Mathematics - 2018-05-18 (page 1 of 2)

Eric Silva

00:00

i wonder how someone who go about writing a book like "examples in analysis" or top or alg and so on

Antonios-Alexandros Robotis

@EricSilva counterexamples in analysis is a thing

Eric Silva

i feel it would either be too short or too big

ik

i was riffin on it

Antonios-Alexandros Robotis

oh haha :P

MatheinBoulomenos

somtimes interesting examples require knowledge from another field

Like you can define class groups in terms of commutative algebra, but I'm not sure you can compute a single nontrivial class group without algberaic number theory or algebraic geometry

Ted Shifrin

That's still algebra :P

anakhronizein

00:04

non-associative, commutative magma ---> rock paper scissors

Eric Silva

@Ted I saw a copy of ur book sitting in a bookstore the other day and bought it

multivariable one

Ted Shifrin

damn, Eric, you shouldn't be wasting your money!

Eric Silva

it was like 10 bucks

the bookstore was closing

anakhronizein

yeah I am waiting for it to get published as a dover book

Ted Shifrin

oh wow

Eric Silva

00:06

and selling everything for ridiculous prices

Ted Shifrin

I doubt that'll ever happen, @anakhronizein ... at least while I'm alive.

anakhronizein

Heh, it's a joke, Ted.

Ted Shifrin

Oh, that's interesting. Who's in charge of my books when I'm gone?

anakhronizein

Your great grand daughter

Ted Shifrin

That's the empty set.

anakhronizein

00:07

Grand daughter????????????

Ted Shifrin

You're barking up the wrong trees.

anakhronizein

I was hoping to.

Ted Shifrin

well, congratulations, Eric. Now you can complain first-hand :P

anakhronizein

Life's too short to be on Ted's good side.

Eric Silva

lol

Ted Shifrin

00:11

There's no need to interact with Ted at all.

anakhronizein

It's okay, Ted. We all love interacting with you.

Ted Shifrin

Um, very false.

anakhronizein

Well I can only speak for myself.

Ted Shifrin

And that's fine with me.

Eric Silva

@Ted why is this book SOOO expensive usually

anakhronizein

00:16

Because he likes making $$$$

Ted Shifrin

I hate it, Eric. I truly do. I tried making Faustian deals with editors and they were never taken seriously.

No, @anakhronizein. It's totally out of my control and I get very little.

anakhronizein

Did you try publishing with Dover?

Did you try publishing freely on the internet?

Ted Shifrin

Notice that I left my diff geo notes in .pdf form for free ... despite about 5 offers from publishers.

Eric Silva

i feel like this price range is like more than id ever pay for a book that isnt like a rare book

Ted Shifrin

Part of the textbook game is to have publishers market your book so that people who wouldn't know otherwise find out about them.

All math books have gone crazy in the last decade, Eric.

For a one-year course it's not that out of the ordinary.

Eric Silva

00:18

ya but that feels bad to me

Ted Shifrin

You've lost touch with what textbooks are like.

anakhronizein

Dover maintains cheap prices.

Eric Silva

it feels v sketchy when books cost this much

Ted Shifrin

Dover does no marketing or advertising, and I'm pretty sure Dover doesn't do first-time publication.

anakhronizein

Dover does first-time publishing

Guaranteed.

Ted Shifrin

00:19

I don't disagree at this point, Eric. As I said, I definitely opted to keep my diff geo text free.

I'm not sure they want to do any typesetting stuff, @anakhronizein. They want pre-produced stuff. Maybe LaTeX has changed that, but originally they worked with the original typeset versions.

anakhronizein

LaTeX has made it very easy.

Eric Silva

a huge portion of dover books are public domain too

anakhronizein

Easy, breezy, beautiful. Dover. (tm)

Ted Shifrin

It's still nontrivial. Trust me. I've given my publishers LaTeX files in all three cases.

It still took a lot of work to get the book produced.

anakhronizein

Yeah it's fine. I am haranguing you needlessly.

Ted Shifrin

00:21

Yeah, you're f***ing obnoxious.

Besides, my 112 lectures are totally free, so screw the book.

anakhronizein

Others seem to publish their textbooks semi-independently fairly easily with low costs, though. I will ask around for you in case your contract expires and you want to change publisher.

