Mathematics - 2021-12-04 (page 1 of 3)

ted: hooray!

A third of the congress should be in prison, @amWhy, but it ain't gonna happen.

amWhy

@TedShifrin I agree whole heartedly.

@TedShifrin And at least a handful of governors, and the former head of the Federal Executive Branch, and ...

And the former titular corrupt, lying idiot in charge

My kingdom for a math question.

ted i have a matrice is it linearly independence?

shintuku

12:07 AM

this all reminds me i need to graduate before ww3 or i'm fodder material

shintuku has brought up the silver lining here. we're well above draft age, ted.

amWhy

@TedShifrin Indeed.

@leslietownes is it in a victorian bodice?

it's a span of the augmented vector

i had an officemate with whom i would trade nonsense linear algebra gibberish word questions. the joke was you'd walk into the office and say hey, can you help me with some question? like it's an actual question. and once they'd stopped what they were doing and were listening you'd say linear algebra gibberish.

humor was hard to come by in grad school.

shintuku

hahaha

this format has potential for most academic topics

12:16 AM

yeah, at least any topic with enough of its own terminology.

i sometimes pose category theory questions like that in here.

geocalc33

How?

A morose definition of (vitreous? Vitriolic?) humor.

12:31 AM

my daughter's funnier than i am. if she's bored she might announce something like "i'm a cat! we're all cats!" and then crawl around meowing at our actual cat.

geocalc33

Can I please get an example of a step function, asymptotic to our natural log(x)?

what do you require by 'asymptotic to'?

does floor(log(x)) work? note that it is within 1 of log(x) for all x?

maybe floor(log(x)) + 1, which has a nice interpretation if log is base 10 and x is a positive integer?

geocalc33

those 2 are good yes

I just graphed them

cool

2:27 AM

did @Govind75 sort out his issue?

@copper.hat Yes, we did.

:-) Cool

We turned induction right-side up.

Been to upper Kensington for me yet?

:-) not yet, i should have driven by today, i went for a 'run' at tilden.

Ah. It’s a serious bike ride up those hills.

2:33 AM

i don't mind hills so much, it is cars that bother me/

Well, that’s worse downtown.

I hope I can come visit when COVID dims. Just not sure my back/neck can take the driving.

let's rent a party bus

I think we figured that the geographic centroid in CA was pitiful.

yep

hence, party bus

If I drive, you can come along, but I’ll stay up there longer to see friends in the greater area.

I wonder if Olivia and munchkin can babysit Screech :)

2:45 AM

one of them would end up babysitting the other two, but i think it's anyone's guess who would come out on top

Yes, that occurred to me.

AMDG

3:54 AM

Ok, I have a recursive algorithm that's pretty simple for computing $1 \bmod x$ which I didn't intend, but it sort of just came up. desmos.com/calculator/9avzkybkt3

However, in it's current form, I don't find it to be very satisfactory. You can see it does not converge very quickly as x goes to zero.

How do I solve recurrence relations generally?

This recurrence relation works such that you add a certain number of terms of $x$ while we are less than zero and is pretty simple, however, I don't know the first thing about solving recurrence relations.

6:06 AM

Hi All,

hyper-neutrino

I am only one person and not All, but hi

i am just All's west coast representative

imgur.com/a/4LWq4ou

Pls see link below.

Three forces P , Q , R act perpendicularly to the sides of a triangle at their middle points and are proportional to the sides. The forces are in equilibrium...

My question is that, these three forces in equilibrium means triangle wont move any where, triangle also in equilibrium or not?

i don't know physics or what it would even mean for these forces to 'act' on a triangle (is it some kind of rigid body? what is holding it together). if the triangle is just a device to define the magnitudes of the forces, maybe it's just a statement that the vector sum of the forces is 0.

I am confused because if these three equilibrium forces act on a particle or concurrent, it means their resultant is zero. But in this case we have three different sides of triangle suppose triangle as has three uniform rods.

Here forces are concurrent the triangle is not a uniform lamina, also centroid/COM of triangle is at different place from the point of concurrency of triangle, in second picture of attachment i have shown that.

