Mathematics - 2015-01-27 (page 2 of 3)

Ramanewbie

09:02

@DavidZhang YEAH I was looking for this one, tks !

Mike Miller

@DavidZhang Got the email. Thanks for the file.

See ya folks

Ramanewbie

@DavidZhang Where are you from ?

David Zhang

@Ramanewbie I'm a high school student in Tennessee, why do you ask?

Ramanewbie

@DavidZhang Just your name ^^

David Zhang

Oh, my parents are Chinese immigrants, if that's what you're getting at.

Ramanewbie

09:09

ok.

Vim

09:44

Hey!

Feel like this site is full of brilliant real mathematicians

~_~

Anybody like me, a college freshman and just began calculus and linear algebra courses?

Huy

Good morning, @MikeMiller: I assume you're already asleep. :D

Vim

@Huy It's afternoon in my zone

Huy

@Vim: Too bad.

Chris's sis

Greetings

Sayan Chattopadhyay

Hi guys

David Zhang

09:54

@Vim I'm in high school, and definitely not a brilliant real mathematician.

Sayan Chattopadhyay

@DavidZhang hi

David Zhang

@SayanChattopadhyay hey

mikeonly

10:21

Hi.

Is there some name for limits of $\frac{\sin(x)}{x}$ at 0 and $(1+\frac{1}{x})^x$ at $\infty$ in English?

In Russian there are generally called “wonderful limits”.

Sayan Chattopadhyay

u r Russian @mikeonly

mikeonly

@SayanChattopadhyay I am taught in Russian. I am Ukrainian though.

Huy

@DanielFischer: Is there a simple way to show that $(\partial_{x^1})_p, \dots, (\partial_{x^n})_p \in T_p M$ are linearly independent? The proof we used introduces some weird cutoff function without any motivation, in my opinion, and seems a bit far-fetched.

mikeonly

@Huy, can you remark on my question concerning those limits?

Huy

@mikeonly: No.

mikeonly

10:31

So there is no way to refer to them except for explicitly defining them?

Huy

@mikeonly: What is the precise definition of a "wonderful limit"?

Vim

Hey guys

Just had dinner

$(\partial_{x^1})_p, \dots, (\partial_{x^n})_p \in T_p M$ well , the code can't be visualized here

mikeonly

@Huy It's just a conventional name only for these ones. I think it's because they are used throughout.

Jan 17 at 21:31, by robjohn

Chat guidelines | $\LaTeX$ in chat

Vim

well I click on start chat Jax but it doesn't work?

Daniel Fischer

@Huy You use a cut-off function so that you have no problem extending your function that is annihilated by all but one of the differential operators to the whole manifold. If that's the proof you're looking at. Of course, if you define $T_pM$ as the space of derivations on the germs of smooth functions, the linear independence is practically built into the definition.

Vim

10:40

just opened a blank page

@mikeonly Would you please show me how? Many thanks!

Huy

@DanielFischer: In GR, we defined vectors as derivations satisfying the Leibniz rule. We did not prove there that they are linearly independent but just assumed it to be given. In DG however, we defined them as directional derivatives. So there's no simple way to avoid the cut-off functions, right?

Najib Idrissi

@Vim You need to add that link as a bookmark, and then when you're on this page (the chat), click on the bookmark

mikeonly

@Vim math.ucla.edu/~robjohn/math/mathjax.html follow these instructions.

Daniel Fischer

@Huy Depends on whether you defined them as acting on locally defined functions, or globally defined. If locally defined, you of course don't need a cut-off. If globally, cut-off is the easiest way to extend to the whole manifold.

Huy

@DanielFischer: Globally, unfortunately.

@DanielFischer: Is there a reason to define the vectors as directional derivatives as opposed to derivations, except for intuition? It seems to me most of the proofs are much easier using the more abstract definition.

Daniel Fischer

10:47

@Huy Each of the definitions has its advantages and disadvantages. Directional derivatives are easier to see the concept of, derivations are easier for algebraic manipulations.

The wise course uses both. And possibly also the third classical definition as the space of equivalence classes of some curves.

Vim

Strangely when I right click the Start ChatJax link (I think you mean this link) there is no such option as "Add as bookmark" in the menu.

Perhaps something wrong with my Chrome..

Huy

@DanielFischer: Something else I noticed: In DG, we defined vector fields as maps from $M$ to $TM$, which to me seems like the most natural definition. In GR however, vector fields were defined to be linear maps $X: C^\infty(M) \to C^\infty(M)$ satisfying the Leibniz rule. How are these two definitions equivalent? Or are they different things?

Vim

but that doesn't matter much, thank you anyhow, @NajibIdrissi and @mikeonly

evinda

Could I ask someone something about red-black trees?

Huy

@Vim: Drag and drop the link to your bookmark bar.

