Mathematics - 2017-08-15 (page 1 of 5)

Semiclassical

00:00

Almost every number between 0 and 1 is transcendental. But if you asked someone to name an example of a real number between 0 and 1, they usually won't pick a transcendental one.

Secret

$\pi / 2047$

mdave16

sometimes i feel like i know certain maths without applications, but then i realise i just don't know that many applications, and i can't rule out existence

Semiclassical

Or 1/e, or 1/pi

Typhon

e/2

mdave16

the same goes for normal numbers, within the reals, we can prove that there is an uncountable number of them, yet not only would a normal number not be picked, but we can't construct one!

Typhon

00:02

duh

Secret

Most transcendental things don't even have names because we only have a finite alphabet to name things

Damn autocorrect

Typhon

@mdave16 normal numbers?

what is a "normal" number?

mdave16

Normal number

In mathematics, a normal number is a real number whose infinite sequence of digits in every base b is distributed uniformly in the sense that each of the b digit values has the same natural density 1/b, also all possible b2 pairs of digits are equally likely with density b−2, all b3 triplets of digits equally likely with density b−3, etc. Intuitively this means that no digit, or (finite) combination of digits, occurs more frequently than any other, and this is true whether the number is written in base 10, binary, or any other base. A normal number can be thought of as an infinite sequence of coin...

Secret

Equal probability for every digit to appear

Semiclassical

Every digit in its decimal expansion occurs with equal frequency

Secret

00:04

$\pi$ is suspected to be one, but nobody had finished proving it

Semiclassical

I'm not sure how many specific numbers are known to be normal

mdave16

i lied, we have some normal numbers

i must have misremembered

Secret

0.123456789101112131415...?

Semiclassical

I think the examples have been specifically constructed so as to be provably normal

Trying to prove it for something like pi is a lot harder

Secret

maa.org/sites/default/files/pdf/upload_library/22/Ford/…

Semiclassical

00:08

Anyways. The point of the analogy is that the set of examples we're familiar with can be a lot smaller than the entire set

Secret

Just like how we can only see the visible spectrum, but not all of the EM

Semiclassical

So too it is with equations. If you've seen it written somewhere, it had a use

Secret

$0=0$ is very useful to me to tell either the expression reduces to a tautology or that something went wrong

mdave16

i'm very good at deriving it

Semiclassical

And $a=0\implies x=x+a$ is something we use all the time

mdave16

00:11

you mean the axiom of additive identity?

Secret

Converse is true only for number systems that has a as additive identity though

Semiclassical

Sure. I have in mind just elementary manipulations

Eg $x/(x-1)=(x+1-1)/(x-1)=((x-1)+1)/(x-1)=1+1/(x-1)$

mdave16

according to comments on the right, apparently you are just procrastinating

i feel bad for distracting you

Semiclassical

Eh, today wasn't a day I was expecting to get much done due to other circumstances

mdave16

what sort of work do you have to do?

we could help if it is vaguely mathematical, or we could provide moral support from outside the ring

Semiclassical

00:20

Writing.

Which is mostly a matter of taking stuff from other stuff I've contributed to and combining it into one manuscript

So the math is pretty much done at this point.

mdave16

ah, writing a book?

you can do it! keep going!

Semiclassical

Thesis.

Secret

Writing is one of those jobs that I want an AI that think like me but had no emotions to do it instead

Semiclassical

So like a book except only a handful of people will ever read it

mdave16

what's it on

i'd be up for reading if it's something i'm into

Semiclassical

00:26

It won't be :P

mdave16

how can you be sure

i'm also not doing anything these days, so i'm up for learning anything i'd need. I'm currently trying to get a group to go through a category theory book with

in comparison, right now, i'm trying to find a good broadband and phone deal, so fun

micsthepick

When I try to count the number of edges in a truncated icosahedron (soccer/football) I keep getting wrong answers

mdave16

how are you counting?

micsthepick

Given that the shape contains 12 regular pentagons and 20 hexagons as faces, I first counted the edges that are adjacent to all pentagons, as they do not share faces, and It appears that there should be 12 remaining edges, giving 72 as the total

attempting to count all edges from every face and dividing by two gave me 110

So the second method will work, but you must make sure that hexagons have 6 edges not 8

The first method works if you count the remaining edges right, 5 neighbours divide by 2 because each edge has two neighbouring pentagons, and multiply by the number of pentagons, 12, giving 30 for the number of remaining edges

Secret

00:44

chat.stackexchange.com/transcript/36/2015/3/25/5-16

mdave16

wait, i was distracted by Sky TV ... So, no pentagon shares a face, so that gives us 60 edges right, where are you getting an extra 12 from?

