Mathematics

12:37 AM

@anon: Well, you know there are plenty of indefinite quadratic forms. You can certainly do linear algebra with them, as opposed to positive definite ones. Of course, you have null vectors, etc. Personally, I've never worked with this, as I've stuck more with Riemannian and Hermitian.

anon

@TedShifrin sure, it's a thing we can do. I was hoping to have some semblance of meaning to it though, like with normal metrics.

like, what does the pseudo-R metric of two vectors actually tell us?

Ted Shifrin

Well, there's time-like and space-like paths, etc. Look at O'Neill's graduate text.

What does using an indefinite inner product tell you? It's the same thing at the tangent space level.

The motivation surely comes from relativity theory, as you already noted.

anon

@TedShifrin it would be the product of lengths and cosine of angle in some coordinates. this can't be the case with pseudo metrics though, since for instance <v,v> can be negative.

I'll check out the book, looks interesting

Ted Shifrin

right ... but I'm pointing out that your issue is with the linear algebra more than with the geometry.

OK, I know a bit of that stuff, but not much.

anon

my issue is with the linear algebra not coming with geometric motivation I would say

Ted Shifrin

12:51 AM

that's why people refer to null vectors, space-like, and time-like, etc. :)

angles only make sense when you restrict ...

anon

what?

Ted Shifrin

when you restrict to pairs of timelike or pairs of spacelike vectors. :)

Will Jagy

@anon, recommend Cassels Rational Quadratic Forms for beginnings, including Witt cancellation. A good deal easier than Lam (2005)

store.doverpublications.com/0486466701.html

@Ted, saw the related question on Main. I suspect a book on just quadratic forms is the thing

Prototank

1:26 AM

I'm having trouble showing that the Vitali Covering Lemma fails if the covering set is made up of open intervals. Something I've tried to do: enumerate the rationals and cover them with 1/2^i radius balls. The problem is that I don't think it is a cover.

Also I am failing to see how the openness of my choice causes the lemma to fail >:(

anon

2:19 AM

@Prototank I am not sure what your question is. the set you describe clearly covers the rationals. What are you trying to do with it?

Prototank

I'm trying to cover [0,1] with open intervals in the sense of vitali

but then show that the covering fails the lemma: that any finite collection of intervals you take from the collection must miss $\epsilon$ much of $[0,1]$.

anon

well, I don't think there ever exists a finite disjoint open covering of [0,1]

so any infinite open covering of it will fail to have such a finite subcollection

Ethan Bierlein

@Agawa001 I didn't design that. It's on the Chrome webstore, and it's a signed extension. What's your beef with Code Review anyways?

anon

suppose there were a finite disjoint open covering. let (a,b) be the leftmost interval, and let (c,d) be the secondmost left interval. then the supposed covering fails to cover (b,c), which has positive measure.

well, unless b=c, suppose I didn't think of that

then {(0,1/2),(1/2,1)} would even be a finite disjoint open covering modulo a set of measure zero, so I suppose this is harder than I just made it out to be

skill patrol

5:03 AM

So three logicians walk into a bar. The bartender says "Do all of you want a beer?" The first one says "I don't know," the second one says "I don't know," the third one says "Yes!"

Huy

6:42 AM

@skillpatrol: An engineer, a physicist and a mathematician are sitting in a train travelling through Scotland. At one stop, they can see a black sheep through the window. The engineer exclaims "all sheep in Scotland are black!". The physicst corrects him "no, from what we've seen all we can say is that some sheep in Scotland are black". Thereupon the mathematician says "we only know that in Scotland, there exists at least one sheep that is black on at least one side"

Rigor

6:55 AM

If there was a philosopher with them he would have said "if we didn't see the sheep would it still be black?" @Huy

Agawa001

8:05 AM

@EthanBierlein i have nothing against CReviewers but you should redefine the true behavior of a troll there, also, try to not downvote math-based suggestions. my point was generalized, I dont trust anything sent to me or proposed for me via internet.

Alec Teal

8:17 AM

@EthanBierlein I also hate code-review because it isn't about helping OP, it's about reviewing code. I gave a code with some python-like pseudocode and a 14 year old (genuinely) was like "OP CANNOT READ IT"

Agawa001

@AlecTeal oh really, you give lot of effort of sharing your own code made by yourself then yu get punched like this ? horrible

CR was much fun combined with SO

Patrick Stevens

8:35 AM

Can someone please verify the following lemma I have for math.stackexchange.com/questions/1460108/…

Wlog $0 \in A$, as stated. For all $\epsilon$ there is $x_1^{\epsilon}, x_2^{\epsilon} \in A$ with $|x_1^{\epsilon - x_2^{\epsilon}| = \epsilon$. There is a rational in the interval $x_1^{\epsilon}, x_2^{\epsilon}$ - say $q_{\epsilon}$.

