Mathematics - 2017-09-01 (page 1 of 5)

Kasmir Khaan

00:01

if we have the right side inverse it does

Leaky Nun

but you can't skip all of those steps as if it is apparent

Kasmir Khaan

ax=x
axx' =xx'
a=e

Leaky Nun

@KasmirKhaan you skipped 100 steps again.

we keep talking in circles

Kasmir Khaan

Argggg

It is good to it step by step like you said

Leaky Nun

yes, it is

Kasmir Khaan

00:02

but I dont think my teacher expect us to do it that way

Leaky Nun

then don't do it that way

Kasmir Khaan

Ill ask to be sure

Leaky Nun

just write whatever you want

Kasmir Khaan

I want to learn the regor way ofc

Leaky Nun

and use fallacies everywhere

Kasmir Khaan

00:03

that is the best way

Leaky Nun

your proof of ex=x is not even wrong.

Kasmir Khaan

noooooooo

not even right you mean ?

Leaky Nun

Not even wrong

The phrase "not even wrong" describes an argument or a theory that purports to be scientific but is based on sloppy logic or speculative premises that cannot be discussed in rigorous scientific sense. The phrase is generally attributed to theoretical physicist Wolfgang Pauli, who was known for his colorful objections to incorrect or sloppy thinking. Rudolf Peierls documents an instance in which "a friend showed Pauli the paper of a young physicist which he suspected was not of great value but on which he wanted Pauli's views. Pauli remarked sadly, 'It is not even wrong'." This is also often quoted...

to be right is good

to be wrong is unfortunate

to be not even wrong...

Kasmir Khaan

okay =P

Ill make it right then :D

Leaky Nun

good luck

you can't just use what's intuitive for you

Kasmir Khaan

00:07

well but overall

was those the only comments that I need to fix?

Leaky Nun

your first part is completely wrong

Kasmir Khaan

part a) ?

Leaky Nun

...

the part where you "proved" ex=x

Kasmir Khaan

i wrote a) b) and c)

Leaky Nun

I'm just talking about a).

Kasmir Khaan

00:08

okay

xe=x
xex' =xx'
xx'=e
xx'x=x
(xx')x=x
ex=x

those are the highlights

I did nto type any justicfication here '

Leaky Nun

I told you what your mistakes were

Kasmir Khaan

but why is that proof wrong?

Leaky Nun

you agreed that those were mistakes

you said you saw what I mean

and then you make them again.

Kasmir Khaan

I mean there is a degree of wrong

it can be all false and it can be some missing details

Leaky Nun

@KasmirKhaan one wrong step makes it all wrong.

A chain is only as strong as its weakest part

A proof is completely annihilated by one wrong step.

Kasmir Khaan

00:11

That is true

But am still in the learning curve

Leaky Nun

line three to line four

you skipped one step

that step is the wrong step

Kasmir Khaan

I multiplied by x

on both sides

should i put x*e

Leaky Nun

which side?

Kasmir Khaan

i mean e*x

right side

Leaky Nun

yes, you should

don't skip that step at all.

Kasmir Khaan

00:14

okay ill fix it now

@LeakyNun how should I define x' ?

just say that x' is the inverse of x?

Leaky Nun

you used a symbol without knowing what it means @_@

@KasmirKhaan no.

Kasmir Khaan

hmm

x' is an element such that xx'=e

better?

Leaky Nun

in words?

Kasmir Khaan

x' is in G , such that xx'=e

I think my brain stopped working ><

Ill fix all of this when i wake up

:D

@LeakyNun Goodnight for real now !

Leaky Nun

ok

Daminark

00:26

Hey

Leaky Nun

@Daminark do you want to see a magic trick?

The Bear of Bad News

00:36

Hi

micsthepick

So if I want to arrange four Green and four Red balls in a circle, such that there is at least one pair of Green balls sitting next to each other, how many ways can I do it?

Leaky Nun

@AkivaWeinberger do you want to see a magic trick?