Eric Silva

i feel like in this era we're at the point where if we want to provide math textbooks to people as cheaply as possible we should just post them on the internet like @Ted has done w his diff geo notes

Ted Shifrin

Contracts don't have a time limit of which I'm aware. With aggressive action on my part, I could probably get rights back.

anakhronizein

It depends on the contract. You would know what you signed though.

I just have had friends who have publishing deals expire, but that was not for math books.

Ted Shifrin

I actually didn't keep copies of the contracts when I moved, sadly. I'm pretty sure they're permanent and specifically say that if I'm incapacitated or dead they can get someone else to be an ex-post-facto coauthor.

anakhronizein

00:24

Hott uses Lulu for their publishing, it seems. lulu.com

Hardcover, $30 canadian, 600+ pages.

Eric Silva

publishers sound scary

anakhronizein

Not bad, I'd say.

Publishers are scary. They take advantage of mathematicians who don't know any alternatives.

Poor mathematicians. :(

Ted Shifrin

Well, some mathematicians actually do try to make serious money from their books. That wasn't the market I went into. But plenty do.

anakhronizein

Some "applied" mathematician is currently making money off of your books instead. :(

Ted Shifrin

OK, I'm done with this.

anakhronizein

00:27

You were done before, but you keep on coming back for more of my jokes.

Yeah, Lulu.com seems to advertise towards teachers who want to make textbooks.

Neat. I might consider binding some notes one day.

Note to self: apologize to Ted when he is more receptive. :(

Faust

01:25

@TedShifrin thanks for the suggestions

DarkRunner

01:57

I'm struggling to understand why the method for finding the centroid of a quadrilateral works.

I've read that the centroid of a system must lie on the segment connecting the centroid of it's individual parts

But I have no idea why this has to be

If anyone could help me understand that would be great

DarkRunner

02:23

I asked my question here:

0

Q: Centroid of quadrilateral on coordinate plane

I'm having trouble understanding the motivation between finding the centroid of a quadrilateral. Q: Find the centroid of a quadrilateral with vertices at (-8,12), (7,15) (13,-9), and (-2,-3). I've solved the problem, using the procedure described @How can I construct the centroid of a quadrilat...

Daminark

03:02

@Mathein do you know any decent sources for Galois theory? I'm not liking D&F all too much

Ted Shifrin

04:09

Demonark: Ian Stewart has a wonderful paperback on Galois theory.

Emil Artin also has a very short paperback book, much older.

MultivaluedPersonality

Does anyone know a result called Schur's lemma about n by n matrices? It says something about maximum number of matrices with a certain property (symmetric something something) being exactly something like $(n+1)(n+3)$ or similar.

What about Field and Galois Theory by Patrick Morandi?

Daminark

Hmm, thanks, I'll look into both of your suggestions!

MultivaluedPersonality

04:25

Ah, found the result I was looking for. It had nothing do with symmetric matrices. xD

Daminark

Lol yeah the only Schur's lemma I've heard of was something about irreps

So I was just like whut

MultivaluedPersonality

That's what Google kept telling me lol.

This one says the maximum number of linearly independent mutually commuting complex matrices of order $n$ is $\lfloor n^2/4\rfloor +1$.

Daminark

04:42

Interesting

Secret

04:53

[Random]

in The Factory Floor, 3 mins ago, by Secret

Consider a timeline with an unusual topology such that each second as measured in the rest frame is like:

Alessandro Codenotti

05:29

@Mathei I noticed that $N(\alpha)=2^3$ and argued with norms and factorisations that it must either $\mathfrak p_1^3$ or $\mathfrak p_2^3$

And since $3$ is prime that must also be the order

MatheinBoulomenos

06:42

@Daminark I know the book by Artin that Ted mentioned and it's very good

I also liked the chapters in Aluffi on fields and Galois theory. For more advanced stuff you can also look in Lang. If you want videos, there's a course on Galois theory on coursera: coursera.org/learn/galois (you have to register, but you don't have to pay if you don't want a certificate)

Daminark

07:00

Nifty, thanks!

