Mathematics - 2017-07-25 (page 3 of 5)

@amWhy The problem is that the people who are actually active in the 'chores' are not frequently there and even when they are they don't seem to wish to act on the posts mentioned there.

SteamyRoot

@Astyx It's summer holidays in France, no? Getting anything done then takes a while.

Astyx

Yeah, but I'm still waiting for results, they should not be in holidays

Ted Shifrin

LOL, almost August!

Astyx

4:10 PM

The concours probably require a lot of paperwork

amWhy

@user21820 But new faces can change things: We can suggest closures and deletions, why we vote as such..... We can also counter faulty requests for reopening/ undeleting, talk it out, etc. The dynamics there will change with more participation there. I'm just suggesting that it is a valid location to include when we think posts should be closed, deleted, discuss duplicates, and keep attuned what reasons others may have for actions they encourage.

Astyx

I think each year they work during the whole summer (there will still be some results on the 9th of september)

SteamyRoot

Most likely they're really understaffed.

Astyx

Probably ... I can't do anything but sit and wait anyway

Ted Shifrin

au contraire ...

Astyx

4:13 PM

I can stand up and wait

rolls 5 eyes

But that's tiring

hi Martin

hi all

@TedShifrin Rolls 13 eyes!

Ted Shifrin

4:14 PM

That's getting up there in the eyeroll count ... But Balarka and DogAteMy have kept track of my maximum number of eyes.

Martin Sleziak

@user21820 Somewhat related discussion with amWhy in c.r.u.d.e: chat.stackexchange.com/transcript/2165/2017/7/25

SteamyRoot

Damn mutants everywhere!

Astyx

chat.stackexchange.com/search?q=eyes&user=76056&room=36

Ted Shifrin

Damn, Astyx, you really are bored.

Astyx

At some point you "[rolled] eyes" ... how are we meant to guess how many that is ?

Eye am bored

Ted Shifrin

4:16 PM

I need to think of a good hard problem for you.

SteamyRoot

Want some homework?

Astyx

Hehe

Mats Granvik

Is there a formula for the Gram points?

amWhy

@TedShifrin You've really done quite and amazing feat, in your leadership here. Someone mentioned tobias as a somewhat frequent visitor. Is that the Tobias as in t.b.?

Adeek

hi @TedShifrin

Astyx

4:18 PM

@amWhy It's Tobias Kildetoft

Ted Shifrin

Nah, I don't think I've done that much, @amWhy. No, Tobias is Tobias K, Danish (I believe) algebraist (who, ironically, did a postdoc at UGA, so we met years ago).

@TedShifrin Aha!

hi Karim

Hi @Adeek

hi @Astyx

4:19 PM

So, @Astyx, what flavor problem should I be thinking of for you?

@Mats: What are the Gram points?

Adeek

@Astyx you should do differential topology

Astyx

The thrilling kind

Adeek

it is very interesting to study

Astyx

I'm already somewhat digesting that @Adeek :)

Ted Shifrin

Um, what field, Astyx?

And please answer specifically and unpunnily.

Adeek

4:20 PM

cool

Mats Granvik

https://oeis.org/A002505
Gram points are: "...points t such that Re(zeta(1/2+i*t)) is not equal to zero and Im(zeta(1/2+i*t))=0."

Ted Shifrin

Have you actually learned transversality, Astyx?

Astyx

No preference really @Ted, perhaps some fun combinatorics if you have that under your sleeve

Kinda, I need to re-read that yet

Ted Shifrin

I'm not combinatorial, but I'll give you something from my probability course in a moment, Astyx.

amWhy

@TedShifrin Oh, yes, I recognize that name. (Actually, t.b. was "Theo".) But Tobias K has been around from the start! Cobwebs now cleared!

Ted Shifrin

4:22 PM

I have no knowledge, Mats. Far from my interests/knowledge, sorry.

Astyx

I've read all first chapter of G&P and re-read the beginning of it (up to transversality, excluded)

Mats Granvik

ok

Ted Shifrin

OK, Astyx, I can always give you good G&P-type problems. But I'll give you a combinatorial type problem that Pedro helped me with years ago. Let me find it.

Astyx

I should probably do the ones in the book first :p

Ted Shifrin

There are some excellent ones in the book. Mostly Guillemin gives too many hints, but occasionally he gives too few.

Semiclassical

4:24 PM

The thing I always find interesting in probability/combinatorics is the asymptotic aspects

Astyx

I definetly am going to do them .. just not now

Semiclassical

Which usually ends up meaning "stuff I can whack with the saddle-point approximation"

Ted Shifrin

OK. The combinatorics question was ... What is $\sum\limits_{k=0}^m\binom mk\binom{m+n}{k+1}^{-1}$?

