Mathematics - 2016-02-07 (page 2 of 3)

6:01 PM

Anyone heard about North Korea's ballistic missile launch today? (happened around today morning)

@AndrewThompson so then we just prove it without using the idea that the function is injective? So then if I say that if $x \in C \cap D$ and $f(x)=y$ then $y \in f(C \cap D)$ which implies that $f(x) \in f(C)$ and $f(x) \in f(D)$ and therefore $f(C \cap D) \subset f(C) \cap f(D)$?

Ted Shifrin

hmm, they say satellite, and you say missile ...

your "which implies" isn't right, @Paradox ... $x\in C\cap D$ says that $x\in C$ and $x\in D$, and that implies $y\in f(C)$ and $y\in f(D)$.

Idle001

maneuvers ?

Andrew Thompson

@Paradox101 You're assuming what you want to prove.

Ted Shifrin

sorry, @AndrewT: I'll stop interrupting :)

r9m

6:05 PM

@TedShifrin satellite was released into the orbit .. but it goes against UN's restriction directives on using ballistics for that purpose ..

Idle001

@r9m that satellite is used to remote-direct missiles ?

robjohn

@TedShifrin Hey, Ted. How are things? It is windy here.

Andrew Thompson

No problem @TedShifrin. @Paradox101 You want to show $f(C \cap D) \subset f(C) \cap f(D)$, so we better start with $y \in f(C \cap D)$. Now we want to deduce $y \in f(C) \cap f(D)$. You were doing fine until the "which implies"

Ted Shifrin

calm here, @robjohn, but I'm getting annoyed that for two weeks in a row the NY Times is hours, hours late delivering my paper.

robjohn

@TedShifrin It's got to come a long way.

r9m

6:07 PM

@Idle001 not sure what that satellite will be used for :P We are talking about Kim's N-Korea here :P

JeSuis

@robjohn did you forget me or are you busy ?:p

Ted Shifrin

Tout le monde t'oublie, @JeSuis :)

Idle001

@r9m and i didnt divert from the topic

JeSuis

@TedShifrin pour Jesuis Là :

Paradox 101

@AndrewThompson so if I instead wrote that $y \in f(C \cap D)$ and there is an $x\in C\cap D$ such that $f(x)=y$ then as $x \in C$ then $y \in f(C)$ and same with the subset $D$ then that proves the conclusion?

r9m

6:11 PM

@Idle001 the rocket launched uses ballistic tech (that's against UN directives .. ) I am not sure if anything is known about the satellite it was carrying ..

Andrew Thompson

@Paradox101 Hurray!

Now, give me an example where the inverse inclusion doesn't hold.

Paradox 101

@AndrewThompson thanks a lot for your help :)

ok

Andrew Thompson

No problem.

Ted Shifrin

<--- thinks @AndrewT has the makings of an excellent teacher.

JeSuis

@TedShifrin J'ai une application différentiable $f$ de $\Bbb{R^2}$ dans $\Bbb{R^2}$ qui est propre $\Vert f(x)\Vert\rightarrow +\infty$ et $Df(x)$ est injective. Je dois montrer que $f$ est surjective... je dois donc trouver un vecteur $a$ tel que $f(x)=a$. Puis-je me placer en $0$? Car si ce n'est pas $0$ je n'ai qu'a translater le graphe, non ?

Andrew Thompson

6:14 PM

@TedShifrin Haha, coming from you I'll take that to heart.

Ted Shifrin

$C^1$, pas seulement différentiable, @JeSuis? Pour qu'on puisse appliquer des théorèmes?

@AndrewT: I was being totally sincere.

@JeSuis: On pourrait bien considérer la fonction $g(x)=f(x)-a$?

Andrew Thompson

Thanks!

JeSuis

@TedShifrin si seulement différentiable, mais je n'ai presque théorème encore... oui je voulais étudier la fonction $g(x)=\Vert f(x)-a\Vert^2$

elle est aussi propre.

Ted Shifrin

Evidemment, oui.

Paradox 101

@AndrewThompson that would be a function that's not injective?

JeSuis

6:16 PM

mais peut-on se placer en $0$ ?@TedShifrin

Ted Shifrin

Alors, on n'a pas de théorème d'application inverse?

Andrew Thompson

@Paradox101 Yes, I suppose that cat is out of the bag. Can you give an explicit example?