Now there are uncountably many $\epsilon$ but only countably many possible $q$, so there is some $q \in \mathbb{Q}$ which appears infinitely often as a $q_{\epsilon}$. Therefore there is a rational limit point of $A$.

Agawa001

@Rigor i would say, poor sheep get itself overlaid with black paint

Chris's sis the artist

9:01 AM

@robjohn OK

robjohn

@Chris'ssistheartist Good morning :-)

Chris's sis the artist

@robjohn Morning! :-)

@robjohn I have a hard time with math these days despite the results I get that is due to the allergies that cannot be controlled in any way.

I suspect all comes from pushing me too much, I mean I'm very tough with myself.

robjohn

@Chris'ssistheartist antihistamines don't help at all?

Chris's sis the artist

@robjohn No

Agawa001

@Chris'ssistheartist dust ?

Chris's sis the artist

9:09 AM

Maybe I should not do math for some days (I cannot believe myself saying that). I do math even in my dreams, how could I even stop?

Agawa001

how can maths provoke allergy lol ?

Chris's sis the artist

@Agawa001 Mental fatigue might be a trigger.

Agawa001

@Chris'ssistheartist you must take some rest meanwhile

Huy

I hope you get better soon, @Chris'ssis!

Chris's sis the artist

@Huy Thanks.

Agawa001

9:11 AM

but maths is having nothing to do there, maybe anther issue

Chris's sis the artist

It's depressing one cannot control the symptoms, with one exception though that I try to avoid (sedatives seem to work).

Agawa001

@Chris'ssistheartist tbh, ido maths as a cure

i doubt such issue can be originated from maths

Chris's sis the artist

@Agawa001 It's not math, it's about the mental effort.

Huy

I try to avoid medication whenever possible. I once was drugged to a point that made me practically unable to do sport for over a year.

Agawa001

@Chris'ssistheartist lol ok.

Chris's sis the artist

9:13 AM

@Agawa001 For some years I didn't know the meaning of "break" although I tried to take a break once in a while. :-)

Agawa001

Sep 27 at 18:48, by anon

2 hours ago, by Daniel Fischer

Medical uses of Mathematics, part I.

Chris's sis the artist

@Agawa001 And extremely hard work for me was every single day, even when doing other activities, I did my best to make a progress on my math.

Agawa001

ask the fisher, he is experienced

Chris's sis the artist

@Agawa001 It's hard to determine someone like me to take breaks, that's the whole point since some longer breaks might help a lot. :-)

I do math to be on top, not to be just another one that does math.

Agawa001

@Chris'ssistheartist yes, thats the point, try to relax dont abuse your brain

Chris's sis the artist

9:18 AM

OK, 10 min break! :D

Agawa001

see ya in better case

Chris's sis the artist

9:41 AM

@Agawa001 People often talk about Wonders of the World but it's such an amazing thing when you discover the simple thing about yourself, that you can excel in anything, and this is available for all (perhaps with some exceptions related to medical issues).

Work extremely hard, with passion, and get anywhere, absolutely anywhere.

Agawa001

yes look at ramajunan, he was struggling for his aims without paying interest to his illness, until he died while trying. (i hope no such thing for you)

Chris's sis the artist

@Agawa001 :D

Agawa001

lot of scientists and researchers have this little deep weakness, but noone was throttled or obstructed by it , they keep going on

while people called em weak, they had to prove later how stronger are they

but this s nothing in comparison to your tiny allergy :D

Chris's sis the artist

:D

Agawa001

gosh, french is invading my english

for the nth time, i mistake a french word for english one

Mary Star

11:15 AM

When we have that $f_1$ and $f_2$ are smooth bijective maps, how can we show thst the composition $f_1 \circ f_2$ is also a smooth bijective map?

Tobias Kildetoft

@MaryStar Just apply the definitions

(plus chain rule I suppose)

Mary Star

A bijective map $f$ is surjective and injective.
A map is surjective if : $\forall y \exists x : f(x)=y$.
A map is injective if : $f(x_1)=f(x_2) \Rightarrow x_1=x_2$.
Is this correct?

Which is the definition of a smooth map? @TobiasKildetoft

Tobias Kildetoft

@MaryStar How can you hope to solve the given problem if you are not familiar with the terms?

3

Mary Star

Is a smooth map $f$ a map such that $f \in C^{\infty}$ ? @TobiasKildetoft

Tobias Kildetoft

@MaryStar Whereever you got the problem from, look up the definition there

Algebra

11:27 AM

I just want to say that this website is awesome and this answer makes me really happy:

http://math.stackexchange.com/a/32416/261481

Moses

11:52 AM

@DanielFischer Yip makes sense.