Tobias Kildetoft

I had to track down the calculation of an example yesterday, since I was missing an argument for whether a certain coefficient should be $1$ or $2$. Ended up finding a reference to Jantzen's 1973 thesis, which was fortunately on the shelves in our library. Then it turned out that the argument in it was not quite thorough enough for me to follow.

But after showing it to Jantzen, it only took him like a few minutes to recall that he had that part covered by an earlier example in the thesis.

That's some impressive remembering

Balarka Sen

How's it going

Daminark

07:24

Yo

Maneesh Narayanan

08:11

Prove that there is an additive function $T: R → R$ (A function $T: V → W$ between vector spaces $V$ and $W$ is called additive
if $T(x + y) = T(x) + T(y)$ for all $x, y ∈ V$.) that is not linear. How to give example. I tried to find the example. I couldn't. How do I find? $T: R → R$ $T(x+y)=T(x)+T(y)$ But we need $T(cx)\neq cT(x)$ for some $x\in R$

@secret do you have some idea?

Secret

Hmm...

Maneesh Narayanan

Can you please help me?

Secret

I am thinking

MatheinBoulomenos

@ManeeshNarayanan choose a $\Bbb Q$-basis for $\Bbb R$, then define a map $\Bbb R \to \Bbb Q$ by sending each element in $\Bbb R$ to the sum of its coefficients when written as a sum in that $\Bbb Q$-basis

Maneesh Narayanan

okay.

What do you meant by $\mathbb Q$-basis?

@MatheinBoulomenos

MatheinBoulomenos

08:15

$\Bbb R$ is a vector space over $\Bbb Q$

so it has a basis as a $\Bbb Q$ vector space

sockevalley

Hi!
Im trying to prove that language $ L_1 = \{ a^i b^j c^k | 0 < i < j < k\} $ is not context free.
Is the string $ a^p b^{p+1} c^{p+2} $ a good choice to pump up or down the 5 cases?

Secret

T(x+0)=T(x)=T(x)+T(0) thus T(0)=0. T(c0)=0=/=cT(0)=0?? Uh...

why is the value of T(0) makes no sense under additive nonlinearity

Uh what?

11

Q: Does a nonlinear additive function on R imply a Hamel basis of R?

A function is additive if $f(x+y) = f(x) + f(y)$. Intuitively, it might seem that an additive function from R to R must be linear, specifically of the form $f(x) = kx$. But assuming the axiom of choice, that is wrong, and the proof is rather simple: you just take a Hamel basis of $\mathbb{R}$ a...

analysis logic set-theory axiom-of-choice

How does axiom of choice allow us to bypass that problematic T(c0) case?

MatheinBoulomenos

@Secret what do you mean?

We only need $T(cv) \neq cT(v)$ for one $v$, not for every $v$

Secret

O, then T(c0)=cT(0)=0 has to hold, I see

Alessandro Codenotti

The function will be linear when restricted to $\Bbb Q$ and as long as you fix $T(1)$ you have it on the whole of $\Bbb Q$

sockevalley

08:25

It seems to me that this guys answer is wrong. Since he doesn't from the start pick a proper string?
https://math.stackexchange.com/questions/566724/using-the-pumping-lemma-to-prove-l-aibjck-mid-i-j-k-is-not-conte

Secret

I should study deeper on how axiom of choice allow the T(cv)=/=cT(v), as in, how these v are produced to introduce nonlinearity, hmm...

Maneesh Narayanan

08:42

@MatheinBoulomenos How do we prove that this vector space is infinite? Can you give one infinite L.I set?

MatheinBoulomenos

I could, but why do you even need that it's infinite for the argument?

note that a finite-dimensional vector space over $\Bbb Q$ is countable

Maneesh Narayanan

I was searching for the dimension of this type of vector space. Actually, I couldn't prove it. That is why I asked.

@MatheinBoulomenos

$\{1,2^{\frac{1}{n}},...,2^{\frac{n-1}{n}}\}$. I am not able to find infinite one

@MatheinBoulomenos

sockevalley

09:00

Anyone know how to apply the pumping lemma to CFL in automata theory?

Balarka Sen

10:28

Flaccid Sheaf

33 3

algebraic-topology differential-geometry differential-topology geometric-topology spin-geometry

what a username

Arrow

Suppose $\overline U\subset \mathring V\subset X$. Will the inclusion $U\subset V$ be a cofibration?