JeSuis

Non @TedShifrin :)

Ted Shifrin

Hmm ... Mais, en tout cas, c'est ce que tu aurais fait en considérant la fonction $g$, n'est-ce pas?

JeSuis

@TedShifrin Je viens de commencer le cours depuis quelques semaines, je ne "connais" pas ce théorème, mais je l'ai déjà lu dans un livre.

Ted Shifrin

6:19 PM

Avec l'hypothèse que $Df(x)$ soit toujours un isomorphisme, on y pense ...

JeSuis

@TedShifrin je pensais surtout montrer que $Dg(x)$ est nulle ^^

r9m

@Khallil who's that girl on your avatar? :-)

Ted Shifrin

Ah, je crois que tu as raison. Il ne le faut pas.

@JeSuis: est nulle pour un $x$ quelconque, oui.

Paradox 101

@AndrewThompson maybe something like the floor function? So that for example if $C$ is a set that just contains 1 and $D$ contains only 1.5 then $f(1)=1=f(1.5)$?

JeSuis

@TedShifrin Mais pour cela j'ai besoin que la borne inférieure soitatteinte...

Andrew Thompson

6:24 PM

@Paradox101 That function is certainly not injective, that is true. So why doesn't the inverse inclusion hold in this case?

Ted Shifrin

En fait, oui, @JeSuis.

JeSuis

$\inf\{g(x):x\in \Bbb{R}^2\}$

Ted Shifrin

Il faut employer l'hypothèse que $f$ soit propre.

Comment arrive-t-on à savoir qu'une application achève son minimum/maximum?

JeSuis

@TedShifrin ah comme est elle propre dès qu'on sera très loin, g sera beaucoup plus grand que l'inf ?

Paradox 101

@AndrewThompson if I take this example then $C \cap D$ in a null set for the function of that is also null. But the intersection of the function of these sets individually is not null so then they won't be equal?

JeSuis

6:26 PM

@TedShifrin Sur un compact par une application continue (elle est différentiable donc continue)

Ted Shifrin

En général, quelle hypothèse va nous garantir un max/min?

Aha ... Un compact.

JeSuis

Ok @TedShifrin donc $\inf\{g(x):\in \Bbb{R}^2\}=\inf\{g(x):x\in\Bbb{R}^2, \Vert x\Vert\le M\}$

Andrew Thompson

@Paradox101 Right, so $f(1.5) \cap f(1) = 1$, while $f(1.5 \cap 1) = \emptyset$. Great.

Ted Shifrin

OK ...

pour un valeur approprié de $M$, oui.

Andrew Thompson

Now assume $f$ is injective. Prove the inverse inclusion.

r9m

6:29 PM

@robjohn do you have any idea how to deal with Jack's hypergeometric series here? I know nothing about $_5F_4$ :(

JeSuis

@TedShifrin c'est la constante dans la définition de la limite..sauf erreur

Idle001

une valeur* (sorry but i feel dutiful to correct vocabulary mistakes)

JeSuis

@TedShifrin mais le minimum est sur $\Bbb{R}^2$ comment j'en déduis que $Dg(x_0)=0$ ($x_0$ est le point où l'inf est atteint) sur $\Bbb{R}$ ?

Ted Shifrin

Thanks, @Idle001. My French is too rusty.

Idle001

sorry again just sidy remark, i know ur english beats mine

Ted Shifrin

6:33 PM

@JeSuis: Il faut que ce soit un point à l'intérieur de $\{x: \|x\|\le M\}$.

@Idle001: Et je sais bien que c'est une valeur appropriée :P

Idle001

ah so a typo

Ted Shifrin

After 40+ years, I still know my grammar, but the genders of words and the slang vocabulary are realllllllly tough, @Idle001.

JeSuis

@TedShifrin hm right, so when I have $Dg(x_0)=0$ it means that je Jacobian of $g$ is zero, but after that, I don't know how can I conclude..

Andrew Thompson

I have to read a paper in French for my undergrad. Hope Google translate has gotten more precise with the years.

Idle001

i feel lazy too for typing terminological punctuations

Ted Shifrin

6:35 PM

But please continue to correct me :)

Google Translate sucks @AndrewT :)

Andrew Thompson

Dang it.

Idle001

@TedShifrin french is so sensitive and strict to genders

Ted Shifrin

@JeSuis: Il faut un calcul de $Dg(x_0)$, alors.

