Mathematics - 2016-10-18 (page 2 of 5)

Balarka Sen

11:03

Hi @Anubhav.

Lozansky

Why does $e^z$ have an essential singularity at infinity?

Balarka Sen

Do you understand what it means to have a singularity at infinity?

Lozansky

Its not bounded but it doesnt approach infinity as $|z| \to \infty$

But $|e^z| = |e^x|$, can't we say this approaches infinity as $|z| = \sqrt{x^2+y^2}$ goes to infinity? Is that because we can't necessarily say how $x$ behaves as $|z| \to \infty$?

DHMO

@Lozansky what above $z\to-\infty$?

Balarka Sen

It's a fuss to talk about "stuff happening at infinity". Can you rephrase your definition in terms of something which does not involve the word infinity?

Lozansky

11:16

@BalarkaSen I can talk about $e^{1/w}$ as $w \to 0$

Balarka Sen

Mhm.

Why does it not have a pole at $0$?

@DHMO I didn't ask you the question :P

But, yes.

DHMO

@BalarkaSen but we sort of went through that yesterday, so...

Balarka Sen

@Lozansky I quoted a result about how a holomorphic function looks like near a pole. What does it tell you?

Lozansky

@BalarkaSen $e^{1/w} = 1+1/w+1/(2!w^2)+...$ so we can't express it in terms of $g(w)/w^k$ for finite $k$ (and analytic $g(w) \neq 0$ at $w=0$)?

DHMO

@Lozansky Didn't we go through yesterday how $e^{1/\omega}$ has an essential singularity at zero?

Lozansky

11:21

@DHMO Yes but now I'm allowed to talk about the point at infinity :P So I wanna do it differently

DHMO

Aren't they equivalent?

Balarka Sen

@Lozansky Yeah, but understand it pictorially though. If a function as a pole at some point, what does it do near an nbhd of that pole?

@DHMO Infinity is not a rigorous notion. One extends hol. functions to the Riemann sphere near poles, and then it becomes rigorous. Even then one needs to change charts to actually do computations.

Lozansky

@DHMO Kind of, but not entirely

DHMO

@BalarkaSen @Lozansky alright

Balarka Sen

The $z \to 1/z$ thing is just changing charts from infinity to the origin.

Lozansky

11:23

@BalarkaSen It certainly becomes unbounded

Balarka Sen

@Lozansky Not only that, but it's norm goes to infinity approached from any direction!

Because the $1/z^k$ term dominates, and that heads straight to infinity as $z \to 0$. No funny business near $0$.

DHMO

Would it be better to let $z=re^{i\theta}$?

And then we would have $r\to0$

Balarka Sen

On the other hand, $e^{1/z}$ doesn't go to infinity as $z \to 0$. Can you give me two directions approaching from which the limit at $0$ are different?

@DHMO That's what "approaching from a direction" means.

Lozansky

@BalarkaSen From left/right on the real axis?

Balarka Sen

That's it, @Lozansky.

DHMO

11:26

@BalarkaSen yes, but would it be better to have that explicit

Balarka Sen

@Lozansky The upshot of this is that near essential singularities a function jumps around too much, instead of heading straight to infinity. There's a theorem out there which says that if $f$ is holomorphic everywhere except $0$, with essential singularity at $0$, image of any deleted ball near $0$ by $f$ is dense in the complex plane.

Lozansky

@BalarkaSen @DHMO Can I claim $cosh(z)$ has an essential singularity at $\infty$ since the power series at $w=0$ for $cosh(1/w)$ is alternating?

Balarka Sen

Alternating power series is not sufficient.

Lozansky

@BalarkaSen What does dense mean in this case?

DHMO

@Lozansky You know, $\cosh$ is just $\dfrac{e^z+e^{-z}}2$

Balarka Sen

11:32

Eg, $1/(1 + z)$ also has an alternating Taylor series.

