Mathematics - 2017-07-15 (page 1 of 3)

Anyways, inserting that integral representation means (in principle) that one can get that sum as a sum of contour integrals, resum that to get a single contour integral, and then hopefully give a closed-form result.

Also, Mathematica does give a closed-form result in terms of incomplete gamma functions

Simply Beautiful Art

Hm, the end result is supposed to include incomplete gamma functions, so it looks like its gonna be quite messy evaluating that integral

Semiclassical

Yeah, agreed.

Simply Beautiful Art

The incomplete gamma functions can be seen as a direct result after applying fractional derivatives

Semiclassical

makes sense

Simply Beautiful Art

1:45 AM

I wonder if there's any other approaches

Kasmir Khaan

2:23 AM

hi chatg

anyone here?

BAYMAX

Hi@KasmirKhaan

Daminark

3:18 AM

Hey there!

BAYMAX

hi

Daminark

3:32 AM

What's up?

Kasmir Khaan

@BAYMAX Helloha ! how are you ?

@Daminark Hi Amin =p

BAYMAX

Sky is always up :)

yeah Casimir

you can search for Casimir function :)

Kasmir Khaan

haha

BAYMAX

here

Kasmir Khaan

Am doing rook polynomials atm >< dont distract me =p

Very sure not made by me

Anyway you know what's the idea behind rook polynomials?

kinda new to me

BAYMAX

3:38 AM

to me too!

Kasmir Khaan

Grrrr I wish Ted was here to help us :D

BAYMAX

Potential users :)

Kasmir Khaan

Donno what you mean by that =p

if its good then thanks if not then shame on you -.-

Daminark

Apparently the idea is that you want to codify how many ways you can place rooks on some variant of a chessboard such that they can't kill each other

Kasmir Khaan

hmm I get that but whats the idea behind the solution of that problem ?

and why do we get a polynomial out of it?

BAYMAX

3:42 AM

i only meant that i wish that there should be some potential users to help you , if you understand it its good!

good luck with Rook polynomial

Kasmir Khaan

@BAYMAX haha okay thanks :)

Lucas Henrique

Hey there

Fast question:

Daminark

I'll have to read more about it, I'll get back to you, but it seems like it's about generating functions, which I know are very much a thing in combo but which I know little about

Kasmir Khaan

@Daminark thanks alot! :) Ill keep reading on it as well and look for some lectures

Lucas Henrique

If I have a polynomial $\mathrm{P}(x) = c + \prod_{j=1}^n x - a_j$, how do I find its roots?

Leaky Nun

3:47 AM

@LucasHenrique you can't

Lucas Henrique

Isn't there any cool manipulation? :/

Leaky Nun

not really

Lucas Henrique

I was trying to build another polynomial or use something like $\mathrm{P}(a_i) = \mathrm{P}(a_j)$

The actual problem is $(x-1)(x+2)(x-3)(x-2) = -3$

Daminark

Well $S_4$ is a solvable group so... Lol jk

BAYMAX

so how you are relating to this problem Daminark!

I am curious!

Daminark

3:53 AM

Oh it was joking that you could theoretically just solve by radicals. It's a thing in Galois theory or something

Lucas Henrique

Beyond $S_5$ there's no general solution, right?

Daminark

Yeah, since each $S_n$ has $A_n$ as a normal subgroup which is simple but not prime order, at least I think that's it

hi Demonark, @Lucas

Hey there!

Hey @Ted!

3:59 AM

I don't know why @Kasmir thinks Ted knows everything.

Demonark: Nonsolvability of the general polynomial isn't just simplicity of $A_n$, $n\ge 5$.

BAYMAX

$A_n$ is simple for $n \geq 3$

is this true>

?

Ted Shifrin

No, $n\ge 5$.

$A_4$ has a nontrivial normal subgroup.

BAYMAX

ok

Ted Shifrin

If you think of $A_4$ as the symmetry group of a regular tetrahedron, the three $180º$ rotations (plus the identity) form a Klein subgroup, and it's normal.

