« first day (849 days earlier)      last day (2736 days later) » 

6:03 AM
Discuss the problem of determining a polynomial of degree at most 2 for which p(0), p(1), and p'($\xi$) are prescribed, $\xi$ being any preassigned point.
 
leo
don't know if it is smooth, but it seems like so
 
Could anyone clarify what the above question is requesting? It seems a little vague.
 
leo
@arete find a polynomial $p(x)=ax^2+bx+c$ so that ...
so th problem reduces to find $a,b,c$
which can be done
as you'll find
 
@arete Problem: given $\xi$ and the values $p(0),p(1),p'(\xi)$ for some unknown polynomial $p$ of degree $\le2$, reconstruct the polynomial from the known information (i.e. its coefficients).
 
Thank both of you.
 
leo
6:07 AM
$\theta^{-1}$ seems smooth as well
:-(
 
Can you restate what it is you want leo?
 
leo
@anon okay, it is a bit large to state
here I go
:
say that 2 immersions $F_i:N_i\to M$ ($N,M$ smooth manifolds) are equivalent $i=1,2$ if there exist a diffeomorphism $G:N_1\to N_2$ such that $F_1=F_2\circ G$
The problem is to find two injective inequivalent immersions $F_i$ from $\Bbb R$ to $\Bbb R^2$ with the same image.
@anon
my try was those $\gamma$ and $\delta$
they have the same image
are injective
but seems to be equivalent
 
okay. how about pick $F_2$ and define $F_1=F_2\circ H$, where $H$ is not smooth?
 
leo
yes, we want they to have the same image
let's pick $F_2:]0,\pi[\to S^1$ $F_2(x)=(\cos x,\sin x)$
its image is the upper half circle
 
6:22 AM
identify $\Bbb R^2$ with $\Bbb C$ for convenience. define $\alpha:(-1,1)\to S^1:\theta\mapsto e^{2\pi i\theta}$. define $\beta:=\alpha\circ f$, where $f:(-1,1)\to(-1,1)$ is a monotone and not smooth, like a stretched version of the cantor function. then $\alpha,\beta$ are inequivalent injective immersions of $(-1,1)$ into $S^1$ with the same image, right?
 
leo
@anon ahh yes, that's it. Because $f$ is biyective.
How many nonsmooth functions are there?
 
most of them
 
leo
I guessed so, but I was thinking on it, all what comes to my mind was smooth
unless exotic things
like the function of Weierstrass
nonexotic nonsmooth $|x|$
 
that's not injective though
just fyi, so it can't be used for the question
 
leo
@anon yep and locally smooth
well, almost locally smooth
nice answer to this
 
leo
6:51 AM
I'll better go to sleep. Thanks for your help @anon
 
 
1 hour later…
8:13 AM
$\color{red}{\Huge\text{>8(}}$

$\color{orange}{\Huge\text{>B(}}$

$\color{yellow}{\Huge\text{>:(}}$

$\color{green}{\Huge\text{>;(}}$

$\color{blue}{\Huge\text{>8(}}$
 
@aDangerousIdea: I thought Marvis was here...
Your gravatar is close to Marvis'
Hasn't been here since June
@aDangerousIdea In that size the 'B' looks bad
 
@robjohn I agree.
$\color{red}{\text{>B(}}$
$\color{red}{\Huge\text{>}}\color{green}{\Huge\text{8}}\color{blue}{\Huge\text{(‌​}}$
$\color{orange}{\Huge\text{>}}
\color{white}{\Huge\text{8}}
\color{orange}{\Huge\text{(}}$
 
8:35 AM
Hi
 
@Chris'ssister Hey, what's up?
 
Hi @aDangerousIdea
Hi @robjohn. A bit around here. :D
 
How are you?@Chris'ssister
 
@aDangerousIdea: Not bad. I think of a problem that seems rather hard.
This one $\int \frac{x^2}{\sqrt{1+x^2}} \cdot e^{\arctan x} \ dx$. I see no straightforward solution.
 
