Mathematics - 2016-05-27 (page 1 of 2)

Forever Mozart

00:09

wow game of thrones is so good I watched 3 seasons in 3 days

no spoilers from the later seasons please

arctic tern

a main character dies

Forever Mozart

that happened already

i have heard it happens a lot

but please dont spoil it

Clarinetist

@BalarkaSen Dang, I was out

Surdz

00:25

does binary relation the same as a Relation?

Forever Mozart

00:42

yes

Mike Miller

00:53

@arctictern off the top of your head, do you know the first few G-invariant symmetric polynomials in S^*(su(2))?

not the dual, note

arctic tern

nope

Mike Miller

darn

know anywhere thwy're written down?

arctic tern

nope

Mike Miller

double darn. thanx

Axoren

I hear they play a game.

Consisting of thrones.

Potentially a metaphor for medieval politics.

Mike Miller

02:02

good question; hope someone might look at it. If I recall right, seems like it might be to @archipelago 's taste.

Eric Stucky

@Pedro: Actually I didn't know a proof of this result; we stated it in my combo class this semester but symmetric function things were kind of rushed at the end.

So it was an honest question; I thought I might have misremembered the statement.

user19405892

02:20

hi

why does $9|2^b-2^a$ imply $6|b-a$?

Eric Stucky

that seems pretty false.

Oh, okay I see why that might be true.

arctic tern

@user19405892 any ideas, user?

user19405892

not really

arctic tern

how would you write 9|(2^b-2^a) using modular arithmetic?

user19405892

$2^b \equiv 2^a \pmod{9}$

arctic tern

02:29

now what can you do to that equation to see b-a in it?

user19405892

so we can divide by 2 repeatedly?

arctic tern

divide by 2^a (assume wlog b> or =a)

user19405892

ok

Eric Stucky

(Careful, you can't always divide with mods; why is that okay here?)

user19405892

since 2 is relatively prime with 9

arctic tern

02:30

mmhmm

Eric Stucky

^.^

arctic tern

now what.

user19405892

$2^{b-a} \equiv 1 \pmod{9}$

ah

which only occurs for multiples of 6

Eric Stucky

Yeah, I guess since you know what the answer should be, you can just kind of go through all of the cases. If b-a=0 okay, if b-a=1 nope, if b-a=2 nope, etc.

Unless you were thinking of something more intelligent?

user19405892

I was thinking of since we know that the powers of 2 modulo 9 are periodic, we have $2,4,8,7,5,1,...$

Eric Stucky

02:34

Ah, that's nice.

Pedro Tamaroff

02:55

You can also think about Euler's totient.

$\varphi(9)=3(3-1)=6$.

arctic tern

well, you would have to compute the order of 2 mod 9, which is known to be a divisor of phi(9)=6. in general the order of x mod n can be a proper divisor d of phi(n), in which case n|(x^a-x^b) implies d|(a-b), which is weaker than phi(n)|(a-b).

Mike Miller

@arctictern turns out I didn't need the explicit expression, tho I think it's not hard to find. SU(2) was also in a talk today about Sato-Tate. hell of a group.

user19405892

hi

I just noticed that $9|2^b-2^a$ and $2^5|2^b-2^a$ implies $7|2^b-2^a$?

why is that

Semiclassical

03:19

you just noticed it, but you don't know why it is?

Akiva Weinberger

Oh, wow

user19405892

yes

Akiva Weinberger

You don't need the $2^5$ thing, do you?

user19405892

im not sure maybe i don't

Semiclassical

if you don't know why it's true, how do you know it's true?

Akiva Weinberger

03:21

The reverse isn't true; $8-1$ is a multiple of $7$ but not $9$.

user19405892

i think you need both statements then

Akiva Weinberger

@Semiclassical Experimentation, I guess. And it is true, anyway.

Still don't think you need the $2^5$ thing. Just, "if it's a multiple of 9 then it's a multiple of 7."

Semiclassical

and you guys poke fun at physicist levels of rigor :p

arctic tern

$9\mid 2^b-2^a$ implies $b-a=6n$ and then $7\mid 2^b-2^a$ is equivalent to $7\mid 2^{6n}-1$. then $2^{6n}-1$ is divisible by $2^{3n}-1=(2^n-1)(2^{2n}+2^n+1)$. the order of $2$ mod $7$ is $3$. notice $2^n-1\equiv 0$ mod $7$ if $n\equiv 0$ mod $3$ and $2^{2n}+2^n+1\equiv0$ mod $7$ if $n\equiv 1,2$ mod $3$.

