Mathematics - 2013-03-31 (page 1 of 3)

Peter Tamaroff

12:00 AM

But yeah, you're right.

user43418

@BrianM.Scott Where is it ?

Peter Tamaroff

@BrianM.Scott The core of the idea is clear tough: if $G$ has nonempty intersection at nbhs of $0$, we can translate it and make it intersect in any other real number.

user19161

@user43418 Done.

user43418

@JasperLoy thanks

Brian M. Scott

@user43418 Here.

user19161

12:02 AM

@BrianM.Scott You should have posted it man!

user19161

@BrianM.Scott OIC that guy posted the same thing, LOL.

user43418

@BrianM.Scott Post it if you want ?

Brian M. Scott

@PeterTamaroff Yes $-$ but it seems to be easier to work backwards from a gap around $x$ to a gap around $0$ than forwards from no gap around $0$ to no gap around any $x$!

@user43418 Nah $-$ it’s identical to the answer that’s already been posted.

user43418

@BrianM.Scott Ok Thank you

Peter Tamaroff

@BrianM.Scott Yeah =)

Brian M. Scott

12:04 AM

@JasperLoy I couldn’t: I wrote it here before the question was posted to the main site, and essentially the same answer was already up there by the time I discovered that the question had been posted.

@user43418 You’re welcome.

amWhy

Hello, @Brian ! Good to see you here!

Brian M. Scott

@amWhy Peter dragged me in, kicking and screaming. :-)

Peter Tamaroff

@BrianM.Scott KICKING AND SCREAMING!?

amWhy

@BrianM.Scott That's Peter for you!!! ;-)

Peter Tamaroff

@BrianM.Scott What does your middle M stand for?

Brian M. Scott

12:06 AM

@PeterTamaroff Maynard.

user19161

@PeterTamaroff I guess it is a Michael?

amWhy

I've got a very short, sweet middle name: Jo

Peter Tamaroff

@BrianM.Scott Do people call you Brian May?

Hehehehe

user19161

@PeterTamaroff I know you are PNT, hehe...

Brian M. Scott

@PeterTamaroff Most people have no idea what the M stands for.

amWhy

12:08 AM

@BrianM.Scott Could you refresh my memory re: the domain of your email address?

Brian M. Scott

@amWhy I gave you <b.scott@csuohio.edu>, I think.

amWhy

Yes, I think that was it...I think you mentioned you still have access to your campus account, as Emeritus...

You can delete it, if you'd like, I've got it down! (again)

Brian M. Scott

@amWhy It’s the second-most useful privilege; the most useful is access to the online library, including journal subscriptions.

amWhy

@BrianM.Scott Don't I know it!!! What a bonus ;-)

user65165

ahoy!

amWhy

12:25 AM

Hi, @anon !!

anon

ello

amWhy

or...yellow?

or L'O

Charlie

@jasper ;)

amWhy

@anon I like your answers, especially the expository-ish ones! I first noticed back when you were explaining the basics of permutations to an OP who was having trouble grasping the different notations. You're explanation could serve as a "reference" entry.

anon

thanks

amWhy

12:30 AM

Of course, you "stole" an accept from me, by doing so, but it was certainly a better answer for the OP than mine, at the time.

Charlie

*your @amWhy

amWhy

@anon I like your answers, especially the expository-ish ones! I first noticed back when you were explaining the basics of permutations to an OP who was having trouble grasping the different notations. Your explanation could serve as a "reference" entry.

anon

:/

amWhy

@Charlie I tried correcting...but it was too late and reposted as a new post!

@anon OP's should feel no loyalty to an earlier upvote if a subsequent post is far more helpful and informative. ;-)

nerdy

Suppose i have a system of equations At=b representing two planes in the euclidian space R^3 where A is 2x3 and t is (x y z). My same system has a 1-dimension NullSpace, that means the intersection of the planes shifted at the origin is a line.If the only possible free variables are x and y for example, does it mean the line is in the x-y plane ? If the only possible free variables are x and z for example, does it mean the line is in the x-z plane?

Charlie

12:39 AM

@anon what happened?

anon

reposting due to a simple typo

Charlie

Indeed

@user43418 No, only I say that!!!