Semiclassical

hiya

Khallil

I think she's in Game of Thrones, @r9m.
I can't remember her name though :-)

JeSuis

6:37 PM

it's $2\langle f'(x_0),f(x)-a\rangle$

Khallil

Loving the Saitama egg btw, @r9m! ^_^

r9m

@Khallil ya! One punch .. !! Do you watch one-piece?

Ted Shifrin

@JeSuis: Ça ne fait pas de sens. $f\colon\Bbb R^2\to\Bbb R^2$.

JeSuis

@TedShifrin oui je n'ai pas écrit la matrice car je ne sais pas avec Latex :D

Khallil

I haven't found the time to start it yet, @r9m.

I'm watching BTOOOM! right now.

It's really good!

Ted Shifrin

6:40 PM

Alors, il faut comprendre ce qui se passe, et puis se demander si $Df(x_0)$ puisse être un isomorphisme.

Idle001

good you have time for that luminous talking box called tv

cant handle watching repetitive sensless scenes and reminding the hours spent vainly next to em makes me feel sorrowful

Ted Shifrin

limunous? !!

r9m

@Khallil yup .. I watched Btooom! awesome anime! :D

Ted Shifrin

oh, luminous ...

Idle001

thank u ted, u hit me back

JeSuis

6:43 PM

@TedShifrin $Df(x_0)$ est une application linéaire injective en dimension finie donc surjective, n'est-ce pas ?

Ted Shifrin

LOL, I had no idea what you meant!

@JeSuis: Oui, c'est pour ça que je disais "isomorphisme" :)

mr eyeglasses

disais isomorphisme

disease isomorphism

Andrew Thompson

@Paradox101 Making any progress?

Ted Shifrin

Stop that, @mreyeglasses :D

Idle001

i meant, watching tv isnt a thing of my quotidian @TedShifrin

JeSuis

6:45 PM

@TedShifrin oui, c'est grâce à toi que j'ai pensé à cela :D

Ted Shifrin

Alors, @JeSuis, tu vois le chemin?

Je t'ai compris enfin, @Idle001 :)

user276387

A fair die is rolled twice. If the two results are the same, a coin is tossed. What's the total number of different possible outcomes of this experiment? Answer: $6 \times 6 = 36$, so $30$ outcomes are a dice roll and $12$ outcomes are a coin toss. $30 + 12 = 42$. Why have they subtracted $6$ from the outcomes of the dice roll?

JeSuis

@TedShifrin non lol

Ted Shifrin

@user276387: They haven't subtracted. They've added two possibilities for each of those 6 outcomes.

Mike Miller

Ah, @Ted, you've never told me that. Am I doomed to be a failure of a teacher?

Ted Shifrin

6:51 PM

Shaddup @MikeM.

@JeSuis: Qu'est-ce que tu sais à propos de $\partial f/\partial x(x_0)$ et $\partial f/\partial y(x_0)$?

Mike Miller

I should send this to the students who didn't want to hear me talk about them last Tuesday... :)

Andrew Thompson

Ohh, you had a seminar for undergraduates or something?

Paradox 101

@AndrewThompson if I prove that by contradiction will that be right? I mean if I first state that there is a function such that $f(x)=f(y)=y$ where $x$ and $y$ are notequal would that be the right start?

Mike Miller

No, I TA the graduate differential geometry course.

Someone asked me why they cared about the Frobenius theorem so I gave a very brief overview of the geometry of foliations.

Ted Shifrin

I have lots of other answers to the "why do I care about Frobenius" question :)

Khallil

6:55 PM

Awesome! I didn't think it was very popular so this is a nice coincidence, @r9m!

Ted Shifrin

More differential geometric answers, as opposed to topological answers, but I bet you guys haven't gotten to any geometry yet.

user276387

@TedShifrin So it's $30$ outcomes for the dice roll, and the $6$ where the results are the same where a coin is tossed, i.e. $6 \times 2 = 12$?

Ted Shifrin

@user276387: It's 36 outcomes for the dice roll, but if both outcomes are the same, you pick up an additional option of H/T. So for those 6 cases, there are two outcomes each.

user276387

Thanks @TedShifrin

Ted Shifrin

Sure :)

r9m

6:58 PM

@Khallil it's popular enough! among the drama-thrillers-action that is ..

