Mathematics - 2016-03-07

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Semiclassical

00:01

not sure it helps. and tbh i'm not really interested in trying to figure out their reasoning

Pallas

hi

if r = <x,y,z> and i have r/||r||, then is it true that \frac{x}{\sqrt{x^2 + y^2 + z^2} \frac{\partial x}{\partial r} = 1?

because e_r_x is normal to dx/dr?

Forever Mozart

00:52

@PVAL I finally finished Six Feet Under... :( incredibly sad

Semiclassical

01:08

@pallas that's equivalent to asking if $\frac{x}{r} \frac{\partial x}{\partial r}=1$. but the second term is just $x/r$ since the only radial dependence of $x=r\sin\phi\cos\theta$ is in the leading term

so that equals $x^2/r^2=x^2$ (since $r^2=1$ here) not 1.

Anthony

@MikeMiller Ugh I just realized I won't be here when you're here. I'm flying back home for spring break.

Semiclassical

a true fact you can learn from that, though, is that $$\frac{x}{r}\frac{\partial x}{\partial r}+\frac{y}{r}\frac{\partial y}{\partial r}+\frac{z}{r}\frac{\partial z}{\partial r}=1$$

Mike Miller

01:37

@anthony toooonyyy

Anthony

02:03

:(

Clement C.

02:14

Hi

I have a very silly question about independence of random variables... but I cannot really figure it out quickly, and it's been a long day

I have iid Poisson(lambda) random variables N_1,..,N_k and M_1,..., M_k

And weights (w_{i,j})_{1\leq i,j\leq k}

Now, I am looking at the sum \sum_{1\leq i,j\leq k} w_{i,j} N_i M_j

1). Are the summands N_i M_j independent?

2). What would be the least painful way to compute Var[\sum_{1\leq i,j\leq k} w_{i,j} N_i M_j]?

(For 1), I'm pretty sure the answer is "no.")

usukidoll

02:36

hiiiiiiiiiii

1 hour later…

Easterly Irk

04:00

Anybody know if a formula exists to generate the gnomon of a figurate number for a n-gon?

1 hour later…

GGG

05:02

-1

Q: Null space of linear transformation of polynomials

The set $\mathcal{P}$of all real polynomials $f(x)$ is a linear subspace of the vector space $\mathcal{C}^{\infty}$ of real differentiable functions on the real line. Find the null space of the following mappings defined on $\mathcal{P}$. $F_2(f(x)) = xf(x)$ $F_3(f(x)) = x^2f''(x)-2xf'(x)$ Wh...

linear-algebra polynomials

Does my question make sense?

2 hours later…

Ted Shifrin

07:28

@IlanAizelmanWS: Hi, great news. I'm glad you're doing well!

Nazaf Anwar

07:43

Hi, all. I have a very basic question. I went to the calculus chat room but didn't find anyone active. I wanted to know which version of Stewart Calculus book people usually talk about while recommending the same for people looking forward to appear for GRE math subject test.
I see two versions mainly. Calculus 7E and Essential Calculus and both having Early
Transcendentals as another sub-versions(?).

Thank you.

Tobias Kildetoft

08:01

@TedShifrin Hey, seems I managed to catch you both sides of my sleep for once (assuming you are still here)

2 hours later…

Tobias Kildetoft

09:38

Does anyone know if it is possible to make a function in GAP create global variables?

3 hours later…

Szabolcs

12:57

For people interested in graph theory software:

Nazaf Anwar

I would highly appreciate any response to my question, guys.

Szabolcs

I created a Mathematica interface for igraph, so Mathematica's builtin graph functions can be easily combined with igraph. I also contributed some patches to the igraph core meaning that now we have access to three different isomorphism algorithms that also work with coloured graphs, through the same uniform interface.

Vrouvrou

13:13

I need help for this please: math.stackexchange.com/questions/1686845/…

2 hours later…

Mike Miller

14:43

I left my window open and it rained through it last night :(

None of my books seem destroyed, luckily

Mary Star

15:25

I want to find how many different necklaces we can create with $m$ symmetric beads if we have $k$ colours.

Burnside's Formula is the following: $$\text{ # orbits } =\frac{1}{|G|}\sum_{g\in G} X (g)$$

$G$ is the group of the permutations of the beads, right?

Do we have that $|G|=m!$ ?

Anonymous

Hello people

Mary Star

Hello @anonymister

Anonymous

Hi

Evinda

16:01

Hey @robjohn

I have a question... We have that $f \in C_c^{k}(\mathbb{R}^n)$. How can we show that $|f(x- \epsilon z)-f(x)|< \delta$ for any $\epsilon> \epsilon_0$ ? @robjohn

robjohn

@Evinda that is not true without some constraints

Evinda

After that there is the following remark:
$f$ is continuous and has a compact support. We know that each continuous function is uniformly continuous in a closed and bounded space of $\mathbb{R}^n$.

Do you mean that we have to consider a closed and bounded space so that the above statement is true? @robjohn

robjohn

16:19

@Evinda No, I mean it is not true. Why should that be true for arbitrarily large $\varepsilon$? what is $z$?

