Mathematics - 2016-10-09 (page 1 of 3)

0celo7

00:03

@MikeMiller When was it proved false?

0celo7

00:55

@Semiclassical What does an "excited bound state" refer to?

Mike Miller

Find tables of embeddings of projective spaces somewhere. That should break it.

PVAL-inactive

03:18

www.lehigh.edu/~dmd1/

0celo7

@PVAL-inactive who that?

user227867

Hey there @robjohn happy anti-belated Halloween!

robjohn

@WillHunting How are you?

user227867

@robjohn I am the same as usual, trying to get better. How are you?

user227867

@Adeek Real strength lies in the heart.

user227867

03:27

@TedShifrin Relax. Have some tea. =)

dsillman2000

04:09

How many of you guys read works by Mark Levi?

I honestly think his ideas are so cool

dsillman2000

05:48

Welcome, @Ka

*@Kaumudi

user228700

Hi everyone :-)

user228700

I have a small question regarding statistics; I keep coming across the term "score" and I'm wondering what it means in the language of statistics..?

dsillman2000

They may be in reference to a $z-\text{score}$

user228700

@dsillman2000 Uh, I dunno what that is :/

dsillman2000

A $z\text{-score}$ is basically the number of standard deviations a value is from the mean of all the values.

So basically: $z=\frac{x - \mu}{\sigma}$

Where $x$ is your given value, $\mu$ is the mean of all the values in your set, and $\sigma$ is your standard deviation for the set.

For example, if your SAT score is 1100, the mean is 1026, and the standard deviation is 209

Then your $z\text{-value}=\frac{1100-1026}{209}=0.354$. This result indicates that your SAT score is 0.354 standard deviations above the mean.

That may or may not be what the "score" you're in reference to is, @Kaumudi

Welcome, @ItachíUchiha

Itachí Uchiha

06:04

Thank you very much. :)
Good morning from India

dsillman2000

Lol, good evening from

California

It's 11:05 here, I'm very tired :)

Itachí Uchiha

@dsillman2000 SLEEP!SLEEP :p

user116211

06:35

@ItachíUchiha deleted the whole SE account, I'm seeing.

Itachí Uchiha

@MAFIA36790 @ You remember me?!

user116211

No, I was just seeing your logged in account, but I'm seeing page not found.

Black

06:57

Hi does anyone know the convolution formula to find the density of Z=aX+bY?

I think it should be integrand of f(x-a*t)g(t) dt

sorry f(x-at)g(b*t)

*no problem found the solution

Itachí Uchiha

@arctictern,Question posted.

@arctictern,math.stackexchange.com/questions/1960359/…

@MAFIA36790 Yup,I deleted all accounts except Anime one.

user116211

Why?

user116211

@ItachíUchiha Anyways, use \cdot instead of * to get $2014\cdot 2013\,.$

user116211

Also, use \ldots to get $\ldots$

user116211

This is enough to show continuation of the series.

Itachí Uchiha

07:14

@MAFIA36790 Thank you very much.

Balarka Sen

07:28

@TedShifrin I think it works fine for noncompact. It'd just be a partition of unity argument.

It's in G-P.

Mike Miller

What does?

Balarka Sen

The Whitney embedding theorem, I meant.

Mike Miller

How do you ensure there are finitely
many charts?

Balarka Sen

Hmm, now that I read back we seem to be talking about topological manifolds. In which case, I do not know.

Mike Miller

You shouldn't know how to do that for smooth manifolds, either :P

The argument needs nontrivial work to fix in the noncompact case.

Andrew Thompson

07:42

@BalarkaSen Bredon has a very readable proof I think.

Balarka Sen

I have forgotten the construction I have read in G-P. I think the crucial point is to construct a proper real valued function on the manifold.

Mike Miller

Both you and Andrew are right.

user227867

I am left handed.

user227867

For the first time in my life, I heard of the term water intoxication, and from a doctor too, about drinking too much water.

Balarka Sen

It's a thing.

user228700

07:48

Has any of u tried and succeeded in finding the sum of squares if the terms of an arithmetic series?

user227867

Alcohol tastes terrible. I wonder who likes it and why. Beer, whiskey, wine, liquor, champagne all suck.

user228700

@WillHunting Hi :-) What are the harmful effects of this? (I'm afraid that I may be drinking too much water everyday)

user227867

@Kaumudi Well, in my case, I just felt some weird kind of dizziness.

