Mathematics - 2017-06-26 (page 5 of 6)

Secret

17:00

Currently, my plan for integrals have multiple pipelines: One of these is to continue to observe and solve integral challenges that pop up and learn about the pattern of each solution pathway, another pipeline is I plan to express integration purely in terms of algebra (and possibly category theory when I got there) so I can highlight the sources on where all those substitutions, symmetries etc. came from and see if there is a pattern that unite them all

For pipeline 1, below is my rough draft:

Waiting

@Hippalectryon It looks nice.

Zee

@Secret you may be interested in algebraic microlocal analysis, D-modules and functorial analysis. I don't know much about them...

Hippalectryon

@Waiting Well, if you find a nice solution, please let me know :D so far none of my attempts have really worked

Secret

Definition

$\textbf{Elementary functions}$: Given $f \in \textrm{L}^p(\Bbb{F})$,$f$ is elementary if it can be expressed as a finite combination of $\textrm{exp}$, $\log$, $c\in \Bbb{F}$, powers and $\frac{p}{q},p,q \in \textrm{P}(\Bbb{F})$.

$\textbf{Standard integral functions}$: Given $f \in \textrm{L}^p(\Bbb{F})$, $a, a_k,b \in \Bbb{F}$, $f$ is a standard integral function if it take only linear arguments $a*\textrm{id}+b$ and is any of the following and its derivatives:

$\sin$, $\cos$, $\tan$, $\csc$, $\sec$, $\cot$,

Semiclassical

I still suspect that differential galois theory has some insight to be had.

Secret

17:03

@Zee Yeah, and steamyroot also recommend complex varieties, semi and typhon recommended differential galois theory, all of which I am still ages from a solid background, but I will find a way in between my chemistry career (jots in notebook)

Semiclassical

@Hippalectryon Madre dios

Is that integrand symmetric under $t\to 1/t$ (apart from the limits) ?

Zee

@Secret how the heck do you have the time :s

Secret

@Semiclassical My naive hunch is that if we can incoporate the special functiosn into the galois group and make an extension, we can make them behave as if they are new elementary functions and incoporate other functional identiies into the framework, thus effectively extending rische algorithm to include special functions

Hippalectryon

@Semiclassical yeah it's the same as $\int_0^1$

Semiclassical

Okay. So you could just as well take it to be $\int_0^\infty$ without the 1/2

Not sure that helps, but worth bearing in mind.

Though I guess one has to be careful about the behavior near $t=1$.

Secret

17:05

@Zee I don't know, but I have a history of being incredibly stubborn and often get what I want 5-10 years later

Hippalectryon

@Semiclassical True :-) it just happened to be in $\int_1^\infty$ when I first encountered it

Semiclassical

Sure.

Akiva Weinberger

@Semiclassical Mother god

Zee

@Secret lololol unless your over 50 you don't have a history

Semiclassical

@Secret Here's a digression for you. Do you know what it means for a spectrum to be gapped?

Secret

17:07

@Semiclassical I heard of that term in some solid states news, but details not sure

Semiclassical

pfff. younger people have history too. for instance, I have a history of procrastination

Secret

Do no underestimate young minds, they see what the old cannot see and they often hold the future

Semiclassical

whereas others have a history of self-serving cynicism.

@Secret quick intro, then.

Secret

ok

Semiclassical

Suppose I've got a Schrodinger equation with a periodic potential i.e. $-y''(x)+V(x)y(x)=\lambda y(x)$.

Astyx

17:09

Rehi chat

Akiva Weinberger

@Secret In any case, every shift-invariant linear operator (shift-invariant means it commutes with $D$) can be written in the form $\sum a_nD^n$

Semiclassical

I'm setting various constants to one and I'm not bothering to give them their physical names, but $\lambda$ would be energy, $V(x)$ would be a potential, and $y(x)$ would be the wavefunction.

Akiva Weinberger

That's why shift-invariant linear operators commute.

Semiclassical

For it to be periodic, we need $V(x+L)=V(x)$ for some $L$.

Zee

@Semiclassical I almost didn't notice what you did there

You can't do math unless your cynical, else you will get duped

Astyx

17:12

Do you really believe that ?

I was wondering about that some days ago

Semiclassical

That would be more convincing if your cynicism weren't solely exhibited towards what other people have done and are trying to do.

Astyx

I was wondering if you could do anything without being cynical actually

Zee

@Semiclassical I don't think that's a fair assessment, I don't see where I critiqued secret

Or any other before him really...