Brian M. Scott

@Charlie <innocent> I thought that it was pretty nearly a unanimous refrain. </innocent>

Charlie

@BrianM.Scott :-O

pourjour

@Charlie hi

Charlie

12:49 AM

@pourjour Soufian!!! Hi, I'm fine and you?

Peter Tamaroff

@BrianM.Scott Are you there?

pourjour

@Charlie pretty good thanks

@anon any help with this :
what is the set of positive interger n for which $(2n+1, 5)=1$

anon

step 1: what is the set of positive integers m for which (m,5)=1? step 2: restrict to odd m.

Charlie

@pourjour excellent!

Brian M. Scott

@PeterTamaroff I am now.

pourjour

12:53 AM

@anon m are those that cannot be divised by 5 am I right?

anon

alternatively, (2n+1,5)=(2n-4,5)=(n-2,5), so whenever n is 2 more than a number coprime to 5

Peter Tamaroff

@BrianM.Scott OK. We assume $B(x,\epsilon)\cap G$ is empty. This means that there are no numbers inside $(g,h)$ for $g=\sup\{g':g'<x-\epsilon\}$ and $h=\inf\{g:g>x+\epsilon\}$.

But before, I said that there were some $g,h\in G$ that worked.

I think it needs to be justified.

pourjour

@anon can't understand this "so whenever n is 2 more than a number coprime to 5"

Brian M. Scott

@PeterTamaroff All that has to be justified is that $G\cap(\leftarrow,x)\ne\varnothing\ne G\cap(x,\to)$.

Peter Tamaroff

@BrianM.Scott OK, but we said that such $g,h$ were such that no $\ell \in G$ satisfied $g<\ell <h$; or did I say it and wasn't corrected?

Brian M. Scott

12:58 AM

@PeterTamaroff See my first comment in the one where I described a sentence as awful.

Peter Tamaroff

@BrianM.Scott =)

kram1032

So markov processes are defined on continuous spaces. Markov chains are the discrete counterpart. Markov chains have a continuous time version. But I can't find anything about continuous time Markov processes.
Is there any reason why Markov processes can't be time-continuous?

Peter Tamaroff

@BrianM.Scott What I'm wondering then is: how do we prove $G\cap (h,g)=\varnothing$?

Brian M. Scott

@PeterTamaroff What is there to prove? Every element of $G$ is either $\le h$ or $\ge g$ by the definition of $h$ and $g$.

Peter Tamaroff

@BrianM.Scott Sorry, how did we define $g,h$?

pourjour

1:05 AM

@anon I think the set of n is $S=\{ n : n=5p+k : p\in \mathbb{N}$ and$ k\in\{0,1,3,4\}\}$

anon

that's correct

Brian M. Scott

@PeterTamaroff You did it about half a dozen comments back.

Peter Tamaroff

@BrianM.Scott the infs and sups?

pourjour

@anon does p must be a prime?

Brian M. Scott

@PeterTamaroff Yes.

anon

1:07 AM

@pourjour $p\in\Bbb N$ is what you wrote, and what you wrote is correct

Peter Tamaroff

@BrianM.Scott Oh, sorry. I thought that didn't work. I think I lost confidence.

pourjour

@anon and k=0 or 1 or3 or 4

??

anon

yes

Brian M. Scott

@PeterTamaroff Oops! My fault: no, those don’t have to belong to $G$, so a little more work is needed.

Peter Tamaroff

@BrianM.Scott That's why.

pourjour

1:08 AM

@anon thanks

@anon is there any other way without noticing this $(n-2,5)=1$

?

Peter Tamaroff

Maybe if I drink more beer a miracle will happen! A leprechaun will give me the solution!

anon

@pourjour yes, 2n+1 must be among 1,2,3,4 mod 5, so solve 2n+1=k mod 5 for n for each of k=1,2,3,4. (note the multiplicative inverse of 2 mod 5 is 3)

pourjour

@anon that's what already did

you're idea was great too

Brian M. Scott

@PeterTamaroff Take sequences of group elements $g_n$ and $h_n$ converging to the sup and inf. Look at the group elements $h_n-g_n$. They’re positive, so they have an infimum $y$. Show that $y>0$ and that $G\cap(0,y)=\varnothing$.