8:39 AM
@Chris'ssister It is a bit quiet here. We had an hour of silence (for what, I don't know).
 
@robjohn: sorry if you took the silence away. :-)
 
@Chris'ssister actually, aDangerousIdea and I were just looking at smileys
 
@robjohn: smileys make us feel better, right? :) I like smileys too.
 
@Chris'ssister Perhaps >8(
 
@robjohn: I never asked you but now I do it: are you a teacher? You explain things nicely and I was thinking at that.
 
8:43 AM
@Chris'ssister I taught at UCLA for a couple of years. But that was a while ago.
 
@robjohn: I see. That's really nice.
 
@Chris'ssister since then, I've been a part of sci.math and now MSE, both of which feel like teaching, too
 
@robjohn: MSE it's one of the greatest things I've seen in the last years in terms of learning mathematics. If you wanna learn then you have a great possibility to do this here. We need guys like you here. :D I also learned a lot of things here.
 
Just curious: is there a forum that's a bit more specialized (to graduate level math, yet not at the level of mathoverlow?)
 
@Sanchez: I suppose that MSE and MathOverflow catch all level in mathematics. (if I'm not wrong)
 
8:53 AM
i think so, but sometimes i feel difficult to navigate around here due to the broad spectrum of level of questions here.
So I would like to know other places too.
 
@Sanchez: I see.
 
hi @iymz
$\color{orange}{\text{>8(}}$
$\color{orange}{\text{)8<}}$
bye iymz
 
9:16 AM
Good morning
 
@OldJohn: morning!
 
$$
\begin{align}
\int\frac{x^2}{\sqrt{1+x^2}}e^{\arctan(x)}\,\mathrm{d}x
&=\int\tan^2(u)\sec(u)e^u\,\mathrm{d}u\\
&=\int(\sec^3(u)-\sec(u))e^u\,\mathrm{d}u
\end{align}
$$
Now consider
$$
\begin{align}
\int\sec(u)e^u\,\mathrm{d}u
&=e^u\sec(u)-\int\sec(u)\tan(u)e^u\,\mathrm{d}u\\
&=e^u\sec(u)-e^u\sec(u)\tan(u)+\int(\sec(u)\tan^2(u)+\sec^3(u))e^u\,\mathrm{d}u\\
&=e^u\sec(u)-e^u\sec(u)\tan(u)+\int(2\sec^3(u)-\sec(u))e^u\,\mathrm{d}u\\
0&=e^u\sec(u)-e^u\sec(u)\tan(u)+\int(2\sec^3(u)-2\sec(u))e^u\,\mathrm{d}u\\
 
@robjohn: awesome! Thank you!
 
@Chris'ssister I just checked and Mathematica agrees.
 
@robjohn: nice and fast
 
user19161
9:27 AM
Ladies and gentlemen, I have capped today. Thank you for your support.
 
@WillHunting you have capped already?!?!
 
@WillHunting Again, already? Wow!
 
On the rare times I have capped, it has always been late in the evening :)
 
user19161
@OldJohn Thanks to low hanging fruit!
 
user19161
@robjohn Nothing compared to your 40 grand!
 
9:29 AM
@WillHunting but I haven't capped for a long time
 
user19161
A note to my supporters: please check if I have capped before upvoting, thanks!
 
@WillHunting Awww, I was going to upvote a lot of your posts now :-p
 
@robjohn That would have been so cruel.
By the way, my imaginary Erdos number is 2.
@robjohn BTW, I guess the cap will be relative of the time zone of the user? Or else it might be a rolling cap? Not more than 200 rep in last 24 hours? So, in which case, Will Hunting has capped into his early evening. @WillHunting @OldJohn
 
@JayeshBadwaik I know how I hate getting upvotes after I've capped.
 
user19161
@JayeshBadwaik Rep is always from 0000 GMT to 2359 GMT.
 