Mike Miller

@Kaumudi Most people don't. Indeed, most people don't drink enough.

user227867

@Kaumudi Nope, that is too hard for me.

user228700

07:49

@WillHunting OK. Apparently, it reduces the sodium content in our blood. That's why the dizziness.

user228700

@WillHunting OK. This person has:

user228700

pballew.blogspot.in/2012/12/…

user228700

And I wanted to know how to go about proving/deriving the formula. I've only just graduated high school, and let's just say that I'm trying to improve my math skills...

user228700

@MikeMiller Yes, I'm inclined to agree with this.

Balarka Sen

@MikeMiller More reasons to get drunk, eh? At least that'll balance the water shortage!

user228700

07:52

@BalarkaSen I think he meant drinking more water. @MikeMiller: No?

Balarka Sen

I figured. I joked about an alternative to drinking lots of water.

user228700

@BalarkaSen OK :-)

user227867

I have been listening to Jacob Sartorius's songs all day long.

user227867

He has so many haters, sad panda.

user227867

@Kaumudi Do you like pizza? I prefer spaghetti carbonara and tiramisu to all forms of pizza.

user228700

07:59

@dsillman2000 Oh, crap, I didn't even check this! Thanks! :-D

user228700

@WillHunting For one second, I became very confused as to why u were asking me this question seemingly out of the blue but then I remembered my avatar and yes, I love pizza!

user228700

user227867

Hmm, he looks cute. That must be Mike.

Mike Miller

It's not me, but it looks startlingly like someone I knew in undergrad.

user228700

Mike? That's John Green!

user227867

08:04

I only know about John Nash and Ben Green. Bye.

user228700

OK. Bye :-)

user228700

@MikeMiller U knew him?

Mike Miller

Not John Green, no.

user228700

Oh, OK :-) 'Cause John Green attended Kenyon College-liberal arts.

Balarka Sen

I should get to work.

DHMO

09:08

why can infinitely strictly decreasing sequence not exist?

i'm talking about transfinite ordinals here

Secret

Baer–Specker group

In mathematics, in the field of group theory, the Baer–Specker group, or Specker group, named after Reinhold Baer and Ernst Specker, is an example of an infinite Abelian group which is a building block in the structure theory of such groups. == Definition == The Baer–Specker group is the group B = ZN of all integer sequences with componentwise addition, that is, the direct product of countably many copies of Z. == Properties == Reinhold Baer proved in 1937 that this group is not free abelian; Specker proved in 1950 that every countable subgroup of B is free abelian. The group of homomor...

Why do axiom of choice cannot be used to ensure modules to always have a basis?

Evinda

09:30

Hello @robjohn
Are you familiar with subharmonic functions ?

Secret

10:14

Is there exists torsion free cyclic groups?

Balarka Sen

@Secret $\Bbb Z$

Secret

O yes, forgot it just keep on going forever starting from the generator {1}

Secret

10:47

Powerful p-group

In mathematics, in the field of group theory, especially in the study of p-groups and pro-p-groups, the concept of powerful p-groups plays an important role. They were introduced in (Lubotzky & Mann 1987), where a number of applications are given, including results on Schur multipliers. Powerful p-groups are used in the study of automorphisms of p-groups (Khukhro 1998), the solution of the restricted Burnside problem (Vaughan-Lee 1993), the classification of finite p-groups via the coclass conjectures (Leedham-Green & McKay 2002), and provided an excellent method of understanding analytic pro-p...

what about even $p\neq 2$?

DHMO

11:21

In tachyos.org/godel/proof.html?= what is "|-"?

Secret

Elementary abelian group

In group theory, an elementary abelian group (or elementary abelian p-group) is an abelian group in which every nontrivial element has order p. The number p must be prime, and the elementary abelian groups are a particular kind of p-group. The case where p = 2, i.e., an elementary abelian 2-group, is sometimes called a Boolean group. Every elementary abelian p-group is a vector space over the prime field with p elements, and conversely every such vector space is an elementary abelian group. By the classification of finitely generated abelian groups, or by the fact that every vector space has a...

