Mathematics - 2021-12-12

Ted Shifrin

00:00

Can’t you solve explicitly?

I mean — I just did it in my head.

Wojciech Kulma

Honestly I don't know how to do it

Ted Shifrin

Do you know how to take derivatives?

leslie townes

i wouldn't shy away from that, if you do. things simplify quite a bit when you set the thing equal to 0.

Wojciech Kulma

So I calculated the derivative: $e^{-n x^2} \left(n-2 n^2 x^2\right)$
and set it to 0: $\left\{\{n\to 0\},\left\{x\to -\frac{1}{\sqrt{2}
\sqrt{n}}\right\},\left\{x\to \frac{1}{\sqrt{2}
\sqrt{n}}\right\}\right\}$

seems like it's indeed converging then

thanks a lot!

Ted Shifrin

Yeah. Easier to get rid of the outside $n$ to start!

Wojciech Kulma

00:13

btw @TedShifrin I envy your skill to calculate that in head :D

Ted Shifrin

with exponentials, you factor them out because $(e^x)’= e^x$.

then it’s just simple product rule and chain rule. Admittedly, I have 50+ years experience.

leslie townes

while i didn't do it in my head before i said "things simplify quite a bit", i did see that it was going to be solving a degree-2 polynomial in x.

my 'mental math' got me that far. i probably would have goofed on factors of 2 if i'd done it in my head.

Ted Shifrin

I was not meaning to be obnoxious. I was indicating you just get your hands dirty and calculate.

Once again, the iOS version fails and I cannot edit

Finally, After 6 tries.

BannedUser

02:20

hello why is electric flux same for any closed surface enclosing an electron?

is there a simple proof that it is $q/\epsilon_0$ for arbitrary closed surface enclosing an electron?

I know for a sphere where electron is at center of sphere it is true.

leslie townes

mm, it might depend on what multivar calculus you know. i forget if the physicists look at this differently.

BannedUser

I already forgot most of the multivariable calculus. Is there other way? If no I will try to understand as much as possible.

leslie townes

if you know the divergence theorem, i think it at least phrases the flux in terms of an integral of the divergence of a field, and the field has divergence 0 away from the place where maybe it isn't defined for a point charge. which allows you to say, okay, for any surface enclosing the charge, i should get the same flux as i would for a small sphere enclosing the charge. where you plug in your knowledge about that case.

there's probably some physicsy explanation of how the flux relates to the divergence but i don't know it. this is broadly known stuff (among people who are into it), surely someone can do a more competent explanation than i can.

robjohn

02:42

@BannedUser Is the potential field you are considering around the electron $k/r$?

BannedUser

@robjohn well both proton or electron is okay but what is k/r?

PM 2Ring

03:01

@BannedUser The electron can be anywhere in the sphere. Do you understand the Explanation using liquid flow from en.wikipedia.org/wiki/Divergence_theorem ?

Ted Shifrin

The flux is the same for any closed surface whatsoever containing the charge and no others.

Physicists give a heuristic argument for spheres (noting that inverse square cancels the square growth of surface area). It’s hard to make that work for general surfaces.

robjohn

So if the force is inverse square, the potential field is inverse distance (times a constant, thus the $k$).

The Laplacian of that field is $0$ away from the origin. Then you can apply the divergence theorem

@BannedUser The electric potential around the electron is inversely proportional to the distance.

BannedUser

03:21

So it seems that I need to relearn vector calculus to understand it ;_;

shintuku

study using anki

never forget anything

Ted Shifrin

03:38

You cannot understand E&M without serious vector calculus.

Osmium

04:27

What is anki?

@BannedUser Do you know Gauss law?

If you know the basics of surface integrals you can prove it for yourself.

Gauss law states that net electric flux linked with a closed surface is equal to the ratio of net charge (add the charges with their conventional sign) with the permeability of free space, it is the same for all mediums. Note that net flux means net flux by all the charges in the universe not just due to the charges enclosed in the surface.

shintuku

@Osmium spaced repetition program

Osmium

This is a fancy name.

What does it mean?

shintuku

Spaced repetition

Spaced repetition is an evidence-based learning technique that is usually performed with flashcards. Newly introduced and more difficult flashcards are shown more frequently, while older and less difficult flashcards are shown less frequently in order to exploit the psychological spacing effect. The use of spaced repetition has been proven to increase rate of learning.Although the principle is useful in many contexts, spaced repetition is commonly applied in contexts in which a learner must acquire many items and retain them indefinitely in memory. It is, therefore, well suited for the problem...

