Mathematics - 2018-01-30 (page 2 of 5)

Silent

13:00

@AkivaWeinberger Thank you so much! will you please let me know why $h$ is permutation iff $\gcd(b-a,n)=1$?

Akiva Weinberger

@user777 I think you're allowed to have the zero clique, the clique with zero vertices and edges. So for k=2 and larger, the answer would be yes.

@Silent Got to run again, but when is that map injective/surjective?

Silent

Ok, thinking.

user777

@AkivaWeinberger Are you saying that the graph that I give with k=4 would return "yes"

Tuki

hello

Silent

13:20

If $k>0$ and $n>2k$ then how to show that $n(n-1)\cdots(n-k+1)>\frac{n^k}{2^k}$?

Tuki

if i have $$ \int_D 5\cdot \cos(y)dA \text{ when} D:0\le x \le \pi \text{ and} |y|\le x $$ hmm how would i evaluate this ?

MatheinBoulomenos

@Adeek Neukirch - Algebraic Number Theory

Tuki

Does someone understand the notation ?

i Understand that $D$ stands for domain of integration

if domain is $0 \le x \le \pi$ and $|y|\le x$ what does this mean ?

Akiva Weinberger

@user777 Also, note that you wrote "at most k cliques"

user777

@AkivaWeinberger I think I understand the definition, once we find one that says yes for such k=c, then it would be always yes for k>c (so we don't care about next)

Tuki

13:33

$\rightarrow |y|\le 0 \le x \le \pi ??$

Akiva Weinberger

@Tuki $\int_{-x}^x\int_0^\pi{\rm stuff}~\mathrm dx~\mathrm dy$

Shobhit

@Tuki draw the range of x, then use $|y| \le x \implies -x \le y \le x$, to see the favourable region over the range of $x$.

user777

@AkivaWeinberger the wrong with k=2 and larger, you mean for general graph? if so, then this would be wrong; since the definition says that we "must cover all edges of graph". It seems that the question says that: the smallest number of cliques that covers all edges of G"

Tuki

also i think the order doesn't matter in this ?

$\int_{-x}^{x}$ or $\int_{0}^{\pi}$ shouldn't matter which one you evaluate first ?

but one of them might be easier than the other

Akiva Weinberger

@Tuki Oh sorry I wrote it wrong

There shouldn't be any variables in the outer limit, because there shouldn't be any variables in the final answer

So it's $\int_0^\pi\int_{-x}^x dydx$

Tuki

13:41

Yes makes sense

Akiva Weinberger

@user777 Your graph only needs 2 cliques, doesn't it? v1v2v3 and v2v3v4

Tuki

so actually the order does matter

It's getting bit confusing

philmcole

Hey does anybody know WolframAlpha a bit?

I'm strugglin to compute the Fourier Coefficients of a function

user777

@AkivaWeinberger no, the smallest number of cliques are 3: which are v1v3v4, v1v3, and v1v2.

philmcole

I want the integral to look like $a_k = \int_0^1 f(x) e^{-2\pi i k x} \, dx$ but I can't see how I tell this to WA.

Meow Mix

13:46

hi

Tuki

Hi @MeowMix

Akiva Weinberger

@user777 v1v3v4 isn't a clique. (v1,v4) isn't an edge

Unless you misrote

Tuki

@philmcole you mean like this ?

wolframalpha.com/input/?i=integrate+(f(x)*e%5E(-2piik*x))+from+0+t‌o+1

hmm how to tell this chat this whole string is link

you write this to wolfram alpha

integrate (f(x)*e^(-2piik*x)) from 0 to 1

it also interprets i as imaginary unit

as intended i guess ?

user777

@AkivaWeinberger sorry you're right!

it is with 2. (I made mistake! sorry for misleading you) But I don't know: Is it ok to share edges?!

philmcole

@Tuki Thanks, that seems to work. I tried it with the FourierSeries command but this has some presets I don't understand yet

Balarka Sen

13:53

@TedShifrin @Eric I have done this problem before. You can construct a subspace $\text{Imm}_{\pitchfork}(M^n, N^{2n}) \subset \text{Imm}(M^n, N^{2n})$ of immersions $f : M \to N$ with only double points such that at every double point $f(p) = f(q) = x$, $df_*(T_p M) \pitchfork df_*(T_q M)$ (i.e, immersions with transverse double point singularities)

This subspace is an open dense subset.

anakhronizein

Hi Balarka

Balarka Sen

The proof is pretty much identical to the transversality theorems and stability theorems in usual transversality theory

Hi @anakhronizein

quallenjäger

If a function $f(t)$ is continuous on the interval $[t,t+\delta]$, is it possible to derive the estimate $\sup_{u\in[t,t+\delta]}|f(u)-f(t)|\leq C\delta$ for some constant $C$.

