Mathematics - 2017-03-10 (page 2 of 5)

Kane

10:00

Is it fair to say this:

$x(\lambda p)'(x) - 2(\lambda p)(x) = \lambda xp'(x) - \lambda p(x)$
$ = \lambda (xp'(x) - 2p(x))$
$ = \lambda 0$
$ = 0$

Secret

' is linear wrt its arguments, thus the relation you wrote holds

(NB you have a typo in missing a 2)

Kane

@Secret thanks. Ohh I think I see what my problem was. I was multiplying the arguments of the function by lambda, where I should have been multiplying the whole function by lambda. Is that correct?

Secret

@DHMO quantum chem stuff were running in the background, things are so far ok. Still monitoring how the calculation goes to ensure it does not calculate garbage

Kane

So I'm basically considering $\lambda p(x)$ not $p(\lambda x)$

DHMO

@Kane correct

Kane

10:06

@DHMO @Secret cheers boys, thanks for the help

DHMO

@Secret why don't I see your activity in chem if your chem is so good

Secret

Because I am not 100% confident in my answers (I only answer if I think my answer perfectly address the question, and any possibel subquestion that arise from it)

Another reason is that I usually cluster in sites I want to ask question, I usually have no + or - drive to answer questions

DHMO

I see

Secret

The trouble of being a perfectionist: Sometimes I overthink

I also answer questions only if I want to, or if someone ask me and I might be able to answer

The general rule is that when people ask me questions, I will try my best to answer it, but I am also honest to my limits

DHMO

but your answer was quite good

your only answer in chem.SE

Secret

10:12

That one is just happens my memory is near 100% intact (barring that definition typo), thus I can immediately answer it

and most importantly, it pops up in PSE thus I saw it

DHMO

well you never know if you don't try

felipa

hi all

My question has had no love.. is it unclear? math.stackexchange.com/questions/2177922/…

or is there a better way or phrasing it?

Sha Vuklia

He guys

I'm struggling with the following question: "Prove that every planar Euler graph has an Euler-cycle that doesn't crosses itself."

I don't understand the idea... A planar graph is defined as a graph that can be drawn without crossing edges...

So why isn't the answer simply: as the graph is planar, its Euler-cycle doesn't cross itself

felipa

math.stackexchange.com/questions/388436/… this question?

SteamyRoot

Hmmm

Sha Vuklia

10:23

yea that's the same question indeed

I mean, I have the solutions to it

but I don't even understand the question:P

SteamyRoot

What is a crossing, though?

Sha Vuklia

maybe that two edges arrive at the same vertex?

felipa

good point

Sha Vuklia

because I initially thought a crossing as in how you draw to lines that cross

SteamyRoot

I think the point is that you run through the same vertex twice or so

Sha Vuklia

10:24

ah right

I'm gonna see if that makes sense with the solution!

SteamyRoot

At least, that's the only kind of crossing that would make a sensible question, I suppose

Sha Vuklia

true

Secret

How do we describe the case where two lines sit on top of each other like an X in a graph?

Sha Vuklia

ah yea, it is indeed how it's supposed to be interpreted

now the solution makes sense :p

well, my book uses the word "crossing" :l so that's unhelpful

maybe you could use intersection

but wikipedia also uses crossing

SteamyRoot

@Secret that is, in fact, called "crossing" as far as I know

Sha Vuklia

10:27

so you just gotta be careful when interpreting i guess

Secret

wow, tricky terminologies

Sha Vuklia

haha

skill patrol

They are synonymous.

SteamyRoot

Well, it's always a good idea to be careful. But in this case I'd say the question is definitely at fault.

Using an ambiguous term like that without further explanation is bad practice.

Balarka Sen

Hi @SteamyRoot.