@micsthepick how many ways are there not to have at least a pair of green balls sitting next to each other?

micsthepick

Well there could be 1, or 4*3!^2

Leaky Nun

what is 4*3!^2?

oh, you're counting them distinct

micsthepick

144

Leaky Nun

00:47

I think they're not distinct

so it's just 1

micsthepick

interesting

Leaky Nun

hmm

I think we should look at them as a line first

as different configurations have different rotational symmetries

which might make counting a bit complicated

Akiva Weinberger

@LeakyNun Sure

Leaky Nun

A simple sphere eversion

►

Akiva Weinberger

Seen it

Leaky Nun

00:49

at first i'm like "ok that's trivial"

Akiva Weinberger

Really cool

Leaky Nun

and then at the last step i'm like UWOTM8

Akiva Weinberger

The bit in the middle where they isolate the ring is confusing but I think I get it

Leaky Nun

@AkivaWeinberger I don't

I just keep looking at how the curve turns and trying to convince myself that it's an invariant

Akiva Weinberger

From like 43 second in to 48 ish

Leaky Nun

00:50

right

Secret

[Random]

Leaky Nun

do you know what the invariant is called? @AkivaWeinberger

Secret

Quotient groups

Leaky Nun

@Secret yay I like those

Akiva Weinberger

@LeakyNun What invariant?

Leaky Nun

00:51

@AkivaWeinberger a topological invariant

I think this is diff.geom right

Akiva Weinberger

At 0:43-0:48, you see a little loop inside a big loop, right? The big loop is essentially a shaped like $\Bbb R\times L$, where $L$ is a 1D loop

Locally

Secret

$\Bbb{R}/\Bbb{Z} = \{n\Bbb{R},n\in \Bbb{Z}\} = $something to do with [0,1) I think?

Akiva Weinberger

@Secret $n+\Bbb Z$, wouldn't it be?

Think of what happens if you were to unwrap that big loop into just a plane

The small loop then doesn't pass through itself at all, it's just rotating

$\Bbb R/\Bbb Z$ is the circle group. @Secret

It's like the reals mod 1

The circle group can be thought of as the unit circle in the complex plane, $\{z\in\Bbb C,|z|=1\}$, under multiplication

Leaky Nun

@AkivaWeinberger why do they need to pass through each other to form a ring?

Secret

@AkivaWeinberger Ah, no wonder I see copies of [0,1) repeated periodically on the real line when I compute its cosets

Akiva Weinberger

00:54

@LeakyNun I don't see what you mean

Leaky Nun

0:35 - 0:39 self intersection

0:49 ring created

0:59 self separation

Akiva Weinberger

Oh, there

I was thinking of when it passes through itself in 0:43-0:48

It's just a cool trick to turn two separate loops into one loop-within-a-loop, I guess

There's a supplementary website for it with more images I think

Leaky Nun

do you know any topological invariants demonstrated in this eversion?

Akiva Weinberger

Found it

usefuldreams.org/sphereev.htm

@LeakyNun No

Secret

Quotient group $\Bbb{R}/\Bbb{Q}$ = ? need to first check whether $\Bbb{Q}$ is a normal subgroup of $\Bbb{R}$

Akiva Weinberger

00:58

@LeakyNun Have you heard of sphere eversion before this?

Leaky Nun

I saw the older video

@Secret that's just real messy

not explocitly constructible

Akiva Weinberger

Outside In?

Leaky Nun

look up Vitali set

right @AkivaWeinberger

Akiva Weinberger

@Secret All subgroups of abelian groups are normal subgroups

That goes over an invariant called "winding number" which works in 2D

and a similar invariant for 3D

Not winding number.

Secret

Ah yes, because for $H \subset G$ the conjugate $ghg^{-1}$ can be rearranged to cancel out the gs thus $h$ stay invariant

Leaky Nun

01:00

yes I'm asking what is that formally called

Akiva Weinberger

Turning number, I think.

Leaky Nun

I think it's theorema egregium

Akiva Weinberger

It's the winding number of the derivative around the origin

Leaky Nun

but I can't be sure

Akiva Weinberger

Hm, that sounds like it's related, yeah

Leaky Nun

01:02

oh right

Akiva Weinberger

In any case, an invariant can only tell you when two things can't be turned into each other

Leaky Nun

the gaussian curvature of a sphere is (1/k)^2

Akiva Weinberger

The video demonstrates two things that can be turned into each other

Leaky Nun

@AkivaWeinberger turns out in 2D they are equivalent

Akiva Weinberger

@LeakyNun Yup. And it's equal when $k=R$ and $k=-R$.