Or do I need more than continuity of $f$=

PVAL-inactive

14:17

@quallenjäger No it should be easy to come up with counterexamples.

quallenjäger

As I thought

but uniform continuity would be sufficient?

Akiva Weinberger

14:32

@quallenjäger What about $\sqrt x$ on $[0,\delta]$?

You want a single $C$ to work for all $\delta$, right?

quallenjäger

Yes

Hmm, I will have $\delta$ in square root right?

What would I need to have this inequality work?

user193319

If $H$ and $K$ are normal subgroups of order 2 and 3, respectively, why is $HK$ an abelian group? I don't see it. I could use a hint.

Akiva Weinberger

@quallenjäger You'd need $\sqrt\delta\le C\delta$ for this problem, or $\frac1{\sqrt\delta}\le C$, which is impossible as $\frac1{\sqrt\delta}\to\infty$ as $\delta\to0$

Perhaps if it's Lipschitz continuous?

quallenjäger

Or continuous differentiable?

Akiva Weinberger

@quallenjäger Actually, you know what, I think your thing is equivalent to Lipschitz continuity, if the inequality holds for all $t$.

Or, no

Locally Lipschitz, maybe

if you just need small enough $\delta$

(ex: $x^2$ is locally Lipschitz but not Lipschitz)

quallenjäger

14:42

The sup is the problem, but I think with lipschitz continuity it might work.

Akiva Weinberger

@quallenjäger Continuous differentiability implies local Lipschitz continuity

quallenjäger

Yes, the thing is I don't know how to proof with Lipschitz continuity:D

*locally

Akiva Weinberger

@user193319 Choose an element $h\in H$ and $k\in K$. Consider $khk^{-1}$. Since $H$ is normal, this is in $H$.

If you prove $khk^{-1}=h$, you're done (as this is equivalent to $kh=hk$). That's your hint

quallenjäger

$sup_{u\in[t,t+\delta]}|f(u)-f(t)|=sup_{u\in[t,t+\delta]}|f'(t_1)(u-0)-f'(t_2)(t‌-0)|= sup_{u\in[t,t+\delta]}|f'(t_1)-f'(t_2)|(u-t) \leq C(t+\delta-t)=C\delta$

Akiva Weinberger

(And remember that $H$ only has two elements)

quallenjäger

14:49

The last inequality follows because $f'(t)$ is continuous and the set is compact

Is that right?

Akiva Weinberger

Looks like it

quallenjäger

@AkivaWeinberger I think you are right, need locally lipschitz

user8469759

if $f : A \to B$ ($A,B$ topological spaces) and $f$is a continuous, and $V$ is a connected open set in $B$, can I say that $f^{-1}(V)$ is also connected in $A$?

Silent

Please someone help here:

2 hours ago, by Silent

If $k>0$ and $n>2k$ then how to show that $n(n-1)\cdots(n-k+1)>\frac{n^k}{2^k}$?

Akiva Weinberger

@user8469759 No. Consider $f(x)=x^2$, $\Bbb R\to\Bbb R$. Let $V=(1,4)$.

Then $f^{-1}(V)=(-2,-1)\cup(1,2)$.

@Silent If $n>2k$ then $k<\frac n2$, and $n-k+1>n-\frac n2+1=\frac n2+1$${}>\frac n2$

Thus $n(n-1)\dotsb(n-k+1)>\frac n2\frac n2\dotsb\frac n2$

${}=(\frac n2)^k=\frac{n^k}{2^k}$

user8469759

15:01

@AkivaWeinberger forgot to add $f$ is a bijection

Silent

@AkivaWeinberger, thank you so much!

Akiva Weinberger

@user8469759 Still no. The classic example of a bijection which is not a homeomorphism is $f:[0,2\pi)\to S^1$ (where $S^1$ is the unit circle) defined by $f(\theta)=(\cos(\theta),\sin(\theta))$.

Or $f(\theta)=e^{i\theta}$ if you think of it as the unit circle in the complex plane

user8469759

and I add the homeomorphism?