Why prime, why it does not work for composites?

user116211

:32810241 It's Hilbert Choice Operator.

DHMO

@Secret then it is not elementary?

@MAFIA36790 I see thanks

In transfinite ordinals, why can infinitely strictly decreasing sequence not exist?

@MAFIA36790 do you know this?

Secret

Ok, makese sense

user116211

@DHMO Yes, I was lately reading Bourbaki Set Theory.

DHMO

11:25

@MAFIA36790 then why?

user116211

I've taken a break from it to concentrate on my main studies.

user116211

@DHMO What?

DHMO

In transfinite ordinals, why can infinitely strictly decreasing sequence not exist?

user116211

I was not talking about that.

DHMO

then do you know this?

user116211

11:27

no.

DHMO

ok thanks

Itachí Uchiha

How to solve this differential equation:$y-dy/dx-x^n=0$

DHMO

@ItachíUchiha integral factor

Secret

What's the problem of having e.g. $C_4 \times C_9$. 4 is relatively prime to 9 (i.e. gcd(4,9=1)), $C_9$ is a p group with p^n=9^1=9. This seemed to fit the requirement of an elementary group?

$C_i$ are the cyclic groups of order i

DHMO

11:45

@Secret 9 is 3^2

Secret

I am actually quite confused about p groups. If I have $C_{12}$ (thus the order is 12 for any elements in the group) and since $12=2^2\times 3$, is $C_{12}$ a p-group, if yes then how to determine what the p is?

o nvm

DHMO

I thought p stands for prime

Secret

I think the question I am trying to ask is: Why do p-groups only restrict p to prime, is it has something to do with the fact that sylow theorems only works for prime orders?

DHMO

@Secret because primes cannot be decomposed?

Secret

That's true, but it is still not very obvious on how that will screw up everything if the p is replaced by any positive integer in the p-group article... I think I might have to reread sylow theorem in more detail again, to see how it all breaks down

DHMO

11:54

Monster group

In the area of modern algebra known as group theory, the Monster group M (also known as the Fischer–Griess Monster, or the Friendly Giant) is the largest sporadic simple group, having order 246 · 320 · 59 · 76 · 112 · 133 · 17 · 19 · 23 · 29 · 31 · 41 · 47 · 59 · 71 = 808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,000 ≈ 8×1053. The finite simple groups have been completely classified. Every such group belongs to one of 18 countably infinite families, or is one of 26 sporadic groups that do not follow such a systematic pattern. The Monster group contains all but six of the...

What the hell is this

@Secret Is (R,+) holomorphic to (R_{>0},x) ?

You see, I'm just starting to learn groups

Secret

12:12

I think you mean homomorphic, cause holomorphic is related to differentiablility in complex functions?
In that case I think you can write a homomorphic function $f:\mathbb{R} \rightarrow \mathbb{R}^+$

by $f(x)= \left\{\begin{matrix}x,x\neq 0\\ 1, x=0\end{matrix}\right.$

DHMO

@Secret you know, my function is exp

Secret

exp will work because exp(a+b)=exp(a)exp(b) and exp(0)=1

DHMO

@Secret your function did not account for the negative numbers?

Secret

oops you are right, I forgot the negative numbers, but anyway, exp is not only homomorphic, it is also isomorphic because f=exp is bijective

DHMO

@Secret isomorphic is to homomorphic as bijective is to surjective?

Secret

12:19

homomorphism is $f(a)\cdot f(b)=f(a * b)$. Isomorphism is when f is bijective

DHMO

eh, alright

Secret

isomorphism basically means the two structures in question can be converted between each other forward and back, thus in a sense they have the same structure (other than labelling)

DHMO

i see

tachyos.org/godel/proof.html

What do the two last rules mean?

Riccardo

does a Kunneth spectral sequence exist for a generalised homology theory (actually I'm interested in MSpin_* as an homology theory)? Can someone give me some reference for it?