Semiclassical

well, if $2^5$ is to divide $2^b-2^a=2^a(2^{b-a}-1)$, it must be the case that $b> a\geq 5$. (the case of $b=a$ is of course trivial)

so there must exist $a',b'$ such that $a'=a-5$ and $b'=b-5$. but $2^b-2^a=2^5(2^{b'}-2^{a'})$, and neither 7 nor 9 can divide $2^5$. so it's sufficient to ask whether $9|(2^{b'}-2^{a'})$ implies $7|(2^{b'}-2^{a'})$

which means that being divisible by 2^5 is superfluous. all that matters is being divisible by 9. that's exactly what @akiva says

Akiva Weinberger

03:41

It's essentailly 'cause $(2^n\mod9)_{n=0}^\infty=(1,2,4,8,7,5,\dots)$ and then it repeats (notation?), while $(2^n\mod7)_{n=0}^\infty=(1,2,4,1,2,4,\dots)$ and then it repeats

I'm not sure if that's the right notation at all

arctic tern

ah, that's much better.

Akiva Weinberger

But they both repeat in (a factor of) every six numbers, because $2^6=64$ is one more than a multiple of both of them

arctic tern

9|2^b-2^a implies 6|b-a since the order of 2 mod 9 is 6. which in turn implies 3|b-a which implies 7|2^b-2^a since the order of 2 mod 7 is 3.

Akiva Weinberger

Ya. And the reverse doesn't work.

'Cause we have $1$ twice every six in the second thing. Or, $3|b-a$ doesn't imply $6|b-a$.

Semiclassical

now i'm curious what that orbit looks like in $Z_{63}\cong Z_7\times Z_9$

Akiva Weinberger

03:46

1,2,4,8,16,32,repeat

Semiclassical

...durr, yeah. though i more meant in terms of Z_7 x Z_9

Akiva Weinberger

11,22,44,18,27,35,repeat, where by 11 I mean $(1,1)\in\Bbb Z_7\times\Bbb Z_9$

Semiclassical

(1,1), (2,2), (4,4), (1,-1), (2,-2), (4,-4)

Akiva Weinberger

Ooh, that's more symmetrical. I like that.

Semiclassical

snerk

Prince M

04:02

Quick random question... if considering the top half of a hallow torus in R3 (as in sliced by z = 0), would its manifold boundary would be 2 concentric circles?

But then shouldnt boundary of that manifold be a n-1 manifold? Would we consider the two circles seperate boundaries then each would be a n-1 manifold

Jim Wilson

math.stackexchange.com/questions/1800524/…

Guy's I'm going to faint. I just cannot figure out part (d) for this question

Mike Miller

If I fainted for everything I didn't understand I would probably be comatose

Akiva Weinberger

05:01

RIP Mike?

Balarka Sen

07:13

@PrinceM Yes. And since dimension of that is 2, the dimension of the boundary is 2 - 1 = 1 because it's disjoint union of two circles and a circle is a 1-manifold.

Rajesh Dachiraju

07:56

Hi

OneRaynyDay

Hi there!

I was wondering if anyone could help me with this problem: Find all a's such that the matrix is diagonalizable: $\begin{bmatrix} 1 & 1\\ a & 1 \end{bmatrix}$

Tobias Kildetoft

@OneRaynyDay Have you computed the characteristic polynomial?

OneRaynyDay

Yes I have. I got the charpoly as: $\lambda^2-2\lambda+1-a$

Tobias Kildetoft

and what is the discriminant of that?

OneRaynyDay

Sorry, typing in LaTeX still feels foreign to me, but trying to keep the formatting pretty for the chat

Tobias Kildetoft

07:59

it is good practice

OneRaynyDay

The discriminant is $b^2-4ac$, and thus it is $4 - 4*1*(1-a)$, which is $4-4+4a$ which is $0+4a$. Gotcha. It has to be any a that is positive.

