Mathematics - 2024-01-17 (page 1 of 2)

Dannyu NDos

03:05

Is there a notable equivalent condition to a fundamental group to be abelian?

Sine of the Time

09:23

why all these messages are starred?

Thomas Finley

09:53

Hey there! I will be recently starting on multivariate calculus. Can someone suggest any good book of it? I mean a book that is good for beginners and provides good explanations.

robjohn

@SineoftheTime Someone went star-happy. I have removed all of these stars.

rogerroger

10:54

can someone tell me if this is a typo and the last term should be $\mathbb{P}\big(G^1 \in \mathcal{G}^{-1}_{\Delta t,x_0}(S)\big)$? here $P_{\Delta t}(x^0,S)=\mathbb{P}\big(x^1 \in S | x^0\big)$

Ryder Rude

12:49

hi
if we have a group-type algebra with closure, unique identity and unique inverses. does this imply associativity?
if we have the first three properties, then each row of the multiplication table is a permutation of elements
and permutations are associative
so do these three properties imply associativity??
then y do we hav associativity as a separate axiom??

Soumik Mukherjee

Xander Henderson

What is a "group-type algebra"?

Ryder Rude

an algebra with the three properties i wrote

Xander Henderson

What is an "algebra"?

Ryder Rude

a bunch of elements and a multiplication

Xander Henderson

12:52

Usually, an algebra is a set over a field (or a ring) with scalar multiplication, addition of elements and multiplication of elements...

Ryder Rude

oh. maybe i just meant a set with a binary operator

if the operator has these three properties, is associativity implied?

Xander Henderson

No.

Consider exponentiation over the integers. Or multiplication in the octonians.

Ryder Rude

im considering discrete and finite grps.

with these properties, is each row of the multiplication table a permutation of elements? @XanderHenderson

let's consider a row

corresponding to element 'a'

ab=ac wud imply b=c

using the properties

so this means all of $ag_i$ are different for all $i$

Soumik Mukherjee

How to reply to my own message in chat?

Ryder Rude

this means $a$ can be identified as a permutation of $g_i$

and permutations are associative

so i think the conclusion only holds for discrete and finite groups?

so this is y we need the associativity axiom?

Jakobian

12:59

@SoumikMukherjee write : followed by last digits in transcript

Ryder Rude

pls help

Soumik Mukherjee

@SoumikMukherjee Ok it works

Ryder Rude

exponentiation over integers wud not be closed for finite sets

Soumik Mukherjee

@Jakobian Thank you

Jakobian

@rogerroger this is a random set of notations

I'm not sure if you're expecting an answer

Xander Henderson

13:06

@RyderRude Why not? Consider exponentiation modulo $n$.

Jakobian

@RyderRude This is called a loop and loops don't have to be associative

Xander Henderson

3

Q: Existence of finite, non-associative group-like structures

Do there exist sets $G$ with a composition such that $G$ is finite. There is a two-sided identity element $e\in G$ such that $eg = ge = g$ for all $g\in G$. Each $g\in G$ has a unique two-sided inverse $g^{-1}$ with $gg^{-1} = g^{-1} g = e$. For all $g$ and $h$ in $G$ there exists a $k$ in $G$ ...

abstract-algebra

rogerroger

ok I drop some of the unnecessary notation: they say $\mathcal{G}(x)= y$, so that $\mathbb{P}(x\in S)= \mathbb{P} (y \in \mathcal{G}^{-1}(S))$

Jakobian

Then yeah some part is wrong. Not sure which one

Quasigroup

In mathematics, especially in abstract algebra, a quasigroup is an algebraic structure resembling a group in the sense that "division" is always possible. Quasigroups differ from groups mainly in that the associative and identity element properties are optional. A quasigroup with an identity element is called a loop. == Definitions == There are at least two structurally equivalent formal definitions of quasigroup. One defines a quasigroup as a set with one binary operation, and the other, from universal algebra, defines a quasigroup as having three primitive operations. The homomorphic image of...

@RyderRude

@XanderHenderson algebra can also mean universal algebra, perphaps thats the intended meaning here

Xander Henderson

Perhaps, but someone asking such a basic question probably doesn't have that in mind.