This arose from the following probability question (to which this can be applied): What is the expected number of cards I must turn over (52-card deck) to get my first ace?

Astyx

Huh ?

Oh right

I probably have done this before

It'll be interresting to do it again

Ted Shifrin

With what probability?

Semiclassical

4:28 PM

lool

Astyx

Like $3\over 5$

(not almost certainly)

Is that a thing in english too by the way ?

Ted Shifrin

Another one of my probability favorites was: I go to a shop where they give me (at random) one of 5 possible prizes each time I go. How many trips do I expect to make to the shop in order to get (at least) one of each prize?

Yes, in English, too.

Astyx

So you also have almost certainly converging sequences of random variables and cool stuff like that

I didn't know the english-speakers were that evolved

Semiclassical

pfffffft

Astyx

easy :p

Anyway I gotta go

Seeya later !

Ted Shifrin

4:32 PM

Well, I gave you two good problems. Bye.

Astyx

The way is not to actually try and compute that sum right ?

well, not directly

Ted Shifrin

No, there's a closed-form formula for that sum.

Secret

::rolls $\aleph_2$ eyes, wait, that's a cyclops with an infinitely wide eye::

Ted Shifrin

hell, Secret, I never got so advanced as to roll 100 eyes, let alone $\aleph_0$ of 'em.

Secret

I wish one day we humans will discover something physical that has size $\aleph_2$ that would be really awesome

Ted Shifrin

4:35 PM

Do we have a physical model of $\aleph_1$?

Secret

well, we tend to treat things like time and temperature a continuum

Semiclassical

eh, temperature isn't a great example since that's defined in terms of statistical mechanics

Ted Shifrin

Everything I think of that's physical or chemical is just a large finite thing.

Semiclassical

I'd put it a little differently than that.

Temperature is genuinely a continuous variable; there's no fundamental unit of it. However, temperature rests on a statistical description of the system, and so is not a fundamental aspect of the universe.

Ted Shifrin

OK, well put.

Greetings, DogAteMy.

Semiclassical

4:41 PM

Entropy, by contrast, really is a "large but finite quantity" in my understanding

Ted Shifrin

Interesting: Secret raised the question but is strangely silent.

Secret

Well, I don't have much to add on that, semi seemed to explain that quite well

(and momentarily I was in physics meta chat doing stuff)

Ted Shifrin

ah, stuff.

Semiclassical

(Though the whole Gibbs paradox / entropy of mixing aspect creates issues as well.)

Secret

Some theoretical models did try to quantise time, though, but it is still not very popular

Ted Shifrin

4:44 PM

I hope we aren't all going to turn into constructivists.

Semiclassical

Lol

Secret

I did not say space because, well you have string theory, loop quantum gravity and many many many others who tried to digitalise spacetime

Semiclassical

Imaginary time is weird enough for me :/

Ted Shifrin

I can imagine.

Secret

whether we can have some fundamental aspect of the universe that has size $\aleph_1$, I am not sure, but my bets put on time

Akiva Weinberger

4:47 PM

Hi

@Secret Yeah but we need CH for that

Secret

(and that is assuming time is not emergent property of dynamics)

Semiclassical

Eh, I think that blurs things a bit. There's a difference between 'digitalizing' space (distances occur in integer multiples of some discrete unit) and space as a concept not being meaningful below certain scale

Akiva Weinberger

There's only like continuum many open sets, right?

Semiclassical

Quantum mechanics is far more in the spirit of the latter, in that the concept of particle trajectories in space cease to have any operational meaning

Akiva Weinberger

I don't know why I'm saying "like," there aren't any error bars on that

Continuum plus or minus 7

Secret

4:49 PM

Semi: That is true, I just mentioned there are popular mainstream models that try to does that, which is why I cannot say space is a continuum anymore when I factor all mainstream views

Semiclassical

I guess my point is that one shouldn't lump string theory in there

Secret

right, in that case we should focus on the more relevant aspects when thinking about this problem

Semiclassical

Whatever its faults with regard to prediction, it does respect both QM and Lorentz invariance

(As I understand it, anyways)

Secret

@AkivaWeinberger At least experimentally, there is no limit to how precise you can theoretically get a number to some physical quantity, thus it is perfectly fine to have something from the continuum (e.g. a real number) and some error to it

Semiclassical

Um, Heisenberg might have something to say there

Though if you mean in a statistical sense I agree

Lucas Henrique

4:54 PM

Hey guys.