Tobias Kildetoft

I mean for this particular polynomial

OneRaynyDay

Thank you! I was trying to check when the eigenvalue was 0. I know that if the eigenvalue is zero then it's not diagonalizable since the initial matrix is singular

I forgot to check the discriminant :)

Tobias Kildetoft

having $0$ as an eigenvalue is not relevant for being diagonalizable

unless both eigenvalues are $0$

OneRaynyDay

@TobiasKildetoft Hmmm, but doesn't the fact that we get the diagonalized matrix to have a row of entire zeros make it singular?

Tobias Kildetoft

08:04

@OneRaynyDay Sure, that does not mean it cannot be diagonalizable

note that a matrix is singular if and only if $0$ is an eigenvalue

whether or not it is diagonalizable

OneRaynyDay

@TobiasKildetoft Ah you're right, I think I'm getting myself confused by doing eigen stuff for 3-4 hours straight. Thank you!

Tobias Kildetoft

Yeah, it is a slightly confusing thing at first, but after a while it will make a lot of sense and start to feel natural

OneRaynyDay

08:19

@TobiasKildetoft right - thank you :)

@TobiasKildetoft Are you not too busy right now Tobias? Could I go over a slightly harder question of the same variety with you just to make sure I'm understanding it correctly?

If it's not too much too ask

Tobias Kildetoft

Sure, I could use some procrastination

OneRaynyDay

Awesome - thank you! So another problem poses this matrix: $\begin{bmatrix}
1 & a & b\\
0 & 2 & c\\
0 & 0 & 1
\end{bmatrix}$

So my first step was to find the eigenvalues and since this is an upper triangular - rather easy to see : $\lambda = 1,2$

So then I started working on the Eigenspace $E_2$

I found that the matrix became: \begin{bmatrix}
-1 & a & b\\
0 & 0 & c\\
0 & 0 & -1
\end{bmatrix}

And we are trying to find the kernel, so naturally I rref'd it and it became:

\begin{bmatrix}
1 & -a & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{bmatrix}

This kernel shows that there's a single eigenvector spanning the nullspace: $\begin{bmatrix}
a \\
-1 \\
0
\end{bmatrix}$

Tobias Kildetoft

right

OneRaynyDay

This meant the $gemu(2) = 1$ and $algmu(2) = 1$. We have 2 more eigenvectors to find from $E_1$

So then doing the same thing to $E1$ I got: $\begin{bmatrix}
0 & a & b\\
0 & 1 & c\\
0 & 0 & 0
\end{bmatrix}$

Tobias Kildetoft

right (you did not actually need to find the eigenvector to know this part, as this always happens when you have algebraic multiplicity $1$)

OneRaynyDay

08:29

Ah - good point. So this doesn't really eliminate any choices of a from knowing $E_2$'s basis?

Tobias Kildetoft

right

OneRaynyDay

But yeah - continuing on, I then rref'd the $E_1$ and got: $\begin{bmatrix}
0 & a & b\\
0 & 0 & ac-b\\
0 & 0 & 0
\end{bmatrix}$

And we know that $gemu \leq almu$, so then we need to find specifically 2 vectors so that $gemu = almu$ for this to be diagonalizable?

Tobias Kildetoft

You need to be a bit careful here, as you seem to have assumed $a\neq 0$ for that rref

OneRaynyDay

@TobiasKildetoft Oh shoot - should I flip rows and do gaussian elimination from there?

Tobias Kildetoft

(but just switch the first and second row to make this easier)

OneRaynyDay

08:33

Gotcha, that makes sense :)

Tobias Kildetoft

always best to work with the specified numbers as much as possible

OneRaynyDay

Good idea - then I'll flip it and I get this: $\begin{bmatrix}
0 & 1 & c\\
0 & a & b\\
0 & 0 & 0
\end{bmatrix}$ to $\begin{bmatrix}
0 & 1 & c\\
0 & 0 & b - ac\\
0 & 0 & 0
\end{bmatrix}$

Tobias Kildetoft

not quite, you should get $b - ac$

OneRaynyDay

@TobiasKildetoft Derp - you're right. Let me edit that, but on a second thought aren't they equivalent?

Tobias Kildetoft

if you want b/a then you need $a\neq 0$ which we wanted to avoid for now

OneRaynyDay

08:38

Ah right, keep forgetting that exception.

Now that we've boiled it down to this, and we need 2 vectors in the nullspace, then we know that the rank of the matrix has to be 1, which means there's only 1 pivot, which means the 1 that was constant has to be the only row that's non-zero after rref.