Vim

10:52

Amazing!! it works

Thank you Huy

Daniel Fischer

@Huy If you have a smooth section $V$ of $TM$, then $X \colon f \mapsto (p\mapsto V(p)f)$ is a vector field in the GR sense. In the other direction, you go by coordinate functions times cut-off functions.

Huy

@DanielFischer: How come I can't find anything about the GR sense vector fields on wikipedia etc.? Is that a very unusual definition?

Daniel Fischer

@Huy Better ask Ted that. I'm not so au courant with matters differential geometry.

Huy

Okay, thanks anyways for your help, @DanielFischer.

Daniel Fischer

De rien.

Huy

10:55

I prefer je t'en prie.

Daniel Fischer

(Da nich für, if you prefer)

Huy

That must be a North German expression.

Daniel Fischer

It is.

Huy

Hence I've never heard of it before.

evinda

@DavidZhang Could I ask you something about red-black trees?

Sayan Chattopadhyay

10:58

@Huy can in deform an object as i like in topology continuously

I*

But then wont that lead a huge array of different kind of objects

and that will be very difficult to study

Can someone answer my doubt

evinda

According to my lecture notes:
Let x be the child of the node that we delete. Let w be its sibling node and p the father of x.
There are four cases:
1. Case 1: w is red. We cahnge the color of w to black and of p to red and we make a left rotation around the father of x. Now we have one of the cases 2,3 or 4. For example the case 2 is this: Both of the children of w are black. We chage the color of w to red, x to black and we transfer the black that we subtracted from w and x to p. If p was red it becomes black and the algorithm terminates. Otherwise, p gets double black and the algorithm c…

Sayan Chattopadhyay

@evinda this a question related to avl tree right

evinda

@SayanChattopadhyay No to red-black trees

Sayan Chattopadhyay

oh....

Well do you like number theory

evinda

@SayanChattopadhyay Me?

Sayan Chattopadhyay

11:10

yes

evinda

Yes, I do

Vim

.... I'm still struggling with chain table, C language

Sayan Chattopadhyay

so can u answer my question

Just have a look at the question if u dont fell like answering dont@evinda

evinda

I will take a look at it later because I have an exam in a few hours

Sayan Chattopadhyay

oh fine i will ping u then

A Beautiful Mind

11:12

Hello @Chris'ssis.

Sayan Chattopadhyay

Hello @ABeautifulMind

A Beautiful Mind

@SayanChattopadhyay Hi, don't ask me to answer your question.

Chris's sis

@ABeautifulMind Hello. How are you doing?

Sayan Chattopadhyay

Sure

A Beautiful Mind

@Chris'ssis Not good. But I will try to get better.

Sayan Chattopadhyay

11:13

Well @ABeautifulMind are u reading any books related to maths currently

A Beautiful Mind

@SayanChattopadhyay I am not. Are you in high school?

Sayan Chattopadhyay

Yes

Chris's sis

@ABeautifulMind That's good.

Sayan Chattopadhyay

@ABeautifulMind why this question about high school

A Beautiful Mind

@SayanChattopadhyay Nothing, just talking.

Sayan Chattopadhyay

11:16

Oh and what are u doing @ABeautifulMind

A Beautiful Mind

@SayanChattopadhyay I am not working. I am trying to recover from my mental illness.

evinda

0

Q: Changements that have to be done in order to delete node of red-black tree

According to my lecture notes: Let $x$ be the child of the node that we delete. Let $w$ be its sibling node and $p$ the father of $x$. There are four cases: At the first case, $w$ is red. We cahnge the color of $w$ to black and of $p$ to red and we make a left rotation around the father of $...

Does anyone have an idea?

Sayan Chattopadhyay

Mental illness?

No sorry @evinda

@ABeautifulMind what kind of mental illness

A Beautiful Mind

@SayanChattopadhyay It's complicated, I won't say more.

Sayan Chattopadhyay

Oh fine then

but before that what did u used to do

A Beautiful Mind

11:20

I have been talking to Ethan via email. He is not in a good situation either, but I won't say more.

@SayanChattopadhyay I used to be a teacher.

Sayan Chattopadhyay

okhay

@Chris'ssis hows your progress on your book

Chris's sis

@SayanChattopadhyay I'm working on some very hard modules. Well, all goes fine. Thanks for asking.

Sayan Chattopadhyay

Can anyone tell me how do u publish a paper in maths about conjectures

Chris's sis

@SayanChattopadhyay I only published articles with full solutions and proposed problems. About conjectures ... I'd like to know more too.

Sayan Chattopadhyay

so if have to propose problems how do i go about it

in articles@Chris'ssis

Chris's sis

11:25

@SayanChattopadhyay It depends on where you wanna send your proposed problems.

