Mathematics - 2022-02-13

copper.hat

00:56

=p

leslie townes

:P

copper.hat

i'm stuck. no witty retort.

Ted Shifrin

;Q

copper.hat

my daughter turned 21 today.

scary how quickly the time goes

Ted Shifrin

01:18

I turned — agh — 70 – 1.

Your daughter joins Lincoln, Darwin, and Ted.

leslie townes

i'm 29

Ted Shifrin

You mean 39.

leslie townes

i'll settle for 39. i like to benefit from the anchoring effect

Ted Shifrin

Wasn’t Jack Benny permanently 39?

leslie townes

yes. one of my idols, and, perpetual contemporaries.

Ted Shifrin

01:31

Your violin playing as good as his?

leslie townes

no.

:D

copper.hat

Happy Birthday Ted!!!

17

One of my brothers flew to London to meet his daughter & my daughter and brought them our for dinner to celebrate.

Ted Shifrin

That was nice of him!

MathIsLife12

02:00

54

Q: Why does the median minimize $E(|X-c|)$?

Suppose $X$ is a real-valued random variable and let $P_X$ denote the distribution of $X$. Then $$ E(|X-c|) = \int_\mathbb{R} |x-c| dP_X(x). $$ The medians of $X$ are defined as any number $m \in \mathbb{R}$ such that $P(X \leq m) \geq \frac{1}{2}$ and $P(X \geq m) \geq \frac{1}{2}$. Why do the ...

Why is the first answer valid? Im still not seeing how it’s differentiable almost everywhere

If this is the wrong chat to do this sorry

copper.hat

02:46

@MathIsLife12 here is my version math.stackexchange.com/a/4342866/27978

@MathIsLife12 The function (of $c$) is convex everywhere, so is locally Lipschitz and hence differentiable ae.

Lucas Henrique

03:01

hey chat

quick question: what does "differentiating twice gives zero" mean in this context? my professor was defining R-module complexes and there's this $\mathrm{im}(d_{i+1}) \subseteq \ker(d_i)$ condition. he said that they are called differentials because "differentiating twice gives zero". I also recall this when listening something about differential forms and de Rham cohomology

MathIsLife12

@copper.hat Beautiful answer. My question is, is this the same thing has saying the expected value of absolute value of x-m is less than the expected value of the absolute value of a for ALL a

leslie townes

lucas, that condition tells you that $d_i \circ d_{i+1} = 0$

MathIsLife12

the countable discontinuities would get in the way

leslie townes

if you think of d_i and d_{i+1} as the same kind of "differentiation" (i'm not sure this is always helpful), then doing one and then the next is "differentiating twice," i guess

copper.hat

03:43

@MathIsLife12 I don't understand your question, what is $a$?

MathIsLife12

@copper.hat any real number. E[ |X-m| ] < E[ | X-a| ] for ANY real number a

@copper.hat that’s what I’m trying to prove. However, if differentiable almost everywhere is the best we can get, then I don’t think my statement is true

copper.hat

Well, $E[|X-m| ] \le E[|X-a|]$ for any $a$. It is $\le$ not $<$.

Definitely differentable ae.

MathIsLife12

@copper.hat Yes ok that’s good. But is this statement the same as the statement you proved?

copper.hat

Which statement?

MathIsLife12

The median minimizes E[ | X- c| ]. Is this statement the same as my statement with the inequality?

copper.hat

03:52

It says that $m$ is a median iff $E[|X-m| ] \le E[|X-a|]$ for all $a$.

There might be many medians.

MathIsLife12

@copper.hat Yes ok ok. I guess I’m just not understanding how differentiable almost everywhere implies this will work for ALL real numbers a

copper.hat

Are you asking about Did's answer?

MathIsLife12

@copper.hat Yes

I understand yours

copper.hat

Let me read his again.

MathIsLife12

Thanks so much. I’m trying to see if his answer also proves the for ALL real numbers a thing

copper.hat

03:57

So, the thing he does not make clear is that $u$ is a convex function and if $u'$ exists, it is non decreasing.

