12:00 PM
I mean like the the 3d space and all can i represent groups over there

I most certainly won't try to visualize a given group by embedding it in some matrix group first and fiddling with the corresponding linear operators.
Too much work, and not interesting.

Ya but it is interesting for a person like me who has never seen cayley graphs

[note : that all groups can be realized as matrix groups is not trivial. it's a good theorem]

unsolved or solved

solved, 'course.
It's insanely hard if one tries to solve this group-theoretically, but the moment one knows about the language of group actions, it almost becomes a tautology.

12:03 PM
But it is fascinating you dont know what kind of a linear operator you might get while changing a group to the linear operator

"changing a group to the linear operator" sounds like nonsense to me. anyway, I don't expect you'll understand the whole beauty of the theorem yet.

user96977
@PaulPlummer thanks, checking...

@BalarkaSen the theorem you mentioned??

yes

y? i get dyslexia when it comes to abbreviations @BalarkaSen
Ha group actions thats after homomorphisms and all second chapter

12:07 PM
Do linear algebra before doing abstract algebra.

I have not touched i was just going through the pages looking at the content
what is there

@AlexClark When are you going to present your talk on central extensions?

user147690
@BalarkaSen Mon, Wed or Fri I may be selected

ah. have you prepared the talk well?

user147690
@BalarkaSen Not as well as I would have liked, complex analysis has sucked up my whole weekend

12:10 PM
off-topic, but can't help it : I'm tired of all these trivial questions about $H_0$ in the alg-top tag.

user147690
0

Let $X$ be a a topological space, $A\subset X$. I've been told that it is "trivial" that if each path component of $X$ contains at most one path component of $A$ then $H_0(A)\overset {i_*}{\to} H_0(X)$ is injective. But honestly I cannot see why it is true. tried to go with the definition, taki...

uh-oh, @AlexClark.

user147690
You are literally asking him the most trivial questions(about $H_0$) possible after he expresses his displeasure lol

lol

user147690
Wow I wish I had a piano, this guy is inspiring wtf: youtube.com/watch?v=wrlID7C8xLk

user147690
12:14 PM

I can play the guitar...@AlexClark

user147690
I can play guitar pretty well also, but this guy is insane

12:46 PM
@robjohn I'm surrounded by a lot of amazing results :D They simply flow ceaselessly ... :-)

Hi@ᴇʏᴇs

@Chris'ssis that's great

@robjohn i have asked this question any takes:

Hi @Rememberme

How are you?@ᴇʏᴇs
@quid why am i not getting an upvote when it is showing an upvote for the question

12:54 PM
@Rememberme what do you mean? Which question.

metamathematics

I see. On meta there are no points @Rememberme There is only "participation" and this does not depend on the votes one receives.

i didnt know that

The points to determine moderaion privileges are inherited from the main site;
It is a common question @Rememberme

@AlexClark That is simply amazing!

user147690
1:01 PM
@robjohn Makes me want to buy a piano

user147690
Although that's clearly years of practice

@AlexClark no one asked if he wrote that himself. I think the incorporation of the Canon in D is superb as well as the playing ability.

@robjohn Indeed.

user147690
1:25 PM
@robjohn Could I hassle you for an answer check to $\int_C e^z - \frac1z dz$ where $C$ is the lower half of the circle with radius $1$ center $0$ negatively oriented? Is it $-2\pi i$, $\frac1e - e +i\pi$ or neither?

1:55 PM
We have the function $g:[0, 2\pi]\rightarrow \mathbb{R}$,
$g(x)=\frac{(x-\pi)^2}{4}, x \in [0, 2 \pi]$.

This function is not $2\pi$ periodic, is it? @robjohn

2:14 PM
Howdy!!@LeGrandDODOM

@Rememberme fine, you ?

fine

@MaryStar how can you tell if it is only defined on $[0,2\pi]$? It is equal at $0$ and $2\pi$, so for the only points that you can check, it is. This also means it can be extended continuously to a $2\pi$-periodic function on $\mathbb{R}$.