Akiva Weinberger

A small neighborhood around the point $(1,0)$ will have a disconnected inverse image.

@user8469759 If it is a homeomorphism then yes, it will be connected

user8469759

proof?

Akiva Weinberger

15:04

Homeomorphism means that the inverse function $f^{-1}$ is continuous as well

user8469759

yes

Akiva Weinberger

This makes $f^{-1}(V)$ the image of a connected set in a continuous function

and the image of a connected set in a continuous function is always connected

user8469759

did not think of that

ok

just out of curiosity

why is the image of a connected set also connected?

in case of continuity

Akiva Weinberger

By way of contradiction, suppose not. This would mean that the image is the disjoint union of two (nonempty) open sets.

Take the preimage of those two open sets.

The preimages would be nonempty, open, and disjoint as well

meaning that the domain would be the disjoint union of two nonempty open sets, i.e. it would be disconnected. Contradiction

user8469759

why the preimage of two disjoint set is disjoint?

Akiva Weinberger

15:11

$f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$ in general

($x$ in the first set iff $f(x)\in A\cap B$. $x$ is in the set iff $f(x)\in A$ and $f(x)\in B$. These are equivalent)

user8469759

ok

makes sense

Akiva Weinberger

If $A$ and $B$ are disjoint, then the we have $f^{-1}(A)\cap f^{-1}(B)=f^{-1}(A\cap B)=f^{-1}(\emptyset)$${}=\emptyset$ so they're disjoint

Silent

@AkivaWeinberger, I still can't see why $\gcd (m,n)=1$ implies that $x\to mx (\pmod n)$ is bijective, where x ranges from 1 to n.

Akiva Weinberger

\pmod looks better in LaTeX

Silent

ok

Akiva Weinberger

15:17

Suppose $mx\equiv my\pmod n$. Then $m(x-y)\equiv0\pmod n$, or $n|m(x-y)$. Since $n$ and $m$ are coordine, we must have $n|x-y$, or $x\equiv y\pmod n$. Thus, it's injective

It's surjective because an injection between two finite sets of the same size is automatically a surjection

Alternatively: You can compute the multiplicative inverse of $m$ mod $n$ when they're coprime. $x\to m^{-1}x\pmod n$ would be the inverse function in that case

and since it has an inverse function it's bijective

*coprime, not coordine

Silent

Thank you so much! @AkivaWeinberger

anakhronizein

@BalarkaSen if I have an embedded hypersurface S, what does it mean exactly for a vector field X to be "transverse to S"?

Balarka Sen

@anakhronizein I suspect it means $X|_S \oplus TS = TM|_S$?

anakhronizein

That's what I was thinking, but I never know exactly what people mean when they say transverse.

Is there a good treatment on transversality and the general theory of it?

Balarka Sen

The point is that the flowlines of $X$ are transverse to $S$. Or rather, $S$ is a section of the foliation given by the flowlines of $X$

anakhronizein

15:31

Yeah, I guess it does work in this case.

Balarka Sen

@anakhronizein I just learnt whatever that is in Guillemin-Pollack

But I picked up various other notions of transversality in different contexts I guess

anakhronizein

I guess it might be a mathematical maturity thing.

Eric Silva

sup nerds

Balarka Sen

could be

anakhronizein

By the way, I have never asked. What exactly do you do, @BalarkaSen ?

Balarka Sen

15:33

hey @Eric

@anakhronizein Ah, er, that's a little complicated.

Let's say that I am not into academia

anakhronizein

You have a lot of points in algebraic topology on math.stackexchange but everything else is a bit cryptic about your user page. ;)

I can see why, academia sucks.

Eric Silva

academia is -2/10

Balarka Sen

I think my first serious exposure to mathematics was learning algebraic topology, so yeah, that's the tag I roam around the most

Mike Miller

academia is garbage

i will probably do it but it’s a machine that crushes people

Eric Silva

what we need is a machine that crushes machines

and a machine to crush that machine

and so on and so forth

Balarka Sen

15:37

sorry I should stop dropping that joke randomly

Tuki

do you know anyone from here that would be from rus ? @BalarkaSen

Balarka Sen

Hm, no, I think we don't have a chat regular who's from Russia

anakhronizein

Can academica be fixed?

Or is it too far gone?

