Mathematics - 2019-08-27 (page 1 of 2)

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12:54 AM

0

Q: The mysterious numbers $ \frac{13}{20} $ and $20$?

Let $g(x) = x^6 - 30 x $ Let $h(x) = x^6 $ Let $f(x) = x^2 - 2 $ Let $r$ be a reduced fraction $0 < \frac{p}{q} < 2 $ with integers $p,q > 1$ Let $f_{n+1}(x) = f(f_n(x)) = f_n(f(x)) , f_0(x) = x$. Now consider for n going to infinity the following ' averages ' : $$ a(r,v) = \lim n^{-1} \su...

convergence dynamical-systems closed-form average complex-dynamics

1:18 AM

@ALannister so, i guess the issue is when a function is only strictly increasing

Hello! I am having some trouble finding the particular solution of $2a_n+5a_{n+1}=0$, $a_0=3$

My teacher says it is $a_n=3(2/5)^n$ but I say it has to be $a_n=3(-5/2)^n$

1:44 AM

I found the characteristic equation: $2k+5=0\implies k=-\frac{5}{2}\implies a_n=C(-\frac{5}{2})^n$

But haven't come up with the solution that the teacher gave

2 hours later…

3:51 AM

@ALannister I'm not totally sure tbh. the fact that J^T*J is singular at the optimal point doesn't bode well but it's not clear if it's fatal for Gauss-Newton or just suboptimal.

@manooooh: Did you try writing out the first few terms explicitly? It looks like $a_1 = -6/5$, so you're both wrong.

4:06 AM

@ALannister one thing I would suggest re-checking is your computation of $J$: I don't get a symmetric matrix, though I do get a singular matrix. If I were to hazard a guess, the symmetric J you report in the post is actually $J^\top J$ (i.e. the Hessian)

So I'd look at that part with fresh eyes.

@TedShifrin thanks!

4:21 AM

any use for an equation that takes in a number and whose solutions are that number and its reciprocal?

4:48 AM

@ALannister doing some numerical testing (in excel, lol), it does seem to be the case that the slow convergence is due to Hessian being singular at the optimal point

no matter what initial condition $\beta=(1,1,1)+(\delta x_1,\delta x_2,\delta x_3)$ I start with, numerical iteration quickly drives it to $\beta\approx (1,1,1)+\delta(1,1,1)$, and (the second) $(1,1,1)$ corresponds to the zero eigenvector in the Hessian

in other words, no matter what initial condition I pick, the convergence to the optimal point seems to be dominated by the behavior along that direction

and convergence along that (1,1,1) direction is sloooow

@TedShifrin hello Ted

The answer gave by the teacher says: $2a_n+5a_{n-1}=0$ with $a_0=3$ $\implies$ $a_n=\left(\dfrac{2}{5}\right)^n\cdot3$. Proof: base case: $n=0$ implies $a_0=(2/5)^0\cdot3=3$

Inductive step: inductive hypothesis: $a_n=(2/5)^{n}\cdot3$ inductive thesis: $a_{n+1}=(2/5)^{n+1}\cdot3$. Proof: $a_{n+1}=\dfrac{2}{5}\cdot a_n=\dfrac{2}{5}\left(\dfrac{2}{5}\right)^n\cdot3=\left(\dfrac{2}{5}\right)^{n+1}\cdot3$

It seems that $5/2\to-\dfrac{1}{5/2}=-2/5$... Coincidence or big typo?

5:06 AM

@manooooh $n=1$ case gives $2a_1+5a_0=2a_1+15=0\implies a_1=-15/2=3(-5/2)^{1}\neq 3(2/5)$

so yeah, that seems wrooong

plus, $2a_n+5a_{n-1}=0\implies a_n=-\frac{5}{2}a_{n-1}$

oh, wait

4 hours ago, by manooooh

Hello! I am having some trouble finding the particular solution of $2a_n+5a_{n+1}=0$, $a_0=3$

11 mins ago, by manooooh

The answer gave by the teacher says: $2a_n+5a_{n-1}=0$ with $a_0=3$ $\implies$ $a_n=\left(\dfrac{2}{5}\right)^n\cdot3$. Proof: base case: $n=0$ implies $a_0=(2/5)^0\cdot3=3$

these are not the same recurrence relation

@Semiclassical sorry wanted to say $5a_{n-1}$. My bad

ok. in that case I return to $a_n=(-5/2)a_{n-1}\implies a_n=(-5/2)^n a_0$

Your profs answer would be correct if the recurrence relation were $5a_n=2a_{n-1}$

@Semiclassical ya. So $a_n=3(2/5)^n$ is not a valid solution since if $n=1$ then $2a_1+5a_0=0\iff2a_1=-15\iff a_1=-15/2=3(-5/2)^1\neq3(2/5)^1$?

right. the base case is right, but any $a_n =a_0 r^n$ will get the base case right

Thank you!!