Astyx

My question is a real one, I'd really appreciate an answer @Zee

Semiclassical

No, critique would imply actually presenting an argument. Instead you just snipe at people.

Zee

17:15

@Astyx your question if people can do math without being cynical?

Astyx

Yes, sorta

My question would be more general though, not just about math

Zee

@Semiclassical I don't see where I did that, can you elaborate? Maybe it's true and I need to examine myself, maybe your being unfair

Astyx

That seems pessimistic but could still be true

Zee

@Astyx well people can do anything without being cynical. Being cynical just makes you more skeptical of vague notions

Astyx

What do you mean exactly by "cynical" then ?

Semiclassical

17:18

Let's see: Western philosophy is a waste, other people are dunderheads when they don't accept your proof, "no source identified = did not happen"

That's setting aside whether or not I agree with you on a given topic.

Balarka Sen

Skepticism is an essential part of proof-checking in mathematics. But that only comes after you have found the argument/proof :P

Semiclassical

^

otherwise, even if you're right...

Balarka Sen

Otherwise you're just going to stare at nothing at all and contradict the nonexisting :P

Akiva Weinberger

What's the thing? "Convince yourself, convince a friend, convince a skeptic"?

Three stages of writing a proof?

Astyx

Some people like to skip step 1

Secret

17:21

I often did my proof checking like a programmer: By ramming it with any counterexamples I can think of. If the proof crumbles, fix that bug in the proof

Akiva Weinberger

That's the two steps to starting a cult @Astyx

Zee

@Semiclassical or perhaps my views are justified in all of those settings? Western philosophy is bull couse I studied it deeply and found it to be nothing more than intellectual masturbation, no source identified follows from the fact that making a strong claim needs a strong source

Akiva Weinberger

Well, convince your friends to convince other people

Zee

And calling people dinderheads comes from them lacking tact when discussing proofs

Semiclassical

Yes, because people's intellectual capacity is so obviously linked with their civility.

Astyx

17:22

Convince skeptics to convince your friends

Akiva Weinberger

This is making me thirsty

Astyx

Or convince skeptics to convince your friend to convince yourself

Zee

@Semiclassical since when are we in a court room?

Astyx

That's what friends are for after all

Balarka Sen

@Astyx I am unconvinced.

Secret

17:24

I have bias that only those who disagree with me can detect, which is why even those people who disagree with me can be useful in some cases

Astyx

Since when are you a skeptic ? Your opinion does not matter @Balarka :p

Balarka Sen

@Astyx True, but my opinions are convincing.

Doesn't matter if they do not.

Zee

Thus far you have failed to back your criticism, so am calling you unfair

Astyx

Nothing really matters anyway

You guys should listen to Mendelssohn's violin concerto played by Hilary Hahn.

Amazing

Balarka Sen

On the contrary, matter really nothings anyway.

Unconvincing, but inconvenient.

Secret

17:25

@Semiclassical uh, so we solve for a periodic wavefunction and it has a discrete spectrum?

Semiclassical

Nothing really matters on the internet, at any rate

Zee

Sure, that's why you all spend your time here

Semiclassical

@secret depends what we mean by periodic.

Akiva Weinberger

$\Bbb R[[x]]$ acts on $\Bbb R[x]$ and $\Bbb R[x]$ acts on $\Bbb R[[x]]$

Semiclassical

Need I point out how circular that comment is?

Secret

17:26

I am guessing our wavefunction will satisfy $y(x+a)=y(x)$ for some constant $a$?

Astyx

@BalarkaSen You lost me :(

@AkivaWeinberger What's $\Bbb R[[x]]$ ?

Akiva Weinberger

@Semiclassical Well, $\Bbb R[[D]]$ acts on $\Bbb R[x]$ and $\Bbb R[D]$ acts on $\Bbb R[[x]]$

Balarka Sen

@Astyx I am learning to speak Joycean.

Semiclassical

@Secret That's one way. The more generic approach is to consider boundary conditions with $y(L)=e^{i k L}y(0)$.

Akiva Weinberger

@Astyx Formal power series (infinitely long), as opposed to $\Bbb R[x]$, which is polynomials (finitely long)

Astyx

17:28

I see ... good luck with that ! @Balarka

Cool, I didn't know that notation @Akiva

Semiclassical

But if you've got $e^{ik L}=-1$ then $y(2L)=-y(L)=y(0)$, so that gives a $2L$-periodic wavefunction.