Lozansky

@DHMO Yes I know, and then it reduces to the case $\frac{e^{1/w}+e^{-1/w}}{2}$ which we covered?

Balarka Sen

@Lozansky Don't worry about it if you are not familiar with that terminology. It means image of it is a "big" subset of $\Bbb C$.

DHMO

I don't believe we have covered that case

since it now approaches infinity from both the + and the - real axes

Lozansky

@DHMO Can we make use of that fact that $cosh(z)$ is an even function?

DHMO

@Lozansky how?

Lozansky

11:37

So $cosh(\infty) = cosh(-\infty)$

Or is that nonsense?

:|

DHMO

@Lozansky so?

Lozansky

@DHMO Nevermind, I was going for a symmetry but the terms cancelled

Balarka Sen

@Lozansky Hint: use the power series expansion (that is the definition of $\cosh(z)$)

Again, you have to show that the singularity is not (1) removable (easy) and (2) not a pole.

DHMO

so we need to find a direction from which it does not approach infinity

Balarka Sen

It's easier to do it by the power series this time, @DHMO.

Lozansky

11:43

@BalarkaSen $cosh(z) = 1+z^2/2!+z^4/4!+...$ which does not have a common factor $z^k$ (i.e no pole of order $k$) and $cosh(1/z) = 1 + x^{-2}/2+x^{-4}/4!+...$ is not analytic at $z=0$ so it does not have a removable singularity at infinity?

DHMO

@BalarkaSen I see

Balarka Sen

@Lozansky What is $x$?

Lozansky

A typo

Balarka Sen

It doesn't have removable singularity at $0$ because it's not bounded near $0$, period.

Lozansky

Okay

DHMO

11:48

@BalarkaSen Couldn't I just use the i direction?

Balarka Sen

@DHMO That's right.

DHMO

and then show that it is bounded @BalarkaSen

Balarka Sen

$\cosh(ix) = \cos(x)$, yes.

Lozansky

@BalarkaSen $cosh(z)$ does not have a pole at $\infty$ because $cosh(1/w) = 1+w^{-2}/2!+w^{-4}/4!+...$ does not have a pole at $w=0$ since there is no integer $k$ such that $w^kcosh(1/w)$ is analytic and $\neq 0$ at $w=0$?

Balarka Sen

@Lozansky Yes, that's it.

But it's also worth understanding the proof DHMO is speaking of.

Antonio Vargas

11:54

does anyone have any experience with publishing part or all of their thesis?

Tobias Kildetoft

@AntonioVargas Sure

Antonio Vargas

@TobiasKildetoft I'm trying to figure out a good way to go about it. It would be a pretty long paper (>80 pgs) if I tried to publish the entire core of it, but I'm not sure publishing the chapters individually is the way to go either. Did you split yours up?

Tobias Kildetoft

@AntonioVargas Everything I published based on my theses (both masters and PhD) was pretty much completely rewritten for the purpose of the publication

The style of writing just needs to be so different between the two formats

Antonio Vargas

ah yeah I suppose so

at least I was more conversational in my thesis than I would have been in a short paper, but I don't think that would be bad in a paper either

Tobias Kildetoft

A thesis just tends to include a lot more detail than makes sense in a paper

For comparison, my PhD thesis was 86 pages, and I recently published a paper containing all the substance (and in fact quite a lot more) which is 25 pages

SteamyRoot

12:07

To quote a colleague's advisor: if $x$ is the length of your paper, then the quality must be proportional to $x^2$ to get it published.

Tobias Kildetoft

My masters thesis was 120 pages (though about 30 of those were computer code) and I have published two papers based on that, totaling less than 20 pages

Antonio Vargas

I don't think any 15 page paper I'd write is worth 225 pages :)

@TobiasKildetoft How'd you manage to cut so much out?

