Mathematics - 2016-06-18 (page 4 of 4)

user174558

20:00

a=a is a logical axiom, lol.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

by reflex :P

user174558

I think studying logic can make one crazy, like Godel.

user174558

But maybe if you are already crazy, it can cure you.

Danu

My professor recently said something that I think reflects the essence: "The only thing worse than thinking about sets is thinking about numbers"

user174558

There is something worse than those two @danu...

user174558

20:02

It is thinking about shapes.

user174558

What is length, area and volume? That is a very deep question.

user 1618033

Back from jogging and also pretty exhausted (4.5 km).

user174558

I have not jogged for ten years.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

I see :-)

user 1618033

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ 4.5 km

Danu

20:03

@JasperLoy I don't think that's nearly as shitty a rabbit hole as rigorous foundations for set theory/numbers.

@user1618033 So how long does it take you?

user174558

@Danu Recently I ate a lot of carrots, so my shit is like rabbit shit.

Monad

Don't want to misuse any terms here but I accept the existence of infinity the same way I do 0.9 recurring or 0.0 recurring 1. :T

Danu

@JasperLoy Neat, carrots are delicious and healthy.

user 1618033

@Danu 40-45 min these days (because of lack of sleep and other kind of efforts).

Monad

If you believe that infinity cannot exist in reality then perfection is impossible unless imperfection is your perfection, otherwise...

user174558

20:06

@Danu I think the thing that confused me the most is thinking about determinism versus free will. At some point, I don't know what I am thinking.

Danu

@JasperLoy Not freely ;)

user174558

@user1618033 Ten years ago, I ran 5 km every day.

user 1618033

@JasperLoy Cool. Why don't you try that again?

Monad

Btw... who said there's no statement of infinity? According to the infamous Wikipedia there's an axiom of infinity, and an axiom... is a statement??? :T

Are they wrong? ._.

user174558

@user1618033 I am now a fat pig. =)

arctic tern

20:08

> who said there's no statement of infinity

Monad

@TobiasKildetoft

arctic tern

okay, you got me, who?

user 1618033

@JasperLoy LOL :-)))

user174558

@Monad That statement does not make sense, so there is no point discussing it.

Monad

@JasperLoy As in... the statement of infinity, or my statement?

user174558

20:09

@Monad Your statement.

arctic tern

what is this "statement of infinity" that you reference?

also, you probably wouldn't forgive me if I didn't tell you this is me @Jasper

yo

Monad

@arctictern en.wikipedia.org/wiki/Axiom_of_infinity

user174558

Who are you @arctic?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Do you agree that 0.999... = 1 @Monad

arctic tern

can you not

Monad

20:10

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ no :T

arctic tern

not sure if troll

user 1618033

@JasperLoy Height/Weight?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Prove it @Monad

Find the number between

arctic tern

Nov 8 '15 at 17:57, by anon

Pirates of the Carribean - Stop Blowing Holes in my Ship

►

user174558

@arctictern Sorry I don't know you. Are you Alex?

Monad

20:12

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ There is none

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

So they are the same

Shadock

Hello I am a bit ashamed but I have a question, i'm stuck on that for one hour...

arctic tern

2 mins ago, by arctic tern

yo

Monad

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ They are not the same, but they do not have a number in-between

arctic tern

^ hint

user174558

20:12

@user1618033 Right now I am 1.65 m and 70 kg, lol.

user174558

@arctictern The problem is that many people on SE chat say YO, LOL.

user 1618033

@JasperLoy well, that is not really bad!

arctic tern

hmm, you're actually right.

well whatever.

user174558

Oh, you are anon?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

@Monad if they do not have a number in-between then 1 - 0.999... = 0, right?

arctic tern

20:14

you win

Monad

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ no

user174558

anon? LOL!

Shadock

In one of my lesson about diffusion theory the teacher wrote $d_i=\frac{\frac{a}{c_i}}{D}=\frac{\frac{a}{z_i}}{cD}$ with $c_i=c*z_i$

arctic tern

I am almost inclined to ban 0.999...=1 discussions from chat.

user174558

I recently watched High Strung, a very nice movie about a violonist and a dancer.

Shadock

20:15

I don't understand how to play with these fractions to get the result...