Evinda

@robjohn I found it at the proof of the following theorem:

Let $f \in C_c^k(\mathbb{R}^n)$. Let $\rho \in C_c^{\infty}(\mathbb{R}^n), \rho \geq 0, supp{\rho} \subset \{ |x| \leq 1\}, \int_{\mathbb{R}^n} \rho(x) dx=1, \epsilon>0$. We consider $f_{\epsilon}(x)= \epsilon^{-n} \int_{\mathbb{R}^n} f(y) \rho{\left( \frac{x-y}{\epsilon}\right)} dy$.
Then $f_{\epsilon} \in C_c^{\infty}, supp f_{\epsilon} \subset \epsilon- \text{ region of } supp f$

$\partial^a{f_{\epsilon}} \to \partial^a f$ uniformly for $|a| \leq k$ while $\epsilon \to 0$.

robjohn

16:36

Since $\left|f(x-\varepsilon z)-f(x)\right|\le\left|\varepsilon z\right|\sup\limits_{x}|f'(x)|$

Julian Rachman

0

Q: Showing $\mathcal{A}^*$ is Well-Quasi-Ordered

I was told that the problem below is supposed to prove that $\mathcal{A}^*$ (defined below in "Attempted Solution") is well-quasi-ordered. Two problems come to my mind: How is this problem proving that $\mathcal{A}^*$ is well-quasi-ordered iff the ordered set $A$ is well-quasi-ordered? I have a...

proof-verification category-theory

Anyone?

Evinda

@robjohn You mean that we use the intermediate value theorem, right?
If so, then we have that there is a $\xi \in (x- \epsilon z,x)$ such that

$f'(\xi)=\frac{f(x)-f(x- \epsilon z)}{\epsilon z} \Rightarrow |f(x- \epsilon z)-f(x)|=|f'(\xi)| |\epsilon z| \leq |\epsilon z| \sup_x |f'(x)| $

But why does the latter tend to zero?

robjohn

@Evinda since $\varepsilon\to0$.

Evinda

A ok... And since $f \in C_c^k(\mathbb{R}^n)$ we know that it is bounded and so $\sup_x |f'(x)| \in \mathbb{R}$, right? @robjohn

Mike Miller

@robjohn: Scary weather this morning. I decided to head to campus later than usual because of the lightning.

robjohn

16:51

@MikeMiller Yeah. I am considering driving to Victorville today. Luckily, things have quieted down some.

@MikeMiller The lightning and thunder was very close here.

Mike Miller

It wasn't so close here but I consider lightning scary enough to be an excuse not to go somewhere.

Evinda

@robjohn Also the fact that $|f(x- \epsilon z)-f(x)| \to 0$ implies that there is a $\delta$ such that $|f(x-\epsilon z)-f(x)|< \delta$ for any $\epsilon> \epsilon_0$, right?

robjohn

@Evinda why do you keep using $\epsilon\gt\epsilon_0$? We are supposed to have $\epsilon\to0$.

Evinda

@robjohn Oh yes, right... So the statement that I should show is wrong, i.e. that $|f(x- \epsilon z)-f(x)|< \delta$ for any $\epsilon>\epsilon_0$, right?

robjohn

@Evinda Since $z\le1$, $\left|f(x-\epsilon z)-f(z)\right| \le\epsilon\sup\limits_x\left|f'(x)\right| \to0$

Evinda

17:02

@robjohn And the latter tends to 0, only for $\epsilon \to 0$, right?

What can we say about $\sup\limits_x\left|f'(x)\right|$ ?

robjohn

@Evinda $f\in C_C^k$. What do you think we can say about $\sup\limits_x\left|f'(x)\right|$?

Evinda

That it is bounded, right? @robjohn

robjohn

@Evinda you think?

Evinda

Yes, since $f\in C_C^k$ , $f$ and its derivatives are non-zero only near a region around 0. Or am I wrong? @robjohn

robjohn

@Evinda why does it matter that the derivatives are non-zero?

Evinda

17:08

Since we have the supremum of f'(x) @robjohn

Aaa you mean that if f is 0 at a region which is not near 0, so the derivative will be, right? @robjohn

robjohn

$f\in C_C^k\implies f'\in C_C^{k-1}$ which means it is continuous on a compact set.

Therefore, $f'$ is bounded

Evinda

17:22

@robjohn Ah I see... Thank you very much!!!

r9m

@DanielFischer Hi! Can you help me with a problem sir? :-) I was wondering how would I characterize the functions $f \in C^{\infty}[0,\infty)$ such that $\displaystyle \int_0^{\infty} f(x)e^{-nx}\,dx \sim \sum\limits_{k=0}^{\infty} a_k \frac{k!}{n^{k+1}}$ where $\displaystyle f(x) = \sum\limits_{k=0}^{\infty} a_k x^k$ (taylor series at $x = 0$)

does $f(x) \to 0$ as $x \to \infty$ suffice, or can we manage with a weaker condition than that? :)

@RandomVariable Hi man! How are you?! :-) I see I&S is up and active again!

@robjohn Helloooo .. :) How are you?

robjohn

@r9m pretty good. How are you?

r9m

17:38

@robjohn doing okay .. somehow alive! :-)

robjohn

@r9m that's promising!

ypercubeᵀᴹ

Why people take commenting so personal? You add a comment or 2 so they can improve their answer and they answer with, well, like you attacked their mother.

2

An identity is an expression that has the same value no matter what x you plug in. (I understand your desire to be creative and find some wow logical inconsistency and something that is worth endless discussion for nothing. Please waste your own time. I will not participate. Thanks.) — alex.peter 5 mins ago

r9m

@ypercubeᵀᴹ 'attacked their mother' .. :P * golden!