Alessandro Codenotti

11:38

Are you here @Mathei?

MatheinBoulomenos

11:54

@AlessandroCodenotti what's up?

Alessandro Codenotti

I have another general question about the ideal class group

I know that its cardinality contains a lot of information about the number field I'm studying, but what can be deduced about a number field from the structure of its ideal class group? For example if it's cyclic or abelian or has some other nice property

MatheinBoulomenos

Hmm, not sure

mercio

the ideal class group is the galoisgroup of the maximal abelian unramified extension

of the field

and it's always abelian

Alessandro Codenotti

Hmmm what's an unramified extension?

mercio

it's when it's not ramified over any prime

maybe also including the infinite place

Balarka Sen

12:07

@AlessandroCodenotti what you are right now, but shortly afterwards you'll become a ramified extension folds sleeve, proceeds aggressively

Alessandro Codenotti

So for all primes $p\in\Bbb Z$ we have that $(p)$ factors into a product of distinct prime ideals in $\mathcal O$, right?

lol, hi @Balarka

loch

that's when it is unramified over $p$ yes

ÍgjøgnumMeg

@AlessandroCodenotti also from what @mercio said, the subgroups of the ideal class group are subgroups of the galois group of the maximal unramified abelian extension of your number field, so you get information about intermediate extensions too by the galois correspondence

Alessandro Codenotti

What do you mean with abelian extension?

ÍgjøgnumMeg

@AlessandroCodenotti an extension whose galois group is abelian

Alessandro Codenotti

12:21

Ah makes sense, thanks

ÍgjøgnumMeg

@AlessandroCodenotti Also I don't know if you already knew this but the ideal class group tells you whether or not the ring of integers of a number field is a UFD

Alessandro Codenotti

Yes, but for that I just need to know how many elements it has, not the actual structure of the group

ÍgjøgnumMeg

Ah okay, just checking

Then yeah, the structure gives you information about extensions of number fields

Silent

12:40

Is this true : Let $f:[a,b]\to\Bbb R$ be continuous. If $\int_c^df(x)\,dx=0$ for all $[c,d]\subset [a,b]$, then $f$ is zero function.

Secret

it has to...? since all possible definite integrals of f in the interval [a,b] will be zero and $\bigcup_{c,d \in [a,b]} [c,d] = [a,b]$?

mercio

it's true

Silent

thank you

Secret

I don't know how to prove that, my guess it might have something to do with nested intervals theorem

I am also thinking about:

Suppose:

Alessandro Codenotti

You can show that it must be 0 almost everywhere and continuous+0 almost everywhere implies that it is the zero function

user193319

12:52

Does anyone know what Ian means by diagonalization in his comment on this question: math.stackexchange.com/questions/2039788/…

?

Secret

right

This question meanwhile, caused me to wonder about the properties of the following discontinuous function:

Silent

@AlessandroCodenotti Is this correct proof: Suppose $f\ne0$, and WLOG, $f(x)>0$ for some $x\in[a,b]$, then $f(y)>0$ for all $y\in (p,q)\subset [a,b]$, for some $(p,q)$, since $f$ continuous, and taking some $[m,n]\subset(p,q)$ would imply $\int_m^nf(x)\,dx>0$, contradiction.

Alessandro Codenotti

Looks good to me

and even simpler than I was suggesting

Silent

thanks

Secret

$$g(x - \frac{c+d}{2}) = \begin{cases} -1, x < \frac{c+d}{2}, x \in \Bbb{Q} \\ 1, x \geq \frac{c+d}{2},x \in \Bbb{Q}\end{cases}$$ for all $c,d \in [a,b]$

such that for any $\int_p^q g(x)dx = 0$ but $g(x) \neq 0$

Silent

13:00

@Secret so, is this one defined only for rationals?

Secret

uh, it's more like stitching step functions together at the rationals

so it is the limit of the following process:

Silent

ok. is this function continuous over $\Bbb Q$? I think its not.

Secret

No $g$ is discontinuous. I am wondering about a question inspired by your question and I am wondering whether $g$ is discontinuous everywhere

Silent

If you want non-continuous function which is not zero but integrates to zero, just give a point some nonzero value in domain.