6:22 AM

magnitudes and directions, i should have said.

maybe it would be helpful to think of the triangle as a solid surface. the point G is where three strings meet, and you've put weights on the end of each string, hanging off the surface, and the statement is that if the masses are proportional to the side lengths then point G will not move.

i don't know, again, not a physics person.

if my physics is the best second opinion on physics that this chat can provide, we are all in a lot of trouble.

@leslietownes :-) Thank you for your input about my question.

Why math and physics is so difficult to understand intuitively. Life is very short to learn these subjects.

Hi Leslie :)

howdy

6:40 AM

Good morning/evening :-)

7:21 AM

good night :-)

@Govind75's matrix $T$ is nilpotent. And he needs to work in $\mathbb{R}^n$ not $\mathbb{C}^n$.

there are treatments for that.

7:36 AM

Good morning! what does END(V) mean?

set of all endomorphismen ?

yes. what 'endomorphism' means might require further elaboration, but if you have an understanding of that from context, that is almost certainly what it means.

its okay i will read itu p thanks i am not familiar with the notation now i know what to look :)

“Now you know. And knowing is half the battle"

:)

7:57 AM

Mar 2 '12 at 23:26, by Asaf Karagila

I cannot believe that people miss my GI Joe reference!

8:09 AM

yeah, i'm not sure how many people watch that anymore.

:D

not me 😉

nudge nudge wink wink

Interjection: nudge nudge wink wink

(idiomatic, humorous, also attributive) A phrase used to hint that the speaker is euphemistically referring to something else.

:D

robjohn

8:26 AM

@user726941 say no more!

these references aren't getting any more recent

delta t is positive, everybody, hate to break it to you

robjohn

@leslietownes really? How do we know that?

^

🤔

huh? sorry, i was just quoting my favorite show from 2050

I saw two hoodies on sale somewhere yesterday. One of them had the message: "For every $\epsilon \gt 0$" and the other one had -"$\exists$ a $\delta\gt 0$ such that $0\lt |x-x_0|\lt \delta\implies |f(x)-L|\lt \epsilon$. I'm sure there must be a third hoodie which had definition of $f$ on it.

8:29 AM

you live in a weird place, koro

maybe it's time to print up some linear algebra t-shirts and retire on the proceeds

Q: For every symmetric matrix exists B not singular matrix s.t. B−1AB is diagonal matrix.

Haha

unit 1991

8:55 AM

Hi .Can anyone help with this problem>

0

Theorem. For every $q_f:V\to P$ quadratic form exists orthogonal basis,meaning basis of $V$ on which quadratic form has canonical form. And my book then says corollary from theorem. For every symmetric matrix exists $B$ not singular matrix s.t. $B^{-1}AB$ is diagonal matrix. I understood proof ...

unit 1991

10:39 AM

$(a_1,a_2,..,a_n)$ be basis for $V$.($e_1,e_2,..,e_n$) different basis for $V$.$(e_1,..,e_n)=(a_1,a_2,..,a_n)\times P$.Then $[x]_e=P\times[x]_a$.Let $[x]_e=X,[x]_a=Y$.
$X=PY$,$X^T=Y^TP^T$.

$q_f(x)=X^TAX=Y^T(P^TAP)Y$=$Y^TBY$=$\sum_{i,j=1}^nb_{ij}y_iy_j$ and from theorem follows that $P^TAP$ is diagonal.

$A=$$\begin{pmatrix}f(e_1,e_1) & f(e_1,e_2) & .. &f(e_1,e_n\\\ f(e_2,e1) & f(e_2,e_2) & .. &f(e_2,e_n) \\\ .. &.. &..& ..\\\ f(e_n,e_1) &f(e_n,e_2) & .. & f(e_n,e_n) \end{pmatrix}$ where $f$ is bilinear form.

This is my work.Is this correct?

PM 2Ring

1:42 PM

@Wolgwang I made a new diagram for the 5 square dissection:

monoidaltransform

2:07 PM

If $V=\cong W_0\oplus W_1\oplus W_2$ where $V$ is a representation and $W_0,W_1,W_2$ are irreducible representations of finite group $G$, how do I view $W_i$, say, as a subrepresentation of $V$?