Secret

Wowzer functions seemed to describe how the epsilon numbers are constructed in terms of $\omega$

micsthepick

there should be 90 edges total

mdave16

i know, theres lots i haven't counted yet

micsthepick

I thought that after the 60, there should be 12 extra because of some flawed logic

mdave16

00:46

so now, the remaining edges i haven't counted don't touch a pentagon,

micsthepick

and I made an error when attempting to double count every edge and divide by two

mdave16

now, each pentagon will have an edge connecting it's tip to another tip of a pentagon (these are the uncounted ones)

each pentagon has 5

micsthepick

yes

mdave16

but, this is double counting

micsthepick

correct

mdave16

00:47

so 5*12/2 = 5*6 = 30

micsthepick

4 mins ago, by micsthepick

The first method works if you count the remaining edges right, 5 neighbours divide by 2 because each edge has two neighbouring pentagons, and multiply by the number of pentagons, 12, giving 30 for the number of remaining edges

mdave16

30 + 60 from earlier gives me 90

i was a bit slow

micsthepick

I figured it out myself already :P

mdave16

in fairness, i was distracted by deals for a while

which is better, sky sports or sky movies?

Daminark

01:12

Um... good lord

Wait Secret can you delete that message

Or edit it still?

Like it's making the screen long

@Secret

And that affects typing

Merp

Secret

02:06

uh...., unfortunately, I was at the seminar moments after typing this (and just returned, hence the lack of responses), thus it has been 1 hr since then, thus the edit had expired.

I am sorry that the single $$ seemed to make it unexpectedly too long, guess I will just use a pastebin in the future if I need to type such long things again

@anon, since you are room owner, perhaps you can trash the above unexpectedly screen stretching message for me as requested by Daminark?

anon

→ 2 messages moved to trash

Secret

cool thanks

anon

you have a tendency to talk to yourself and use the chatroom as a livejournal

Secret

yeah, that's still an issue to be dealt with. I have started blogging now earlier this week, thus it should be less...

Jasper Loy

07:09

@Secret I do enjoy listening to your random talk though. =)

Secret

lol thanks

(NB Screen stretching is not obvious in my mac and windows, which is probably why I failed to detect it in the first place)

LLL

07:36

math.stackexchange.com/questions/2393463/…. Any comment on this further formal or informal

Tobias Kildetoft

@LLL Your edit only addressed the second thing mentioned in a comment, so it is still not very clear what is being assumed.

Martin Sleziak

08:13

@anon and @Secret If it is problematic here, one solution would be a separate room. (Maybe user interested in the stuff would visit that room. And you can occasionally add a link here. Like - I am currently thinking about strongly algebraic ordinals, I tried to write a bit about their relation to modular abelian categories here - followed by a link to the relevant part of the transcript of the other chat room.)

Needless to say, both names are made up - I don't think such things really exist.

Personally I do not visit the main chatroom very frequently, so I'm not able to judge whether or not it's a problem.

Secret

I actually would have posted in my room more often had SE don't have the system of autofreezing chat rooms after 15 days, thus the pressure of upkeep is annoying in fact, my current chat room is kept alive only because I have set up autofeeds to numerous SE with topics I am interested in

Martin Sleziak

Feeds do not prevent room from freezing.

@Secret Do you mean SecretLabs room?

Secret

yup

That's my current hub for my notebooks in the SE portion of my internet footprint

Martin Sleziak

It seems it hasn't been frozen so far.

But if I notice that it's close to 14 days limit, I'll post something there :-)

Secret

I have numerous such notebooks and time capsules throughout the whole internet, and currently I am kinda starting to condense them all in one location, which takes some time because of the sheer amount of stuff I have posted and the number of domains involved

Tobias Kildetoft

08:20

@Secret Martin is pretty good at keeping chat rooms alive

Martin Sleziak

I think I've failed a few times. (And I guess as soon as my first suspension comes, some of the rooms might go away.)