Rrj

Metric Spaces book : Kumaresan or Searcoid? Which do you suggest

arctic tern

@Shadock put \displaystyle in front to make it bigger

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Write down the difference 0.999... - 1 = ? @Monad

user174558

@Rrj I have heard of neither of them.

Monad

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ 1/∞

arctic tern

20:16

also $$\frac{\displaystyle\left(\frac{A}{B}\right)}{C}=\frac{A}{B}\cdot\frac{1}{C}= \frac{A}{BC} =\frac{ \displaystyle \left(\frac{A}{C}\right)}{B}$$ not sure if I should comment about having an algebra I question in a diffusion class...

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Not a number @Monad

Real number

Monad

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ then 0.999... is neither

Rrj

@JasperLoy howdy, I used to be Nick.

user174558

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ There is no point subtracting the numbers when they have not even been defined.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Is 1/3 = 0.333... A real number @Monad

user174558

20:17

@Rrj I see. These two authors are unknown to me. Sutherland's Introduction to Metric and Topological Spaces is good.

Shadock

$$\frac{\displaystyle\left(\frac{a}{c_i}\right)}{D}=\frac{\displaystyle\left(\fr‌ac{a}{z_i}\right)}{c \cdot D}$$

user174558

@Rrj If you want to study topology, I suggest a full blown book on general topology, such as Willard's General Topology, which is a cheap Dover book.

Rrj

publisher?

Chinatsu-creepy-chan

$\frac{1}{3} = 0.333\dots \implies 3\frac{1}{3} = 0.999\dots = 1$

Monad

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Because 0.333... * 3 = 0.999..., assuming it is... then no

arctic tern

20:19

@Shadock put spaces between random things if it's not parsing correctly (even though it should)

Rrj

I love dover books, cheap and most of them are little treasures

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

8/9 = 0.888... @Monad

user174558

@Rrj Really depends on what topics you want, but a general topology book covers the metric spaces stuff as well.

Rrj

Looking for a neat book for my library, that's all

Shadock

$$\frac{\displaystyle\left(\frac{a}{c_i}\right)}{D}=\frac{\displaystyle \lef( \fr‌ac{a}{z_i} \right)}{c \cdot D}$$

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

20:20

9/9 = 0.999... = 1 @Monad

Shadock

ugh

user174558

@Rrj Then I recommend Willard. Will last you till PhD.

Monad

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ 9/9 =/= 0.999...

Shadock

oops i forget a t

$$\frac{\displaystyle \left(\frac{a}{c_i}\right)}{D}=\frac{\displaystyle \left(\fr‌ac{a}{z_i}\right)}{c \cdot D}$$

Chinatsu-creepy-chan

why did the second dots in my message show up wrong

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

20:21

Ok, what is it? @Monad

Chinatsu-creepy-chan

D:

user174558

Hi @anon. I forgot your favourite topic in math.

Rrj

@JasperLoy does that include metric spaces?

Monad

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ 9/9 = 1

user174558

@Rrj I think so, but I cannot recall. You can take a look at it first from you know where.

Shadock

20:22

$$\frac{\displaystyle\left(\frac{a}{c_i}\right)}{D} = \frac{\displaystyle \left( \fr‌ac{a}{z_i} \right)}{c*D}$$

what's the hell

Well i will take a picture it will be easier i will not take one hour more to explain where i'm stuck lol

anon

@Shadock do you understand that (A/B)/C equals (A/B)(1/C) equals A/(BC) ?

Shadock

Yes

But if ci=c*zi

anon

that means $(a/c_i)/D=(a/(cz_i))/D=a/(cz_iD)=(a/z_i)/(cD)$

Rrj

@JasperLoy I forgot, you are graduate?

Shadock

Let me write it on paper

user174558

20:25

@Rrj I finished my degree. Are you the Nick from India or another Nick?

Rrj

another nick

user174558

@Rrj The physics Nick?

Rrj

no, like evinda

user174558

Ah, I recall!

Shadock

@anon thank you !

exams exams exams... my brain becomes to be on fire

user174558

20:28

@Rrj amazon.com/General-Topology-Dover-Books-Mathematics/dp/…

user174558

@Rrj Hold on, let me take a look at that book...