9:37 AM
@WillHunting That is so crude.
 
@JayeshBadwaik No, caps are always UTC
 
user19161
@robjohn Better written with the comma.
 
user19161
Eats shoots and leaves? No. Eats, shoots and leaves!
 
@WillHunting GN alert!
2
 
user19161
I think the very high rep users have a lot of rep wasted due to capping.
 
9:41 AM
 
user19161
I am beginning to fall in love with my steelblue square. It just looks so beautiful!
 
I've always liked blue colour (my favorite one). :)
 
user19161
Hehe, that's how I hypnotize people! =)
 
my favorite colors are black and dark maroon.
 
@JayeshBadwaik: blue and dark here.
 
9:50 AM
My laptop bag by Liviya (which is I use everywhere) is maroon and black (random color chosen by the shipper), My Dell XPS M1530 is maroon and black (my choice), my portable external hard disk drive is maroon and black (random choice by the shipper), there was a time in my college when everything I owned was black and maroon.
 
I wonder why companies discontinue a perfectly excellent product and in its place bring out a completely crappy product.
 
10:16 AM
@OldJohn Just answered two questions on algebraic topology :D
 
@BenjaLim well done - I steer clear of those :)
 
@OldJohn :D
 
I remember once when I think I just about understood Mayer-Vietoris ... but that was long ago
 
@OldJohn The minute you understand that every ses of chain complexes gives rise to an LES in homology
Mayer Vietoris
is
just another LES
 
I actually got most of my understanding of AT from Fraleigh's book on abstract algebra!
I also remember spending hours chasing elements round a diagram (5-lemma?) - and not really understanding the point of it :P
 
10:20 AM
hahahahahahahahhahahahahahahaha
It's quite useful sometimes
I've used it a few times
It's handy
If you two Long exact sequences say that of the pair $(X,A)$ and that of the pair $(Y,B)$ with a map of pairs $f:(X,A) \to (Y,B)$
 
:)
 
By naturality you will get that the squares in the commutative diagrams of the two LESs' placed next together commuting
and then you want to ask if certain vertical maps are isomorphisms
will then a particular one sandwiched in between be
that's where the five lemma comes in
@OldJohn AT
Just lot's of diagrams and long exact sequences
@OldJohn Like in one assignment we used Mayer Vietoris like 50 million times
and then
in my AT final
I could state the MVS with my eyes closed and even remember how the boundary map was defined
 
I think one of my number theory books has some Galois cohomology in it somewhere, but I am sure I am never going to get to that chapter :)
 
is it neukirch?
 
@BenjaLim no - I don't own that one - it probably one that I picked up cheaply in a sale in the vague hope that it might be useful one day
 
10:25 AM
@OldJohn ah
@OldJohn I'm reading algebraic number theory now
are you familiar with those sort of topics
just got of with trace, norm , discriminant
 
@BenjaLim I know a bit - but only really just scratching the surface at the moment
 
Why don't start reading up a bit on it?
 
I think the book was Husemollers book on Elliptic curves
 
We can learn together :D
@OldJohn That sounds really advanced
@OldJohn Oh btw
 
@BenjaLim Not a bad idea - but you will get through it much quicker than I will
 
10:27 AM
@OldJohn rubbish
@OldJohn In australia fyi we like to call our mates "cunt"
and you only use it among your peers
I know it sounds really rude
But I would call a friend of mine a good cunt
 
@BenjaLim Yes - I have heard of that quaint terminology before :)))
 
and someone I hate a shit cunt
 
@BenjaLim I hope there are no flaggers around!
I base my knowledge on pages like this
 
Cunt () is a word that primarily describes the female genitalia, particularly the vulva, and is widely considered to be vulgar. The earliest citation of this usage in the 1972 Oxford English Dictionary, c 1230, refers to the London street known as Gropecunt Lane. Scholar Germaine Greer has said that "it is one of the few remaining words in the English language with a genuine power to shock." Cunt is also used as a derogatory epithet referring to people of either sex. This usage is relatively recent, dating from the late nineteenth century. Reflecting different national usages, cun...
@OldJohn you have facebook?
 