Secret

I do have another room planned in the pipeline: Repository of Unnatural Algebraic Structures. However, that room cannot be set up unless I get my solid algebra background, otherwise upkeep with relevant topics can be difficult there

But now that I am on my way to set up a blog, I think it will be better to just do it there

The algebra note is also the reason why I discuss a bit less algebra recently in the maths chat, which does kinda help on my procrastination issue as a lol side effect

Daminark

Hey everyone, I've been wondering how reasonable it is to attempt the proof that every manifold (smooth?) is a CW complex

Balarka Sen

08:35

Not reasonable. For non-smooth guys it's actually wide open in dimension 4.

Daminark

Hmm, is there at least some intuition about it?

(Also how about triangulability?)

Balarka Sen

Yes, so smooth manifolds are actually stronger than CW, they are triangulable.

What's an intuitive proof, hm.

Usually you triangulate by giving it a Riemannian metric and then using the convex geodesic neighborhoods carefully

@Daminark Maybe you'll like the Morse theory idea?

Daminark

I dunno any Morse theory

Balarka Sen

Let me tell you about it a bit.

Take a torus on $\Bbb R^3$, embedded not in the usual way but from ground-up. So it touches the xy-plane at a point, and the donut-hole lies parallel to the yz-plane.

Daminark

Sweet

Tobias Kildetoft

08:40

Does anyone here know anything about Hodge theory? I would really like to understand the recent ideas by some authors which are supposed to be an algebraic version of it, and it might help if I had some idea what the "usual" Hodge theory is all about.

Balarka Sen

@TobiasKildetoft I can tell you a bit about Hodge decomposition, but surely Mike and Ted knows Hodge theory.

@Daminark Ok, here is a nice picture.

Daminark

Ah

I think that'll get you, at least orientable surfaces, yeah?

Like as a connected sum of tori

Balarka Sen

We're just working with the torus for now.

In any case, take the "height function" on the torus. That's a map $f : S \to \Bbb R$ such that $f(x)$ is the "height of $x$ in that picture".

This is just the projection map of that torus onto the z-axis.

Notice that there are four critical points of $f$; the bottom, the first saddle, the second saddle, and then the top.

Daminark

True dat

Leaky Nun

How difficult is it to prove that the dual space of an infinite-dimensional vector space is never isomorphic to itself? I tried using cardinalities but it wouldn't work

@BalarkaSen

Tobias Kildetoft

08:50

@LeakyNun Why didn't it work?

Balarka Sen

@Daminark Well, let's say image of $f$ is $[a, b]$ and the critical points are $a, x, y, b$.

Consider $f^{-1}[a, x)$. That's a disk in the torus of height smaller than the first saddle.

Leaky Nun

@TobiasKildetoft because R and R^N have the same cardinality

Tobias Kildetoft

@LeakyNun Right, and the dual space has strictly larger cardinality

Leaky Nun

@TobiasKildetoft I don't think so

consider the real sequences with finite support. Its cardinality is c. So is its dual space's.

Daminark

@BalarkaSen Would it be a disk? From the picture it seems open

LLL

08:53

math.stackexchange.com/questions/2393463/…. I have again edited the question.

Tobias Kildetoft

@LeakyNun Hmm

Balarka Sen

@Daminark Let's move to DC, it's too crowded here

Tobias Kildetoft

@LeakyNun Right, so instead we need to just find sufficiently many linearly independent maps that we exceed the dimension of the original space

Leaky Nun

@TobiasKildetoft I had some progress when restricting the coefficients to 0 and 1, but not much.

Tobias Kildetoft

Consider a basis and the corresponding duals. Now imagine adding infinitely many of these in any way you want. Make this precise to get a larger set of independent maps

Martin Sleziak

09:01

I'll also add a link to this: Why are vector spaces not isomorphic to their duals? Other posts which are linked there might be of interest, too.

And this MO question is linked in comments: Slick proof?: A vector space has the same dimension as its dual if and only if it is finite dimensional

Secret

A weird extended question, must a mathematical object A be isomorphic to itself because A = A is a tautology?

Leaky Nun

@MartinSleziak how long is that?

@Secret I suggest that you look up the terms you are trying to use before asking questions about them...

in this case, "isomorphic".

Martin Sleziak

@LeakyNun I'm not sure I understand the question...?

Tobias Kildetoft

@Secret There are "places" where this does not hold, but we don't go there :)

Leaky Nun

@MartinSleziak how long is the answer... so long

Martin Sleziak

09:08

So it was a rhetorical question, right? (And as such, I should not have answered it.)