Ethan Bierlein

@Agawa001 I don't specifically downvote math-based answers. I downvote answers of they're unclear, and don't seem to provide an useful information to the OP.

@AlecTeal There are a few things wrong there. By reviewing the code, and providing a review, you are helping the OP. Secondly, pseudocode is off-topic. It always has been. Finally, are you referring to me when you say "some 14 year old"?

You two seem like folks with a grudge. Next time read the help center a little more carefully. Code Review is not some "unwelcoming" community. We love it when new users become active members, we love new contributions, and we certainly love it when new users become active in our chatroom.

N3buchadnezzar

12:16 PM

heya

Moses

1:12 PM

@DanielFischer If you have a chance please look at this post. I prove part of the Riesz decompostion theorem without doing the embedding but I also show that we can get $\sigma_{\mathcal{A}}(\sigma(a)) = \sigma_{k}$. Let me know where the problem is. Thanks.

Agawa001

@EthanBierlein i think you are exagerating facts here, there is nothing in rapport with hate or virtual points, its just i might be in the wrong place all that time.

Ethan Bierlein

@Agawa001 No, you weren't in the wrong place. You were in the right place. Code Review and it's chat rooms are great communities to be a part of, but you decided to be a troll and annoy the hell out of everyone. You're the misguided one here.

Agawa001

lol, ok i ll not go further with this but try to not type CR literally, i dont want this converstaion to port in your chatroom thru this duga thing

Ethan Bierlein

Duga only scrapes comments off Stack Overflow.

This conversation won't be picked up.

Agawa001

CR is wonderful place, no doubt; thanks

skill patrol

1:51 PM

@N3buchadnezzar hi pal

N3buchadnezzar

Does anyone know how I can find the n'th term of the continued fraction expansion of e?

skill patrol

Which continued fraction expansion?

N3buchadnezzar

List of representations of e

The mathematical constant e can be represented in a variety of ways as a real number. Since e is an irrational number (see proof that e is irrational), it cannot be represented as a fraction, but it can be represented as a continued fraction. Using calculus, e may also be represented as an infinite series, infinite product, or other sort of limit of a sequence. == As a continued fraction == Euler proved that the number e is represented as the infinite simple continued fraction (sequence A003417 in OEIS): Its convergence can be tripled by allowing just one fractional number: Here are some infinite...

First one here

Mary Star

2:23 PM

When we have that $t^m=0$ where $m$ is a positve integer, do we conclude that $t=0$ ?

Chandan

Hello can some some explain the below sequence
A1 = 4: 4, 7, 19, 49 ...

Agawa001

3:24 PM

@Chandan that would fit for a puzzle, but you can use OEIS

oeis.org/A133264

Chandan

i see that it fits the puzzle

i do not see any equation or any explanation

Agawa001

2 mins ago, by Agawa001

https://oeis.org/A133264

Smallest number whose sum of digits is 3n+1.

Mary Star

When we have that $t^m=0$ where $m$ is a positve integer, do we conclude that $t=0$ ? @DanielFischer @robjohn @SamuelYusim

robjohn

@MaryStar If you are in an Integral Domain, yes

Agawa001

@robjohn you last answer for derivating trigonometic functions was really cool

Mary Star

3:38 PM

I am looking at this exercise math.stackexchange.com/questions/1461063/… @robjohn So in this case it stand that $t=0$, right?

robjohn

@Agawa001 Thanks. It took me a while to dig up that old answer, but I knew I wanted to use that image.

@MaryStar Is it not the case that $t\in\mathbb{R}$?

@MaryStar Isn't $\mathbb{R}$ an integral domain?

Balarka Sen

@Alizter You figured out the thing about finite subgroups of $S^1$ from yesterday?

I was actually getting you to classify all discrete subgroups of $S^1$, by classifying discrete subgroups of $\Bbb R$ in turn.

Huy

3:57 PM

@MaryStar: look at the backwards direction first if you don't see the mistake you've made...

robjohn

@MaryStar $\left(t^2,t^3\right)''=(2,6t)$ and $\left(t^2,t^3\right)'''=(0,6)$. Plug in $t=0$.

Agawa001

@robjohn do you think there is a mathematical proof of this

Balarka Sen

@robjohn More generally, if and only if you are in a reduced ring.

Agawa001

mathematically saying, i find this way far to apply

robjohn

@BalarkaSen don't know that I've ever heard of that. When I think of "no zero divisors", I think "integral domain".

