two lines?

lines i meant

the intersection of two lines

Hello, can i ask for help here?

two planes would have been easier

a plane would be some weird infinite cube-like thing

(in projective space, that is)

yes, meow, you want to use linear algebra, vector algebra, whatever ... in $\Bbb R^3$.

alright :P

No, @meow, $\Bbb P^2$ is the plane. To talk about various planes, you need to move up to $\Bbb P^3$, which you will a few sections hence.

@Ted French 8th graders got the worst result out of all European countries that took the test. 12th grader also did very poorly compared to 20 years ago. This got the media all worked up. But when you listen to our retarded Ministre de l'Education, the results are no wonder

@ted yes i know it is the only "plane", just like how the only plane in $\mathbb{R}^2$ is the only plane in itself

@LeGrandDODOM which test are you talking about? Was Italy among the countries taking it? I'm pretty curious

Well, things will only get worse with our upcoming president in the US. But I have to say I work with little kids every week (going in a few hours today), and when a 10-year old has to use his fingers to find $2+5$ or $3\times 4$, I know we're in big trouble.

right @meow

Could someone help me with this math question?

its just weird to think about infinite cube things in $4$-space :P

In $\Bbb P^3$, a plane is like a plane in $\Bbb R^3$, but with some extra points at infinity.

No cube things.

hyperplane

or something

A plane in $\Bbb P^3$ corresponds to a $3$-dimensional subspace of $\Bbb R^4$, just as you've done points and lines ...

Yup.

yes i know

i was calling the $3$-dimensional subspace of $\mathbb{R}^n$ an "infinite cube thing" instead of a "Hyperplane" on acciident

Ah ...

@Alessandro nces.ed.gov/timss I've taken a cursory glance at the results of the 12th graders' results, and Italy performed very poorly

Mathematics terminology exists to make things precise :)

Did you think about/understand the discussion about the Möbius strip?

@Rakso Is there more information?

uhhh not really

i should probably read that

@LeGrandDODOM our school system is known to be pretty bad so I'm (sadly) not surprised

over again

@Brody I need to get the probability in % to win. And my record is 87 wins and 10000 games played.

@Ted look what she says from 3:00 dailymotion.com/video/… It's pathetic

@Rakso, they want you to say 87/10.000 is the answer, but it isn't really. :) It's just all the information you have.

Hi.

good evening @Balarka

@TedShifrin That is the part that confuses me. I did wrote 87/10.000 but it doesnt solve the question. Its just the win ratio not the probability to win one game.

i found another kid in my school who's interested in math

Good night @Balarka.

@Rakso There's not enough info I think. Your observations may approximate the expected probability of winning a game, but it doesn't help derive a more concrete answer without more stuff.

@Alessandro Ted corrected you there.

he's trying to learn trigonometry, but i don't think he really understands, so today at lunch i gave him a mini-lesson on the unit circle, and how it related to right triangles

@meow: There may be lots of kids interested in math. You don't need to want to read graduate math books at the age of 14 to be interested in math!

@Brody Could you mention what more information i would need in order to calculate it? Im interested in learning.

@LeGrandDodo: So the minster of education pokes fun at math because she doesn't know it.

yeah I always greet with my timezone because I keep forgetting where are you all from. Maybe I should start just using greetings that don't reference time :P

Not uncommon.

@TedShifrin i have observed many kids interested in math; many participated in this year's math competition

@Rakso As far as the problem goes, you might not need anything else. It's the "format" they want the answer in: win-to-lose odds.

I wasn't correcting @Alessandro. I was telling @Balarka to go to bed.

about 20 or 30, to be precise, in a grade of 200-300 students

Uhm I think you were answering to @LeGrandDODOM instead of me? @Ted

At least that's the impression I have about the problem so far...

@TedShifrin Yeah, it was a joke.

@Brody The exact task translated to english is as follows: "Write an application called Play123Main that lays the solitaire 1-2-3 10.000 times and then calculates the probability (%) for the solitaire to win."

@Rakso: Basically the law of large numbers tells you that if you repeat an experiment a large number $N$ of times, if the probability of success is $p$, then you expect approximately $pN$ successes ...