Leaky Nun

01:03

same turning number <=> homeomorphic

@AkivaWeinberger exactly!

Akiva Weinberger

Yup @LeakyNun

And in 3D, the point is moot anyway since it turns out all immersions of the sphere can be turned into each other

Leaky Nun

what is an immersion?

Akiva Weinberger

Like, any frame of the sphere eversion is an immersion

It's a way to put the sphere in $\Bbb R^3$ without creases

I think formally it's a smooth map $S^2\to\Bbb R^3$ where the derivative matrix of that map is always invertible

Leaky Nun

btw are those manifolds?

Akiva Weinberger

Just a sec

A manifold is just an $n$-dimensional surface

A 2-manifold is a surface

Leaky Nun

01:06

because a circle with turning number 1 is homeomorphic to a circle with turning number -1 as topological manifolds

Akiva Weinberger

Doesn't really matter how it's immersed in space

So like a torus or a sphere

Leaky Nun

so what is that?

diff.geom surface?

Akiva Weinberger

Differential geometry, to my understanding, studies smooth manifolds

So manifolds with a smooth structure on them

"Smooth structure" means, uh,

I think you assign a tangent space to each point? Or something

Ask like Balarka or Daminark, they know diff geo I think

Leaky Nun

so the two circles are distinguished under what discipline?

Akiva Weinberger

I think the phrase is "there's no regular homotopy between them"

"between the two immersions"

Leaky Nun

01:10

what is an immersion, formally?

shoild I think of it as a kind of transformation between $f(x) = \exp(2\pi x)$ and $g(x) = \exp(-2\pi x)$?

Akiva Weinberger

I'm back

Leaky Nun

welcome back

Akiva Weinberger

Say the circle is thought of as $\{(x,y)\in\Bbb R^2:x^2+y^2=1\}$

The first immersion $f_1:S^1\to\Bbb R^2$ is just the identity map

The second immersion $f_2:S^1\to\Bbb R^2$ is the map $(x,y)\mapsto(-x,-y)$

Leaky Nun

oh, like an embedding

Akiva Weinberger

The antipodal map

@LeakyNun Yeah, except these are allowed to cross themselves (i.e. not be injective)

Leaky Nun

01:20

something like you can only embed some 2-manifolds in 4 dimension

@AkivaWeinberger oh

Akiva Weinberger

and also they have to be smooth (no creases or corners)

So like the figure-eight immersion is also a thing

Leaky Nun

I'm thoroughly confused

at some point you seem to have no knowledge of this

at other points you seem to have complete knowledge

Akiva Weinberger

Uh

Hm

thinks of a response

Leaky Nun

then what is invariant under regular homotopy of 2-manifolds?

Akiva Weinberger

All I know is that every immersion of the sphere can be turned into every other immersion through regular homotopy

I've seen tori turn inside-out as well, so I'm guessing the same is true for tori

Leaky Nun

01:24

hmm

Akiva Weinberger

There's a weird immersion of the Klein bottle that might not be connected to the usual one via regular homotopy

Leaky Nun

I should learn some diff.geom some time

HO-mo-to-py :o

Akiva Weinberger

Oh, is it?

Damn

That's the weird one

Just a sec

Leaky Nun

I'm blind

Akiva Weinberger

Leaky Nun

01:26

@AkivaWeinberger u wot

Akiva Weinberger

If you take the figure-eight thing and rotate it around with a twist, you get a new immersion of the Klein bottle

anon

huh

Leaky Nun

@AkivaWeinberger is it even closed?

@anon μnμ

Akiva Weinberger

@LeakyNun That picture is half of it

Leaky Nun

oh

Akiva Weinberger

01:27

It's the thing cut in half

The one on top is the full one but it's kinda hard to see

anon

ɥnɥ-ɥn

Akiva Weinberger

So I'd be surprised if you could regularly homotope(?) that into the usual picture of the Klein bottle

Ooh look a paper on Google

Leaky Nun

@AkivaWeinberger I'd be surprised if I could do any regular homotopy at all

Akiva Weinberger

maths.ed.ac.uk/~aar/papers/pinkall2.pdf

"Regular Homotopy Classes of Immersed Surfaces"

I didn't read it yet but it seems relevant?