Lucas Henrique

@LucasHenrique by "symmetry", it may be interesting to use the $y = x - \mathrm{AM}(3,2,1-2) = x - 1$. Then the problem becomes $y(y-1)(y-2)(y+3) = -3$... which is not interesting at all

Daminark

4:07 AM

Oh whoops @Ted, what else do you need?

Ted Shifrin

It's called solvability :)

Daminark

Well yeah, I thought S_n isn't solvable because A_n is simple

That gives you a composition series where factors aren't of prime order

BAYMAX

I am thinking how to get exertise in abstract algebra,

Daminark

Which by J-H means you're done in general

BAYMAX

its too wide field for me to get hands on,i usually forget the definitions

><

Ted Shifrin

4:10 AM

Well, OK, you have to prove more to get to that point, but OK, Demonark.

Lucas Henrique

As a highschooler, everything I know about abstract algebra is the definition and basic properties of rings

BAYMAX

great

Lucas Henrique

I can't even get the meaning of ideals and quotient rings... smh

Daminark

@Baymax I'm not in too deep, I've only got some basic group theory and a few random facts here and there, but a lot of it boils down to practice and time

BAYMAX

oh hmm

Daminark

4:12 AM

Find a book that suits your style with good exercises and work through it

BAYMAX

thanks@Daminark

will try to do

Ted Shifrin

Lucas, you might find my algebra book more concrete and helpful (if you can find it in a library).

BAYMAX

you have an algebra book @TedShifrin

I thought it was linear algebra

Ted Shifrin

yeah, it was the first book I wrote

That was second.

Daminark

@Ted I was gonna recommend Lang!

Lol jk

Ted Shifrin

4:14 AM

@Lucas: FYI, I started with integers, then polynomials, then general commutative rings ... groups after that.

Kasmir Khaan

@TedShifrin Heyyyy Ted :D

Lucas Henrique

@Daminark "every joke is half the truth"

Ted Shifrin

hi @Kasmir

Kasmir Khaan

@TedShifrin I dont only think you know everything , you DO know everything :D

Ted Shifrin

definitely NO.

Kasmir Khaan

4:15 AM

ermm

Modest man -.-

What is the idea behind generating functions?

Ted Shifrin

I'm not the right person for that.

Kasmir Khaan

I mean like basic idea ( did not read about it yet )

ermm okay thanks anyway :)

Daminark

@LucasHenrique I mean I don't know Lang much, it's along similar lines to recommending Hartshorne for alg geo or something

I used Herstein a bit and Rotman a bit, that was mostly it

Lucas Henrique

@KasmirKhaan I only know the empirical idea from what I've already used

Kasmir Khaan

@LucasHenrique anything you think its usefull just tell me :D

Lucas Henrique

4:23 AM

Generating functions are "infinite" polynomials of the form $f(x) = \sum_{j=0}^\infty a_j x^j$ that usually converge and are used in two ways:

Ted Shifrin

I don't think they have to converge at all :)

Just formal power series

Lucas Henrique

1. To find the coefficients given a closed form
2. To find the closed form given coefficients

But I'm also not a pro in gen functions, I usually use them to determine the closed form of the terms of a sequence

@KasmirKhaan the Wikipedia page has plenty of content in this subject

Generating function

In mathematics, the term generating function is used to describe an infinite sequence of numbers (an) by treating them as the coefficients of a series expansion. The sum of this infinite series is the generating function. Unlike an ordinary series, this formal series is allowed to diverge, meaning that the generating function is not always a true function and the "variable" is actually an indeterminate. Generating functions were first introduced by Abraham de Moivre in 1730, in order to solve the general linear recurrence problem. One can generalize to formal series in more than one indeterminate...

Kasmir Khaan

Thanks alot!:)

Ehm I know about that >< but I prefer allways to find ideas from others to make it "softer"for me to understand new topics =p

Lucas Henrique

@TedShifrin "Unlike an ordinary series, this formal series is allowed to diverge, meaning that the generating function is not always a true function and the "variable" is actually an indeterminate."