Tobias Kildetoft

More precisely, those places are called semicategories

Leaky Nun

@MartinSleziak you can say so

Martin Sleziak

Is this a rhetorical question (Y/N)? :-)

Leaky Nun

@MartinSleziak :p

Martin Sleziak

A joke stolen from QI. (At least that's where I heard it.)

Tobias Kildetoft

09:12

I actually spent some time trying to find the correct notion of a fiat $2$-semicategory, but that turned out to be more trouble than it was worth

Leaky Nun

> Now let $V^*$ be the dual of $V$. Since $V^* = \mathcal{L}(V,F)$ (where $\mathcal{L}(V,W)$ is the vector space of all $F$-linear maps from $V$ to $W$), and $V=\mathop{\oplus}\limits_{i\in\kappa}F$, then again from abstract nonsense we know that
$$V^*\cong \prod_{i\in\kappa}\mathcal{L}(F,F) \cong \prod_{i\in\kappa}F.$$
Therefore, $|V^*| = |F|^{\kappa}$.

I know that "abstract nonsense" is a reference to category theory, but could somebody explain?

@MartinSleziak

Tobias Kildetoft

@LeakyNun Hom sends direct sums (on the left) to direct products.

Martin Sleziak

To add a bit of context for others, thet above quote is from Arturo Magidin's answer.

Leaky Nun

@TobiasKildetoft can it be proved?

Tobias Kildetoft

@LeakyNun Sure, it is just a matter of writing up the definitions (and making sure you are working in the correct categories for each object, which is not a problem here)

Leaky Nun

09:19

@TobiasKildetoft without categories?

Tobias Kildetoft

@LeakyNun Yes, but it takes a bit more work, writing up what everything actually does

Leaky Nun

@TobiasKildetoft in other words can I prove that an element in $\operatorname{Hom}(V,F)$ is uniquely determined by its value in the standard basis of $V$?

Tobias Kildetoft

@LeakyNun That is a standard linear algebra thing

linear maps are uniquely determined by their value on a basis

Martin Sleziak

Basically, what you're saying Tobias is this? $$\mathcal L (\prod\limits_{i\in I} V_i,W) \cong \prod\limits_{i\in I} \mathcal L (V_i,W)$$

Leaky Nun

@MartinSleziak no, sum

$\bigoplus$

$\displaystyle \mathcal L \left( \bigoplus_{i\in I} V_i,W \right) \cong \prod_{i\in I} \mathcal L (V_i,W)$

Tobias Kildetoft

09:23

Because $\operatorname{Hom}(-,W)$ is contravariant, so it turns direct sums into direct products (well, given that it turns them into something)

Leaky Nun

I like to think of $\Bbb N$ as a vector space, over $F_2$

with addition being $\oplus$

7+9 = 14

Secret

From a very brief skim read of the first few lines in nlab, semicategories kinda reminds of semirings and semigroups without identies a bit. Interesting. However will save that for later after the foundation is solid

Tobias Kildetoft

@Secret semicategories are terrible things where nothing makes sense any more.

Secret

I like the weirdness and so far I am not afraid of them yet. But to explore the weird, one needs to be well prepared

Tobias Kildetoft

Certainly not much has been written about them that I could find. Though it is interesting to note that the forgetful functor from categories to semicategories has both a left- and a right- adjoint, where one is the "naive" additions of identities, and the other is the Karoubi envelope (which I had never considered before in this context)

Leaky Nun

09:54

Let $\{v_a : a \in A\}$ be vectors indexed by infinite set $A$. What does it mean that they are linearly independent?

@AkivaWeinberger buenos dias

Akiva Weinberger

Buenos días

Tobias Kildetoft

@LeakyNun Same as if they were indexed by a finite set

Leaky Nun

@TobiasKildetoft but infinite linear combination may not be well-defined

Tobias Kildetoft

i.e. no linear combination gives $0$ except the one

Right, linear combinations are still finite

Leaky Nun

alright, thanks

@TobiasKildetoft @AkivaWeinberger consider the vector space $\displaystyle \bigoplus_{r \in \Bbb R} F_2$ over the field $F_2$...

