Mathematics - 2017-10-07 (page 3 of 6)

RE60K

16:03

How to prove SL2(F3) is generated by (1,1;0,1) and (1,0;1,1). First of all what is F3?

Kasmir Khaan

@TedShifrin hi Ted :)

Ted Shifrin

hi Kasmir

Kasmir Khaan

:D

Ted Shifrin

Just dropped by for a second

Kasmir Khaan

was waiting when you gonna show up :D

grr damn it :(

how are you btw?

after the operation ?

Ted Shifrin

16:07

LOL ... oh, it went fine, thanks

Kasmir Khaan

=p

Ted Shifrin

I'll be back later.

Kasmir Khaan

that is good

okay then ill keep working on problems :D

RE60K

Well both of the matrices have determinant 1, so the group generated by them will be subgroup of SL2(F3), now how to show that all matrices of SL2(F3) can be written as product of these matrices or their inverses?

Kasmir Khaan

if N is normal in G , and H is any subgroup of G , then N intersect H is a normal subgroup of H

Leaky Nun

16:09

@RE60K well there are only that many elements of SL2(F3) lol

Kasmir Khaan

why do we need N to be normal here ?

Leaky Nun

@KasmirKhaan try proving it for general N

Kasmir Khaan

without N being normal the statement is still true right?

Leaky Nun

@KasmirKhaan prove it

Kasmir Khaan

well it is one line proof

Leaky Nun

16:10

show me

Kasmir Khaan

let a be in the intersection

so a is in H and in N

any element of H normalize H

RE60K

I found the order of both of them is 3 if i am not wrong... Also it says "You may assume this subgroup has order 24 - this will be an exercise in Section 3.2." so I think what you are saying makes sense...

Kasmir Khaan

@LeakyNun i need 3 mins and ill come back , but tell me if am missing something

brb

Leaky Nun

you're missing the rest of your proof?

nbro

Meaning of this notation $y \in \{0, 1\}^C$?

RE60K

16:13

complement?

nbro

$y$ in this case should be either $0$ or $1$, if I'm not wrong, since it's the output of a classification algorithm.

So, it should not be the complement.

Danu

o/

Leaky Nun

@nbro more context?

@Danu how was the thesis?

nbro

@LeakyNun ? classification?

Leaky Nun

@nbro like, where do you see it? the whole sentence? the whole paragraph?

RE60K

16:18

@LeakyNun how would I say that there are 24 of the generated elements, hand-brute-force?

Leaky Nun

@RE60K sure

for two elements of F3, 5 pairs have product 0, 2 pairs have product 1, and 2 pairs have product 2

nbro

$y \in \{ 0, 1\}^C$ means that $y$ should either be $0$ or $1$, as I said above. This is strange because $A^C$ is usually used to denote the complement of set $A$.

Leaky Nun

fo the determinant to be 1, it's either 1-0, 2-1, or 0-2

so 2x5 + 2x2 + 5x2 = 24

@nbro whole paragraph.

nbro

So, I asked this question to understand if someone has ever seen it.

That's the whole paragraph.

Leaky Nun

the whole paragraph states $y \in \{0,1\}^C$ without context?

come on

you're asking us for help and you're hiding the text

nbro

16:21

Omg, I told you that the context is "classification"!

Statistical classification

In machine learning and statistics, classification is the problem of identifying to which of a set of categories (sub-populations) a new observation belongs, on the basis of a training set of data containing observations (or instances) whose category membership is known. An example would be assigning a given email into "spam" or "non-spam" classes or assigning a diagnosis to a given patient as described by observed characteristics of the patient (gender, blood pressure, presence or absence of certain symptoms, etc.). Classification is an example of pattern recognition. In the terminology of machine...

RE60K

decision trees? neural networks? bayes' classifier? linear regression? k-nn?

nbro

Read something about it.

After that, come back!

Leaky Nun

did you see that notation in a text? Include that text here.

nbro

Ok, never mind.