@robjohn :-)

Semiclassical

@ypercubeᵀᴹ 'attacked their own child' might be the more appropriate metaphor here; they created the question, after all, not the other way around :)

Random Variable

@r9m I've had a cold for the past couple weeks that just won't go away completely. But other than that, I'm doing OK.

r9m

17:46

@RandomVariable colds can be annoyingly sticky .. it's hard to shake 'em off completely. I see Shobhit's back on action! :-)

Julian Rachman

Anyone?

0

Q: Showing $\mathcal{A}^*$ is Well-Quasi-Ordered

I was told that the problem below is supposed to prove that $\mathcal{A}^*$ (defined below in "Attempted Solution") is well-quasi-ordered. Two problems come to my mind: How is this problem proving that $\mathcal{A}^*$ is well-quasi-ordered iff the ordered set $A$ is well-quasi-ordered? I have a...

proof-verification category-theory

Random Variable

@r9m I thought he had abandoned the site.

r9m

@RandomVariable apparently not .. he's back and the site has gotten more active again! :) Btw what you think of this one? :-)

Random Variable

@r9m I'm more interested in this one. There is a paper about it here, but I've been trying to come up with an alternative approach.

r9m

@RandomVariable ooh yeah! that's a nice problem .. I have a way in mind but been too lazy to write up .. lemme check the paper though .. thanks!!

r9m

18:13

oh! Weierstrass factorization theorem there too .. 'kay

s.harp

18:31

Is the empty intersection of subsets of $X$ usually defined to be $X$? And would the empty intersection be a "finite intersection"?

Daniel Fischer

@r9m I don't expect there is a useful characterisation. $C^{\infty}$ can be very ugly. The series need not converge. If it converges, it need not have anything to do with the behaviour of $f$ away from $0$, and it need not have anything to do with the behaviour of $\int_0^{\infty} f(x) e^{-nx}\,dx$. If you require $f$ to be real-analytic, you might have a chance to get a characterisation.

Mike Miller

Morning @DanielF.

@s.harp: I would say so, yes.

Daniel Fischer

@s.harp Yes. The empty set is a finite set, so $\bigcap \varnothing$ is an intersection of finitely many sets. And if the universe of discourse is $U$, then $\bigcap \varnothing = U$ is common convention.

Mike Miller

@DanielF: I'm quite fond of $C^\infty$. I won't have you besmirching its good name.

Daniel Fischer

@MikeMiller Happy whatever time of day.

Evinda

18:40

@robjohn Let $E(t,x)= \frac{H(t)}{(2 \sqrt{\pi t})^n} e^{-\frac{|x|^2}{4t}}, x \in \mathbb{R}^n$. I want to show that $\frac{\partial{E}}{\partial{t}}- \Delta{E}= \delta(t,x)$.

So it suffices to show that $\langle \frac{\partial{E}}{\partial{t}}-\Delta E, \phi \rangle= \langle \delta, \phi \rangle$.

So firstly we have to show that $E$ is integrable.

In order to do this I thought to show that $E$ is continuous.
It holds that $E(t,x)=\left\{\begin{matrix}
\frac{e^{-\frac{|x|^2}{4t}}}{(2 \sqrt{\pi t})^n} &, x>0 \\ \\

@robjohn Sorry I meant: $E(t,x)=\left\{\begin{matrix}
\frac{e^{-\frac{|x|^2}{4t}}}{(2 \sqrt{\pi t})^n} &, t>0 \\ \\
0 &, t \leq 0
\end{matrix}\right.$

robjohn

@Evinda So $H(t)=1$

Evinda

$H(t)=\left\{\begin{matrix}
1 &, t>0 \\
0& , \text{otherwise}
\end{matrix}\right.$

It's the Heaviside function @robjohn

Random Variable

18:57

@DanielFischer Isn't $f(z) = \exp \left(i z \frac{z^{2}-b^{2}}{z^{2}-c^{2}} \right) $ bounded in the upper half-plane for positive values of $b$ greater than $c$? My reasoning is that $f(z)$ is essentially $e^{iz}$ when $|z|$ is large in magnitude and essentially $\exp \left(\frac{i}{2} \frac{c^{2}-b^{2}}{z \mp c} \right) $ near $z = \pm c$. But are those estimates accurate enough?

Andrew Thompson

I just got a 'welcome to math.SE-chat, keep some things in mind ... be nice ...'-bubble. Did everyone get that or did I do something bad?

s.harp

I got it too

Does anybody know how to write fraktur letters without boldface?

nevermind sorry you can google that quite easily :S

Mike Miller

I will never forgive you

2

Andrew Thompson

19:12

I think that bubble we got was meant for you, Mike.

s.harp

D:

Andrew Thompson

They just sent it to everyone so it wouldn't be awkward.

Evinda

@robjohn How else could we show that E(t,x) is integrable?

r9m

@DanielFischer 'kay .. so does the result hold if $f$ is uniformly approximable by $p(x)e^{-x}$ for polynomials $p$ on $[0,\infty)$ (I know $f \to 0$ as $x \to \infty$ is a sufficient condition for that)

Daniel Fischer

@RandomVariable Hmm, you should go into more detail when writing. As a handwaving argument in speech it works, but in writing one should be more precise. Write $f(z) = e^{iz} \exp\Bigl(-iz\frac{b^2-c^2}{z^2-c^2}\Bigr)$ to deal with large $\lvert z\rvert$, and similar for $z$ close to $\pm c$, then you can explicitly show the boundedness.