3:33 PM

Hello, i have been disconnected from linear algebra for a while, could someone tell me what i need to look up when i read this senteance?
IE why is it true......

let L be a linear function, $ L: \mathbb{R^n}\rightarrow \mathbb{R} $ and let $<.,.>$ be the eucledian scalarproduct on $\mathbb{R^n}$ then it is known from linear algebra that there exists only one $ v_L $ such that $L(x) = <x,v_L>$ for all $x$ ,$ x, v_L \in \mathbb{R^n}$

Thorgott

4:01 PM

@monoidaltransform what do you mean

@MadSpaces try proving this yourself, express everything in terms of the standard basis of $\mathbb{R}^n$ and see what has to happen

I did try. i do not see how we can have one vector which scalar product gives the same result for all x .. it doesnt make sense to me

I usually try to search and work myself before i ask any question to be honest.

Regardless.. we know that a linear function can be expressed by a transformation matrix.. thus $L= A * x $..

is it the same matrix? which is meant by this vector.. since it has the dimensions of a vector.

Thorgott

@MadSpaces it doesn't, $L(x)$ depends on $x$ just like $\langle x,v_L\rangle$ does

(well, unless $L=0$, in which case $v_L=0$)

Astyx

4:17 PM

If you want a more abstract approach, first prove that every n-1-dimensional subspace of $\mathbb R^n$ is the orthogonal complement of a vector

I dont want to prove it to be honest i just want to see the proof since i am doing Analysis right now and i am not interested in linear algebra.
But this has come up in the lecture and i am interested to see the proof, but really not in trying to prove it myself

:D

@MadSpaces the vector is $\sum_{i=1}^n L(e_i) e_i$

I knew it has to do with the transformation matrix. Let me play with this one around to assure its truth for myself @LeakyNun 5q+5q :)

what is 5q?

what is five q plus five q ?

ten q : )))

"thank you"

4:22 PM

I see

4:33 PM

Is this only true for $L \rightarrow \mathbb{R^1}$? because if not than we wont have scalars!

5:01 PM

Review how you make the standard matrix for a linear transformation. The column vectors are ….

the picture of unit vectors of f

Um, wrong.

Maybe look it up rather than random guesses.

whatsup ted?

my answer was correct :D

The picture of? Oh, the image.

Sorry my german brain no speaken english

5:08 PM

So if the images are in $\Bbb R^m$ you get an $m\times n$ matrix.

True statement.

@MadSpaces woher kommen Sie?

Kartoffelfresserland, @LeakyNun und Sie?

@MadSpaces welche Teil?

ich komme aus HK

Hessen )

I do not want to say which city since i talk alot of smack about my professors

hahahaha

5:12 PM

You should only hear what they say about you.

ahahahahahah

i am pretty sure its not that bad :)

Of course, we know that in Europe professors hardly know the students.

Well, actually my department is small, so all professors know us personally :)

Ah, good.

Welp.. thats debatable, since you literally kind of get scolded if you miss a lecture or get asked " ah so you decided to not write teh exam.. i see!"

:D

But its nice : )

5:16 PM

Decided not to write the exam? There’s a choice?

i am not sure if you are sarcastic

I scolded my students frequently for missing class or bad homeworks.

Why : (

Thats not nice Ted :(

Not sarcastic. I’m not used to giving students an option. No make-up tests unless death or violently ill.

Jesus christ i am glad i wasnt your student

This seem too hardcore.

But its your style so its okay.

5:18 PM

I’m nicer than plenty.

In introductory classes I post solutions, so it’s not fair to anyone to give options.

I do not understand this mindset of making it like school, aka punishing students and making them commit to this. i understand that you might feel that alot of them are immature and need a good strong figure to guide them, but plenty as well are well matured and have many life problems other than studying. Such rules only become a handicap to them.

I know this too well.... sadly.

And students talk to one another, even if you tell them not to.

In the US most of my students were 18-20.

Yea.. these were kids...

But anyway, each professor has his or her own way. If you think its a good method, than you keep at it. i do not believe there exists a one good method.. each has ups and downs.

The methods are quite universal in universities in the US. What is not universal is that I had up to 8 office hours a week and most profs had 1-3.