Leaky Nun

don't tell me that sentence $y \in \{0,1\}^C$ popped up in your dream

nbro

16:23

@RE60K It does not matter which algorithms you use to classify. Classification always output either $0$ or $1$: either you belong to a class or not.

Kasmir Khaan

back

@LeakyNun am missing something ?

Leaky Nun

@KasmirKhaan the proof, of course

RE60K

but every classifier works differently, hence its analysis will be different and the notation may appear depending upon what you are using

Kasmir Khaan

the element in N intersect H are by definition in Hh

and H is normal in H

Leaky Nun

@KasmirKhaan so?

Kasmir Khaan

16:24

so H intersect N is normal in H

Leaky Nun

@KasmirKhaan why?

what is the definition of "H intersect N is normal in H"?

Kasmir Khaan

because H intersect N is a subset of N

RE60K

@LeakyNun sym(4) also has order 24, so does SL2(F3) isomorphic to S4?

Leaky Nun

@RE60K no idea

you can google it

Kasmir Khaan

means that elements of the intersection are closed under conjugation by H

Leaky Nun

16:25

groupprops.subwiki.org/wiki/Special_linear_group:SL(2,3)

@RE60K no, it isn't

@KasmirKhaan and have you proved that?

no, you've only proved that the element a in H intersect N is in H under conjugation by H

Kasmir Khaan

hmm

what do i need to prove more?

and why is N assuemd to be normal here

Leaky Nun

@KasmirKhaan you need to prove that an element in H intersect N is still in H intersect N under conjugation by H

do you see the difference?

Kasmir Khaan

okay yeah i see it

Leaky Nun

I don't see how that counts as "I have already tried" when you haven't even understood the question

Kasmir Khaan

I came here to ask about why N had to be normal

Leaky Nun

16:29

@KasmirKhaan read the question first

before you doubt the hypothesis of the question

Kasmir Khaan

hmm

since N is normal in G

N is normal also in any subset of G

let a be in the intersection of N and H

so hah' is also in the intersection of N and H

Leaky Nun

@KasmirKhaan why?

Kasmir Khaan

because of the fact N being normal

If N is normal in the group G

it is also normal in a subset of that group

Leaky Nun

I can’t point out where you are wrong if you won’t stop skipping steps

@KasmirKhaan that’s plainly nonsense

Kasmir Khaan

okay gNg' = N

for all g in G

hmm

i see why that is wrong statement now

Leaky Nun

16:37

good

Kasmir Khaan

so if we let A be the intersection of H and N

i need to show that hAh' = A

Leaky Nun

yes

Kasmir Khaan

now using N being normal

A is a subset of N

ag=ga

so in the equation hah'

i can change it ahh'

=a

thus closed under conjugation by elements of H

Leaky Nun

@KasmirKhaan ?!

Kasmir Khaan

I used that fact that all element of N have the structure that

ng =gn

Leaky Nun

16:43

no they don’t

Kasmir Khaan

gNg' = N

definition of being normal

Leaky Nun

yes

Kasmir Khaan

in the expression hah' = ahh' = a no ?

Leaky Nun

why hah’ = ahh’?

Kasmir Khaan

i feel like am mixing defintions

oh well

since N intersect H is a subgroup of both H and Nn

h ( H cap N ) h' = h (element that is in both H and N ) h' , by closure

Leaky Nun

16:50

you’re equating a set with an element, but continue

@KasmirKhaan by closure of what?

Kasmir Khaan

closure under multiplication in a subgroup

Felix.C

Is the vector gradient simply spoken the standard inner product of the gradient operator and a vector? Being in the $\mathbb{R}^n$

Kasmir Khaan

okay let me rethink

i drew a picture and what i wrote does not make sense

if we take an element in H \ N

there is no reason ( so far ) why that product still in the intersection

@LeakyNun hi

Leaky Nun

hi

Kasmir Khaan

so i drew a picture

N is normal in G

means that gNg' = N

now we take A subset of N

accually subgrup on N

this means that gAg' = A

Leaky Nun

17:02

why?