Secret

Silent

13:06

e,g, f(x)=0 for all [0,1] except 1/2, and f(1/2)=1

Secret

Put it simply, $g$ is a function that flip sign in a step function like manner at every rational

but since it does not quite resemble the dirichlet function, I am not sure how to analyse the severity of its discontinuity

ÍgjøgnumMeg

Is there a notation for some kind of unordered cartesian product?

feynhat

hello

When we speak of the quotient topology generated on $Y$ by a map $f: X \to Y$, is it implied that $f$ is surjective?

Alessandro Codenotti

13:22

@feynhat Not necessarily, but $f$ surjective is the interesting case

Because the topology on $Y\setminus f(X)$ is discrete

(I guess some authors might require $f$ to be surjective, but that's not strictly needed)

feynhat

@AlessandroCodenotti How do you define quotient topology?

Alessandro Codenotti

$U$ is open in the quotient topology on $Y$ with respect to $f$ iff $f^{-1}(U)$ is open in $X$

feynhat

Or in other words, it is the largest topology on $Y$ such that $f$ is continuous, right?

Alessandro Codenotti

Right, that's equivalent

googling I see that requiring $f$ to be surjective is pretty common, so you should which convention your professor/book is using (my professor didn't ask for $f$ to be surjective, but all the interesting examples have $f$ surjective anyway)

feynhat

But this may not necessarily imply that $f$ is surjective. Munkres defines quotient topology as the topology on $Y$ with respect to which $f$ is a quotient map. Now, for $f$ to be a quotient map $f$ must be surjective. We have conflicting definitions.

Szabolcs

13:30

Is there any theory for describing the class of convex 2D objects that can intersect a circle in at most 4 points?

Alessandro Codenotti

It's a matter of convention, Munkres probably assumes that $f$ is surjective then

feynhat

@AlessandroCodenotti You said that topology on $Y \setminus f(X)$ is discrete because $f(U) = \phi \ \ \ \ \forall U \subset Y \setminus f(X)$, so we can include all the subsets of $Y \setminus f(X)$ without disturbing the continuity of $f$, right?

Alessandro Codenotti

It's not that we can, but we must include them if we want $f$ to be continuous since $\varnothing$ is open in $X$

feynhat

If we remove some of them from the topology, $f$ will still be continuous right?

Its just that it will no longer be the largest topology.

Alessandro Codenotti

Ah, yes sorry you're right

but we want the finest topology

Secret

13:42

My previous question is now organised into a MSE:

0

Q: A highly pathological discontinuous function. Trying to analyse its value at irrational points and continuity behaviour

Consider the following pathological function $g : [-1,1) \to \{-1,1\}$ constructed in stages: \begin{align} g_0 (x) & = \begin{cases}-1, x < 0\\ 1, x \geq 0\end{cases}\\ g_1 (x) & = \begin{cases}1, x \in [-1,-\frac{1}{2})\\-1, x \in [-\frac{1}{2},0)\\1, x \in [0,\frac{1}{2})\\-1, x \in [\frac{1}...

limits discontinuous-functions

Alessandro Codenotti

I don't think $g_n$ converges, even pointwise, to a function $g$

Secret

hmm...

I wonder how we can approach proving the nonexistence of a discontinuous function with discontinuity a dense set of the reals, such that despite it is either -1 or 1, any integral of a subinterval (with rational end points) of its domain is zero

Alessandro Codenotti

Just make it $1$ or $-1$ on the rationals and $0$ everywhere else

mercio

what is $g$

Secret

13:58

Best described as a discontinuous function with alternating 1 and -1 such that it has a jump at every rational and that for p,q rational, $\int_p^q g(x)dx = 0$

Its integer analogue will be the following function:

mercio

best described as "doesn't exist"

Secret

so the integer analogue has the property that it is rotationally symmetric about every integer point

mercio

o..o

Secret

I am wondering whether we can do the same if the jumps are at the rationals

and that is what g is

mercio

what is a jump

Secret

14:01

jump discontinuity, such as at x=0 in the step function

mercio

what is a jump discontinuity

Secret

I don't know if you will call discontinuities in a dense set jump discontinuities though....

for any function $h$, $p$ is a jump discontinuity if $\lim_{x \to p^+} h(x) \neq \lim_{x \to p^-} h(x)$

mercio

so those limits have to exist ?

