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12:00 AM
But yeah, you're right.
 
@BrianM.Scott Where is it ?
 
@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.
 
@JasperLoy thanks
 
@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.
 
@BrianM.Scott Post it if you want ?
 
@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.
 
@BrianM.Scott Ok Thank you
 
@BrianM.Scott Yeah =)
 
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.
 
Hello, @Brian ! Good to see you here!
 
@amWhy Peter dragged me in, kicking and screaming. :-)
 
@BrianM.Scott KICKING AND SCREAMING!?
 
@BrianM.Scott That's Peter for you!!! ;-)
 
@BrianM.Scott What does your middle M stand for?
 
12:06 AM
@PeterTamaroff Maynard.
 
user19161
@PeterTamaroff I guess it is a Michael?
 
I've got a very short, sweet middle name: Jo
 
@BrianM.Scott Do people call you Brian May?
Hehehehe
 
user19161
@PeterTamaroff I know you are PNT, hehe...
 
@PeterTamaroff Most people have no idea what the M stands for.
 
12:08 AM
@BrianM.Scott Could you refresh my memory re: the domain of your email address?
 
@amWhy I gave you <b.scott@csuohio.edu>, I think.
 
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)
 
@amWhy It’s the second-most useful privilege; the most useful is access to the online library, including journal subscriptions.
 
@BrianM.Scott Don't I know it!!! What a bonus ;-)
 
ahoy!
 
12:25 AM
Hi, @anon !!
 
ello
 
or...yellow?
or L'O
 
@jasper ;)
 
@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.
 
thanks
 
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.
 
*your @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.
 
:/
 
@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. ;-)
 
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?
 
12:39 AM
@anon what happened?
 
reposting due to a simple typo
 
Indeed
@user43418 No, only I say that!!!
 
@Charlie <innocent> I thought that it was pretty nearly a unanimous refrain. </innocent>
 
@BrianM.Scott :-O
 
@Charlie hi
 
12:49 AM
@pourjour Soufian!!! Hi, I'm fine and you?
 
@BrianM.Scott Are you there?
 
@Charlie pretty good thanks
@anon any help with this :
what is the set of positive interger n for which $(2n+1, 5)=1$
 
step 1: what is the set of positive integers m for which (m,5)=1? step 2: restrict to odd m.
 
@pourjour excellent!
 
@PeterTamaroff I am now.
 
12:53 AM
@anon m are those that cannot be divised by 5 am I right?
 
alternatively, (2n+1,5)=(2n-4,5)=(n-2,5), so whenever n is 2 more than a number coprime to 5
 
@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.
 
@anon can't understand this "so whenever n is 2 more than a number coprime to 5"
 
@PeterTamaroff All that has to be justified is that $G\cap(\leftarrow,x)\ne\varnothing\ne G\cap(x,\to)$.
 
@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?
 
12:58 AM
@PeterTamaroff See my first comment in the one where I described a sentence as awful.
 
@BrianM.Scott =)
 
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?
 
@BrianM.Scott What I'm wondering then is: how do we prove $G\cap (h,g)=\varnothing$?
 
@PeterTamaroff What is there to prove? Every element of $G$ is either $\le h$ or $\ge g$ by the definition of $h$ and $g$.
 
@BrianM.Scott Sorry, how did we define $g,h$?
 
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\}\}$
 
that's correct
 
@PeterTamaroff You did it about half a dozen comments back.
 
@BrianM.Scott the infs and sups?
 
@anon does p must be a prime?
 
@PeterTamaroff Yes.
 
1:07 AM
@pourjour $p\in\Bbb N$ is what you wrote, and what you wrote is correct
 
@BrianM.Scott Oh, sorry. I thought that didn't work. I think I lost confidence.
 
@anon and k=0 or 1 or3 or 4
??
 
yes
 
@PeterTamaroff Oops! My fault: no, those don’t have to belong to $G$, so a little more work is needed.
 
@BrianM.Scott That's why.
 
1:08 AM
@anon thanks
@anon is there any other way without noticing this $(n-2,5)=1$
?
 
Maybe if I drink more beer a miracle will happen! A leprechaun will give me the solution!
 
@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)
 
@anon that's what already did
you're idea was great too
 
@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 $G\cap(0,y)=\varnothing$.
 
@BrianM.Scott You win the mathwebs Brian! You win!
 
1:15 AM
@PeterTamaroff :-)!!
 
@anon BTW, do you have any experience with complex numbers?
 
yes
 
@pourjour They are good guys, I swear.
 
hahahahaha^
 
@PeterTamaroff are you drunk?
 
1:19 AM
@pourjour No. It takes a lot of alcohol to get me drunk.
I barely had two glasses of beer.
 