Osmium

04:43

Flux through a surface is proportional to number of field lines passing through it. If you have a closed surface containing an electron at any point, it will be constant. This is because even if the charge moves inside the surface the net field lines due to it passing through the surface will be the same.

Ted Shifrin

04:55

The number of field lines is infinite. This stuff doesn’t really quite make sense.

copper.hat

engineering speak

Osmium

It is a convention to draw same number of field lines on paper through two surfaces if they have same net flux.

leslie townes

sounds kinda circular but i get it

in magnetism you can just shake some iron filings onto a paper under the thing in question and count 'em

Osmium

05:26

You're right it is ciruclar. I took the converse of the convention to be true; which is false. Ted Shifrin is right that it doesn't make any sense.

Which type of enginner are you?

Prithu biswas

05:42

@shintuku I can't find gamelin.

shintuku

it's on genesis library

there are also lectures on youtube that use gamelin as the main textbook

Prithu biswas

When I download it is saying it is a djvu file.

shintuku

yeah i love djvu, the viewers are cheaper on RAM compared to pdf

Prithu biswas

But I have nothing to read djvu.

shintuku

i use windjview.sourceforge.io/en

many math textbooks i have are in djvu

what's up koro

what are you studying lately

Prithu biswas

05:55

nvm I converted it to pdf.

Koro

Hey shin! what's up

I'm currently studying linear algebra.

shintuku

argh i have to finish it up too soon enough

so little time

you need it for something else?

Koro

how fast you finish depends on your syllabus too. :)

shintuku

heheh

copper.hat

@Koro did you sort out the svd issue?

Prithu biswas

06:07

@shintuku In metric spaces (gamelin) , ∀x,y ∈ X (d(x,y) ≥ 0) is stated as an axiom. But on wiki it is a lemma?

Wiki on metric spaces

copper.hat

no it is not. it is a consequence of the first two axioms.

Prithu biswas

leslie townes

yes, some books take it as an axiom

Prithu biswas

@copper.hat This is what I am seeing in gamelin.

Koro

copper: I understood svd but I'm yet to know about many of its applications.

leslie townes

06:16

there's no law that says books have to use minimal sets of axioms, and many don't

rudin takes an equivalent approach as gamelin

copper.hat

my son wanted to learn stick shift today. after an hour he was cruising my aging wrx over the hills past inspiration point. i was surprised & impressed.

Prithu biswas

@leslietownes why?? if it is such a direct consequence , shouldn't I use white correction fluid to just remove that part?

leslie townes

when i was learning i found 'real' driving fairly easy on stick. hard is an hour of stop & go traffic.

shintuku

@Prithubiswas typically to save space and for practicality

you can derive it yourself if you want

leslie townes

prithu, if you care, go right ahead

copper.hat

06:19

yeah, we did quite a bit of slow stop/start/reverse stuff at the start.

leslie townes

i wonder if any of the axiom-folks out there have come up with single-axiom axiomatization of metric spaces

shintuku

can you actually single-axiom anything

copper.hat

@Prithubiswas focus on the point of the axioms rather than their minimality.

the one i hate is universal properties.

Prithu biswas

@shintuku well , you can just take the conjunction of all of the axioms :P

leslie townes

shin: do the general theorems spit out reasonable stuff for stuff like metric spaces? i had the impression that some of this was done best by hand

Koro

06:20

@Prithubiswas if d satisfies only 1.2 and $d(x,y)\le d(x,z)+d(y,z)$ for all x,y and z in X then also d is a metric space.

so you can actually replace 4 numbers of axioms by only 2.

leslie townes

don't let lists of axioms fool you, the precise choices made in axiomatizing metric spaces are not the point of anything in gamelin

shintuku

at leslie: i just went with gamelin's axioms so I wouldn't know heh

copper.hat

this is like going to a burger joint and ordering a salad

leslie townes

particularly if you plan on doing analysis beyond the basics, almost no theorem or concept is phrased in maximal generality

which drives some people crazy the way redundant axioms drive some people crazy

but almost all of the interesting stuff isn't at that level of generality

copper.hat

i hate, i mean really hate, redundancy over and over again.

Prithu biswas

06:23

@Koro That is sort of different from only removing one axiom.

leslie townes

redundancy kills me. it literally makes me die. also imprecision in language.

copper.hat

i find myself using literally more and more each day

i think it is early brain rot

Koro

@Prithubiswas If d satisfies only the properties mentioned in my earlier comment, then you can show that all 4 axioms of the metric space in the image you sent are also satisfied and converse is also true of course.

copper.hat

@Prithubiswas why are you concerned about minimality?

shintuku

since the day I had to write 70 lines to prove there exists a function between the natural numbers and the integers, i got tired worrying about axioms

copper.hat

06:26

no convex psqs :-( must be end of term time/

shintuku

pastebin.com/02h30nWG

Koro

copper: I think Prithu is studying logics and is trying to put the definitions in line with that.

copper.hat

whoa. that sounds like my kid's highschool maths torture

@Koro thanks!