Tobias Kildetoft

@AntonioVargas Well, part of the thesis was based on an already published paper with a coauthor which I did not count here (17 pages, though that includes quite a bit extra compared to the part in my thesis) and a major part was background stuff that would take way too much space in a journal paper

Antonio Vargas

I see

@TobiasKildetoft Thanks for the advice, I'm going to think on it

Ramanujan

@DHMO I have a question

DHMO

12:22

@Ramanujan ask

HiHello

any idea lim (x, y) -> (0, 0) sin(u)/sin^2(sqrt(u)?

DHMO

@HiHello what is u?

HiHello

x^2 + y^2

Ramanujan

How can one say “product of r successive integers is divisible by r!”

DHMO

that is difficult to prove...

SteamyRoot

12:26

so you're saying that $r! | \frac{(r+k)!}{k!}$ ?

Ramanujan

@DHMO please

Balarka Sen

You're literally trying to prove the binomial coefficients are integers.

0celo7

Hi

Ramanujan

@BalarkaSen I till now don't know what binomial coefficient(even expansion)is😥

SteamyRoot

What @BalarkaSen said. In the expression I posted, let $n = r+k$, and hence $r = n-k$. Then you're trying to prove that $(n-k)! \; | \; \frac{n!}{k!}$, or thus that $\frac{n!}{k!(n-k)!}$ is an integer

Perhaps math.stackexchange.com/questions/11601/… is of use?

DHMO

12:30

@HiHello use power series lol

Ramanujan

@DHMO explain my question

DHMO

@Ramanujan click that link?

HiHello

@DHMO Thanks. I`lll try

Ramanujan

@DHMO out of my reach,i can't see any relation with my question

DHMO

@Ramanujan alright, here:

lets look at how 6x7x8x9 is divisible by 1x2x3x4

are you familiar with prime factoring?

Ramanujan

12:37

Yes,do you mean every no. Can be written in the form of multiple of prime numbers?

DHMO

yes

so, if we say x is divisible by y, that means that every prime factor in y is also in x

right?

Ramanujan

At least one?

DHMO

every

like, if x is 2x2x3x5 and y is 2x3

for each prime factor in y, you can also find in x

Ramanujan

Will y also contain 5?@DHMO

DHMO

it can, but it is not necessary

x is still divisible by y

Ramanujan

12:41

I got it,next?

DHMO

nice.

now, let us only focus on the prime 2

(other primes can be used similarly)

Ramanujan

Ok

DHMO

now, we need a good way to count the number of 2s in (the prime factoring of) 1x2x3x4

Ramanujan

There are three 2(s)

Gridley Quayle

Hi.

I am confused on a linear algebra problem. I've been asked to find the JCF of a 4x4 matrix $A$. I worked out the characteristic polynomial to be $x^3(x+1)$ and the minimal polynomial to be the same. When I work out the nullity of $A-0I$ however I get 2. I thought since we have an $x^3$ term that there would be a $J_{0,3}$ Jordan block but since the nullity is 2 we should have 2 Jordan blocks which points to $J_{0,2} \oplus J_{0,1} $ Where am I getting confused.

DHMO

12:47

@Ramanujan yes. however, there is a more systematic way

we find that for every 2 numbers, one of them is divisible by 2

Gridley Quayle

$A=\begin{pmatrix}
1 & -2 & 1 &0\\
1& -2 & 1&0\\
1& -1& 0&0\\
1& -2& 1&0
\end{pmatrix} $ for reference

Ramanujan

!?!?!?

DHMO

@Ramanujan in the first two numbers (1 and 2), one of them is divisible by 2

Ramanujan

Yep

DHMO

and in the next two numbers, we also have one number divisible by 2

Ramanujan

12:52

OK but 1×2×3×4×5 so in this we should take next 3 numbers?

For 3?

DHMO

@Ramanujan you are trying to count the occurrences of 3?

Ramanujan

Yes

DHMO

so we group them into {1,2,3} and {4,5}

the last group does not contain any number divisible by 3

Ramanujan

So in the next 3 numbers,we also have one number divisible by 3 and so on?