Fast question: does $x^2 - 2y^2 = -1$ have infinite integral solutions?

Secret

I think heisenberg only works for conjugate observables, so if you are measuring two commmuting observables, there should be no limit on how precise it can get.

Even with heisenberg, I suppose one can measure one of these infinitely accurately and get a completely uncertain value for the conjugate variable, though I don't think heiseberg's equality will like having alephs plug into it

Lucas Henrique

Pell's equation guarantees that $x^2 - 2y^2 = 1$ does, but the $-1$ makes the problem different.

Akiva Weinberger

@Secret No you misunderstood me

Ted Shifrin

Do you have any solutions, @Lucas?

Akiva Weinberger

I meant like $|\Bbb R|\pm7$, as a joke

Lucas Henrique

4:56 PM

@TedShifrin yes, $(7; 5)$

Akiva Weinberger

@TedShifrin Well $(1,1)$ certainly works

Secret

Akiva: O, in that case, I don't know, if the error is finite or even just countably infinite, it will make no difference because of cardinal arithmetic

Ted Shifrin

Yes, DogAteMy, that's the one I'd found.

Lucas Henrique

:39009955 yes, hahaha

Akiva Weinberger

You know the standard solution to the *regular version?

I have no idea why I said "formal"

I think I was thinking of "former"

Ted Shifrin

4:58 PM

Here's a hint, perhaps, @Lucas. $(1,1)$ corresponds to the number $1+1\sqrt2$, perhaps.

Lucas Henrique

what does "correspond" mean?

Ted Shifrin

I mean we're thinking about the ring $\Bbb Z[\sqrt2]$.

Akiva Weinberger

@LucasHenrique Do you know how the other version, $x^2-2y^2=1$, is solved?

Lucas Henrique

I tried to use the factorization but... I'm not used of using $\Bbb Z[\sqrt{2}]$

@TedShifrin exactly

Akiva Weinberger

The usual solution is to know that, in $\Bbb Z[\sqrt2]$, if you take the power of something of norm 1, you get something else of norm 1

So now you can multiply any of those with $1+\sqrt2$ to get something of norm $-1$

Anonymous

5:01 PM

In my book they write that: Any relation $F(y,y',...,y^{n},x)=0$ is called an ordinary differential equation. After that they say $F:\Bbb R^{n+1} \times \Bbb R \rightarrow \Bbb R$. What does the second part i.e. $F:\Bbb R^{n+1} \times \Bbb R \rightarrow \Bbb R$ mean? Could someone explain it to me in simple english?

Ted Shifrin

Do you mean $f$ or $F$, @Blue?

Anonymous

@TedShifrin $F$, corrected now

Ted Shifrin

Is $f$ a solution of that differential equation, @Blue, or are we talking about $F$?

Akiva Weinberger

@Blue $F$ has $n+2$ inputs and $1$ output

Mats Granvik

I have an iterative formula that finds the points "t" such that:
1) Re[Zeta[1/2+I*t]]=0 and Im[Zeta[1/2+I*t]] not equal to 0
2) Re[Zeta[1/2+I*t]] not equal to 0 and Im[Zeta[1/2+I*t]] = 0
3) Re[Zeta[1/2+I*t]] not equal to 0 and Im[Zeta[1/2+I*t]] not equal to 0 and Re[Zeta[1/2+I*t]]=Im[Zeta[1/2+I*t]]
4) Re[Zeta[1/2+I*t]] not equal to 0 and Im[Zeta[1/2+I*t]] not equal to 0 and Re[Zeta[1/2+I*t]]=-Im[Zeta[1/2+I*t]]
But the same points "t" can be found with the RiemannSiegelTheta function and the FindRoot[] command in Mathematica.

Ted Shifrin

5:02 PM

Note that there are $n+1$ inputs coming from $y,y',\dots,y^{(n)}$, and then one more input from the independent variable $x$. That gives us $\Bbb R^{n+1}\times\Bbb R$.

So, for example, @Blue, the differential equation $y'^2 + 3yy'' = x^2$ would correspond to $F(a,b,c,x) = b^2+3ac-x^2$.

Akiva Weinberger

@LucasHenrique In any case, the answer is yes.

Anonymous

@TedShifrin Umm, what does $a,b,c$ stand for?