Prithu biswas

@copper.hat It is kind of pointless for me to make 1 lemma an axiom.

Like if we have axioms A,B,C,D and it turns out D can be derived from A,B,C ; then it is pointless to me to let D be an axiom anymore.

leslie townes

well, as a sneak preview, many hypotheses of many theorems in analysis can be relaxed in about 20 directions. this is the beginning of a series of expository choices to avoid getting into that stuff

even if it's sometimes not strictly necessary

at least here you can be sure that you aren't losing interesting examples by adding the extra thing :)

Prithu biswas

06:30

@Koro I think you are replacing A,B,C,D with A,E. =)

And I am replacing A,B,C,D with A,B,C.

Koro

@Prithubiswas this is not possible in the case of metric spaces though (considering A,B,C,D as axioms you sent in the image.)

Prithu biswas

@Koro The numbering are note the same , I just used an random example.

Koro

And yes, you rightly pointed out I replaced A,B,C and D by B and E.

copper.hat

@Prithubiswas it depends on what you are trying to do and where you want to spend your time. redundant axioms are not an issue unless they are inconsistent.

Koro

Prithu: Do you think that at least one of the axioms in metric space definition (in the image you sent) is redundant?

Prithu biswas

06:36

Koro : (1.1),(1.2),(1,3),(1.4) ⇔ (1.2),(new)
Me : (1.1),(1.2),(1.3),(1.4) ⇔ (1.2),(1.3),(1.4)

@Koro Here.

@Koro The first one is redundant.

leslie townes

a lot of the theorems about metric spaces can be repeated, with similar proofs, without the metric at all, and the assumption of a metric actually excludes a lot of useful examples. still, nobody has to start with more general spaces first, and then only specialize when they "actually need" the metric

it's a similar vibe

Koro

Prithu: do you have any reason to believe why (1.1) is redundant? Do you have any proof of that? Would you like to share that please?

Prithu biswas

30 mins ago, by Prithu biswas

Wiki on metric spaces

Koro

thanks, I'll take a look at that.

shintuku

koro i think it is because you can derive it from identity of indiscernibles and triangle inequality

Koro

06:41

Oh yes, of course!!

Prithu: I think you are right. :)

I never noticed that before :(

leslie townes

commutativity of addition is a redundant ring axiom, if you assume a ring to have a multiplicative identity (which many books do). in particular in a book on field theory it is a redundant field axiom

Prithu biswas

@shintuku can you show me that proof?

copper.hat

06:57

here is an example of why i have no intuition for elementary modular arithmetic: show that for positive integers $m,n$ that $3^m+3^n+1$ cannot be a perfect square. it is straightforward if one follows the hint (mod 8), but without that hint i would be up all night.
am i missing something huge, is there something in the problem (other than the hint) that would suggest $8$ as an appropriate modulus?

leslie townes

copper, to my knowledge you are not missing anything huge

this might be a case where 20 minutes of somebody's cleverness could supply why it's a 'natural' choice (after 20 minutes and being clever)

copper.hat

thanks. that helps. (seriously)

leslie townes

from my POV, looking at small moduli in beginning number theory is just a natural move because it can be brute forced

if there's a layer beyond that, i don't know what it is

copper.hat

part of me was hoping i was missing something...

leslie townes

someone with enough number theory background could say more, maybe even at an elementary level

but intro textbooks tend to stop before pulling back whatever wizard of oz curtain is hiding us from any deeper truth

copper.hat

07:02

if it is down to a large collection of unrelated observations i might as well quit now :-)

leslie townes

i don't think it's such a huge collection. look modulo n for small n, or n connected to the problem in some obvious way.

at a high level though i think of almost anything about diophantine equations as like that

no coherent anything, just lotsa cases

Koro

imagine a number num=0.axby.... such that whatever combination of numbers 1,2,...,9,0 you can think of (repetition is allowed), it's all there in num. Can such a number be called "well-defined"?

leslie townes

and 'although some minor variation of this problem might be hard or unsolved, i wouldn't be asked about this if it were hard'

koro: the property you mention doesn't seem to define a number? it's just a property a number might or might not have?

like champernownes constant

i'm not sure i understand the condition

shintuku

@Koro i think you can with decimal expansion

Koro

I recall having a discussion in another chat room, on denseness of {10^n a} in [0,1] where a is irrational.