Secret

I am not famialr with the Hilbert Choice operator thus I have no idea

DHMO

12:35

@Secret After some thinking I think it is not the Hilbert Choice operator

it just means "then it can be deduced that"

the first one means "if WF1 is true, and if it is true that WF1 implies WF2, then it can be deduced that WF2 is true"

And of course MP means Modus Ponens

Secret

WF |- ∀V WF might mean If WF is true, then it can be deduced that WF is true for all V

DHMO

bingo

Secret

Gen might mean "generalising"?

DHMO

@Secret I believe so

What might WF mean?

Secret

Well I am not really a logic person and right now I am in the middle of group theory started by suddenly want to read all algebraic structures starting from magmas from 4 days ago

WF is just some formula, it is stated at the paragraph at the top

DHMO

12:39

@Secret what might it stand for?

tachyos.org/godel/1+1=2.html

I just figured how this can be shorter

by combining Ax4 and MP together in the beginning to create a new lemma

Secret

en.wikipedia.org/wiki/Well-formed_formula A skim read of this suggest WF is basically a proposition thus it can be assigned a boolean value to it

DHMO

@Secret nice, thanks

11 lines to prove x=x...

the proof would be better if it had formal ways to reference the lemmas

Daniel Cortild

14:18

What can I say about $f(x)$ if $f(x)=f(f(x)+9x)-33x$?

robjohn

@Evinda what about them?

Ramanewbie

@ted They're dead.
Anyway "naughty" is "méchant" while "mean" is more like "cheapskate".

Simple Art

@Ramanewbie Lol

Mike Miller

14:38

@Balarka Did you get to work?

Tom.A

Hi all

there's a recursive series I am trying to build without success what so ever. Let (an) be the number of ways to pave a path of length n with 3 different bricks. Red - 3cm long, Blue - 2 cm long, Green- 1cm long. there shouldn't be any blue bricks beside green bricks. can someone please suggest an hint? ( I have tried to use 3 different subsets and what so ever could not come up with an independent (an)

Balarka Sen

15:00

@MikeMiller Yes, though I spent most of the time thinking about something unrelated. But I proved what you wanted me to prove: given a nonorientable surface $S$, $TS$ has a line bundle as a complement (immerse in $\Bbb R^3$, pullback the trivial 3-plane bundle to $TS$, take it's orthogonal complement). So $w(TM \oplus \xi) = 1$, giving $w_1(TM) = w_1(\xi)$ and $w_2(TM) + w_1(TM)w_1(\xi) + w_2(\xi) = 0$ implying $w_2(TM) = w_1(TM)w_1(\xi)= w_1(TM)^2$.

Mike Miller

And you also know something more about that.

Balarka Sen

Hmm, yes, I guess for any vector bundle $E/M$ s.t. $E \oplus \xi = \Bbb{1}$ for some line bundle $\xi$, $w(E \oplus \xi) = 1$ so $w_1(E) = w_1(\xi)$, $w_2(E) = w_1(\xi)^2$, $w_3(E) = w_1(\xi)^3$ etc.

Mike Miller

I meant about what $w_2(TS)$ was.

Balarka Sen

Ah, well, it's $0$ if $S$ has odd number of cross caps and $1$ if even number of cross caps.

$1$ meaning the generator of $H^2(S; \Bbb Z/2)$.

Mike Miller

Sure. It would probably have been easier to say it's $\chi(X)$ mod 2.

You asked a while back about what the SW classes had to do with cobordisms. What do you conclude?

Balarka Sen

15:06

Yes, I did prove that (I wanted to ping/e-mail you with all of that but procrastinated on writing).

@MikeMiller It seems two surfaces are cobordant iff their SW classes have the same "expression". Eg $\Bbb{RP}^2$ is not nullcobordant, so it's $w = 1 + \alpha + \alpha^2$ but $\Bbb{RP}^2 \# \Bbb{RP}^2$ is so $w = 1 + \alpha$. I am still unclear on how to write that down.

Two nonorientable surfaces, I guess. $S^2$ is nullcobordant, but $w = 1$.

Mike Miller

I really don't know what you mean by 'expression'.

Balarka Sen

RP^2 and RP^2 # RP^2 # RP^2 are cobordant, so w = 1 + a + a^2 for both. Of course I can't say these are the same in any sense, because the two a's live in different rings. But as formal expressions... I am not sure if this is helpful because S^2 and RP^2 # RP^2 are cobordant too but w = 1 for the first and w = 1 + a for the latter.

