Hi @BalarkaSen @Astyx

I can't show it. We know that $p^2-1$ is even. We also know that it holds for $p=5$. I tried some form of induction, by setting $r=p_2-p$, where $p_2$ is the subsequent prime number. Then we get $(p+r)^2-1=p^2+2pr+r^2-1$. We know that $r$ is even, so $pr$ and $r^2$ are even. But I don't know how to continue

@ShaVuklia You will never make induction work for primes

Well, instead of +1, we do +r as defined, but I know it's not working :d

No, I really mean never

Would you mind explaining?

in this case as mentioned, you just need that $p$ is odd, so you could do induction

I don't see how I can use induction, when $p+1$ doesn't help here?

$p+2$

oh

alright, that was stupid of me:P

now I see it

What is it about the primes that makes you say "never"?

@skullpetrol Ok, so never might be a bit strong. But it is practically impossible to make an induction argument work for the primes which would not work for all (odd) numbers

I see.

@BalarkaSen ah, right, that works too

hi chat

Quick question what is the mathjax for hyperbolic inverse functions, can't seem to find something working. I tried \argth, \arctanh, \atanh for instance, they still go italic.

hi @Semiclassical

@pilko I doubt there is a preset one for those

just use \operatorname

$\tanh^{-1}$ is what I do.

^

how to show a rational link has either one or two component

Hey, I'm looking for the derivative of tr(F(p)^{-1}A) with respect to the k-th element of p, i.e., d tr(F(p)^{-1}A) / d p_k . F(p), A are a square matrices where F(p) = A + \sum_{k=1}^K exp(p_k)S_k. I thought that the solution is d tr(F(p)^{-1}A) / d p_k = tr ( d tr(F(p)^{-1}A) / d F(p) * d F(p) / d p_k ), i.e., -exp(p_k) F(p)^{-1}AF(p)^{-1}S_k but if I compare this with numerical approximation the results are different.

any idea

ok, thanks. I'll continue to use my fav then : french argth. :p

@ShaVuklia As pointed out, you can use induction. But it's possible to do without induction too: let $p = 2n+ 1$; the $p^2 - 1 = 4n^2 + 4n = 4n(n+1)$. $n(n+1)$ is even because given two consecutive naturals at least one of them is div. by $2$. So the whole thing is div. by $4 \cdot 2 = 8$

Hi @TobiasKildetoft.

hey guys btw :)

@SylentNyte I would say that the results of mean, range, variance, ... give exactly the same results are calculated for both data-sets

@Druss2k Im asking for a reason why that is the case

I know that they give the same results of means and etc, but I'm having difficulty to explain why that is the case

ooh wait

What do you mean by "why"? You can eg, have different kurtosis.

@SylentNyte you could just generate some toy examples where this is the case. It still does not answer why but it is just possible that this may happen by chance.

The question was "can you explain why your answers to the two data sets are identical". I wrote that they "the values in the data set are spread out equally from their respective means" however I dont think that is a suitable enough explanation

I think if all the standard moments of two random variables are the same, then they're drawn from the same distribution (up to some measure-zero nonsense).

@Semiclassical I am pretty sure you can generate pathological counterexamples :) But yeah, for nice enough dudes I think so

ooh nice

That's not an answer to your question either...

Could be right. I'm sure there's a relevant theorem.

ooh

sorry, the mean is different

yet the range is the same

and everything else is the same

What you wrote is perfectly fine, although I have no idea what kind explanation you are looking for; s.d. measures exactly how scattered the data is from it's measure of central tendency (in your case, mean)

(By measure-zero nonsense I mean, for instance, that there's no difference in moments between the uniform distribution on [0,1] and the restriction of that to (0,1])

@BalarkaSen I just thought that what I wrote was not enough. Thanks though

I don't know why you thought so. That's what I am asking.

@BalarkaSen Actually, look here: mathoverflow.net/questions/3525/…

Evidently the moments doesn't determine the distribution if the moments grow too fast. That's beyond just measure-zero stuff.

Does $|F_5[X]/(x^3+2x^2-x-2)|$ mean $5^3$? so the answer is $125$?

Fair enough.

@BalarkaSen Because I'm not doing well in statistics and I want to improve tbh

How do I calculate the derivative of tr(F(x)^{-1}A) where F(x), A are square matrices?