So we were trying to characterise such irrationals for which the denseness condition will hold.

So I was suggested a number similar to the construction I gave above.

shintuku

07:06

koro is there any reason why you couldn't define it with decimal expansions?

Koro

If we take any open interval (p,q) in [0,1] then by the virtue of num defined above there will be a pattern such that 10^n *num will have fractional part in (p,q) so such "num" makes fractional part of 10^n *num dense in [0,1].

shin: Isn't that decimal expansion only? We take 0.axbycd.... something like that.

But not sure if such a num is well defined in the sense that -if shin asks, what is 100th decimal place in the num-I wouldn't know or would I.

copper.hat

can't say i am following...

shintuku

Let an arbitrary combination of numbers of arbitrary size be given and an arbitrary decimal place be given. then you can write a number that has that number combination starting at that decimal place

i guess you can prove it by contradiction

Koro

copper: for a background here's how the question arose

brilliant.org/wiki/dense-set

there's one try yourself problem: S={{2^n a}: n is natural} and a is an irrational number. {.} is fractional part. Then is S dense in [0,1]?

The answer to that seems no, not in general. Then my question was for what irrational a we can make S dense.

For that I was suggested that "num" kind of construction I gave above. But today I thought -how is it well -defined?

What is 100th decimal place there for example?

leslie townes

koro the recipe appears to be a garbled description of a property of a number that is intended (apparently) to make the property hold

one hoped to be equivalent to that property

a property of a number doesn't have a 100th decimal place. i'm confused by the question

Koro

07:18

Oh you mean since there are infinitely many ways to construct such a number, then we can choose one of those many ways to construct the number. And yes, with that "the question about 100th decimal place of num" is not a question anymore.

leslie townes

yep

Koro

:)

copper.hat

choose $a$ to be $ 0.110010001...$ then you can see that this not dense in $[0,1]$.

Koro

One stupid question: I know I am missing something. Consider a pattern 1234. Now as per our assumption (that is the way num was constructed above), num has repetition of 1234 and hence num could be rational also as there is repetition.

So num could be rational also.

nvm, I'll think about that.

copper.hat

what assumption?

Koro

07:25

copper, I had difficulty thinking in terms of binaries so I switched 2^n to 10^n.

copper.hat

i see things in black & white

Koro

here's the construction: imagine a number num=0.axby.... such that whatever combination of numbers 1,2,...,9,0 you can think of (repetition is allowed), it's all there in decimal expansion of num.

copper.hat

imgur.com/JOyEY2H

leslie townes

it might help to introduce some notations. e.g S_n(x) for the truncation of the decimal part of x to n places, thought of as a sequence of decimal digits (where you will need to make a choice to make this well defined when x has two decimal representations). S_n is a map from positive reals to {0,...,9}^n. if x is such that for every n, the restriction of S_n to {10^k x: k a positive integer} is surjective, then {10^k x: k a positive integer} will be dense mod 1

and you can write down examples of x with this property

the question is whether the converse holds

copper.hat

@Koro you can't put an uncountable collection into a countable one?

Koro

07:28

(Leslie understands my questions so well). :)

leslie townes

it might really help to talk about maps with definitions, and not "combinations" "being there" in a number

Koro

@copper.hat for that, we allow repetition.

leslie townes

the converse would follow if you could arrange arbitrary but fixed length decimal truncations of two numbers to be the same by just making sure the two numbers were close enough

this doesn't sound like a very interesting problem, whatever the outcome is

more linear algebra please

Koro

@leslietownes The construction of S_n makes sense. Thank you so much :).

Converse (if the set S ={ {10^n a}: n is natural} is dense in [0,1] then a is of the form num) also seems interesting but I'll think about that some other time. :)

unit 1991

Hi.Can anyone help with this problem?

0

Q: Question about doing $f$ on basis vectors.

We have $f:V_1\to V_2$ linear map and let's $(e_1,..,e_n)$ be basis for $V_1$ and $(c_1,..,c_n)$ be different basis for $V_1$ $(e_1,..,e_n)=(c_1,..,c_n)\Gamma_1$ where $\Gamma_1$ is change of basis matrix from $(c_1,..,c_n)$ to $(e_1,..,e_n)$. Then book says that from here follows that $(f(e_1),...

