Mathematics

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

arete

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

anon

@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).

arete

Thank both of you.

leo

6:07 AM

$\theta^{-1}$ seems smooth as well

:-(

anon

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

anon

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

anon

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?

anon

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 $|x|$

anon

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

a Dangerous Idea

8:13 AM

$\color{red}{\Huge\text{>8(}}$

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

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

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

$\color{blue}{\Huge\text{>8(}}$

robjohn

@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

a Dangerous Idea

@robjohn I agree.

$\color{red}{\text{>B(}}$

$\color{red}{\Huge\text{>}}\color{green}{\Huge\text{8}}\color{blue}{\Huge\text{(‌}}$

$\color{orange}{\Huge\text{>}}
\color{white}{\Huge\text{8}}
\color{orange}{\Huge\text{(}}$

Chris's sister

8:35 AM

Hi

a Dangerous Idea

hi

robjohn

@Chris'ssister Hey, what's up?

Chris's sister

Hi @aDangerousIdea

Hi @robjohn. A bit around here. :D

a Dangerous Idea

How are you?@Chris'ssister

Chris's sister

@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.

robjohn

8:39 AM

@Chris'ssister It is a bit quiet here. We had an hour of silence (for what, I don't know).

Chris's sister

@robjohn: sorry if you took the silence away. :-)

robjohn

@Chris'ssister actually, aDangerousIdea and I were just looking at smileys

Chris's sister

@robjohn: smileys make us feel better, right? :) I like smileys too.

robjohn

@Chris'ssister Perhaps >8(

Chris's sister

@robjohn: I never asked you but now I do it: are you a teacher? You explain things nicely and I was thinking at that.

robjohn

8:43 AM

@Chris'ssister I taught at UCLA for a couple of years. But that was a while ago.

Chris's sister

@robjohn: I see. That's really nice.

robjohn

@Chris'ssister since then, I've been a part of sci.math and now MSE, both of which feel like teaching, too

Chris's sister

@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.

Sanchez

Just curious: is there a forum that's a bit more specialized (to graduate level math, yet not at the level of mathoverlow?)

Chris's sister

@Sanchez: I suppose that MSE and MathOverflow catch all level in mathematics. (if I'm not wrong)

Sanchez

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.

Chris's sister

@Sanchez: I see.

a Dangerous Idea

hi @iymz

$\color{orange}{\text{>8(}}$

$\color{orange}{\text{)8<}}$

bye iymz

Old John

9:16 AM

Good morning

Chris's sister

@OldJohn: morning!

robjohn

$$
\begin{align}
\int\frac{x^2}{\sqrt{1+x^2}}e^{\arctan(x)}\,\mathrm{d}x
&=\int\tan^2(u)\sec(u)e^u\,\mathrm{d}u\\
&=\int(\sec^3(u)-\sec(u))e^u\,\mathrm{d}u
\end{align}
$$
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\\

Chris's sister

@robjohn: awesome! Thank you!

robjohn

@Chris'ssister I just checked and Mathematica agrees.

Chris's sister

@robjohn: nice and fast

user19161

9:27 AM

Ladies and gentlemen, I have capped today. Thank you for your support.

Old John

@WillHunting you have capped already?!?!

robjohn