Alessandro Codenotti

they're just variables

Ted Shifrin

They're my variables in $\Bbb R^3$ ... you're going to put in $y, y', y''$ ...

heya @Alessandro

Alessandro Codenotti

Hi @Ted

and everyone else as well

Anonymous

5:07 PM

@TedShifrin Oh, now that makes sense. $a,b,c$ are the sample values of $y,y',y"$ respectively.

Anonymous

I think I got it

Ted Shifrin

yes, you can think of it that way

Anonymous

Thanks :)

Ted Shifrin

Sure :)

Akiva Weinberger

@AlessandroCodenotti Helslo

Hlelo

Ted Shifrin

5:11 PM

damn, you're confuzling today, DogAteMy.

Akiva Weinberger

A while ago I had fallen into the *habit of pronouncing "hello" weirdly

and I hadn't really paid much attention to it until a friend imitated me

Ted Shifrin

habbit, hobbit ...

Akiva Weinberger

I also tend to say "food" more like "füd" 'cause why not

Eric Silva

Hi chat

Ted Shifrin

hi Eric

Alessandro Codenotti

5:14 PM

hi

Akiva Weinberger

hʌj

Akiva Weinberger

5:27 PM

bäː

mathiu_lady

hello...can anybody post some link or give some name of books where I can get details proof of RADEMACHER's theorem. I'm studying it from Evans, Gariepy's book but I'm facing a lot of problems there

please read my messege fully...thank you

Lucas Henrique

I'm sorry, but I can't see how this fact trivializes the question

What's the construction that we can do with this fact?

Hi chat!

Hey!

hey Daminark

5:33 PM

Hey everyone!

Lucas Henrique

hello :)

BAYMAX

Why a line through origin is not a maximal susbspace of $\Bbb{R}^{3}$ where as a plane passing through origin is a maximal subspace!

for plane we can take the plane and any point in $\Bbb{R}^{3}/$ Plane and they together span whole of $R^3$

Daminark

How's it going?

BAYMAX

but what about the line

dont the line and a point in $\Bbb{R}^3$ / Line don't span the whole of $\Bbb{R}^3$ ?

skullpatrol

@amWhy I am now.

What's up?

SteamyRoot

5:42 PM

@BAYMAX $\dim U + \dim V = \dim (U+V) + \dim (U \cap V)$

For vector subspaces $U,V$ of some vector space.

Actually, wait, nevermind.

That doesn't do much in this case, since you care about a point outside the origin.

Eitehr way, a line and a point outside that line will only ever span a plane.

BAYMAX

yeah

Daminark

@Baymax A line is generated by one point, so a line and a point will generate 2 dimensions

BAYMAX

that's like it

Daminark

Which isn't enough to generate $\mathbb{R}^3$

BAYMAX

yes

Daminark

5:46 PM

The other way to think about it is that it's not maximal because you can always fit a line in a plane, right?

So it can't be maximal

BAYMAX

yeah because there is plane which is maximal

and we are claiming that a line isa maximal subspace

and the line fits in the plane

so we cannot say that line is a maximal susbspace

Lucas Henrique

@AkivaWeinberger so if $a = x + \sqrt{2}y$ where $a\overline{a} = -1$, then... ?

I mean, I could construct $a^{2k+1}\overline{a}^{2k+1} = -1$. That would not necessarily mean that the number will be of the form $n^2 - 2m^2$

Lucas Henrique

6:04 PM

3

Mike Miller

funny

Akiva Weinberger

6:40 PM

@LucasHenrique Yes it will. All powers of $a$ are of the form $n+m\sqrt2$.

Specifically, $a^{2k+1}$ is of the form $n+m\sqrt2$, and thus $a^{2k+1}\bar a^{2k+1}=(n+m\sqrt2)\overline{(n+m\sqrt2)}$

${}=(n+m\sqrt2)(n-m\sqrt2)=n^2-2m^2$.

Take $a=1+\sqrt2$.

As an example:$$(1+\sqrt2)^5=41+29\sqrt2$$

$$41^2-2(29)^2=1681-2(841)=-1$$

(On a related note, $\sqrt2-\frac{41}{29}$ is around 0.00042046)

Akiva Weinberger

7:11 PM

@user21820 @Adeek An uncountable $\Bbb Q$-linearly independent set is given in another answer: math.stackexchange.com/a/67751/166353

It uses the fact that there is an uncountable chain of subsets in $\Bbb N$, and then uses that to construct things of the form $\sum_{n\in A_r}\frac1{n!}$ with the desired property

Akiva Weinberger

7:32 PM

Is nobody here?