CowperKettle

09:42

> I went for a job interview and they asked me to state my biggest weakness in three words.
'Not very good at maths' I replied

BannedUser

09:59

@Osmium I can only prove for a sphere

I don't know about other surface

show me if you can

i think it's impossible to prove with basic surface integration

Osmium

10:16

See this: physics.stackexchange.com/questions/38404/…

BannedUser

10:28

Thanks I will try to digest that

soupless

11:38

This may be too simple, but what is the name of the theorem where $$\int_a^b \int_u^v f(x) g(y)\,dx\,dy = \left(\int_a^b f(x)\,dx\right)\left(\int_u^v g(y)\,dy\right)?$$

I noticed that $f(x)\,dx$ is constant when integrating in terms of $dy$ and $g(y)\,dy$ is constant when integrating with respect to $dx$.

BannedUser

14:40

@Osmium Thank you so much for the proof. Thank god there was a proof by geometry.

but the proof is bit sus...

steradian part is bit sus

Osmium

15:47

Glad I was able to help you. And it makes me more happy that I helped someone in physics, my field of interest. But my advice is that you learn multivariable calculus. You won't be able to understand the definition of potential, its relation with the field and many more things.

Osmium

16:16

Have you read the article on the wiki: google.com/url?sa=t&source=web&rct=j&url=https://…

Xander Henderson

steradian: short for radian on steroids.

But in French, which is why the word order is reversed (as with the natural logarithm, which is abbreviated $\ln$).

user193319

16:30

In the context of complex analysis, what are bounded and unbounded components?

leslie townes

of what?

for a subset A of C, "A is bounded" means there's M with the property that |z| < M for all z in A, and "A is unbounded" means "A is not bounded" in the sense just defined

with more context maybe we can say more or address a specific setting

user193319

Yes, a subset. In the theorem I am reading, I think the subset is taken to be open.

leslie townes

often in complex analysis one considers 'simple closed curves' in the plane. they have an open 'inside' that is bounded and an open 'outside' that is not

actually proving this from the definitions can be fairly difficult but it is easy to see in specific examples (e.g. circles)

for just an entirely arbitrary curve there is no reason why the complement of the curve has to have any bounded components, or have any unbounded components, or why it couldn't have more than one of both type, but you still might classify them according to whether or not they are bounded or unbounded

porridgemathematics

16:47

hi, does anyone know why in this answer, it is claimed that $\sum_{w} \frac{1}{|w|^3}$ converges, where $w = n_1 + n_2 \tau$ and $n_1,n_2$ run through integers such that $(n_1,n_2) \neq (0,0)$? Apparently showing $|w|^2 \geq C(n_1^2 + n_2^2)$ for some $C > 0$ is enough, but I do not see why

3

Q: Why is the modular $\lambda$ function a quotient of two meromorphic functions in the U.H.P.?

In Ahlfors' complex analysis text, page 278 it says: It is quite clear from (9) that $\lambda(\tau)$ is the quotient of two analytic functions in the upper half plane $\text{Im} \tau > 0$. Formula (9) is the definition of the Weierstrass $\wp$ function $$\wp(z;\omega_1, \omega_2)=\frac{1}{z...

complex-analysis modular-forms elliptic-functions

as far as I can tell, this just tells us this series is dominated by $\sum_{(n_1,n_2) \neq (0,0)} \frac{1}{(n_1^2 + n_2^2)^{\frac{3}{2}}}$ and I don't see why this converges

oh never mind, I see why it converges now :/

leslie townes

some counting and some estimates

yeah

if you have ahlfors around i think he does at least a few examples although maybe nothing in general

soupless

@leslietownes
Do you mind if I ask this question to you? https://chat.stackexchange.com/transcript/message/59858082#59858082

porridgemathematics

yeah im using ahlfors, he doesn't prove this at all though, he covers the convergence of a kind of similar series for the p function, but that just involves showing $\sum_{(n,m) \neq (0,0)} \frac{1}{(|n| + |m|)^3}$ converges, which is imo much easier than this

because the counting you need to do is simpler, and then you just compare to $\sum_{n=1}^{\infty} \frac{1}{n^2}$

oh wait. i am an imbecile , $\sqrt{n_1^2 + n_2^2} \geq C_0 (|n_1|+|n_2|)$ by equivalence of norms... then just use the 'easier' result I mentioned

where the counting is just there are $4k$ pairs $(n_1,n_2)$ s.t. $|n_1|+|n_2| = k$, hence our sum is basically $\sum_{k=1}^{\infty} \frac{4k}{k^3}$

the way I originally saw it was by comparing that sum to an integral..