SteamyRoot

10:31

Ohi

Sha Vuklia

ohh

based on my solutions

a crossing is thing such that

say we have a cycle $(v_0,v_1,\dots,v_k)$

and say $v_i=v_j$ ($i<j$)

then we have $\{v_{i-1},v_i\},\{v_{j-1},v_j\},\{v_i,v_{i+1}\},\{v_j,v_{j+1}\}$

Vrouvrou

Hello

Sha Vuklia

and these edges are clockwise or anti-clockwise, so you really make a 'cross'

Vrouvrou

@DHMO bonjour

DHMO

@Vrouvrou bonjour

Vrouvrou

10:37

désolé pour hier j'ai eu une coupure

DHMO

pas de probleme

felipa

anyone have any suggestions for how I can improve my question?

there is clearly something wrong with given the lack of interest

Vrouvrou

Alors j'ai $\liminf_{n\rightarrow \infty} \sup_{y\in\mathbb{R}^N}\int_{B_{\rho}(y)}\Phi(|u_n|) dx$@DHMO

LeGrandDODOM

@Vrouvrou une coupure de Dedekind ?

Vrouvrou

??? une coupure d'internet

Secret

10:40

Who are the combinitorics guys in this chat?

DHMO

@Secret ?

Secret

felipa's question is some combinitorics information theoric question, those guys will be best bet for him in working that out

LeGrandDODOM

@MickLH WUB WUB WUB WUB WUB WUB

DHMO

@Vrouvrou utilise la definition de liminf par lim et inf qui se donne ici

felipa

@Secret thanks!

are there any combinatorics/information theory people on chat?

DHMO

10:43

et tu peux utiliser epsilon-delta ensuite

Vrouvrou

@DHMO par la definition c'est égale à $$\lim_{n\rightarrow\infty} \left[\inf_{k\geq n} \sup_{y\in\mathbb{R}^N}\int_{B_{\rho}(y)}\Phi(|u_k|) dx\right]$$

Secret

[Meanwhile thinking about PhinotPi's question]
$(a+b)+c=a+(b+c)$

$a_x+b_y=a_{x\oplus y}$

$a-a=I_0$

$I_0$ + id

$a_1=-a_1$

$I_0=-I_0$

$a+b=-(-b+-a)$

R = group ...?

$a_1b_1=c_1$

$a_1a_1=a_1$

$a_1b_0=c_1$

$a_0b_1=-a_0$

$a_0b_0=a_0$

$aI_0=a$

$I_0a=I_0$

$(ab)a=a(ba)$

$a(b(-a))=a(ba)$

$a(b+c)=(ab)c$

$(b+c)a=ba+ca$

$(-a)b=-(ab)$

$(-a)+b+a=ba$

$ab=c, c(-b)=a$

DHMO

@Vrouvrou $\Phi$ veut dire quoi?

Vrouvrou

$\phi$ c'est une fonction continue convexe et positive

felipa

@Secret It would be interesting to see the main interests of the people on chat aggregated

This could be done by counting the tags of their answers/questions

Secret

10:47

The interesting thing about this chat is that the math field is weakly correlated to timezones until recently this year

You will find the algebra guys mostly awake in the afternoon and morning in sydney time

felipa

@Secret but only until recently?

did something change?

Secret

Well, you can say DHMO, I and many others broke the pattern for the algebra guys for being on chat more than half of the day almost 24/7 (although I am not terribly good at algebra, I still need to work on it)

Vrouvrou

comment continuer @DHMO?

DHMO

@Vrouvrou c'est tres complique et je n'ai pas le temps, desole

felipa

@Secret aha.. so you are in a good position to know who is into combinatorics/information theory

Secret

10:51

mick also tend to have a lot of trouble with combinitorics and infinite series closed form questions. But as far I know, I don't recall the small group of regulars I knew are particularly good at combinitorics. Perhaps, that group is not that frequent on this chat

felipa

@Secret ah.. that's a shame

DHMO

@Secret could you help @Vrouvrou?