Secret

yeah (and finite)

user193319

14:19

Is $f_n : [0,1] \to \Bbb{R}$ defined by $f_n = n 1_{[0,1]}$ a cauchy sequence with respect to the $L^1$ norm? If my calculations are right, it seems to reduce to determining whether $\epsilon > 0$, there is $N \in \Bbb{N}$ s.t. if $s \in \Bbb{N}$ and $n \ge N$, then $\frac{2s}{n+s} < \epsilon$...But I'm having trouble showing that this is true...

Secret

16:30

ok cr** I am screwed

I forgot the rationals are not well ordered in the usual ordering, thus there is actually no anchor to ensure the function is of certain form

user193319

16:48

Simple question: given $a,b > 0$, does there exist $x \in \Bbb{R}$ such that $|1-x| < a$ and $|x| < b$?

Martin Sleziak

@user193319 From triangle inequality you get $1\le |1-x| + |x| < a+b$.

So $a+b>1$ is a necessary conditions. I think it is also sufficient - if $a+b>1$ it is possible to find such $x$.

user193319

wait...can't I just choose $x \in (0, \min \{a+1,b\})$?

Does that work?

Martin Sleziak

Such choice definitely gives $|x|=x<b$.

But what happens, for example, if $a=b=1/2$.

You choose $x$ such that $0<x<1/2$ and you have $|1-x|>1/2=a$.

user193319

Ah...This is annoying. I'm just trying to show that $\mathcal{C}[0,1]$ is dense in $L^P[0,1]$, it seemed to just reduce to this problem, but I guess I'm wrong.

Secret

desmos.com/calculator/bqbvqlxnwz

if only there's a more efficient way to generate those intervals...

Martin Sleziak

17:01

@user193319 What about $x=\frac{b-a+1}2=b-\frac{a+b-1}2$. (Assuming $a+b>1$.)

You have $x<b$. You also have $1-x=\frac{a-b+1}2=a-\frac{a+b-1}2<a$.

This still probably needs some fine tuning to avoid negative values and avoid $x>1$.

user193319

@MartinSleziak That's possible, but these calculations are being to get too fine-grained to solve this problem, so I'm being to think that my approach to proving the density of $\mathcal{C}[0,1]$ is wrong.

Balarka Sen

17:17

Man it feels good to be doing math again

Daminark

@Balarka college application business is finally over?

Balarka Sen

Yup

Daminark

Woot

Balarka Sen

Truly

Nûr

17:32

Hello. If someone here can help... math.stackexchange.com/questions/2784875/…

That's solved.

user193319

17:49

How does one show that $C[0,1]$, the step of all continuous real-valued functions on $[0,1]$, is dense in $L^p[0,1]$? Any hints or MSE link would be appreciated.

Martin Sleziak

@user193319 I have asked in the searching chatroom. On the off-chance that somebody has a bit of spare time and is willing to spend some time searching.

Jon

18:07

I have question regarding the geometric effect of a linear transformation for example the linear transformation $R^2$ to $R^2$ which is $L(x) = (-x_1,x_2)^t$ my main question about this question is I know that if you say$x_1 =1 , x_2 = 1$ you can plot the first point but what about the second what exactly is a reflection? Why is the reflect $(x_1,x_2)^t$?

feynhat

18:26

@Jon "you can plot the first point but what about the second"
What exactly are these "first" and "second"?

Jon

First point is $L(x) =(-x_1,x_2)$ second point $(x_1,x_2)$

So $(-1,1)$ and $(1,1)$ are the points

feynhat

@Jon So you want to know why the image of the first point $(-x_1, x_2)$ is $(x_1, x_2)$. Did I understand your question correctly?

Jon

Yeah wouldn't a reflection be the opposite like $(-x_1,-x_2)$?

feynhat

Your reflection transformation is $L$. So apply $L$ on $(-x_1, x_2)$. What do you get?

Jon

I don't understand. L is the linear transformation?

feynhat

18:34

Which axis are you taking the reflection about?

Jon

$x_2 $ axis.

feynhat

Yes. And the reflection about $x_2$ axis is given by the transformation $L$.