MathIsLife12

Yeah. I guess my question boils down to, if we have a function f and we want to prove f(c) \leq f(a) did all

for all a

Then does differentiable almost everywhere still guarantee that can happen

copper.hat

So, if $c$ is a minimiser, then if $u'$ exists for $x<c$ we must have $u'(x) \le 0$ and for $x>c$ we have $u'(x) \ge 0$.

MathIsLife12

ahhhhhhh

Thank you so much

copper.hat

In the other direction, if $c$ satisfies $u'(x) \le 0$ for $x<c$ and $u'(x) \ge 0$ for $x>c$ (where it exists) then $c$ is a minimiser.

This is where convexity is crucial.

I like convexity :-).

MathIsLife12

No wonder convex optimization studied so much more than concave optimization …..

lol

copper.hat

04:03

:-)

concave is studied more in Austrailia

MathIsLife12

That was a good one

copper.hat

I don't really like my answer, in particular the little l,e,g algebra at the end.

the answer is straighfroward is you are comfortable with the convex subgradient.

But many folks are not used to computing the subgradient of an integral function, so I basically did that in my answer, but did not mention subgradient.

MathIsLife12

Yeah I used to want to go into convex optimization but I quickly changed my mind

copper.hat

i like it. but if you learn that before normal optimisation then it can throw you a littte, sort of like learning in an automatic before a manual

Koro

@TedShifrin wishing you a very happy Birthday professor Ted. :)

copper.hat

04:08

it was my daughter's 21st today as well :-)

Koro

@copper.hat Very happy birthday to your daughter as well :).

copper.hat

thanks Koro :-)

i sent her a bottle of champagne for her birthday

Koro

Is that alcohol?

I'm sorry I don't know about drinks :(. I never wished to have one.

copper.hat

yes, a sparkling wine, typically used for special events

since i can't be with her...

Koro

copper, what is implicit function theorem?

It's there in multivariable calculus and I haven't yet studied this.

I plan to do it soon.

Just quickly wanted to know what we do with it.

Only maximization and minimization?

copper.hat

04:20

no, it is a very useful tool

it is used a lot in the study of surfaces.

for example, on the unit circle, we have $\phi(x,y) = x^2+y^2-1 = 0$, so if $\phi_x(x^*,y^*) \neq 0$ then we can find a (locally defined near $y^*$) function $x$ such that $\phi(x(y),y) = 0$.

In this case we can write an explicit formula, so the IFT is not needed, but in general it is useful.

for intuition, use a linear function like $\phi(x,y) = ax+by = 0$. we see that we can write $x$ as a function of $y $ if $a = \phi_x(x,y) \neq 0$, that is $x=-{1 \over b} y$.

Koro

Thanks a lot.

Its name supplements what it does!!

copper.hat

05:09

i should have written $x=-{b \over a} y$.

Abhishek Ghosh

06:11

Consider a simple undirected weighted graph G, all of whose edge weights are distinct. Is the following statement true or false:
"One or both the edges with the third smallest and the fourth smallest edges are part of any MST of G."
-----------------
Nothing else is given in the above question. Now is this statement true or false? Does having a connected graph with only 2 edges, make this statement false? I am not actually getting it. If the graph itself has only two edges, then no point of having the third or fourth minimum edge weight in the MST, as they do not exist.

It is like saying:

One or both of Mr. Robert's daughters go to school.

Now if Mr Robert has no girl child in the first place, then will this statement be true or false.

Personally I feel it is false..

S.M.T

07:44

@robjohn Thank you yesterday for your help. I solved my the question

S.M.T

12:13

Q: Why is it that expanding a determinant along any row or column always give the same value.

I really want to understand it. It will be grateful if you could share any link to it as well.