@r9m Oy. Since we're integrating over a set of infinite measure, uniform approximation doesn't really give you very much. Suppose you have $f(x) = \frac{1}{x}$ for $x > 1$. Then whatever polynomial $p$ you take, you have $f(x) - p(x)e^{-x} > \frac{1}{2x}$ for $x > x_p$. Then, as $n\to \infty$ you have $\int_{x_p}^{\infty} \bigl(f(x) - p(x)e^{-x}\bigr)e^{-nx}\,dx \in \Theta (n^{-1})$, so I wouldn't be too optimistic.

What is the context for this?

r9m

19:38

@DanielFischer I see your point now .. I was actually trying $\int_0^{\infty} \frac{e^{-nx}}{1+e^x}\,dx$ and once I expanded $\frac{1}{1+e^{x}}$ in it's taylor series at $x = 0$ and integrated term by term I got the exact asymptotics :o is that a coincidence?

Daniel Fischer

@r9m That's probably not a coincidence. That function is analytic.

s.harp

If $F \subset \mathfrak P(X)$, $A \subset X$ how is the trace of $F$ on $A$ defined?

r9m

@DanielFischer yes .. I am having a hard time pinpointing why this is happening .. I tried it for a couple of other functions too (I had alternative ways of verifying exact asymptotics .. and it checks out :o .. )

Daniel Fischer

@r9m Hmm, wait. The radius of convergence of the MacLaurin series of $\frac{1}{1+e^z}$ is $\pi$. Then $$\sum_{k = 0}^{\infty} a_k \frac{k!}{n^{k+1}}$$ doesn't converge if $\frac{1}{1+e^z} = \sum a_k z^k$.

Mike Miller

@s.harp: How's it defined if $A=X$?

Daniel Fischer

19:46

@s.harp $\{ A \cap M : M \in F\}$

Usually.

Mike Miller

(That was not a pedagogical comment, I'm asking for myself.)

s.harp

@MikeMiller I don't know

Mike Miller

@DanielFischer: How does that parse? $M$ is an operator, right?

Daniel Fischer

@MikeMiller No, $M$ is a subset of $X$.

s.harp

@DanielFischer thank you, but in the book I am reading it says if $F$ is a filter then the trace of $F$ on $A$ is a filter iff every set of $F$ meets $A$

Daniel Fischer

19:48

$F$ is a family of subsets of $X$.

s.harp

but the definition there does not allow the trace to be a filter, since it does not contain ie the whole set $X$

Mike Miller

Oh, that's a mathfrak P. I thought it was a B, for bounded operators.

Sorry.

Daniel Fischer

@s.harp Yes. If not all elements of $F$ meet $A$, then $\varnothing$ is in the trace.

@s.harp A filter on $A$, and $A$ is in the trace.

s.harp

I meant, if you take for example the principal filter of $A$, then every set of $F$ meets $A$. But the definition above makes the trace to be just $\{A\}$ which is not a filter on $X$

Daniel Fischer

@s.harp The trace in $A$ of a filter on $X$ is a filter on $A$.

(If all filter sets meet $A$)

s.harp

19:51

Ok thank you

Evinda

Also it holds that $\langle E, -\frac{\partial{\phi}}{\partial{t}}-\Delta{\phi} \rangle= -\int_{t \geq 0} \int_{\mathbb{R}^n} E \left( \frac{\partial{\phi}}{\partial{t}}+ \Delta{\phi}\right) dxdt=\int_{t \geq 0 \int_{\mathbb{R}^n}} \frac{1}{(2 \sqrt{\pi t})^n} e^{-\frac{|x|^2}{4t}} dxdt$ . Is it right so far? How could we continue? @robjohn

s.harp

But I am shocked that Bourbaki just write: Let $F$ be a filter of a set $X$ and $A$ a subset of $X$. Then the trace $F_A$ of $F$ on $A$ is a filter iff each set of $F$ meets $A$

from the grammar I expect $F_A$ to be a filter on $X$ :(

Random Variable

20:08

@DanielFischer So $\left|e^{iz} \right| \le 1$ in the upper half-plane, and $\exp\Bigl(-iz\frac{b^2-c^2}{z^2-c^2}\Bigr)$ is tending to $1$ as $|z| \to \infty$. Is that what you meant about being more precise?

Daniel Fischer

@RandomVariable Yes. The case $z \approx \pm c$ is the one that takes a bit more work to break down.

r9m

@DanielFischer hmm .. I see! But how do I frame the question if I wanna ask that over main?

Daniel Fischer

@r9m I probably misinterpreted your series. It's meant as an asymptotic expansion, $$\int_0^{\infty} f(x) e^{-nx}\,dx = \sum_{k = 0}^N a_k \frac{k!}{n^{k+1}} + o(n^{-N-1}),$$ is it?

r9m

@DanielFischer yes .. that's what I meant (:P my life depends on it .. I have used it to prove something and now I can't explain why this is happening :P)

Ted Shifrin

hi @DanielF, @r9m, DogAteMy :)

Daniel Fischer

20:19

Hi @Ted.