@PeterTamaroff hhh
 
I don't need alcohol to get drunk
 
@pourjour "hhh"?
 
@PeterTamaroff laugh
 
@PeterTamaroff nevermind
 
1:20 AM
yeah, don't you laugh with your mouth closed like a normal person?
 
@Charlie That is a laughter?
 
@PeterTamaroff yes
 
@anon When I'm on the internet, I just exhale fast through my nose.
Rare cases make me LOL
 
@PeterTamaroff like kkkk rsrsrsrsr
@PeterTamaroff i laugh all the time
the easiest thing is to make me laugh
 
I found this hilarious for example.
 
1:23 AM
@anon 8750393 I'm trying to find fixed point of this transformation $g$in a complex plan :as $g:z\rightarrow(1+i)\overline z -1+3i$
 
@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 $a,b$
 
@PeterTamaroff yeah but I couldn't
because of the bar over z
 
$\bar{z}=a-bi$
 
I will retry and show you the result
 
@anon seems that you are always in a bad mood, because of your gravatar
 
1:28 AM
other way around
oh, you mean seems to others
 
@anon just like me every morning
 
exactly
 
so I think $z=1-i$
is that correct
 
no
 
@anon so z=?
ahh $z=-1+i$
am I wrong again?
 
1:40 AM
 
@PeterTamaroff that's a good one!
 
@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)
 
thanks :D
 
@Charlie It's hilarious.
 
@PeterTamaroff so much!!!!
 
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!!!
 
@JasperLoy :D
 
@anon any ideas?
 
@pourjour what's a homotecy?
 
@anon I mean homothecy
 
1:54 AM
homothety
 
either or
 
user19161
@Charlie That smiley is OK, I hate this one ;-)
 
@JasperLoy aah
 
user19161
@Charlie ;-) looks wicked!
 
1:55 AM
@JasperLoy you change the mewaning of things when you blink
 
user19161
@Charlie I like using =)
 
@JasperLoy I see
 
user19161
@Charlie I learnt it from the great Pedro.
 
@JasperLoy I know
 
user19161
@Charlie What we just said in email, HAHAHAHAHAHAHA
 
1:58 AM
@JasperLoy }:)
 
ok people I've got to sleep
 
@JasperLoy did you like that cute bears i used to send you?
@pourjour good night, sleep tight!
 
@Charlie good night! :D
 
user19161
@Charlie Hmm, some of them.
 
@pourjour :D
@JasperLoy ow
@JasperLoy :DDD
 
user19161
2:06 AM
@Charlie Goodnight!
 
@JasperLoy goodnight
 
user19161
@Charlie Yes, going to sleep now...
 
@JasperLoy sleep tight! :)
 
user19161
It's Easter here already, lol.
 
@JasperLoy HAPPY EASTER!!!!
 
2:31 AM
 
2:51 AM
are you Russian charlie
 
@Ethan No.
 
oh, but you have seen swan lake
 
@Ethan of course :)
@κρανίοπεριπολία hello
Good night everyone!
 
3:48 AM
@JasperLoy 哥哥
 
@Karl'sstudents what kind of math do you do
 
@Ethan differential geometry :D
 
4:01 AM
(removed)
 
$\Huge\text{(}$removed$\Huge\text{)}$
@Karl'sstudents Hi honey.
 
@κρανίοπεριπολία hey bee :-)
 
@Karl'sstudents :-D
bzzz...
 
@κρανίοπεριπολία Please up vote it for me. :-)
 
 
1 hour later…
5:25 AM
@Karl'sstudents Do you mean star it?
0.999...hours later...
 
@κρανίοπεριπολία Yes. :D
 
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!
 
is swaping the rows of a matrix then doing RREF the same as doing the RREF of a matrix then swaping its rows ?
 
6:28 AM
@awllower yeah it's one of my favorite books
 
Oh!
Glad to know that!
It just enables us to visualize the quadratic forms!
:D
 
first time I read it I wrote programs to test equality of forms and things like that
 
Indeed.
It is quite fun to test the result of this diagram!
 
I also understood the value of Zoltarevs reciprocity proof from that book
I'd seen it before but not realized its importance
 
@caveman hi
@caveman how are you?
:-D
 
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!
 
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 ?
 
7:52 AM
@nerdy Post the question on main.
 
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?
 
8:14 AM
$$\frac{1}{8} (\pi +\log (4)) \Im(\rho _1)=8.00005480321887696343731402153...$$
$$\Im(\rho _1)=14.1347251417346937904572519836$$
 
what are those zeros of
 
the Riemann zeta function
 
almost 8 lol
 
yes almost 8
 
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 ?
 
8:33 AM
(removed)
 
(removed) hours later...
 