Balarka Sen

i think the proportion of students vs jobs is a serious issue that would take a nontrivial idea to fix

anakhronizein

Because almost everyone I have actually talked to has more or less told me that academia sucks, yet majority of these people are in academia.

user8469759

15:44

please give me your insight about this

0

Q: PDF of a point in a curve and of a point in a surface.

Let $\gamma : (\alpha,\beta) \to \mathcal{C} \subset \mathbb{R}^3$ a bijection (or maybe stronger an homeomorphism) representing a parametrized curve. Suppose $t$ is random variable distributed in $(\alpha,\beta)$, I want to derive $p_{X_1,X_2,X_3}(x_1,x_2,x_3)$ (namely the pdf of a point $(X_1...

what do you reckon?

Eric Silva

@anakhronizein tbh at least in the US i think the system would just break before it could be fixed

Tuki

another thing is how you would try to fix it ?

anakhronizein

Can we speed up the breaking of academia, perhaps so that it fixes itself by the time I get my Ph.D.?????????

Tuki

your doing PhD right now ? @anakhronizein

anakhronizein

16:06

No. :P

I am not even sure I want to do my Ph.D. I feel like I suck at math.

Trey

how to solve $\int \frac{sin^3(\sqrt{x}}{\sqrt{x}})$ ?

anakhronizein

Well first try substitution with $u=\sqrt{x}$

Mr. Xcoder

Is it just me or derivative calculus is far easier than integral calculus?

Trey

@anakhronizein I can't see how that would help

Akiva Weinberger

@Trey What would $\mathrm du$ be?

anakhronizein

16:20

Well did you try it?

Akiva Weinberger

@Mr.Xcoder Differentiation is mechanical

Trey

oh, screw what I said, it worked beautifully

Mr. Xcoder

@AkivaWeinberger mechanical?

Tuki

@Mr.Xcoder usually derivative is the easier direction

anakhronizein

#1 thing you should try when you are doing an integral is to substitute $u=\text{uglything}$ when there happens to be an ugly thing.

As I tell some students: $u$-substitution stands for "$\text{ugly}$-substitution".

Tuki

16:25

also @AkivaWeinberger what do you mean with differentiation is mechanical ?

anakhronizein

@Tuki he probably means that the rules of differentiation allow you to just algorithmically differentiate a function.

Whereas the rules we teach for integration do not allow you to do this.

Tuki

but these rules don't apply to all possible functions

anakhronizein

No. But for the vast majority of them they do.

And you can tell when you are able to do so easily. Whereas with integration you have some guess work or ingenuity you must use.

Tuki

probably

do you have example of function that is easy to compute derivative but almost impossible to integrate ?

numerical integration is usually option too ^^

user777

@AkivaWeinberger I just want to point out that this problem is Clique Edge Cover, so you cannot take the edge twice, therefore we cannot take k=2; because we will get two triangle sharing one edge between them!

Tuki

16:30

I mean that is there function that is easy to derivate but it would be extremely hard to inverse the derivate i mean "antiderivate" == indefinite integral

Xander Henderson

$\frac{1}{\sqrt{1+x^2}}$ ?

I mean, differentiation is generally comparatively easy

antiderivatives are quite hard in general, and most functions don't even have elementary antiderivatives

Tuki

This reminds me of how Diffie-Hellman key exchange algorithm works

It's easy to compute in one direction but hard to reverse

I don't know if you have read about this @XanderHenderson ?

Xander Henderson

@Tuki I don't know that specific algorithm, but I think I know the general idea. Unfortunately, I think that the question of antiderivatives is more subtle.

user193319

0

Q: Prove that every non-abelian group of order 6 has a non normal subgroup of order 2.

I tried using the result that if p is the least prime dividing the order of G, then any subgroup of index p is normal in G. So a subgroup H of order 3 is normal in G. Then, if there was a normal subgroup K of order 2 then G=HK. But how do I get a contradiction?

abstract-algebra group-theory finite-groups

Xander Henderson

In the case of a encryption/decryption scheme, there is a right answer

Tuki

16:40

yes i agree

Xander Henderson

in the case of integration, most functions don't actually have elementary antiderivatives

for example, $\int t^t \,\mathrm{d}t$

or $\int \mathrm{e}^{|t|^2},\mathrm{d}t$

Eric Silva

differentiation making sense is a way stronger regularity condition on functions

integration making sense is basically true for any function that isnt bullshit

Xander Henderson

@EricSilva But even when you have as much regularity as you could ever want, integration can be impossible

like, why is $\int \sqrt{1-t^4}, \mathrm{d}t$ so hard?