But that is why I asked the question "What is an 'algebra'." :D

Jakobian

13:19

Its not as uncommon as you think

Its just a set with operations on it after all

Xander Henderson

Again, that is why I asked the question.

Soumik Mukherjee

@SoumikMukherjee Any hint on this anyone?

user123234

does someone know about simply connectedness?

Xander Henderson

@user123234 Nope. Only complex connectedness.

user123234

this was a joke right? @XanderHenderson

I have namely a question. I have given a simply connected group $G$. And I could show that some property holds on $H\subset G$. Why can I then say that it holds for $G$ because of simply connectednes? @XanderHenderson

nickbros123

13:26

@ThomasFinley serge Lang I would say, for a gentle intro

Jakobian

@SoumikMukherjee Write $D$ as image of $\mathbb{R}^3$ using Vieta's formulas

Thorgott

@user123234 that depends very much on the property

and probably also the subgroup $H$

user123234

let me write it more explicitly

We have given two simply connected closed matrix groups $G_1,G_2$ and let $g_1:=\{X: \exp(tX)\in G_1~\forall t\in \Bbb{R}\},g_2$ be the corresponding lie algebras. Then we have shown that there exists two smooth group homomorphisms $\phi:G_1\rightarrow G_2$ and $\psi:G_2\rightarrow G_2$ s.t. $\phi\circ \psi=id$ on $\exp(g_1)$ and $\psi\circ \phi=id$ on $\exp(g_2)$. But then they wrote that by simply connectedness we know $\phi\circ \psi=id$ on $G_1$ and $\psi\circ \phi=id$ on $G_2$ @Thorgott

Ryder Rude

13:43

@XanderHenderson @Jakobian thankss

i cant understand something. if the multi table is like a sudoku, then each element can be identified with a permutation group element

and permutations associate??

Jakobian

Not true

Ryder Rude

permuatations dont associate?

Jakobian

Fixing $x$, $y\mapsto xy$ and $y\mapsto yx$ are permutations in a quasigroup. But there's no obvious way to identify $x$ with a permutation

Xander Henderson

Well, then it must be true...

Jakobian

What must be true

Ryder Rude

13:53

can the pre multi and post multi separately be identified with one permutation each?

i can create a permutation matrix

thats what i meant by identification

Jakobian

They are permutations

nickbros123

@ThomasFinley you seem to be quite well equipped with the math prerequisites, so you can skip serge lang and look into quite a lot of options: i personally liked $\text{hubbard}^2$. shifrin is also great (I proved a problem i had in the back of my head for a few months by an intuition developed by shifrin regarding block matrices. problems are great too) unfortunately, i havent really read much into the other general recommendations

Ryder Rude

@Jakobian ok so we can make permutation matrices for x, y and z. and these matrices associate

Jakobian

You can try to identify each element $x$ with pair $(\rho(x), \lambda(x))$ of right and left multiplication and that might give you some description similar to "groups are permutations" theorem, but it won't be a group in general, because it can't

Ryder Rude

if i identify them $x,y,z$ with matrices, then x(yz) g_i= (xy) z g_i for all g_i. does this part hold?

we can think of g_i as a list of grp elements and we r applying permutations from the left multiplication

Jakobian

13:58

Uh...

Ryder Rude

[g_i] is the list, then (xy) z [g_i]=x(yz) [g_i]

Jakobian

You're making no sense

Ryder Rude

and then we can right multiply by the inverse of g_i to get (xy)z=x(yz)

Thorgott

14:11

@user123234 I don't think that's needed. The sets $\exp(g_1)$ and $\exp(g_2)$ are neighborhoods of the respective identities, hence are generating sets (this needs only connectedness). But two group homomorphisms that agree on a generating set are equal.

user123234

@Thorgott I see that $\exp(g_i)$ are neigborhoods of the identities, but why are they generating sets? So what does this mean?

Thorgott

that's a general fact. take a symmetric neighborhood $U$ of the identity in a Lie group $G$ (symmetric meaning $U=U^{-1}$). try showing that the subgroup generated by $U$ is both open and closed in $G$, hence equal to $G$ if $G$ is connected.

user123234

ah okey perfect thanks a lot

Ryder Rude

ooh i think you r right. i may map each element to a permutation matrix. but the multiplication of these matrices does not correspond to the multiplication of the elements @Jakobian

so i have not really represented the group

thanks

Soumik Mukherjee

14:35

@Jakobian okay so $(x,y,z) \mapsto (-x-y-z, xy+yz+zx, -xyz)$. If this map is surjective then $D$ is continuous image of a connected set, hence connected. But how to show surjectivity?