@robjohn this one has a very nice closed form $$\int_0^{\pi/4} \frac{\cos (2 x) \log (\cos (x))}{1+\sin ^2(2 x)} \, dx$$

@Chris'ssis the $\cos(x)$ can be eliminated easily and that simplifies it greatly
@AlexClark Since it is not a closed contour, I would need to actually compute the contour integral without using residues. I have to go for a bit, I can look at it later.

2:23 PM
@robjohn What do you mean? Not sure I got the point.

user147690
@robjohn Thanks

@robjohn some of my recent question may interest you, anyways I can't gurantee, see them for yourself.

How did you find it or where did you get it from??

@Rememberme which, one?
@MatsGranvik hello

The closed form of the summation one

2:35 PM
Tried some questions on internet, couldn't proceed so asked it.
BTW what is this parity word in english mean?
sign?

hi@MatsGranvik

@MatsGranvik what are you currently doing?

So I am relearning trigonometry for the first time in 7 years. Can someone help me out? You have this boat that sails 70 miles in the direction ${139}^{\circ}{41}^{\prime}$. How far south has it sailed to the nearest mile?

@ADG Not much, I have been to my mothers place to today. Yesterday I spent the whole day seeking patterns in the Andre LeClair approximation of the zeta zeros. I found nothing.

2:39 PM
@MatsGranvik oh you'r up for reimann hypothesis?

@MatsGranvik then we have the same goal!!!!!!

@ADG Yes, I have been for a while now. Oh you too?

Yes!!!!!!!!!

@MatsGranvik Nah. maybe later

@MatsGranvik what developments do you have in RH?
i really want to know...

2:42 PM
I assume the problem I have above has something to do with bearings... and I think I understand the concept...

@Rememberme Well I am an amateur.

same here
YOu have found any weird combination in the zeroes i mean some weird pattern....

@Rememberme The Riemann hypothesis is either a massively unsolved problem in algebra and polynomial roots, or there is a short cut via Andre LeClairs approximation of the zeta zeros.

hmm @MatsGranvik i want your email.....

The problem with Andre LeClairs approximation is that it is an approximation and therefore inexact.

2:45 PM
so we have got an approximation to the zeroes
how does the zeroes go......any combination

You can find my email here:
@Rememberme

https://matsgranvik.wordpress.com/cv-english/

The picture not the abo fi

@MatsGranvik IMO it is something that is very unexplored, and a potential research topic, a new branch is poised to be created in the near future.
Another thing will be that we will be able to find a pattern [maybe] to prime numbers.

user96977
what kind of CV is that

user96977
hehe, ok :)

2:48 PM
@TruthSerum Not much of a CV I know.

@MatsGranvik @ADG we can continue our talk in the new chat room

There is the number theory chat room.
$$2 \pi \exp (1) \exp \left(W\left(\frac{n-\frac{11}{8}}{\exp (1)}\right)\right)$$

lets talk in the place i made

?? what is this expression

what is it??

2:50 PM
@ADG That is the Andre LeClair approximation of the Zeta Zeros.
Mathematica:
N[Table[2*Pi*Exp[1]*Exp[ProductLog[(n - 11/8)/Exp[1]]], {n, 1, 12}]]

oh its this one....

user96977
where does the imaginary part come from?

@TruthSerum What do you mean imaginary?
W=LambertW= ProductLog

user96977
do the complex roots not have imaginary parts?

yes they do

2:52 PM
@TruthSerum Yes

user96977
maybe you get i when W(x) with x<0 ?

$$1/2+i*2 \pi \exp (1) \exp \left(W\left(\frac{n-\frac{11}{8}}{\exp (1)}\right)\right)$$

user96977
exp(1) is some shorthand for e to avoid confusion i guess

@MatsGranvik Is there any other approximation

user96977
@MatsGranvik i see now

2:54 PM
got it...?@TruthSerum

@Rememberme I don't know what they look like, but I have seen people commenting that there are other approximations than this one.

user96977
yeh, that whole thing was just the imaginary part

{-0.137955, 0.229925, 0.597804, 0.965684, 1.33356, 1.70144, 2.06932, \
2.4372, 2.80508, 3.17296, 3.54084, 3.90872}

user96977
how is that sequence related to the distribution of primes?

$$\frac{n-\frac{11}{8}}{\exp (1)}$$

2:56 PM
zeroes...huh....

user96977
thanks, but i don't have mathematica

Or roughly @TruthSerum
$$\Lambda (k) very approximately \sum _{n=1}^{\infty } n^{\rho _n}$$

@MatsGranvik If i think of the Riemann hypothesis more carefully what happens if i assume that what it says is false what will happen?
I know this is a big assumption but still what might happen

@Rememberme That I am not able to answer, other than what other people have said.
It is fully possible since Log(Log(Billionth zeta zero)) is about 30 something.