Andrew Thompson

@MikeMiller Nice, which book are you using? @Paradox101 Hm, I think a direct argument is better. I think you can do the exact same argument as you did before, in fact, however as you now want to prove $f(C) \cap f(D) \subset f(C \cap D)$, you have to start with $y \in f(C) \cap f(D)$. Try to carry out the same argument, but note that you will need two (a priori distinct) $x$'s, using the notation from before. Then injectivity will save you.

mr eyeglasses

Lang's differential geometry looks so hard

Ted Shifrin

@mreyeglasses: Not a good place to start.

I seriously recommend starting with the undergraduate curves/surfaces stuff before you go off the deep end to $n$-dimensional manifolds. Lang is doing it all with Banach manifolds, even worse.

Andrew Thompson

I'm suffering the backlash from having a 'top-down' education in diffgeo. At this point I can only think of the tangent space as equivalence classes of curves.

Ted Shifrin

I'm not sure that's the worst symptom you're enjoying :D

Andrew Thompson

7:04 PM

Well, when you start writing $[\gamma]$ for elements in a Lie algebra you might need a trip to rehab.

Ted Shifrin

um, yeah, that's bad. Let's use vector notation (or matrix).

Andrew Thompson

Yesh. Will eventually be cured, taking a Lie theory course this semester.

Ted Shifrin

Spankings are always another option :D

Andrew Thompson

A lecturer of mine seriously considered bringing a spankingthingy (switch?) to the oral exam as a joke.

Ted Shifrin

:)

Mike Miller

7:07 PM

@AndrewT: Morita's "Geometry of differential forms".

Andrew Thompson

oh the wrong morita.

Not heard of.

trilolil

Hello everybody! :)

I posted a question on se but it hasn't been viewed a lot the past days

only 10 times

it is about calculating the orientation of a satellite

I was wondering if any of you guys could give me a few hints?

0

Q: satellites attitude determination TRIAD - how are orbital reference frame vectors constructed?

I posted this same question on space.stackexchange but never received any answer. So I am posting here hoping to get an answer as this is a quite mathematical topic. I am trying to fully understand the TRIAD algorithm for attitude determination of a satellite. https://en.wikipedia.org/wiki/Tria...

Ted Shifrin

Hi dogatemyhomework

mr eyeglasses

lol

dog named columbus

Mike Miller

@Ted: They did Moser's trick, so they've done some geometry. I gave them two examples of things they probably already care about where Frobenius is useful: integrability of almost complex structures and the Lie group - Lie algebra correspondence.

Ted Shifrin

7:11 PM

I was thinking of Maurer-Cartan forms and the fundamental theorem for surfaces/hypersurfaces.

This is Cartan's trick :)

Paradox 101

@AndrewThompson so then we start with $y \in f(C) \cap f(D)$ which means that $y \in f(C)$ and $y \in f(D)$ then if we also assume that $x \in C$ and $x \in D$ then $x \in C\cap D$ so from this point we assume an injective function $f(x)=y$?

Khallil

I was hooked when I saw the opening sequence on YouTube, @r9m!

It's really catchy. "The game has only just begun"

mr eyeglasses

@Khallil you like nano?

Khallil

At some point, I also wanna check out "The Garden of Words".
This AMV makes it look so good. The animation looks like the budget was unlimited.
https://www.youtube.com/watch?v=pfBhOcjhowA&index=2&list=WL

Andrew Thompson

@Paradox101 What you write is either correct but extremely unclear or gibberish. '...then if we also assume that" Why can you make such an assumption? (Regardless you are on the right track.)

Khallil

7:13 PM

I've only heard that one song as far as I know but it's good, @r9m!

Andrew Thompson

@Paradox101 But the key here is $y \in f(C)$ and $y \in f(D)$. This should tell you two things.

Mike Miller

@Ted: I'm an ignoramus and know nothing about hypersurfaces.

I'm an artist

@r9m hey. Only serious research can help there. I've never ever seen anyone calculating anything like that.

Paradox 101

@AndrewThompson if I instead write after the $y$ parts that let $y=f(x)$ where the function is injective and then say that $f(x) \in f(C)$ and same with $D$ and then state that this implies that $x \in C$ and same with D would that make sense?

I'm an artist

Very tired here due to the excessive work.

Andrew Thompson

7:20 PM

@Paradox101 Now it seems you are on the right track (and might be done.) Write me the full argument and I'll look at it.