Secret

@DHMO You need to translate vrouvrou's to english cause I only knew chinese and english. Even then form the look of it it looks like banach space stuff, thus I am not sure if I can handle that

DHMO

> Use epsilon-delta to write $\displaystyle \liminf_{n\rightarrow \infty} \sup_{y\in\mathbb{R}^N}\int_{B_{\rho}(y)}\Phi(|u_n|) dx$

That's all @Secret

@Secret that's a strange tense

Vrouvrou

11:05

@Secret i speak english

i have that

Kane

Say I have a set: $w=\left \{ \left \{ a_{n} \right \}\in \mathbb{R}:a_{n+2}=a_{n+1}+a_{n} \ \forall n\geq 0 \right \}$

How can I show the $0\in \mathbb{R}$ is in w?

Secret

@Vrouvrou I know, hence the 's as in referring to the above specific question

DHMO

@Kane I don't think you can.

Kane

I understand 0 is the first term of the fibonacci sequence

but they want me to show this is a subspace of R

Obviously it is, but I mean, shouldnt I show the 0 of the reals is in the set?

DHMO

you didn't say that it is the fibonacci sequence

Vrouvrou

11:09

@Secret $\liminf_{n\rightarrow \infty} \sup_{y\in\mathbb{R}^N}\int_{B_{\rho}(y)}\Phi(|u_n|) dx>\tau>0$ what are the steps to find that $\forall \varepsilon>0, \exists z\in \mathbb{R}^N, \int_{B_{\rho}(z)}\Phi(|u_n|} dx>\tau-\varepsilon>0$

Kane

@DHMO Well since it is, can I just say that by definition, when n = 0, the term of the fibonacci sequence is 0?

DHMO

@Kane yes

Kane

@DHMO sorry I didn't mention that I was talking about the finonacci sequence. Thanks for that

Gomez

Hello. I will be glad if anyone could please help me with this problem.

What is the number of subsets of $\{1,2,3, \cdots,n\} containing k elements?

Thanks for helping

DHMO

@Gomez nCk

Gomez

11:18

Thanks

Vrouvrou

@Secret have you an idea please

Secret

I am typing and thinking

Given
$\liminf_{n\rightarrow \infty} \sup_{y\in\mathbb{R}^N}\int_{B_{\rho}(y)}\Phi(|u_n|) dx>\tau>0$

$\forall n, \exists N, \sup_{y\in\mathbb{R}^N}\int_{B_{\rho}(y)}\Phi(|u_n|) dx>\sup_{y\in\mathbb{R}^N}\int_{B_{\rho}(y)}\Phi(|u_N|) dx>\tau>0$

$\forall n,y , \exists N, Y, \int_{B_{\rho}(Y)}\Phi(|u_n|) dx>\int_{B_{\rho}(Y)}\Phi(|u_N|) dx>\int_{B_{\rho}(y)}\Phi(|u_n|) dx>\int_{B_{\rho}(y)}\Phi(|u_N|) dx>\tau>0$

Therefore

$\forall \varepsilon >0 , \exists z, \int_{B_{\rho}(z)}\Phi(|u_n|) dx>\tau>0$

Not sure if that's the correct way to think, given I have zero experience in analysis courses as I mentioend earlier

Little Rookie

Hello all, please help me with this question: Motivation for forming a specific bound for a recursively defined sequence.

http://math.stackexchange.com/questions/2179430/motivation-for-forming-a-specific-bound-for-a-recursively-defined-sequence

Please help me with my question or upvote it, so more people will notice the question.

Thanks.

MickLH

11:43

@LeGrandDODOM yus.

Secret

12:04

[Nerd test] I pretty sure I completely missed the context

is this actually a triple integral at all?

DHMO

sure

Secret

NB: I actually knew LITERALLY NOTHING about computer data structures

Or more generally, I knew nothing about how to fix or assemble a computer

Secret

12:26

Understanding computer data structures and understanding maths are two different things. The former requires quite a bit of memorisation

s.harp

12:48

and mathematics doesnt?

Akiva Weinberger

@Secret Is there an option for "ask Wolfram Alpha"

Secret

@AkivaWeinberger I wish, but there isn't

SteamyRoot

13:01

Is it bad that I scored over 75% in every single category on that test?