Tobias Kildetoft

right

OneRaynyDay

So therefore b-ac is equal to 0(?), and we have two vectors like: $\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}$ and $\begin{bmatrix}
0 \\
c \\
-1
\end{bmatrix}$

Tobias Kildetoft

yep, looks good

OneRaynyDay

So then we have the conclusion that it works for any $a,b,c$ such that $b-ac = 0$ :) right?

Martin Sleziak

Wouldn't a quicker way be checking when $A-I=\begin{pmatrix}
0 & a & b \\
0 & 1 & c \\
0 & 0 & 0 \\
\end{pmatrix}$ has rank equal to 1?

Tobias Kildetoft

08:42

yes, it is diagonalizable precisely when $b - ac = 0$

Martin Sleziak

Which is equivalent to the determinat $\begin{vmatrix}
a & b \\
1 & c \\
\end{vmatrix}=ac-b$ being non-zero.

Tobias Kildetoft

@MartinSleziak Sure, basically the same calculation (I am not sure if the question also asked for a basis of eigenvectors in case the matrix was diagonalizable)

OneRaynyDay

@MartinSleziak Ooh, I didn't really quite get this step you're proposing - How does this submatrix's determinant correlate to its rank?

Martin Sleziak

First rank of $\begin{pmatrix}
0 & a & b \\
0 & 1 & c \\
0 & 0 & 0 \\
\end{pmatrix}$ and rank of $\begin{pmatrix}
a & b \\
1 & c \\
\end{pmatrix}$ is the same, right?

OneRaynyDay

I mean - I understand that the matrix is degenerate when its determinant is zero - but we don't know how many ranks it really has if its determinant is zero right? We just know it has a nontrivial kernel

Martin Sleziak

08:44

If you agree to this the rest is easy.

Let us denote $B=\begin{pmatrix}
a & b \\
1 & c \\
\end{pmatrix}$.

Clearly $\operatorname{rank} B\ne 0$.

So we have only two possibilities.

Either $\operatorname{rank} B=1$, which means it is singular and $\det B=0$.

OneRaynyDay

Ah - because of the 1 holding it to having at least $dim(image) \geq 1$

Martin Sleziak

Or $\operatorname{rank}=2$, which means it is invertible and $\det B\ne0$.

OneRaynyDay

right okie

Martin Sleziak

@OneRaynyDay Exactly. Since $B\ne0$, we know that $\operatorname{rank} B\ne 0$.

OneRaynyDay

Cool - thanks so much @MartinSleziak and @TobiasKildetoft you guys are the best :^)

Also learned how to write matrices in LaTeX as well haha

Martin Sleziak

08:47

Exactly as you say, determinant distinguishes only between full rank and smaller rank. This case was simpler, because the only possibilities were 1 and 2.

OneRaynyDay

@MartinSleziak Gotcha :)

wing

Hello Guys, sorry for the off-topic, I have an active question and I was wondering If anyone of you could comment or answer.
[title](http://math.stackexchange.com/questions/1799681/order-size-estimation-of-converging-sum-used-for-approximation-of-logarithm)

Slereah

09:04

Hey is it a good idea to try to span a manifold with two center bipolar coordinates

Two-center bipolar coordinates

For related concepts, see Bipolar coordinates. In mathematics, two-center bipolar coordinates is a coordinate system, based on two coordinates which give distances from two fixed centers, and . This system is very useful in some scientific applications (e.g. calculating the electric field of a dipole on a plane). == Transformation to Cartesian coordinates == The transformation to Cartesian coordinates from two-center bipolar coordinates is where the centers of this coordinate system are at and . == Transformation to polar coordinates == When x>0 the transformation to polar coordinates from...

They span the entire manifold and run over $\Bbb R$, but they have this weird condition that they have to form a triangle so they aren't totally independant

Will that have any important consequences on manifold business

Like trying to solve the Laplace equation on them or whatnot

I can't find a lot of ressources on them so it's hard to judge

Benjamin R

09:22

Hi All, when we say "fix a vector $\bar{v} \in V$"... or "fix a basis $\beta$", why don't people just say "Let" or "Define"?