Jakobian

12:40

If $g:X\to\mathbb{R}$ is upper semi-continuous, $f:X\to\mathbb{R}$ is lower semi-continuous with $g\leq f$, then we can define $F(x) = \begin{cases} \{g(x)\} & g(x) = f(x) \\ (g(x), f(x)) & g(x)\neq f(x) \end{cases}$

I think that by using Micheal selection theorem now, we obtain a continuous function $h:X\to\mathbb{R}$ such that $g\leq h\leq f$ and $g(x) < h(x) < f(x)$ whenever $g(x) < f(x)$.

they asked me to prove a weaker version of this by setting $F(x) = [g(x), f(x)]$, but I think this multi-function gives us a stronger result

yep, it does

it's pretty cute

Jakobian

14:07

actually, this doesn't work because $F(x)$ has to be closed for every $x$.

Koro

How would one feel if they knew how to solve every question in an exam but they spent too much time on few questions and therefore couldn't even take a look at all the questions in the given time?

So eventually they could complete almost slightly more than half of the questions and didn't even have time to verify/cross-check whether what they did was wrong or right.

leslie townes

18:32

koro: i think there's only one answer to that question, which is "not great"

geocalc33

$\lim\limits_{n\to \infty; n\in \mathbb R^+}\text{surface area} \{(x,y): y= x^n(1-x)^n,x\in[0,1]\}=\pi$

how would you prove this?

oh

I forgot to mention the surface is formed by revolving about x=1/2

I think the problem is equivalent to $\lim\limits_{n\to \infty; n\in \mathbb R^+}\text{length} \{(x,y): y= x^n(1-x)^n,x\in[0,1]\}=1$

and then revolving that unit length to sweep out the $\pi$

copper.hat

18:49

sometimes i add what i think is a cute short answer but the accepted one is more complex and i feel slightly cheated as a result. i need to return to my search for a life.

leslie townes

copper: the answers to math.stackexchange.com/questions/4380704/… are both cool but the "cute short answer" i was very tempted to provide was "It's complicated"

in that both answers are basically a textbook dump

i don't mean to insult them, a lot of analysis is like that. "how is that motivated?" "you have the luck of being born several hundred years after gauss and euler and when you read enough of what they did, this comes naturally"

a huge amount of analysis, particularly complex analysis, is just people figuring out general theory to justify what they had done in special cases

books make it look so much cleaner than it is or was

copper.hat

@leslietownes yep, i think that much of teaching looses that perspective.

in the particular whine above, i think the cute short answer is a better intuition.

leslie townes

can i see? i promise not to downvote it.

actually i change my mind. i make no promises. or i promise that i might not downvote it.

copper.hat

just a mo. really it was just a whine.

robjohn

@geocalc33 that would seem to limit to a disc of radius $\frac12$, which would have surface area $\frac\pi4$. Are you sure that is the axis of rotation?

copper.hat

19:03

math.stackexchange.com/q/4380762/27978

i'm still working on my mse jumpsuit.

that would be sooo cool

leslie townes

it's definitely more conceptual, although the OP in the position of asking the question might not be in the position of knowing the definition of support function or its properties. B-.

also, the small matter of the second half of my drive

copper.hat

its a new look, you'll find it in lots of places now, the more work we do the more we see it

leslie townes

hahaha

copper.hat

true, but if they don't know what the support function is, i suspect they may have issues showing that a compact convex set is the convex hull of its boundary.

thanks for the sympathy vote :-)

my highest voted answer is truly mundane.

i need to work on my narrative.

this morning a buddy was admonishing me to bill for my thinking time, not just end product.

leslie townes

i realized that the accepted answer is a less conceptual version of your answer. it seems like more is going on, but it kind of isn't.

it's also a gasps PSQ.

copper.hat

19:10

i have a tough time doing that

i know, i'm living on the edge, thinking of defecting to china

as a skater

geocalc33

say $n \mapsto e^n$ for $n$ a natural number. Does it make sense to ask about the number $f(n)?$

copper.hat

[for the cia, that was a joke]

leslie townes

i generally don't have a tough time doing that, although there are certainly gray areas and i think relative to my colleagues i underbill.

geocalc: your first sentence defines the rule of a function without naming the function. is the function f?

