Mathematics - 2023-01-30

Ted Shifrin

12:17 AM

Huh?

ペガサス Seiya

12:37 AM

nani desu ka?

Wojciech Kulma

@TedShifrin which part requires clarification?

Ted Shifrin

It sounds like garbage to me. How does an open ball become closed?

Wojciech Kulma

you take out the boundary of it

I got an open ball wit center Theta and a positive radius r. I make the radius a bit smaller and get a closed ball with center Theta and radius r-epsilon

Ted Shifrin

It doesn’t contain its boundary. That’s the whole problem.

Wojciech Kulma

thanks!

Ted Shifrin

12:50 AM

Yes, of course, it contains smaller closed balls.

Wojciech Kulma

you're right, I just reduce a radius. I turn (-10, 10) into [-9.99999999999, 9.9999999999999]

Ted Shifrin

But that may destroy what you’re trying to do.

Wojciech Kulma

I got a sequence theta_n converging pointwise towards the center theta in the original ball

I want it to work in the smaller one

Ted Shifrin

For example, a continuous map on the closed unit ball must have a fixed point, but it may not any longer if you shrink.

Thorgott

3:10 AM

@ShaVuklia isn't that just a splitting of $TM$

copper.hat

5:02 AM

yellow

Ajay

5:57 AM

@copper.hat is not a primary color

Koro

6:40 AM

I think that this is wrong.

21 hours ago, by Koro

Leslie: Suppose that I take $y_0\in$ X**- JX. I define $f:span (JX\cup \{y_0\})\to F$ as follows: $f(J(x)+ \lambda y_0)= \lambda$, then I can extend it to all of X** by Hahn Banach.

There is no reason to believe that f is bounded

So Hahn Banach may not be applicable.

Koro

6:54 AM

@leslietownes: I think we do need closedness as we discussed yesterday.

PM 2Ring

7:34 AM

@Ajay Yellow is a primary colour in printing, which uses a subtractive colour model. en.wikipedia.org/wiki/CMYK_color_model Of course, there are an infinite number of colour space basis vector sets, but some sets are more useful than others. There a few different basis vector sets used on modern screens, but none of them are perfect, so there are plenty of colours that lie outside their gamut.

@amWhy Philosophers have spent a lot of time discussing that stuff. See en.wikipedia.org/wiki/Qualia

It's also possible that you do perceive differences in colours that most people with (so-called) normal colour vision cannot detect. It's known that some women have 4 kinds of colour-sensitive cones in their eyes, which may give them enhanced colour discrimination abilities. See en.wikipedia.org/wiki/Tetrachromacy

user193319

12:03 PM

Problem: Two black bags contain the following number of balls: (a) Bag 1: 6 red balls and 2 blue balls (b) Bag 2: 6 red balls and $x$ blue balls.

It is known that the probability of drawing $2$ balls of the same color is $0.4375$. How many blue balls are in bag 2?

So, in total there are $6+x$ balls in bag $2$. Hence, $$Pr(\mbox{two balls of same color}) = Pr(\mbox{two red ball}) + Pr({\mbox{2 blue balls})$$

$$Pr(\mbox{two balls of same color}) = Pr(\mbox{two red ball}) + Pr(\mbox{2 blue balls})$$

$$Pr(\mbox{two balls of same color}) = \frac{6}{6+x} \cdot \frac{5}{6+x-1} + \frac{x}{6+x} + \frac{x-1}{6+x-1}$$

So, set that equal to $0.4375$ and solve for $x$, right? I had wolframalpha do that and I got $x$ has a complex number...Where did I go wrong in my calculations?

Koro

1:09 PM

@Jakobian ok

geocalc33

1:37 PM

given two cylinders that intersect orthogonally what is the volume of the intersection?

robjohn

2:17 PM

When I view this answer on my laptop, it looks fine. When I view it on my phone, the MathJax does not render and the numeric answer turns into a link.

Evidently, this was a problem in 2017 as well, but I deleted the comment. I assume the problem was fixed.

Alessandro Codenotti

It's working on my phone

robjohn

Very strange.

@AlessandroCodenotti what kind of phone?