Akiva Weinberger

But we could have $D:f(x)\mapsto f'(x)$, and then $D_1:f(D)\mapsto f'(D)$, etc.?

Astyx

I mainly act on instinct

Akiva Weinberger

So $\Bbb R[D_1]$ acts on $\Bbb R[[D]]$ acts on $\Bbb R[x]$?

Semiclassical

What's important here is that the Hamiltonian operator and the translation operator both commute, so every eigenstate has well-defined quantum numbers $\lambda,k$.

Astyx

17:30

I've fallen in love with this music

Semiclassical

Physically: each state has a well-defined energy and quasi-momentum.

Akiva Weinberger

Also, for $A\in\Bbb R[[D]]$, and for $\underline x:\Bbb R[x]\to\Bbb R[x]$ meaning multiplication by $x$,

then $D_1(A)=A\underline x-\underline xA$

Balarka Sen

@Astyx In what context?

Semiclassical

But these two quantities are not independent; if I specify $k$, then there's a discrete spectrum, and vice versa (with $k$ taken mod 2pi/L ) @secret

Astyx

In all context I guess, how else do you expect me to act ?

Semiclassical

17:31

That relation is basically just the dispersion relation for the potential.

Balarka Sen

I dunno, my "subconscious" instincts frequently conflict with my cerebral choices and moral principles.

Semiclassical

What's important for my purposes is that, for a typical potential, there will be 'gaps' in the spectrum.

Balarka Sen

I almost never act on instincts.

Akiva Weinberger

Carpe jugulum

Secret

ah, so that's why I heard of that in solid states stuff, because semiconductors

Astyx

17:35

I've reached a point where my moral principles define my instinct, and my brain doesn't work mostly

Semiclassical

What that means is that there are energies such that, if I were to send in a plane wave with that energy, there's no transmission: the scattering solutions don't propogate.

Astyx

@Akiva Terry Pratchett ?

Secret

yes and I recall in photovoltaics they worry about band gaps a lot, since it determines what wavelength of the sunlight they can capture

Semiclassical

right.

typically one would start talking about the fermi level at this point

But I'm only interested in the scattering problem.

The main question is: How many gaps do you get for a given potential?

(setting aside any physical concerns)

Balarka Sen

I don't think it's particularly possible to have moral principles defining instincts, but maybe that's true vice-versa. I don't trust in human instincts, in any case. They are too basic to be meaningful (whatever that means).

s.harp

17:38

are you doing old quantum theory again

Semiclassical

It turns out that, typically, there's an infinitude of such gaps. However, as you go up in energy they're exponentially narrow. (There's a tunneling calculation you can do via WKB for which that comes out.)

Astyx

I'm not too sure what you mean by instinct then @Balarka

Semiclassical

But when I say 'typically' I really mean "If I write down some explicit V(x)" function as a trigonometric polynomial.

Which is a natural thing to do, because if you write out the Fourier series for $V(x)$ and truncate it, then in the Fourier basis the action of $V(x)$ multiplying $y(x)$ is just multiplication by a matrix with constant diagonals.

Secret

btw why are $k$ called quasi momentum. From what you wrote, $k$ forms a dispersion relation, which is exactly what we expect for momentum of particle states in k space, what makes them quasi?

What happens at the limit where the exponential relation forces all the gaps to bunch up together and merge at very high energy, do we get a continuous spectrum (I am guessing that's where we hit the free propagating case)?

Semiclassical

So you end up being able to make good approximations of the spectrum in that way, at least to leading order for low gaps.

Balarka Sen

17:41

@Astyx It's hard to put words to thought, but what I assume instinct to mean is a mixture, on some proportions, of biological instincts and the mystical "free will".

Semiclassical

@Secret It's quasi-momentum because it's only defined mod $2\pi/L$.

Balarka Sen

What is it to you?

Secret

ah, it's quantised

Semiclassical

It's really $e^{i k L}$ that matters for the boundary condition, after all.

Eh, I wouldn't say even quantised. Just that there's no physical difference between $k$ and $k+2\pi/L$.

Secret

ok I see

Semiclassical

17:43

As for what goes on at high energy, it turns out that you still don't get a truly continuous spectrum.

No matter how high I go in energy, the spectrum at any particular $k$ is still discrete.

Or, at least, that's the typical story.

Astyx

Simply put, what I mean by instinct is "the first thing that comes to my mind"

Balarka Sen

Yeah that's a very very dangerous thing to act on in my opinion.