There's only a 4h30m difference between here and India, I don't know why but I believed it to be bigger

@Alessandro really? from italy?

yep

@Rakso I think your repeated runs of the game would count as a "calculation". In that case, take your proportion $p$ (wins / total games) and then $(100\times p)$% is you probability.

by plane i assume

No, by time zones!

no math pun intended

> how does a 2-dimensional affine subspace of $\mathbb{R}^n$ get to work?

> by plane!

Some commute.

Let's blow a plane up.

@Brody That is what i've done so farm but i was not sure it was right since it doesnt really answer the quetsion in my oppinion.

@BalarkaSen no thank you

inb4 discretion notice against potentially offensive joke

I was just referencing the blowing-up construction.

you could condense the plane, however, by using a linear transformation of rank $< 2$

@TedShifrin So basically we are using division laws $q = n * p$

@BalarkaSen Of course :P

oh boy, now i gotta calculate some cross products

and construct some normal vectors

that's definitely one way to do it, @meow

@Rakso I don't know what the problem is supposed to entail exactly. Do they want your program to calculate the exact probability of winning under certain conditions? Or just get a point estimate from repeated simulations?

@Rakso: So it sounds like they want you to tally the winning percentage for large numbers of games (say 8000-10.000) and look at those percentages and draw some conclusions. Not just the one 10.000 result.

@Brody: I suspect the latter.

Yeah, probably @Ted

hi chat

It has to be the latter @Brody

Hallo Semiclassical

I think i got it then. Thanks to the both of you! @Brody @TedShifrin

Have fun, @Rakso.

@TedShifrin Always fun with math ;)

Ted did the helping :) Good luck

who's excited for the snow? :P

Where is snow? Georgia is having tornados, and it's sunny as usual here in San Diego :)

I am actually thinking about switching majors from Computer Science to math because i feel it is more interesting for me

Can you do both, @Rakso?

I'm up in the northern nj area

Ah, well, it is almost December. I remember snow in Boston the second or third week in October many years ago.

@TedShifrin I would have to study both programs at once. I am going to try it after new years since i applied for all courses for both Math and CS, will be challenging but interesting

It should have been colder than it is in this part of the world too.

It will be hard, @Rakso, but it will make you more powerful for either a job or graduate school.

We're just having a pile of smog instead of snow though.

A quick and funny problem: count the number of one-to-one polynomials with coefficients in $\mathbb F_p$ and degree $\leq p-1$.

Let's build some smogmen

@TedShifrin im mostly interested in Math over CS since i dont understand it as well as CS. I have worked as a programmer for 1 year now before starting my university studied.

Fair enough, @Rakso. If you already have programming experience and know data structures, etc., that plus the math might be enough. I don't know ...

One-to-one, huh, @LeGrand?

Not too difficult.

What is $\mathbb{F}_p$?

Also, is @TedShifrin a spider?

I suspect so.

$\mathbb{Z}/p\mathbb{Z}$ for the first and probably not for the second @Brody

Are you now petrified of me, as well, @Brody?

No, the anthropomorphic ones are fine @Ted. It's the natural ones that are weird and alien.

So why am I a spider?

Spiders are fine.

@TedShifrin Because you have eight and a half eyes.

Probably.

@TedShifrin You have eight, sometimes six, eyes. No? Maybe other animals have this.

Oh ... But not nearly enough legs.

Sometimes nine, even, @Brody.

chat.stackexchange.com/…

well that's some evidence

LOL, so you should compute the probability density function for the number of eyes :P

LOL

yikes

I hope that's an outlier

LOL

That seems to be the mode, anyhow.

I'm glad Balarka has nothing better to do in life than count my eyes.

Yeah, this is much better than studying for exams.

Weird thing about this distribution is that some are proportions (e.g. 6 out of 8) and not just straight counts

Most were proportions, if I remember me well enough.

The total number of eyes is not fixed apparently

Indeed.

So Ted's not a spider. Probably just some Lovecraftian beast

Phew

Now I understand why you're scared to meet me. Makes perfect sense.

Well, I'm imagining the boiler operator from Spirited Away. Pretty friendly guy, harmless

Oops. 20 seems to be the mode then.

You're misremembering what mode means.

Maximum observation. The mode has the maximal frequency

Not maximum observation.

No?