Agawa001

4:06 PM

i could give it a try if the op asks for conditions, but listing all the range ! i just cant

Balarka Sen

@robjohn Having no zero divisors is stronger than having no nilpotents.

robjohn

@BalarkaSen well, of course (I assume that is what "reduced" means). However, I'd have needed to know what "reduced" meant beforehand.

Balarka Sen

yes, that's what reduced means.

robjohn

@BalarkaSen in any case, I believe that $t\in\mathbb{R}$ and that is an integral domain, so her statement is true.

Balarka Sen

I agree. I was merely pointing out that it's sufficient for a ring to be reduced for $x^m = 0 \implies x = 0$ to hold, but all the same.

robjohn

4:13 PM

@BalarkaSen Well, I learned what a reduced ring is.

Alizter

@BalarkaSen no

Balarka Sen

Just a terminology though :) The thing is, you can find lots of interesting rings which are not domains, yet reduced. A buckload of such rings come as ring of functions over algebraic varieties, say.

@Alizter Do you know what the discrete subgroups of $\Bbb R$ are?

Mary Star

We have that $\gamma ''(t)=(m(m-1)t^{m-2}, n(n-1)t^{n-2})$ and $\gamma '''(t)=(m(m-1)(m-2)t^{m-3}, n(n-1)(n-2)t^{n-3})$.

So that $\gamma ''(0)$ is non-zero it must stand: $m=2$ or $n=2$ but not $m=n=2$.

If $m=2$ it must be $n=3$, so that $\gamma '''(0)$ is non-zero.
If $n=2$ it must be $m=3$, so that $\gamma '''(0)$ is non-zero.

Is this correct? @robjohn @Huy

Alizter

@BalarkaSen No idea

robjohn

@MaryStar so that $\gamma'''(0)$ is non-zero and not linearly dependent with $\gamma''(0)$

Balarka Sen

4:19 PM

Yikes. Then that is going to be a problem.

Alizter

I never studied R as a group

Always permutations and symmetries

Mary Star

4:32 PM

Ah yes... Thank you very much!! :-) @robjohn

Kelsey

6:10 PM

What is 940 + sqrt(10) divided by 12321 plus pi^2

Razieh Noori

how is some matrix

0

Q: A question on pseudospectra of matrix polynomial

Suppose: ${A_j} \in {\mathbb{C}^{n \times n}},0<{w_j}\in \mathbb{R} (j = 0,1,2....m)$ ${\rm{P(}}\lambda {\rm{) = }}{{\rm{A}}_m}{\lambda ^m} + .....{A_1}\lambda + {A_0}$ is a matrix polynomial, and $\lambda $ is a complex variable. ${\rm{q(}}\lambda {\rm{) = }}{{\rm{w}}_m}{\lambda ^m} + ........

who here has good knowledge about operator theory?

sorry i found answer

PVAL

6:35 PM

Huh On arxiv, apparently Lueck has 64 papers under Lueck and 14 under Luck with an umlaut over the u.

Huy

bad luck

PVAL

@Huy Your topology department is chaired by his student right?

Huy

@PVAL if you tell me who his student is, maybe I can answer

@PVAL I'm not too much into topology so I don't know by heart

PVAL

Err, someone on here said their department was chaired by a student of Luecke. It probably isn't you then.

Huy

no, probably not

I don't even know who's a topologist at my uni

I only know the analysts and the algebraists and the numerical guys

PVAL

6:40 PM

Salamon is

Huy

really?

he was never my teacher but I know him

he plays the guitar too

I thought he was some analyst/geometer

PVAL

symplectic toplogist/geometer

Huy

cool

how do you know him?

PVAL

Think its a matter of what he wants to call himself as to what side he falls on

ive read books by him.

I don't know him personally.

Huy

ic

user153330

6:57 PM

@BalarkaSen terminology is very important

@MaryStar i think you should revise bro

Mary Star

@user153330 What have I done wrong?

skill patrol

7:38 PM

@user153330 but don't let it get in the way of your train of inquiry, right?

Agawa001

^ what does this mean ?

means dont be inquisitive about english terminology ?

skill patrol

No, I'm referring to terms getting in the way of thinking

Chris's sis the artist

8:07 PM

I was trying to explain to some kid how to think about the problem with the arrangement of n persons around a circular table. I was also thinking that often it happens that the text of the problem might not be clear for a kid.

Of course, I can judge things in different ways, but I think the text should be put in a very clear way such that there is no doubt about the way to go.

Hippalectryon

@Chris'ssistheartist What's the original text ?

Chris's sis the artist

I mean math problems should not be a puzzle, but a math problem.

Balarka Sen

The problem you mention is definitely a math problem, not a puzzle.

Chez Cosette

Transcript for

$Mathematics$ Mathematics