Mathematics - 2018-09-24 (page 4 of 5)

2:15 PM

Does it make sense to talk about a point (in $\mathbb{C}$) being encircled for an open path?

an open path ?

Not closed

I have a picture

I wouldn't call that open. Also, no.

Here's my path

And the point of concern is $-1$

By the way here is the full talk: "The Riemann Hypthotesis" https://youtu.be/jXugkzFW5qY

2:17 PM

We know.

groans

Talk about this again and I will start flagging you for spam.

what makes -1 "enclosed" by your path there ?

The question is kinda like: how much must we shrink the path before $-1$ no longer is encircled?

@MikeMiller ahaha

2:18 PM

@mercio Lol that's what's bothering me

so the question uses "encircled" and doesn't explain what it means ?

(and also "shrink")

@mercio This is a question on Nyquist criterion, and the method is based on Cauchy's argument principle

dunno what that is

Kari

@mercio Probably homotopy of paths

Like, you have a contour plot of some transfer function

2:21 PM

homotopy doesn't really tell a notion of "how much"

And you're looking at the number of poles and zeros inside contour

There's just nothing to say right now. The notion only makes sense for closed paths.

the most relevant thing I can say is that if you add [0;2] to the contour (so how much = of length 2), then -1 goes from not enclosed to enclosed, which is the opposite of what you asked

Kari

Whether or not a point is enclosed depends wildly on your choice of enclosing path.

Kari

2:24 PM

Somewhat random question, but does anybody know what this symbol is?

maybe some cyrillic letter ?

Xi?

It's [zh]

no

@Mercio @Kari it's zh

oh

2:25 PM

(it's not xi)

sniped

Or close to it :P

it's rz

so it's from the phonetic alphabet ?

ж

ÍgjøgnumMeg

2:25 PM

@mercio no it's cyrillic like you said

ah

x)

ÍgjøgnumMeg

has a similar sound to the "s" in "measure"

it looks like two K standing back to back

Rithaniel

Yeah, I should look things up before I spout out an answer

@ÍgjøgnumMeg servus

ÍgjøgnumMeg

2:26 PM

@Leaky grueass di

@MikeMiller Yeah I don't get what they mean by a point being encircled by a non-closed path

It's nonsense to my eye.

^

What's the precise statement being made?

The way it's described in the book is like this: you have a curve $\gamma$ in the complex plane which is a half circle $Re^{i \phi}$ for $\phi \in [-\pi/2, \pi/2]$ with origin cut out by a smaller half circle $re^{i \phi}$, traversing ccw. Then you plot the contour $\gamma' = G_o(\gamma)$ for some transfer function $G_o(s)$ and count the number of loops $\gamma'$ does around $-1$

@Semiclassical I can link picture

2:37 PM

so you have two half circles ?

I don't see any half circle

hmm. does "closed loop system" refer to Figure 3.16a or 3.16b

my guess is that it's actually the latter

I'm pretty confident the problem is that this is not a complex analysis problem

And none of us know Nyquist better than NyQuil

@mercio Yeah like this

2:40 PM

that's more than two half circles

What?

that's two half circles and two line segments

and they make a closed loop

@MikeMiller well, I do. But not much more

@Semiclassical Yes, 3.16b describes the closed loop system in a block scheme

@mercio Yeah true :P

3.16b does not describe a closed loop from complex analysis

2:43 PM

@Semiclassical Ah, so we can leave it to you

Thanks for so kindly volunteering!

I guess what I'd note is that the usual argument principle in complex analysis requires a closed contour: $\oint_C \frac{f'(z)}{f(z)}\,dz = 2\pi i (N-P)$

@mercio It's a closed loop system

@Lozansky in the sense that the feedback loop is closed

it's a statement about the electrical system which is giving rise to the transfer function, not about the image of the transfer function itself

However, suppose you split the closed contour $C$ into two parts i.e. $C=C_1-C_2$, where $C_1,C_2$ are open contours with the same endpoints

But also my main goal was to make a NyQuil joke

Then you will have $\int_{C_1} \frac{f'(z)}{f(z)}\,dz = 2\pi i (N-P)+\int_{C_2} \frac{f'(z)}{f(z)}\,dz$

And if $f'(z)/f(z)$ happens to vanish along $C_2$, then that second integral drops

in which case you're able to appeal to the argument principle by referencing the integral along C_1 alone

But something still seems off

Rithaniel

2:49 PM

It was a top tier NyQuil joke, no worries.

lol

Thanks mate

@Lozansky To clarify, what exactly is being plotted in Figure 3.16a?