Sayan Chattopadhyay

i want to send my problem

can u tell me some ways i can do that@Chris'ssis

Chris's sis

@SayanChattopadhyay Sure. For AMM, you can go here maa.org/publications/periodicals/submissions-to-maa-periodicals

Sayan Chattopadhyay

i mean i have to go with more details right

Chris's sis

@SayanChattopadhyay An important thing: proposed problems and the research articles cannot be sent in the same way. For the latter ones you need to regitser on their site and provide with more information.

Sayan Chattopadhyay

okay and is my problem of that level that i can send it there

Chris's sis

11:28

For proposed problems, use [email protected]

Sayan Chattopadhyay

@Chris'ssis is my question of that standard or level that i can send it there or solve it in myself

Chris's sis

@SayanChattopadhyay I don't know what you mean. When you send things to them you need to follow some rules described on site.

Sayan Chattopadhyay

i mean that will the question be thought about by people or is my question just a wastage of time

Chris's sis

@SayanChattopadhyay I didn't see such things so far. All questions are precious, from each one you can learn precious lessons.

Sayan Chattopadhyay

Thanks for the advice

and will the charge u for sending a question

Chris's sis

11:34

@SayanChattopadhyay You are the creator of the question, so you need to send it to them. :-)

Sayan Chattopadhyay

i mean any fee for sending the question

Chris's sis

@SayanChattopadhyay No, there is no such a fee.

Sayan Chattopadhyay

thanks then for the knowledge

@Chris'ssis can u provide me with the link

Chris's sis

@SayanChattopadhyay In general, when you send a question with a solution, put all in latex, follow their guidelines, and I'm sure that if what you proposed is considered proper to be published, it will be published.

Sayan Chattopadhyay

PLS can u give me the link

Chris's sis

11:37

@SayanChattopadhyay maa.org/publications/periodicals/submissions-to-maa-periodicals

Proposed Problems and Solutions should be sent to:

Doug Hensley, Monthly Problems
Department of Mathematics
Texas A&M University
College Station, TX 77840

In lieu of duplicate hardcopy, authors may submit PDF’s to [email protected].

Martin Sleziak

in Tagging, 2 mins ago, by Martin Sleziak

To summarize the discussion in the comment to the post discussed above: graph-symmetries combines both question from graph theory and questions. Does it seem reasonable to remove this tag and use symmetry in combination with either graphing-functions or graph-theory insted?

(Maybe more users will notice a tag-relationed issue when mentioned here rather than on meta or in the tagging chatroom.)

Ramanewbie

@DanielFischer do you speakl French ?

Daniel Fischer

@Ramanewbie Un peu. Pas trop, parce que j'oublie beaucoup de la langage quand je ne suis pas en France.

Ramanewbie

@DanielFischer have you been in France then ? Where ?

Daniel Fischer

11:54

@Ramanewbie Mostly in the Pyrenées and in Savoie, but also a bit on the Cote d'Argent and in the Massif Central, plus once in the Normandie. And a bit here-and-there occasionally.

Ramanewbie

What about Arthur @daniel

Daniel Fischer

@Ramanewbie J' sais pas.

Ramanewbie

ok

Huy

12:33

@DanielFischer: After having defined the tangent bundle, we show that it has a unique naturally induced structure of a $2n$-dimensional smooth manifold. Is that of any relevance? I don't see it explicitly being used in the later course anymore.

Daniel Fischer

@Huy Without the manifold structure, you couldn't even say that the differential of a smooth function is smooth, and define higher derivatives.

Huy

@DanielFischer: Hm. I have to think about that.

@DanielFischer: By "differential of a smooth function" do you mean $df: TM \to TN$ of $f: M \to N$?

Daniel Fischer

Yes, that.

Huy

(I'm sorry for those trivial questions but different notions being used than what I've seen in GR confuses me a lot)

Ok, that makes sense.

It doesn't seem like we say the differential is smooth any time soon. Let's see what happens in the lecture notes next.

The Artist

4

Q: A plan to defeat a betting game where the odds of winning are 50/50. Help me understand why it's flawed.

My friend has this plan where he implies that it's impossible to lose, as long as the odds of winning are 50/50 on each bet. His idea is that basically you keep doubling your bet until you win and then start over again. So for example, you bet 1 dollar and you lose, your net profit is now -1 dol...

probability gambling

A Beautiful Mind

12:48

@huy How is the game theory class?

The Artist

^^^^ a very interesting question

@ABeautifulMind how's antartica ?

Huy

@ABeautifulMind: It hasn't started yet. I'll start preparing for it some time next week, and it will start in about 3-4 weeks.

A Beautiful Mind

@TheArtist It sucks, as usual.

The Artist

@ABeautifulMind how much does a banana cost their?

A Beautiful Mind

@TheArtist You should ask how I am instead, silly.