5:12 AM

np

Yup, I had a feeling there was a typo in the problem somewhere, @manooooh.

doesn't absolve the prof if it was reported right, though. if you've got $pa_n+qa_{n-1}=0$ then the solution had better have a minus sign in it

but meh. it's easy to miss these things in transcribing

I usually try to put the indices in increasing order when I write these recurrence equations.

same

I like when it's in the form $a_n=...$

MR. @Erico!

5:18 AM

(or $a_{n+2}=...$. not too picky about that)

@TedShifrin you mean $n,n+1,n+2,\ldots$?

Yeah, just to minimize confusion. And when you use linear algebra to solve these, you need to pay careful attention (just like with your solution).

actually, come to think of it, I usually do mine in descending order. not sure why

i guess because I write polynomials like x^n+... and not 1+...

Ok, thank you! You both helps a lot

I'm almost surely not as consistent as I claim to be.

5:22 AM

i'd quote the emerson line about 'foolish consistency' but I don't remember it off the top of my head right

oh, I think I came up with a tidy way to understand that one problem with awful notation

The matrix/subspace one?

yeah.

OK.

if one can establish $V=A\oplus B^\perp$ under the proper assumptions, then one has $v=a+b_\perp$ and one can verify that $Iv=v$ for their $I$

so once one gets the subspace stuff right, then the formula itself is just trivial

still haven't written up the solution tho

That seems logical to me.

5:27 AM

yeah

I just gave a solution to a question earlier today that was based on a sum decomposition and orthogonal complements. (Of course, it was one where the "given" answer to an exercise was wrong in the first place.)

lol

Isn't the only reasonable hypothesis to get that that their intersection is trivial? Or are you trying to state it in terms of $A$ and $B$?

basically, yes. in the original matrix language, it's that $\alpha_\bot'\beta_\bot$ is full rank

Wait, both with perps?

5:31 AM

yeah.

So no nontrivial element of $B^\perp$ is orthogonal to somebody in $A^\perp$.

right. and $A$ itself is the subspace of vectors orthogonal to $A^\perp$, since it's just finite dimensional

So it'd have to be a nontrivial element in both $A$ and $B^\perp$

Gotcha. Right. So that does it.

I still have issues with the non-uniqueness despite the notation ...

no disagreement there. it's annoying

Probably an engineer or statistician :P

5:34 AM

worse: econometrics

LOL

That's stat :)

yeah

for context, the book the OP cited was "Likelihood-based inference in cointegrated vector autoregressive models"

so that's certainly stats, albeit probably with a very financial context?

Hell if I know. :)

I guess the right way to think of $\alpha_p$ is some full column rank 'representative' of $A^\perp$, under the equivalence $\alpha\sim \beta$ when $\text{Col}(\alpha)=\text{Col}(\beta)$

Right. In my writing, I just said things like "a matrix of appropriate rank whose column space is the subspace $V$" ... or whose columns give a basis for $V$, better.

5:39 AM

It's the latter here, since it's got full column rank

not necessarily an orthonormal basis but definitely a basis

Well, "appropriate" meant maximal for the given shape.

ahhh

everything in here has more rows than columns, so full column rank is maximal

I would have said "an $m\times n$ matrix of rank $n$."

But, anyhow .... enough.

quite

the weird part is that, if you go back to the literature on this (which I did) they'll actually denote $\alpha_\perp$ as an orthogonal complement of $\alpha$.

like, they make a point of that at the end of the intro to the paper I looked at

but yeah, no point jawing about it

Orthogonal complement of a matrix?