Paradox 101

$y \in f(C) \cap f(D)$ and therefore $y \in f(C)$ and $y \in f(D)$. Let $y=f(x)$ where the function is injective, then, $f(x) \in f(C)$ and $f(x) \in f(D)$ . Then $x \in C$ and $x \in D$ and therefore $x \in C \cap D$ which implies $f(x) \in f(C \cap D)$ i.e $y \in f(C \cap D)$? @AndrewThompson

Andrew Thompson

@Paradox101 That is correct. Can you tell me where you used injectivity?

Paradox 101

@AndrewThompson when stating that $x \in C \cap D$

Andrew Thompson

Right, so in fact when you wrote "Then $x \in C$ and $x \in D$." Here's how I would've written it:

Let $y \in f(C) \cap f(D)$. Since $y \in f(C)$, there is an $x \in C$ such that $f(x) = y$. Since $y \in f(D)$, there is an $x' \in D$ such that $f(x') = y$. Since $f$ is injective, $x = x'$.Hence $x \in C \cap D$, so $f(x) = y \in f(C \cap D)$.

But your way of writing it is fine, no worries, I'm just being pedantic.

mr eyeglasses

I don't think it's pedantic

Andrew Thompson

7:33 PM

Well, what (s)he is writing is completely correct, however it might suggest (s)he doesn't understand it completely. Had a person written that where I didn't know the author I wouldn't have any suspicions.

Which is the problem of teaching people stuff over the internet: you don't know if they understand what they're writing.

Paradox 101

@AndrewThompson your way is far better, it's concise and conveys everything clearly. Mine is all over the place. Thanks a lot for being so patient :)

Andrew Thompson

@Paradox101 No problem.

guest

I agree, typing words takes no understanding :)

Ted Shifrin

Yes, skull is the master of that :P

guest

:P

trilolil

7:36 PM

0

Q: satellites attitude determination TRIAD - how are orbital reference frame vectors constructed?

I posted this same question on space.stackexchange but never received any answer. So I am posting here hoping to get an answer as this is a quite mathematical topic. I am trying to fully understand the TRIAD algorithm for attitude determination of a satellite. https://en.wikipedia.org/wiki/Tria...

matrices vector-spaces systems-of-equations orientation

I am just leaving that here

praying for something to happen...

JeSuis

@TedShifrin désolé j'étais parti manger... Elles sont injectives, en gros c'est une base de $\Bbb{R}^2$, si j'ai bien compris

Ted Shifrin

@MikeM: I won't offer to send notes, because you'll say you can't read them. Basically, for surfaces in $\Bbb R^3$, having the Gauss and Codazzi equations hold is the obstruction to having a given first and second fundamental form determine an immersed surface. The nicest way to see this is the Cartan game.

trilolil

mais ne sois pas desole mon grand...

Idle001

second most perfect moment of a coder right beside when bying a new computer, seeing his new mathematic formula working out its programatical fruits . toast of the moment

Ted Shifrin

Mais, vis-à-vis du vecteur $f(x_0)-a$, elles sont ... ? @JeSuis

Mike Miller

7:39 PM

@Ted: Oh well. :)

JeSuis

orthogonal ? @TedShifrin Je ne comprends le caractère de la question

Ted Shifrin

Oui, c'est ça, et donc ... ?

@MikeM: Or you could look at section 6.3 of Chern/Chen/Lam Lectures on Differential Geometry. Or a Griffiths paper I could dig up.

JeSuis

@TedShifrin ahhh, le vecteur $f(x_0)-a$ projeté sur les "dérivées partielles" sont nulles

Mike Miller

I probably won't. I'm answering a question then getting back to work.

Andrew Thompson

@MikeMiller What does TAing entail in the US? Are you lecturing a lot? For us its just waiting for someone to ask for help.

Ted Shifrin

7:41 PM

@JeSuis, et puisqu'elles te donnent une base ...

@AndrewT: We're glad to ask for help.

JeSuis

@TedShifrin excellent!!

Mike Miller

@AndrewThompson I write a lecture the day before Tuesday, give it, and hold two office hours a week, and they're free to email me questions.

For some reason, nobody does.

mr eyeglasses

Are you intimidating @MikeMiller

JeSuis

@TedShifrin quand tu m'as dit et donc.. j'ai fait : ahhh oui, cool! :D

Ted Shifrin

@MikeM: I'll have my ex-students who keep bugging me with questions bug you instead.

JeSuis

7:42 PM

Merci beaucoup

Mike Miller

No, I think they're just not thinking about the material enough to have questions to ask.