Eric Silva

integration in the sense of finding antiderivatives i guess

Tuki

I am not sure how elementary function is defined ?

Eric Silva

16:45

it's weird tho cuz numerically integrating is easy

MatheinBoulomenos

Integration in the sense of finding definite integrals can also be quite hard

Eric Silva

@MatheinBoulomenos i mean symbolically this is very frequently the problem of either finding a trick or finding an antiderivative

both of which can be arbitrarily hard

quallenjäger

Can I actually think ODE as kind of discretization scheme of the solution, which converges to the originial solution?

Balarka Sen

i dont understand what that means

quallenjäger

For example, $x'(t)=f(x)$, then the left hand side can be thought of $lim \frac{x(t+\delta)-x(t)}{\delta}$ and this is kind of discretization scheme of $x(t)$ as $x(\delta)=x(t)+\delta f(x(t))$

MatheinBoulomenos

16:54

@BalarkaSen I heard a talk about Quillen K-theory, I thought you might find that interesting

Balarka Sen

@quallenjäger Ehh sure I guess

You're really thinking of a difference equation, not a ODE, then. I mean that's like a discrete dynamical version of an ODE

@MatheinBoulomenos What's a Quillen K-theory?

quallenjäger

Ok but in limit, this difference equation converges to the ODE?

pointwise or uniform?

MatheinBoulomenos

@BalarkaSen it's basically algebraic K-theory, but the construction is more general than that

it doesn't only work for rings

Balarka Sen

@Mathein I see

MatheinBoulomenos

What you do at least for rings is that you start with a ring R, then you look at the category of finitely generated projective modules over R, then you turn that into another category which is complicated to define, then you take the nerve of that category, which gives you a simplical set, then you take the geometric realization of that which gives you a (giant) CW-complex, then you take the homotopy groups of that and that's the K-groups of R for you

Balarka Sen

16:59

Oh that is a cool construction

MatheinBoulomenos

it's quite remarkable that you start with an algebraic object and end with an algebraic object, but you pass through a topological space at some point

Balarka Sen

Well the category of f.g. projective modules over R is like the category of vector bundles over a space

I'd like to see the intermediate algebraic object (the category you say is complicated to define)

MatheinBoulomenos

Okay suppose that we have an exact category C, i.e. an additive full subcategory of an abelian category that is closed under extensions

that basically allows us to talk about exact sequences in C, although C itself doesn't have to be abelian

exact sequences in C are exact sequences in the surrounding abelian category such that all the terms are in C

Akiva Weinberger

@Tuki https://en.wikipedia.org/wiki/Elementary_function

MatheinBoulomenos

we call a monomorphism/epimorphisms admissable if we have an exact sequence in C such that they are the first or the last term, respectively

Balarka Sen

17:04

@MatheinBoulomenos mmk

Akiva Weinberger

Basically, combinations of +,-,x,/,exponentition,logarithms,trig

Mike Miller

@Balarka Keyword Q-construction

Akiva Weinberger

@Tuki I read about this recently. It is cool.

MatheinBoulomenos

yes, that's basically what I'm describing

In QC you take the objects of C but a morphism from X to Y is a triangle X <- Z -> Y, where X <- Z is an admissable epi and Z -> Y is an amissable mono

and then you compose these triangles by taking pullbacks

Akiva Weinberger

Have you read about RSA as well? It's a very similar idea and also cool @Tuki

Tuki

17:06

yes i have @AkivaWeinberger

Balarka Sen

thats a strange category

anakhronizein

"transversally orientable" <--- what might this mean (when describing a surface)?

BAYMAX

Hi chat!

MatheinBoulomenos

Hi @BAYMAX

Balarka Sen

@anakhronizein Prolly that it's co-orientable (ie admits an orientable normal bundle)

BAYMAX

17:09

I had a question which I ma not sure of answer exists or not:)

quallenjäger

Does bounded derivative imply continuous derivative?

BAYMAX

like the average of the distance of any two points inside a square?

quallenjäger

Because, by Darboux-Theorem, if the derivative is not continuous, at least one side limit does not exists.

Is there any counter example to bounded derivative but not continuous derivative?