Peter Tamaroff

@BrianM.Scott You win the mathwebs Brian! You win!

Brian M. Scott

1:15 AM

@PeterTamaroff :-)!!

pourjour

@anon BTW, do you have any experience with complex numbers?

anon

yes

Peter Tamaroff

@pourjour They are good guys, I swear.

Charlie

hahahahaha^

pourjour

@PeterTamaroff are you drunk?

Peter Tamaroff

1:19 AM

@pourjour No. It takes a lot of alcohol to get me drunk.

I barely had two glasses of beer.

pourjour

@PeterTamaroff hhh

Charlie

I don't need alcohol to get drunk

Peter Tamaroff

@pourjour "hhh"?

Charlie

@PeterTamaroff laugh

pourjour

@PeterTamaroff nevermind

anon

1:20 AM

yeah, don't you laugh with your mouth closed like a normal person?

Peter Tamaroff

@Charlie That is a laughter?

Charlie

@PeterTamaroff yes

Peter Tamaroff

@anon When I'm on the internet, I just exhale fast through my nose.

Rare cases make me LOL

Charlie

@PeterTamaroff like kkkk rsrsrsrsr

@PeterTamaroff i laugh all the time

the easiest thing is to make me laugh

Peter Tamaroff

I found this hilarious for example.

pourjour

1:23 AM

@anon 8750393 I'm trying to find fixed point of this transformation $g$in a complex plan :as $g:z\rightarrow(1+i)\overline z -1+3i$

Peter Tamaroff

@pourjour So you have $(1+i)\bar z-1+3i=z$.

Solve that.

It will be easier if you put $z=a+bi$ and solve for $a,b$

pourjour

@PeterTamaroff yeah but I couldn't

because of the bar over z

anon

$\bar{z}=a-bi$

pourjour

I will retry and show you the result

Charlie

@anon seems that you are always in a bad mood, because of your gravatar

anon

1:28 AM

other way around

oh, you mean seems to others

Charlie

@anon just like me every morning

anon

exactly

pourjour

so I think $z=1-i$

is that correct

anon

no

pourjour

@anon so z=?

ahh $z=-1+i$

am I wrong again?

Peter Tamaroff

1:40 AM

►

Charlie

@PeterTamaroff that's a good one!

anon

@pourjour firstly, expand (1+i)(a-bi)-1+3i=z to be (a+b-1)+(a-b+3)i=a+bi, so a+b-1=a and a-b+3=b, so b=1 and a=-1, i.e. z=-1+i (correct)

pourjour

thanks :D

Peter Tamaroff

@Charlie It's hilarious.

Charlie

@PeterTamaroff so much!!!!

pourjour

1:44 AM

now I'm struggling to prove that $z'=2i\overline z -5 +5i$ is an homotecy

user19161

I now have the Copy Editor Badge, yay!!!

Charlie

@JasperLoy :D

pourjour

@anon any ideas?

anon

@pourjour what's a homotecy?

pourjour

@anon I mean homothecy

Charlie

1:54 AM

homothety

anon

either or

user19161

@Charlie That smiley is OK, I hate this one ;-)

Charlie

@JasperLoy aah

user19161

@Charlie ;-) looks wicked!

pourjour

Charlie

1:55 AM

@JasperLoy you change the mewaning of things when you blink

user19161

@Charlie I like using =)

Charlie

@JasperLoy I see

user19161

@Charlie I learnt it from the great Pedro.

Charlie

@JasperLoy I know

user19161

@Charlie What we just said in email, HAHAHAHAHAHAHA

Charlie

1:58 AM

@JasperLoy }:)

pourjour

ok people I've got to sleep

Charlie

@JasperLoy did you like that cute bears i used to send you?

@pourjour good night, sleep tight!

pourjour

@Charlie good night! :D

user19161

@Charlie Hmm, some of them.

Charlie

@pourjour :D

@JasperLoy ow

@JasperLoy :DDD

user19161

2:06 AM

@Charlie Goodnight!

Charlie

@JasperLoy goodnight

user19161

@Charlie Yes, going to sleep now...

Charlie

@JasperLoy sleep tight! :)

user19161

It's Easter here already, lol.

Charlie

@JasperLoy HAPPY EASTER!!!!