Ted Shifrin

Well, grading midterm exams is depressing even after I've retired :P

r9m

@TedShifrin Hello professor! :-)

Ted Shifrin

I doubt your life depends on anything mathematical, @r9m, but it's a common expression :P

Mambo

@DanielFischer Sorry to bother. I am trying to find an example of special kind of maps on Banach algebras called derivations

Daniel Fischer

@r9m Okay, then we have a much better chance. We integrate by parts and get a remainder $$\pm \frac{1}{n^k} \int_0^{\infty} f^{(k)}(x)e^{-nx}\,dx.$$ So all we need is that $f^{(k)}(x)$ stays civil.

Mambo

20:23

Would you help please

Ted Shifrin

@AndrewT: Clearly the bubble was meant solely for me.

Andrew Thompson

stands up and slams hand on table The bubble was mine!

Ted Shifrin

I haven't seen you misbehave ... ever.

Well, maybe just a bit of temper, like now ... :D

robjohn

@Evinda $$ \int_{\mathbb{R}^n}\frac{e^{-\frac{|x|^2}{4t}}}{(2 \sqrt{\pi t})^n}\,\mathrm{d}x $$ is independent of $t$

r9m

@DanielFischer ooh!! By parts by parts ... dear me .,. that's right! :D I forgot about that ..

Ted Shifrin

20:25

hi @robjohn

Andrew Thompson

Haha. Maybe the closest I've been to being worthy of such a bubble is nagging Mike with diffgeo-questions or asking an artist critical questions about their art.

Mambo

@robjohn hello

Ted Shifrin

I do occasionally lose patience with people who repeatedly refuse to listen/read/think ... or who want us to do every step of every problem for them.

I haven't seen you learning any of what I call differential geometry, @AndrewT :D

robjohn

@TedShifrin How goes?

Daniel Fischer

@Mambo Hmm, the typical derivation is a differential operator. But I can't think of a Banach algebra of infinitely differentiable functions that is closed under differentiation.

Ted Shifrin

20:27

Other than dental work and pains, I'm doing well, thanks, @robjohn. Looking forward to more blustery weather today.

Mambo

@DanielFischer the definition is something else here

Andrew Thompson

We didn't learn much in general. It was a stundentarranged course, we were all supermotivated in the beginning of the semester, and that faded as time went.

robjohn

@TedShifrin we had very blustery and thunderous weather this morning.

Ted Shifrin

I'm sorry to hear you aren't still super-motivated, @AndrewT

Andrew Thompson

We all passed, the clever ones in a respectable manner, but I doubt any of us truly enjoyed that course.

Daniel Fischer

20:28

@Mambo A linear map satisfying the product rule, $D(a\cdot b) = a\cdot D(b) + D(a)\cdot b$, right?

Andrew Thompson

@TedShifrin Well, I am, for things not differential geometry.

Ted Shifrin

@AndrewT: When I was an undergraduate, I think one of my top favorite courses was Guillemin and Pollack. And in the numerous times I've taught it, my students, too, have really liked it, even though it was hard.

Mambo

Yes. here that $\cdot$ is module operation

Ted Shifrin

I still don't think you guys did differential geometry. Learning basics about manifolds isn't differential geometry per se.

Daniel Fischer

@Mambo What is "module operation"?

Ted Shifrin

20:30

Yes, @AndrewT, I know you (and mr @Pedro) are becoming totally renegade algebraists :(

Mike Miller

@TedShifrin: Unfortunately one of our first years absolutely despised that book. More algebraic/symbolic thinking than geometric. Takes all types.

Andrew Thompson

@TedShifrin I agree, and so does a good diffgeo friend of mine.

Ted Shifrin

Well, the book is a bit chatty for grad students, @MikeM. And it annoys me how Guillemin gives away too much with hints (that's his teaching style), so I've always written a lot of my own exercises. Even when I've taught algebra, I've tried to encourage geometry/conceptual thinking, as opposed to symbolic manipulation, but, sadly, most algebraists seem to like to teach more symbolically :(

Mike Miller

It just happens that for some people, that's an easier or more helpful way to think. I don't blame them for that.

Ted Shifrin

So we all got that bubble to be politically correct/nice because I finally snapped when someone repeatedly ignored what I said ...

Mike Miller

20:32

I certainly can't cope that way myself.

Mambo

@DanielFischer Suppose $A$ is a Banach algebra. A Banach space $B$, a left $A$ - module is called a left Banach $A$ module if there exists $K > 0$ such that $$\norm{a.x} \leq K \norm{a}\norm{x} (a \in A , x \in B)$$

Ted Shifrin

@MikeM: It makes me sad how few algebra students understand what the point of conjugation/normality actually is.

Mike Miller

When you say algebra students, you mean undergrad?

Ted Shifrin

Most books just settle for saying it's what's needed to make the quotient be a well-defined group and leave it at that.

Mambo

It is representation of algebra $A$ on linear operators of $B$

Ted Shifrin

20:33

I might mean larger scope than that, @MikeM :P

Mambo

@DanielFischer *continuous representation

Daniel Fischer

@Mambo But then the product rule makes no sense. So please give a precise definition of what a derivation should be in that context.

Mike Miller

@Ted: I would have trouble finding more than one algebra grad student here who didn't understand the point.

Ted Shifrin

Well, perhaps I have a unique viewpoint on the point :P

Mambo

A bounded linear map $D : A \to B$ is called derivation if it satisfies the product rule.