Or at least for me.

Astyx

Nietzsche would say we're all weaklings obeying the slave's moral cause of Christianity

Secret

@Astyx You can draw trees if you try to plot my instinct, it grows very fast and erratically

Semiclassical

To see where this is coming from, if you had no potential but still demanded the quasi-periodic boundary conditions, then you'd have eigenfunctions $y(x;k)=e^{i k x}$ with energy $\lambda=k^2$.

Astyx

17:45

@BalarkaSen Depends what the first thing that comes into your mind is, really

Semiclassical

However, since $k$ only matters mod $2\pi/L$, there's also going to be branches of solutions given by $\lambda = (k-2\pi/L)^2,(k+2\pi /L)^2,$ etc.

Balarka Sen

That is why I added "for me". The first things which come to my mind, given a situation in time and place, is generally contradictory to my rational (or let's put it as "filtered") thoughts and choices.

It's a perfectly fine life-choice to act on your first-thought. I do not believe in it.

Akiva Weinberger

@Secret The most accessible proof of Morley's Miracle in geometry is "backwards" in a sense

(There's a Mathologer video on it, as a sequel to the fun "Illuminati" episode he did)

Semiclassical

And these branches cross, giving rise to 2-fold degeneracies at $k=0,\pm \pi/L,\pm 2\pi/L$, etc

Balarka Sen

I strongly believe that if I were to exterminate all rational ("filtered") thoughts (as Burroughs would say :P), the thoughts which remain would be basic, degenerate and destructive. I think this is true for every intelligent being but that is just my belief.

Secret

17:49

Here's a snipplet on how "first thought came to mind"works for me:

Jun 6 at 14:08, by Secret

The inspiration pathway on how I came up with this scenario is as follows:
My old computer have not been turned on for a long time$\rightarrow$ suspect there might be spiders living in the CPU due to long disuse$\rightarrow$imagining what the spiders will see when the computer is turned on again: blue highways of light lit in a complex metallic catacomb as if something is going to be activated $\rightarrow$ switching the spiders into people and imagine a futuristic scenario where humanity set up a very primitive civillization because of them forgetting the technological advancement of a lon…

Semiclassical

If I now perturb the problem by adding a potential, each crossing of levels would become an avoided crossing i.e. a gap

Astyx

@BalarkaSen Then I'm not an intelligent being :(

Perhaps we're not meaning the same thing by "first choice" really

Balarka Sen

Or, maybe your first thoughts are not really first thoughts, just subconciously filtered.

Yeah that is what I am guessing.

Astyx

I guess there is a lot of subconcious work I'm not aware of in my head

Or I hope there is at least

Balarka Sen

We are thinking of the same things, yup.

Astyx

17:51

Sniped

Semiclassical

At least, a generic perturbation will do so

But it turns out that there's a special class of periodic potentials for which only the first n gaps open

Secret

Do they look like bessels? (my favourite guess whenever weird periodic stuff came up)

Astyx

That's what I get for being a slow typer

Semiclassical

In which case there will be exactly n gaps in the spectrum

Nah. It's something weirder

Balarka Sen

@Astyx I was glad we are agreeing at that point, not being happy for writing my thought out before you :)

Semiclassical

17:53

Weierstrass elliptic functions provide the simplest cases, I think?

Astyx

I know, I'm kidding around :p

Secret

o, that's quite bumpy, I wonder if the poles have phsyical meaning in this potential

Semiclassical

(This is somehow related to a Pierls transition of the lattice when you include the lattice energy under deformations but I really don't remember )

Secret

ah, I see, then it kinda make sense to have all those poles forming a lattice like pattern

Semiclassical

Well, I think the poles would be off the real line? Potential should still be continuous on the real line

But it is doubly periodic

Balarka Sen

17:56

I think a lot of psychoanalysts from the 19th century have thought about the subconscious process involved in thinking and acting, Freud being one of the primary ones of course. I have not read his works rigorously, but peeked at his tome on dream analysis and some classical Freudian interpretation of Dostoyevsky. I think it was very insightful.

Secret

http://mathworld.wolfram.com/WeierstrassEllipticFunction.html

(Need to zoom in to see whether the red stuff are poles in the Re plots)

Balarka Sen

"Between the conception and the creation
Between the emotion and the response
Falls the Shadow

Life is very long"

Semiclassical

for thine is life is for thine is the

Astyx

Winter is coming

Wait ..