Jakobian

@SoumikMukherjee by definition

Soumik Mukherjee

14:51

Oh right

Thank you @Jakobian

Ryder Rude

en.m.wikipedia.org/wiki/Cayley's_theorem in fact the proof of this theorem uses the associativity of the group as an assumption

after mapping the elements to permutations, we have to use associativity to show that the permutation matrix multiplication is a representation of the group

the book i was reading handwaved this proof by omitting this part @Jakobian

Jakobian

@RyderRude yeah. If you don't then you need to introduce additional factors in your multiplication

Thomas Finley

15:39

@nickbros123 Thanks a lot for your suggestions! They were indeed helpful.

Ryder Rude

16:10

what do u do in free time

robjohn

16:35

free time? time is money!

Ryder Rude

@robjohn yeah. it's like a resource. do u try to use all of it in productivity?

Jakobian

16:52

I don't want to think of time as a resource

Ryder Rude

whats ur view

psie

Basic question. Suppose I'm given a periodic function and the definition of the function is only given for some interval to the right of the origin, say $[0,1]$ with period $1$. How can I then determine whether or not the function is even or odd?

robjohn

$f(x)=f(1-x)$ is even; $f(x)=-f(1-x)$ is odd (to spelll it out)

Xander Henderson

@psie What @robjohn said.

Sine of the Time

@psie @xander confirmed what @robjohn said

leslie townes

16:59

@psie what sine of the time said

psie

jesus

said nothing

leslie townes

no, he was more into weeping

Xander Henderson

@robjohn I like spellling.

robjohn

@XanderHenderson that may bee

leslie townes

psie so stepping back just slightly, the question is directly about a property of the [unique] periodic extension of f, not f itself. robjohn's formulas implicitly invoke what the formula for that extension would have to be on [-1,0] (if you call it g to preserve the distinction from f, g(t) for t in that range is forced by periodicity to be f(t+1))

its also very important that you said "with period 1" there, the extension would not be unique if you allowed for a different period. it's not unusual for fourier series treatments to also consider extending a function defined initially on [0,L] to a 2L-periodic function on [-L,L], where you suddenly have choices in how you do this, and some room to make "the" [now more appropriately "an"] extension either even or odd (or at least as close to odd as would matter for the fourier series)

psie

17:10

good points

CroCo

Dear all, I would appreciate some clarification regarding this notation in convex optimization. What is the mathematical meaning of "argmin" in this notation
$$
x = \text{argmin } f(x) \text{ subject to } Ax=b
$$
I'm familiar with this notation though
$$
\min_{x} f(x) \text{ subject to } Ax=b.
$$

leslie townes

it is a bad notational idea to denote both that argmin and the implicit variable on the RHS by the same letter "x"

the "min" is the smallest value of f subject to that constraint, the "argmin" is the value of x that realizes the smallest value of f subject to that constraint

neither of these notations making much sense when these things don't exist or aren't unique

croco as an example, replacing the linear constraint with a simpler constraint, if f(x) = (x - 2)^2 then min_{x in [0,10]} f(x) is 0, while argmin_{x in [0,10]} f(x) = 2, and f(2) = 0

CroCo

17:26

@leslietownes thank you so much for this informative answer. In a paper I'm reading, this notation shows up

Ted Shifrin

Note that the arg in argmin means argument (which is the input variable)

leslie townes

ooh, there it seems like they're maybe using argmin to denote the set of all places where the minimum is realized, to allow for some non uniqueness

Ted Shifrin

It does seem.

CroCo

where $u$ is the control input to the robot. My supervisor wasn't happy about it and he said why you're using this notation. Is the solution mulivalued?

I don't have proper answer for his question.