$A$ - algebra and $B$ is Banach $A$-bimodule

Andrew Thompson

20:36

Regarding symbolpushing, off to prove that something is a pushout using the simplicial identities. Buh-bye!

Ted Shifrin

Much as I complain about the algebra books not making the point, I didn't really make it clearly enough in my own algebra book, @MikeM ... although in lecturing I made a big point of it.

Push well, @AndrewT.

@DanielF: Is your point that in most norms, differentiation won't be bounded?

Mambo

For each $b \in B$, $a \to a.b - b.a$ is a derivation.

But other than these !!

@DanielFischer Any problem with the def ?

Daniel Fischer

@TedShifrin Yeah, a sensible norm can only consider finitely many derivatives, so you can't make $C^{\infty}(X)$ a Banach algebra with any sensible norm.

Ted Shifrin

@MikeM: Sadly, neither of my topology students figured out what the one-point compactification of the line with two origins is ... because they haven't understood what the one-point compactification of $\Bbb R$ is. Oh well, ...

Mike Miller

@Ted: I find the latter surprising.

Ted Shifrin

20:40

I suspect they hadn't read the section in Munkres where he talks about one-point compactification and gives $\Bbb R$ as an example.

Mike Miller

It's obvious what the 2-point compactification is...

Daniel Fischer

@Mambo No, that definition works, if we have a two-sided module.

Another.Chemist

Good day to all!

Ted Shifrin

Somehow they're thinking of $\infty$ as an isolated point :(

Another.Chemist

I was wondering if someone can help me out with: math.stackexchange.com/questions/1686507/…

Ted Shifrin

20:41

Good day, Chemist.

Mambo

Those derivations like adjoint operation are called inner derivations.

Ted Shifrin

Your question is impossible to decipher, @Another.Chemist. Part of it is an English issue, part of it isn't. What do you mean by "Probe"?

Mambo

I am trying to find a derivation which is not inner

Mike Miller

@Ted: I think a nice hint might be to ask them to find a 2-pt compactification first.

Ted Shifrin

Do you mean "Prove"?

@MikeM: First they have to understand what the correct neighborhoods of $\infty$ are.

Another.Chemist

20:43

Thank you so much Mambo. But My chemist knowledge tried to solve those matrices problems.... this is why I'm asking for help :)

Ted Shifrin

I think they've both seen the Riemann sphere, so ... really ...

Another.Chemist

mmm... good point... I write it thinking in spanish....- what I mean is to demonstrate

Mambo

@robjohn Do you have any idea to check where?

Mike Miller

See, I'm not even thinking of nbhds. Just... a compact space with an open dense wmbedding of $\Bbb R$ and 1/2 extra points.

Daniel Fischer

@MikeMiller Gonna be mean and say the line with two origins doesn't have a compactification, since it's not a Hausdorff space ;)

Ted Shifrin

20:45

Fair enough, @MikeM.

Another.Chemist

corrected

Ted Shifrin

@DanielF: Why does it need to be Hausdorff? It needs to be locally compact, which it is.

Daniel Fischer

@TedShifrin Because any subspace of a Hausdorff space is Hausdorff, and compact = quasicompact + Hausdorff.

Mambo

@DanielFischer Do you have any idea where to check?

Ted Shifrin

I don't define compact to be Hausdorff, @DanielF. Maybe that's a European custom.

Daniel Fischer

20:47

@Mambo Sorry, no idea.

Ted Shifrin

@Another.Chemist: What you have makes no sense. To say that $\hat U = BY$ means that $BX\hat B = O$.

Daniel Fischer

@TedShifrin That's how Bourbaki did it, and HE knew right from wrong.

Ted Shifrin

Well, ha ha ... Not the custom in this country.

Bourbaki is thinking more about algebraic geometry terminology, and Zariski topology is never Hausdorff, of course.

robjohn

@Mambo nope, sorry.

Another.Chemist

O?

Mambo

20:49

@DanielFischer Okay np. I will try. Do you think set of inner derivations be closed in set of all linear operators from $A$ to $B$ ?

Ted Shifrin

I mean the zero matrix. Seems highly unlikely.

Daniel Fischer

@Mambo A very cautious "I guess so".

user19405892

hi

Mambo

Why do you think so?

Another.Chemist

sigh... My maths are not so advanced ... and to be honest, I get lost :/

Mambo

20:51

Why should the limit be like adjoint at some point?

Daniel Fischer

@Mambo I don't think either way. I very cautiously guessed. For no particular reason, just a hint of a soupcon of a gut feeling.

Ted Shifrin

Are you sure your definitions are right, @Another.Chemist. Shapes of matrices aren't making sense in some cases. I suspect the definition of $\hat B$ may be wrong.

Mambo

@DanielFischer The set of derivations is closed. It is fine. But for the inner derivations !!

Mike Miller

@DanielF: Ah, but I never said a double origin once. That was all Mr. Shifrin's doing.

Mambo

@DanielFischer There are certain algebras where the inner derivations are only derivations.

Another.Chemist

20:55

I had the same doubt. But, I focused on the definitions ... and it worked, at least with the tried that I posted with the question

Ted Shifrin

@Another.Chemist. On one hand you say $B$ is a $k\times 1$ matrix, but then you have an equation $B=I-A$, which makes it an $n\times n$ matrix. Something is just nonsense here.

And then I don't know where $BY=BU$ comes from.