leslie townes

soupless i don't know if that has a name. as you've phrased it it's just the fact that int a..b c f(t) dt = c int a..b f(t) dt applied twice. if you were referring to something like \int_D f(x) g(y) dA, where D is a rectangular region in the plane and the thing is defined as a double integral, then fubini's theorem is what lets you express that thing in terms of iterated single variable integrals

but if you've got it in iterated integral form already it's just the linearity of the integral

soupless

@leslietownes It's a weaker version of Fubini's Theorem, isn't it?

leslie townes

17:03

yes, even the thing you see in calculus books is a weaker version of fubini's theorem

it's one of the many consequences of results commonly called fubini's theorem

the iterated-integral thing in fubini makes sense even if your f(x,y) isn't of the form u(x) v(y)

soupless

I asked you about the average mass from the Cafe last night. Here is my take: let $x$ and $y$ be the BMI and height in meters, respectively. The average of the mass $(xy^2)$ where $16.9 \leq x \leq 29.9$ and $1.2192 \leq y \leq 1.778$ is, if I am right, $$\frac{1}{(29.9 - 16.9)(1.778 - 1.2192)}\int_{16.9}^{29.9}\int_{1.2192}^{1.778}xy^2\,dx\,dy$$

Wait, that's too much. I could've just taken the average of $x$ and $y^2$ independently

leslie townes

for a discrete data set it could just be a sum. if you had a continuous thing you'd have an integral like that for the average value of xy^2

soupless

@leslietownes I used an integral since mass and height are continuous. not the mass and height themselves, you know what i mean :)

porridgemathematics

17:26

its funny you mention u(x)v(y), it turns out that fubini-tonelli holds in arbitrary measure spaces as long as your functions are of this form (you can drop the sigma finiteness assumption )

so you get something even stronger

leslie townes

do you even need countable additivity of the measures?

i wasn't gonna go there, but let's go

porridgemathematics

havent thought about that

i believe you may still need it

but thats just because the proof in my heads uses approximation by simple functions.. and im guessing the limit theorems wont play as nicely with non countably additive 'measures'

leslie townes

some of that still works although sometimes you need an epsilon of room that won't go away

porridgemathematics

ah well, if you have monotone convergence, then it sounds like it could work?

leslie townes

i don't think you do have MCT, in general

Ted Shifrin

19:55

@XanderHenderson In French, most adjectives follow the nouns they modify. There are a few exceptions.

leslie townes

20:16

or so the teds shifrin would have us believe

Xander Henderson

@TedShifrin Sure, but I don't speak French, and was making s*** up. :P

Ted Shifrin

20:34

Well, you did inquire. I merely responded ... je n'ai que répondu.

UnderMathUate

20:54

I've been looking at this for about 10 min, and I'm stuck: Let $a, b \in \mathbb Z$ with $a \neq 0$. Prove that if $a | b$, then $a| (-b)$ and $(-a) | b$.

Ted Shifrin

You should not be stuck.

UnderMathUate

I know this should be very simple.

I'm not sure why I'm not seeing it.

I'll try rubber ducking

Ted Shifrin

Write definitions.

UnderMathUate

Assume $a|b$ such that $b = ac$ for some $c \in \mathbb Z$.

Ted Shifrin

What is rubber ducking?

Not such that. Say this means

UnderMathUate

20:58

Where you say your thought-process out loud to see where you're misunderstanding. I think the original idea was for it to be an inanimate object such as a rubber duck.

@TedShifrin Hm?

ok

Ted Shifrin

Oh, talking to the rubber duck.

UnderMathUate

$a | b$ means that $b = ac$ for some $c \in \mathbb Z$.

Ted Shifrin

Ok, that’s all you need. Look at the things you must show.

leslie townes

ted a great-aunt gave us binoculars. this has increased the amount of birding we do at the duck pond.

Ted Shifrin

Oh, very cool. Does munchkin binoculate?

UnderMathUate

21:00

First, I need to show that $a | (-b)$. This means that $-b = al$ for some $l \in \mathbb Z$.

Wait, let $l= -1$?

Ted Shifrin

Don’t use the same letter, @Under. $c$ already is in play.

UnderMathUate

Whoops

leslie townes

she does binoculate. they're a small pair, just her size. we used them to see a group of northern shovelers swim in a circle to create turbulence that brought food to the surface at the duck pond.

i didn't know that northern shovelers did this until much later in life. what a head start she is getting

Ted Shifrin

LOL

I still have no idea, earthly or otherwise.

leslie townes

under: this is nitpicky but for things yet proven it may help to adopt language that reflects that. "I need to show that a | (-b). This would require me to show that there is some integer d with -b = ad."