I restarted my iPhone, and things still don't render

Alessandro Codenotti

Android, it's an old xiaomi of some kind, I don't remember the model

ペガサス Seiya

2:35 PM

@robjohn Sometimes it bugs on my iPhone too

user193319

2:54 PM

Whoops...the last indented line should be $$Pr(\mbox{two balls of same color}) = \frac{6}{6+x} \cdot \frac{5}{6+x-1} + \frac{x}{6+x} \cdot \frac{x-1}{6+x-1}$$

The corrected version I had initially put into wolframalpha...that still gives complex solutions for $x$...Not sure where my reasoning went wrong. I'm assuming we're selecting the balls without replacement, hence the $-1$'s everywhere. Even if you select the balls with replacement, the corresponding equation yields complex solutions... :(

Sha Vuklia

3:10 PM

@Thorgott yea, such that the distribution is right-invariant

I had a question, but I think I've solved it (modulo maybe a few details)

you know how a connection on a vector bundle $E\to M$ is a map $\Gamma(E)\to\Omega^1(M)\otimes\Gamma(E)$? there's also a way to arrive at that map via the Ehresmann connection on the frame bundle, which in turn has a covariant derivative

so I wanted to see that this covariant derivative was the same as the original connection, and I think I see that now

had to make some identifications explicit which had been a bit obscure to me until now

Thorgott

3:57 PM

ok good, I wouldn't know this stuff off the top of my head

Ted Shifrin

4:14 PM

@robjohn No good on my iPad now either!

Koro

@robjohn not rendering properly on my iphone either.

ペガサス Seiya

So I'm not alone?

Obliv

4:32 PM

is there a simpler way to take this derivative or do i have to do two product rules $e^{3x}(3\cos(2x)-2\sin(2x))$

geocalc33

4:44 PM

I stumped the experts with an easy to state but hard to prove question

Ted Shifrin

4:58 PM

@Obliv why two product rules? Just one.

Koro

5:22 PM

A convex subset A of $R^n$ is the unit ball for a norm associated with an inner
product iff A is a solid ellipsoid.

What is the meaning of this statement exactly?

Isn't the unit ball always convex?

Unit ball is a unit ball w.r.t. to the dot product norm but still not an ellipsoid?

the statement looks wrong to me.

Thorgott

every inner product induces a norm, every norm induces a unit ball and the claim is that the subsets of $\mathbb{R}^n$ that can be obtained that way are precisely the solid ellipsoids

Koro

so the unit ball is also a solid ellipsoid?

Thorgott

sure

Koro

ohh

Thorgott

$x_1^2+\dotsc+x_n^2\le1$

a much more interesting question is which subsets of $\mathbb{R}^n$ arise as unit balls of some norm (not necessarily induced by an inner product)

the answer is quite cool

Koro

5:29 PM

not sure how to prove the statement though.

Thorgott

the point is, it's impossible to write down all the norms on $\mathbb{R}^n$, but it's very easy to write down all inner products

Koro

So suppose that a convex set A is the unit ball associated to an inner product norm on R^n, so I have to show that A is a solid ellipsoid, right?

Thorgott

that's the claim, yes

Koro

@Thorgott hmm yes. Riesz representation theorem

Thorgott

you might have to quote that one theorem from linear algebra

Koro

5:33 PM

so do I have to suppose 'the unit ball' to be centered at the origin?

or that it doesn't matter?

I guess 'the' refers to the one centred at the origin.

Thorgott

the unit ball is $\{x\colon\langle x,x\rangle\le1\}$

Koro

Suppose that A is convex, and that A={x in R^n, (x,x)<1}

does it matter if I take open ball or the closed ball?

I guess no as far as the convexity is concerned?

Ted Shifrin

@Thorgott Well, we certainly get rectangular boxes, too.

@Thorgott I see the spectre on the wall.

Koro

hmm, I think I got it now. Every inner product can be associated to a matrix.

Ted Shifrin

What sort of matrix?

Thorgott

5:39 PM

@TedShifrin right on the nose

@Koro you get analogous statements in either case

though I wouldn't call an open unit ball a solid ellipsoid

well, I guess it sounds less wrong on paper than it did in my head

Koro

@TedShifrin So I write the matrix $A= [a_{ij}],$ where $a_{ij}= (e_j,e_i)$.

Ted Shifrin

Why do you reverse $i$ and $j$?

What sort of matrix will this be?

Koro

It is symmetric.

positive definite too.

I reversed i and j as I got used to it when I learnt inner product matrix the first time.

Hoffman and Kunze do it.

Ted Shifrin

I can see the reversal showing up because $Ae_i = \sum a_{ji}e_j$, but why not write $Ae_j = \sum a_{ij}e_i$ instead? Shrug.

Right. Positive definite symmetric.