Leaky Nun

@_@ not going to read it

Akiva Weinberger

01:31

Oh crap homology yeah

You don't know that yet (I think)

Leaky Nun

I don't know anything

I'm just randomly throwing terms like "manifold" as if they mean something

Akiva Weinberger

@LeakyNun Oh OK so

I think what it's saying is

You know that weird picture of the Klein bottle, the figure eight rotated and swung around?

So that's with a half-twist, and you get a Klein bottle

Leaky Nun

hmm?

Akiva Weinberger

If you do it with a full-twist, you get another torus

And it seems to be saying that this new torus isn't regularly homotopic to the regular torus

but all tori are regularly homotopic to one of those two

Leaky Nun

mm..

Akiva Weinberger

01:38

I'm trying to work out how that makes sense

'cause you could image doing it with no twist at all, just taking a figure eight and taking the surface of revolution of it 'round some axis that doesn't intersect it

Leaky Nun

I could

Akiva Weinberger

And theoretically that's regularly homotopic to one of the other two

Leaky Nun

I see

Akiva Weinberger

@LeakyNun On the link to the paper

On like page 11 there's Fig. 4, two weird tori

Apparently the one on the left (the less-weird one) can be turned into the normal one

and the one on the right can't

Don't see how that's done

Leaky Nun

I fail to imagine a normal round figure-eight

I know how its cross-sections look like (figure eights of course)

I fail to understand the whole picture

@AkivaWeinberger you mean the normal torus?

Akiva Weinberger

01:47

What on earth is going on here 'cause I have no idea

jp-petit.org/science/maths_f/Retournement_sphere/…

Leaky Nun

Smale's inside out paradox

►

this is much better

Prototank

Any discrete mathematics fans in here?

Leaky Nun

@Prototank have some knowledge but mostly none

Prototank

I have a kind of a wonky problem that I'm working on. I'm counting binary strings of length $n$ that have the same number of occurrences of the substrings $00$ and $11$. I started by saying "Suppose we have $k$ occurrences of each substring", anticipating a sum over all $k$ after we're done. Then I tried to do something with compositions of $k$

So for example 000111 has 2 occurances of each substring

Akiva Weinberger

02:06

And 011001 has one of each, but 0110 has one 11 and no 00

Prototank

yeah

Akiva Weinberger

Hm. If it starts with 0 and ends with 1 (so it's like 0…1), and it has the same amount of 0s as 1s… will it work?

Salt

you could try counting with the "digits" "00" "11", you'd miss some '000' bits though

Akiva Weinberger

Like I just randomly mashed 0001110101110001

That has eight zeros and eight ones, and starts on opposite digits

4 occurrences of 00, 4 occurrences of 11

Prototank

That's pretty cool, I have to understand why that works combinatorially (I can try that)

and then I'd have to understand how this example lends itself to different cases, like 00110

Akiva Weinberger

02:09

That has one more 0 than 1, but it also starts and ends on 0

Random: 00100111100

Six 0s, five 1s, starts and ends on 0

Prototank

3 of each

Akiva Weinberger

3 occurrences of 00, 3 occurrences of 11

Yeah

Prototank

Hah this is strange

Leaky Nun

000011100

Akiva Weinberger

That doesn't work. Also there are six 0s and three 1s

Leaky Nun

02:14

one could map 00 to 1, 01 to 0, 10 to 0, and 11 to -1

and sum each frame

I'm thinking in terms of CS not combinatorics :P

Salt

02:31

you notice that you have to balance out long sequences of one digit with lots of short sequences of another? starting sequences with consecutive 0's longer than any subsequence of consecutive 1's: 00-impossible 00011011 0000111011 0000110111 000001111011 000001110111 000001101111

Martin Sleziak

Some user mentioned a few times ago some kind of VoIP program they use to discuss mathematics. Do you remember the name of the software?

Salt

was it discord?

Martin Sleziak

Yes, it was discord. Wikipedia: Discord (software)

Thanks Salt!