That is way too much

Let me quote you einstein on this one when he was in prague and was annoyed by his lectures (which he always prepared poorly) and hated giving
Die Tintenscheißerei ist endlos
"the ink shittery is endless!"

5:25 PM

Hard courses, challenging homework.

Refering to the university work he had to do

Dont you have helping students to do that.. in germany,p rofessors hire students to do these work for them

aka tutorials and correcting homework and so on.

UnderMathUate

One of my professors is like this. He has 8 official office hours, but he tends to keep an "open door" policy. Meaning, it's not uncommon to find him working with students after 6 pm on concepts and assignments.

Very minimal grading help. I graded all the homeworks in my upper-level courses, but typically got one of my former top students to grade for the Calculus with proofs courses I taught. No help with tutorials, office hours, or exam grading.

I'm not counting hours of preparation and grading.

Most graders do a terrible job.

i am sorry you have so much work.
Maybe you should complain or ask for a raise..

I gotta go catch my bus. maybe talk later!

5:41 PM

I retired 6+ years ago. No worries.

And I did the work I loved.

5:56 PM

the wheel rim on my bike split this morning curtailing a morning ride. luckily i was only a mile away from home.

i wouldn't let that stop you from asking for a raise, ted. the worst they can do is say no.

copper: the excuses for not checking on ted's house are piling up.

Yikes. I've never heard of a rim split.

:-) this was in the opposite direction with a friend.

@leslie When I got the university highest teaching award (permanent $5K raise), I noticed that my departmental raises dwindled accordingly. Great award.

hah. they don't even try to be subtle about it.

6:00 PM

imgur.com/ZacdJOC

I did complain ... but "research" runs the university.

robjohn

It gets the grants

Wow, @copper. That could have been dangerous if you were zooming down Cutting.

at my wife's school (liberal arts college) the university-wide teaching award is a one-time $1000, up from $500 a few years ago.

you could earn more dressed as spiderman on hollywood boulevard for a few hours a week for a year.

@leslie This is the highest. A maximum of 5 university-wide per year. I served on the selection committee for 4 years subsequently.

6:01 PM

@TedShifrin i have had many hardware failures over the years, luckily none caused injury and equally luckily none left me in an awkward situation.

my wife's is 2 a year. the one positive thing i will say about it is that it's open to instructors without tenure.

this is a common exclusion from those types of awards.

the advantages are pensions, health care, etc.

Yes, although UGA has lots of college-wide awards which do allow instructors.

@copper I guess I never rode my bike as viciously as you.

i had a hardware failure once on colby ave in oakland. someone flung out their car door so widely and so fast that i ran into it.

That is super dangerous.

It happened to me in Cambridge, MA.

6:03 PM

@TedShifrin i am really not a mad mountain biker.

it ruined my front tire and gave me a nice door shaped bruise, but on the plus side, i damaged the car so much that the driver couldn't close the door afterward.

Serves the a**hole right.

enjoy parking your car with one door open in oakland.

And driving?

yeah. haha

6:05 PM

i used to have an office in downtown berkeley and used to cycle up solano through the tunnel. about a third of the trips i narrowly avoided being doored (ugg, new verb) or someone reversing out of those angled spaces.

I would be scared to bicycle through that tunnel. I never did, although admittedly I didn't bicycle around there much.

i just walked the bike on the sidewalk, the tunnel road is outside my risk profile

Ah, OK.

people really underestimate the danger of driving both to themselves & others.

@TedShifrin It amazes me how many people I hear from who aspire to do research at places like Google, FB, etc. yet haven't heard of the peer-review process.

6:06 PM

and overestimate other less frequent risks.

@Clarinetist getting into the peer review cycle is not that easy.

@copper.hat Sure, but at least having an awareness of it would be a good idea before committing yourself to a PhD in ML, thinking that you'll work for Google Brain some day, no?

clarinetist, i don't know. in some areas it's possible to do 'research' oriented work where it just becomes a 'white paper' that just goes into the company's document management system. or is published as patent prior art.

i certainly would never have published anything if it were not for my advisor's connections.

the idea of going into a phd program with no understanding of peer review is a little weird.

that is true.