5:42 AM

yeeeep

Um, NO. I forbid that crap.

quotation from another paper by the same author:

"From Johansen (1996) we adopt the following definitions: for an $m \times n$ matrix $a$ with full column rank $n$, we define $\bar{a} = a(a'a)^{-1}$ and let the orthogonal complement of $a$ be the full rank $m \times (m - n)$ matrix $a$ that has $a'_\perp a=0$." @TedShifrin

where Johansen 1996 is the book I mentioned

sooooo yeah

You have a typo in $\bar a$, I think. This should be the projector matrix?

also, lol at "the" orthogonal complement, as though it's unique

No, that's just horrendous.

5:48 AM

i'm on a pc without easy chatjax atm, so probably

yeah, it's pretty awful

Anyhow, I would excoriate the author(s).

On that note, g'night :)

night

6:41 AM

Can anybody help me in my math problem

I know that sinx is may one function ,but f(x)=2x+ greatest integer x+sinx cosx is onto

7:26 AM

Morning all

@yuvrajsingh If you draw the graph, you should see that g(x)=2x+floor(x) has "gaps" of height one. wolframalpha.com/input/…

Since sin(x)*cos(x) has values between -1/2 and 1/2, it is too small to "close" the gaps.

So using this intuition I would say that it's not onto - and graph in WA seems to confirm it.

But I do not see immediately what would be the best strategy to actually prove that's not onto.

Maybe showing first that the function is increasing would help. For the function without floor, you can simply differentiate.

As a side note, there are also rooms called basic mathematics and calculus and analysis - in case you would find them useful. You can find some other chatrooms here: List of chatrooms.

8:07 AM

@MartinSleziak it make difficult for me to prove function to be onto ,how to prove function to be onto ,yiu refer in your answer that sinxcosx value to small to fill gap what you want to imply by this

I suppose you can sketch the graph of g(x)=2x+floor(x). (And the graph is also linked above.)

sinxcosx value to small to fill gap what you want to imply by this

sinxcosx value to small to fill gap what you want to imply by this

@yuvrajsingh When you look at the graph, you see that this function is increasing and that at zero there is a "jump". You have g(0)=0 but g(x)<-1 for negative values, right?

So the function g(x) is not onto, the values from the interval (-1,0) are not attained.

Do you agree so far?

Yes

@MartinSleziak

Ok.

And your function is f(x)=g(x)+sin(x)*cos(x).

If you can show that f(x) is non-decreasing, you can use the same argument before. (Had I noticed that, I would have started with this argument.)

Anyway, even without monotonicity, at least we have some intuition why f(x) is not surjective.

You know that values in (-1,0) are not attained by g(x).

Does sin(x)*cos(x) help us to get these values?

8:18 AM

No

You can notice that |sin(x)*cos(x)|<=1/2.

So for positive x you have g(x)>=0 and f(x)>=-1/2.

Yes

For negative x you have g(x)<0 and f(x)<-1/2.

Yes

In fact, I should have written that for positive x you have g(x)>0 and f(x)>-1/2.

So -1/2 can only by attained for x=0. But we have f(0)!=-1/2.

In any case, if you show first that f(x) is non-decreasing, then you get a more straighforward argument.

8:22 AM

Sir i differentiate it it is greater than zero

Yes, and therefore increasing.

Hence nit onto

I have one more question

So if we know that f(x) is increasing and if we notice that f(0)=0 but f(x)<-1 for x<0, we can see that it's not surjective.

Can i ask

(1/2)^modx =x^2-a maximum possible value of a for this to be solution i sketch the graph

@MartinSleziak

Sorry i have written wrong value of a for which it has maximum solution @MartinSleziak

Since I have linked to some chatrooms, I will mention that I have seen some people from India around the rooms JEE Maths zone and CSIR-TIFR-ISI-NBHM. Just in case you want to check those roomse.

@yuvrajsingh I do not understand what you mean by "(1/2)^modx".

In any case, since I have some stuff to do IRL, I'll leave this for somebody else.

Have a nice day!

8:56 AM

15 hours ago, by MadSpaceMemer

To be honest I dont even know why I bother so much with these stupid futile questions, since I dont even study mathematics. :D

Flagged as stuck up troll, violating be nice policy

9:54 AM

Let V denote a homogeneous system of m linear
equations in n unknowns over the field Q of rational numbers. Suppose solution set over Q is infinite and suppose there is another field F for which Q is subfield of F then show that if solution set over Q infinite iff solution set over F is infinite.

Can anybody help?