Koro

positive synthetic matrix, haha

but I have one problem yet: I didn't use convexity of A anywhere.

Thorgott

5:46 PM

why not write $Ae_j=a_j^ie_i$ :P

Ted Shifrin

@Thorgott I do, sometimes, but I always put the summation symbols.

Thorgott

I never drop summation symbols, but I did actually start arranging my indices like that at some point

nothing to say in my defense

Koro

proving $\phi(m)\phi(n)=\phi([m,n])\phi((m,n)),\phi$ is Totient function requires very intensive use of indices.

Ted Shifrin

Only in teaching graduate differential geometry did I do upper and lower indices. And bars over indices in hermitian geometry. Research, too, of course.

Thorgott

oh, bars over indices is a new one to me

Koro

5:50 PM

is it OK to use symbols from different languages?

Ted Shifrin

It helps you keep track of well-definedness just as upper/lower does in the Riemannian case. Yeah, the hermitian metric is $\sum h_{j\bar k}dz^j\otimes dz^{\bar k}$. $dz^{\bar k} = d\bar z^k$.

Koro

suppose we have already used English alphabets in a proof, and let's say we have used all the Greek alphabets too, can I use say Russian alphabets ?

Thorgott

haven't seen that, but makes sense

Koro

Defining solid ellipsoid as the set {x in R^n: $x^T Ax\le 1$}, where A is a positive definite matrix, the converse to my statement also follows just by creating an inner product using A, I think.

Ted Shifrin

You still need to quote a famous theorem to see the ellipsoid.

Koro

5:55 PM

ohh

But forward direction of the proof is complete, or did I misunderstand something?

Do you mean for the opposite direction ? (from solid ellipsoid to the unit ball with inner product norm)

Thorgott

well, the question is what your definition of a solid ellipsoid is

Ted Shifrin

Why is that set an ellipsoid, yes?

leslie townes

@Koro if you are using that many symbols, i think you need a rewrite and not a different alphabet. :D

Ted Shifrin

I hate to agree with leslie, but I must.

Koro

here is the definition: the set {x in R^n: $x^T Ax\le 1$} is called solid ellipsoid, where A is positive definite matrix.

Ted Shifrin

6:02 PM

That's not what the world would recognize as an ellipsoid.

We have a common notion generalizing the ellipse $x^2/a^2 + y^2/b^2 = 1$ from high school.

Koro

yeah, I know right. But I saw that set definition the first time.

leslie townes

it's not uncommon for people to use different letter types to denote mathematical objects of different types, even if they don't use all of the letters first. but usually that's stuff like, uppercase vs. lowercase roman, lowercase greek vs. uppercase greek, sometimes fraktur lettering. sometimes hebrew (set theory folks use hebrew for at least certain cardinals).

Ted Shifrin

It is a horrid definition.

Koro

it has 'cross terms' like $x_ix_j$ apart from just the square terms.

leslie townes

i haven't seen russian letters brought into english math, except for Sha for the shafarevich group of whatever you take the shafarevich group of.

Thorgott

6:06 PM

yeah, so the actual content right here is to understand why this is no more general than what Ted is proposing

up to a change of coordinates, of course

Koro

@TedShifrin our first functional analysis class started with Hahn Banach theorem, then next week it was category and functors, then Banach Alouglu theorem, today uniform boundedness principal. When asked to the teacher why he does that (not following any pattern), he said -'fun'.

robjohn

@TedShifrin The other MathJax on that page renders properly. I've checked the MathJax there, and it seems fine.

Ted Shifrin

principle :) ... Yeah, uniform boundedness is important.

Koro

:( I tend to misspell this one, and another one is center vs centre.

Ted Shifrin

@robjohn Bizarre.

center vs centre is purely American vs British

principal vs principle is two totally different words.

Koro

6:10 PM

I'm comfortable with 'stationery' vs 'stationary' via the following mnemonic: pap$\color{blue}{e}$r is 'station$\color{blue}{e}$ry'

Ted Shifrin

Yes, that might help.

You can remember that the school principal is not your pal.

Koro

ohh cool. Thanks. I won't forget it now.

Ted Shifrin

You then have to remember that that is the principal principle.

I just made this stuff up :D

But people who write (even in books) principle curvatures make me want to hit them.

leslie townes

ted: how do you feel about Reimannian geometry

Ted Shifrin

vomits

There's neither reime nor reeson.