I don't think $f'(z)/f(z)$ can be identically zero on $C_2$

Nyquist stability criterion

It's evidently the image of some complex function as $\omega$ is varied along the positive real line, but which one?

for context, I think it is worth explicitly referencing this:

In control theory and stability theory, the Nyquist stability criterion, discovered by Swedish-American electrical engineer Harry Nyquist at Bell Telephone Laboratories in 1932, is a graphical technique for determining the stability of a dynamical system. Because it only looks at the Nyquist plot of the open loop systems, it can be applied without explicitly computing the poles and zeros of either the closed-loop or open-loop system (although the number of each type of right-half-plane singularities must be known). As a result, it can be applied to systems defined by non-rational functions, such...

2:52 PM

@Semiclassical It's the Nyquist curve of $G(s)$, i.e. the image of $G(i \omega)$ on the path described by the two half circles and line segments I guess :P

Yeah that's the one

I hope I missed the Riemann hypothesis buzz

And the image there is a clue: You get a closed curve when you consider not just $\omega=0$ to $\infty$ but also $\omega =-\infty$ to $0$

but the image of a closed path by a holomorphic function is supposed to be a closed path

Sure. But $\omega=0$ to $\omega=\infty$ isn't a closed path

On the other hand, $\omega=-\infty$ to $\infty$ is

yep

2:54 PM

@Semiclassical Ah, but then you superposition the mirror image on the real axis?

Right

However, the Nyquist theorem is stated in terms of a Nyquist contour

"a contour that encompasses the right-half of the complex plane"

Which means that you'd have $\omega=0$ to $\omega=\infty$, sure

where $s=i\omega$

Yes

but you'd also have $s=+\infty$ to $s=0$ as well

Because the instability is for poles in the rhp

And presumably that $s=\infty$ to $s=0$ path is precisely the line connecting $G(\omega=\infty)$ to $G(\omega =0)$

2:57 PM

$G(i \omega)$ where $\omega: 0 \to \infty$ maybe

Eh, point at infinity

:>

But I think you solved the mystery Semi. I was confused, because all previous examples were of closed paths :P

right. the point is that you need to include that line segment from G(0) to G(\infty) for the purpose of counting

otherwise it's indeed nonsense

I guess one thing that's not obvious to me is that $G(s)$ is positive real for any positive real $s$

But that's presumably understood on the basis of how $G(s)$ is obtained

right, $G(s)$ is just a Laplace transform

Yes

yeah, see the caption to their Nyquist plot

". Although the frequencies are not indicated on the curve, it can be inferred that the zero-frequency point is on the right, and the curve spirals toward the origin at high frequency. This is because gain at zero frequency must be purely real (on the X axis) and is commonly non-zero, while most physical processes have some amount of low-pass filtering, so the high-frequency response is zero."

that last line is the key one

3:14 PM

Makes sense I guess. Not sure how I would interpret an imaginary static gain :P

Yeah

3:26 PM

Does anybody have example of ring R such that R/I is notherian but R is not and I is finitely generated?

3:39 PM

What's up

the ceiling

oO

So the navier Stokes problem is pretty difficult

It's a set of coupled partial differential equations

@Ninjahatori what about $I=R$ - if you want a proper ideal then you can take R to be a product of a noetherian and non-noetherian ring and mod out the non-noetherian part

but if we mod out that mod out part is not finitely genrated I guess

3:48 PM

If $R=R_1\times R_2$, then the ideal $R_2$ is generated by $(0,1)$

hi @loch

hi @LeakyNun

does there any example of of finitely generated non-notherian ring

if you mean finitely generated in the sense of modules, then every ring is a free rank 1 module over itself..