DHMO

@DanielCortild that f(x) has a fixed point

Daniel Cortild

15:21

@DHMO But what would it be? When $x=0$? That would give me $f(x)=f(f(x))$...

DHMO

@DanielCortild it would be f(0).

Daniel Cortild

Yea so $f(0)=f(f(0))$, so that means that $f(0)=0$? But to conclude that, I have to prove that $f(x)$ is injective no?

DHMO

@DanielCortild i didn't say f(0)=0

I said f(0) is a fix point of f

Daniel Cortild

Ohh... But what does that help than?

DHMO

x is a fix point of f <=> x = f(x)

Mike Miller

15:23

@BalarkaSen It seems like your latter statement shows that this notion is not helpful.

DHMO

@DanielCortild how do i know what you are trying to find

Balarka Sen

I guess.

Mike Miller

You already found full cobordism invariants for surfaces! But you haven't talked about them yet...

Daniel Cortild

Ohh yea, wait i'll put the original problem...

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ so that $f(f(x)+9y)=f(y)+9x+24y \ \text{for all } x,y \in \mathbb{R}.$

DHMO

if 9x+24y = 0, then $f\left(f(x)-\dfrac{27}8x\right) = f(x)$

Balarka Sen

15:28

@MikeMiller $w_1^2$, I guess?

Mike Miller

@BalarkaSen Both $w_1^2$ and $w_2$ seem to be cobordism invariants here.

Or, to be more precise in the case of disconnected things, the evaluations of $w_1^2$ and $w_2$ against the fundamental class.

Balarka Sen

Yeah. They are not distinct invariants here, unfortunately.

DHMO

if y=0, then f(f(x)) = f(0) + 9 x

@DanielCortild this is interesting

Mike Miller

Now prove that $\langle W(M), [M]\rangle$ is a cobordism invariant, where $W(M)$ is one of $w_1^n$ or $w_n$, in the spirit of surfaces.

Daniel Cortild

@DHMO Yes... But I don't see where that leads...

Mike Miller

15:31

Once you've done that, tell me if your proof applies to anything else.

DHMO

@DanielCortild so f(f(x)) is basically a straight line

Daniel Cortild

Ohh yes... But what would the answers than be?

DHMO

idk, i'm just exploring

Oct 3 at 16:49, by Mike Miller

I sent my advisor a reasonably long email and he cut it up into four paragraphs, and responded with "no, no, yes, not sure how you'd do that".

How did this get 11 stars?

Mike Miller

Presumably 11 people clicked the star button next to it.

If that's not how it got 11 stars, then that would be interesting.

Balarka Sen

lol

DHMO

15:36

if x=0, then f(f(0)+9y) = f(y)+24y

Balarka Sen

@MikeMiller Alright, I'll think about it. Do you want me to send you the proofs for w_1 = 0 iff orientable and w_n = chi mod 2?

They are not very interesting though :) The former is an easy consequence of classifying maps lifting to universal cover of the Grassmannian (which I was talking about yesterday), and the latter is Yoneda's lemma.

So was wondering if you'd care enough to read.

Mike Miller

No.

Balarka Sen

Thought so.

Mike Miller

I do wonder if you can guess what $w_2 = 0$ means.

I don't really know what you mean by saying the latter is Yoneda, though.

Balarka Sen

I'll have to think about that. Ted already asked what $w_1$ is Poincare dual to (I'll ponder on that, can't be that hard)

Mike Miller

15:44

You actually already know that.

Balarka Sen

@MikeMiller $w_n$ gives a map $[X, G_n] \to H^n(X; \Bbb Z/2)$ (pullback the universal bundle, take the SW class) and euler class gives a map $[X, G_n] \to H_0(X; \Bbb Z/2)$ (pullback the universal bundle, take the zero set of a generic section). These are natural transformations; Poincare dualize to get both their values in $H^n(X; \Bbb Z/2)$. They agree on a point, so by Yoneda lemma are the same natural transforms. So $w_n$ is the same as Euler class.