DHMO

if we have eight numbers, then {1,2,3} and {4,5,6} and {7,8}

we can discard the last group

Ramanujan

12:56

OK next procedure?

DHMO

so we have 2 occurrences of "3" in the prime factoring of 8!

Ramanujan

Yes

DHMO

it isn't that simple

let's look at 10!

the above would give you 3 occurrences of "3"

Gridley Quayle

I've sorted it out, the minimal poly was actually $x^2(x+1)$

Ramanujan

@DHMO so what do we do now?

DHMO

13:00

@Ramanujan that is because "9" is doubly divisible by 3

but we only counted it once

Ramanujan

Then…

DHMO

so for every 9 numbers we need to count again

so this gives us one extra

making the total count 3+1 = 4

Ramanujan

OK,then?

DHMO

now, we would establish that this is the minimum possible count for 10 consecutive numbers

that means, for example, let us consider another 10 consecutive numbers starting from 3: 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Ramanujan

Are we going for divisible of 3

DHMO

13:04

yes

Ramanujan

Then there are 5 counts divisible by

DHMO

yes

@Ramanujan let us just count every 3 numbers, and forget that we need to count every 9 numbers, just for the moment (just for now)

Ramanujan

But it dont hold product of r successive inteegers is divisible by r!

DHMO

@Ramanujan i'm building it step by step

i said the proof would be difficult

Ramanujan

OK,iam Waiting

DHMO

13:09

so, that would be 4 counts, right?

Ramanujan

Yes but again it dont hold product of r successive inteegers is divisible by r!

DHMO

@Ramanujan not so fast...

{1,2,3,4,5,6,7,8,9,10}: 3 counts
{2,3,4,5,6,7,8,9,10,11}: 3 counts
{3,4,5,6,7,8,9,10,11,12}: 4 counts
{4,5,6,7,8,9,10,11,12,13}: 3 counts
{5,6,7,8,9,10,11,12,13,14}: 3 counts
{6,7,8,9,10,11,12,13,14,15}: 4 counts
So we can see that {1,2,3,4,5,6,7,8,9,10} is minimum, right?

Ramanujan

Do you mean 1,2,3,4,5,6,7,8,9,10 is minimum in case of 1×2×3…×10

DHMO

yes

1x2x3x4x5x6x7x8x9x10: 3 counts
2x3x4x5x6x7x8x9x10x11: 3 counts
3x4x5x6x7x8x9x10x11x12: 4 counts
etc.

Ramanujan

Then?

DHMO

13:16

so, if we count the multiples of 9, it would also be minimum

it is quite easy for intuition: since 0 is a multiple of 9, if you count directly after 0, it would take a while for you to meet another multiple of 9

that applies to every number, because 0 is a multiple of every number

I wish I could make an animation to show you this

Ramanujan

Eager to see animation

DHMO

something like this

imagine that the green part is moving as shown

from 1-10 to 2-11 to 3-12

the red squares are multiples of 3

if you count from a multiple of 3 (3-12) you would get maximum

if you count just after a multiple of 3 (1-10) you would need to wait for a while before reaching the first multiple of 3

Ramanujan

What do you mean to say by ~you would need to wait for a while before reaching the first multiple of 3

DHMO

well, if you count from 1

you would need to pass two numbers (2 and 3) before reaching the first multiple of 3

whereas if you count from 3, you immediately have a multiple of 3

Ramanujan

Do you mean counting a number from 3 : including first no. To be 3?

DHMO

13:24

yes

could you see that the last green part includes the most red squares?