@BalarkaSen ahh right! Thank you, that is indeed also a clever way to do it:) But I guess it's easier to use induction in the case of 24 XD

Nah, once you know it's divisible by $8$ what remains is to prove it's div. by $3$. That's because $p - 1, p, p+1$ are a consecutive set of 3 integers; so some of them must be div. by $3$. But $p$ is prime, so it's not. So $p \pm 1$ is div by $3$.

@jublikon It equals that, yes

So $(p - 1)(p+1) = p^2 - 1$ is div by $3$

Collecting up all the results, it's div by $3 \cdot 8 = 24$

@TobiasKildetoft thanx

hi guys I have a quick question if e^x=u what does e^2x= and e^-2x= interns of u

Hey everyone!

hey

hi

@MikeMiller Random question out of nowhere. Suppose you take a foliated manifold; do the "generic" - whatever that means - leaves have the same topological type? I think this is a consequence of generic leaves being of trivial holonomy, so nearby leaves also have the same topological type. That gives a comeagre set in the space of leaves of the same topological type, right?

is variance normally used for a population or for a sample?

@Balarka have you seen this question? I wouldn't be too surprised if this set of points can be very ugly in some sense

can you take $A_n$ to be increasingly fatter cantor sets

that would really be crazy

Ok Baire says that's not possible. [0,1] is complete and our fat cantor sets are all nowhere dense closed sets...

mutters joke about /bairely/ being able to understand that theorem

@Adamapple $e^x=u$, $e^{2x}=(e^x)^2=u^2$...

Hi, I have a doubt, the plane tangent to the graph of $f(x,y) = e^{x^2} + y^2$ at $(0,1,e)$ Is $ f_x(0,1)(x - 0) + f_y(0,1)(y - 1) = z - e$ ??

Wow! Exciting race for first place for overall rep.

Andre 424,461; Brian 424,453.

Andre 424,651

Damn

Also heh @Balarka that's a good one

Guys, my book on graph theory asks the following: Say we have a graph $G$ with $V=\{1,2,3,4,5,6\}$. If we know that vertex $i$ has degree $i$ for $i=1,2,3,4,5$, does that mean we know the degree of vertex 6? Well, I tried giving a counter example. I drew the two graphs, with vertices {{1,5},{5,6},{1,6}} and {{1,5}}. In this case, we know the degree for $i=1,2,3,4,5$, yet vertex 6 can have either degree 0 or 2.

However, as this is a 'general' counter example, does this mean that it holds for all $n$?

@BalarkaSen yeah I was thinking about fat cantor sets too

couldn't conclude anything interesting tho

Oh wait... I used degrees 0 in my counter example

Well, we could connect the remaining vertices, and not connecting them with vertices 1, 5, and 6

What I'm trying to say is; would the following work as a general counter example: For the first graph, connect 1 with $n$ and $n-1$, and connect the other vertices without connecting them with 1, $n-1$, or $n$. The second graph would be: connect 1 with $n-1$, and connect $2,3,...,n-2$ without connecting them with 1, $n-1$, or $n$.

@ShaVuklia You are not giving the vertices the required degrees

How not? All vertices but the $n$th have degree $\geq 1$?

@ShaVuklia It specifically said that vertex $i$ should have degree $i$

Oh, like that

not sure where you got the requirement of $\geq 1$ from

@TobiasKildetoft hi can you help me with complex analysis ?

@KasmirKhaan Probably not. Why do you think I can?

you helped me before @TobiasKildetoft but with other field of math

I just tried out all kinds of graphs, and it turns out that degree 6$\in\{2,3\}$ (maybe degree 6=2 or degree 6=3, but I stopped there). Even if I get the answer this way, I don't feel like this is a very economical approach. Is there a smart way of doing this?

@ShaVuklia Start by noting that vertex 5 is connected to all the others

Yes I did that

then that 4 is connected to all but 1

And vertex 1 is only connected to vertex 5

and so on

How do I use the fact that 4 is connected to all but 1? Because what I did was: I noted that 5 is connected to all, and 1 is only connected to 5. And then I started my cases: consider 2 connected to 6, and 2 not connected to 6. And so on. But what do you do with the fact that 4 is connected to all but 1?

@ShaVuklia Continue that way to see precisely which are connected to which

Oh okay, that's what I was doing. I thought it was a cumbersome approach

one more step and you are done

Ah yea, you're right. I conclude that 2 cannot be connected to 3, and then I'm there indeed

I winged it on the first analysis exam and feel rather confident. Oddly, most of the test was less difficult than exam questions from the prerequisite course.