Tobias Kildetoft

@BenjaminR "define" would be incorrect if you don't actually define it

but "let" is synonymous with "fix" in this context (as long as you restructure the sentence a bit)

it is just a stylistic choice

Benjamin R

@Tobias Okay, true... I mean, it's math, not an imperative/stateful programming language, the value of $x$ never changes

"Fix" just f**ks up my head.

Slereah

I think it's to convey that the choice of basis isn't important

You're just picking a particular basis from the set of all basis

Benjamin R

But it's never important unless we are talking about the "standard basis"

I think "Let" is clearer from an English-language perspective. But yeah, thanks for clarifying!

I get a bit autistic about these sorts of things.

Unnecessary ambiguity of terms gives me chest pains

Have to do groaning.

Balarka Sen

10:46

Hi @AlexClark

Tobias Kildetoft

@BalarkaSen Hey, what are you up to at the moment?

Balarka Sen

Hi @TobiasKildetoft. I haven't gotten much of my math done for the past couple days - have been busy with school work (stat & physics). My current project (which I need to get time to work on) is to study differential topology from Guillemin-Pollack.

Tobias Kildetoft

neat

Balarka Sen

What about you?

Tobias Kildetoft

Writing up some stuff. Just finished writing up an example of why a certain assumption cannot be dropped from some earlier results of others, together with what does happen when one drops it

Balarka Sen

10:56

Ah. I won't ask what you are writing on though, since I guess it would be too technical and foreign to me.

:)

Tobias Kildetoft

And now I am going to write up why the assumptions I replaced it by cannot be avoided (at least completely)

Well, the assumption that cannot be removed is just that $p\geq 2h-2$ where $p$ is the characterisic of the field one is working over and $h$ is the Coxeter number of the root system associated to the algebraic group

And usually one can remove that by instead assuming a certain conjecture, but not in this case (though the modified statement becomes equivalent to that conjecture)

the statement in question is a reciprocity between two types of modules (or rather on multiplicities of filtrations of those modules)

Balarka Sen

Reciprocity as in?

Leaky Nun

(latex testing) $\{n:n\ge2\land n\in\mathbb N\}$

Tobias Kildetoft

as in the multiplicity of the module $M$ in the module $N$ is the same as the multiplity of the module $L$ in the module $M$ (where $L$ and $N$ are related in a specified way)

Leaky Nun

(Do I use $:$ or $|$?)

Balarka Sen

11:01

What's multiplicity of a submodule?

Tobias Kildetoft

Written in symbols $[M:N]_{\nabla} = [N:L]_G$ (the subscripts are not important)

@BalarkaSen so if the module is simple (which $L$ is in this case) then it is the number of times it occurs in a composition series, i.e. the number of subfactors isomorphic to it in a filtration where the subfactors are simple

Balarka Sen

ah, gotcha

Tobias Kildetoft

@LeakyNun either can be used, but it would be more common to write the $\in \mathbb{N}$ in the first part

@BalarkaSen the thing is that here the module $M$ also has a different type of filtration where the subfactors are not simple (if $N$ was simple this would not be very interesting)

Leaky Nun

@TobiasKildetoft Oh, thanks

Balarka Sen

12:16

@Clarinetist I saw that you were away when I pinged you yesterday. I'll be here for a while, so if you come here please shoot me a ping. I got a couple questions to ask you.

Regarding statistics of course.

jordan178

12:50

Hi guys

Semiclassical

morning chat

jordan178

$P(R=n+i)=\begin{pmatrix}n+i-1\\
i
\end{pmatrix}(p)^{n}(q)^{i}$ and $q := (1-p)$ and$i = 0,1,2,..$ How can I show that this is true?

We throw a coin with success probabilty p and R is the amount of coin tosses we need untill we have n successes

Semiclassical

...i really want to yell at mathematica, except i know i'm the one who screwed up :/

meant for it to do two sweeps of data overnight, but i did the setup wrong and it only did the first one

mutter mutter mutter

jordan178

mutter is german and means mum

Semiclassical

i see

Mike Miller

13:00

morning

Semiclassical

morning @mike

@MikeMiller how're you?

jordan178

Im fine thx

user19405892

hi

what is the period defined as in modular arithmetic?

for example the powers of 2 mod 10

$2,4,8,6,\ldots$

what if we start with $2^0$? Then it is not periodic or do we not consider $2^0$ when considering the period?