Ah, you were using the wrong word. Just as I said.

Right. Mode is the most commonly occurring data value.

Maximum... value.

Statistics is not interested in vast outliers ...

Ya, I just don't know what it's called for usual distributions.

other than how they affect standard deviations ...

it's not a meaningful thing, really, @Balarka.

Unless it has very high probability of occurring.

I agree, I agree.

I've only done intro stats stuff @Ted. Do the endpoints of a confidence interval converge to the true parameter as the sample size tends to infinity?

I'm no expert, @Brody. I've never taken stat, and I did finally teach probability just before I retired. But this is what the law of large numbers is about ...

Funnily enough, the first question I ever asked here was a probability question on median v. mean

It never actually got answered >:/

Hard to believe, @Semiclassic.

@TedShifrin Interesting. And a bit surprising tbh

Something I learned from an exercise in Spivak (but saw nowhere else), @Semiclassic: If $a_1,\dots,a_n\in\Bbb R$, are data points, what are the minimizing points for $f(x)=\sum (x-a_i)^2$ and $g(x)=\sum |x-a_i|$?

It's a beautiful theorem, @Brody.

Hello, perhaps a silly question, but the following relation R on the set of integers Z, R = {(a, b) ∈ Z^2 ∶ a^2 = b^2 }, this is simply {(1,1),(4,4),(9,9),(16,16)....}, right?

The mean for the former and the median for the latter IIRC

It's not that surprising, if I'm honest. The initial version of it was quickly seen to be trivial, and I left it up to anyone else to provide a precise interesting version

For reference: math.stackexchange.com/q/725319/137524

Yup @Balarka ... Did I give you that before, I think?

@devilius you also have (-1,1),(1,-1),(-1,-1) and so on

What does 'data points' mean in this context?

@Ale

@Alessandro ah yes, of course. Thanks!

Typically when I hear that phrase I think of elements of R^2, not R^1.

I think the one with the $g(x)$ is in your multivariable book. I proved it in the stat course I took this year.

My first question was about integer valued polynomials apparently

@Semiclassical Just a bunch of real numbers.

No, I have $f(x)$. You can't really do $g(x)$ with calculus, @Balarka.

@TedShifrin Not really following the question.

Yeah, fair enough.

Minimizing what, exactly?

Minimizing $f(x)$ and $g(x)$.

The respective functions $f$ and $g$ !

Oh, you mean what value of $x$ minimizes each?

Is the first easy? Seems like a Calc I application

It can be done without calculus, @Brody.

First should be the mean value, just by eyeballing it

Totally easy, @Brody. For extra credit, do it totally without calculus :) Then do it in $\Bbb R^n$ when you get to chapter 5 of my book :P

lol, anticipated

Are you going to roll some eyes at me?

I think I can see the easy approach. (spoiler removed)

Yup @Semiclassic.

$g(x)$ seems like a much bigger pain in the butt.

It is worth thinking about, actually. I think it's cool.

It takes some case-by-case inequality trickery to get there.

Bleh, cases.

$\sqrt{u^2}$, but sums of radicals are no better

No cases needed. Just thinking.

Think about the graph.

I think you and I have the same technique in mind.

Oh, duh.

LOL

Yeah, piecewise linear things are easy to minimize :)

Simplest case of the simplex method? :D

hah.

A completely unserious approach:

Minimize $F=\sum_{k=1}^n y_k$ subject to the $k$ constraints $y_k^2=x-a_k$ :)

You're missing something.

Am I? I could imagine the Lagrangian multiplier method not actually working here because of reasons, but formally it seems right.

You mean $y_k^2 = (x-a_k)^2$? But how do you force the nonnegative solutions?

Oh. Need $y_k\geq 0$.

Yeah, that makes more sense.

Hrm.

Nuts.

Hey guys

Go do some serious work, @Semiclassic :)

Rehi, DogAteMy.

"Hi DogAteMy"

I found this

ugh, i dun wanna

https://i.imgur.com/wWyRSfR.jpg

Colin Adams, DogAteMy?

Yes

Nice book.

Old friend of mine ...