Ted Shifrin

Well, the minimum can occur multiple times, unless you have some sort of convexity or something else to prohibit that.

leslie townes

17:29

if there's more than one potential minimizer, or at least no clear reason why there could only be only one minimizer, then "arg min" would indeed be (potentially) multivalued

people definitely abuse this notation all the time (e.g. writing "=" instead of "\in" when argmin is set valued)

Ted Shifrin

Ugh, they do?

leslie townes

yes ted. i tried to stop them

CroCo

@TedShifrin isn't the quadratic function convex?

leslie townes

the further you get from the math department, the less they care :(

Ted Shifrin

Do you know $\beta>0$?

CroCo

17:32

yes

Ted Shifrin

Then you win.

Sine of the Time

what's the award?

Ted Shifrin

@CroCo I assume the independent variable is $t$?

What are you offering, Sine?

Sine of the Time

3k rep points

Ted Shifrin

That's a lot of winning.

CroCo

17:34

In robotics, $t$ is time and yes it is independent

@TedShifrin What do you mean by "unless you have some sort of convexity"

Ted Shifrin

Nonconvex functions can certainly have multiple minima.

CroCo

My question why my supervisor raised this question "is the solution multivalued?"

Hasini

Hi

Ted Shifrin

Actually, being quadratic in this case will not make it strictly convex.

What if $\dot q=0$ on an interval, for example?

Hasini

Usually we regard set union, intersection, difference and complement as set operations. Can we also say that "subset", "set equality" are operations?

Ted Shifrin

17:37

No, those are not operations.

Jakobian

no, they're relations

Ted Shifrin

An operation is a function from the set (perhaps two or more copies of the set) to itself.

Jakobian

there is a division between an $n$-ary relation and $n$-ary operation

Hasini

Okay thanks a lot @TedShifrin and @Jakobian

But complement is an operation right?

Jakobian

an $n$-ary relation is a subset of $A^n$, and an $n$-ary operation is a function from $A^n$ to $A$

Ted Shifrin

17:38

Yes. Check the definition I just gave you.

Jakobian

yes, complement is an unary operation, meaning $n = 1$

Ted Shifrin

Or that Jakobian just gave you formally.

Oh wait.

Hasini

Okay. Two or more copies means - product of sets?

Jakobian

to contrast with binary operations, meaning $n = 2$, that you gave examples of

@Hasini disjoint union?

Ted Shifrin

Complement is a map from the subsets of $A$ to subsets of $A$.

That doesn't fit our definition.

CroCo

17:39

@TedShifrin the control inputs must be zero then (i.e. $u=0$)

Jakobian

$A = \mathcal{P}(X)$ for a set $X$

Ted Shifrin

But is that an operation on $X$?

Jakobian

complement is an unary operation on $A$

Hasini

@Jakobian Disjoint union of copies?

Ted Shifrin

But that is not an operation on the underlying set $X$.

Jakobian

17:40

@Hasini disjoint union of sets is a way to make copies of those sets

Ted Shifrin

Be careful here.

Jakobian

I'm not saying that it is

the same thing applies to all the other operations Hasini mentioned

Ted Shifrin

I think Hasini intended the set to be the set he started with.

Oh, right.

Hasini

@Jakobian Okayy, thanks a lot again!!

Ted Shifrin

Very confusing, actually.

So I did not respond right at all.

You have to specify that $A=\mathscr P(X)$ where $A$ is the set you started with.

Hasini

17:42

Some relations are functions. Like one to one relation. then there can be instances where equality and subset can be operations too?

Jakobian

@Hasini disjoint union of sets $A_i$ can be realized as the set $\sqcup_i A_i = \bigcup_i A_i\times\{i\}$

Ted Shifrin

How is equality or subset an operation? What operation are you performing?

Jakobian

You can check that $A_i\times \{i\}$ are all disjoint, even if $A_i$ themselves aren't

@TedShifrin no, relation

Ted Shifrin

I was asking Hasini, not you.

Hasini

:)

Jakobian

17:43

ah

Hasini

But thanks a lot @Jakobian

Pls forgive for not being able to answer at once @TedShifrin

But @Jakobian's response has been helpful as I couldn't answer at once

:)

Jakobian

@Hasini For a relation $R$ to be a function we need for all $x$ to exist unique $y$ such that $xRy$. For subset relation this means that each subset $y\subseteq X$ would have to have a unique subset $x\subseteq y$.

When can this happen?

Same question applies but for equality

Hasini

Okayy, I think I get it now.