@Ninjahatori no, by Hilbert Basis

3:53 PM

@loch then how can we take I=R since there is no finitely generated non-notherian ring

I was using a different interpretation

i don't understand what you're saying lol
$I=R$ is finitely generated as a $R$-module, generated by $1$
$R/I$ is the zero ring (so noetheiran)

and so you can take $R$ to be any non-noetherian ring

for your statement

4:19 PM

@Ninjahatori Check your question. I was already typing at it before I noticed you talking here about it.

4:32 PM

Short sanity check: If a function defined on an open subset of $\Bbb C$ is a polynomial when restricted to compact convex sets, then it follows that this function is a polynomial on any connected subset of the domain. Right?

@s.harp is this related to Atiyah?

I've seen people say "his definition of weakly analytic is just analytic", when the definition seems to actually be just polynomial. So I'm wondering if I forgot basic complex analysis here

I take it as a yes

CaptainAmerica16

@s.harp That's what I'm trying to figure out right now as well. Some of his statements referring to the function seem to come out of no where.

How many users are removed here??

I keep getting -5 because "user was removed"

Common site here on Math.SE.

4:43 PM

@CaptainAmerica16 Its not about his paper, but about what people are saying about his paper. ie somebody says "X is obvious nonsense because Y", but in my opinion "X is obvious nonsense because Z", where Z is a lot more nonsense than Y and a lot easier of a statement

@Abcd I'd say it's not uncommon for that to happen once a month for me.

@rschwieb why are users removed?

@Abcd In fact as I look, it happened twice to me in september

@Abcd It could happen for various reasons: admins delete an account violating the rules, or the user requested removal

@Abcd checking my history, in september i got -20, in august -10 and in july -10 due to removals

@rschwieb Same here, 2wice in september

CaptainAmerica16

4:50 PM

@s.harp I see.

@s.harp yes.

but clearly either is a problem.

So I'm looking at standing waves in a box according to the first fundamental harmonic. What is the equation for a string (standing wave) with length $L$ fixed at two points.

^^^^^^ Does anyone know why some questions are yellow-colorer?

You have one of those tags set as a favorite tag

5:00 PM

Is it just of the form $y=A\sin(x)$

It highlights questions with your favorite tags

where $A$ is amplitude

@MikeMiller :) many thanks

@Ultradark No, that wouldn't satisfy the boundary condition at $x=L$

@Lozansky okay

I'm trying to put together a good question. So each side of the box has boundary condition equal to zero

Pythonista

5:14 PM

Haven't found a good resource discussing what it means that a Lie algebra captures "most" of the information of the Lie group. Why does it capture "most" of the information can anyone expand on this or give resources.

@rschwieb the example you given why R/I is F2 I don't understand it?

can anyone help me for a few minutes

Because $F\times F\times F\times\ldots /\{0\}\times F\times F\times F\ldots\cong F$.

@Ultradark Not sure what you mean. What box? For a string in one dimension, the solution to the wave equation $u_{tt}-c^2u_{xx} = 0$ with boundary conditions $u(0,t)=u(L,t) = 0$ is $u(x,t) = \sum_{k=1}^{\infty} (a_k \cos \dfrac{k\pi ct}{L}+b_k \sin \dfrac{k \pi c t}{L}) \sin \dfrac{k \pi x}{L}$. The coeffiecients are determined by initial conditions $u(x,0) = g(x), u_t(x,0) = h(x)$

Jasper Loy

@MikeMiller Oh dear, it looks like someone close to Atiyah should have a good talk with him.

5:21 PM

@Lozansky the box has dimension $1$ by $1$. I just want to use the simplest possible equations. The string has length $/sqrt(2)$ Let me see if I can find a picture to help

@rschwieb Sir, can you explain me 2ℤ(𝟚)×ℚ why this is not -notherian and why ideal of this is principal? I guess non noetherian beacause since it is not finitely generated but how to write it properly I don't know? It is another example

@Ninjahatori I have no idea what $2\mathbb {Z(2)}$ means. What ideal are you talking about?

localization of Z at prime ideal 2Z

Can thee speak again freely in my deepest regrets for not believing everything I read above a certain number of times by "different" "individuals"

?

what?