EnjoysMath

prime numbers: math.stackexchange.com/questions/2786678/…

Jon

I see so $x_2$ is the only one to change because the reflection is just on the $x_2$ axis not on the $x_1$ axis.

EnjoysMath

Questions?

T_T

Mikhail

19:55

Is there some analytic form for an integral like $$\int_{x_i=-a}^{+a} f(x-x_i) dx_i$$

Mikhail

20:10

Its like I'm looking at the function through a window that is 2*a

But also for example at $x=0$ the "windows" overlap the most

Jon

There is Simpsons double integral math.stackexchange.com/questions/2554573/… @Mikhail

Mikhail

Not really sure how that's related

Jon

Never mind then I was thinking as an algorithm.

Sebastiano

20:30

Good evening. Is there somebody to help me for this question?

0

Q: Snell's law in relativity: a clarity on the notes

I am following a training course and updating of fundamentals of modern physics at my university. I had any photocopies of a colleague (see images) I think they are not very clear without a comment and without a detailed explanation. I kindly ask if there is anyone who can help me to understa...

special-relativity reference-frames refraction

nbro

20:43

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

A little of mathematical comedy due to mathematical terminology and possibly inferrable meanings of terms

ÍgjøgnumMeg

l e l

Balarka Sen

latus rectum is damn good

Poline Sandra

hi

Leaky Nun

@BalarkaSen rectum?

Alessandro Codenotti

Hi

Balarka Sen

20:48

@LeakyNun Not just any regular rectum, latus rectum.

nbro

Anyway, a few weeks ago I had shared here an algorithm to find the tangents which need to be computed during Chan's algorithm. That algorithm was incorrect, because it was not handling a few cases...

Poline Sandra

i want to prove that $f_n(x)=\begin{cases} nx,~ 0\leq x\leq\frac1n\\ 1, ~\frac1n\leq x\leq1\end{cases}$ is not a Cauchy sequence in $(\mathcal{C}([0,1],\mathbb{R}),d_{\infty})$

i found $d_{\infty}(f_p,f_q)=1-\frac{q}{p}$

how to continue

?

Leaky Nun

@PolineSandra is that the supremum (maximum) norm?

nbro

Hopefully, nobody used it for some mission-critical application :)

Poline Sandra

@LeakyNun yes

Leaky Nun

20:53

@PolineSandra well note that $d_\infty(f_p,f_{2p}) = -1$

by your calculations

(I didn't verify

btw why is your $d_\infty$ not symmetric?

Poline Sandra

$d_{\infty}(f,g)=sup_{x}|f(x)-g(x)|$

Leaky Nun

right, but your result is not symmetric

Poline Sandra

i choose $p>q$

Leaky Nun

oh

...

Poline Sandra

i'm in cauchy

Leaky Nun

20:56

well then note that $d_\infty(f_{2p},f_p) = 0.5$

so just choose $\varepsilon = 0.5$ to derive a contradiction

Poline Sandra

ok thak you

Sha Vuklia

21:46

hey @Leaky, mind helping me out with sth in algebra?

Leaky Nun

?

Sha Vuklia

so it's about groups;

let $G$ be a group of order $pq$, where $p>q$ are prime, $q$ doesn't divide $p-1$. We know that $G$ contains a cyclic subgroup $H$ of order $p$ (by Cauchy’s thm), and since $q=[G:H]$ is the smallest prime number that divides $pq$, we know that $H$ is normal. Now let $a\in G, a\notin H$. My book says that $aH$ generates $G/H$.

I don’t really see why. I know that the order of $a$ must be $p,q,$ or $pq$, but apparently it can’t be $p$. Not sure if that would even be sufficient, but otherwise I have no idea how to show taht $\langle aH\rangle=G/H$.

Leaky Nun

pass to the quotient

$aH$ is not the identity

ÍgjøgnumMeg

The order of $G/H$ is $q$ which is prime so it is generated by any non-identity element, and since $a \notin H$, this is such an element

Leaky Nun

$|G/H| = q$

...

so who got sniped?

ÍgjøgnumMeg

21:51

lol

apologies

Leaky Nun

:P

Sha Vuklia

why is it generated by any non-identity element though?