Rrj

@JasperLoy thanks, promising reviews...

@JasperLoy the "counterxamples" books by dover are also cool

user174558

@Rrj Yes, I think you should just buy Willard. It is very, very good. I studied out of it ten years ago.

anon

@JasperLoy I read around. Currently I am reading a couple things by Qiaochu about applications of groupoid cardinality.
(1) One can prove if $a_n=\#\{H\le G:[G:H]=n\}$ then $$\sum_{n=0}^\infty\frac{\#\hom(G,S_n)}{n!}z^n=\exp\left(\sum_{n=1}\frac{a_n}{n}z^n\right).$$ (2) One can prove Mednykh's formula for surfaces $\Sigma$ $$\frac{\#\hom(\pi_1(\Sigma),G)}{|G|}=\sum_{\rm irrep}\left(\frac{\dim V}{|G|}\right)^{\chi(\Sigma)}$$ inspired by topological quantum field theory.

user174558

@anon Are you going to grad school soon?

anon

20:33

Still working on understaing (2). Particularly enjoyed classifying principal $G$-bundles over $\Bbb S^1$.

Maybe in a year.

user174558

For me, at most 5 more years.

user174558

If I don't make it then, I will forget it.

Mike Miller

@anon when you understand it, can you explain it to me?

anon

sure

Mike Miller

you might also like classifying principal G-bundles over $S^3$

anon

20:35

oh boy

user174558

I think I would like to work in algebraic geometry and differential geometry.

anon

the groupoid of G-bundles over S^1 with automorphisms turns out to be equivalent to the action groupoid of G acting on itself by conjugation

Rrj

@JasperLoy just downloaded it, will take a look

user174558

@Rrj Some copies are searchable, some are not.

Mike Miller

You'll have a tougher time with the groupoid of bundles over $S^3$ but I think you'll like what the set of bundles is, at least

anon

20:37

@Mike I assume it has something to do with recognizing S^3 has a group structure just as S^1 does, and translating the arguments using it

more generally what I don't quite get is why the groupoid of G-bundles over M with automorphisms is equivalent to the action groupoid of G acting on hom(pi1(M),G) by pointwise conjugation. given pi1(S^3)=Z this would mean it is essentially the same groupoid as with S^1.

user174558

I am leaving this chat now. Sorry to bother Mike with my book discussions. You may ignore me if needed.

Mike Miller

I'm a little skeptical of that.

That would be true if you said flat G-bundles

anon

sorry: principal G-bundles, G a finite group, M compact oriented without boundary

dunno what flat means

Mike Miller

Ahh, those are automatically flat. My question about $S^3$ is interesting for G a compact Lie group

anon

well, finite implies compact, so whatever the answer is it must restrict to the answer I gave for finite G...

Mike Miller

20:41

I can go into a little detail about the differences, not sure if you want to see though.

anon

go for it

Mike Miller

are you familiar with the idea of a structure group?

anon

nope

something about a group acting on fibers attached to every point of the base space?

Mike Miller

let me know if this is too general for your taste. so one of the most general descriptions of "bundles with structure" is a fiber bundle $E \to M$ with fiber $F$ and a class of local trivializations $p^{-1}(U) \cong U \times F$ such that the transition functions between these local trivializations lies in some fixed group $G \subset \text{Homeo}(F)$ ($F$ here should probably be compact but you can salvage it with other assumptions)

these give a cocycle in Cech cohomology $Z^1(M;G)$. we say that two cocycles are equivalent if they differ by a boundary in Cech cohomology. there is a better way of phrasing this but I'm momentarily blanking

anon

> these give a cocycle in Cech cohomology

noped out

Mike Miller

20:48

I want to say what flat is without saying curvature :)

I can fix this, gimme a second

Danu

I would like to learn about Cech cohom.

bolbteppa

8 + 5 = 3 with cocycle 1 right?

anon

heh

Danu

Any decent book that covers it?

Mike Miller

forster ;)

OK, a bundle $E$ with structure group $G \subset \text{Homeo}(F)$ is a fiber bundle with a maximal atlas of local trivializations w/ transition maps in $G$

anon

20:50

So like a vector bundle should have a GL() structure group?