@BenjaLim yep
 
10:30 AM
@OldJohn add me!!!
 
@BenjaLim Add meeeeeeeeeeeeeeeeeeeee. :P
 
@JayeshBadwaik What is your facebook?
 
@BenjaLim jayesh.badwaik
 
@BenjaLim done
 
10:32 AM
Done!
Oh wow
I can believe that we have friends in common
Peter Tamaroff
Mariano
And Jonas
 
@BenjaLim Yep
@BenjaLim But I have to go - got to get the kitchen ready for a total re-build next week :(
 
@BenjaLim About that jewish problems thing, I did not know that non-mathematical things could be put up on arxiv.
 
speak later, folks
 
@JayeshBadwaik huh?
@JayeshBadwaik @OldJohn facebook.com/…
 
10:35 AM
@BenjaLim Nevermind, I just found problems below the story.
 
@JayeshBadwaik link doesn't open
@JayeshBadwaik arxiv jewish thing?
 
ah
 
Is that you in the photo?
the one you linked?
 
@JayeshBadwaik No that's my lecturer mathoverflow.net/users/910/vigleik-angeltveit
 
10:38 AM
@BenjaLim okay...
:7073460 hmm.
 
must go - back later!
 
@OldJohn later!
 
11:00 AM
@JayeshBadwaik A oi thar captian
 
11:13 AM
hi
 
How are thy female dogs, doing in this splendid afternoon?
@aDangerousIdea That looks like a couple of adult movies I have watched.
 
11:30 AM
@N3buchadnezzar It's so much better with the "(removed)" mystery, no?
 
@aDangerousIdea Indeed
Are you good with laplace?
 
Not really, sorry.
I prefer my lap dancers to not wear any lace ;-)
 
@N3buchadnezzar Hola. Wassup?
@N3buchadnezzar Laplace? Transforms?
 
good morning BenjaLim
 
@JayeshBadwaik Let $$ F(s) = e^{-as} \cdot \frac{1}{(s+b)^2}$$
 
11:40 AM
@N3buchadnezzar Okay...
 
bye
 
Find $f(t)$, as in $\mathcal{L}^{-1}\left\{ F(s) \right\}$.
Looks like convolution, after a bit of trouble I figured that
 
$e^{-as}$ is for a Heaviside function right?
Hmm, so a convolution of a heaviside function with the inverse of $1/(s+b)^2$
 
$$\mathcal{L}^{-1} \left\{e^{-as} \right\}= u(t-a)$$ and $$\mathcal{L}^{-1}\left\{ \frac{1}{(s+b)^2}\right\} = t \cdot e^{-b t}$$
 
Yup, its convolution then.
 
11:44 AM
@JayeshBadwaik Seems like quite a cumbersome way to find its inverse, is there not something easier?
 
@N3buchadnezzar its not as hard as it looks, the unit step function makes it quite simple.
 
Well I used some time finding the inverse of $1/(s+b)^2$... :p
So
 
:P
Also, with inverse laplace transforms, I follow the policy of integrals, don't spend awful amount of time looking for the best method, pick the one that works and get ahead with it.
 
$$ f(t) = \int_0^t u(t - (\tau+a)) \cdot \tau \cdot e^{-b t}\,\mathrm{d}\tau $$
@JayeshBadwaik Is this correct?
 
@N3buchadnezzar yeah yeah.
 
11:47 AM
Fixed it, I think :p
 
wait, there should be one more $t-\tau$
yup, now this is better.
Ahh wait, you missed a sign I suppose
 
Where?
 