copper.hat

i have a hard time justifying why a relatively simple solution took 8 hrs

geocalc33

@leslietownes oh yeah, $f$ is the function respecting that map

leslie townes

19:13

we have the cover that doing something relatively simple might require consulting 5-10 long and detail ridden documents so there is a general understanding that even if the finished product is one page it probably did take a long time. harder with technical work where maybe it's just a question of when a lightbulb goes on.

geo: if you have f defined on natural numbers by whatever rule, it does make sense to ask about $f(n)$ for any function. you say 'the number' $f(n)$ and i wonder if the fact that here, the codomain is a set of numbers has any importance.

e.g. your rule spits out real numbers that aren't integers or rational numbers. there are questions that it makes sense to ask about n that it wouldn't make sense to ask about f(n).

copper.hat

for some reason (i have my suspicions) convex psqs seem to escape. that said, the mad rush to close seems to have abated.

leslie townes

i am leery of responsing to PSQs simply because i don't want to give an answer and then have the OP ask me about it in comments in a way that shows they don't understand the problem. that brings me down. i'd much rather ask for supplemental info in comment, and not answer if they don't respond.

by then someone else has usually answered, so all is well :D

copper.hat

sometimes i answer a psq just because the comment would be too long

and you know living life on the edge, like a skater

geocalc33

@leslietownes $f: \Bbb N \to \Bbb R_{\gt 0}$ so $f(n)=e^n.$ I want to ask about a function $g(\cdot)$ whose domain is the image under $f$ YET instead of inputting elements from the image of $f$ you forcefully require inputs of $g$ to be from the set $\Bbb N$

but it's a contradiction because numbers $n$ are not defined in $\Bbb R_{\gt 0}$ prescribed by $f$

leslie townes

any g from f(N) to wherever is going to induce a function from N to wherever by composition. you might notationally distinguish the two (maybe call g circ f by "g hat" or something)

geocalc33

19:35

okay thank you

copper.hat

i got my social security yearly update. they always admonish to check that your income is reported correctly, the latest year always seems to be zero which sends me off wondering what to do, then the faq indicated that it may take more than a year to update. one wonders what they actually do if it takes that long.

of course it would help if i had a long term memory...

leslie townes

haha. i have a 3-year gap in grad school when i was definitely earning money and somehow it was not reported. i've never bothered to correct it because it wasn't that much money

also, fairly sure that if the planet still exists by the time i retire, social security won't

they might as well tell me my lesliecoin balance

geocalc33

might start geocalccoin33

copper.hat

yeah, my grad student earnings do not show.

i suspect that crypto will undergo a dotcom sort of event in the next few months

leslie townes

19:51

it's weird because it does show my grad school earnings in the first 2 years. then radio silence.

copper: except for lesliecoin

copper.hat

i am working still because of my financial naivety early on

leslie townes

nobody's going to retire in the future except for people who hold lesliecoin.

everyone else will be toiling away in the salt mines until they expire

copper.hat

the first was when they redid a sales performance based options agreement in my first few months of working. it would have hit target in 2 years, but ended up being aligned with rest of the company on a 4 year standard. i should have renegotiated at that point, but was truly naive.

since i was on an f-1 that was a big deal then

leslie townes

what's this, mr. gordy? i sign right here and get $500 and a record on motown?????

copper.hat

i want it, i want it all and i want it now

Hasini

20:10

Hi

A sequence is a list of things (usually numbers) written in a definite order. There, is it required that it has a pattern? I thought it should, since usually in a sequence we should be able to guess/obtain the next term...

Am I right?

Thanks a lot in advance.

leslie townes

hasini: to define a sequence, you need some way of specifying the nth term, for every n. if you haven't been given that, you haven't been given a sequence.

e.g. "1, 2, 4, 8, ???, ???, ???" is a guessing game and not a sequence.

but, there are lots of ways of defining sequences that do not give very much of an idea of what the nth term is, or whether there is any pattern.

e.g. a_n = "the nth digit after the decimal in the decimal expansion of pi" is a perfectly good sequence. it doesn't shed much light on what a_328754243865 is, or how to compute it, or what patterns might exist in a_n. but it does define a sequence.

copper.hat

20:34

evans hall is to be razed

leslie townes

do they have a date? it has been planned for demolition since at least the early 2000s.