Mike Miller

an isomorphism of bundles over $M$ is a fiberwise homeomorphism $E \to E'$ such that in each trivialization, the homeomorphism is given by an element of $G$. that's what I wanted, ok.

yeah, exactly

or principle $G$-bundle and structure group the set of left-multiplications by elements of $G$

or fiber bundles and the whole structure group

bolbteppa

I guess the thinking is that the numbers cycle through $\{0,1,...,9\}$ but then you have this extra set of numbers $\{0,1\}$ to cycle through

Mike Miller

so if you have a fiber bundle with structure group $G$, a reduction to structure group $G' \subset G$ is picking a maximal subatlas such that the transition maps live in $G'$ instead

anon

wait pi_1(S^3) doesn't equal Z what was I even saying up there

Monad

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ sorry I was busy talking with relatives... I don't have time to continue discussing and I'm not just trying to disprove you for the sake of winning regardless of cliche this excuse is going to sound... but my grandfather is a lecturer for mathematics in University and 0.999 =/= 1, whether it's a real number though is another story.

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ bye

of how cliche*

bolbteppa

20:52

@Monad $0.999... = 1$

Monad

nope... but anyway

I gtg sorry man...

anon

if we start that again I'm moving all the messages to a different room

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Thanks for trying @Monad

anon

let people be wrong on the internet

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:-)

Mike Miller

20:54

really I should say that in the appropriate generality: $G'$ is just a group equipped with a homomorphism $\varphi: G' \to G$, and then a reduction is a choice of atlas with trivializations etc in $G'$ such that when you send them to $G$ with $\varphi$ you get (a subset of) your original atlas

Semiclassical

eh, let's just agree to work in base 12 and leave it at that

bolbteppa

@Monad It has an easy proof en.wikipedia.org/wiki/0.999...#Infinite_series_and_sequences

Semiclassical

@bolbteppa "Forget it, Jake. It's Chinatown."

2

Mike Miller

i'm not sure if this is parsing very well. the best example I have is that a complex structure on a real $(2n)$-dim vector bundle is a reduction of the structure group from $GL(2n,\Bbb R)$ to $GL(n,\Bbb C)$; a Riemannian metric on each fiber is a reduction of the structure group from $GL(n,\Bbb R)$ to $O(n,\Bbb R)$; an orientation is a reduction to $GL^+(n,\Bbb R)$, etc

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

@Semiclassical classic

:)

Semiclassical

20:55

heh

bolbteppa

$1 + 2+ 3 + ... = -1/12$ on the other hand :p

Semiclassical

ahh, zeta

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:P

Mike Miller

@anon should I keep going or no?

anon

it's interesting but I'm mostly filing it away for later

Semiclassical

20:57

@bolbteppa makes me think of this: smbc-comics.com/index.php?id=3777

bolbteppa

I remember some insane conformal field theory thing where $-1/12$ showed up as part of a central charge and plugging Zeta in as not being obvious to me was a high source of shame :p

Mike Miller

alright, I'll just say what I wanted to, I guess

then a flat structure on a $G$-bundle is a "reduction of the structure group" to $G^\delta$, $G$ with the discrete topology. equivalently, it's picking a maximal atlas of charts such that the transition functions are constant. so it's clear why if $G$ was a discrete group to begin with, a flat structure is god-given

user 1618033

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I perceive you as one of the wisest people here undoubtedly. Why do you think people here don't star the pictures of Ramanujan? I posted more pictures in a row, and no star.

Mike Miller

then your classification theorem actually extends to flat bundles with arbitrary structure group, where isomorphisms are demanded to preserve the flat structure

bolbteppa

That's pretty good, always think of modelling the formation of mountains and trees with fractals in that Mandlebrot documentary when I hear the word fractal, there is magic there

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

21:00

@user1618033 different people have different heroes

2

user 1618033

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ one of the explanations was that some are scared to death by the stuff of this guy, but I feel this is not the best explanation. I think it's more.

Mike Miller

but it doesn't survive for arbitrary bundles and arbitrary isomorphisms. as an explicit example, there are $\Bbb Z$-many $U(1)$-bundles on $S^2$ (you can probably think of three off the top of your head), but only one of those (the trivial one) has a flat structure, which is more-or-less unique, as predicted by your groupoid

user 1618033

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Great answers! Really!