Semiclassical

I remember the presentation in here being pretty okay as far as finite gap potentials: imath.kiev.ua/~symmetry/Symmetry2001/Belokolos273-280.pdf

The connection to pierls, if I'm remembering right, is that if you include both the electronic and the lattice energy, and demand that the system be allowed to deform in order to minimize the energy

Then the one-gap potential pops right out

Secret

18:03

That's a really backward proof. Conway literally act like a businessman mindset handling a project (instead of an academic) by assuming the project (the end goal of the proof) works, and then derive what happened

This is often the pathway that we use in proof by contradiction, except here it lead to a reverse engineering of the problem itself by building backwards from the goal, and thus showed that the proof works

I think I might consider this proof pathway more in my future proofs, seems pretty useful

Balarka Sen

@Astyx When's your next exam? :)

Astyx

Tomorrow

Math and physics, in that order

Semiclassical

And this reflects a Pierls transition from the cis- state of a 1d lattice to a trans- state

Astyx

9am and 1pm

Balarka Sen

Gotcha. When does this ordeal end?

Astyx

18:06

For the Mines, tomorrow. For the whole thing on the 23rd of july

Semiclassical

Eg polyacetylene

Astyx

(The Mines is a school, or rather a group of schools)

Akiva Weinberger

@Secret It's like how you prove the converse of the Pythagorean theorem as well (given a triangle with sides $a$, $b$, and $c$ with $a^2+b^2=c^2$, prove it's right).

Balarka Sen

Ah ok. Still a long time to go :(

Akiva Weinberger

The idea is that you can build a second triangle with legs $a$ and $b$ and a right angle between them

Astyx

18:07

Yeah, I'm not sure I'll attend the Centrale one though

Akiva Weinberger

Normal Pythagoras says the third side is $c$

Astyx

That's the last one, from the 17th to the 23rd

Akiva Weinberger

From the SSS rule for congruence, it's congruent to our first triangle, and so the first triangle is right.

Semiclassical

Anyways. The reason I'm remembering this stuff is that I dimly recall a connection to differential Galois theory

Akiva Weinberger

What Conway's done is similar; build another diagram and show it's similar to the original

Balarka Sen

18:08

When you're all done and dusted, remind me and I'll recommend you good movies to watch.

:P

Semiclassical

From this paper I think: arxiv.org/abs/1011.1642

With the finite-gap potentials being interesting as the only ones which permit an exact solution by quadrature

Akiva Weinberger

@Secret Normally, "working backwards" works because you have a sequence of equivalences (rather than of implications). That's essentially what happened in that old question of yours that you linked to earlier. But Conway's thing seems to be a completely different style…

Astyx

@BalarkaSen Will do ! I've been longing for good movies

Balarka Sen

thumbs

Semiclassical

Anyways, that might open up a new range of special function stuff for you to play with

Secret

18:12

cool thanks

@AkivaWeinberger That kind of thinking is not easy for me because I often reluctant to work with duplicated objects, but I guess I need to start changing that mindset a bit and start to accept duplicated objects

Ok guys I am heading to sleep. Tomorrow I might not be on that much cause I really need to catch up on my chemistry literature review. any interesting questions or challenges, I will answer later

Waiting

18:35

@Hippalectryon I think one can rearrange all and apply Watson’s lemma. On the other hand, I also think it is possible to do it pretty elementarily (I have more ideas to develop further).

Referring to $$\int_1^\infty\sqrt{\frac1{(1-t^2)^2}-\frac{(n+1)^2t^{2n}}{(1-t^{2n+2})^2}}\sim \log(n)/2$$

user193319

Is the following true: Let $(x_n)$ be a sequence of nonnegative numbers diverging to positive infinity. Then every subsequence of $(x_{n})$ diverges to positive infinity.

Astyx

Yes, you don't even need them to be nonnegative

user193319

@Astyx Oh! I see. So if $(x_{k_n})$ is some subsequence, proving the theorem would just be matter of noting that $k_n \ge n$ for every $n \in \Bbb{N}$?

Astyx

If you're thinking what I think you're thinking, yes

user193319

@Astyx Haha Hopefully we are thinking the same thing. Thanks for the help!

Astyx

18:46

Glad, as always

In $M_2(\Bbb C)$, if $A$ is diagonal and $B$ any matrix such that for all $t\in\Bbb C$ $B+tA$ is diagonalizable, how can I show B is diagonal ?