@TheArtist Very cheap. I eat bananas all the time.

The Artist

12:49

@ABeautifulMind how are u

@ABeautifulMind how much does a mango cost their?

A Beautiful Mind

@TheArtist I am not good. I am giving myself another 6 years maximum to get well, study and enter grad school.

@TheArtist I don't know about mangoes but they are more expensive than bananas.

The Artist

@ABeautifulMind oh

@ABeautifulMind hope you get well soon! Which grad school are you planning to go to

A Beautiful Mind

@TheArtist Any which accepts me. I would be lucky to get into even one, when people consider my history.

The Artist

@ABeautifulMind do they look into medical history ? I don't think so...

A Beautiful Mind

@TheArtist Well, never mind. How are you?

The Artist

12:53

@ABeautifulMind I'm good...do you use AIR CONDITIONING in Antartica?

A Beautiful Mind

@TheArtist Yes, I do. I have it in my room.

The Artist

@ABeautifulMind cool (y)

Huy

@DanielFischer: After introducing smooth vector fields, we look at the example $TS^1 \cong S^1 \times \mathbb{R}$ and afterwards it is stated that the analogous does not hold for the $2$-sphere, in words "there is no non-zero vector field". This is supposed to state "no non-zero smooth vector field", right? That would be the hairy ball theorem, correct?

Daniel Fischer

@Huy You can relax "smooth" to "continuous", yes, known as the hairy ball theorem, or in German, der Igelsatz: "Ein stetig gekämmter Igel hat mindestens einen Glatzpunkt."

Huy

Ok.

@DanielFischer: We used smooth vector fields in this context because we used the fact that the tangent bundle of $M$ is diffeomorphic to $M \times \mathbb{R}^n$ iff $M$ possesses $n$ smooth vector fields that are linearly independent at each point.

Can I relax that statement with continuity as well?

Daniel Fischer

13:01

@Huy Yes. If the fields are linearly independent at a point, all close enough fields will be too, and you can approximate a zero-free continuous vector field as good as you want by a (zero-free) smooth field.

r9m

13:13

@Chris'ssis have you seen this one ? $\displaystyle \sum_{n=1}^{\infty}(-1)^{\frac{n(n+1)}{2}}\frac1{n!}\int_0^1x(x+1)\cdots(x+n-1) dx$

Chris's sis

@r9m Possibly. Why that $(-1)^{\frac{n(n+1)}{2}}$? It looks annoying ... maybe you wanna remove it.

r9m

@Chris'ssis oh! I just read Olivier's answer on main to this Q :) nice

Chris's sis

@r9m hmmm, let me see ...

@r9m I might have used a similar way, at least for the beginning.

r9m

@Chris'ssis okay :)

Chris's sis

@r9m Soon it's going to appear a generalization I did with a mathematician on a problem where I used that Stirling's formula along with a clever use of the symmetry.

Lembik

13:22

can anyone suggest tags for math.stackexchange.com/questions/1121774/… ?

Chris's sis

@r9m don't worry, I'll show it you. Just to remember when it's published.

r9m

@Chris'ssis okay :D

Chris's sis

@r9m by the way, did you make some progress on my problem posted yesterday? :D At least tell me if you know any approaching idea.

r9m

@Chris'ssis I keep forgetting .. sorry is it the one with the radicals ?

Chris's sis

@r9m I tested it on some mathematicians. Some asked me: "is this a joke?" I replied: "no, it's not". I admit it blows up minds by its very nice good-looking. :-)

@r9m Yeap.

r9m

13:28

@Chris'ssis lol .... okay :P

Chris's sis

@r9m actually, I know no one that can do it. I like to create stuff with unusual approaches like Ramanujan.

r9m

@Chris'ssis would you decapitate me if I told you I don't remember the problem statement ? :P (I thought I had copied it to a text file .. seems I haven't :( .. )

Chris's sis

@r9m hahaha, if you don't have it then maybe you didn't like it. No worry. :-)

Lembik

I've improved it now math.stackexchange.com/questions/1121774/…

sorry for the earlier version

r9m

@Chris'ssis what kind of inference is that ? -_-

Chris's sis

13:34

@r9m Is there something posted on internet you don't have? :-)))

r9m

@Chris'ssis are you accusing me of being a database ? -_-

Chris's sis

@r9m lol, the largest one I know :-))))))))

r9m

@Chris'ssis You'd be surprised to see how little I have in comparison to what I want to have :P lol

Chris's sis

@r9m lolll, I'm sure of that! :D

r9m

:P rofl

Chris's sis

13:38

:D

r9m

@Chris'ssis I'll bbl .. :)

I've run out of cigs ...

Chris's sis

@r9m Do you smoke? OK

Huy

Smoking is not only bad for your health but also disgusting, in my opinion.

