ are terrible symbols to stick side by side, I get it the follow alphabetically but the glyphs look way too similar, I'm gonna use other ones in the future, $j$ and $k$ seam way better!

And because I was moaning about glyphs I forgot the $dx$ at the end and missed the edit window

Ted Shifrin

8:54 PM

Did you notice my product rule comment earlier? That's all that’s going on.

You have to be careful with infinite limits to make sure things converge appropriately.

Chairman Meow

So I have a problem. Here it is:
Let $g(x)$ be the generator polynomial of a binary cyclic code $C$. Let $h(x)=\left(x^{n}-1\right)/g(x)$ be the parity check polynomial.
Show that if $a(x)$ is a polynomial satisfying gcd$\left(a(x), h(x)\right)=1$, then $a(x)g(x)$ is a generator of $C$.

Now what I am not understanding is why I must show that $a\left(x\right)$ must satisfy gcd$\left(a(x), h(x)\right)=1$.
As long as $a(x)$ is of least degree and a divisor of $x^n -1$, the product $a(x)g(x)$ should be a generator for $C$, correct?

There is something obvious my tiny brain is probably missing.

Andrew Micallef

I did, $f^* f = |f|^2$ which is a product, so we need to use integration by parts to undo. I'm about to run off to work. I guess the bit that is troubling me is how that works when I need to differentiate one of those functions first.
I'll keep thinking about this / puzzling over it in my head. and will try to resist the overwhelming urge to write down where I'm at on a scrap of paper at lunch :P

Ted Shifrin

9:23 PM

@Andrew: Ultimately, you should perhaps explain to picky @robjohn and me what your goal is here.

robjohn

9:40 PM

@TedShifrin Perhaps my pickiness is too much for him.

Ted Shifrin

Well, I was wrong, and you were right! But I'm not sure what he's trying to do with the Schrödinger equation.

Astyx

Hi robjohn, Ted

Ted Shifrin

Ah, Astyx gave robjohn top billing. Surely that'll placate him for a while.

Hi, @Astyx.

Astyx

@TedShifrin A very strategic decision

Ted Shifrin

Hi, @Rithaniel

Deadcode

9:47 PM

@robjohn Yep, that's cool, but not really similar. Its exponents can grow without bound, go either up or down, and loop without bound (i.e. a program has the option of never halting).

robjohn

@Deadcode Yes, old west programming. (1)

@Astyx hi there.

Rithaniel

10:03 PM

@TedShifrin Hey Ted

Ted Shifrin

Hope you're doing ok, @Rithaniel. Haven't seen you in a while.

user863565

10:31 PM

hey does anyone know what The general form of n-th order ODE is given as F(x, y, y’,…., yn ) = 0

what is function F()?

Ted Shifrin

Anything you want.

Typically a continuous or many times differentiable function.

user863565

so I can say x+y+y'=0

sin(y')=0?

Ted Shifrin

That last one wasn't very interesting, but you certainly could have $yy'' + \sin(x)y^2 + e^y = 0$.

user863565

Ted Shifrin

Now that's a general partial differential equation.

user863565

10:34 PM

ah wrong picture

I was quite confused about linear partial differential equation

Ted Shifrin

Just as with ordinary differential equations. $y$ and all its derivatives appear linearly, no products, no powers.

user863565

I was confused about those partial differentiation in terms of some variable in suppose function y

Ted Shifrin

I do not understand.

user863565

$\partial^{(2)} y(x,y)/\partial^{(2)}(x)+\partial^{(2)} y(x,y)/\partial^{(2)}(y)+\partial^{(1)} y(x,y)/\partial^{(1)}(y)+\partial^{(1)} y(x,y)/\partial^{(1)}(x)=f(x,y)$ is second degree partial differential equation right?

because it got many variable unlike ordinary differentiation and linear ordinary differential equation only differentiate in terms of one variable instead of multiple variable

so I am curious how does general linear partial differential equation looks like

sorry I found the définition online

Ted Shifrin

No, what you're writing doesn't make sense.

user863565

10:46 PM

@TedShifrin ah i get it now thank you

@TedShifrin yeah I admit that. 😅

mshwf

Hi all, I have a question,
Isn't all quadrilateral shapes with the same side lengths, have the same area?

robjohn

@mshwf no

rhombuses can have varying areas, depending on their angles.

Ted Shifrin

It's true for rectangles, though, @mshwf.

But if we take crazy quadrilaterals, there's not even a formula with just the side lengths.

mshwf

I'm imagining having 4 sticks (with different lengths) configured in quadrilateral shape, I think shrinking an angle stretch another maintaining the same area

Ted Shifrin

Well, many shapes won't even be able to be changed, depending on the lengths.

But area generally changes. Robjohn's suggestion about rhombuses is the best thing to think about.

mshwf

10:55 PM

@TedShifrin so I'll need to get two angles to be able to calc the area?

Ted Shifrin

Two opposite angles will do, yes.

mshwf

thanks

robjohn

@TedShifrin if the opposite angles are supplementary, one can use Brahmagupta's formula.

Rithaniel

@TedShifrin Eh, I'm okay. My mom recently got a feeding tube put in, so she's been able to get food in her, which she wasn't able to for a while. Myself, I've been wandering about, distracted as I usually am

copper.hat

11:14 PM

that's tough for everyone. best wishes.

Ted Shifrin

My best wishes to your mom and family. I’m sorry! 😢

Rithaniel

Not to worry, she's actually able to walk around because she has energy. It's a great improvement

Also, we (or, rather, I) have been saying she's a cyborg, now

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$Mathematics$ Mathematics