Kasmir Khaan

r_0 (C) = 1 , any idea why placing 0 non-attacking rooks can be done in one way on any chess board?

Better now :)

Daminark

4:31 AM

You're not making any choices, right?

Leaky Nun

@KasmirKhaan there's only one way: not to place anything

Kasmir Khaan

hmm makes sense now when I think about it =p

but sometimes with the same logic it can be infinity as the answer

I never know which is which =p

Leaky Nun

@KasmirKhaan why?

Kasmir Khaan

I did not mean it in this case , but sometimes they define things without telling us full story

Lucas Henrique

@KasmirKhaan with the same logic, in my understanding, it could be 0

Kasmir Khaan

4:33 AM

I think in this case its clear =p

Leaky Nun

I think logic is sometimes overrated

I mean, what people refer to as "logic" is merely their own intuition.

Kasmir Khaan

@LeakyNun true ><

Lucas Henrique

@LeakyNun or underrated, maybe?

I usually use the term logical with the same sense of "reasonable thinking", but I know it's wrong

The Raiders of Las Vegas

5:27 AM

logic: : a proper or reasonable way of thinking about or understanding something. : a particular way of thinking about something. : the science that studies the formal processes used in thinking and reasoning.

Daminark

7:44 AM

Hey @s.harp and @Liad!

Liad

@Daminark hi, how are you ?

Daminark

I'm well, how's it going with you?

Liad

im fine, doing exams this month

did you finished the semester already ?

Daminark

Yup, I've been done for about a month now, actually

Liad

No exams ?

Daminark

7:49 AM

I did have exams, it was all in a week

Liad

Wow..

You had like one or two courses or it is just like that in your university?

Daminark

Well, in this case both were sorta true

Like, I had only 2 classes give exams during exam week

One class finished up stuff in 9th week, and another had a paper that was due the Saturday after finals week

But in general there's just one week when finals happen

Liad

Amazing. at my university it is a whole month, and if you need the second test, another month

When does the next semester begins ?

Daminark

Not sure what to think about that. On one hand it's nice to have multiple chances, and not to have everything crushed into a week, but it's also nice to be done in a week

We do quarters instead of semesters, so we'll be starting, I think September 25 $\pm \epsilon$

Liad

Hehe

Where are you studying ?

Daminark

7:59 AM

Next quarter? I'm between doing various things

Liad

No i mean which university

Daminark

Oh where, I'm at Chicago

You?

Huji

In Jerusalem?

Yea

8:00 AM

Cool!

Liad

There is a TA here that i heard will be at Chicago next year

Daminark

If only to ask what I thought you were asking, what are you studying?

Hey @Steamy!

SteamyRoot

Ohi

Daminark

And @Liad, nice, like a current grad student who's coming here as a postdoc?

Steamy, now's a good time for the raid btw, I'm in the barn and there's some Hagoromo laying around

Liad

His name is Asaf Katz

SteamyRoot

8:02 AM

:O

Liad

And im studying Math-computer science. you ?

Daminark

Math-theoretical computer science

A lot of stuff in the coding side of the CS department is done in C which isn't quite my speed

Liad

:-)

I took C/C++ courses last year , it is better than Java :P

Daminark

I haven't ever used Java, actually. First quarter of the intro sequence was Racket, second was C

Somehow, I think C just involves too much micromanaging/ways to screw up

Segfault this, code dump that, mem leak over there

Liad

This is why i liked it :P it felt like talking with the computer

but i agree there is a lot of places to screw up there

because of that i made the exercises on the school's computers

Daminark

8:08 AM

But yeah I mean, next year I'll probably focus on the more theoretical side of compsci. Sitting in on discrete, thinking of do information theory and error-correcting codes, algorithms, combinatorics, all that

Liad

Have you done computability theory course?

Daminark

Not yet, I'll do that too though

Second part of logic

Liad

Great. im gonna go learn topology :P got exam soon . see you letter

Daminark

See you!