Mike Miller

not ready for my meeting

been messing up this computation all week. want to get it right before we talk

Semiclassical

@user19405892 oeis.org/wiki/Eventually_periodic_sequences

jordan178

13:06

Mike Miller just tell them that you found an elegant proof but your page is too small to write it down

Semiclassical

tired joke is tired

Balarka Sen

also, not the ideal excuse to have

Semiclassical

morning @balarka

Balarka Sen

good morning.

Mike Miller

i also need to tell him that the thing I proved to him in our meeting last week was completely wrong

Semiclassical

13:07

@MikeMiller that's exactly how I've been feeling this week

i had a nice computational result on tuesday, but all the other attempts this week have gone wrong

jordan178

how can a proof be wrong

Semiclassical

"the thing i thought i proved to him", i imagine

jordan178

;)

Semiclassical

including this last calculation, with the consequence that it won't finish until 5pm instead of 1pm

Mike Miller

He didn't tell me it was wrong then so I can be glad I know it was subtle

Semiclassical

13:09

which is just greeeeat

Balarka Sen

tell him that you wrote down a proof and put it in a klein bottle, and couldn't get it out anymore.

or maybe don't.

Mike Miller

none of this has been funny

jordan178

lmao

it is funny

Balarka Sen

i tried

Semiclassical

not cynical enough to be gallows humor, which is about the only kind of humor one can find in a situation like this

Mike Miller

13:11

@SemiC I think I'm honestly close now, just want to get to campus and start writing and have finished up when I see him

Semiclassical

gotcha

i think this last calculation would have worked, but this four-hour delay is just really really inconvenient

especially as this is the last day the person to whom i'd want to show this (a visiting prof) will actually be here

1pm on his last day? not great, but not horrible

5pm on his last day? substantially worse.

jordan178

maybe there is a complete proof on the internet

Semiclassical

if there's a complete proof on the internet, then someone else has already done it. not a great improvement in that case.

jordan178

it seems weird to me that you are just about to proof sth no one ever has proven

Semiclassical

that's why they call it research.

academic research: making people grumpy and/or anxious since forever

jordan178

13:14

lol

Balarka Sen

here's a probably better joke from the book i am reading: "They say that the famous mathematician d'Alembert, giving lessons in mathematics to a very noble and very stupid student, couldn't explain to him the proof of a theorem and desperately exclaimed: "Upon my word, sir, this theorem is correct!". His pupil replied: "Sir, why didn't you say so earlier? You are a noble man and so am I. Your word is quite enough for me!"".

well, not really joke but anecdote.

jordan178

that one really isn't funny :D

Do you guys wish sometimes that you did something else than maths?

Semiclassical

technically I do physics, but that basically equals applied math for me

Balarka Sen

@jordan178 no

but life gets in the way

jordan178

Im only a pure maths bsc but still sometimes I think I should have done sth else

modern physics seems to be so crazy

Semiclassical

13:18

and it depends on what you mean by that. i don't regret being a math/physics guy, but i'm not planning to stay in academia (at least not research level)

jordan178

like I can't even believe it

I won't go into academia because I don't think that there is a slightest chance that I could find something out that terrytao.wordpress.com couldnt find

Mike Miller

I like to watch movies as well as do math

jordan178

Whats your all-time-fav?

Mike Miller

i enjoy the math and my hobbies and volunteerish work

user 1618033

@MikeMiller movies are a waste of time

Balarka Sen

13:23

I am not a movie fanatic but I like watching them occasionally.

jordan178

movies are a part of human culture

Mike Miller

I don't believe in favorites, but three I really enjoy ("three of my top 5, in no particular order") are stalker, the man who fell to earth, and synecdoche New York

Balarka Sen

I also read occasionally.

jordan178

I have some volunteerish work for you math.stackexchange.com/questions/1802126/…

Balarka Sen

also, stalker is my all time favorite.

user 1618033

13:26

@MikeMiller but don't you also enjoy going out in the city with some girlfriend?

No one said anything about this option.

Math, movies, but some dates? No one cares about some dates?

Mike Miller

@jordan178 ]i'm alright, thanks though

jordan178

lol

user 1618033

OK

Back to my work (trying to finalize an EPIC article)

jordan178

gl hf

Clarinetist

@BalarkaSen What's up?

user19405892

13:40

are the powers of $2$ periodic modulo 10?

Leaky Nun

Yes, $2,4,8,6$.

user19405892

but what about $2^0$?

we don't consider that?