Ramanujan

Yes

DHMO

and the first green part includes the least red squares

Ramanujan

Can say,yes

DHMO

and can you see that it would be true for any size of the green part as well as any distance between two red squares (but fixed distance between two red squares)

Soham Chowdhury

a little notational question

(pinging @Balarka)

DHMO

13:26

@Ramanujan so it is minimum right

Ramanujan

Yes

DHMO

now if you apply that for every prime, you can conclude that for every prime the count is minimum

Soham Chowdhury

does $F^\times/\Bbb A_F^\times$ mean the same thing as $\Bbb A_F^\times/F^\times$?

(i.e. a group quotient)

Balarka Sen

Hi @Soham. I think so.

Soham Chowdhury

(note that everything is abelian here)

long time no see, friend

Ramanujan

13:27

Ok

Balarka Sen

Weird notation tho. Maybe he's doing left actions.

Ramanujan

When we start from that prime

DHMO

@Ramanujan therefore, if we have another r consecutive integers, their count would be equal or larger

Balarka Sen

Yeah, haven't seen you around in a while. How's things?

arctic tern

@SohamChowdhury is it $F^\times\backslash\Bbb A_F^\times$ perhaps?

Soham Chowdhury

13:28

Ramanujan

Yes

arctic tern

H\G is same as G/H

Soham Chowdhury

you mean \setminus? I do not think so, he's explicitly talking about quotients

DHMO

@Ramanujan and so your theorem is proved

arctic tern

@SohamChowdhury no, \backslash, not \setminus

Soham Chowdhury

13:29

oh, okay

I've never encountered that notation before.

Balarka Sen

That's the notation for left-quotients

Soham Chowdhury

Yeah, but for abelian everything it's unfamiliar to me.

Thanks though @arctictern

arctic tern

not sure why they'd use that for an abelian situation, unless they use it generically for all left actions

usually its only for groups acting on sets/spaces, not quotient groups too

Soham Chowdhury

coset spaces of a group action?

as in, the orbits

Balarka Sen

sure

arctic tern

13:31

"coset spaces" are orbits in the regular action, we don't use the word "coset" for orbits otherwise

for instance if H,K are subgroups of G, then H left-acts on G and K right-acts on G from multiplication, in which case G/K is the set of right K-orbits (which will be the left cosets of K), H\G will be the set of left H-orbits (which will be the right cosets of H), and both (H\G)/K and H(G/K) may be canonically identified with the set of H-K double cosets (of the form HgK), which is notated H\G/K.

Balarka Sen

@arctictern adeles admit a valuation though, and I think that action preserves the topology induced by the valuation. so it's probably not just a quotient as groups; there's more structures involved

I think so, at the very least

arctic tern

true

DHMO

@Ramanujan ok?

Soham Chowdhury

I very strongly hope everything stays abelian, if not, I'm bailing

Balarka Sen

@Soham did you see the recent noncommutativegeometry post?

Ramanujan

13:41

What about r! ? Any example @DHMO

Balarka Sen

or maybe it was neverendingbooks, I never remember which one is the active blog

DHMO

@Ramanujan r! is just the product of the first r numbers

Ramanujan

Give me any quick problem on that as HW

DHMO

for example, $10! = 2^8 \times 3^4 \times 5^2 \times 7^1$

and we just proved that for every prime, the exponent is minimum

Ramanujan

WHAT?

DHMO

13:43

...

alright, let us go through what we did

if we have 10 consecutive numbers, and we want to count how many numbers are divisible by 3, we proved that the count would be minimum if we start from 1

right?

Ramanujan

From 1?

Oh,so minimum 1 right

DHMO

meaning, 1x2x3x4x5x6x7x8x9x10 is the minimum

if you start from any other number, the count would be equal to or greater than this count

for example, 3x4x5x6x7x8x9x10x11x12 has a larger count

because we start from 3 instead of starting from 1

Ramanujan

OK,till now everything is fine

DHMO

@Ramanujan now, let us establish how to prove that one number is divisible by the other

Ramanujan

Ok

DHMO

13:48

for example, $x = 2^a \times 3^b \times 5^c \times 7^d \times \cdots$ and $y = 2^e \times 3^f \times 5^g \times 7^h \times \cdots$

Ramanujan

And {a,b,c…} are greater than {f,g,h,…} then @DHMO

DHMO

@Ramanujan you mean {a,b,c,d...} are greater than {e,f,g,h,...}

more rigorously, $a \ge e$ and $b \ge f$ and $c \ge g$ and $d \ge h$ and $\cdots$

Ramanujan

Yeah,sorry,then?