@Steamy Seems like the summer is always the perfect time for it. I've scored a stick already (there were a few laying around so like c'mon)

Balarka Sen

hi chat

SteamyRoot

8:18 AM

@Daminark Hrmf. Nobody at my campus has any, so I'll end up ordering a bunch myself.

But I'll probably have to fit my office with extra locks, then...

ohi

Daminark

Hey @Balarka

@Steamy Kek

Astyx

Hi chat

Daminark

Hey @Astyx!

Balarka Sen

what will I do now

what will I ever do

I'll walk the street with my hair down, so

Astyx

I'm facing the same dilemna

Balarka Sen

8:25 AM

The hot water at ten.
And if it rains, a closed car at four.
And we shall play a game of chess,
Pressing lidless eyes and waiting for a knock upon the door.

Daminark

m8w0t?

Also jfc Titchmarsh's proof of the Cauchy integral theorem is just ridiculously long

Balarka Sen

it starts with Goursats lemma doesnt it

Daminark

Huh? Which is that?

Balarka Sen

Cauchy integral theorem for a triangle contour

Daminark

I partially zoned out during that proof but it does some stuff with squares

Like it uses the whole, $|\int_C f(z)dz| \le ML$ where $|f(z)| \le M$ and $L$ is the length of $C$

Balarka Sen

8:29 AM

I don't like that proof. I would much rather prove it for C^1 holomorphic functions

using the fact that $f(z)dz$ is a closed form

Daminark

Yeah same

Balarka Sen

all the hassle is to do it for just complex differentiable guys

Daminark

I don't get why the lecturer did the proof of Titchmarsh, we were literally told to just use the Stein-Shakarchi proof since we also wanted to get to the proof of the integral formula, holomorphic functions have power series, and Liouville

Wait what do you mean?

Balarka Sen

Stein-Shakarchi proof uses Goursat's lemma

What do you mean what do I mean?

Daminark

@BalarkaSen This, I'm not sure what you mean by "just complex differentiable", and what the hassle is

Balarka Sen

8:33 AM

The proof by seeing $f(z)dz$ is a closed form works only if $f$ is not only complex differentiable, but continuously complex differentiable

Because Green's theorem works for C^1 functions only

The usual Goursat trick (which eg SS does) does not use that extra continuity of $f'$ hypothesis

Daminark

Wait hold on though if something is holomorphic you should already know it's analytic, I don't see why there's a problem

Balarka Sen

That complex differentiable implies C1 C2 blergh blergh blergh C^omega is itself a consequence of Cauchy integral theorem :P

you can't use Cauchy integral theorem to prove Cauchy integral theorem

that's the thing

Daminark

Oh, well shit

Balarka Sen

:)

Daminark

And are there any shortcuts for just holomorphic functions

Astyx

8:37 AM

How can I find an reccurence relation on $$\int_{0}^{+\infty} {dt\over(1+t^a)^n}$$ ?

($a\gt 1$)

Balarka Sen

@Daminark Not that I'm aware of. The only proof I have ever seen proves it for a triangle, and then triangulates the interior of any given contour.

Or, well, explicitly comes up with an antiderivative (which does require the triangle thing) of $f(z)$

Daminark

I see

Balarka Sen

By setting $F(x) = \int_{a}^{x} f(t) dt$, say, where integration is over a contour which moves from $a$ to $x$ by (1) a path parallel to x-axis and then (2) a path parallel to y-axis.

So it's an "L" shaped path

Daminark

Hmm

I mean I suppose it's convenient to know things in full generality

Zeta.Investigator

8:58 AM

Maryam Mirzakhani, winner of Fields medal died today due to cancer :(

26

ifpnews.com/exclusive/iran-math-genius-die-cancer

Secret

$$\int_0^{\infty}\frac{dt}{(1+t^a)^n}$$

Let $u=t^a \implies du= at^{a-1}dt$

$$\frac{1}{a}\int_0^{\infty}\frac{du}{u^{1-\frac{1}{a}}(1+u)^n}$$

and partial frac the resulting integrand?

Astyx

What do you mean by that last sentence ?