Secret

not really, it just means you might be a supernerd

SteamyRoot

I see.

Well, I guess the test did call me a "Highly Dorky Nerd God"

PhiNotPi

@Secret oh, hey

Secret

there's nothing new up there, I just copy the equations in that question and see if they can be combined in ways to show something else

The type 1 elements together with the identity form a type of reflection group as all elements are involutive (i.e. its own inverse) and they obey the axioms of groups.

and this equation then constrains on the type of reflection group we are considering
$a+b=-(-b+-a)$

Don't disturb

I also have some nerdtest to give you if any interested. It's consisted of only one question.

The scores you got are really nice.

Secret

13:16

ok

Don't disturb

Please find the asymptotic behaviour of

$$\sum_{k=1}^N \left(\frac{H_k}{k}\right)^5$$ as $N\to \infty$

This is my nerdtest.

Secret

What special function is $H_k$ ?

Don't disturb

@Secret harmonic number, $H_k=1+1/2+...+1/k$

Secret

Hmm, I am not terribly good at series. Let's see what the partial sums told us:
N=1
$1^5$

N=2
$\left[\frac{1}{2}\left(1+\frac{1}{2}\right)\right]^5+1^5$

N=3
$\left[\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}\right)\right]^5+\left[\frac{1}{2}\left(1+\frac{1}{2}\right)\right]^5+1^5$

...

now to check the identities of harmonic numbers multiplied to some constants

if we can condense these terms up, it will become a series of N of these terms, which might be easier to manage

Akiva Weinberger

I find that integrals and series are a specific type of taste

Some people love them, some people don't care for them

Secret

13:23

They are hard to solve, but they have symmetry properties that are important in understanding some aspect of functional space as what that (forgot name) conjecture told us

SteamyRoot

Hmmmm

Akiva Weinberger

@Secret ???

Elaborate

SteamyRoot

That better not converge to $\sqrt{2}$...

Secret

@AkivaWeinberger :

Jan 21 at 16:06, by Secret

[Integral symmetries] Cauchy Riemann theorem like analogues in reals:

$$\frac{d^n}{dx^n}f=\lambda\int^{(m)}fdx$$

(where $f=\sin, \cos,e^x$ (possibly more?),$\lambda \in \mathbb{R}$)

For example

$$\int x^n \sin x e^x dx$$
Consider

$(uv)''=u''v+2u'v'+uv''$

Plug $u=\sin x$, $v=e^x$. Then

$(uv)''=-uv+2u'v+uv$

Note how the antisymmetry of $\sin x$ wrt the functional $()''$ results in cancellation. Hence

$(uv)''=2u'v$

Now it is easy to check that $\cos x$ is also antisymmetric wrt $()''$ thus we could have start with $u' =\sin x\implies u = -\cos x$. Hence

Of course, those involving special functions are not that simple, but the symmetry is there, it is just very hard to find. That's why we need complex varieties to abstract them

SteamyRoot

@Don'tdisturb Do you want us to actually calculate the exact limit?

(if the limit is finite, of course)

or is just convergent/divergent enough?

Don't disturb

13:27

@SteamyRoot No, find the asymptotic behaviour (which also include an approach of the limit)

@SteamyRoot It's easy to see at a first glance the series converges.

PhiNotPi

@Secret $(ab)c = (-b + a + b)c = ((-b)c + ac + bc) = (-(bc) + ac + bc) = (ac)(bc) = a(c + bc) = a(c + -c + b + c) = a(b+c)$

not sure if anything there is useful or not... I was actually expecting to reach some kind of contradiction but I'm happy it's self-consistent.

SteamyRoot

Yeah, can't be bothered with that sum.

Secret

@PhiNotPi The fact you get it from generalising a geometric manipulation suggest it is very hard to be inconsistent because it has to reduce to that geometric case

SteamyRoot

Not the kind of math I want to spend the time I don't have, on.