Akiva Weinberger

09:57

@LeakyNun Similarly, the span of $\{1,x,x^2,\dots,x^n,\dots\}$ is the set of finite sums, i.e. it's the polynomials rather than the formal series

Leaky Nun

@AkivaWeinberger right

Akiva Weinberger

The idea is that it's the smallest ring that contains all of them @LeakyNun

And the polynomials is smaller than the the formal series

Leaky Nun

And then consider the span of $\left\{\dfrac1{1-x},1,x,x^2,\cdots,x^n,\cdots\right\}$ lol

Akiva Weinberger

That would be linearly independent :)

And still not the entirety of formal series, I think that requires an uncountable number of things to span it

Leaky Nun

I know

Akiva Weinberger

10:02

and you probably need choice to get a minimal spanning set? A basis?

Leaky Nun

@AkivaWeinberger I think so

How would one define a field of cardinality $2^\mathfrak c$?

Kirill

I have an equation $x(1-\gamma_1x)(1-\gamma_2x)=(\beta_1(1-\gamma_2x)+\beta_2(1-\gamma_1x))(1-x-x^2‌)$ (partial fraction decomposition of the Fibonacci function). The text says: "the comparsion of constants implies $\beta_2=-\beta_1$". What comparsion of what constants, how did the author do that?

Leaky Nun

@Kirill of the constant term

Kirill

@LeakyNun please, concrete

Leaky Nun

@Kirill the coefficient of $x^0$

In other words, let $x=0$.

Kirill

10:08

aha

so, it would be not to bad multiply all the terms out on the both sides? And, "comparsion" is the comparsion of the coefficients of the polynomials?

Leaky Nun

@Kirill yes to the latter

Kirill

so, $\neg $yes to the first, means you don't like the first idea. Ok, I will follow the text, thanks.

Leaky Nun

@Kirill no, I didn't understand the former

Kirill

@LeakyNun I mean, in order to compare the coeffitiens I need to see the polynomial, to multiply out the brackets

Huy

@Kirill: multiplying all terms out on both sides would be much more work than required here, since setting $x=0$ already yields $0 = \beta_1 + \beta_2$.

Leaky Nun

10:16

@Kirill of course you don't need

Kirill

@Huy I thought I could get some nice equations for the rest gammas!

Huy

what are you studying? @Kirill

Kirill

@Huy mathematics?..:)

Huy

@Kirill: more specifically?

Kirill

@Huy at the moment I am looking at the generating functions on the example of Fibonacci. They use formal power series, partial fraction decomposition and a good portion of luck.

Leaky Nun

10:39

@Kirill just use power series lol

Kirill

@LeakyNun what you mean?

Kirill

10:55

I don' understand. The text says, "we can interprete $\frac{1}{1-\gamma_ix}$ as the geometric series $\sum_{n=0}^{\infty}(\gamma_ix)^n$". But we can do that only if $\gamma_ix < 1$, and for $\gamma_i$ being $\frac{1+ \sqrt{5}}{2}$ that is not the case. On the other hand, we are looking at the formal power series . That means, we are not interested in the convergence, but at the same time we took the limit of this series for our calculations. Is that not cheating, at all?

Leaky Nun

$\begin{array}{rcl}
f(x) &=& \displaystyle \sum_{n=0}^\infty a_n x^n \\
f(x) &=& \displaystyle 0 + x + \sum_{n=2}^\infty a_n x^n \\
f(x) &=& \displaystyle 0 + x + \sum_{n=2}^\infty a_{n-1} x^n + \sum_{n=2}^\infty a_{n-2} x^n \\
f(x) &=& \displaystyle 0 + x + \sum_{n=1}^\infty a_n x^{n+1} + \sum_{n=0}^\infty a_n x^{n+2} \\
f(x) &=& \displaystyle 0 + x + x \sum_{n=1}^\infty a_n x^n + x^2 \sum_{n=0}^\infty a_n x^n \\
f(x) &=& \displaystyle 0 + x + x \sum_{n=0}^\infty a_n x^n + x^2 \sum_{n=0}^\infty a_n x^n \\

@Kirill ^

Kirill

@LeakyNun I've made that, that is not what I or the text want.

Leaky Nun

@Kirill that's just not rigorous, lol

Kirill

@LeakyNun the aim is to get a formula for $a_n$ - that is that with the square roots of 5.

Leaky Nun

@Kirill oh, I use matrix for that :P

Transcript for

$Mathematics$ Mathematics