I was just thinking about the longitude/meridian chat we had a while back, is all

So that's which one is which

One serious thing I am trying to do is get this Griffiths-Dwork stuff to make sense.

meridian is the one which is tight

I can follow the example calculation now, which I'm proud of, and even reproduce the needed computations in Macaulay2.

(I just finished the book, by the way. I agree, it's very nice)

I use the vocabulary differently. I'm sticking with mine.

Topologists are weird, anyhow.

But making it work for the example I want to use it for is harder :/

It's not an intrinsic difference, anyway

Lines of longitude on the earth are inconsistent with that terminology.

I think latitude should be perpendicular to the axis of revolution (i.e., HEIGHT).

I don't know how to minimize functions w/o calculus :( @Ted

Do you know how to find the vertex of a parabola without calculus?

@Brody: You knew how to minimize quadratics when you took Algebra I in high school. You just weren't asked to do it. (I would have asked you.)

Well, can't exactly minimize $f(x)=-x^2$ :P

@Brody It's a fun exercise trying to minimize the function $x^x$, using only the fact that $e^x\ge x+1$ for all $x$ (and with equality on when $x=0$) and without calculus

I won't even take the energy to smack you, @Semiclassic.

lol

Hmm, let me think then. There's definitely more than memorizing the formula for the vertex :p

@AkivaWeinberger Let's annoy Ted by calling the (1, 1)-curve the longitude.

DogAteMy: Let him do the easy one first.

Actually, semi-serious comment on that.

Where did that formula come from, @Brody? This is actually a very important thing.

puts @Balarka on ignore

I know the spoiler. In the process of remembering why/how though...

You had to do this for methods of integration in Calc II, also, @Brody. And it does show up several times in my book. (Yeah, yeah, I know.)

I was reading a research paper over the summer, and formally it made sense; it used a Legendre transform to give the leading term of a saddle-point integral. So far, so good

Oh yeah, I remember your bringing that up, @Semiclassic.

The problem I realized, though, is that at some point their function stops being concave up everywhere

Steepest descent stuff?

And once that happens, Legendre transform gets baaad.

Nah @Balarka

@Brody What are you trying to solve, out of curiosity?

Hint @Balarka: $LDL^\top$.

Eh, I'd call it steepest descent stuff in the broadest sense.

Though I tend to fold those things together lazily

Oh, if that was aimed at you, @Semiclassic, I withdraw my nah.

@TedShifrin I don't know what you're hinting at.

Honestly, I don't think we were taught how to minimize functions. We were just given the general vertex coordinates, and I recognized it can be obtained by easy calculus

Shouldn't call it steepest descent if you're not working in the complex plane.

Where completing the square showed up ...

Agreed, @Semiclassical.

Oh.

Sure.

@Brody. Here's a hint: When $x\in\Bbb R$, what's the minimum of $x^2$?

What this was was: Suppose you've got an integral $\int_{-\infty}^\infty e^{-\lambda f(x)}\,dx$ with $\lambda\gg 0$.

I recall using CTS for deriving the quadratic formula and solving roots and such, but not minimization

Bad on your teacher :P

CTS = completing the square?

It's how you prove the quadratic formula, of course.

Yep @Akiva

Ted's hint + CTS should work

Suppose you've got a quadratic in squared form, i.e. $y=a(x-b)^2+c$

DogAteMy is just too impatient for me to be Socratic.

Sorry

So is @Semiclassic. So I'll just leave. I have to get going soon anyhow.

pah

Bye @Ted.

Anyways, where I was going before

lol. Bye

Bbye

I will think on this

*am thinking

if $f(x)$ is concave down, then it's got a single local maximum

and therefore the asymptotic behavior of that integral w/r/t $\lambda$ is $e^{\lambda f(x_*)}$ where $x^*$ is the unique local minimum

where do you buy books from on the internet?

libgen!

ducks

Minimum:minimize :: maximum:maximize :: extremum:??

Extremize?

Extremalize

@Sophie Amazon

@BalarkaSen me too thanks

Or just Google the book and see what comes up

Actually, I'm forgetting how the story worked entirely. Bottom line was that it went bad for certain parameter values

Oh duh, vertex form of the quadratic

Yah. Alternatively, if $y=a(x-b)^2+c$, what happens if I start with $x=b$ and then change it slightly?

Obtain the extreme value?