Secret

uh, I am not very sure if we can do a partial fraction on that to split up the two terms in the denominator since $a$ is real

Astyx

Oh I get what you mean

But yeah, that seems unlikely

Secret

Wolfram alpha give a nice closed form which I don't have enough background to understand

wolframalpha.com/input/?i=integrate+1%2F(1%2Bx%5Epi)

hypergeometric functions

(that's for n=1, though)

Astyx

9:09 AM

I'll look into that

Secret

wolframalpha.com/input/?i=integrate+1%2F(1%2Bx%5Ea)%5En

Astyx

Thanks

Secret

General case here

Alessandro Codenotti

@BalarkaSen Goursat was on rectangles in my complex analysis course, weird

Astyx

Oh but that's the indefinite integral

Alessandro Codenotti

9:11 AM

Not that it makes any difference

Balarka Sen

@Alessandro Sounds fair. That should neither be harder nor easier than triangles

Astyx

@Secret According to Mathematica, you get $$\Gamma(1+{1/a})\Gamma(n-{1\over a})\over \Gamma(n)$$

So I guess the induction formula is $I_{n+1} = (1-{1\over na})I_n$

Secret

I got that kind of expression too once you use the definite integral

wolframalpha.com/input/…

(Wolfram alpha refuses to do anything if I use a generic a)

Daminark

What is up with this gamma function?

Astyx

Right I got it

Oh clever actually

I'll type it in a sec

Danu

9:33 AM

@MikeMiller Source?

Astyx

$$\begin{align}
I_n&=\int_0^{+\infty}{dt\over (1+t^a)^n} \\
&= \underbrace{\left[{t\over (1+t^a)^n}\right]_0^{+\infty}}_{=0\text{ since }a>1} + \int_0^{+\infty}{na t^a\over (1+t^a)^{n+1}}dt \\
&= \int_0^{+\infty}{na(1+ t^a-1)\over (1+t^a)^{n+1}}dt \\
&= na(I_n - I_{n+1})
\end{align}$$

@Secret

Secret

O wow, straight out integration by parts. I guess I need to be more comfortable handling integral functions as it is...

Astyx

Me too :/

Mary Star

Hello!! We have three lineswith equations $a_{i1}x+a_{i2}y+a_{i3}=0$. I want to show that $\det((a_{ij}))=0$ iff the lines are pairwise parallel or they have a common point.

So, we have the system with 3 equations and 2 unknowns:
$\left\{\begin{matrix}a_{11}x+a_{12}y=-a_{13} \\ a_{21}x+a_{22}y=-a_{23} \\ a_{31}x+a_{32}y=-a_{33}\end{matrix}\right.$

To check the existence of solutions we consider the respective matrix $A'=\begin{pmatrix}a_{11}&a_{12}&-a_{13} \\ a_{21}&a_{22}&-a_{23} \\ a_{31}&a_{32}&-a_{33}\end{pmatrix}$.

user84215

Why do the letters in the box not appear in math mode?
$$\fbox{gwsrgwergertg}$$

user84215

9:49 AM

with code \fbox

Daminark

(removed)

Secret

@MaryStar Well, if you row reduce (or compute via submatrix expansion), that determinant is indeed zero

Secret

[Unrelated]

Mary Star

@Secret Ah ok!! So, then we have tha when the three lines are parallel that the determinant is 0. When can we say if the three lines have a common point?

Secret

It will mean the system of linear equation have a unique solution, that is only possible if the coefficient matrix is invertible

Mary Star

10:04 AM

@Secret Why s this the only case? And isn't the coefficient matrix is invertible when the detreminant is non-zero?

Secret

O wait a sec, your lines are in $\Bbb{R}^2$. That means you only need two lines to intersect at a point and the third line can be of any slope as long it intersect that common point. I need to think what that means in terms of matrix because we are having a non square coefficient matrix here...

user84215

When you define a new command, how long does it last?