ÍgjøgnumMeg

I guess the thing to note is that the order of any element divides the order of the group by Lagrange's theorem (order of element is order of subgroup it generates) and so if your group has prime order $q$, then any non-identity element has order $q$

Sha Vuklia

if we have $Z/10Z$, then, then $\bar 2$ doesn't generate the group

Leaky Nun

10 is not prime

Sha Vuklia

21:53

oh right

ÍgjøgnumMeg

$\Bbb Z/10\Bbb Z$ doesn't have prime order

Leaky Nun

yay

ÍgjøgnumMeg

lol

Sha Vuklia

haha thanks

Leaky Nun

@ÍgjøgnumMeg can Ext be defined for groups? does it still correspond to the equivalence class of extensions?

ÍgjøgnumMeg

21:54

absolutely no idea

lol

Sha Vuklia

hm, how does it follow then that $q$ divides the order of $a$? I understand that the order of $a$ must be at least $q$, but why the divisibility property?

ÍgjøgnumMeg

Lagrange's Theorem

Sha Vuklia

o shit again

let me think

ÍgjøgnumMeg

sorry

I mean

if $G$ is a group and $a \in G$ then the order of $a$ divides the order of $G$

Sha Vuklia

yes, so the order of $G$ is $pq$

but we have to exclude that the order of $a$ is $p$

ÍgjøgnumMeg

21:59

Yes but the group you're talking about is $G/H$

which has order $q$

Sha Vuklia

right, but I'm talking about $a$ now, and not $\overline a$

ÍgjøgnumMeg

I see

so what's your issue?

Sha Vuklia

well, why can't the order of $a$ be $p$?

ÍgjøgnumMeg

er

Leaky Nun

because $aH$ generates $G/H$

there is a surjective homomorphism $\varphi : G \to G/H$ sending $a$ to $aH$

so $q = \operatorname{ord}(aH) \mid \operatorname{ord}(a)$

Sha Vuklia

22:05

oh, I had forgotten about that

ÍgjøgnumMeg

ah ye

Daminark

22:17

Hey there nerds

Leaky Nun

hi

ÍgjøgnumMeg

gwarn

geocalc33

22:34

anybody know how to find the roots of (f(z))^(1/log(z))?

Leaky Nun

22:44

Dec 17 '17 at 10:02, by Balarka Sen

@LeakyNun Find a holomorphic covering of $\Bbb C \setminus \{1, -1\}$ by the unit complex disk.

Dec 17 '17 at 17:11, by Balarka Sen

But this by no means is a proof; I only constructed a topological covering map $D^2 \to \Bbb C \setminus \{-1, 1\}$ modulo all details. I want a holomorphic covering map.

@BalarkaSen what you asked for is the modular lambda function

and is not very easy to define elementarily

and it's also used in the proof of little picard theorem

Balarka Sen

Correct, that's why I asked you.

Leaky Nun

do you suppose I could have come up with a proof of little picard myself...

Balarka Sen

I feel like I can do a trickster argument by saying $\Bbb C \setminus \{1, -1\}$ admits a metric of constant negative curvature, so it's covered by $\Bbb H^2$ by invoking uniformization theorem lol but that's just ass

@LeakyNun Sure why not.

You're a smart person

Antonios-Alexandros Robotis

@BalarkaSen how are you

Balarka Sen

Much better. The admissions are over now, so I feel a lot better.

Antonios-Alexandros Robotis

22:54

when will you hear?

Balarka Sen

A month later.

Antonios-Alexandros Robotis

good luck my friend

Balarka Sen

Thanks!

Antonios-Alexandros Robotis

@TedShifrin indeed I am now being advised by bogomolov

Mary Star

Hello!!

Let $g:\mathbb{R}\rightarrow \mathbb{R}$ be arbitrary and let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ be defined by $f(x,y)=yg(x)$.
I want to prove that $f$ is differentiable in the origin iff $g$ is continuous in $x=0$.

Could you give me a hint how we could show that?
Do we maybe have to use the definition of derivative with the limit?

Leaky Nun

22:59

@MaryStar how do you define differentiability for $\Bbb R^2 \to \Bbb R$?

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$Mathematics$ Mathematics