SemiC means, what happens if you look at $x = b + \epsilon$ or $b - \epsilon$ for some small $\epsilon > 0$?

^

assume $a>0$ to make that definite.

anyone versed in stats/gambling theory?

very simple question

"Just ask, don't ask to ask."

Are you looking for $0<|y(b)-y(b\pm \varepsilon)|$?

Yeah. More simply, if you move $x$ away from $b$ in either direction then $y$ must increase

@AkivaWeinberger You'd get more out of Rolfsen's book IMO.

So it's a local minimum.

Oh yeah, $a>0$

Adams' book is written below your level.

So i'm reading about the martingale betting system (keep doubling your bet until you win, and you will always end up winning your initial bet). I understand the logic that previous coin flips (or roulette spins) have no bearing on future ones. And I understand that eventually one giant string of losses will happen and lose the rest. And logically, it seems that no betting system should ever be able to beat the law of expected returns.

However, the one caveat everyone writes about with this theory, is that it does seem to work if you had an infinite bankroll. And logically it seems true. You would never lose. If this is true, how can the amount of money somebody has possibly break basic math theory?

Alternatively, $y(x)-y(b)=a(x-b)^2\geq 0$ with equality only if $x=b$.

@Semiclassical Right, ah...

So $x=b$ is the unique global minimum.

Hmm, guess that's a step toward formalizing extrema. They never taught me anything besides equating the first derivative to zero and cranking out a solution

Not that it's nontrivial. Just glad to see something concrete

Anyways, you'll want to now apply this to $f(x)=\sum_k (x-a_k)^2$.

@MikeMiller It is. But I had it on my shelf unfinished for a few years and I recently re-noticed it, and I was about to travel for Thanksgiving so I thought I'd bring a book

Is "Rolfsen's book" this?: maths.ed.ac.uk/~aar/papers/rolfsen.pdf @MikeMiller

Not loading but probably. It's the book for a geometric topologist to learn knot theory.

I don't think I know too much geometric topology, but I'll look at the PDF

Huh, $f(x)=\sum_{i} (x-a_i)^2$ is a quadratic itself

Ding!

Do we care what the constant term is

Shouldn't, applying what we know generally about quadratics

Eh, all you're trying to do is minimize it

And sliding a quadratic up/down won't change that.

Yeah, for $i\in\{1,2,\ldots, n\}$ we have $f(x)=n\left(x-\dfrac{1}{n}\sum_i a_i\right)+\text{constant}$

Neat, thanks chat (esp. @Semiclassical)

@Brody Never mind. But you're gonna want a ${}^2$ somewhere.

You're right, I forgot to include that

Mistype

$f(x)=n\left(x-\dfrac{1}{n}\sum_i a_i\right)^2+\text{const.}$---Hoping there's a pen-to-paper method for $g(x)=\sum_i |x-a_i|$ too

So $f$ is minimized at the mean

I know the answer, but not (yet) how to obtain it analytically

Ah

Speaking about minimizing $g(x)$, that is

@MikeMiller Do you know an easy description for the canonical Heegard diagram of a knot given a Seifert surface?

No.

Darn.

Does complex analytic imply real analytic?

@PVAL-inactive But I've never thought about it.

@Simeon Yes. Prove it.

Hello

I am Bulbasaur

Hey

Hi

Hello

Hi

Just found out that my assignment due tomorrow has been pushed back

But I want to finish most parts anyway because I don't have that much more

:D

Just the rest of 4c from last night, 4d, 3a and 2

But I feel so drowsy

Is 0 in the ideal? Is 1 in the ideal? For the every division ring is simple

0 is a member of every ideal

in any ring

So by closure under subtraction

0-0 in I?

Sorry on a phone

$0-0 /in I $

closed under outside multiplication

Other slash

$0 \cdot r = 0$

I have done closed under multiplication already

It's just the closed under subtraction part

Zero ideal is closed under addition

0 +0 like this??

$0 \pm 0 = 0$

I KO'd a cockroach in a hand-to-hand combat a while ago.

OK I understand that code that's 0+0=0 and 0-0=0

@BalarkaSen has reached Level 2

What about for r?

It's like if I do r +r that's 2r not r x. X

r-r

Oh wait $r_{1} - r_{2}$