Secret

[Meanwhile] Looking up derivatives of harmonic numbers

$\lim_{k\to \infty} \frac{H_k}{k}$ has the form $\frac{\infty}{\infty}$. L'Hopital the term gives $\lim_{k\to \infty}(H_k)'=(computing)$

Don't disturb

13:38

@Secret $H_k$ behaves like $\log(k)$ when $k$ is large (one may see that using Stolz-Cesaro theorem).

Secret

So using linearity of derivatvies, the asymptotic behaviour of $H_k'$ should behave like 1/k

SteamyRoot

$\ln(k) + \gamma + \frac{1}{2k} \cdots$

If I recall correctly

Meh, if there's no closed form, it's no fun :P

SoumyoB

HOLY CRAP

I got accepted into the University of Delaware's PhD program

9

with a GPA of 2.72

and others with a far better GPA than me had applied to far more universities than I did and still are getting rejections after rejections

Secret

$\cdots +\left[\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}\right)\right]^5+\left[\frac{1}‌{2}\left(1+\frac{1}{2}\right)\right]^5+1^5$ I am currently trying to see if I can use the do nothing technique to set up some kind of recursion relation

SoumyoB

I feel like this guy

SteamyRoot

13:44

Well, perhaps they aimed higher (and too high) ?

Either way, congratulations!

SoumyoB

thanks @SteamyRoot

I think it was very much a stroke of luck combined with good recommendations

Compulsive Mathurbator

@SoumyoB Congratulations

SoumyoB

thanks!

Secret

$\{\frac{1}{k^5}\}$ this is convergent by p test, so hmm...

DHMO

@Secret wt da hell is dat latex

Secret

13:46

@DHMO I have no idea, somehow the 2 get stuck up there

DHMO

@Secret u need some spaces in ur code

or else SE will arbitrarily insert spaces

and those spaces (U+200A/B/C/D) aren't the spaces that you want

SteamyRoot

@SoumyoB So, what area/subject are you planning to do research in?

SoumyoB

stochastic processes

and a little bit of combinatorics, although I know UD perhaps isn't the best place for the latter

but it is, for the former

Little Rookie

cool, are u a stats major?

SteamyRoot

Heh, shame, that means I'll probably never end up meeting you at a conference :P

SoumyoB

13:51

I'm a math major

@SteamyRoot wait are you in UD too?

Little Rookie

well stats is one of the many track in maths

didnt u apply for a specific track?

SteamyRoot

No, not at all

SoumyoB

no just a mathematics PhD

Little Rookie

i mean during ur undergrad period

SteamyRoot

I'm in Europe; but there's always a few US PhD students at conferences held in Europe

Little Rookie

13:53

maybe your university dont have that system

SoumyoB

@LittleRookie indeed, my university doesn't offer a specialization, we just have some compulsory courses and a minimum number of credits, but beyond that we're free to choose courses as we wish

I chose mostly from probability area

@SteamyRoot being in which area then could we have met?

SteamyRoot

Group theory

SoumyoB

I see...

too bad I've not had much interest in it until recently

Little Rookie

@SoumyoB thats cool, in my university, if i were to study a course offered in other track, i might have to take them as an unrestricted elective instead of a maths module.

SoumyoB

@LittleRookie is your uni in EU too?

Little Rookie

13:56

Nope, asia

Thus different system

SteamyRoot

Well, non-UK Europe is very different in general

Secret

$$\sum_{k=1}^N \left(\frac{H_k}{k}\right)^5=\left[\begin{pmatrix}H_1^5 \\ H_2^5 \\ H_3^5 \\ \vdots\end{pmatrix}\cdot \begin{pmatrix}1^5 \\ \frac{1}{2^5} \\ \frac{1}{3^5} \\ \vdots\end{pmatrix}\right]=$$

SoumyoB

@SteamyRoot I reckon so, more so after Brexit :P

SteamyRoot

Looks like you forgot some $5$th power on the LHS?