Which is the same as $\chi(X)$ mod 2 for the tangent bundle.

Mike Miller

Sorry, what category is your domain?

Balarka Sen

Can't the category of smooth $n$-manifolds work?

Mike Miller

OK, I see what you're doing now. That's almost OK, except that you've written down one functor that's covariant and another that's contravariant.

With what maps?

Also, what is $G_n$ in the category of smooth manifolds?

Balarka Sen

Smooth maps, as usual. Ah, yikes, good point. But you can cellular approximate to the $n$-skeleton, I believe, which will be an $n$-manifold.

Mike Miller

15:53

Since when is a skeleton a manifold? :P

Anyway, your result is true. But the proof is not.

Balarka Sen

Hm. Can I fix it?

Without working on the category of infinite dimensional manifolds like you did for line bundles previously.

Perturbative

Hey everyone, quick question : Does $\lim_{n \to \infty} \sup |c_n|^{\frac{1}{n}} = \sup (\lim_{n \to \infty} |c_n|^{\frac{1}{n}})$, where $c_n$ is some arbitrary sequence?

Mike Miller

@BalarkaSen I don't think that would fix it, though, because it was important to you that a dimension $n$ be chosen.

Balarka Sen

Oh, right, Poincare duality.

Well, crap.

DHMO

@Perturbative what is sup(x)?

Perturbative

16:03

If $\{x\}$ is a subset of some set with a least upper bound property then $\sup(x)$ would be $x$ I'd say, am I correct?

DHMO

but lim ... is not a set

i'm referring to the RHS

you treated sup as a function of a number

Perturbative

Ah, so that expression on the RHS isn't even defined

DHMO

exactly

Perturbative

@DHMO, thanks!

DHMO

@Perturbative, you're welcome!

alan2here

16:09

Hello all :0

I mean :)

I've just discovered,

f[n] = The number of 1s in the binary representation of (f[n-1]^3 - 1)

Mike Miller

@BalarkaSen Anyway, you'll be able to prove later in the book that there is a characteristic class of oriented vector bundles called the Euler class, and a characteristic class of vector bundles called the "$\Bbb Z/2$-Euler class". You'll be able to prove that the latter agrees with SW, is equal to the Euler class itself mod 2 when the bundle is orientable, and that the Euler class of the tangent bundle is $\chi(M)$ times the fundamental class.

Balarka Sen

Hrm, alright.

alan2here

Seems to produce very long sequences

for given small integer starting values

DHMO

@alan2here for example?

alan2here

when you repeat and earlier value, subtract 1 again
and stop at 1, this being the end of the sequence

DHMO

16:13

but f[n] < log_2(f[n-1]^3-1) < log_2(f[n-1]^3) = 3log_2(f[n-1]) ...

alan2here

ok, one moment

DHMO

I don't see how long it can be

judging from the fact that it is limited by 3log(x)

so it must keep decreasing

alan2here

ahh, I mean
f[n] = (The number of 1s in the binary representation of f[n-1])^3 - 1

sorry, I had the brackets in the wrong place

DHMO

well, (log_2(n))^3 don't really take you very far

@alan2here well... one moment is quite long

Are all groups the product of some cyclic groups?

(I'm all new to group theory)

arctic tern

16:29

very no

all finite abelian groups are though

DHMO

interesting

what about infinite a. groups?

arctic tern

(Q,+) and Q/Z are not cyclic or products of cyclic groups (although they are both "locally" cyclic)

DHMO

what is Q/Z?

(sorry I'm new)

arctic tern

Q (rationals) and Z (integers) are groups under addition, Q/Z is the quotient group

DHMO

what does it mean?

what does locally cyclic mean?

sorry for many questions

arctic tern

16:31

don't worry about "locally cyclic", that's too advanced

are you familiar with "mod"?

DHMO

yes

arctic tern

well, Q/Z is like rational numbers (mod 1)

DHMO

so 5/3 = 2/3?

arctic tern

so for instance 1/2+2/3 = 7/6 = 1/6 (mod 1)

right

DHMO

yay

what do you call a group like this where there is always an element between two?

arctic tern

16:33

what does "an element between two" mean?

