Mathematics - 2017-02-12 (page 4 of 4)

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barto

21:01

@PaulPlummer Okay thanks; probably it was just an "oubli". After all, "homeomorphism on its image" wouldn't be as useful a notion if it depended on the space surrounding the image :D

SteamyRoot

21:19

Meeeh. I finally got a desktop for my office; and I want to install GAP...

But they're probably going to release version 4.8.7 any day now and the only way to update is pretty much a reinstall afaik

user243301

@DanielFischer Please, and sorry by asking here, but if can you answer about an example of a complex-valued square integrable function $f:\mathbb{R}\to\mathbb{C}$? I believe that it means the following, I should be able to show that $\int_{\mathbb{C}}\left|f(s)\right|^2ds=\int_{\mathbb{C}}f(s)\overline{f(s)}ds$ is finite . Thanks.

Secret

Last night dream: A challenge on finding an uncountable pairwise ordered sequence of integers. 5 days' later, the challenge host reveal an unexpected answer:

6,5,4,3,1,0,0..,5,3,1,0,0..,4,3,2,1,0,0,0..,16,5,2,1,0,0…

He then said that the lesson he want to get through to the students is that while he requires consecutive terms to be ordered in a way such that the preceding term is larger than the next term, he never said that every possible pairs in the sequence must have the same ordering, hence the answer need not be a chain (CR). The answer he given is actually a general ordering (GO).

Reality check: That stuff, if it ever makes sense, will be some weird poset consists of union of many posets with different types of ordering. The resulting sequence is then one which have many types of ordered relations chained together in one expression

even then, 0 > 0 does not make sense for any ordering relations

also in maths, there is no such thing called "general ordering"

Daniel Fischer

@user243301 The domain of the function is $\mathbb{R}$, so you integrate over $\mathbb{R}$, not over $\mathbb{C}$. Apart from that, you're right, supposing the measurability of the function is clear. Then you have to show that $$\int_{-\infty}^{+\infty} \lvert f(x)\rvert^2\,dx < +\infty.$$

user243301

21:49

@DanielFischer thanks and sorry I was very wrong integrating over $\mathbb{C}$, it was the main problem to find examples...

Paradox 101

Can anyone please clarify something in this question? Given vector $A= (2,-1,-1)$ and vector $C=(0,1,1)$, $C$ rotates about $A$ with an angular velocity of 2 rad/s . Find the velocity of the head of $C$. As velocity is the cross product of angular velocity $w$ and $r$ and $w$ is in the direction of $A$, so the velocity should be a cross product of $A$ and $C$? If so, where does the angular velocity 2 rad/s fit in?

Secret

22:13

The tip of the vector C (which if it is a position vector, is (0,1,1)) is actually rotating in a circular motion with vector A as the rotation axis. Therefore your r is actually the perpendicular distance of the tip of C from the line spanned from A (which can be obtained by computing C-proj(C)_A). Now the angular velocity vector is mutually orthogonal to the r vector, thus the direction is clear (indeed w has the same direction as A). Hence multiplying r by w gives the velocity of the tip

Paradox 101

22:28

@Secret Can you explain what you mean by C-proj(C)_A? I don't understand the notation.

Astyx

@Akiva They do indeed

On what conditions is a real function a derivative of another ? Is it enough for it to satisfy the IVT ?

And hi @Ted

Ted Shifrin

salut @Astyx

Astyx

Comment vas-tu ?

Ted Shifrin

Maintenant je me trouve un an âgé de plus :P

Astyx

Joyeux anniversaire !

Ted Shifrin

22:38

Merci bien :)

Astyx

Tu as soufflé les bougies ce midi ?

Ted Shifrin

Rien d'intéressant se passe ici?

Pas de bougies, merci.

Astyx

Ici où ?

Ted Shifrin

Dans le chat, bien sûr?

Astyx

Je ne sais pas je viens d'arriver

Paradox 101

22:40

@Secret I still don't get where the angular velocity 2 rad/s comes in, given that vectors A and C are omega and r

Astyx

Enfin si, j'ai posé une question super intéressante sur les dérivées de fonctions :p

Ted Shifrin

Tu es sûr qu'elle soit tellement intéressante?

Astyx

Assez pour que je me la sois posée disons (ce qui n'est évidemment pas suffisant)

Après les questions que je me pose à 23h30 sont pas forcément les meilleures

Ted Shifrin

You need an absolutely continuous function for the Lebesgue version of the first fundamental theorem of calculus to work, I believe.

$f(x)=\begin{cases} \sin(1/x), & x\ne 0 \\ 0, & x=0\end{cases}$ satisfies the intermediate value property. Is it a derivative?

Astyx

You're missing a $

Ted Shifrin

22:47

LOL, thanks.

Astyx

Sorry, I lost my internet connection

Ted Shifrin

Likely story :D

Astyx

Anyways, why isn't this function a derivative ?

Ted Shifrin

Well, give me the function.

Astyx

But a derivative does not have to be continuous does it ?

I see the problem with the function you defined above

Ted Shifrin

22:57

No, a derivative does not have to be continuous.

Astyx

And doesn't absolute continuity imply continuity ? I think I misunderstood what you said

Ted Shifrin

Yeah, absolute continuity implies continuity. The Lebesgue stuff tells you that it has to be absolutely continuous a.e., I guess.

But I'm just thinking about old-style Riemann here.

Astyx

"a.e." ?

Ted Shifrin

almost everywhere

Astyx

Oh right

Anyways, I'm not going to waste too much of your time on your birthday :)

Ted Shifrin

23:00

If you define $g(x)=\int_a^x f(t)\,dt$, then $g$ can be differentiable but have $g'(b)\ne f(b)$, of course, for certain $b$.

Astyx

Yup

Ted Shifrin

I'm back from a hike and lunch and now have to get ready to go out to drinks and dinner with other friends. It is a tough life.

Astyx

It sucks to have friends, doesn't it ? :p

Ted Shifrin

Ne sois pas jaloux, enfin :P

Astyx

Au moins moi je peut rester dormir dans mon lit pendant les prochaines heures ... na !

Ted Shifrin

23:04

Il faut te dire bonne nuit, alors :)

Astyx

Oui, merci :) Bonne fin de journée à toi ! Et merci pour l'aide :)

À la prochaine !

Ted Shifrin

À la prochaine.

Daminark

Hey!

Paul Plummer

@Astyx I remember one of my analysis teachers (one of the best classes I ever had) gave us a paper that talked about how it is difficult to classify which functions are derivatives

Paul Plummer

23:23

@Astyx I found the paper. It is Derivatives why they elude classification by Andrew Bruckner