Kasmir Khaan

well if all the elements in N

have the property that gNg' = N

means also that a subgroup of N also have that property

Leaky Nun

@KasmirKhaan why?

Kasmir Khaan

honestly i dont know now

i thought i had it

doesnt A inherit that property ?

from N

it is a subgroup of N afterall

Leaky Nun

@KasmirKhaan please

stop asserting things without proof

Kasmir Khaan

@LeakyNun is what am saying at least right?

i mean so i can continue

Leaky Nun

17:08

no, it’s not even wrong

if N < H and H < G we do not necessarily have N < G

so your weaker version is obviously false

Kasmir Khaan

okay i see

Leaky Nun

a chain of reasoning is a sequence of steps where each step clearly follows from the previous

if you will stop inherting bad habits from your teacher

and start doing real proofs

then you will be able to have a more solid understanding of the subject

Kasmir Khaan

@LeakyNun i dont know what is am missing?

should i find a surjecitve home such that the intersection is the kernel of this hom ?

Leaky Nun

@KasmirKhaan you’re complicating things

Kasmir Khaan

what am sure of so far

is that N intersect H is a subgroup of H

this much is clear

closed under conjugation by elements of H

i dont know how to get that

Leaky Nun

17:18

I won’t give you any hints

except that you should keep asking yourself “what does this word mean?” until the statement you want to prove is in the purest language of group theory

Kasmir Khaan

i know what i want to prove

Leaky Nun

really?

Kasmir Khaan

h ( N int H ) h' is in N int H

Leaky Nun

continue expanding the statement

Kasmir Khaan

that thing being in H is clear

but being in N is not

Leaky Nun

17:22

I don’t want to see “int” here

expand that one

Kasmir Khaan

how should i write it

element that are in both H and N

Leaky Nun

@KasmirKhaan yes, use symbols

Kasmir Khaan

i dont know how to make that symbol here

can you make it so i copy it?

Leaky Nun

I don’t mean literally symbols

just not words

Kasmir Khaan

okay

using previous notation

Leaky Nun

17:23

or words but the whole statement

words is probably better

Kasmir Khaan

hah' is in A

Leaky Nun

@KasmirKhaan we’re still rewriting the statement you need to prove

Kasmir Khaan

yes i know

but did not use N being normal

ahh

i think i got it

using N being normal

hah' is in N

and that complete the proof :D

Leaky Nun

good

Kasmir Khaan

:D

it was easiear showing it in H and N seperate

Leaky Nun

17:26

yes

remember my technique

Kasmir Khaan

Yes i will try to keep doing that :D

simplifiying the problem into small pieces

Leaky Nun

good

Kasmir Khaan

17:47

okay , prove that a subgroup N of G is normal iff gNg' is a subset of N , for all g in G

the first implication is easy

assuming N is normal , so gNg' = N , and N is a subset of N

assuming that gNg' is a subset of N

Tobias Kildetoft

Next show that the subgroup is normall iff the set of left cosets equals the set of right cosets.

Kasmir Khaan

hmm hi @TobiasKildetoft

Tobias Kildetoft

(and hi)

Kasmir Khaan

my idea was different

Tobias Kildetoft

This was not meant as a hint for this problem, but as a different problem

Kasmir Khaan

17:50

oh okay ><

assuming gNg' is a subset of N

if i take x ,y in N

need to show product is still in N

oh wait

I dont know what is the point of this exercice

can it happend that gNg' is in N but N is not normal ?

Tobias Kildetoft

@KasmirKhaan It can't happen for all $g$ (that is the point)

Kasmir Khaan

aha okay =p

Tobias Kildetoft

but it is possible for non-normal subgroups to have $gNg^{-1}$ be a proper subset of $N$

Kasmir Khaan

so being normal means that gNg' = N , we get whole N

for all g in G

did not encounter something like that

gng' is proper subset ie

Tobias Kildetoft

right, and you are trying to show that it suffices to assume that you get a subset for all $g$

Kasmir Khaan

17:56

i dont know how to do that yet

Tobias Kildetoft

So one direction is obvious, right?