$u(t-(\tau - a))$
 
I want to pint some stuff out but I don't know if the library lets me the math building is too far away
 
user19161
@cassandra0 A walk might do you good.
 
11:52 AM
@JayeshBadwaik I think my laplace transform of $e^{-as}$ is correct though
 
@cassandra0 This might help. Don't take it too seriously, even I am a victim sometimes.
;-)
 
user19161
I think I will go out for a walk tonight.
 
no it's too far because I have an appointment somewhere else
 
@N3buchadnezzar Ahh, double mistake
@cassandra0 print later. :-)
 
@JayeshBadwaik Need to think this over some more
$$F(s) = G(t) \cdot H(t) = \mathcal{L}\{g(t)\} \cdot \mathcal{L}\{h(t)\} $$
 
11:56 AM
@N3buchadnezzar you messing up variables again.
 
@JayeshBadwaik What is correct then ?
 
$F(s) = G(s) \cdot H(s) = \mathcal{L}\{g(t)\} \cdot \mathcal{L}\{h(t)\}$
sorry for being pedantic, but it can mess up the understanding sometimes.
hahahha
 
Okay, phew. I thought I had made some grave mistake, and needed to reconsider my entire existence.
 
hmm, anyway, proceed
 
And then
@JayeshBadwaik $$\mathcal{L}^{-1}\{ F(s) \} = \mathcal{L}\{g(t)\} * \mathcal{L}\{ h(t)\}$$ ?
 
12:02 PM
@N3buchadnezzar Yes.
 
Okay, I just wanted to make sure my intuition was correct
@JayeshBadwaik Hmm gives, me an integral I am not able to solve :p
 
@N3buchadnezzar Ehh? You cannot integrate $te^{-st}$ over some interval?
 
Gimme a minute to tex what I have tried
$$ \begin{align*} \mathcal{L}^{-1}\{ F(s) \} & = \mathcal{L}\{g(t)\} * \mathcal{L}\{ h(t)\} \\ & = \int_0^t \frac{1}{a+\tau}\cdot (\tau - t) e^{-a(t-\tau)}\mathrm{d}\tau\end{align*}$$
 
25 mins ago, by N3buchadnezzar
$$ f(t) = \int_0^t u(t - (\tau+a)) \cdot \tau \cdot e^{-b t}\,\mathrm{d}\tau $$
from here, you got
42 secs ago, by N3buchadnezzar
$$ \begin{align*} \mathcal{L}^{-1}\{ F(s) \} & = \mathcal{L}\{g(t)\} * \mathcal{L}\{ h(t)\} \\ & = \int_0^t \frac{1}{1+\tau}\cdot (\tau - t) e^{-a(t-\tau)}\mathrm{d}\tau\end{align*}$$
?
 
Made a mistake $\mathcal{L}^{-1} \{ e^{-as} \} = 1/(s+a)$
 
12:14 PM
Ohh
Naah, you are mistaken
you are doing it reverse
$\mathcal{L} \{ e^{-at} \} = 1/(s+a)$
$\mathcal{L}^{-1} \{ e^{-as} \} = \delta(t+a)$
 
Ah!
Then it gets easier
 
yes..
 
@JayeshBadwaik And the dirac delta is zero everywhere except at $t = -a$
 
@N3buchadnezzar Yeah.
 
@JayeshBadwaik It should be $\delta(t-a)$ btw ;)
 
12:23 PM
@N3buchadnezzar hmm, right.
 
And then the integral windles down into
$(t-a) e^{-b(t-a)}$
 
hmm
I have to restart my XServer after the update
see you later
 
12:53 PM
hello
 
@robjohn Hello?
 
hi @PeterTamaroff
 
@cassandra0 Yello.
 
did you see Anons answer to my question it's so good
 
@PeterTamaroff Hey there
 
12:59 PM
good day robjohn
 
@robjohn I have a problem. I mean, proofwriting.
 

« first day (849 days earlier)      last day (2736 days later) »