Leaky Nun

Well, $1\to(2\to4\to8\to6\to)$

Balarka Sen

@Clarinetist Hi.

user19405892

but then we have $1,2,4,8,6,2,4,8,6,\ldots$

Clarinetist

13:42

Hello @BalarkaSen

user19405892

oh so it is periodic from a point

Balarka Sen

So, I was wondering if you can explain me what's up with the fourth moment of given distribution. Apparently it measures how much "peaked" the frequency distribution is. How can I see this?

I have a vague explanation, but I am not quite convinced.

Leaky Nun

@user19405892 Yep.

Clarinetist

@BalarkaSen Oh wow. Well, you just happened to ask a really good question that I wish I knew the answer to. I've wanted to know the answer myself

@BalarkaSen Usually, however, the fourth moment isn't used, but rather, the (normalized) fourth central moment

@BalarkaSen It's also known as "kurtosis"

Leaky Nun

@user19405892

Balarka Sen

13:45

@Clarinetist Yep, I am aware.

@Clarinetist Right, one divides out by $\sigma^4$ to normalize the order (and get rid of units), I suppose.

Clarinetist

@BalarkaSen I will find out and ping you, how's that sound? That question has been on my mind every so often, but I forget to ask it

Balarka Sen

That'd be great, thanks very much.

Clarinetist

No problem

Balarka Sen

Thanks for the time, by the way. I have been studying some stat myself, so plan to ask more questions to the resident statistician (aka, you) here more often, if it's fine by you :)

Clarinetist

@BalarkaSen Definitely. I'd like to see more probability and stats questions in chat

Balarka Sen

13:51

Great. Hope to see you around more often.

Clarinetist

@BalarkaSen Yeah, I will be for the summer. Grad school kept me REALLY busy during last fall and spring

Balarka Sen

I have never studied probability and/or stat before, but I think both are quite interesting.

@Clarinetist Ah, I see.

Clarinetist

@BalarkaSen If you're familiar with linear algebra, I've learned that grad-level stats is essentially undergrad probability and stats with matrices. So, a lot of working with squared norms and quadratic forms

Balarka Sen

Ah, yes, I have heard of things called "random matrix theory" and "free probability theory" and whatnot. (I am familiar with linear algebra, yes).

Clarinetist

@BalarkaSen By the way, on the Wikipedia page for Kurtosis (en.wikipedia.org/wiki/Kurtosis#Interpretation)

@BalarkaSen

> The exact interpretation of the Pearson measure of kurtosis (or excess kurtosis) used to be disputed, but is now settled. As Westfall (2014) notes, "...its only unambiguous interpretation is in terms of tail extremity; i.e., either existing outliers (for the sample kurtosis) or propensity to produce outliers (for the kurtosis of a probability distribution)." The logic is simple: Kurtosis is the average (or expected value) of the standardized data raised to the fourth power. Any standardized values that are less than 1 (i.e., data within one standard deviation of the mean, whe…

@BalarkaSen So I would assume you're right about dividing by $\sigma^4$: it's just to make it unitless

Balarka Sen

13:58

I am not quite sure what measuring outliers actually mean.

Clarinetist

@BalarkaSen Outliers are quite arbitrary, unfortunately. The general idea of an "outlier" is that it's a set of data points which are clearly very far from most of the data, and could be caused by any number of things

@BalarkaSen For example, consider five points: 1, 1, 2, 2, 500

@BalarkaSen I think any reasonable person would think 500 is an outlier in this data set

@BalarkaSen As you might expect, this has a rather high kurtosis

Balarka Sen

Right. I suppose one could define them as data points being very far away from the central tendency, whatever that might mean.

Clarinetist

@BalarkaSen Basically

Albas

@BalarkaSen this stuff that you talk about in statistics it is your school stuff or self study?

Balarka Sen

@Clarinetist I agree.

@Albas Schoolwork.

Clarinetist

14:02

@BalarkaSen Ah, what textbook?

Albas

Hmm... That is something very shocking . For me random variables will be introduced this year. Your school is pretty ahead with stuff

Balarka Sen

I do not think you'd know the name of the textbook: it's not particularly famous. On a related note, if you can recommend something to read up from, I'd like to know.