DHMO

like

$\forall A,B\in S:\exists X\in S:A\le X\le B$

Balarka Sen

a group doesn't come with any natural order

arctic tern

groups are generally not ordered by some $\le$ relation

also, you probably want $a< x<b$, so $x$ is actually in between $a$ and $b$ (otherwise $x=a$ and $x=b$ would satisfy the inequality $a\le x\le b$)

alan2here

@DHMO ok, here you go

(1s count)^3 - 1
while recuring, -1 again
stop at 1

11111
b((5^3) - 1)
1111100
b((5^3) - 1)
recurs
b((5^3) - 2)
1111011
b((6^3) - 1)
11010111
b((6^3) - 1)
recurs
b((6^3) - 2)
11010110
b((5^3) - 1)
recurs
b((5^3) - 2)
recurs
b((5^3) - 3)
1111010
b((5^3) - 1)
recurs
b((5^3) - 2)
recurs
b((5^3) - 3)
recurs
b((5^3) - 4)
1111001
b((5^3) - 1)
recurs
b((5^3) - 2)
recurs
b((5^3) - 3)
recurs
b((5^3) - 4)
recurs
b((5^3) - 5)
1111000
b((4^3) - 1)
111111
b((6^3) - 1)
recurs
b((6^3) - 2)
recurs
b((6^3) - 3)

DHMO

that's a new rule

alan2here

16:36

sorry, I thought I put it in the description earlier, before I took a moment to write up this table

without it you just get loops all the time

DHMO

@BalarkaSen but it does come with order?

sorry if this is stupid question

I'm all new to this

Mike Miller

@DHMO No, it does not.

If you look at the definition of a group, there's no order in it anywhere. Some groups you know have a way of ordering them, but that doesn't mean that's something you can do to a group.

alan2here

without all the working, so you can just see the sequence

DHMO

does order mean something different?

alan2here

11111
1111100
1111011
11010111
11010110
1111010
1111001
1111000
111111
11010101
1110111

DHMO

16:38

like 1/2 < 2/3

arctic tern

no, "order" does not appear anywhere in the definition of a group

for instance, there is no order in the group of permutations of {1,2,3}

DHMO

well

alright

by the way I thought of a group involving Q that is cyclic

alan2here

groups and the such took me quite a bit of thinking about to remember all the stuff, and I'm still very rusty myself

DHMO

well

how do i multiply two groups together?

Balarka Sen

Hi @TedShifrin

Ted Shifrin

16:46

Hi @Balarka, tern.

Balarka Sen

The proof of $w_n = \chi(M)$ mod 2 proof I told you yesterday had a subtle/dumb flaw.

Which, unfortunately, seems unfixable.

Ted Shifrin

Ah ... so back to the axioms?

Balarka Sen

I don't know how to prove it from axioms.

Mike says it's later in Milnor-Stasheff. I flipped through a bit and see bunch of Steenrod squares etc :S

Ted Shifrin

Mike will yell at me, but you might pretend that the bundle is a direct sum of line bundles. Can you do it then?

(This is called the splitting principle.)

Mike Miller

@TedShifrin I know the splitting principal, but I'm unsure if that argument will work.

Ted Shifrin

16:52

Hmm, interesting. It seems to work for complex bundles, but I don't see the real case yet.

Mike Miller

@TedShifrin No, the problem si that you need to pullback to a larger-dimensional manifold, so the Euler class is no longer a number.

Ted Shifrin

I am not pulling back anywhere.

I guess it's true that $e(L\oplus L') = e(L)\cup e(L')$ even for real line bundles?

Balarka Sen

In my definition, it is.

Ted Shifrin

I'm not used to thinking about the euler class except for oriented bundles of even rank.

alan2here

cool, found a really short one that converges to 0 instead of chaoticly jumping around larger values, that'll do I think, as there is probably not much use to these sequences :¬P

111
11010
11000
110
101
100
0

DHMO

16:57

maybe you should make a tree diagram

or just a map

alan2here

I'd have to write a program to do it I think

DHMO

well, you only have a few numbers to work with

Balarka Sen

@TedShifrin Interesting fact though, that principle.

alan2here

lol, It's all just 0s and 1s

Transcript for

$Mathematics$ Mathematics