10:10 AM

@LeakyNun math.stackexchange.com/questions/3335723/…

my attempt

10:20 AM

@Mathein Venjakob hat mir grad auch den Minikurs empfohlen lol

MatheinBoulomenos

10:49 AM

@ÍgjøgnumMeg ah, hattest du schon Kontakt mit Venjakob?

11:01 AM

@MatheinBoulomenos hab i di gsagt, dass i bin nach USA gomme

MatheinBoulomenos

@LeakyNun cool! I don't recognize the language though

I'm just trying to speak in your dialect lol

MatheinBoulomenos

I ko scho a bissle Schwäbisch schwätze, aber net gscheit, mei mudder is deitschlehrarin un es heißt muddersproch...

ich rede eigentlich nur Hochdeutsch, höchstens ironisch Dialekt

11:17 AM

@Leaky i ka dialekt! Aber ned da gliche als wia @Mathein

@Mathein und ja ich hab ihm eine mail geschrieben

@Mathein quick question: I wanna prove that $2^{p-1} \equiv 1 \bmod p^2$ iff $2$ splits completely in the degree $p$ subfield of $\Bbb Q(\zeta_{p^2})$. I think I can do it easily with the Artin symbol but I wonder if there's an obvious way that I'm missing?

MatheinBoulomenos

11:33 AM

@ÍgjøgnumMeg I don't see an easy way without the Artin symbol

@ÍgjøgnumMeg ANT1 is going to be pretty easy for you, I think

Okay good lol

yey

@Mathein the local fields stuff is really what I need from it

but it's good to get a refresher anyway

MatheinBoulomenos

@ÍgjøgnumMeg don't worry, ANT2 is going to be pretty tough (not for you specifically, but in general)

since you do proofs of LCFT

Yeahhhhh

and GCFT?

MatheinBoulomenos

we did that in "ANT3"

but that's not always offered

ah fair :)

MatheinBoulomenos

11:47 AM

but there's going to be some follow-up to ANT2, they always have a specialization module in NT following that

can't wait to start scheming!

MatheinBoulomenos

@LeakyNun what are you going to take at MIT?

AG1, LA, LG, NT1, AT1; AG2, LGR, NTS, NT2, AT2; ich hoffe

where L is Lie and R is Representation and S is Seminar

noch net registiert

MatheinBoulomenos

sounds really cool!

sounds really hard!

MatheinBoulomenos

11:53 AM

yeah, that too

@MatheinBoulomenos does "des" mean "therefore"?

MatheinBoulomenos

no

> Christ lag in Todesbanden,
für unsre Sünd gegeben,
der ist wieder erstanden
und hat uns bracht das Leben.
Des wir sollen fröhlich sein,
Gott loben und dankbar sein
und singen Halleluja.
Halleluja.

fifth line?

MatheinBoulomenos

I'd interpret that as dependent on "fröhlich". We shall be happy about that

what does "des" mean there?

MatheinBoulomenos

11:58 AM

in modern German it would be "dessen", it's a demonstative pronoun in the genitive

hmm

MatheinBoulomenos

apparently, "des" can mean "therefore" according to Grimm, but it's mostly middle high German, which that song isn't

Grimm is giving "des laszt uns alle fröhlich sein" as an example of fröhlich + genitive

woerterbuchnetz.de/cgi-bin/WBNetz/…

12:18 PM

Des is how I say das

ha

Akiva Weinberger

1:04 PM

Take any right triangle. Draw the altitude, splitting it into two smaller right triangles. Erase the smaller one. Repeat, drawing the altitude and erasing the smaller resulting triangle. What happens in the limit?

Do you eventually erase the entire triangle?

Also I kinda wanna see this illustrated

i would say so, yeah. after every step, the right triangle you get is similar to the original one but scaled down by a factor

and since the scale factor $r$ won't change, the area will go like $r^n\to 0$

@ALannister rerunning my code in Mathematica instead of Excel, I get much faster convergence. not sure why that's going on but I trust MMA > Excel (I probably had a typo in the latter)

Akiva Weinberger

Oh right yeah

so this isn't actually the picture I want

I'm trying to make a visualization of $x+x^2+x^3+\dots$

but this is actually $x+x(1-x)+x(1-x)^2+\dots$ isn't it

one notable thing about your function, btw: if you approach $(1,1,1)$ along the (1,1,1) direction (i.e. $(x_1,x_2,x_3)=(1,1,1)+t(1,1,1)$) then you can check that $f(1+t,1+t,1+t)=t^4$