You have to get the exercise precisely correct.

Mambo

@DanielFischer Do you know anything about group algebras of free groups?

Another.Chemist

@TedShifrin don't forget the hat on the definitions

Ted Shifrin

You need to proofread very carefully. It makes no sense as it is now.

Another.Chemist

It is literally what I receive

Ted Shifrin

20:57

Well, it's wrong.

Daniel Fischer

@Mambo No, not really.

Ted Shifrin

Go to your professor and ask him to explain what I just wrote up there ^^^^ ...

Another.Chemist

ok, thank you so much ! :)

Ted Shifrin

Good luck!

Mambo

@DanielFischer Thank you Sorry for bothering too much

Daniel Fischer

21:00

@Mambo No problem. Sorry I couldn't help.

Random Variable

@DanielFischer Perhaps it's an inefficient approach, but by substituting $x+iy$ for $z$, I get that $ \left|\exp \left(- i z \frac{b^{2}-c^{2}}{z^{2}-c^{2}} \right) \right| \le 1 $ if $y >0$. I could have made a mistake, though.

Daniel Fischer

@RandomVariable We want $\operatorname{Im} \frac{z}{z^2-c^2} \leqslant 0$, or equivalently $\operatorname{Im} \bigl(z(\overline{z}^2-c^2)\bigr) \leqslant 0$. Hrrmbl hrmmbl, $x\cdot (-2xy) + y(x^2-y^2-c^2) = - (x^2y + y^3 + yc^2)$, yup looks non-positive.

Evinda

21:21

0

Q: Shoe equality of distribution

Let $E(t,x)= \frac{H(t)}{(2 \sqrt{\pi t})^n} e^{-\frac{|x|^2}{4t}}, x \in \mathbb{R}^n$. I want to show that $\frac{\partial{E}}{\partial{t}}- \Delta{E}= \delta(t,x)$. So it suffices to show that $\langle \frac{\partial{E}}{\partial{t}}-\Delta E, \phi \rangle= \langle \delta, \phi \rangle$. S...

pde weak-derivatives

@robjohn Oh, now I saw your answer... Why is it independent of t? Isn't it a function of it?

Ted Shifrin

@DanielF: Oops. Upon further checking of the textbook, it does require Hausdorff and locally compact to get a Hausdorff and compact one-point compactification. Mea culpa.

Semiclassical

afternoon

Random Variable

@DanielFischer I didn't actually think I made a mistake. I don't know why I say stuff like that.

Semiclassical

hey @ted. here's something small (and not actually too surprising) which you might find amusing.

Another.Chemist

math.stackexchange.com/questions/1686507/… I improve it ^.^

Semiclassical

21:28

namely, i was checking books out of our math library and saw a book on the return cart with a name that looked familiar

"Theodore Shifrin" :)

Mike Miller

was it "Geometry: A geometric approach"?

Semiclassical

couldn't tell. all i could see was the name.

JeSuis

Salut @TedShifrin

robjohn

21:42

@Evinda substitute $x\mapsto x\sqrt{t}$

Hippalectryon

@JeSuis o/

JeSuis

ca va @Hippalectryon ?

Hippalectryon

@JeSuis Oui :D et toi ?

Akiva Weinberger

¿Cómo estáis?

Evinda

@robjohn If we substitute $x$ with $y \sqrt{t}$, then the integral is equal to $\int_{\mathbb{R}^n} \frac{e^{-\frac{y^2}{4}}}{(2 \sqrt{\pi t})^n} \sqrt{t} dy$.

Or am I wrong?

Daniel Fischer

21:47

@RandomVariable It's good practice to be cautious.

JeSuis

ça va grâce à la grève national j'ai pas d'examen mercredi_

Hippalectryon

@JeSuis Haha :P

JeSuis

nationale*

@Hippalectryon tu connais le livre Geometry and the Imagination ?

Hippalectryon

@JeSuis Pas vraiment... laisse moi regarder de quoi ça parle :D

@JeSuis Celui de 99 (Conway) ou de 52 (Hilbert) ?

JeSuis

Hilbert

dommage il n'est pas traduit...

Hippalectryon

22:01

@JeSuis Ca a l'air simpa

JeSuis

@Hippalectryon en tout cas je vais aller voir cela demain à la bu... bon je vais dormir, bonne soirée et bon travail!

Hippalectryon

@JeSuis Bonne nuit :-)

robjohn

@Evinda No, it is not. Remember that $\mathrm{d}x$ is really $\mathrm{d}x_1 \,\mathrm{d}x_2 \dots\mathrm{d}x_n$

r9m

@DanielFischer Looks like I hit a new benchmark in stupidity today .. :|

Daniel Fischer

@r9m ?

robjohn

22:12

Wow! I just got this message. I guess I'd better watch what I say.

Daniel Fischer

@robjohn Yeah, you're a well-known trouble-maker, aren't you?

robjohn

@DanielFischer I just looked, and this comment was flagged.

r9m

@DanielFischer nvm ..

Daniel Fischer

@robjohn Errr, wot? Who, why???

r9m

@robjohn ?! Hey I didn't do it .. looks like it was addressed to me .. :o

robjohn

22:16

@DanielFischer I have no idea. I was just looking to see if something had been flagged recently, and that was why i got that message.

Daniel Fischer

Chat guidelines | $\LaTeX$ in chat

7

robjohn

@r9m it was. I don't know who flagged it. It is probably one of the three downvoters I've had in the last two days.