@DanielFischer: Any idea what the symbol $\sqcap$ usually denotes? I've seen it for the first time, our prof denotes the set of orientations on a vector space as $\sqcap(V)$, but I've never seen that before.

Daniel Fischer

13:54

@Huy It's rarely used, no idea what the most un-rare use is.

David Zhang

@Huy I've seen it used to denote the greatest lower bound, but I agree, it is not commonly used and does not really have a "usual" meaning

Huy

@DanielFischer @DavidZhang: Any idea how you would "say" it? Our prof. said he didn't know what the name of the symbol was so he would just say "table V".

David Zhang

@Huy No, I have no idea. If I had to pronounce an expression containing it, I would just say what it stands for, e.g. "the set of orientations on V"

Daniel Fischer

How to say it would depend on what it's used for. Here, "orientations" could be an option.

Chris's sis

@Huy It's weird that while I was smoking I was even more creative than now. But, well, it had other undesired side-effects.

Huy

14:04

@Chris'ssis: Doesn't have to be a causal link between the two things.

@Chris'ssis: My father was forced to stop smoking when he started going out with my mother, and he become much more creative then. :P

Chris's sis

@Huy That's good! :-) However, it's known the nicotine stimulates (some processes in) the brain.

A Beautiful Mind

14:22

@Chris'ssis I know a professor who smokes a lot a lot.

Huy

@DanielFischer: We want to check for $n=1,2$ whether $\mathbb{R}P^n$ is orientable. The argument given in the notes is a bit unclear to me. It starts with a double cover $\pi: S^n \to \mathbb{R}P^n = S^n / \sim$, and then concludes $\mathbb{R}P^1 = S^1 / \sim \cong S^1$ is orientable whereas, since the Mobius strip $\subseteq \mathbb{R}P^2$, $\mathbb{R}P^2$ is not orientable. How can I see that the Mobius strip is contained in $\mathbb{R}P^2$?

@DanielFischer: Also, how is $S^1 / \sim \cong S^1$?

Daniel Fischer

@Huy Restrict your attention to a closed hemisphere. Hammer it flat to become a closed disk in your mind if you want. Think a "straight" strip on the closed hemisphere along a great circle passing through the pole. The identification of antipodes on the ends of the strip lying on the equator of the sphere glues the strip to be a Möbius strip.

@Huy Take a closed semicircle and glue the endpoints together.

Huy

I see.

@DanielFischer: We defined an immersion as a smooth map $f: M \to N$ with $df(p)$ being injective for all $p \in M$. All of this definition is completely clear to me but I have a really hard time visualising what an immersion might look like, or how I can imagine $f$ if I know its differential is injective. Do you have any advice for that?

(@DanielFischer: Since we defined vectors through directional derivatives, we defined the differential by mapping some path inducing the vector via $f$ on $M$ and then the result is the corresponding vector on $T_{f(p)}N$, if that is of any relevance)

(the path is mapped on $N$, not $M$, of course)

Also, in this context, what is the difference between the differential of a function and the push forward of the tangent map? The former is used in DG, the latter in GR, in my courses. They seem to be the same, right? @DanielFischer

Daniel Fischer

14:43

@Huy Each point of $M$ has a neighbourhood $U$ such that $f(U)$ looks like the graph of a function $\mathbb{R}^{\dim M} \to \mathbb{R}^{\dim N - \dim M}$. That's not necessarily particularly helpful, though. Globally, images of immersions can be less nice, $8$ is an immersed $S^1$ for example (But $8$ isn't bad, just one easy self-intersection. However, it's the most-easily typed example.)

@Huy "push forward of the tangent map"? The differential is also often called the tangent map. But also $df(p) \colon T_pM \to T_{f(p)}N$ is often called the tangent map. Not sure which one is meant.

Huy

@DanielFischer: Sorry, I confused something. The tangent map is also called push-forward, in GR, and its adjoint the pull-back. So push-forward is the same as differential, correct?

I'll try to make use of your description of immersions.

Daniel Fischer

@Huy Yes, one can also use push-forward and pull-back. While calling the tangent map push-forward is somewhat rare, it's very common to call the map on the differential forms the pull-back (by $f$).

Huy

Ok, good to know.

A Beautiful Mind

15:08

Hello @DonLarynx, don't say goodbye, you are my friend.

Mike Miller

Morning.

Don Larynx

Hello

Huy

Morning, @MikeMiller: I watched you sleep.

Mike Miller

Thanks for looking out for me.

Gato

@robjohn using that $\pi$ is irrational we have that (and your answer) $\left|e^{ip}-1\right|=\left|e^{i(p-2\pi q)}-1\right|\le|p-2\pi q|<\frac{1}{q}<\varepsilon$ so there are integers $p$ for which
$e^{ip}$ is arbitrarily close to ( but not equal to ) $1$.