Alessandro Codenotti

1:30 PM

I have a Radon measure $\mu$ on a topological space $X$ and I know that the support of $\mu$ is a discrete subspace of $X$. What are nontrivial examples of such situations? (Where by trivial I mean a sum of Dirac measures of points in the support)

I'm trying to understand how nice must $\mu$ be under this hypothesis

Actually I think there are no other examples

something thats bothering me, if I have a metric on $U\times P$ (which is _not_ a product metric), then is
$$d( [x_0, p_0] , [x_1, p_1] ) ≤ \inf_{x\in U} d([x, p_0], [x, p_1])$$
(everything is path connected)? A simpler abstraction would be if I had two curves $\gamma_0, \gamma_1$ in a metric space and the question would be:
$$d(\gamma_0(0), \gamma_1(1)) ≤ \inf_{t\in [0,1]}d(\gamma_0(t), \gamma_1(t))$$

(with $\gamma_1$ and $\gamma_0$ never intersecting, otherwise its trivially false)

(ok the abstracted problem is almost always false, lol)

Alessandro Codenotti

What do you mean with "is not a product metric"?

And with the square brackets?

its not of the form $d_U + d_P$ or $\max(d_U, d_P)$, the topology induced by it is the product topology tho

square brackets are just to distinguish points from the brackets of the metric

too many brackets of the same type screw notation badly

Alessandro Codenotti

Oh ok, so they're just ordered pairs

This seems very false to me

MatheinBoulomenos

1:48 PM

@AlessandroCodenotti is it sufficient to show that on a discrete space all Radon measures are (infinite) linear combinations of Dirac measures? This follows directly from inner regularity

since an infinite sum of non-negative terms is the supremum over all finite "subsums"

Alessandro Codenotti

Like look at the usual metric on $\Bbb R^2$, which is not a sum or a max and the points $[0,0]$ and $[1,1]$, then the distance is sqrt(2), while the inf is 1

@MatheinBoulomenos I agree that this is true, but I don't know whether it's enough

That's precisely my doubt

@AlessandroCodenotti oh man i meant $≥\inf$

Alessandro Codenotti

Oh ok, seems believable now, but I don't know whether it is actually true

$≤\inf$ is gonna be false for almost every metric

2:14 PM

decided to ask it on the main site math.stackexchange.com/questions/3335947/…

1 hour later…

3:23 PM

(so the result is false as you can see by deforming the unit square in an appropriate way, maybe once I have time I'll answer my own question.. hope the result I thought I needed this for isn't actually false..)

3:53 PM

user image

This is a symmetric representation of the first few primes relative to the even numbers

the primes are white lines, and the evens are blue lines

the divergence is $0$ around a small circle at the center of the diagram

some things are beautiful but meaningless

3

4:38 PM

example: Plato's Model of the Universe

google.com/…

kepler's polyhedra for the solar system:

entirely silly as a model, but still pretty

Let $H$ be a subgroup of $G$. How to show that $Ha^{-1}=Hb^{-1}$ implies $Ha=Hb$?

Is there any easier way that something like multiplying cosets $HaHb=Hab$?

The sensible starting point for a problem like this is to write out what it means to be a member of $Ha^{-1}$ and $Hb^{-1}$ respectively

that said, I'm not totally convinced this is true in general

i mean, one does have $Ha^{-1}=b^{-1}H\iff bH=Ha$

But non-commutativity seems like it should create issues

(maybe if $H$ is a normal subgroup?)

4:59 PM

@Semiclassical Ok. I have been thinking counterexample, but nothing occurs to me

anyone know how to find the order of the set of all 2 by 2 matrices with positive determinant?

well, on the assumption that one needs a non-normal subgroup

simplest example would be the order-2 subgroups of S3 (i looked it up)

ok.

checking...

0

Q: Isomorphic sets?

I want to define a homotopic flow on a countable set $S\subset\Bbb N$. I'm thinking of the set $S\subset\Bbb N$ as being continuously mapped s.t. after a countable amount of time, the sets completely overlap and have exactly the same elements. Q: Given an arbitrary set $S,$ how can I define ...

calculus reference-request transformation

how can I improve this question? I wrote it pretty quickly..