Kasmir Khaan

yes N being normal means that gNg' = N , subset of N

Tobias Kildetoft

right

Kasmir Khaan

but assuming gNg' is a subset of N

means that gNg' =N

is harder to prove

Tobias Kildetoft

now you want to show that $N$ is a subset of $gNg^{-1}$

(hint: conjugate again, now with a new but suitably chosen element)

Kasmir Khaan

17:58

ah that is neat

i dont get the hint but i got the first idea

proving one set is contained in the other

Tobias Kildetoft

For examples of what happens for non-normal subgroups, see math.stackexchange.com/questions/107862/…

Kasmir Khaan

hmm gonna need some time to get what u wrote =p

@TobiasKildetoft can you clarify some terms for me ?

i mean in general case

i have some concepts that i dont feel i got them right

like , join of 2 subgroups, normalizer of a sugroup

N_G ( H) , does it mean those element such that gHg' = H

so we say that g "normalize " H ?

Tobias Kildetoft

18:26

@KasmirKhaan Right

Also, the join of two subgroups is the smallest subgroup containing both of those subgroups (I hardly ever see that called the join)

Note that when defining the normalizer we really do need to require $gHg^{-1} = H$, rather than $gHg^{-1}\subseteq H$ as demonstrated by those example I linked.

Jeff

19:03

Hi folks, I have a few "name" questions does anyone know if there's a name for the point (or value) where a piecewise function changes from one piece to another? Is there a name for each of the different formulas (you know, like "pieces")?

I refer to that point in class often and have no name for it.

Tobias Kildetoft

@Jeff That is not really well-defined mathematically, so I doubt there is an "official" name for it.

Jeff

@TobiasKildetoft Well, in class I call it "the cutover point". So I guess that will have to do.

Do we have the authority in here to declare that the official name? :D I say yes.

Tobias Kildetoft

@Jeff I would actually advise you to not put too much emphasis on it, as it really is not a well-defined notion.

Jeff

@TobiasKildetoft well, the problem is that in teaching precalc, I often refer to that point in class. I just need something to call it, no matter how informal.

Tobias Kildetoft

True, in practice, it does matter as a point to check continuity at

Lucas Henrique

19:10

Hey, guys. :)

Jeff

@TobiasKildetoft Yup, that, too. The textbook in one of the calc classes I teach calls them a "suspicious point".

Leaky Nun

528

Q: Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$

As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem) $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}.$$ However, Euler was Euler and he gave other proofs. I believe many of you know some nice proofs of this, can you please sh...

This is dank

Daminark

19:30

Hey there

Leaky Nun

19:43

The most shocking fact I’ve learnt today:

there is an integer polynomial [in quite a lot of variables] that it is not provable if it has any integer solution under ZFC.

@Secret @user21820

Tobias Kildetoft

@LeakyNun Even worse, there is one which includes a parameter, where the parameter can represent any sentence in ZFC and the polynomial has integer roots iff that sentence is true in ZFC

Balarka Sen

lol

and that is why, kids, you don't mess around with logic and set theory

Leaky Nun

@TobiasKildetoft Even worse, you can do it in any reasonable axiomatic system, and the polynomial is actually here (K is the parameter):

15

A: Does anyone know a polynomial whose lack of roots can't be proved?

For every consistent recursively axiomatizable theory $T$ there exists (and one can effectively compute it from the axioms of $T$) an integer number $K$ such that the following Diophantine equation (where all letters except $K$ are variables) has no solutions over non-negative integers, but this ...

Daminark

@Balarka to be fair, these types of things are extremely wacky but more fun than computing $\lim_{n\to\infty} \int_0^n (1-\frac{x}{n})^n\log(2+\cos(\frac{x}{n})) dx$

Balarka Sen

I would not compute that integral

but I would not be keen to study set theory either

Daminark

19:59

Ah so you want to study scheme theory?

:P

Balarka Sen

I would much rather

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$Mathematics$ Mathematics