3:01 PM
So there would be zero which wont have a real part equal to 1/2

and 30 is not a big number.

yes it is not
I think we can look at this way also

Exp(MangoldtLambda(n)) = 1, 2, 3, 2, 5, 1, 7, 2, 3, 1, 11, 1...

@MatsGranvik can you come to the number theory room?

3:10 PM
@MatsGranvik out of curiosity, why are you taking log(log( )) of a zeta zero to make that statement?

@anon It is just something that Ian Stewart said about the Riemann hypothesis in his book chapter about the Riemann hypothesis, where he said that many things in number theory are often about Log(Log()) size when given as a bound or asymptotic for some sequence.

@Rememberme it means primes will appear a little more frequently than expected. RH is equivalent (IIRC) to an error term in the prime number theorem, which (again IIRC) derives from the so-called explicit formula. if you look at the explicit formula for the reweighted counting function $\psi(x)$ in terms of zeta zeros, it's like a Fourier expansion where imaginary parts of nontrivial zeros control local oscillations, and real parts control the order of magnitude of the individual oscillations.
a nontrivial zero would advantage one frequency over the others
@MatsGranvik I've seen log(log(n)) more than I've seen log(log(f(n))), so I'm not sure that's quite applicable. In any case infinite is a big number, and we've had numerical evidences over long stretches in number theory that later turned out false before (although not with as many implications and connections as the RH).

primes occurring more frequently @anon that is very interesting can i find a place from which it will start occurring more frequently if i assume RH to be wrong

@Remember the whole point is that the explicit formula provides you a knowledge of the "positions" of the primes.
and life is very easy for the explicit formula once you know that RH is true.

@Rememberme it's a global phenomena, not a local thing. you'd have to compute the empirical error in the prime number theorem and see if it fits the profile of there being a nontrivial zero. I don't think it does to our knowledge, but if RH is false all that means is the deviation of a nontrivial zero from the critical line is very small.

3:19 PM
otherwise you'll pick up residues from anywhere else, which will mess up the error term.
@anon You never know what the deviation is. I don't think there has been work on distribution of zeros outside the critical line.
(AFAIK, so i might be wrong)

or maybe it also means the deviation occurs high up, it's been awhile since I've looked at the stuff

me too. i wrote up loads of notes when i was studying PNT and posted it on a forum, i think, but never studied much about it.

PNT is very amazing isnt it?
you can find number of primes occurring

no, you can't.

@Rememberme yes. I'd like to see a conceptual or heuristic proof of it at some point.

3:24 PM
you can only find a (very good, i am sure) asymptotic.
@anon i like the proof which uses Perron's formula.
it's intuitive.

@BalarkaSen If the proof of FLT was itself so difficult how difficult would the proof of RH be??

well, the proof of RH over finite fields has been way, way difficult (weil cohomologies, etale stuff...) than the proof of FLT...

yes But we have still the RH left over

yes, and that left over is believed (again, afaik) to be more difficult than RH over finite fields.
now use transitivity...
:p

@BalarkaSen I'm thinking way more elementary than that. For instance, $\pi(n)\sim \frac{n}{H_n}$ which perhaps can be reworked using combinatorics. One can interpret $H_n$ as the collective rate of $n$ people working together on a task, when it would take the first $1$ unit of time to complete it alone, the second $2$ units of time, and so on. Maybe some kind of sieve thinking can heuristically justify the result, is what I'd hope.

3:28 PM
if you mean justification, there are a lot of probabilistic stuff.
you can find the probabilistic interpretation in my notes.

all of the heuristics (e.g. Cramer) I'm aware of are based off of the PNT
but I'll take a look

BTW @Balarka, I saw your earlier remark. The change of basis formula has a lot more consequence than your group theory comment. In particular, it's important for understanding what the invariants of a linear map (as opposed to just a matrix representation) are.

sure, i wasn't trying to list all the possibilities. of course, it tells us about the invariant subspaces.