Secret

So, for the scenario where 3 lines intersect at a point in $\Bbb{R}^2$, the row echelon form of the coefficient matrix is will have a row of zeros, thus in order for there be a solution, there will be a constriant on the coefficients of the 3rd line in terms of the first 2 lines so that the final row evaluates to zero

Viewing the augmented matrix as a 3x3 matrix, the determinant will be zero thus the determinant is insufficient to tell you any more information about whether 3 lines meet at a point

In general, if you have n linear equations in n unknowns (i.e. n lines in $\Bbb{R}^n$) the determinant will be able to tell you whether there is a unique solution, or whether you have infinite or no solutions

but for the more general $n$ lines in $\Bbb{R}^m$, the determinant tells you little about the nature of the solutions

Secret

10:23 AM

[Unrelated]

$$\overset{+}{\left.\begin{matrix}
\\
\\
\\
\end{matrix} \right|} \begin{matrix} * & * & * \\0 & 0 & * \\0 & 0 & *\end{matrix} \overset{+}{\left.\begin{matrix}
\\
\\
\\
\end{matrix} \right|}=0$$

user84215

I have asked two question; neither of them has received any response.

Secret

4

Q: How to put a character inside a square

In a math formula, is there a way to put a number inside a square, as a decoration? I.e., instead of $\bar{1}$ or $\widehat{1}$ I would like to write something like $\insquare{1}$ and have the “1” inside a small square. Is this possible?

> There are three simple options. One is \fbox{}, the content of which is typeset in text mode, but can handle math mode as well. Loading the amsmath package provides \boxed{}, the content of which is typeset in math mode.

For /newcommand, I am not sure, it probably works for ther rest of your document

user84215

When I refresh the page, it still works.

user84215

that is, it remains forever?

Secret

10:38 AM

For the SE chat, the defaults will be resetted once you reload the page and the newcommand is no longer in your screen, but in a latex document, well, it will just stays, that is why using newcommand is quite dangerous if you are not careful. I personally don't use it much thus I am nto very famialr with the way to restore the defaults

user84215

But when I refreshed the page, it still works well.

user84215

in SE chat.

user84215

How can I cancel new commands?

Secret

where is your last /newcommand?

user84215

\ses

Secret

10:44 AM

no, I mean the permanent link to your last message that contain the /newcommand

user84215

in practicing mathjax room

Secret

link ?

user84215

in Practicing MathJax, 12 mins ago, by aminliverpool

$$\ses{A}{B}{C}$$

Secret

O, your newcommand is still visible on the screen, thus when you run mathjax it will be executed and thus you cannot undo it.

Try practicing other mathjax so that the message move past the screen, then things should go back to normal

user84215

there is no command to undo new commands?

Secret

10:50 AM

not that i know of

user84215

Thanks.

Simply Beautiful Art

11:29 AM

@Secret Much easier to just do a quick arc shaped contour with an angle above the positive real lines of $2\pi/a$.

Then its just residue theorem and symmetry

Secret

Well, my complex analysis skill is not as good, thus that does not quite occured to me

Simply Beautiful Art

@Astyx The formula should hold for non-integer $n$, hence using induction is not going to be worth while

Astyx

I'm not too sure what you mean by that

Simply Beautiful Art

To prove it by induction for any positive real $n$ and $a$ (such that the integral converges), you'd have to solve it for all $0<n\le1$ and apply induction on that, but solving it for that is about as hard as not using induction in the first place (you'll probably solve the general case if you've solved for $0<n\le1$)

Astyx

To prove what ?

Simply Beautiful Art

11:46 AM

To prove the formula from Mathematica for real values of $a,n$

Astyx

Oh right, yeah

That was not my goal

Simply Beautiful Art

@Secret Say, you interested in fractional calculus?

Secret

I have read about it (and I have once tried to do a 3rd year undergrad project on that, but I was not admitted into the program) but I know little about it except it describes stochastic processes

Simply Beautiful Art

12:02 PM

Hm, okay. Just seemed like something you might be interested in

Secret

fractional calculus are often defined in terms of integrals, so integral equations and integrodifferential equations are common things in this domain

Simply Beautiful Art

Yeah