6:08 PM

of course in US law academia, there is no peer review. so maybe an exception there.

there go my dreams of working in legal academics at facebook.

@leslietownes From hanging out on stats websites a lot lately, I meet tons of people who (1) aspire to become research scientists at Google, FB, and Twitter; (2) realize they need a PhD for it; and (3) don't know anything about the peer-review process. It's concerning.

I actually had to mention to some people that work that sounds fascinating doesn't necessarily mean the work is great.

do you think the interest is serious? or is it like going on a sports forum and meeting people who want to be on the NBA? is google/fb/twitter just a placeholder for "i wanna earn a lot of money with X"

@leslietownes It's more like "I don't want a boring DS job where I'm doing dashboards, so I'd rather get into more novel work, but only FB, Twitter, and Google have that, so I'll aim for those research jobs"

i do think there are some areas where i'm not sure what you'd publish in. recommendation algorithms for websites like amazon, or targeted advertising, are huge $$ but usually trade secrets and all of the academic work i've seen on them is terrible. i don't know where you'd publish good research on that.

if someone knows, let me know, by the way.

clarinetist: i think i'm with you on this: good luck with that, folks.

let's all think about how depressing it is to aspire to one day work in advertising, and not even the fun kind of advertising with jingles and concepts and martinis at lunch.

Martinis?

6:14 PM

negronis? i never watched 'mad men,' i'm extrapolating.

Just saying ... my ears perked up.

oh, right. right.

@leslietownes I'm glad I got a taste of research in my last position, but I'm glad I'm no longer doing that for a living. It's ridiculous how slow peer review is. The very first article I worked on nearly 2.5 years ago just got published about a month ago.

Especially for us stats people. In many social/medical science journals, reviewers are used to using certain statistical methods for which assumptions don't reasonably hold, so when a stats person comes in and uses a new fancy method, the journals will often say "use this simpler method"

the newsboys ran out in the street shouting "extra! extra! hot off the presses! results from nearly 2.5 years ago!"

I had to settle that debate as reviewer #3 for a journal last month

6:18 PM

ha! yes. that is not a huge thing in my wife's area because new fancy methods are hard to come by. but a standard referee report will be like "do this using the model i used in my phd in 1995 instead of the model your advisor used in his phd in 1995"

"you know, so people can understand it"

Better than the NSF rejection I got because one reviewer said the results we sought were in an obscure Italian article from the late 19th century. We went and read it. They were not.

there are wide communities of academic consumers of statistical information for whom statistical methods have no real meaning, outside of 'thing you see a lot, and thing do when you write your own papers, and thing you get suspicious of and maybe criticize if it's not a carbon copy of something done in 50 other papers'

ted: at least because of the timeline, it's likely that the reviewer wasn't citing his own work? :D looking for a silver lining here

Yeah, great silver lining.

a reviewer clearly self-cited something he said we were re-proving, and was wrong, but it wasn't the stakes of an NSF thing. the editor just ignored the reviewer after we wrote an email comparing the two results. and the crummy paper got published anyway.

the high stakes of NSF make it a pretty good example for a lot of phenomena in game theory.

Well, one of my great embarrassments is that Robert Bryant pointed out after one of my papers was published that Cartan had a counterexample to one of my claims. I no longer remember the error I'd made (differential systems stuff).

It was not one of my greatest papers, regardless, but sorta interesting pieces to it.

6:27 PM

great timing, robert.

Too geometric a description but I offer it here anyhow — complex surfaces in $\Bbb CP^5$ with the property that their second osculating spaces are everywhere four-dimensional.

(Generically it's always a 5-dimensional space.)

So it turned into PDE and linear systems in algebraic geometry.

Pls see link

imgur.com/a/4LWq4ou

One of my few solo papers in my career. There's a moral in that story.

since vatican ii the church has opened up a little, but not to the point of allowing second osculating spaces.

Three forces P , Q , R act perpendicularly to the sides of a triangle at their middle points and are proportional to the sides. The forces are in equilibrium...

I am confused because if these three equilibrium forces act on a particle or concurrent, it means their resultant is zero. But in this case we have three different sides of triangle suppose triangle as has three uniform rods.