MatheinBoulomenos

when you take R to be the ring of (real or complex) continuous functions on a compact Hausdorff space, you get the topological K-theory. (This is probably an application of Serre-Swan at some point)

Balarka Sen

Yup, category of fg projective C(X)-modules is isomorphic to the category of vbs over X

where X is cpt hsdrf

il spk lk ths frm nw on

Daminark

17:14

@BalarkaSen my health deteriorates by the moment

anakhronizein

@BalarkaSen thanks, that was my first guess but then he kept on using it despite co-orientable being a thing.

Balarka Sen

@dmnrk i hp u gt wl sn

MatheinBoulomenos

Hi @Daminark

For Dedekind domains you get some exact sequences that arise from a bunch of category theory that is done just to give you some homotopy fibrations at the point before you take the homotopy groups

Daminark

How's it going Mathein?

MatheinBoulomenos

these exact sequences are really non-obvious even for the low-degree K-groups which you can define algebraically

Fine, thanks. I'm having exams soon. How are you doing?

Balarka Sen

17:18

@MikeMiller So I proved that for differentiable functions $f : S^1 \to S^1$ with $f'$ of bounded variation, if the suspension foliation of $f$ in $S^1 \times S^1$ has no compact leaves, has to be topologically conjugate to rotation by an irrational angle.

Daminark

Midterm season is coming but otherwise well

Balarka Sen

This shows that the Denjoy blowup cannot be done $C^1$-ly in particular

MatheinBoulomenos

in the seminar, it takes like over 3 hours of lemmas about category theory to get to these fibrations

Balarka Sen

Which we wondered about a year ago

Mike Miller

ooooh cool

GPhys

17:23

hey @AkivaWeinberger

Are you doing well? Applying to Uni yet?

Tuki

in fact i am applying to uni

or at least trying ^^

hmm can you compute derivatives numerically ?

or more like compute limits numerically ?

MatheinBoulomenos

17:40

Hi @Ted

Ted Shifrin

Howdy

Balarka Sen

Hey @Ted

Daminark

Hey Ted!

Balarka Sen

waves vigorously

Ted Shifrin

@quallenjäger: Bounded derivative needn't imply $C^1$, no. Take the usual $x^2\sin(1/x)$ example on $[-1,1]$.

Tuki

17:41

hey @Ted

Ted Shifrin

Howdy @Balarka, Demonark, Tuki.

The Great Duck

@Tuki not really

Ted Shifrin

you mean rigorously, @Balarka?

The Great Duck

consider $\sin{x}$ as $x$ increases without bound

Balarka Sen

I only handwave

Not rigorously

Abcd

17:42

Please tell me the mathjax code for single bond on Chemistry.se.

The Great Duck

not only does it not exist, but it is not computable either

@Abcd no

go ask on chemistry

:p

Abcd

hmm

Tuki

yes this is what i thought

The Great Duck

@Tuki a naive model of a hypercomputer would likely be a computer that can compute any limit.

Abcd

@TheGreatDuck no body's online there... that's the problem

The Great Duck

17:44

@Abcd well then google latex for that.

Shobhit

hi @TedShifrin

The Great Duck

mathjax is just latex formatting

in your browser

Shobhit

as soon as u come, we all bombard you with questions, XD @TedShifrin

Ted Shifrin

I'm hiding.

Hi, DogAteMy.

Akiva Weinberger

Hi

The Great Duck

17:46

if two functions' derivatives are linearly independent, then are the original functions?

Akiva Weinberger

@GPhys Yeah, I got into Yale

How have you been?

GPhys

nope, but I'm nearby (I'm at NYU now for my PhD)

Akiva Weinberger

Ah

I'm in Manhattan right now

waves

MatheinBoulomenos

@TedShifrin apparently there was a misunderstanding or a change of topics on the projective geometry seminar. They changed the title from "classical projective algebraic geometry" to "classical projective geometry" and the announcement says that they're doing stuff like quadrics, cross-ratios, Pappus's theorem, Desargues's theorem and "weak Bezout theoerem", whatever that means

Balarka Sen

@Akiva Hm, why are you still in high school if you got into Yale?

MatheinBoulomenos

17:47

No varieties, I'm disappointed

The Great Duck

@AkivaWeinberger light waves back. Note: digital signals are light.

Balarka Sen

Is it some phase difference between the first semester?

Ted Shifrin

Ah, that's far more elementary, Mathein. I have most of that stuff in Chapter 8 of my Algebra book, and have taught a good deal of it the last month of my algebra course.