A Beautiful Mind

15:20

@Huy Are you in love with him?

@MikeMiller Although you don't talk much to me in chat, I am glad you emailed me.

Mike Miller

@Huy You might prove that if a group $G$ acts by orientation preserving diffeomorphisms, then $M/G$ is orientable.

Sure.

Huy

@MikeMiller: Sorry, I have to waste some more time with trivial examples first, because they're too difficult for me.

Mike Miller

@Huy But this is your trivial example.

Huy

I'm looking at other trivial examples.

Mike Miller

Oh, I see. How trivial.

Don Larynx

15:24

Anyone remember how to find a perpendicular line to $y = mx + b$? Good times :D

I haven't had to do this since 9th grade

Huy

@DonLarynx: Negative reciprocal slope gives you a right angle.

A Beautiful Mind

@DonLarynx The product of the gradients is -1.

Don Larynx

Indeed, indeed. Thanks all.

Ilya_Gazman

Good morning people

A Beautiful Mind

That would be 1 million USD.

Ilya_Gazman

15:28

There is so much snow here in NY

Don Larynx

would be, should be, could be @A

A Beautiful Mind

That makes no sense @don.

Don Larynx

@A You should look into Number Theory by Andrews. I like it.

Mike Miller

@Ilya_Gazman I've heard predicted snowfall is between seventy and eighty feet.

A Beautiful Mind

When did I become @A @don?

Ilya_Gazman

15:30

@MikeMiller you probably meant inches. It covers the wheels of the cars

A Beautiful Mind

@Ilya_Gazman LOL, inches and feet are so different!

Mike Miller

No, @Ilya, I did not.

Ilya_Gazman

@ABeautifulMind, @MikeMiller Hey amy be you can help me with a question that I posted?

3

Q: Integer Factorization: Possible progress

I build an algorithm for solving Integer Factorization problem when the number to factor is selected by multiplication of two prime numbers with the same bit count. So far I only been able to factor 60 bit numbers. I want to explain to you how it works and perhaps you be able to suggest me ways ...

factoring

Huy

@MikeMiller: I'm looking at the statement that a curve is an immersion iff it is regular. How can I imagine $d \gamma(p)$ for some curve $\gamma$, or more importantly the tangent space of an interval? It's just $\mathbb{R}$, I guess, but somehow I get all confused trying to imagine this setup.

A Beautiful Mind

@Ilya_Gazman I am only a broken mind.

Ilya_Gazman

15:35

@ABeautifulMind look at it, it might be the fix that you need

@MikeMiller at the place where I leave they only been speaking about 1-2 feet of snow.

Mike Miller

@Ilya You must be lucky. You don't have to deal with the yetis when there's so little.

Huy

@ABeautifulMind: Yes, I am. Please don't tell my girlfriend.

Mike Miller

@Huy: The tangent space of an interval is just the interval times $\Bbb R$. I just imagine it not very tangently at all, as $I \times \Bbb R$, each copy of $\Bbb R$ perpendicular to $I$.

A Beautiful Mind

@Huy Are you breaking up with her? Many people in this chat have broken up recently.

Ilya_Gazman

@MikeMiller who are the yetis? lol

Huy

15:37

@ABeautifulMind: I didn't plan to do that anytime soon. I'll go snowboarding with her, this weekend, if you must know.

@MikeMiller: And how can I think about $d \gamma(p)$? The basis of the tangent space of the interval is just $\partial_t|t=p$, right? Which vector in $T_{\gamma(p)}M$ will this be mapped to?

Mike Miller

I don't know if you're going to get anything more satisfying than chasing the definition.

robjohn

@Gato Then, using that $e^{ip}$, you can make a partition of the circle by arcs no bigger than $1/q$.

Mike Miller

If you pick coordinates, it's the actual derivative of the curve at that point.

robjohn

@Ilya_Gazman we had some much needed rain here last night.

Huy

This is all too trivial for me to understand, @MikeMiller.

Ilya_Gazman

15:42

@robjohn last night I saw how snow going bottom up, instead just falling down, that was crazy.

Mike Miller

@Ilya_Gazman When the snow hits around 10 feet, around 5% of the population has latent genes triggered, ala X-men, and morph into yeti like beasts. These beasts are mostly friendly, and you can make a valuable friend if you are willing to offer some hunks of meat.

Huy

@MikeMiller: Too confusing if we're suddenly in $\mathbb{R}^n$, or even subsets of it. I can't think properly anymore. But I already had a cup of coffee. :(

robjohn

@Ilya_Gazman wind?

Balarka Sen

@PedroTamaroff Fair enough, but stuff like homological algebra is definitely interesting on it's own.

Ilya_Gazman

@robjohn yep, more like blizzard.