$b=13$

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

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

$\Im(\rho _{16}) \cdot a = 34.098073858319456895...$
 
8:52 AM
hi
is there anyone please
 
9:14 AM
@Karl'sstudents do you find him annoying?
 
9:36 AM
graveyard....
 
hello
 
10:06 AM
the radius of convergence of z^2/(e^z+1) is \pi?
 
hi ,can i ask a question ?
 
what is a differential manifold in dimension 1
?
 
@κρανίοπεριπολία No. He is a handsome guy.
 
11:05 AM
There is a new joke in LaTeX:
\let\ea\expandafter
\ea\sports\to\the\game
 
@Karl'sstudents what;s that?
 
@Ilya a joke.
 
I mean, what;s the point?
shall I complile it with Math Jax?
 
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.
 
12:02 PM
@Karl'sstudents why would you want me to star that?
 
@κρανίοπεριπολία just 4 fun
 
@Karl'sstudents But he can be very sensitive sometimes...
 
@κρανίοπεριπολία does kranioperipolia have any meaning?
oh whoops
wrong one
 
@κρανίοπεριπολία :-)
 
fixed
 
12:05 PM
@kram1032 skullpatrol
 
heh. Does that explain the HUGE Skullpatrol in the favorites on the right?
 
nice
 
hi
 
user19161
12:32 PM
(removed)
 
the black square is back
 
user19161
This room should be renamed as (removed).
 
Is $\cos(\frac{\pi}{7})$ irrational ? and if no is $\cos^3(\frac{\pi}{7})$ irrational ?
 
user19161
@DominicMichaelis Huhu.
 
user19161
12:42 PM
I like it that I have 4 gold badges now.
 
this one is a nice question even though the op really shouldn't get reputation for that
 
1:14 PM
I suppose I should pick a username.
 
oh what a waste i gave such a nice bijection and he only wants one for natural numbers
 
@user67848 hastings will be nice... :-)
 
my function is a bijection from $\mathbb{Q} \to (100,200) \cap \mathbb{Q}$ ;)
 
1:35 PM
3
Q: Property regarding partial derivatives

Jean-Francois RossignolLet $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ 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...

 
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.
 
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...
 
i gonna fix that ;)
 
user19161
Actually, I think my answer is the most elegant. =)
 
a banananda :D
 
user19161
1:55 PM
I have upvoted the question and the other 2 answers. =)
 
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.
 
yeah yours is good
but arctan is a stupid function ^^
 
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. =)
 
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 -.-
 
3:01 PM
As we know, an invertible matrix could be rewritten into the product of several elementary matrices.
 
As we know.
 
Can we discover an analogous one for 1-1 continuously differentiable mapping?
 
regards to where you are mapping
 
For example, $f\colon\mathbb R^n\to\mathbb R^n$.
 
you know what lie algebras are ?
sry lie groups
 
3:10 PM
I hope if there's some result locally at, say, $0$.
No.
Is that necessary?
 
i guess that is what you are searching for ;)
 
I want to summarize theorems in multi-variable calculus, say, reverse/implicit function theorem, rank theorem, etc.
 
Happy easter everyone!!!!
@κρανίοπεριπολία Hi :)
 
Goodbye!
 
@dominic hi
 
3:18 PM
hi :)
 
@DominicMichaelis how are you?
 
tired :D
 
@DominicMichaelis oh
 
user19161
@Charlie Nice bird.
 
@JasperLoy thanks! It's a swan :)
 
user19161
3:30 PM
@Charlie Is a swan a bird?
 
@JasperLoy I think so...
 
user19161
These days, not many people vote for lhf...
 
@JasperLoy it's a "black swan"
 
user19161
@Charlie A black swan is a swan and a swan is a bird. QED.
 
@JasperLoy lhf?
 
3:33 PM
low hanging fruit
easy questions
 
@JasperLoy precisely
 
user19161
@user1 Haha, funny name user1.
 
i do have such a lot answers with 8 upvotes ...
 
@JasperLoy Glad you like it. I am not all that creative. :)
 
user19161
@user1 What led you to this site?
 
3:36 PM
@JasperLoy mathoverflow has a link in its faq, I think.
And I have no idea what led me to MO.
 
user19161
@user1 Ah, I see. I am not on MO. I only know simple things like 1+1=2.
 
hi
I need a diagram to solve differential equation
can someone please help me
 
user19161
Yes, post on the site.
 
I think it's simple
 
you should be able to solve simple tasks ;)
Gee that preimages of closed sets under continuous functions are closed is not liked on this site
here i bet when i would use am > gm i would have 9000 upvotes now
but i like the preimage definition
 
3:47 PM
nah I need just clear manner
 

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