6:29 PM

Your sentence makes no sense.

Here forces are concurrent the triangle is not a uniform lamina, also centroid/COM of triangle is at different place from the point of concurrency of triangle, in second picture of attachment i have shown that.

123, i do hope someone has better ideas than i did. i'm still guessing that thinking of the triangle as a rigid body made out of rods is not the right approach.

Why is the centroid even relevant?

@leslietownes Do i think it as triangular lamina?

Are you supposed to check that three vectors add up to $0$?

No lamina in view.

We're balancing the three rods, not the actual triangular lamina.

6:31 PM

@TedShifrin Okay pls clear this topic..

So you're not questioning that the three vectors add up to $0$? Then we're done.

@TedShifrin My question is something else. I know and calculate the answer these vectors are 0.

this came up last night. it wasn't clear if the statement was about, if the forces are this, they sum to zero, or: the forces are this, the sum is zero, and [?? what is the problem]

My confusion is about the intuition about this question. Calculation i have done.

Well, I answered your other confusion.

Why should this have anything to do with the centroid?

6:35 PM

@TedShifrin Because i think object is balanced by its center of mass. That's why i mentioned centroid is at different place from point of concurrency of forces.

Balanced if you have gravity acting perpendicular to this plane. Totally not relevant.

The forces are not concurrent. The perpendicular bisectors intersect at the circumcenter.

@TedShifrin I assumed there is no gravity.

But that's what the centroid is about. Go think about that.

Random horizontal forces have nothing to do with the centroid.

I am thinking about this from whole day from different point of view but still don't get the answer. That's why i asked here.

In book question there is no discussion of centroid.

Nor should there be.

Your thinking is just wrong.

6:39 PM

My confusion is that, If i have triangle with uniform rod and i apply force on each side which is perpendicular to their mid point and proportional the sides.

Does this triangle also in equilibrium?

If those vectors add up to $0$, what is the answer?

Because sum of all forces adds up to zero.

Aye Sir answer is zero...

So why are we not done? Why do you keep asking?

But these forces having different point of application.

Correct.

But to add the vectors, you move them head to tail. These are free vectors. The net force is obtained by moving the vectors wherever you want. If you want to compute torques, that's something else.

Maybe that's what you're really thinking. There is no net translation to this system. If you want to ask about rotation of the system, you need to compute torques and consider where the forces are acting.

6:44 PM

@TedShifrin Thank you. I was thinking of shape of triangle and point of application of force, which confused me.

@TedShifrin Thanks , exactly what i wanted to ask.

OK, so calculate the torques about a fixed axis perpendicular to this plane passing through whatever point you want. I would use the circumcenter (since that's easiest to relate to the points of application of the forces).

@TedShifrin Okay... It clear me complete picture.

You will find in an instant that the net torque about that axis is obviously $0$ (no matter what the strengths of the forces are).

But the object is likely to rotate about other points.

We have two formulas of torque $\tau = r \times F = I \alpha$ , in relation $r \times F$ how it take care of rotational inertia of body?

Forget $I$. Compute from the definition.

What is $r\times F$?

6:49 PM

$r$ perpendicular distance from the reference point (may be axis of rotation) O to the force $F$.

to the point at which $F$ is exerted, not to $F$ itself.

I don't know why mechanics books don't clear the reference point confusion. Does it axis of rotation or it can be any other point wherever we want.

@TedShifrin to the line of action of force.

One always talks about torque about an axis (or in the 2D case, a point, the axis being understood to be orthogonal to the plane).

No, to the POINT of contact.

Otherwise, $r$ makes no sense.

@TedShifrin Mechanics book use word moment of force not torque.

Your book, perhaps, not mine.

Anyhow, in your case, using the origin I specified, what is $r\times F$ for each of your three forces?

6:54 PM

1drv.ms/u/s!AozWlUoG8z4tiHJcBDouxLT3N3CV

Pls see link above, page of the book and topic moment of force.

I am not going to look. I do not care.

@TedShifrin Okay.. But in the whole chapter. They used reference point O but they don't clear the reader does this reference point always axis of rotation of not.

i'd focus on ted's question. for each force, we have this r, and this F, and what's r times F.