Sha Vuklia

Guys, I want to show Gauss’ formula. The proof in my book starts as follows: Let $d$ be a positive divisor of $n$. We show that the number of elements $\overline a\in\mathbb Z/n\mathbb Z$ with $\operatorname{order}(\overline a)=d$ equals $\phi(d)$ (where $\phi$ is Euler’s function). I’ve already shown that $\operatorname{order}(\overline a)=n/\gcd(a,n)$. So we can write $\gcd(a,b)=n/d$. Now my book says this is equivalent to saying
$$
a=b\cdot\frac{n}{d}\text{ where }\gcd(b,d)=1\text{ and }1\leq b\leq d.

Astyx

19:09

$A$ is not scalar !

Perturbative

Hey everyone

SteamyRoot

@ShaVuklia "I’ve already shown that $\operatorname{order}(\bar{a})=n/\gcd(a,n)$. So we can write $\gcd(a,b)=n/d$.

What is $b$ there?

Sha Vuklia

yea sorry that's a mistake

it should be $\gcd(a,n)=n/d$ @Steamy

never mind i got it

Astyx

19:30

Sanity check : $\begin{pmatrix}1& \alpha\\0&1\end{pmatrix}$ is not diagonalizable when $\alpha \ne 0$ ?

SteamyRoot

@Astyx It's easy to check the dimension of the eigenspace of $1$ there :P

Astyx

Yeah, I'm braindead at the moment

SteamyRoot

@ShaVuklia Well done!

Astyx

Any idea about my question up there ?

Sha Vuklia

@Steamy lol well I got help, though I was already almost there :P

would you have time for a follow up question tho?

on this same proof

SteamyRoot

19:34

Just ask, I'll try :P

Doing two questions at the same time is a bit tough tho

Sha Vuklia

My book proceeds to say that the number of $b$ for which $a=bn/d$ equals $\phi(d)$. I don't see why this is true, because
$$
\phi(d)=\#\{\overline b\in\mathbb Z/d\mathbb Z\vert \gcd(b,d)=1\text{ and }1\leq b\leq d\}.
$$
I see that the conditions on $\phi(d)$ are exactly the conditions for $b$, but in the case of $b$, we are working in $\mathbb Z$, and in the case of $\phi(d)$, we work in $\mathbb Z/d\mathbb Z$. How do we know that those numbers match up?

maybe it has to do with the fact that $1\leq b\leq d$

so it doesn't matter that we work in the cyclic group

we won't "lose" any elements modulo $d$

that must be it I think

ugh I hate myself :P I could have answered both questions myself, but I just didn't see it straight away

SteamyRoot

It's normal not to see things straight away. Just need to have a little patience - and sometimes asking the question and looking at what you just asked, helps you realise how to solve it.

Sha Vuklia

that's true

Astyx

Oh I guess I found a solution

Not sure it's optimal though

SteamyRoot

@Astyx Go on :O

Astyx

19:44

Since $A$ is not scalar, that's the same as saying $B + tE_{1,1}$ is diagonalisable. You can find $t$ such that the caracteristic polynomial is $0$, meaning that $B+tE_{1,1}$ is diagonal (even scalar) and thus $B$ is diagonal

SteamyRoot

What's $E_{1,1}$?

Astyx

First vector of the canonical basis of $M_2(\Bbb C)$

$\begin{pmatrix}1&0\\0&0\end{pmatrix}$

diagonalisable or diagonalizable ? Or none of these ?

Sha Vuklia

hm, I still have one question left. How do we know for sure that $\phi(d)$ equals the number of $b$'s? To me it seems all we've shown is that this number of $b$'s cannot exceed $\phi(d)$ because of the conditions on $b$. However, assume this number is smaller than $\phi(d)$. That means there would be a $b$ such that $\gcd(b,d)=1$ and $1\leq b\leq d$, yet $a\neq b\cdot\dfrac{n}{d}$. This should be a contradiction, but I don't see how.