Balarka Sen

15:44

Even though the motivation comes from homology in algebraic topology.

robjohn

@Ilya_Gazman Ah... I experienced some of those when I was in grad school.

Mike Miller

@Huy The problem, I think, is that beause $I$ is in $\Bbb R$, you're automatically taking it in coordinates (and thinking of the basis as $\partial_t$). This makes it harder to think abstractly, which you need to do when your codomain is a manifold.

Huy

How do I think properly then? I can't control the way my brain works, apparently.

Ilya_Gazman

@robjohn may be you can help with the bounty that I posted?

robjohn

@Ilya_Gazman Let me look...

Don Larynx

15:47

@Ilya_Gazman it seems unclear what you are asking.

Mike Miller

@Huy The definition of the tangent space at a point on a manifold is the space of functionals on the space of functionals on the space of functions at that point etc

Do you know a definition that looks like that?

Huy

@MikeMiller: I do, but we used directional derivatives to define vectors in DG. I know the other definition from GR.

I'm trying to think in the sense our DG lecturer wanted to.

Mike Miller

So in DG all your manifolds live in some $\Bbb R^n$?

Ilya_Gazman

@DonLarynx I want to find a good representation for $delta(x)$, so far I only succeeded represent a short part of it where it act like a linear line

Huy

@MikeMiller: I don't think (we ever assumed that).

Mike Miller

15:50

You have to be in a coordinate chart to talk about directional derivatives, or at least embedded in $\Bbb R^n$. This works out OK if you know what a vector bundle is, but I doubt that.

Otherwise, what are the directions?

Huy

@MikeMiller: Yes, we defined it using a chart.

Mike Miller

Bah.

Huy

My thoughts exactly.

Mike Miller

OK, then your question has its answer immediate: what's $d\gamma$? The literal derivative of $\gamma$.

Don Larynx

Hi everyone!! I bring you another problem with an elegant solution.

Find the perpendicular distance to the origin (0, 0) from the line defined by the equation $ax - by = 1$.

Huy

15:52

But then, the literal derivative doesn't really map vectors from $T_{p}[0,1]$ to $T_{f(p)}M$, does it? If it does, how so?

robjohn

@Ilya_Gazman I don't think I can offer much. I have not investigated fast factoring algorithms.

Mike Miller

@Huy It maps the basis vector to the derivative of the curve at that point.

robjohn

@DonLarynx $\frac1{\sqrt{a^2+b^2}}$

Ilya_Gazman

@robjohn do you think you can explain part of the experiments conclusion that I found?

Don Larynx

@robjohn correct! How did you figure it out so quickly? I believe it took me 30-35 minutes to solve.

Huy

15:58

@DonLarynx: Solve for $y$, then the perpendicular line is $y = - \frac{b}{a}x$ and now you can find the intersection point and compute the distance from the origin using Pythagoras, would be the straightforward approach.

robjohn

@DonLarynx Your points are those for which $(a,-b)\cdot(x,y)=1$ write that as $$|(a,-b)||(x,y)|\cos(\theta)=1$$

Don Larynx

That's what I did, too @Huy.

Huy

@DonLarynx: Not sure how that took you 30 minutes.

robjohn

@DonLarynx Divide by $|(a,-b)|\cos(\theta)$ and the minimum happens at $\theta=0$

Don Larynx

@Huy: I don't know. @robjohn that's interesting.

Huy

16:03

@DonLarynx: The intersection can be found setting $-\frac{b}{a} x = \frac{1}{b} (ax-1)$, i.e. $\frac{b}{a} x + \frac{a}{b} x = \frac{1}{b}$. The LHS is $(\frac{a^2+b^2}{ab})x$ and thus $x = \frac{a}{a^2+b^2}$. Correspondingly, you plug in to find $y = -\frac{b}{a^2+b^2}$ and then Pythagoras gives you $l = \sqrt{\frac{a^2+b^2}{(a^2+b^2)^2}} = \frac{1}{\sqrt{a^2+b^2}}$.

Sayan Chattopadhyay

Hi @DonLarynx

Andrew Thompson

Anyone up for answering a likely very simple question regarding submanifolds?

Don Larynx

Hi @SayanChattopadhyay

Sayan Chattopadhyay

U wrote a very awesome question about the perpendicular distance from origin

Don Larynx

I have another interesting problem, this one only took me 5 minutes to solve using @robjohn's method.

Find the shortest possible distance between two lattice points on the line defined by $ax - by = c$, where $a, b, c \in \Bbb{Z}$

@SayanChattopadhyay: Attempt this one.

Sayan Chattopadhyay

16:10

Won't the shortest distance be 1

Don Larynx

Prove it.

Sayan Chattopadhyay

Yup prove ......

anorton

"I don't like the word 'integers,' so I replaced it with 'numbers.'" The justifications that people give for edit summaries can be quite surprising...