I wrote a program that prints odd-step OEIS collatz sequences recursively, but I don't know how to simplify and optimize my code or prove it works (although the first terms seems to work)

5:02 PM

did you solve the collatz conjecuter?

no sorry

sorry

@silent i think that does give rise to a counterexample, but you should try for yourself. (you're looking for $a,b$ such that $Ha^{-1}=Hb^{-1}$ and seeing if $Ha=Hb$ as well)

here is my code gist.github.com/Prince-Polka/f18139318d525a6c8f123ec1134f641a

@Semiclassical no.

5:09 PM

thank you for that entirely helpful and revealing statement.

@Silent can you convert "$Ha = Hb$" into $\underline{~} \in H$ where _ is a blank you should fill in?

@LeakyNun In that blank, $ba^{-1}$ must be there

and what is $Ha^{-1} = Hb^{-1}$ equivalent to?

$b^{-1}a\in H$

does one imply the other?

i.e. does $ba^{-1} \in H$ imply $b^{-1} a \in H$?

hint: let $b=ca$

5:13 PM

no. thank you very much

Emanuel Frátrik

6:04 PM

hi i need help with proof

lim as n-> inf [n*(An)!=0] as n-> inf => sum (An) diverges

(An) is sequence

@EmanuelFrátrik: So write down what it means to say $\inf na_n\ne 0$. Do we know $a_n\ge 0$ for all $n$, by the way?

@Semiclassical The conjugate subgroups are the three generated by the reflections across vertices of the equilateral triangle.

Emanuel Frátrik

a_n is just general sequence

Are you sure? If there are negative terms, then the inf will always be negative.

And then you can certainly write down a convergent series.

Emanuel Frátrik

6:19 PM

we know that lim n*a_n != 0 a from that we should prove that sum of a_n is divergent

I just told you that's false if you don't assume $a_n\ge 0$.

I'm not going to say it a third time.

Emanuel Frátrik

ok thanks

Oh wait. I misread because you're typing without ChatJax. I read infimum originally. You're assuming $\lim_{n\to\infty} na_n$ exists and is nonzero?

Then it is correct that the series diverges. Sorry I misread your typing.

Do you know the limit comparison test?

Emanuel Frátrik

yes

i know it

So write down $\lim na_n = c\ne 0$, and what does limit comparison tell you?

6:25 PM

In general “life of semiclassical” news, I’m going back to working for my uni

Emanuel Frátrik

ok now i know how to prove it thank u very much

As a “teaching specialist” rather than teaching assistant

You're welcome. Sorry I misread. Please learn a little ChatJax to type math in here :P

Well, that's better pay, @Semiclassic !

Actual lectures rather than labs?

Translation: full-time TA work rather than half-time

Emanuel Frátrik

of course but i was little confused of CatJax

6:26 PM

@Emanuel: If you know any LaTeX, it's basically the same.

@TedShifrin did you see my one question like a week ago about the linguistics of Green's functions?

Well, @anakhro, we also have Green's identities. Same thing, I assume.

No, it’s just full-time TA. (Grad student TAs are half-time)

Oh, that's crummy, @Semiclassic. You should be more ambitious about job-hunting.

yeah, I’m not proud of it

6:28 PM

But it's good for the short term.

Yeah

This is one problem with you guys not having any teaching experience as grad students.

Lab TA is just not the same as doing some problem sessions or lectures.

Agreed.

Emanuel Frátrik

I used to use LaTeX but when u don't work often in it, it goes out of your head :D

It isn't too hard to remember dollar signs and backslash lim, @Emanuel :P

6:29 PM

One fortunate thing is that it looks like all four sections will be for the same prof + course

So that reduces the headache. (Being a servant of multiple profs is not fun)

@Semiclassic: I wonder why they don't have enough grad students to enslave. Too much grant support?

MatheinBoulomenos

I'm curious about the language. In Germany we have "tutors" which hold exercise sessions and grade homework and we have "teaching assistants" (typically just one per lecture) which help the lecturer make the exercise sheets etc.
But I guess in English you call what we call "tutors" or officially "Übungsgruppenleiter" also "TA"?

Yeah, I dunno. Seems like they had a crunch this year

It's Green's identities, but that's closer to like "Euclid's axioms", rather than calling any number of functions the same name "Green's function" but in a different context.

In the US, TAs typically run problem sessions for large lectures, @Mathein. The instructor writes the problems (more like picks problems out of the book or out of an electronic problem bank).