Ted Shifrin

Pas de quoi, @JeSuis. Pas mal :)

@mreyeglasses: I'm intimidating and it doesn't stop them :D

JeSuis

@TedShifrin demain je passerai au tableau alors!

mr eyeglasses

@TedShifrin I feel like you are quite approachable just from your internet persona

Ted Shifrin

@AndrewT: Often algebra is too much about manipulating letters.

Félicitations, @JeSuis :)

JeSuis

7:45 PM

:))

Andrew Thompson

@TedShifrin Indeed. Figured I shouldn't make fun of students on the internet. But I do believe they should do concrete stuff as to be able to manipulate letters cleverly :)

Michael

Holy crap I just watched the most boring lecture of my entire life

Idle001

@JeSuis tu es en quelle annee ?

Ted Shifrin

Gee, thanks for that compliment, @Michael!

mr eyeglasses

lol

JeSuis

7:46 PM

@Idle001 tu es français ? Je suis en L3 maths

Michael

@TedShifrin ;). You're lectures are actually pretty nice and interesting. ONly issue is I can't always see the blackboard

Ted Shifrin

@Michael: They improved at that after a few times.

It was amateurs (students) volunteering to do the videos.

Idle001

ca voudrais dire tu seras au master l'annee prochaine @JeSuis

mr eyeglasses

We need @TedShifrin's lectures in 1080p HD

Mike Miller

Cool, 20k.

Ted Shifrin

7:47 PM

Too late for that, @mreyeglasses.

mr eyeglasses

What if TedShifrin started making YouTube videos

Michael

@TedShifrin What is a good way to practice proofing? I feel like I am cheating whenever I google the proofs...

Andrew Thompson

@Michael Because you are. :)

Ted Shifrin

The only way to practice is to do lots, @Michael, and get someone to criticize what you write. (That's why I'm such a believer in graded homework, ignoring cheating.)

I'm an artist

@TedShifrin how does it seem to you if a student told you he had over 20.000 results in mathematics? Is that much? Just curious if this volume is a promising one for a student or out there are greater expectations.

Ted Shifrin

7:49 PM

As with everything else, you need to start with examples in lectures and texts as models and try to understand them 1000%, then do similar ones yourself, then harder ones, etc.

Michael

Ok give me a proof that a first year undergraduate is expected to be able to find. I'll come back to u in a day

I promise I won't cheat

Ted Shifrin

Here's an easy one at the beginning of linear algebra. Suppose $x$ is a vector in $\Bbb R^n$ and $x\cdot y = 0$ for all $y\in\Bbb R^n$. Prove that $x$ must be the zero vector.

There are lots of possible ways to proceed. See how many proofs you can find and which is simplest.

JeSuis

@TedShifrin Une question que j'ai pas réussi : J'ai une application linéaire $A:\Bbb{R}[X]\mapsto \Bbb{R}$ such that $A(P)=P(1)$, pour la norme $\Vert P\Vert=\bigl(\sum_{i=1}^n\vert a_i\vert^p\bigr)^{1/p}$, j'ai montré que pour $p=1$ elle est continue. Mais pour $p>1$ je dois montrer que non. Donc je dois trouver une suite qui tend vers une limite tel que l'image par $A$ de celle-ci ne tende pas vers l'image de la limite. Mais je ne trouve pas une telle suite de polynômes..

Idle001

people are googling their car-keys these days

JeSuis

@Idle001 oui tout à fait

Ted Shifrin

7:51 PM

Intéressante, @JeSuis.

LOL @Idle001 ...

I'm an artist

@Semiclassical is 20.000-25.000 results (say, discoveries) a reasonable amount of math stuff for a student?

Ted Shifrin

Well, getting help here is a generalized weak form of googling.

@JeSuis, c'est évident pour $p=\infty$?

user43418

Hi

Question: I am trying to prove using convexity that $e^{\lambda X} \leq X+ (1-X)e^{\lambda}$

but I am not sure what expression for $\lambda X$ I should be using

Ted Shifrin

What's your definition of convexity?

user43418

and what definition of convexity

well

the one that I know is $f(ax+(1-a)y) \leq af(x)+(1-a)f(y)$

Ted Shifrin

7:55 PM

@JeSuis: $p\in\Bbb N$ ou plus général?

user43418

but I have a feeling it might be $f(ax+(1-a)y) \leq af(x)-(1-a)f(y)$

here

I'm an artist

Around with some research. BBL

Ted Shifrin

Good, @user43418. Apply that immediately for appropriate $x$, $y$, and $a$.