Thorgott

6:16 PM

anyway, you should spend some time thinking about the more complicated problem I posed earlier

leslie townes

i like how there's a famous gelfand, and a famous gelfond, and as far as i can tell it's transliteration of the same name (but written differently in both russian and english)

Ted Shifrin

leslie should? I should?

Almost as confusing as Weil (French) and Weyl (German).

leslie townes

it's like how actors have to add an initial or something if there's someone in SAG who already goes by their name

gelfand's agent must have said "ok, from now on we're gonna spell this differently"

Tobias Kildetoft

@leslietownes Well, two famous Gelfand(s) in fact

Koro

@Thorgott yeah, okay.

leslie townes

6:19 PM

@TobiasKildetoft if you say so, but i've never personally seen "both" of those guys in the same place

Koro

but for any norm p on R^n, the set {x in R^n: p(x)<=1} is the unit ball, right? @Thorgott

Tobias Kildetoft

@leslietownes The two Gelfands have a category named after them together with Bernstein

Thorgott

it's the unit ball of that norm

different norms yield different unit balls, of course

(additional question: can two different norms have the same unit ball?)

Koro

squares, diamonds, balls can be such sets.

you want complete characterisation of such sets.

Ted Shifrin

There are all the $\ell^p$ shapes, too.

Koro

6:27 PM

ohh yes, we aren't confined to inner products anymore.

leslie townes

that's another question like thorgott's, which sets can be the unit ball of a norm.

Thorgott

that was my original question, leslie

leslie townes

oh, it was the 'more complicated problem' from earlier. cool.

Thorgott

when I tutored analysis 2 a couple years ago, this was an exercise in the first week

it was probably the hardest exercise in the entire course

leslie townes

yeah. i've seen it in some books, but usually broken down as a series of exercises or with ample hints.

just dumping it on people unprepared, haha. i guess it's one of those things "well, an answer doesn't require any deep theory, so they can figure it out."

Thorgott

6:35 PM

to be fair, there is some nice geometric intuition

Ted Shifrin

I have never seen or worked on this one.

leslie townes

i.e., there's no good reason why you'd need to see a ton of other stuff first, before attacking that problem. so if the question is just, where do you put it? i guess the beginning is as good a place as any.

Thorgott

it's a good exercise, but it is tough

ペガサス Seiya

Can't sleep again

robjohn

6:52 PM

@TedShifrin I tried loading the desktop site on my iPhone, and the rendering did not improve.

Nothing in the MathJax header of that page has changed.

Alessandro Codenotti

Another nice (but easier) question is which open sets are the unit ball of a metric (compatible with the topology)

Obliv

7:07 PM

I totally forgot how would I differentiate $\frac{1}{4-x^2}$

I can factor out $\frac{1}{-1(x+2)(x-2)}$ then do a product rule of $\frac{1}{x+2} \cdot \frac{1}{x-2}$?

just wondering if there is a faster way

Thorgott

why not use the quotient rule?

Obliv

you can use the 1 as a function?

Obliv

7:25 PM

would this be considered explicit form for a d.e $\frac{2X-1}{X-1} = e^t$

I don't think I could separate anything further

geocalc33

orthogonal intersection of 2 double cones. Q1: closed form volume? related to steinmetz solid?

orthogonal intersection between 2 "ted's icons" Q2: closed form volume?

double cone^

Obliv

7:45 PM

for $\frac{dP}{dt}=P(1-P); P = \frac{C_1 e^t}{1+C_1 e^t}$ I found $\frac{dP}{dt} = \frac{C_1 e^t}{(1+C_1 e^t)^2}$ subbing into the equation I got $P(1-P) = \frac{C_1 e^t}{1+C_1 e^t} (1-\frac{C_1 e^t}{1+C_1 e^t}) = \frac{C_1 e^t}{1+C_1 e^t}-\frac{C_1 e^{2t}}{(1+C_1 e^t)^2}$ is any of my arithmetic wrong

okay yeah and to finalize the expression you multiply top and bottom by $C_1 e^t$ which reduces the numerator of the sum to 0 meaning the solution doesn't exist I think

robjohn

8:27 PM

@geocalc33 also called a bicone.

geocalc33

thanks @robjohn

Gwyn

9:43 PM

I need to evaluate $\lim_{n\to+\infty} \int_1^2 \frac{x^n}{1+x+x^2+\dots+x^{2n}}dx$. I noticed that, from the inequality AM-GM, for $x>0$ it is $\frac{x^n}{1+x+x^2+\dots+x^{2n}} \le \frac{1}{2n+1}$ and so, since the integral is non-negative, the limit is $0$ by the squeeze theorem. Is this correct? Are there other elementary techniques to evaluate this limit?