6:31 PM

Well, this is a physics TA position so a bit different beast

At UGA they don't have large lectures, so TAs actually get trained to teach their own small sections of precalculus and calculus and then teach them.

@anakhro: But it's a Green's function for a given differential operator.

Context.

MatheinBoulomenos

@TedShifrin I don't know what to put in an English CV, then. "Tutor" to me sounds more like one-on-one informal tutoring

Yes, @Mathein, that is what it suggests. I would recommend you just put "ran problem session and graded homework for [name of course]."

Indeed. Usually the nomenclature tends towards calling it something like "a Green function"

MatheinBoulomenos

@Ted thanks, that is the most precise

6:33 PM

I understand your complaint, @anakhro, but that just sounds horrible :P

Tobias Kildetoft

@MatheinBoulomenos A common piece of advice is to write the official title with an explanation of what it covers

For us it breaks down as: the prof lectures four times a week. Students have weekly 1h group problem work and one 2h group lab work, both in their assigned section of (at most) 20 students

@Mathein: You can put how many students, too, if you are so inclined.

MatheinBoulomenos

@TobiasKildetoft the official title translates to "problem session conductor" or something clunky like that

Is that the problem then, that "Green" is not a good name to treat with this convention?

6:35 PM

So four sections translates 70-80 students, with 4hrs of discussion time and 8 hrs of lab time throughout the week

Tobias Kildetoft

@MatheinBoulomenos I would probably write it as something like "TA ([German title]) - [explanation]"

So, definitely going to have to treat this as my job for the time being (aka no big research projects)

MatheinBoulomenos

@TobiasKildetoft thanks

Tobias Kildetoft

@Semiclassical Well, presumably is will be your job.

Yeah. Nothing figurative about it

6:37 PM

Let $f$ be a linear functional from $\Bbb R$, a vector space over $\Bbb Q$ to $\Bbb Q$. Does there exist any such non-zero functional?

I went like this: if $f(x)=r$ for some $r\ne0$ and $r\in\Bbb Q$, then $f(cx)=cr$ for $c\in \Bbb Q$. Hence, $f$ maps to a dense subset in $\Bbb R$. I don't know how to proceed further.

@Tobias: Is it really so customary in Europe for undergraduates to be given TA duties?

Senior undergrads here can work as intro TAs

Tobias Kildetoft

@TedShifrin Yes. I started in the beginning of my second year

It's relatively rare in the US, Semiclassic.

Though they’d be quarter-time

MatheinBoulomenos

6:38 PM

@Silent yes, there is a non-zero such functional

@Silent: Does it have to be continuous?

Tobias Kildetoft

Though it seems less common in Aarhus than in Copenhagen for example, so it is very dependent on the location

MatheinBoulomenos

you can just pick a basis for $\Bbb R$ and then define it on the basis

I TA'ed as a sophomore

@TedShifrin no

@MatheinBoulomenos omg! :)

MatheinBoulomenos

I think it depends on the location here in Germany as well. But here in Heidelberg we have a lot of people taking intro LA and intro real analysis (all the physics undergrads have to take that + math majors + some CS majors), they couldn't handle that with all masters students.

Emanuel Frátrik

6:43 PM

Sorry for disturbing guys but i need help with next proof :D

Seems like there should be some pretty trivial non-zero maps from rationals to rationals, though maybe I’m missing something

Oh, maybe I parsed the question wrong

MatheinBoulomenos

we're looking at maps from reals to rationals

Oh, rational combinations of reals

Emanuel Frátrik

we have decreasing sequence a_n , a_n>0 for all n , sum a_n is convergent...i have to prove that lim n*a_n = 0 ... Excuse my confusing interpretation but its for me harder to use ChatJax

i know that sum a_n converge => lim a_n=0 but don't know how to use it

MatheinBoulomenos

@EmanuelFrátrik do you know proof by contraposition?

Emanuel Frátrik

6:50 PM

hmmm sounds helpful, ima try it

thanks for advice it works :D

$f:E\to R$ linear, what is f(B(a,r)) please

$f(B(a,r))=\{f(x), ||x-a||<r\}$

can i go more than this ?

@Ted yeah I TA'd in undergrad too

lol

Also hey @Ted @Mathein et al.

i should change my name to "et al"

Or al

MatheinBoulomenos

hi @ÍgjøgnumMeg

wus gooood y'all

6:58 PM

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