Thanks a million, both of you!!!

Have a nice day!

Ted Shifrin

Writing $x$ for a subset of $X$ is unacceptable.

Jakobian

its acceptable in my mathematical circles

if we're thinking of $x, y$ as elements of $A$, then we might as well use lower case

Ted Shifrin

17:50

When there is a set $X$, writing a subset as $x$ is unacceptable notation. It will confuse everyone (except you, of course).

CroCo

@leslietownes could you please elaborate on "uniqueness" in this context?

Ted Shifrin

Similarly, students who map $X\to Y$ and send $y\in X$ and $x\in Y$ are asking for trouble.

Jakobian

I don't think its confusing

Ted Shifrin

@CroCo This response doesn't help me. How does that say you cannot have more than one minimizing value of $t$?

Jakobian

the latter, yes, the former, no

Ted Shifrin

17:52

We will have to agree to disagree strongly.

This comes back to clarity when explaining things to others — i.e., teaching.

Jakobian

there's a difference between $y\in X$ and $x\in Y$ to using lower case for sets

Thorgott

honestly, I think that is less bad

lower case for sets in such a context is indeed unacceptable

Jakobian

everything is a set, doesn't matter

its a psychological obstruction

Ted Shifrin

I'm fine with letting $x,y\in X$ when there is no underlying set $Y$ in the picture.

Thorgott

yes, it's a matter of clarity

Ted Shifrin

17:55

My experience with students over 50 years indicates that choosing incompatible letters invariably makes them mess things up.

Soumik Mukherjee

@Jakobian proper class left the chat

Ted Shifrin

We're not talking about perfectly pedantic advanced thinkers like Jakobian. But in this room we do not always communicate with Jakobians.

Thorgott

set theory texts especially are careful with using lower/upper case, caligraphy, fraktur font, etc.

to distinguish at which "level" of containment one is operating

Jakobian

maybe the ones that allow atoms

Ted Shifrin

But most humans struggle to write (some) Greek letters and (most of) fraktur :D

Xander Henderson

17:57

A function is $1$-periodic if $f(x) = f(x+1)$. It is odd if $f(x) = -f(-x)$. Combining these gives $f(x) = -f(-x) = -f(-x+1) = -f(-(x-1)) = f(x-1)$. So an odd, $1$-periodic function satisfies $f(x) = f(x-1)$. Or, if you prefer, $f(x) = -f(-x+1) = -f(1-x)$ (which is what @robjohn wrote).

Thorgott

no, no one does atoms

Jakobian

some people like to think of set theory that way, I met a few

Xander Henderson

Huh... I hit enter on that comment ages ago...

Jakobian

maybe they don't study them

Xander Henderson

Chat is borked.

@SoumikMukherjee This room has never had any class, anyway---proper or not.

@Jakobian You are clearly not the typical student.

Soumik Mukherjee

18:00

@XanderHenderson Maybe gently press the comment instead of hitting

Jakobian

@XanderHenderson a typical student might think $x$ and $X$ denote the same object

I'm not really concerned

Xander Henderson

You seem to be someone who has excelled in mathematics throughout your academic career, thus it seems difficulty for you to empathize with those who struggle with these ideas. Writing $x \in Y$ and $y \in X$ is absolutely going to confuse people. Writing $x \subseteq X$ is bad notation for most purposes.

Thorgott

no they don't

Ted Shifrin

@Thorgott Huh?

Oh, the thinking of set theory comment?

Xander Henderson

@TedShifrin Fraktur is the devil. Burn it.

Burn it to the ground.

Thorgott

18:03

the typical student does not think $x$ and $X$ might denote the same object*

Ted Shifrin

A very old German person may erase you for that, @Xander.

Jakobian

Might think isn't the same as, does

Xander Henderson

Then bury the ashes at the bottom of the Mariana Trench.

Ted Shifrin

I'm not sure that Jakobian has any idea what a "typical" mathematics student is or can grok.

In fact, I'm quite sure he hasn't.

Xander Henderson

@Thorgott Honestly, a lot of my students don't seem to understand that $a$ and $A$ are different symbols. That is usually one of the first discussions I have to have with them after grading their first written assignments.