Astyx

I wrote something I didn't mean

SteamyRoot

@Astyx Depends, whether you prefer British English or some ex-colony's bastardisation ;)

Astyx

19:49

You can find $t$ such that the determinant of the caracteristic polynomial is 0

British English for the win

So s

That only works if $B_{2,2} \ne 0$ though

But you can add the identity so you're good

Works

Or I'm stupid

Sha Vuklia

(oh, oh, maybe I see it)

I've already shown that $b\dfrac{n}{d}\leq a$

Astyx

Does any of you know enough about passivation to teach me ?

lattice

20:23

Hey

Astyx

Hi

lattice

Sorry, I don't know what that is^^

I'm also rather looking for help

we shall describe certain affine schemes and the topological space that they're based on

the first two examples are $\mathbb{Z}_{(p)}:=\{\frac{a}{b}|a,b\in\mathbb{Z},p\nmid b\}$ and $\mathbb{C}\times\mathbb{C}$

So first I should find the prime ideals in those rings and determine the open/closed sets, right?

heather

hello, I'm trying to simplify the expression $\frac{c^n}{c(c-1)^{n-1}}$ and was looking for a little assistance.

I was thinking I could divide by $c$ to get $\frac{c^{n-1}}{(c-1)^{n-1}}$

but I'm not sure what else I could do.

Astyx

Regroup the terms so you get only one exponent

heather

how do you mean @Astyx?

Astyx

20:36

${a^n\over b^n} = \left({a\over b}\right)^n$

heather

ah, I see - so $\frac{c}{c-1}^{n-1}$

so...would that mean I could do $-1^{n-1}$?

Astyx

I don't get what you mean

lattice

No^^

heather

@Astyx dividing by c in the fraction

lattice

$(\frac{a}{a+b})^n\neq b^n$ in general

Astyx

20:39

You get $1\over(1-1/c)^n$

heather

um...why not? @lattice c/c = 1, so 1/-1, so -1...you can't simplify before you apply the exponent?

Astyx

You're not being careful enough

Write it down step by step

lattice

Also $(\frac{a}{a+b})^n\neq b^{-n}$ in general if you mean this

try with $a=b=1$

If we consider $\mathbb{C}\times\mathbb{C}$ as a ring, is the multiplication meant to be coordinatewise?

Astyx

Yes

Dodsy

Hey guys

Astyx

20:48

$(a,b)(c,d) = (ac,bd)$

Hi @Dodsy, how goes ?

Dodsy

and gals

Astyx

gals ?

Dodsy

very well thanks :)

Astyx

Glad to hear so

Dodsy

@Astyx Meaning women, or ladies.

Astyx

20:48

Yeah, but are there any

Dodsy

Heather.

Astyx

Oh right

lattice

then the only ideals are $\{(0,0)\}$, $C\times\{0\}$, $\{0\}\times\mathbb{C}$ and $\mathbb{C}\times\mathbb{C}$, right?

Astyx

I think so @lattice

lattice

okay good

Dodsy

20:50

hm.

lattice

that makes it easier to determine the Zariski topology^^

Astyx

I don't know much about it (meaning I know nothing about it)

lattice

well it's some topology on the spectrum (i.e. the set of prime ideals)

if there are only 4 ideals (3 of which are prime I guess), then there is not much options for the open/closed sets of prime ideals^^

Dodsy

omg haha

I just almost pooped myself.

Was on some forum reading about collatz conjecture variations (turns out even more exist)

and a popup came up

scared me stupid.

Astyx

Heh :p

Dodsy

21:04

it's very sad, I still cannot enrol into courses :C

Balarka Sen

21:42

Cute but easy geometry problem for all of you: Suppose $C$ is a given circle and $AB$ a diameter of it that's drawn on it.

$X$ some random point in the exterior of the circle which lies in the strip given by the tangents to the circle at $A$ and $B$. Can you drop a perpendicular on $AB$ from $X$ just by using a straightedge?

Arrow

Hello
Let $F$ be algebraically closed and $G$ a group. Consider two finite dimensional $F$-linear representations of $G$ which have the same traces. If both these reps are simple, they're isomorphic. If only one of them is simple but they're of different dimension, they need not be isomorphic. But what if one of them is simple and they have the same dimension? Are they isomorphic?

Balarka Sen

Oh, actually there's a second part which tells me to find out if the possible position of $X$ can be more general than "in between the strip given by $T_A S^1$ and $T_B S^1$". I think my solution immediately generalizes to the interior of the circle.

Mike Miller

@Arrow There aren't many people here who are qualified to talk about modular representation theory

Semiclassical

I might be missing something but isn't it enough to draw a line through X which is parallel to the tangents?

Balarka Sen

Ok, yes, pretty sure it extends to everywhere except when $X$ is right on $AB$ (or rather, the extended straightline)

Semiclassical

21:53

Though I guess the point may be that there's no compass available

Balarka Sen

@Semiclassical Technically, you're not given the tangents; $X$ is just positioned that way. But even then, can you draw parallels with just straightedges?