UnderMathUate

21:03

Ha, yes. I see why that helps.

leslie townes

even for purposes of talking to a rubber duck.

we have two rubber ducks. one is a child's duck and another is a piece of swag i got at a professional event with the name of a law firm printed on it. i don't know why they thought this was a good tchotchke to represent their law firm.

Ted Shifrin

I feel left out. I have cat toys, but no rubber duck.

leslie townes

maybe the law firm swag store was out of everything. "i'm sorry. no stationery. no, no tiny single-use umbrellas. no tide pens, no ballpoint pens. no crappy rubiks cubes. all we have left is the rubber duck."

UnderMathUate

It's ok, I don't have an actual rubber duck either. Just action figures.

leslie townes

the rubber duck can double as a cat toy. it's a little heavier than your average cat toy, but because of its odd shape and center of gravity, it rolls unpredictably, which makes it ideal.

ugh, this reminds me that today is a bath day for the daughter. these can go smoothly or very unpleasantly.

Ted Shifrin

21:12

Still easier than trimming Screech’s claws. It’s impossible.

leslie townes

we can do that only on fridays when we watch tv or a movie. after livvy sleeps about 30 minutes in my lap, i can grab livvy's paws and trim them without her waking up.

on weekdays she doesn't sit in my lap for longer than 5 or 10 minutes and never goes into a deep sleep. we have to be watching something.

Ted Shifrin

I have no solution.

leslie townes

get the vet to do it at yearly checkups, i guess? you get a few weeks of slightly duller claws, anyway.

they make those claw caps that can go on and stay on for longer, but olivia would never forgive me if i put her in those

Ted Shifrin

The vet wants her sedated if I take her in. Such trouble…

So, @Under, you got those proofs?

Balarka Sen

21:28

So if $|G|$ is coprime to characteristic of $k$, $kG$ is semisimple. What's an orthogonal family of idempotents, explicitly? Probably, for every conjugacy class $\mathcal{O}$ of $G$, you take $e = \sum_{g \in \mathcal{O}} g$

Ted Shifrin

Wow, this is like Balarka 1.5 days.

amWhy

@TedShifrin I like the name! That would be fitting, at times, for my cat; though he's grown into a "howler"!

Balarka Sen

?

Didn't understand the reference

Ted Shifrin

Balarka 1.0-1.5 was doing so much algebra.

Balarka Sen

I have an exam. I can't afford to be a 100 years if I want to pass.

Ted Shifrin

21:32

@amWhy I named her Screech when I thought she was a he, but the name still fits.

What does orthogonal even mean in that question, @Balarka?

amWhy

@TedShifrin Sweet!

Balarka Sen

Assume $k$ alg. closed. You know $kG$ is semisimple, $kG \cong \prod_{i = 1}^r M_{n_i}(k)$, the number of factors $r$ is number of conjugacy classes. Pick the identity matrix from each factor, call them $e_1, \cdots, e_r$. These satisfy $e_i^2 = e_i$, $e_i e_j = 0$

That's what I mean by an orthogonal set of idempotents

leslie townes

balarka that sounds right by loose analogy to stuff i understand that is not this.

Balarka Sen

I want to explicitly understand these inside $kG$

copper.hat

i hate unexplained downvotes.

Balarka Sen

21:34

@leslietownes It sounds like the right thing but I am getting confused trying to compute $e^2$

Maybe I am dumb

Ted Shifrin

Oh, thanks, @Balarka.

Balarka Sen

Maybe think dually, $kG$ is the space of functions $G \to k$, multiplication is convolution. I just look at the characteristic function of each conjugacy class $\mathcal{O} \subset G$

Ted Shifrin

You're saying that if you add the elements of a fixed conjugacy class that should be an idempotent? Can we do an example?

Balarka Sen

It sounds suspicious that $\chi_{\mathcal{O}} * \chi_{\mathcal{O}} = \chi_{\mathcal{O}}$

I don't think that works

Ted Shifrin

We probably need @Thor again.

I never knew this stuff at all.

Balarka Sen

21:42

I think we need to bring in characters or something

Akiva Weinberger

Wait - is the brachistochrone problem the same as finding the geodesic in a weird metric? 'Cause the speed of the particle is solely determined by the vertical position

(from conservation of potential+kinetic energy)

leslie townes

berkeley just phoned me asking for donations. who gave them my number?

Ted Shifrin

I don't quite see it, @DogAteMy.