@DanielFischer oops. I missed that that expired again.

r9m

@robjohn what would a downvoter gain by flagging a message in chat? :o

robjohn

@r9m Who says that flags and downvotes are logical? I don't think any of the downvotes I've received recently are logical anyway.

These seem to be more a personal attack

Perhaps they don't like my avatar

r9m

@robjohn sigh ..

Hippalectryon

22:21

Who doesn't like cute little mean squares :(

r9m

@robjohn I see Chris'ssis isn't/hasn't been around lately .. moreover he seems to have changed his user name once again ..

Vrouvrou

Hello, can we find an example where $\cap F_n=\emptyset$ :math.stackexchange.com/questions/1685896/…

user2692669

gre.kaomanfen.com/question/635 : can someone explain why they say that they answer is D (The relationship cannot be determined from the information given.). They say that Q could be >90 OR <90 but doesn;t messing with that angle mess with the constraint that RPQ angle is 30 degrees\?

Hippalectryon

@Vrouvrou Take closed balls of radius $1/n$

robjohn

@r9m Yes. She was around for quite a while after the last name change. However, she hasn't been to chat this year.

Dan Rust

22:29

@Vrouvrou Take $E=\mathbb{Q}$ and let $F_n = [\sqrt{2},\sqrt{2}+1/n]\cap\mathbb{Q}$.

robjohn

@r9m Oh, I didn't see that she was here a bit in February

and that her name has changed again.

r9m

@robjohn I see .. odd .. must be busy (with her book probably)..

Hippalectryon

@user2692669 The constrain does act on the rectangles, but it is not constraining enough that it determines Q

Dan Rust

This might be a long shot. Does anyone here have experience with publishing papers that describe new mathematical software? Having trouble finding a journal which (i) is not just looking for papers implementing new algorithms (our program only implements published algorithms which haven't been coded before) and (ii) doesn't mind the underlying mathematics being very pure (algebraic topology in my case)

user2692669

@Hippalectryon if we fiddle with RQ line, wouldn't it change length? I can't base the answer on a theorem

Ted Shifrin

22:41

@robjohn @DanielF: I'm assuming that "be on your best behavior" note was entirely because I lost patience with someone a few days ago. :P

salut @JeSuis, @Hippa.

Mike Miller

@DanRust: Look at Ben Burton and Ryan Budney's papers about Regina and see where those went. (I think such papers exist.)

Hippalectryon

@TedShifrin Hi :-) JeSuis is already gone

Ted Shifrin

done = parti?

Hippalectryon

to bed. Yes.

Ted Shifrin

Ah ... tu t'endors maintenant?

Hippalectryon

22:43

Not yet :P

Ted Shifrin

@MikeMiller You've used that one before. :)

Daniel Fischer

@TedShifrin No, everybody gets that message once, it's not personal.

Ted Shifrin

Well, @DanielF, there must be some reason we're all getting it now.

Mike Miller

@Ted ? I've never actually used Regina.

Ted Shifrin

Oops, did I mis-link?

Hippalectryon

22:44

@user2692669 Let me find a concrete example, 6 mins

Ted Shifrin

Nope, I did not mis-link. :P

Daniel Fischer

@TedShifrin Yes, it's a new thing. Couldn't have got it earlier.

Ted Shifrin

OK, @DanielF ... Thanks.

Dan Rust

@MikeMiller cheers for the suggestion.... looks like they haven't published regina yet, as they say to cite the program only by its download URL.

Ted Shifrin

I wonder if @Balarka has figured out sets of discontinuity for pointwise limits of continuous functions.

Hippalectryon

22:50

@user2692669 i.imgur.com/WBOCl3V.png, i.imgur.com/h8MbQ8k.png

user2692669

@Hippalectryon I cringed a little when I saw that it wasn't exactly 30 degrees xD

Hippalectryon

:P

user2692669

@Hippalectryon I thought of something but I'm not sure if it's right:

Mike Miller

@Ted: On my phone there are no arrows.

user2692669

@Hippalectryon Line RQ, if we "play" with it, forms a circle. So if we want the radius NOT to be perpendicular to the line then the line (PQ) must be crossing exactly two points and no more.

thus making it possible to be >90 OR <90 but with one solution per case (my head hurts :P )

Mike Miller

22:59

@DanRust: HM, maybe try looking at SnapPea then.

Hippalectryon

yep

Jessy Cat

23:32

0

Q: Without Cauchy Integral Formula or Residues: $\int_{L}\frac{dz}{z^{2}+1}$

I am currently working on the following: Prove that $\displaystyle \int_{L}\frac{dz}{z^{2}+1} = 0$ if $L$ is any closed rectifiable simple curve on the outside of the unit disc, i.e., $L$ is contained in the region $|z|>1$. Show that the equality is in general false for arbitrary closed rectifia...

complex-analysis complex-numbers complex-integration

Please help! I'm having so much trouble...

Akiva Weinberger

@TedShifrin I keep on missing when you're on

@robjohn How does on determined which of their comments are flagged?

Mike Miller

@AkivaWeinberger: A mod does.

Akiva Weinberger

Ah

'Cause I got that message recently

robjohn

@AkivaWeinberger one doesn't. A moderator does.

Akiva Weinberger

and I'm not sure what I could've said that got me flagged

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