Akiva Weinberger

…I have to graduate first?

Ted Shifrin

DogAteMy: Most probably not, actually :P

Balarka Sen

17:48

Oh you can apply to schools before graduating high school?

MatheinBoulomenos

Yeah, I might not take the seminar

Akiva Weinberger

@BalarkaSen Everyone does

Ted Shifrin

You always apply a year early, Balarka. Same for grad school.

The Great Duck

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Q: Are two functions $f$ and $g$ with linear independent derivatives neccessarily linearly independent and vice versa?

Suppose that there exists functions $f$ and $g$ defined on the real numbers and differentiable everywhere. If their derivatives $f'$ and $g'$ are linearly independent on some nonzero interval then are $f$ and $g$ linearly independent on the same interval? Similarly if $f$ and $g$ are linearly ind...

linear-algebra differential-equations functions

Balarka Sen

That's not how it works over here

Ted Shifrin

17:48

I mean, you apply December to start the following September.

So you have a dead year?

The Great Duck

the answer was in an earlier version but I changed it to better capture what i intended

MatheinBoulomenos

Here you don't apply (or can't apply) before you finished high school

The Great Duck

i think the answer still works but im having a hard time interpreting it.

MatheinBoulomenos

for the bachelor, every uni here takes math students anyway, assuming your high school grades are not terrible

Balarka Sen

Well no the high school finals are in like the end of March/beginning of April, the university/college admissions are a month of two after that, and the first semester starts at August/September

Ted Shifrin

17:49

So is acceptance immediate within a month? Here it takes months, and then you have to get ready to move miles away.

Oh, so things happen way fast. But in the US acceptance drags on for months (some students have to wait 'til the very deadline to hear from the school they want).

Balarka Sen

Yeah it's too fast

MatheinBoulomenos

There are certain majors/university combinations that just automatically accept everyone who meets the criteria, math is typically one of them, at least for the bachelor

Akiva Weinberger

There is the phenomenon of "senioritis" where after getting accepted students kinda stop caring about school

MatheinBoulomenos

for medicine or psychology, it can be difficult to get accepted

Akiva Weinberger

I mean, we still have to not fail, but still

Ted Shifrin

17:52

US is just totally different. Most students don't know their major when they apply to college. Those that do most often change a few times.

Akiva Weinberger

We don't have to try über-hard

Ted Shifrin

(My experience teaching the calculus theory courses was that typically the kids who came in saying they wanted to major in math were not as strong as the ones who ultimately switched to math during or after my courses.)

The Great Duck

@AaronHall hello. I'm not causing any problems, am I?

by posting my question?

MatheinBoulomenos

There are quite a bit of students switching subjects here, too

especially in the first years between physics and math or something like that

The Great Duck

@TedShifrin I added a Math major but I've never switched from my original. It does amaze me how many people have been here for at least 6 or 7 years and still not graduating.

Ted Shifrin

17:54

Senioritis is not uncommon, DogAteMy. I expect it's worse than usual at your school because it's such a pressure cooker.

BTW, @Balarka, do you know how to do this correctly?

The Great Duck

Whether I actually care about the grades themselves is irrelevant. I like the material to such a degree that I'd probably have to intentionally fail to do bad.

well aside from the one class where the professor gave a 200 problem exam (an exaggeration but it was ridiculously overkill; only one student managed to finish it with more than a 70%).

and almost got fired

MatheinBoulomenos

one of my friends started studying in math in Heidelberg and then his plan was literally to transfer exactly to that university where his girlfriend got accepted for French-German studies, because you can get into math anywhere here, unlike French-German studies apparently (also his grades were really good)

Ted Shifrin

But I assume some unis are way stronger in math than others, Mathein.

The Great Duck

^^^^

MatheinBoulomenos

yeah, but he wanted to live with his girlfriend

The Great Duck

17:58

uuuh

MatheinBoulomenos

and it doesn't matter that much for the bachelor

Daminark

I guess that's fair if there's not a notable quality difference

The Great Duck

if he is doing very good then he should stay where he will do the best he can

happyEddie

We have a variety V in the projective plane defined by the polynomial Y^2Z - X^3 - Z^3, is the mapping from V to the projective plane defined by [x:y:z] -> [x^2:xy:z^2] a morphism? The only troubling point is [0:1:0]...

The Great Duck

sacrificing his education for that is a bad idea

unless they're getting married or something

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$Mathematics$ Mathematics