5:27 PM

My latex is being annoying again why isn't it putting the summation bounds above and below the sigma like it should

@Ninjahatori $\mathbb Z_{(2)}\times \mathbb Q$ is a Noetherian ring.

when I do the double money sign things either side

$$\frac{\sum_{i=\pi (n )+1}^{\pi(n+1 )}\sum _{j=1}^{{\bigl\lfloor\frac {\ln (n +1) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1}\Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln
( p_{{i}})}{\ln(n+1)}=\delta(f(n),g(n))$$
$${\{f(n),g(n)}\} \subset \mathbb N$$
Where $\delta$ is the Kronecker delta function.

@rschwieb math.stackexchange.com/questions/183199/… in this link above ring is given.

@Ultradark Plug in $k=1$ if you want the solution for the first fundamental standing wave. But you still need to find coefficients $a_1, b_1$ to satisfy your initial condition.

@Ninjahatori OK, well that is completely different, and is the reason you have to give COMPLETE CONTEXT instead of random snippets of information. It's not Noetherian because the ideal $\{0\}\times \mathbb Q$ has infinite chains of $\mathbb Z_{(2)}$ submodules.

5:29 PM

okay

And how do you increase the font of certain symbols to accommodate for the relative size in a particular algebraic expression its just dumb if we cant control font size why cant I do that

I know you all know what im saying and it will take u like 5 seconds to tell me

fine

@rschwieb thanks a lot sir

@Ninjahatori You should follow the link to where the author explains

@rschwieb @rschwieb are you familiar with Witt rings and quadratic forms?

Fine I will look at it

@Ninjahatori Quadratic forms, yes, Witt rings no.

5:38 PM

I have certain doubts in Quadratic forms over abelian groups so can I discuss it with you whenever you have free time?

@Lozansky can I share a diagram to help you get a visual of what I'm trying to describe mathematically

@Ultradark Shoot

@rschwieb I have certain doubts in Quadratic forms over abelian groups. So can I discuss it with you whenever you have free time?

@Lozansky so these are basically sine waves, varying the amplitude parameter

@Ninjahatori I've never thought about quadratic forms over abelian groups, just fields (and local rings.)

5:43 PM

and I want to view them as standing waves, fixed at (0,0) and (1,1)

@Ninjahatori You may as well post here in case I don't know.

@rschwieb It can be seen as just map b from MM TO Q/Z just like in case of vector space we have map from VV to K

cartesian product of M and V twice

@Lozansky so I just have to solve the wave equation in this box?

I realize that you have to do a change of coordinates which is a bit annoying

@Abcd ?

@LeakyNun Can you please see 1 program? Are you free for 5 minutes?

5:50 PM

@Abcd is it related to Atiyah proof

@Ninjahatori I am a high school student. I dont even properly know whats Riemann Hypothesis :P

@Abcd ?

@LeakyNun At least say "yes" please so that I can send man. "?" doesn't mean anything.

yes

@LeakyNun Thanks, I have sent it to you in the other room.

5:54 PM

can someone help me formulate this problem

@Ultradark Sorry, I don't see how this is different from the one-dimensional case. Your sines are symmetric wrt $y=x$ so just rotate your coordinate system $\pi/4$?

vzn

6:37 PM

Sir Michael Atiyah's Proof of the Rie…

For all those who are interested in Sir Michael Atiyah's proof...

Q: Fixed field and prime field

7:00 PM

0

I know that F is subset of field K. We know Any $\sigma$ in $Aut(K)$ if $\alpha$ is in F then $\alpha^{-1}$ is in F since identity goes to identity and $\sigma(\alpha.\alpha^{-1})$= $\sigma(1) =1 $ ; $\sigma(\alpha).\sigma(\alpha^{-1})$ = $\alpha.\sigma(\alpha^{-1})$ =1 so $\sigma(\alpha^{-1}...

7:25 PM

is there a simple way to show that there are $n^{n-3}n!$ ways to decompose an $n$-cycle into a product of $n-1$ transpositions (assuming I didn't get it completely wrong) ?

I think I did get it wrong

... make that $n^{n-2}$

7:54 PM

@mercio most obvious thing to do is induction, i.e. count how many new decompositions you get upon increasing the cycle length by 1

(note: 'most obvious' need not be the best/simplest way)

@LeakyNun what is splitting field of x^3-x-1 in Q?

sorry x^3-3x-1 splitting field?