Nah, don't think the Brexit will change anything

UK/US and mainland Europe have always been rather different in our university systems

Little Rookie

Hello all, please help me with this question: Motivation for forming a specific bound for a recursively defined sequence.

http://math.stackexchange.com/questions/2179430/motivation-for-forming-a-specific-bound-for-a-recursively-defined-sequence

SoumyoB

13:59

I didn't apply at all to any UK universities since they are much more particular about GPA, or so I've heard

SteamyRoot

Well, unless your advisor knows your potential advisor there, you'll need good grades, yes :P

But, the main difference is that here in Europe, you traditionally go Bachelor -> Master (-> PhD)

SoumyoB

yeah I have a Master

SteamyRoot

Whereas in the US, you go Bachelor -> PhD or Master

SoumyoB

in fact I have indeed applied to a program in combinatorics in Vienna

SFB F50, if you've heard of it

well bachelors in the EU is only of 3 years, so I guess some more time is needed in course work before pursuing a PhD?

SteamyRoot

Depending on the country, bachelors is 3 or 4 years; and master is 1 or 2

My university being 3 - 2

SoumyoB

14:04

I see

SteamyRoot

Haven't heard of that program, but I do know a professor in Vienna

(in fact, an MSE regular)

SoumyoB

well I've got to do my assignment now, see you later!

SteamyRoot

Enjoy

Secret

14:21

ugh why are all my fractions misbehaving

$$\sum_{k=1}^N \left(\frac{H_k}{k}\right)^5=\sum_{k=1}^N \left(\frac{\sum_{m=1}^k\frac{1}{m}}{k}\right)^5=\sum_{k=1}^N\frac{1}{k^5} \left(\sum_{m=1}^k\frac{1}{m}\right)^5=\sum_{k=1}^N\frac{1}{k^5} \left(\sum_{m_1=1}^k\sum_{m_2=1}^k\sum_{m_3=1}^k\sum_{m_4=1}^k\sum_{m_5=1}^k \frac{1} {m_1m_2m_3m_4m_5} \right)$$

SteamyRoot

I guess the formula was so ugly that even mathjax refused to render it?

Secret

nested sums are nothing compared to recursive sums

and I am not even done yet

THERE

Now this part:
$\sum_{m_1=1}^k\sum_{m_2=1}^k\sum_{m_3=1}^k\sum_{m_4=1}^k\sum_{m_5=1}^k\frac{1}{m_1m_2m_3m_4m_5}$
reminds of multinomial related stuff...

Semiclassical

oh lawd.

SteamyRoot

Well, it renders correctly now.

Still ugly, though.

Semiclassical

^

Secret

14:28

Are there any formulae for products of sums where the summand are a single power of x?

Semiclassical

Harmonic-type sums have shown up a lot on the main site, so you miiiight be able to find somethere there.

@Secret An obvious thing to do if you want the asymptotics of that sum is Euler-Maclaurin.

That won't be so useful if you want the exact value, of course.

Secret

Got something... slightly nicer...?
$$\sum_{k=1}^N \left(\frac{H_k}{k}\right)^5=\sum_{k=1}^N \left(H_k\frac{1}{k}\right)^5=\sum_{k=1}^N \left(H_k(H_k-H_{k-1})\right)^5=\sum_{k=1}^N \left(H_k^2-H_kH_{k-1}\right)^5$$

Akiva Weinberger

12 hours ago, by Akiva Weinberger

Interesting geometry theorem: Consider three circles on the plane, no two of the same radius. Each pair of circles has four common tangents; draw the two outer ones, and consider the point of intersection of those two tangents. Repeat for each pair of circles; this gives us three points. The theorem is that these points are collinear.

12 hours ago, by Akiva Weinberger

And here's a proof without words: https://youtu.be/LE3gQKeIyLM The proof goes through the third dimension!

Secret

(using the identity $H_n=H_{n-1}+\frac{1}{n}$)

Akiva Weinberger

Does the same proof work for four spheres, such that the six vertices of the cones determined by the pairs of spheres lie on a plane?