Xander Henderson

9:59 PM

@geocalc33 That is not the figure I usually associate to the phrase "double cone".

@robjohn That is the term I would use.

A double cone is the surface (with a singularity) swept out by rotating one line about another, e.g. rotate the line given by $y=x$ in the $xy$-plane about the $y$-axis.

Ted Shifrin

Actually, that is a cone, quite simply. We abuse language all the time when we call one nappe of a cone a cone.

Otherwise, we'd have to call conic sections double-conic sections. shrug

Xander Henderson

@TedShifrin I agree, and that is the way I teach it in my courses.

But the Google results for "double cone" mostly give that image.

Ted Shifrin

I'm not surprised. After all, Google is smarter than all of us put together :P

Xander Henderson

@TedShifrin Well, no. But I am making a descriptivist (rather than prescriptivist) observation here.

geocalc33

yeah i think bicone is better than double cone

or reverse cone

finite doubly capped cone

ペガサス Seiya

10:12 PM

@TedShifrin Oh yeah? I'd like to see Google solve my hice problem

Xander Henderson

@geocalc33 Bicone seems fine.

geocalc33

yes

based cone

ペガサス Seiya

Based on what?

geocalc33

well the cones share a common circular base

Xander Henderson

I understand a "based cone" to be the solid swept out by rotating a triangle about one of its heights. So "double based cone" might be okay.

geocalc33

10:16 PM

i will search google and try to find the volume of two perpendicularly intersecting bicones

ペガサス Seiya

How much ice cream can I fit inside a cone?

geocalc33

inside a bicone?

probably a good amount

Ted Shifrin

Good luck eating it, though.

ペガサス Seiya

I can fold my tongue

Like Bugs Bunny or something

Ted Shifrin

Nope. Not the issue. It's all sealed safely inside, impervious to your tongues.

Xander Henderson

10:21 PM

I have a four dimensional tongue.

That has to help, right?

Ted Shifrin

I don't think it does.

Well, actually I take it back. If the bicone lives in $\Bbb R^4$, then the inside is no longer protected from wily tongues.

ペガサス Seiya

I have a tongue made of titanium. That'll work

Xander Henderson

@ペガサスSeiya All bicones are made of unobtanium. Your titanium tongue is of no use.

ペガサス Seiya

@XanderHenderson I didn't wanna do this but fine. My tongue is made of Lead-206 that was made by the ACCELERATED $\beta$-decay of U-238

geocalc33

wow

Xander Henderson

10:28 PM

@ペガサスSeiya Again, the bicone is made of unobtainium. Your tongue loses.

ペガサス Seiya

@XanderHenderson Fine, I'll bring my tricone then

Xander Henderson

I'd prefer a tricorn.

Tricorn (mathematics)

In mathematics, the tricorn, sometimes called the Mandelbar set, is a fractal defined in a similar way to the Mandelbrot set, but using the mapping z ↦ z ¯ 2 + c {\displaystyle z\mapsto {\bar {z}}^{2}+c} instead of z ↦ z 2 + c {\displaystyle z\mapsto...

parz

hello

ペガサス Seiya

@XanderHenderson So a shuriken?

parz

impossible ice cream?

geocalc33

10:32 PM

galbatorix

Sine of the Time

hi

ペガサス Seiya

Pegasasu Ryusei Ken!

ペガサス Seiya

10:45 PM

I saw a tiny car that was shaped like a jellybean

I wanna drive it

geocalc33

11:27 PM

what is a motivation of analytic continuation?

tells you more about the function, on an extended domain, complex plane?

are there other motivations for it?

leslie townes

that's a pretty broad question and hard to answer at that level of generality

the geometric series formula is arguably one motivation. 1 + x + x^2 + ... = 1/(1-x) whenever |x| < 1. as it happens the condition |x| < 1 is not needed for RHS to make sense (it makes sense for x except 1)

so at least in the series context, sometimes it's clearly possible to identify a function that a series represents, where the function also makes sense on a larger domain than the series does

you can think of analytic continuation as a kind of generalization of that (although you don't have to)

geocalc33

okay yeah I think i get that part

just can't get an answer to my one question i posted. seems like it requires a lot of thought even from experts

I'm just looking to understand the answer if someone posts one

maybe i have to motivate it better i don't know

I'll try constructive feedback

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$Mathematics$ Mathematics