Random thing I found: mathforum.org/kb/servlet/JiveServlet/download/…

An assignment to "clean up" a proof that $e$ is irrational

Ted Shifrin

That is the outline of the proof I always gave in my Calc Theory class (following Spivak's book).

Eric Silva

isn't this proof in rudin

Ted Shifrin

Yeah, I think it's the universal proof.

Eric Silva

it's an oldie but a goodie

this reminds me that i dont know how to prove pi is irrational just off the top of my head

Daminark

My calc prof did some weird stuff

I remember that day we proved both, and he was like "Yeah I'm not gonna do Spivak's proof, the only thing that accomplishes is scaring you", after which he did some integration stuff

Like I think he was integrating some expression involving cosine, by parts, and some witchcraft happened after that

Ted Shifrin

7:46 PM

The $\pi$ irrationality proof in Spivak is pretty straightforward. The transcendentality of $e$ is a god-awful mess.

Daminark

It was recursive integration by parts, in fact

Lucas Henrique

@AkivaWeinberger oh, so

Eric Silva

isn't transcendality usually just a very difficult thing

Daminark

Like, $\frac{x^n(\pi - x)^n}{n!}{\cos(x)}$

That or sine, I forget

Lucas Henrique

$\overline{a^n} = \overline{a}^n$?

SteamyRoot

7:48 PM

@Daminark sine

and then you integrate it from $0$ to $\pi$, if I'm not mistaken

Daminark

@Ted I don't think he meant straightforward as much as, it felt like the sort of thing you just couldn't come up with

Alessandro Codenotti

@TedShifrin we had 4 hours of non mandatort lectures devoted to proving that $e$ and $\pi$ are trascendental last semester, it was quite interesting

Daminark

He said the proof he gave us, maybe we could come up with it, even if it took... a couple years

Eric Silva

@Daminark integration by parts is literally most of my life honestly

Perturbative

Integration by parts is love, integration by parts is life

Daminark

7:49 PM

@Steamy yeah that sounds right

And like, you sorta assume that $\pi = a/b$, and then define this $f(x) = x^n(a - bx)^n/n!$, which is a polynomial of degree 2n

Ted Shifrin

Well, Munkres makes the point in his point set topology book that the first proof that the best student wouldn't come up with (given enough time) is that of the Urysohn Lemma. I have a hard time deciding what I might have come up with, given how much I've learned and already seen.

Eric Silva

when people ask me what my job is this summer i usually respond "I integrate by parts"

Ted Shifrin

= Stokes's Theorem ;P

Eric Silva

Stoke;s's theorem

Akiva Weinberger

@LucasHenrique Yes. Try to prove it.

Daminark

7:52 PM

So if you integrate that by parts 2n+1 times, $f$ will die off because of derivatives, and then things work out nicely

Akiva Weinberger

In fact, it's easier if we generalize it a bit first @LucasHenrique

Daminark

The theorem of Stoke

Akiva Weinberger

Try to show that $\overline{ab}=\bar{a\vphantom b}\bar b$ @LucasHenrique

Ted Shifrin

BTW, Eric, I'm still waiting for you to think about that $\int \tau/\kappa\,ds$ theorem :)

Eric Silva

oh yeah i forgot about that

Daminark

7:53 PM

@Ted Lol Soug went the opposite way with that equality

He was saying "Yeah so next quarter Schlag will start you off by talking about differential forms, which make the proof of integration by parts easier"

Ted Shifrin

what equality?

oh.

Akiva Weinberger

Oh I spent a good amount of time thinking of that proof (Niven's proof of $\pi$'s irrationality) but I've forgotten most of it

Ted Shifrin

You still haven't learned nearly enough about forms, Demonark.

Akiva Weinberger

Someone should make a similar "clean up this proof" sheet for me to work through

Daminark

Oh yeah it was Bourbaki's proof I think

Eric Silva

7:55 PM

forms are so great honestly

maybe my favorite mathematical objects

SteamyRoot

@Daminark Both Niven and Bourbaki use that function, more or less

Ted Shifrin

I just read an article about how Alzheimers is essentially type-3 diabetes and that it's a matter of insulin resistance. Sadly, according to the formula she gives (in terms of glucose and triglycerides), I'm just over the danger threshold. :(

Akiva Weinberger

Oh I've only seen Niven's one-page proof

Eric Silva

uh oh

Akiva Weinberger

I mean, I've read books that have had other proofs in them

I just never bothered to look at them because they just looked so awful

Daminark

7:56 PM

Yeah that's true @Ted, I do hope to go deeper into those at some point

Ted Shifrin

GRR.