DHMO

If the above condition holds (if the above condition is true), then we can conclude that $x$ is divisible by $y$, right?

Ramanujan

Surely

DHMO

13:52

And we proved that the count is minimum for every prime, if we start from 1?

Ramanujan

Yep

DHMO

now, the count is actually the exponent (index), right?

Ramanujan

Ok

DHMO

now, let $x$ be the product of $r$ consecutive integers, and let $y$ be $r!$, ok?

Ramanujan

Ok

DHMO

14:02

so we just said that for every prime, the count for $x$ is larger than the count for $y$, right? @Ramanujan

Ramanujan

$14^{th}$ one@DHMO

@DHMO I was to solve 14 th problem with that,yes

DHMO

and count is just exponent

so $x$ must be divisible by $y$

Ramanujan

Thanks@DHMO got some idea,going to watch some lecture master to (not shure) in it, any lecture on YouTube you know?

DHMO

@Ramanujan khan academy?

Ramanujan

I don't think he upload these topics

DHMO

14:10

no idea then

Ramanujan

BTW what should I search to get it? (In which type it comes?)

DHMO

@Ramanujan number theory?

Mary Star

14:27

We have that $f(x)=x^4-2x-1$ is irreduccible in $\mathbb{Q}[x]$. Let $a\in \mathbb{Q}$ be a root of $f$. How could we check if $\sqrt{3}\in \mathbb{Q}[a]$ ?

user243301

I hope that you are well @parisa . Best resgards!

Antonio Vargas

@MaryStar, you can't let $a\in \mathbb Q$ be a root of $f$ because $f$ has no roots in $\mathbb Q$

Mary Star

Sorry... I meant $a\in \mathbb{C}$. @AntonioVargas

Antonio Vargas

ah

Mary Star

Do you maybe have an idea how we could check it? @AntonioVargas

Mike Miller

14:44

@BalarkaSen It's certainly not true that trivial over 2-skeleton = w_2 is zero. See TS^2.

DHMO

@MikeMiller do you have any idea about the problem above asked by @MaryStar? I'm intrigued also

Balarka Sen

@MikeMiller Thanks. I should fiddle with examples more before posing a conjecture and/or trying to prove it.

Lozansky

Can Cauchy's residue theorem be applied to singular points on the contour?

DHMO

@Lozansky no

Lozansky

@DHMO Thanks

Balarka Sen

14:55

It doesn't even make any sense. The function has to be defined on the contour to compute the line integral in the first place.

Mike Miller

@Balarka In any case, there's a geometric interpretation of w2 vanishing.

Lozansky

@BalarkaSen In my book it says "If $\Gamma$ is a simple closed positively oriented contour and $f$ is analytic inside and on $\Gamma$ except at the points $z_1, z_2,...z_n$ inside $\Gamma$, then bla bla bla"

Balarka Sen

Curious. Do I have the tools to find that interpretation out?

Mike Miller

Sure, why not? What did it mean that w1 was zero?

Balarka Sen

The bundle is orientable, of course.

robjohn

14:58

@Lozansky The integral along the curve will not converge. So I'd say no.

Mike Miller

Can you tell me in terms of structure groups/classifying spaces?

robjohn

@Lozansky The trick is to avoid the singularity with a small detour around the singularity.

Balarka Sen

The classifying map lifts from $BO(n)$ to $BSO(n)$. (I mentioned that in the bunch of messages I pinged you)

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