Clarinetist

@BalarkaSen That's an easy one: Wackerly et al., Mathematical Statistics with Applications

Balarka Sen

Great, thanks.

I'll note it down.

Clarinetist

@BalarkaSen Expensive, but so worth it. I bought it in reviewing for mathematical statistics in grad school, and it is SUCH a nice read

Balarka Sen

14:05

@Albas I don't know random variables yet but the probability book I am reading up from has a chapter on random variables at the very end. I plan to read it anyway.

Clarinetist

@BalarkaSen After that, if you're wanting a graduate-level treatment, Casella and Berger's Statistical Inference

@BalarkaSen You're going to feel like random variables are not going to be treated very rigorously, just a FYI

Balarka Sen

Thanks. I'll have a look at that one too, but I still prefer to study mathematics than statistics :)

Clarinetist

@BalarkaSen I completely get that. However, Casella and Berger's treatment is still very mathematical. From there, though, there's a lot of applied stats books which just, quite frankly, don't care about the math. I stay away from those

@BalarkaSen Measure Theory + Statistics is something I'd like to study in the future

Balarka Sen

@Clarinetist Eh, most of what is in my probability book (more like a booklet: it's by E.S.Wentzel) is not treated too rigorously. But I have a companion book by Lipschutz which apparently treats stuff rigorously.

Albas

Hmm.. That is nice you guys go into such detail with stuff. I have been only introduced to stuff such as mean deviation, standard deviation , variance

Balarka Sen

14:08

@Clarinetist Measure theory is cool beans, I'd want to learn some at some point.

Albas

Which is good, but we have been taught in a more ad-hoc manner. Not rigorous at all

Clarinetist

@Albas That's usually how people start. It's very slow in the beginning, but once you start bringing in multivariable calculus, it's quite neat

Albas

There is multivariable calc in it???

Clarinetist

@BalarkaSen Yeah, I'm reading the Wentzel. It's not very rigorous, but there's some really, really neat math happening behind what is explained

@Albas Why, of course

@Albas Unless you're taught the really basic, undergraduate-freshman-level statistics, you'll be bringing in multivariable calc

Balarka Sen

@Clarinetist Are you reading the small book? It's a first course on probability, thus asking.

Clarinetist

14:12

@BalarkaSen Probability Theory (First Steps)

Balarka Sen

Oh, great, that's it.

Albas

That sounds interesting

Balarka Sen

Yeah, I agree there's much more going on behind what is being explained.

Clarinetist

@BalarkaSen Hmm... are you familiar with inverse images? I imagine you are, just making sure

Balarka Sen

i.e., I like the analytic form of law of large numbers that's being written out. But the proof's in the last chapter I think.

@Clarinetist Yes. Why do you ask?

Clarinetist

14:13

@BalarkaSen I've been working on a document for a friend of mine who knows analysis but knows no probability. Are you interested in reading it?

Balarka Sen

What is the document about?

Clarinetist

@BalarkaSen Intro to probability, basically, with an analysis point-of-view

Most probability books assume you've never seen analysis

Balarka Sen

Well, I am just starting on probability, so not sure how much I would understand though! But sure.

Clarinetist

@BalarkaSen Let me know when you have this downloaded: dropbox.com/s/xhl4kotkcswzgnu/Mathematical_Statistics.pdf?dl=0

Balarka Sen

Got it.

Clarinetist

14:19

@BalarkaSen There's the document. Let me know if you have any questions on it. It's primarily focused on the theory. If you're looking for applications, I'd get a textbook

Balarka Sen

Thanks, I'll definitely read it.

Clarinetist

@BalarkaSen And let me know if you have any questions/comments

Balarka Sen

Absolutely.

Samuel Yusim

14:32

yo

Eric Stucky

ye

Samuel Yusim

yu

Eric Stucky

me :D

Hey robjohn

I don't see you around as frequently

Samuel Yusim

so I'm reading about some randomized algorithm, and the proof of termination uses techniques from some dudes in the 1850's or something who created a probability model for how last names propogate and die out

kind of weird that something like that would have anything to do with this

Eric Stucky

yeah that's nifty :)

Semiclassical

14:42

Yves's comment here is the understatement of the day: math.stackexchange.com/q/1802208/137524

Samuel Yusim

also interesting: one of the guys who invented this probability model I mentioned also invented eugenics

"whoops"

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$Mathematics$ Mathematics