Semiclassical

Arguably, the math translation of the one re: Kempner's series is just "If you're a really big number, you probably contain a nine."

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:-)

Mike Miller

21:01

what I was thinking of earlier was the classification of $G$-bundles on $S^3$, not necessarily flat. there's only one flat bundle up to isomorphism with $G$ fixed. but one can ask to classify any $G$-bundle (where here I would like to demand $G$ a Lie group, or you have a little trouble classifying).

that's all

bolbteppa

"Homology is everywhere, it measures defects/obstructions" - 100 years from now

Semiclassical

I wonder a bit about that.

SAWblade

Is anyone here experienced with Mathematica? :0

user 1618033

@Semiclassical did you try my series above?

Semiclassical

can't say i have. not really feeling it today.

@SAWblade That's a pretty broad question. Better to ask about what features/functions you're trying to use.

SAWblade

21:13

Fair enough. I'd like to know how I could go about writing code to get Mathematica to compute this: i.imgur.com/BwK7RzI.png

Semiclassical

Hmm. For the most part, that'll just be use of the Sum function

For the product...hmm.

SAWblade

I can't figure out how to make it compute a variable number of indices, though.

And the product is mean as well I agree. xD

Semiclassical

Are you trying to get it to simplify that sum, or to compute it for some particular combination of a_i's and n?

SAWblade

Simplify the sum. :0

Semiclassical

Then I wouldn't hold out much hope.

SAWblade

21:16

Mm, figured.

Semiclassical

I do remember how to make it do a product, though.

SAWblade

Now to instead try to write code for input values of $m,n$.

Semiclassical

Times@@Table[a[i],{i,1,3}] would give a[1]a[2]a[3]

The With function might be useful for that

SAWblade

With?

Semiclassical

e.g. With[{m=3,n=2},f[m,n]] outputs f[3,2]

it gives you a way to use specific values within an environment without actually defining them

SAWblade

21:19

Ah, sweet.

Semiclassical

There's also the Module and Block functions, but they're a bit weirder

SAWblade

How do they work? :0

Semiclassical

I don't know, hence why I say they're weirder :P

SAWblade

Okey doke. xD

Thanks for the advice, by the way, I appreciate it. :)

Semiclassical

np

user 1618033

21:30

@Semiclassical No worry, you only miss a cool problem. :D

Alan Russel

hello everyone

I am trying to figuring out how many nodes we have in a complete tic tac toe game tree. I am not sure if my solution is correct though

(I am trying to calculate a simple upper bound)

I have this
1 node(s) in level 0
=> overall nodes after round 0 : 1 node(s)

9 node(s) in level 1
=> overall nodes after round 1 : 10 node(s)

72 node(s) in level 2
=> overall nodes after round 2 : 82 node(s)

504 node(s) in level 3
=> overall nodes after round 3 : 586 node(s)

3024 node(s) in level 4
=> overall nodes after round 4 : 3610 node(s)

15120 node(s) in level 5
=> overall nodes after round 5 : 18730 node(s)

60480 node(s) in level 6
=> overall nodes after round 6 : 79210 node(s)

So I think there are in total (as an upper bound) 623530. Is this right?

\begin{equation}
\begin{split}
\text{Total number of nodes} & = $1 + 9 + 9 * 8 + 9 * 8 * 7 + ...$\\
& = \sum^{8}_{i=0} \frac{9!}{(9-i)!}
\end{split}
\end{equation}

Please tell me if you need more information to help me out :) thx

Hiroto Takahashi

22:17

Hi

someone know what is the geometrical interpretation of Cartier divisors?

Mike Miller

I suspect there are other things more deserving of your attention right now.

Andrew L

22:48

I'm not sure how to solve this problem. Teacher skipped the whole section. How will I approximate the solutions of 'tan theta = 0.78' in the interval [0,2pi)

arctan of 0.78 0.662426294 radians, but I'm not sure what to do from there.

tangent is positive in quadrants 1 and 3

Jake1234

23:09

@TedShifrin Thanks a lot.

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$Mathematics$ Mathematics