Salt

also if you switch the places of any two consecutive digits, either the digit's are the same and you don't change the number of 00's and 11's or they are different and wlog you get 1,01,1 -> 1101, 0,01,1-> 0101 neither of which change the number of 00's minus the number of 11's

Prototank

@Salt that was helpful

Salt

02:44

seems like it would prove weinbergers claim quite easily

Leaky Nun

but if you can swap two consecutive digits, then you can permute the digits :O

however this cannot happen at the beginning, as 101 -> 011 doesn't work

Prototank

right, we are talking about the digits inside of the single element buffers

Salt

right, that was meant to be implied...the swap has to occur after first and before the last

element

Prototank

I take that as a proof of @AkivaWeinberger 's claim. Surely I can count such situations.

Salt

but, you're right, it does allow one to permute in any way the digits between the first, say, 0 and last, 1

Prototank

02:49

However I have a hard time explaining why these sequences are precisely the sequences that have the same number of 00's and 11's

I think I believe it now, though

Salt

well, if you take one of these 0...1 sequences and do this -> 10....1 it still has the same number of 00's and 11's, so the 0....1's are quite all of them

aren't*

Prototank

Right

Salt

but i think that pretty much covers the rest of them up to symmetry

Prototank

and similarly one could do 0...10

yeah

Akiva Weinberger

03:12

I was thinking of it differently

You can separate each string into 'runs' of a single digit

So like think of 0001110101110001 as

000-111-0-1-0-111-000-1

The nice thing is that if you start with 0 and end with 1, you will always have the same number of 0-runs and 1-runs

because you can group it in groups of a 0-run and a 1-run

(000-111)-(0-1)-(0-111)-(000-1)

Another thing is, in each 0-run, the number of 00s is one less than the length of the run

(000 is length 3, has 2 instances of 00)

So the number of 00s is the total number of 0s minus the amount of 0-runs ('cause each run has one fewer than it "should")

Similarly, the number of 11s is the total number of 1s minus the amount of 1-runs

So if there's the same amount of 0-runs as 1-runs, and there's the same amount of 0s as 1s, then they're gonna be the same.

That means if it starts with a 0 and ends with a 1, and there's the same amount of 0s as 1s, they're gonna be the same.

@Prototank

@Salt

Prototank

Thanks a lot

This has been pretty fruitful, I will think more on this

Salt

03:28

i feel like i've seen the swapping trick before, but i don't remember where

user84215

Please vote to reopen the following post (only one vote is needed): math.stackexchange.com/questions/2410749/… . If this post is off-topic, then why are there two tags "learning" and "education" ?

user84215

The description of the tag "learning": "Questions about the process of learning mathematics, both inside and outside a formal environment, including learning strategies, recommendations for learning particular subjects, and studying habits"

skullpatrol

::yawn::

Salt

it seems pretty moot, honestly

user84215

The description of the tag "education": "For questions related to the teaching and learning of mathematics".

user84215

03:37

.

user84215

I asked a moderator to see whether my question is off-topic or not, and he/she replied: "Anyway, even if we come to the conclusion the question is not off topic here (I'm not sure about whether it is)".

anon

do you realistically think someone would say "no"?

Secret

The question sounds very meta, but it is not meta either because its ties with MSE is not very strong (any SE chat room can potentially did that). My issue with such proposal is once again the problem of schedule and the unstable attendances. The existing chat room infrastructure works ok because it is spontaneous, but trying to do a tutorial in a chat room setting is going to be difficult unless it is some important topic with significant applications. Not to mention the time allocations of

experts are not stable either as we researchers have commitments in our own institution or industrial researches

Talks may be possible (such as the AMA of h bar and reddit), but I vaguely recall that MSE community said that people are rarely free enough to do AMA in maths

Oliver K

03:56

Hey guys! if v = vertices, e = edges and m = minimum degrees of vertices; in a simple graph; why does 2e/v >= m?

user84215

First, I want to know why my question is off-topic while there are at least two tags related to learning and similar posts which have not been considered off-topic.

Leaky Nun

@OliverK because mv/2 <= e

user84215

Second, it is not necessary that all people participate in a workshop simultaneously, they can do the same things as they may do in other chat rooms. @Secret

skullpatrol

Don't be so impatient @MathematicsAminPhysics Rome was not built in a day :-)

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$Mathematics$ Mathematics