They call me incessantly, and I never answer. I just give money mid-December every year. MIT doesn't usually harass me.

leslie townes

they didn't have my number for a long time. they must have bought some data somewhere.

Akiva Weinberger

Something like $ds^2=-\frac1{y}(dx^2+dy^2)$

Ted Shifrin

21:46

On the new iOS on my iPhone, I activated the option to send all unknown numbers to voicemail. It works wonderfully.

Akiva Weinberger

which differs from the metric of the half-plane mode in that the $y$ isn't squared

Ted Shifrin

You don't mean the $-$, DogAteMy.

Balarka Sen

Yeah OK this is the classic confusion. The indicator functions $\mathbf{1}_{\mathcal{O}}$ of the conjugacy classes of $G$ forms a basis of $Z(kG)$, but also the characters of irreps form a basis of $Z(kG)$. It's the latter that's an orthogonal family of idempotents.

The inner product $\langle \chi_1, \chi_2 \rangle$ of characters is actually convolution.

Ted Shifrin

So the usual Euler-Lagrange turns into the geodesic Euler-Lagrange?

It's additionally confusing that the brachistochrone is also a tautochrone, DogAteMy.

Balarka Sen

Or rather, $\langle \chi_1, \chi_2 \rangle = (\chi_1 * \chi_2)(1)$

Ted Shifrin

21:50

Physically it seems suspicious, DogAteMy. Because gravity isn't the only force acting. There's the normal force on the bead exerted to keep it on the wire.

Balarka Sen

One side is $1/|G| \sum \chi_1(g^{-1}) \chi_2(g)$, the other side is $1/|G| \sum \chi_1(xg^{-1}) \chi_2(g)$ evaluated at $x = 1$

This stuff is so hard to organize in my head and nobody writes a good account

Akiva Weinberger

22:06

@TedShifrin I do. This acts on the lower-half plane, where $y<0$

Ted Shifrin

Oh, with starting the bead at $y=0$?

Akiva Weinberger

Yeah

Ted Shifrin

Sort of a singular situation.

Akiva Weinberger

Yeah, we're drawing a geodesic from an ideal point

Balarka Sen

22:18

Better perspective: Let $(V, \rho)$ be a rep of $G$ and $(V^*, \rho^*)$ the dual rep, where $\rho^*(g)$ is the dual of the operator $\rho(g^{-1})$, the inverse is there to keep $G$-actions on the left.

Then $V \otimes V^*$ is a representation of $G \times G$.

There's a $G\times G$-equivariant map $V \otimes V^* \to k[G]$ where $k[G]$ is naturally a $G\times G$-module by left & right actions, given by $v \otimes \xi \mapsto \sum_{g \in G} \xi(\rho(g) x) \mathbf{1}_g$.

unit 1991

Hi.Where I can find proof of $cos^2mx=\frac{1+cos2mx}{2}$ ?

Ted Shifrin

Google double angle formulas.

Addition formula for $\cos$ is the obvious thing to do.

Balarka Sen

I meant $v$ in place of $x$

unit 1991

@TedShifrin Can you give some hints?

Balarka Sen

In other words, the function $G \to k$ corresponding to $v \otimes \xi$ just eats $g \in G$ and spits out $\xi(\rho(g) v)$, the $\xi$-th component of the vector$\rho(g)v$

Akiva Weinberger

22:28

@unit1991 Do you recall the formula for $\cos(A+B)$?

unit 1991

@AkivaWeinberger Yes

Akiva Weinberger

Try $A=mx$ and $B=mx$

Also useful: $\cos^2+\sin^2=1$ (and therefore $\sin^2=1-\cos^2$)

Ted Shifrin

Just do it with $m=1$ and then substitute when you're done.

unit 1991

@TedShifrin @AkivaWeinberger Thank you

Leaky Nun

23:43

@BalarkaSen why are they idempotent?

UnderMathUate

This is what I had come up with: Assume $a | b$ such that $b = ac$ for some $c \in \mathbb Z$.
i) Observe that $(-1) b = ac(-1) = a(-c)$. As $-c \in \mathbb Z$, so $a | -b$.
ii) Observe also that $(-1)b = -a(c)$

I'm not sure what to do about the second one. This would be stating $-a|-b$, right?

leslie townes

ha! yes, but you might need to cite a proof that (-1)b = -b. or this might not be the point. depends on the class and how deep down the integer rabbit hole they went.

or rewrite the proof so you don't need to compute (-1)b

UnderMathUate

23:58

I doubt I would need to go that far for this particular class. At least at this point -- at the beginning, we had a discussion about why those kind of manipulations were ok, but that was pretty much it.

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