Jakobian

18:05

@Thorgott "does" should be "might" and "might" should be "does"

Thorgott

well, I don't think "typical student" should mean "typical student in the first week of college"

Xander Henderson

They get really spooked when I show them the notational index of my thesis, which includes five or six different versions of the letter "D".

Jakobian

then you've said what I meant

robjohn

$\mathfrak{\text{The Devil}}$

Ted Shifrin

That makes the devil far more devilish.

Hasini

18:07

If this is because of the question I asked, I apologize. Please don't argue

Xander Henderson

@Thorgott I don't know that "typical student" is a good standard, is all I'm saying. What is "typical" is going to depend a lot on experience, as well as how successful those experiences are. But my first year students here are significantly weaker (coming through the door) than my first year students in grad school.

Hasini

Both @Jakobian's and @TedShifrin's comments have been useful

Thanks a lot!!!

leslie townes

@CroCo a good example might be min_{x in R} cos(x) = -1, but this minimum value is realized for many different values of x (any odd multiple of pi). so in a context the notation is only defined when there is a unique minimizer, "argmin_{x in R} cos(x)" is not defined, and in a context where "argmin" can be set-valued, it's the set of odd multiples of pi

Ted Shifrin

Sorry about all the static for you, @Hasini. :)

Soumik Mukherjee

@XanderHenderson are DX and XD included?

Jakobian

18:07

@Hasini Its just that people here seem to have an obsession with "not confusing the student"

and apparently what I wrote seems to fall into the category of "is confusing to a student"

Hasini

Hmm, well it was the answer given to me

And I was not confused

And in fact I got very useful info

Jakobian

good enough for me

Hasini

From both of you

So then I hope everything is alright?

:)

Ted Shifrin

Oh, everything's fine, Hasini. Don't worry about it.

leslie townes

@CroCo and in the latter case it would not make sense to do arithmetic or perform any analysis on "argmin_{x in R} cos(x)" that potentially depended on which odd multiple of pi you were talking about

Hasini

18:09

Hee hee :)

Okayyy :)

Xander Henderson

Oh, sh*t... I just noticed that there is fraktur in my thesis. What have I done?!

Hasini

cool

Ted Shifrin

I think @CroCo has a good question (from his adviser). Can we or can we not argue that there is a unique minimum point?

Xander Henderson

There are also a couple of commutative diagrams in there. They make me feel dirty.

Ted Shifrin

There certainly was in mine, @Xander, cuz of Lie algebras.

Xander Henderson

18:11

$\mathfrak{M}$ is for Minkowski content. :/

Ted Shifrin

Well, it's either that or $\mathscr M$.

Xander Henderson

robjohn

@XanderHenderson Evil Thesis

Xander Henderson

Better burn it.

Ted Shifrin

There's a particularly Chern-like page in my thesis which will make Xander faint or worse.

robjohn

18:13

@TedShifrin depends on the function; did I miss the function?

Soumik Mukherjee

@XanderHenderson Is there a typo? s instead of q?

Xander Henderson

@SoumikMukherjee Why $s$?

Ted Shifrin

@robjohn This.

leslie townes

@TedShifrin the question was from a usage in a paper croco was reading. i think more context would be needed (the argmin was over "u" and the symbolic expression that the argmin was taken over didn't expressly involve u at all, but other variables)

Xander Henderson

Oh, yes. There is a typo. Damnit... another one.

Soumik Mukherjee

18:15

In the "The $s$ dimensional Upper Minkowski content" line.

Ted Shifrin

Oh, that's why I asked if $t$ was the independent variable, @leslie.

Xander Henderson

That's two that I've spotted in the last 10 minutes.

This is why I never look at this document... :(

YOU KNOW WHAT I MEANT!

leslie townes

well, i don't think it's possible to reverse engineer what the t-dependence is from that notation alone

we could probably say more with a link to the paper in question

Xander Henderson

One of FIVE commutative diagrams in my thesis.

Thorgott

I have more

Xander Henderson

18:18

Five is too many. More should earn jail time.

Ted Shifrin

I have lots lots more. I'm trying to post the page for Xander. I'm having technical difficulties.

leslie townes

i was gonna say, your advisor kinda dropped the ball on "could we maybe do this without any commutative diagrams, we're so close"

Ted Shifrin

Xander Henderson

@leslietownes Yeah, but I spent so much fckng time trying to read Tate's thesis and other related papers that it kind of seemed like a waste if I didn't get to discuss those constructions. :(

Ted Shifrin

Expressly for Xander.