No, it's the first. Do you know what that sentence means geometrically?

JeSuis

@TedShifrin on a $p\in[1,\infty]$

Ted Shifrin

OK, @JeSuis: $p=2k$ et $p=\infty$ sont pas très difficiles.

Considère le cas des polynômes quadratiques.

JeSuis

7:58 PM

pour la norme infinie on peut prendre $1+X+X^2+\cdots+X^n$ sauf erreur.

Polynômes quadratiques : en x^2 ?@TedShifrin

user43418

@TedShifrin Le problème est que je tombe toujours sur: $\lambda X=X*0+(1-X)*(-\lambda)$

Ted Shifrin

Si $a_n^2+b_n^2 \to a_0^2+b_0^2$, faut-il que $a_n+b_n\to a_0+b_0$?

Tout le monde est français ?

JeSuis

LOL, @TedShifrin

je me posais cette question

user43418

est donc $e^{\lambda X} \leq X+(X-1)e^{\lambda}$ et non pas $e^{\lambda X} \leq X+(1-X)e^{\lambda}$

Michael

J'aime la baguette

JeSuis

8:00 PM

:D

la*

Ted Shifrin

@user43418: Il faut prendre $a=X$, et puis $x=0$ et $y=?$.

user43418

c'est pour ca que je pensais que peut-être que c'est la definition qui était fausse

Ted Shifrin

@Michael: You working on the problem I gave you?

user43418

y= \lambda ?

Ted Shifrin

Evidemment.

Michael

8:02 PM

@TedShifrin Yes sir I am. I just need to reveiw some concepts that you taught about $\Bbb R^{n}$ in your lectures

Ted Shifrin

Oh, do $\Bbb R^2$ or wherever you've already had vectors.

JeSuis

@TedShifrin non il ne faut pas :D

Ted Shifrin

Oh, @user43418, peut-être qu'il faut $x=\lambda$ et $y=0$?

@JeSuis: Alors, il faut jouer comme ça.

OK, lunch time for moi. See you all later.

user43418

@TedShifrin non...

JeSuis

@TedShifrin Bon appétit.

mr eyeglasses

8:10 PM

Croissant

guest

French vanilla

I'm an artist

I'm done with the work today. I'm ridiculously tired.

Is any done with my integral?

Michael

@I'manartist what is Ei?

I'm an artist

@Michael a possible mystery

Michael

@I'manartist How mysterious is it? o-0

I'm an artist

8:17 PM

Let me listening now to some Elvis Presley songs.

JeSuis

@I'manartist I can only say that the numerator is $2E_i(x^2y^2/x^2+y^2)-()-()$ :p

I'm an artist

Exponential integral

In mathematics, the exponential integral Ei is a special function on the complex plane. It is defined as one particular definite integral of the ratio between an exponential function and its argument. == DefinitionsEdit == For real non zero values of x, the exponential integral Ei(x) is defined as The Risch algorithm shows that Ei is not an elementary function. The definition above can be used for positive values of x, but the integral has to be understood in terms of the Cauchy principal value due to the singularity of the integrand at zero. For complex values of the argument, the definition...

@JeSuis :D

Michael

I saw complex, then I'm like nah

I'm an artist

8:29 PM

In case you miss the point (some of you) I want to tell you that the tiny area of calculating integrals and series is extremely vast. Then, to master this area, one needs to work for years, and a lot of hours a day. Just read this once in a while, especially before opening a contradictory discussion with me.

Let me find some good music (now).

I'm an artist

8:50 PM

Let me put a challenge that will also tell a lot about one can do in my area. I'll ofer a free copy of my book for a solution to this probem. Defining $\varphi=\sum_{k=1}^{\infty} \frac{1}{k^2 4^k}$, calculate in closed-form $$\sum_{k=1}^{\infty} \frac{H_k}{k^3 2^k}$$ without using special functions at all (no polylogarithms, no beta function)

Idle001

@I'manartist what is the secret behind that picture u continuously post

I'm an artist

It was posted on MSE, and it appeare din many problem, but no one did it according to the requirements above.

I let you also cheat! Here is the point: I let you have talks with colleagues, professors with anyone you want, but don't post it on sites where thousands of people can work on it.

Idle001

i cant calculate this even with a pen and paper