Howdy@TedShifrin...

well, canonical forms, yes ... but I'm thinking about coefficients of the characteristic polynomial being independent of matrix representation ... this is important even in topology/geometry with characteristic classes represented by invariants of the curvature matrix of $2$-forms

3:35 PM
ok, now you've got me interested.

hi @Remember

@TedShifrin What do you mean here when you say to take the derivative using a sequence? math.stackexchange.com/questions/913582/…

I answered you a day or tow ago, mr eyeglasses :P

Oh I didn't see it
Let me search for it

Take the definition of the derivative as a limit, and use the sequence of fixed points $x_n\to a$ to evaluate $f'(a)$.

3:36 PM
actually, i don't know much about the eigenpolynomial to say anything about it. most of the book i saw just says that the $z^{n-1}$-th coefficient is $\tr(A)$ and the constant coefficient is $\det(A)$

@TedShifrin since you read the earlier transcript what do you think the way i visualize a group geometrically

I didn't read it all, @Remember
ah, the others are the sums of the $k\times k$ principal minors, @Balarka.
They're the invariant polynomials in the eigenvalues.

i'll believe you.

How magnanimous of you.

@TedShifrin I don't understand what it means to use a sequence $x_n \to a$ to evaluate a derivative. We want to evaluate the derivative at $a$, but how does the sequence come into play when using the definition of the derivative?

3:38 PM
:p

Because you can compute $$\lim_{x\to a} \frac{f(x)-f(a)}{x-a} = \lim_{n\to\infty}\frac{f(x_n)-f(a)}{x_n-a}.$$

hmmm, I just finished another amazing integral I developed today - it's going to be added to my book too.

Oh, I didn't know you were allowed to do that

Well, mr eyeglasses, sit down and prove to yourself that it is valid.

@BalarkaSen when computing the det(A-x*I) via Leibniz, the c^k coefficient will include all terms with n-k entries from the matrix, which collected together will give the leibniz expansions of the principal minors

3:39 PM
Ok

This is the sequential definition of limits.
g'morning, @anon

@anon ok. i didn't try to compute it with leibniz. maddening amount of work.

morning
don't actually write it out, just convince yourself that's how everything will go

Exterior algebra explains that you can do expansion in cofactors of arbitrary size, @Balarka.

sounds cool. i'll have to do exterior algebras at some point of time, i guess.
i.e., when doing forms.

3:41 PM
Yup, glad I can help motivate you :D

If we have the Fourier series of $g:$

$g \sim \sum_{k=1}^{\infty} b_k \sin (kx)$

and we want to find the Fourier series of $g'$ can we just say

$g' \sim \sum_{k=1}^{\infty} k b_k \cos (kx)$ ?

Or do we have to write the general formula of a Fourier series and find the coefficents? @robjohn

@MaryStar, that's correct, but you have to worry some about whether it makes sense for a general function. But if you know $g$ is smooth enough, it's fine.

@TedShifrin It is given that $g$ is $C^{\infty}$. So I can do that, or not?

yes, then it's fine

@TedShifrin Ok... Thanks!! :-)

3:48 PM
Hey

hi @n3b

@TedShifrin I dislike fourier analysis :p

I suppose you may dislike anything you wish.

I like the topic, but not preping for the exam I suppose

I'm sure a number of my students disliked differential geometry, too. There's no accounting for taste :)

4:01 PM
@robjohn Why can it be extended continuously to a $2\pi$-periodic function on $\mathbb{R}$?

@MaryStar Isn't what you've written the full Fourier series for $g(x)$?

@robjohn What does this mean? I got stuck right now...

morning

@MaryStar what did you mean by the "general formula of a Fourier series"? I thought that you had given the Fourier series.
@MaryStar since $f(0)=f(2\pi)$, and $f$ is continuous on $[0,2\pi]$, just define $f(x+2\pi)=f(x)$ and it will be continuous on $\mathbb{R}$.