@Balarka Absolutely not. Reeb stability only works on compact leaves.

Have you seen Huy lately @MikeMiller?

does anyone mind helping me with an olympiad geometry problerm

Hi @Zach

hi @AlessandroCodenotti

anyone?

shouldnt olympians be able to do their own geometry problems? :P

True Olympians.

:-)

help anyone?

I won't help other than say it would probably be a good first step to expand it

See you all later

Cya

@KasmirKhaan the radius of convergence of 1/(z-z0) is what?

hint: of 1/(z-1) is the unit open circle

@DHMO at the point z= 3+2i

... I am guiding you!

I am not blind!

@Kasmir: Where are the poles of your function $f(z)$? If you start at $3+2i$, do you encounter one of those poles first? Does that tell you something?

Hi @TedShifrin ! I did great on the re -exam :D i think I can get 25 / 30

I hope you are right :)

@DHMO I know ! am just new on this topic so need a bti warm up !

Hi prof.

Hi skull.

tomoow i get the paper

so I can tell you tomorrow how much i got =P

Hi @Ted

we had to prove cauchy theorem and 2 small line integral theorems

BTW, @Kasmir, I answered a cool Stokes's Theorem question on main yesterday. You should try doing it!

Hi @Alessandro

@TedShifrin Oh thanks ! I will save that question and see it other time , now am doing complex analysis

Ted are poles the removable discontinuity?

I mean like them in a way

No, not at all a removable singularity, @Kasmir.

Hmm I have no idea how to approch this problem

Should I do the power serie expansion of it firsT?

NOOOOOOO.

Hi guys/gals

I missed the lecture because I had the rexam on same time

hi

that is why am so lost =p

It's a theorem that a power series centered at $z_0$ will converge on the largest disk possible until a singularity of the function appears on the boundary of the disk.

Hi @Astyx

Hi @ted

@Kasmir: Are you now taking an actual complex analysis course?

@TedShifrin Yes we started right after we did the first exam

So you should know what a singularity is. There are three kinds — removable, poles (of various orders), and essential singularities.

@TedShifrin did not put alot of time on it because was busy with the reexam

That's a big problem with re-exams. They stop you from learning well the courses you're now taking.

Yes removable is like the line (x^2-1 ) (x+1)

Hi chat

Hi @Semi

Yepp and lots of us took the re-exam

Hi

You have a typo there, Kasmir.

where?

/

oh yes yes thanks

Those are the only singularities to worry about in one complex variable, at least

We did in class an example sinz /z

@skull: Does comprehensive question quality factor against the blatant homework askers who show no effort?

OK, @Kasmir. So have you talked about Laurent series yet?

(You really need to take 10 minutes to learn the basics of ChatJax, finally.)

(I say that mostly because I've been running into Du Val singularies in stuff I read lately)

We did it in class but I had re -exam so I was not there

Same time , unlucky for me

Are there some good videos on complex analysis online for undergrad?

Second verse, same as the first

@Semiclassic: Singularity theory in higher dimensions is a totally different animal.

@TedShifrin don't really know.

Yeah

You should read your textbook, @Kasmir.

Time to become an adult math student. Not just a lowly calculus student.

Haha yes yes :D

My teacher btw does some research on that field

And read the suggestion I gave you as my first comment. It tells you the answer if you sit and think.

multivarible complex analysis

ah, several complex variables ...

good stuff.

I took a graduate course in that many moons ago.

Actually, two.

Is it something new?

I mean relative

:35489083

Oh. Math vocabulary question of the day: universal vs versal vs miniversal

Where are the poles of your function $f(z)$? If you start at $3+2i$, do you encounter one of those poles first? ( this hint ) @TedShifrin ?

Is there a dumb explanation of what those mean?

(Also, miniversal is a fun word)

It is.

Yes, that hint, @Kasmir. I went on to explain why it's relevant.

h

H mm

Okay Ill save those in a file

and come back to them when i read the book

Those are standard terms in singularity theory, @Semiclassic, coming from whether your deformation of a given singularity (mapping) is sufficiently generic.

Should I spend time on first chapter ? its alot of the algebra of complex numbers

Probably best to get caught up quickly, @Kasmir, and then go back and fill in things more carefully. You should know this by now.