I believe it should (the proof goes through the fourth dimension but is essentially unchanged), but I can't visualize any specific examples to make sure that it works.

Semiclassical

14:37

Wowee-zowee

In other words, can Monge's theorem be generalized to higher dimensions?

(spoiler: I have no earthly idea.)

(not much of a spoiler, really.)

Akiva Weinberger

@Semiclassical You know Mathematica. If it interests you, would you mind trying to see if you could generate an illustration of the four spheres, six cones and points, and (hopefully) the plane containing the points?

I have no idea how easy/hard this is

Semiclassical

Hmm.

SteamyRoot

All I can say is that question reminds me of theorems in projective geometry like Ceva and Menelaos etc

Akiva Weinberger

Last night I went through formalizing the informal argument given in the video, to make sure it works. I see no reason it shouldn't generalize to all dimensions.

Semiclassical

Probably the main work in Mathematica is to relate the cones to the spheres.

DHMO

14:42

yesterday, by Alessandro Codenotti

Take a $3\times 3$ magic square represented as a matrix with all rows, columns and diagonals summing up to $N$. Show that $3N$ divides its determinant

Does anyone have any idea about this?

Semiclassical

I've seen that question before.

Hmmm

Akiva Weinberger

@SteamyRoot It's very projective. If two spheres are allowed to be the same radius, the point ends up on the plane at infinity.

SteamyRoot

Probably that shouldn't change the collinearity.

Semiclassical

The fact that all rows/columns sum to N means that $(1,1,1)$ is an eigenvector both as a column vector and as a row vector.

DHMO

@Semiclassical wow

PhiNotPi

14:43

@Secret I think I found something that answers my question from earlier: my algebraic structure is a quandle based on the group of reflections/translations with composition.

Semiclassical

What this doesn't use is that the diagonals sum to N as well.

Though I guess from that you at least know that the trace of the matrix is N.

DHMO

@AkivaWeinberger no comprendo la "prueba". Los conos van al infinito entonces los puntos son colineal?

Semiclassical

So you've got that the trace is N and N is an eigenvalue. Hence the remaining eigenvalues are of the form $\pm \lambda$.

With the determinant therefore being $-\lambda^2 N$. So we'd need to show that $\lambda$ is a multiple of 3.

hmm

Akiva Weinberger

@DHMO The non-projective version has the spheres of distinct radii

so the points are all finite

They're the vertices of the cones.

Adeek

hey @AkivaWeinberger

want to discuss something ?

in topology

Akiva Weinberger

14:47

Got to go, sorry

Adeek

about cofibrations ?

DHMO

@AkivaWeinberger how is it proved that the vertices are collinear?

Adeek

oh ok

Alessandro Codenotti

15:14

@Semiclassical I think I posted it in chat before

Semiclassical

Probably.

Akiva Weinberger

@DHMO Ah, sorry, I misunderstood what you meant.

(I confused "prueba" for "teorema")

What's going on in the proof of the 2D case is that they're associating each circle with a sphere of radius $1$. The smaller the circle, the further away the sphere.

Consider the plane through the centers of these spheres. The "horizon" of this plane, when projected back onto the plane of the problem, gives you the desired line.

So there aren't cones at all in the video; those are cylinders. Cones are only relevant in the statement of the 3D version.

Timothy Chow expressed a similar doubt in this comment. Over there, I wrote a more detailed explanation.

Secret

35

Q: A closed form for the sum $\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$

How can I find a closed form for the following sum? $$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$ ($H_n=\sum_{k=1}^n\frac{1}{k}$).

I strongly suspect a way towards that question is to use $Li_5$ instead of $Li_2$

imgur.com/a/5Re7R

What on earth is this alien language

I guess I need to be more familar with the algebraic properties of summations...

Alessandro Codenotti

15:34

@Akiva do you have 10 minutes to talk about topology?