@Ninjahatori $\Bbb Q(\sqrt{-23})[X]/(X^3-X-1)$

wolframalpha.com/input/?i=disc(x%5E3-x-1)

Thank you

and what about x^3-3x-1

Hi guys, just tuning in for short: Anyone aware of any educated opinion on Atiyah?

or even interested?

@LeakyNun for x^3-3x-1 is it just Q[X]/(X^3-3X-1) right

8:03 PM

that's what you just asked me

@Rudi_Birnbaum chat.stackexchange.com/rooms/83556/…

please keep discussion of atiyah out of this chatroom

is it right or there is much simpler form?

@Semiclassical hmm I was dubious but I might see something working

@LeakyNun sorry didn't want to offend anyone. also did not see your link yet. I'm on travel and can just check in for short

@Ninjahatori no, yours isn't right

I changed the polynomial second coefficient you calculated for x^3-x-1

8:07 PM

oh

yes, that's right

is there much simpler form than this?

no

fine thanks

@LeakyNun Now I understand there a separate chatroom, thanks!

@mercio did you get ma email?

probably, I haven't checked it

@Semiclassical alright so do you have an easy way to show that $n^{n-2} = (n/2) (\sum_{k=1}^{n-1} k^{k-2} (n-k)^{n-k-2} (n-2)!/(k-1)!(n-k-1)!)$

I am not sure this is a step in the right direction

@Rudi_Birnbaum I'm reading the thing

8:26 PM

Dont worry too much about it, its just a talk.

do you think they have a chance of knowing the $\bigwedge$ notation for exterior product ?

though I still don't like putting that next to $Sym^n$

I know there is one person who knows about it, a good friend

but not the rest

i found it out in a discussion today

"for an irrep E of a group blablabla, then Gamma_R_z |H = ..." you are doing it again

you are telling a story with a group G, a subgroup H, and an irrep E of G

and then all of a sudden

Gamma_R_z comes in the story

kills E

kills the audience

and goes away

I can go with Alt^2 E

yes you should do that

8:31 PM

@mercio that looks like a binomial theorem

except it is not

but then I need to show that its id to rotz

@mercio yeah, those factors of k^k etc make that not work

bphi

A homework question says "show that G/H has a natural group structure", does that mean "normal subgroup"?

you need to explain at one point that if your representation stabilizes the xy plane and the z axis, then Rotz is the action on the (x^y) axis

which is obtained by taking the exterior product of the plane representation with itself

@bphi not necessarily

8:35 PM

@mercio ok

bphi

@mercio G=S3, H=A3

what is the size of G/H

@Rudi_Birnbaum when you set the story with just a group G, a subgroup H, and a 2dim irrep E, you are not talking about a representation into O(3) so you are not talking about a Gamma Rot z, not yet anyway

you could mention that E is the xy plane, it would help

bphi

3?

= [(A ⊗ A) ⊕ (B ⊗ B) ⊕ (A ⊗ B)]|H is wrong

@bphi no

@mercio what then?

8:38 PM

I don't like the bracket notation

Its used all over in chem ...

the bracket notation only makes sense when there is a tensor square inside

so when you start decomposing it it looks weird

I think its simply the anti symmetric part of a tensor

and you should have two (A * B) terms

bit they span the same space

8:39 PM

see

confusing !

maybe one A * B and one B * A ?

so adding them doesn't dp anything I thought

oh

F^n(A+B) = sum F^k(A) tensor F^(n-k)(B) is pretty much well known for F^n = nth symmetric power or nth exterior power

aside from that little inaccuracy it should be okay

I have to sleep now, tomorrow is another tough day from 7:30 to 23:00 program ...

CU

good night

@mercio as confirmation of your suspicion earlier: oeis.org/A000272

"a(n) is also the number of ways of expressing an n-cycle in the symmetric group S_n as a product of n-1 transpositions"

8:52 PM

mhm

my usual approach when doing permutation stuff is to try to get insight from a string diagram

it's always surprising how many interpretations of combinatorial sequences there are

ya

If $G_c(i \omega) = \dfrac{G_o(i\omega)}{1+G_o(i\omega)}$ and $|G_c(iw)| \leq \dfrac{1}{\sqrt{2}}$ then does it follow $|G_o(i\omega)| \leq \sqrt{2}$?