Charlie

2:31 AM

►

Ethan

2:51 AM

are you Russian charlie

Charlie

@Ethan No.

Ethan

oh, but you have seen swan lake

Charlie

@Ethan of course :)

@κρανίοπεριπολία hello

Good night everyone!

Karl's students

3:48 AM

@JasperLoy 哥哥

Ethan

@Karl'sstudents what kind of math do you do

Karl's students

@Ethan differential geometry :D

Karl's students

4:01 AM

(removed)

κρανίοπεριπολία

$\Huge\text{(}$removed$\Huge\text{)}$

@Karl'sstudents Hi honey.

Karl's students

@κρανίοπεριπολία hey bee :-)

κρανίοπεριπολία

@Karl'sstudents :-D

bzzz...

Karl's students

@κρανίοπεριπολία Please up vote it for me. :-)

κρανίοπεριπολία

5:25 AM

@Karl'sstudents Do you mean star it?

0.999...hours later...

Karl's students

@κρανίοπεριπολία Yes. :D

awllower

5:43 AM

@JasperLoy I see. Thanks for pointing it out.

BTW, has anyone read the book sensual quadratic forms, by Conway?

I think of it as the most elementary approach to this subject!

nerdy

is swaping the rows of a matrix then doing RREF the same as doing the RREF of a matrix then swaping its rows ?

caveman

6:28 AM

@awllower yeah it's one of my favorite books

awllower

Oh!
Glad to know that!

It just enables us to visualize the quadratic forms!
:D

caveman

first time I read it I wrote programs to test equality of forms and things like that

awllower

Indeed.
It is quite fun to test the result of this diagram!

caveman

I also understood the value of Zoltarevs reciprocity proof from that book

I'd seen it before but not realized its importance

κρανίοπεριπολία

@caveman hi

@caveman how are you?

:-D

awllower

6:44 AM

@caveman Does the book contain thi topic?

I have not seen it?

Anyway, let us continue later, as I am a little occupied now.
Bye!

nerdy

lets suppose i have a system Ax=b with x being ( x y z w ).Suppose i also know that the intersection of all graphs is 2-dimensional, that is, the nullspace has 2 free variables and 2 lead variables.Suppose i know that z and w could be made lead variables and i want a RREF matrix such that the pivots have z and w as corresponding variables.What should i do ?

κρανίοπεριπολία

7:52 AM

@nerdy Post the question on main.

Jonas Teuwen

He yes, how about you use the main site for that? It's the purpose.

Not that you can't ask mathematical questions in chat; it's just that you should just bloody wait like anybody else that wants an answer from strangers.

Ben

Has anybody read Stephen Smale's biography?

κρανίοπεριπολία

@nerdy +1 :-D

Mats Granvik

8:14 AM

$$\frac{1}{8} (\pi +\log (4)) \Im(\rho _1)=8.00005480321887696343731402153...$$
$$\Im(\rho _1)=14.1347251417346937904572519836$$

Ethan

what are those zeros of

Mats Granvik

the Riemann zeta function

Ethan

almost 8 lol

Mats Granvik

yes almost 8

nerdy

saa

Yo, does the idea of a pivot in linear algebra has only to do with coeficient 1

or it laos has to do with leftmost element ?

does a pivot need only to be 1, or does it need also to be the leftmost element and 1 ?

(removed)
(removed) ?

wtf ?

Karl's students

8:33 AM

(removed)

anon

(removed) hours later...

Mats Granvik

$b=13$

$a=\sum _{n=0}^{\infty } \left(\frac{1}{b n+1}-\frac{1}{b n+2}\right)$

$\Im(\rho _{15}) \cdot a = 33.098071066828152722...$

$\Im(\rho _{16}) \cdot a = 34.098073858319456895...$

Vrouvrou

8:52 AM

hi

is there anyone please

κρανίοπεριπολία

9:14 AM

@Karl'sstudents do you find him annoying?

Jayesh Badwaik

9:36 AM

graveyard....

Vrouvrou

hello

Mathematician

10:06 AM

the radius of convergence of z^2/(e^z+1) is \pi?

Vrouvrou

hi ,can i ask a question ?

κρανίοπεριπολία

askaway

Vrouvrou

what is a differential manifold in dimension 1

?