Ted Shifrin

Aww ... my old home.

Balarka Sen

Oh, many happy returns, @TedShifrin.

Ted Shifrin

Ah, thanks, a Balarka :)

leslie townes

i will be sad to see it go. there was a point in my life where i computed that i'd spent a nonzero percentage of my life inside of it.

Ted Shifrin

20:37

Yes, leslie; you did double duty.

copper.hat

i think evans library was my favorite place on campus

leslie townes

wouldn't surprise me if they already closed that. a lot of subject libraries vanished between my undergrad and the end of my postdoc.

no, we don't have that book, it's at offsite storage in livermore. we can get it to you next tuesday.

copper.hat

i think the inability to browse is a huge loss

codys, libraries

leslie townes

browsing is mostly how i'd find books. i'd look for one, and then look at the ones around it, 10 minutes later walk out with a better book than the one i was looking for.

Ted Shifrin

@copper.hat Agreed.

leslie townes

20:44

when i was at iowa they closed the math library. i enjoyed it for a few months. then everything was offsite and it was multiple days to get literally anything other than a calculus book.

Ted Shifrin

Wow. That's crazy. As far as I know, the UGA science library (which was in the same building as the math department) is still going strong.

copper.hat

it is incredulous to me that the loss is not fully appreciated by management

Lukas Heger

21:20

I agree that closing a library is a great loss

@Ted Heartily and belatedly, I wish you a happy birthday!

robjohn

21:37

@TedShifrin: I just noticed. Congrats on missing 70 ;-)

Ted Shifrin

Thanks, @Lukas @robjohn! But not for long :(

robjohn

@TedShifrin The last couple of years just seem to have vaporized.

maths researcher

If $f:\mathbb{S}^{n-1}\rightarrow X$ is continuous and $F:\mathbb{S}^{n-1}\times I\rightarrow X$ is a continuous map such that $F(a,0)=f(a)$ for all $a\in \mathbb{S}^{n-1}$, what can we say about $X\cup_{f} D^n$ and $X\cup_{F} (D^n\times 0)$?

geocalc33

21:53

favorite Mellin pair?

maths researcher

can we say that they are homeomorphic?

Lucas Henrique

22:07

@mathsresearcher what's $\cup_{f}$ again?

Jakobian

adjoin

maths researcher

attaching space where we identify $f(a)$ with $a$

Jakobian

@mathsresearcher I'm pretty sure they are homeomorphic and it should be trivial

maths researcher

Yeah that's my thought just asking as a sanity check

Jakobian

Actually, aren't they exactly the same space?

Lucas Henrique

22:09

the universal property of the quotient topology should do the trick. hmm...

Jakobian

$D^n\cup_f X$ is the space obtained from $D^n$ by considering the decomposition $\{f^{-1}(x) : x\in X\}\cup \{\{x\} : x\in D^n\setminus \mathbb{S}^{n-1}\}$
$(D^n\times \{0\})\cup_F X$ is obtained from $D^n\times \{0\}$ by decomposition $\{(F\restriction S^{n-1}\times \{0\})^{-1}(x) : x\in X\}\cup \{\{x\} : x\in (D^{n-1}\times \{0\})\setminus (\mathbb{S}^{n-1}\times \{0\})\}$

if we use the homeomorphism $f:D^n\times\{0\}\to D^n$ given by $f(x, 0) = x$ then I think it's clear that those are the exact same constructions

because we have a homeomorphism $f$ of two spaces $A, B$ with equivalence relation $\sim_A$ and $\sim_B$ such that $f$ maps equivalence classes of $\sim_A$ to equivalence classes of $\sim_B$

geocalc33

23:05

consider $f$ as the presentation of $g.$How is the presentation of $g$ in $\Bbb R^3$ related to the presentation of $f$ in $\Bbb C$?

geocalc33

23:25

can you rotate $\Bbb C$ euclideanly within $\Bbb C^2$?

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$Mathematics$ Mathematics