Xander Henderson

18:22

@TedShifrin ... lovely.

Thorgott

beauty

Ted Shifrin

That page was done long, long before we had LaTeX ourselves. I think I even carefully proofread it when it was published in Transactions AMS.

Xander Henderson

@Thorgott BURN IT WITH FIRE!

leslie townes

ted: to think some people call that geometry. where's the triangles?

Thorgott

Ted Shifrin

18:23

Yeah, there sure aren't triangles in that!

Thorgott

oh wait, this one was bigger

Xander Henderson

(Yes, there is a typo.)

Ted Shifrin

@Thor For things like that Spivak invented LAMS-TeX. LaTeX can't do anything at all like that.

Thorgott

I use tikzcd for diagrams, works like a charm

Ted Shifrin

Ah, I never learned tikzcd.

I think I won't ever have to, now.

Thorgott

18:25

I wish it was supported on MSE

Ted Shifrin

@XanderHenderson You mean the missing $^2$?

Xander Henderson

@TedShifrin Yup.

Thorgott

any time you have to draw diagrams on MSE is a nightmare

Ted Shifrin

Yup, it is.

leslie townes

xander: who drew the drum? you?

Ted Shifrin

18:27

Surely he stole it.

Xander Henderson

@leslietownes My mother.

She is a freelance technical illustrator.

Ted Shifrin

Oh wow.

Xander Henderson

She illustrated this book, for example: amazon.com/Life-Pueblo-Understanding-Through-Archaeology/dp/…

(That's my sister on the cover.)

Ted Shifrin

Very cool.

leslie townes

it looks great. could see it in a patent, if people were still patenting drums

psie

18:36

Can the integral $\int_0^\pi \frac{\sin x}{x} dx$ be solved analytically? This integral times $2/ \pi$ appears as the limit of the partial sums of the Fourier series $S_n$, evaluated at the maximum, of an (odd) square wave function. I just want to see if I could possibly verify its value...

Xander Henderson

@psie What do you mean by "solved analytically"?

robjohn

@psie Do you mean is there a closed form?

Xander Henderson

Typically, integrals can be evaluated, but not solved. I presume that you are asking for some closed form in terms of elementary functions?

psie

yeah, not numerically.

Sine of the Time

Trigonometric integral

In mathematics, trigonometric integrals are a family of nonelementary integrals involving trigonometric functions. == Sine integral == The different sine integral definitions are Note that the integrand sin ⁡ ( t ) t {\displaystyle {\frac {\sin(t)}{t}}} is the sinc function, and also the zeroth spherical Bessel function. Since sinc is an even entire function (holomorphic over the entire complex plane), Si is entire...

Xander Henderson

18:39

@psie It can be expressed in terms of the $\operatorname{Si}$ function, but, like, what do you mean by an "analytic" expression?

robjohn

@psie I evaluate things like that in this answer where I look at the overshoot in Gibb's Phenomenon

Xander Henderson

@robjohn Where, in that answer, does the bass drop?

CroCo

@leslietownes @TedShifrin this is the part of the paper that talks about the optimization I've mentioned earlier.

Since I'm using same controller in my work, my supervisor raised the question about "why $u \in arg min$ and he said is the solution multivalued?"

I don't have proper answer but I feel yes we have multiple minimizers since the robot is redundant (i.e. more degrees of freedom) and there are some control inputs that drive the robot to the desired configuration. I'm not sure about this answer.

I would like to prepare for the answer for our next meeting.

But I feel I'm lacking some fundamental knowledge in the optimization.

The same authors use $\min_u$ in their other papers for the same controller. :<

leslie townes

18:59

croco: i can't even guess at how the function being minimized depends on the "control vector," this must be implicit from context other than this excerpt. i note that for a moment (the use of "a" in "a control vector . .. that ensures . . ."), the author does not sound as though there is any claim made about uniqueness of the minimizer, although other language (the use of "the" in "the solution to (2)") is maybe suggestive that there is only one minimizer

and i would never infer anything about authorial intent from this kind of prose alone

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$Mathematics$ Mathematics