The Fourier series of $g$ is $g \sim \sum_{k=1}^{\infty} b_k \sin (kx)$.

With general formula I mean $g' \sim \frac{A_0}{2}+\sum_{k=1}^{\infty}A_k \cos (kx)+B_k \sin (kx)$.

Can we just say, since $g \sim \sum_{k=1}^{\infty} b_k \sin (kx)$ we have that $g' \sim \sum_{k=1}^{\infty} k b_k \cos (kx)$ or do we have to prove this by using the general formula?
@robjohn So that we can say that $f \in C^1 (\mathbb{R})$ do we have first to define $f(x+2\pi)=f(x)$ ?

4:38 PM
why do they pull out only 1/n^2 here
goo.gl/vRVWts but 6/n^2 here
goo.gl/61Z1fo?

4:54 PM
Ugh. Tag creation. It's generally a terrible idea, right?
Like if a question was 4 years old, and you decided it deserved to be the first question tagged 'dodecahedron' a few minutes ago?

ugh

ugh

tag, you're it!

Don't get me wrong! I quite enjoy dodecahedra, and all their various forms :)

I prefer them sauteed

4:56 PM
Weird, it became tag synonym'd, I think. How unusual!

oh, it's that guy.

dodecahedra are cool objects, @pjs36

@JoeStavitsky do you see anything else they can pull out in the first one besides 1/n^2 ?
if not, your question is invalid

@anon, 2/n^2?

@JoeStavitsky you mean like 2i+1=2(i+1/2)?

5:00 PM
They are indeed, @BalarkaSen, I concur

@anon, yea I did screw up with the parentheses, but are you saying that doesnt make things simpler?

one usually doesn't "factor out" integers if that means creating fractions where there weren't any before
unless there's a good reason to

@anon got it ty
@anon took me embarrassingly long to see the issue was the paren

@pjs36 i'm mostly interested on them because it gives me cool geometric intuition on the galois theory of quintic extensions of $\Bbb Q$

Have you seen Klein's essay on that, Balarka? I've tried to look it over, but could never get far
But as far as I know, that's one of the highlights

5:05 PM
yeah, only a bit.
I have read King's book, which is far more accessible.

Ah, I see
My favorite part about it is that the alternating group $A_5$ has an irreducible representation of degree $3$, and its character values involve $\phi = \frac{1+\sqrt{5}}{2}$, which just strikes me as incredible

i find the theory developed by Klein quite cool. for example, knowledge of the symmetries of the dodecahedron and it's dual icosahedron tells me that $A_5 \cong \text{PSL}_2(\Bbb F_5)$, which explains a lot about the sextic invariant of the quintic foudn by Lagrange, and also why it's solvable in terms of the Weierstrass $\wp$ function.

I don't know anything about the latter two, but it sounds interesting

it most definitely is
in fact, iirc, icosahedral invariants can be used to solve a quintic, i think.
The whole point is that the icosahedral invariants satisfy a fundamental relation, $X^5 + Y^3 + Z^5 = 0$, which is something like the modular surface $X(5)$ or it's quotient, which of course is related to the quintic.
hello @Ted

5:22 PM
Can we just let it be known that cars are stupid?
I'm supposed to fill with transmission fluid until it runs out the bolthole! How weird is that?!

idk man the Tesla is pretty smart
lol bolthole
A quadric surface has cross-sections that are parabolas for constant x and for constant z, ellipses for constant y>0, a single point for y=0, and the empty set for y<0. What could be an equation for the surface, also identify the surface
I know it is elliptic paraboloid
why doesn't $x^2/9 + y - z^2/4 = 0$ work
Oh $x^2/9-z^2/4=0$ gives a hyperbola not ellipse

@MaryStar what would be the general fourier series for $g$?
@MaryStar if a function is continuous on $[0,2\pi]$ and its value at $0$ is the same as its value at $2\pi$, why wouldn't it be continuous if we extended it by $f(x+2\pi)=f(x)$?

5:49 PM
It's too bad that good questions get closed. I admit that I answered this question, but I think it is a good question with a pretty basic answer whose methods are applicable elsewhere. If it has been asked before, I can see it being closed for that, but I'd hate to see it deleted because the OP could not do enough on the problem.