Karl's students

@κρανίοπεριπολία No. He is a handsome guy.

Karl's students

11:05 AM

There is a new joke in LaTeX:

\let\ea\expandafter

\ea\sports\to\the\game

Ilya

@Karl'sstudents what;s that?

Karl's students

@Ilya a joke.

Ilya

I mean, what;s the point?

shall I complile it with Math Jax?

Karl's students

@Ilya: It is better to ask chat.stackexchange.com/rooms/41/tex-latex-and-friends. :-)

kram1032

11:36 AM

Hmm.

What kind of distribution do you get if you uniformly sample a finite interval in projective space?

E.g. you generate pairs of random variables, of which one lies in an interval $X=[a,b]$ and the other in an interval $Y=[c,d]$ and you divide the two $\frac{X}{Y}$

I see. That stuff is called ratio distrubitons.

κρανίοπεριπολία

12:02 PM

@Karl'sstudents why would you want me to star that?

Karl's students

@κρανίοπεριπολία just 4 fun

κρανίοπεριπολία

@Karl'sstudents But he can be very sensitive sometimes...

kram1032

@κρανίοπεριπολία does kranioperipolia have any meaning?

oh whoops

wrong one

Karl's students

@κρανίοπεριπολία :-)

kram1032

fixed

κρανίοπεριπολία

12:05 PM

@kram1032 skullpatrol

kram1032

heh. Does that explain the HUGE Skullpatrol in the favorites on the right?

yip

nice

hi

hi

12:16 PM

lo

user19161

12:32 PM

(removed)

kram1032

the black square is back

user19161

This room should be renamed as (removed).

Dominic Michaelis

Is $\cos(\frac{\pi}{7})$ irrational ? and if no is $\cos^3(\frac{\pi}{7})$ irrational ?

user19161

@DominicMichaelis Huhu.

Dominic Michaelis

huhu

user19161

12:42 PM

I like it that I have 4 gold badges now.

Dominic Michaelis

this one is a nice question even though the op really shouldn't get reputation for that

user67848

1:14 PM

I suppose I should pick a username.

Dominic Michaelis

oh what a waste i gave such a nice bijection and he only wants one for natural numbers

here

Jayesh Badwaik

@user67848 hastings will be nice... :-)

Dominic Michaelis

my function is a bijection from $\mathbb{Q} \to (100,200) \cap \mathbb{Q}$ ;)

Carpediem

1:35 PM

3

Q: Property regarding partial derivatives

Let $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function. For each $x \in \mathbb{R}$, define a function $g_x: \mathbb{R} \rightarrow \mathbb{R}$ by $g_x(y)=f(x,y)$. Suppose that for each $x$, there is a unique y such that $g_x'(y)=0$; let $c(x)$ be this $y$. Su...

multivariable-calculus

user19161

1:51 PM

@DominicMichaelis I gave another answer on the formula question you answered between 100 and 200 and also edited it, I hope I have not changed the meaning of the question.

Dominic Michaelis

mh i guess it is how the op meant it although i unterstood it a bit different

but my answer is neater cause of it sends rationals to rationals and irrationals to irrationals :)

user19161

OK, but still no upvote for my answer, sad panda...

Dominic Michaelis

i gonna fix that ;)

user19161

Actually, I think my answer is the most elegant. =)

Dominic Michaelis

a banananda :D

user19161

1:55 PM

I have upvoted the question and the other 2 answers. =)

Dominic Michaelis

i didn't upvote the other answer, he posted imho a worse answer then mine 10 minutes after me

user19161

Ah, I think mine is quite simple and also different, so I posted it.

Dominic Michaelis

yeah yours is good

but arctan is a stupid function ^^

Carpediem

Let $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function. For each $x \in \mathbb{R}$, define a function $g_x: \mathbb{R} \rightarrow \mathbb{R}$ by $g_x(y)=f(x,y)$. Suppose that for each $x$, there is a unique y such that $g_x'(y)=0$; let $c(x)$ be this $y$.

How do I show that c is differentiable ?

user19161

You should post on the main site. =)

Dominic Michaelis

2:12 PM

i hate when someone don't know an approximation and i need to prove it -.-

This reminds me to much on numerics -.-