Yes but wanted more exprienced answer ,

I think I will start with complex integration

then series representations of analytic functions

Hmm, sounds right. (Like, for instance, the various ways an An singularity can resolve into others of lower singularity)

Because the first 3 chapters are about Elementary functions and algebra of complex numbers

The phrase I remember is 'miniversal unfolding' which quite apart from anything else is fun to say

This stuff is basically due to John Mather, @Semiclassic.

Hmmkay

There are various books/papers in the 70s and 80s, but this is some seriously complicated math.

The issue I run into with a lot of this stuff is the language. Anything that's in terms of sheaf theory is inaccessible to me

Well, you don't need that much sheaf stuff. But there are jet spaces and stratifications involved, for sure.

Yeah.

Several of my (joint) papers used this kind of stuff to prove very geometric stuff was quite pedantically generic.

Here's a silly physics word in return: diabolical point

LOL ... un point diabolique? :)

Actually, it derives from the Italian word diavolo

Almost as good as pasta alla puttanesca :P

Heh. A diavolo is apparently a kind of double-cone yo-yo

Of course, except in sloppy parlance, cones are always double-cones.

The silliness of it is that a diabolical point is just a point where two energy levels in a spectrum cross as a function of some external parameter

Doesn't need to be a cone point at all, then ...

Well, I think they have in mind what it looks like in the complex plane

Hey @Ted

Heya @Daminark

How's it going?

@TedShifrin that has a different origin though :P

Hey @Alessandro

Hi @Danimark

Gewiß, @Alessandro :P

Oh so funny story

@Semiclassical funnily I always heard that thing being called a diablo, same origin but a spanish name

A friend of mine is doing all reading courses this quarter

Ugh, @Daminark.

He's mainly looking at the interaction of discrete math/theoretical compsci with linguistics

Among other things, he's working on what I think he called higher order Fourier analysis on the Boolean hypercube

Now, he's also head of math club, and next Tuesday we're having a talk on Fourier analysis

It'll be analysis style, and while some $L^p$ stuff might get dropped around here and there, it's supposed to be pretty generally accessible

Fourier is cool. It can tell you how deep to dig your wine cellar ... or about distribution of primes.

Now, as it turns out, the guy I was talking to knows absolutely nothing of normal Fourier analysis

Which somewhat surprised me, like I guess Fourier analysis on the hypercube is probably different enough from that which you'd encounter in analysis that you don't need one before the other, but it feels somewhat odd

Checking a source it looks like the Italian is actually 'diabolo' which is even closer to tgat

What precisely is the Boolean hypercube, @Daminark?

$\{0,1\}^n$

Yeah I think @Semi has it right

Oh, so a finite approximation of the Cantor set :P

Yeah

Haha

Basically it's just a finite abelian group Fourier transform.

There's a picture that emphasizes that, lemme find it

Ah, that makes sense

I mean

Insofar as something that I have no experience with can "make sense" to me

Did you get that quotient Banach thingy?

I stopped thinking about it, but I was pretty sure I could do it.

Not just yet, it was taking a good deal of time and I still had more problems to do

So I used a different approach instead

If we assume $V\cap W = \{0\}$, we could consider the projection operator $P:V+W\to W$

Proved that it was bounded if $W$ is finite dimensional

So that if we had a Cauchy sequence $\{x_n\}_{n=1}^{\infty} \in V+W$, we express it as $x_n = v_n + w_n$, and then $w_n = P(x_n)$ would also be a Cauchy sequence by Lipschitz continuity

Then $v_n$ would be as well, and then since $V$ and $W$ are closed, we could send each of them to a limit so that $x_n\to v+w \in V+W$

Therefore it was closed

Well, even the existence of a complementary subspace is not obvious when you don't have an inner product.

Hey Ted

what is the idea of every analytic function can be represented by power series

Or something along those lines

Can you please give me some insight on that ?

I mean, the problem was that if $W$ is finite dimensional and $V$ is closed, then $V+W$ is closed. You could consider a basis for $V\cap W$, extend it to a basis for $W$. The remaining vectors for some space $W'$ such that $V+W = V \oplus W'$

Is that what you meant?

Oh, I was talking about why the quotient is Banach.

OH

Sorry, I wasn't sure what you meant

@Kasmir: I told you this once already. You need the Cauchy integral formula and then you plug in the power series for $1/(z-\zeta)$.

blargh. I can't find the image I remember.

Tragic, I know.

Okay thanks Ted :)