Akiva Weinberger

@AlessandroCodenotti No, I'm in class

SteamyRoot

That's the perfect moment to talk about topology ;)

Alessandro Codenotti

Ah, I see, a math class?

DHMO

@AlessandroCodenotti what topology?

Alessandro Codenotti

@SteamyRoot especially if it's a topology class I guess

@DHMO algebraic topology

SteamyRoot

15:37

Algebraic topology is best topology :D

DHMO

@AlessandroCodenotti well, mind to state the actual thing you wanna talk about?

Semiclassical

@AkivaWeinberger That's what I've gotten so far using Mathematica.

Alessandro Codenotti

@Balarka (pinging him just in case he's around) gave me the fundamental group of a connected sum of 2 tori to calculate, and I'm trying to adapt a method @Akiva showed me for a sphere to this case

Semiclassical

It's a bit of a cheat; rather than find the cone explicitly, I picked a random distribution of points on two different spheres and then had Mathematica find the convex hull.

That's why it doesn't extend beyond the two spheres.

And holy sh** where'd this pain in my back come from.

Alessandro Codenotti

@Semiclassical isn't the laziest solution the best solution?

Semiclassical

15:42

Only when it doesn't lead to problems later.

SteamyRoot

connected sum of $2$ tori... so a torus of genus $2$ ?

Felix.C

Hey guys I'm a bit confused about the wirtinger derivates, as I understand they define $df/dz\bar := 1/2 (df/dx +idf/dy)$
As of the meaning of "Definition" I conclude that they are just given too us, I saw somewhere on Wikipedia some explaining why you'd choose them like that but I wasn't sure if I understood it right.
In any case is there actually a way to derive a holomorph function with $d/ \bar dz$, as of the limes definition of the derivate?
Edit: I'm sry for the \bar thing I don't get it

Semiclassical

This is actually rather alarming. I feel like someone's pushed a spear into my right lung.

SteamyRoot

You may want to get medical aid... :s

Compulsive Mathurbator

@Semiclassical Angina?

Alessandro Codenotti

15:45

@SteamyRoot yes, I'd call that $T_2$, Balarka calls it $\Sigma_2$, I don't know if either is standard notation

Compulsive Mathurbator

Although that would present in the left

Alessandro Codenotti

@Semiclassical out of nowhere? That's worrying

SteamyRoot

I've seen both used.

Semiclassical

yeah. I'm going to walk over to the health center.

Alessandro Codenotti

@Felix.C what were you trying to do with the \bar?

SteamyRoot

15:47

Take your phone.

Akiva Weinberger

@AlessandroCodenotti No. English class, currently

It's 10:47am here

Felix.C

Alessandro Codenotti

you need d\bar z then $d\bar z$

SteamyRoot

@AlessandroCodenotti Well, I know two different approaches. One is Van Kampen, the other is working with the fundamental polygon.

Alessandro Codenotti

@SteamyRoot Akiva showed me how to prove that the sphere is simply connected by covering it with $2$ open sets intersecting in an annulus, which I think is basically Van Kampen (we should see that theorem next week in class). I'm trying to use the same approach with a $T_2$

LeGrandDODOM

15:53

math.stackexchange.com/questions/2180737/a-hard-integral asked 7 minutes ago and viewed 163 times

lmao

arctic tern

>163

largest heegner number

LeGrandDODOM

most questions asked 1 hour ago have less than 60 views

so the clickbait title is the culprit ?

SteamyRoot

Look at the OP's profile, more specifically the badges

Adeek

hey @MikeMiller are you around ?

just want to verify with you something

LeGrandDODOM

math.stackexchange.com/questions/2027844/… 1600 views

this guy is fishy

SteamyRoot

15:58

@Alessandro Yeah, that's exactly Seifert-Van Kampen

Akiva Weinberger

How do you tell how many views something has?

SteamyRoot

It says so on the right

LeGrandDODOM

@AkivaWeinberger top right corner

Transcript for

$Mathematics$ Mathematics