Mike Miller

Integers offend me

Sayan Chattopadhyay

Hey don can these two points be called as x1 y1 and x2y2 and then use distance formulae

Then put these val of X and Y from the equation into the formula and get the distance

Hippalectryon

16:22

-4

Q: Prove that.. Please quick.

If the length of perpendicular from the point (1,1)to the line ax-by+c=o be 1, show that 1/c+1/a-1/b=c/2ab. if p and p' be the length of the perpendiculars from the origin upon the straight line whose eqn are xsecA+ycosecA=a and xcosA-ysinA=acos^2A, prove that 4p^2+(2p')^2=4a^2cosA^2. Please ...

geometry

^ gg title

Huy

PLEASE QUICK

DONT IGNORE ME

PLS

Don Larynx

uses @TedShifrin's smack technique on @Huy

4

@Sayan: Write it out explicitly.

Huy

@DonLarynx: That didn't hurt remotely as much as Ted's. Do you even lift?

Don Larynx

leaves room @Huy

Sayan Chattopadhyay

See let me name the points as x1 and y1 and the other set of points as x2 and y2 . now find the values ofx and y from the equation and then put iron the distance between the points which is sqrt of x2-x1 whole square +y2-y1 whole square

Did u think about my question @DonLarynx

Is it fine.....@DonLarynx

Pls answer is my method right

Mike Miller

16:34

@AndrewT What's the question?

zed111

hey everyone, this is not fit for a full question/post on the site so I ask here: In a group if $(ab)^n=a^n b^n \forall a,b \in G $ and for all n in integers, then prove that G is abelian. I can prove ab = ba using only n = 2 case, then why is more info given?

Andrew Thompson

Decided to post it instead, @MikeMiller. math.stackexchange.com/questions/1122013/…

Sayan Chattopadhyay

@MikeMiller hi

Mike Miller

Because more is necessary, @zed. There are groups with $(ab)^n=a^nb^n$ for all $a,b$ for some fixed $n$ that are not abelian. If it's true for three consecutive integers $n$ then your group is Abelian.

A Beautiful Mind

@MikeMiller You have vast knowledge.

Sayan Chattopadhyay

16:36

@MikeMiller hi

zed111

@MikeMiller can you show me whats wrong here:

Mike Miller

Not really.

Sayan Chattopadhyay

What is an abelian

zed111

$$(ab)^2=a^2b^2$$, then premultiply with a^-1 and postmultiply with b^-1 to get ba = ab

Sayan Chattopadhyay

What is an abelian pps answer

Pls

Mike Miller

16:39

No, that's correct. The trouble is when $n$ is bigger than 2.

Kevin Driscoll

@SayanChattopadhyay It means that elements commute. Just Google it.

A Beautiful Mind

@SayanChattopadhyay Please don't ask random questions based on random messages.

Sayan Chattopadhyay

Ok. Fine

A Beautiful Mind

Also, please don't ping random people to answer your questions.

Hippalectryon

@ABeautifulMind Ding ding ding

@ABeautifulMind ding ding

A Beautiful Mind

16:42

@Hippalectryon Ding.

Hippalectryon

:D

Huy

@MikeMiller: On wikipedia, an embedding is defined as an injective immersion which is homeomorphic onto its image. Isn't stating it is injective redundant, since a homeomorphism is bijective?

Mike Miller

Sure.

Huy

Oh, wait, is it possible that it is only injective onto its image, no?

Mike Miller

Yes, injective immersion is not necessarily an embedding. But as you mentioned, they could nuke the word injective if so desired

A Beautiful Mind

16:46

Sometimes, they use redundancy for aesthetic purposes, or whatever.

Huy

Oh, ok, I see now.

I don't find it very aesthetically appealing.

A Beautiful Mind

Well, if injective immersions are common, that could be one reason.

Mike Miller

The canonical example of an injective immersion that is not an embedding is an irrational line in the torus: $t \mapsto (e^{it},e^{i\alpha t})$, where $\alpha$ is irrational.

That image is dense in the torus.

A Beautiful Mind

For example, I am redundant in this room, but my colour serves an aesthetic purpose.

Huy

@MikeMiller You have vast knowledge.

A Beautiful Mind

16:48

@Huy Sounds familiar, huh.

@Huy @mike may win the Fields medal one day. I hope he puts in a good word for me then if I need to get some position.

Mike Miller

Literally anybody under the age of 40 on Jan 1, 2018 may get a fields medal one day. Please stop saying that of users in this room.

A Beautiful Mind

I am sorry if I pissed you off. I just think highly of you.

Anyway, I don't say that of every user in this room. =)

Mike Miller

You didn't piss me off, I just think it's silly.

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$Mathematics$ Mathematics