I don't think so.

Semiclassical

Right

Arrow

@MikeMiller ok. I will ask on the main site.

Semiclassical

Will think about it

Balarka Sen

it's acute one

Semiclassical

21:54

I'll try to get it right, then

Balarka Sen

don't come up with a circular proof though

Semiclassical

Pfft, as if I'd be so obtuse

Balarka Sen

Fair and square, patron

Semiclassical

...damn, I lose

Perturbative

Does anyone know a proof that shows that the two definitions of a derivative (as used in diff topology) are equivalent, the first definition is where the derivative is defined as the linear transformation between tangent spaces and the other being the definition of the derivative as a limit?

Balarka Sen

21:58

@Semiclassical Straighten up, no hard feelings.

Semiclassical

Right, I'll be on point next time

Akiva Weinberger

@BalarkaSen Draw from $X$ to $A$, intersecting the circle at $A'$. Draw from $X$ to $B$, intersecting the circle at $B'$. Call the intersection between $A'B$ and $AB'$, $X'$; the intersection between $XX'$ and $AB$ is the desired point

Balarka Sen

Ta-da.

Mike Miller

@Perturbative The second definition is never used outside of single variable calculus (and directional derivatives, once you know your function is differentiable).

Perturbative

@MikeMiller Ah okay, I was wondering because two of the books on Differential Topology that I'm using mention the second definition and then drop it after that

Akiva Weinberger

22:10

@BalarkaSen Oh I have a proof

Assume the standard circle $x^2+y^2=1$ where the diameter is the $x$-axis

For every point $p=(x,y)\in\Bbb R^2$, define $f_1(p )=\frac{x+1}y$ and $f_2(p )=\frac y{1-x}$.

Note that, by similar triangles, $f_1(p )=f_2(p )$ when $p$ is on the unit circle.

Using the same notation as my solution above, I want to prove that $f_1(X)=f_2(X')$ and $f_2(X)=f_1(X')$—that is, they switch.

$f_1(X)=f_1(A')$ (since the value of $f_1$ is constant on lines through $A$) $f_1(A')=f_2(A')$ (since it's on the circle) $f_2(A')=f_2(X')$ (since it's constant on lines through $B$). Thus, $f_1(X)=f_2(X')$.

Similarly, $f_2(X)=f_1(X')$.

Balarka Sen

@AkivaWeinberger I'm sorry, what are you giving a proof of? Uniqueness of orthocenter?

Akiva Weinberger

@BalarkaSen The projection onto the diameter using only straightedge

Thus, $f_1(X)f_2(X)=f_1(X')f_2(X')$.

Balarka Sen

I don't get it. $X'$ is the orthocenter right there, by unqiueness, so $XX'$ is perpendicular to $AB$ by Euclidean geometry.

Akiva Weinberger

That means the value of $\frac{x+1}y\cdot\frac y{1-x}=\frac{x+1}{1-x}$ is equal for both $X$ and $X'$; solving, they have the same $x$-coordinate, so joining them gives a vertical line to the diameter.

@BalarkaSen Orthocenter of what?

Also, remind me which one an orthocenter is?

…Oh, intersection of altitudes

I see. 'Cause $AA'B$ and $AB'B$ are right angles…

…That's a much simpler proof…

Balarka Sen

Right, it's the orthocenter of $XAB$.

Akiva Weinberger

22:19

As you can tell, I'm not good at Euclidean geometry

Balarka Sen

You did the right construction, nonetheless, so cookie for you.

Akiva Weinberger

I still like my proof

Maybe I'm biased 'cause I came up with it :P

Balarka Sen

Let me bookmark it so I can come back tomorrow and read it.

nah, I always learn something from your approaches. Better read it

It's just that right now I'm too busy looking at dank internet memes

Akiva Weinberger

@BalarkaSen Oh, you flatter me ^_^

Balarka Sen

Just reminding myself I should try my best not to vaporize on my admission tests next year. Reading clever proofs is part of that.

Semiclassical

22:32

My inclination is to do a proof using complex numbers, lol

Whether that's a good inclination is another matter entirely

Avantgarde

hello math people

Balarka Sen

hi

Avantgarde

seems like people aren't home

Akiva Weinberger

Is chat home?

Balarka Sen

is homer chat?

good, i am successfully speaking joycean now

Avantgarde

22:44

home sweet home

Transcript for

$Mathematics$ Mathematics