thanks ok @TedShifrin thanks :)

Nice, somebody knows what a trema is: Angular momentum cylindrical coördinates

Which edition, @Sab? Look in the table of contents: Significance of the Derivative.

3Rd Ed

ein Umlaut, @DanielF?

Yes, chapter 11, @Sab.

It's chapter 11

OK, I'm outta here. Y'all misbehave without me.

It's kind of different from Stewart

@TedShifrin No, not an Umlaut.

@TedShifrin See you in your dreams.

Oh, because umlauts are for missing "e" ...

I'm not asleep, @Jasper.

It;s too early to be sleeping, right @Ted

@TedShifrin I ran out of things to say.

I'm sure Ted is going to walk the dog

Probably about time for you, @Sab.

@JasperLoy don't forget your Totem

I haven't a dog. Time for a martini, however :D

:P

@G.T.R I do not know what totem is, lol.

I'll go have a tea and eat something then crunch some questions till 2

@JasperLoy Inception

then take a nap and start again

@cc if you don't mind, what does the covariance and p(X,Y) -actually- mean? I can get Variance and Expectation, but not these..

What does $p(X,Y)$ mean, @Studentmath?

@G.T.R Comment on montre que $x^4=x\Rightarrow (x^n=0\Rightarrow x=0)$ ?

@robjohn I CRY HERE FOR HAPPINESS!!!

@Chris'ssis Could you give a hint on your big sum with $2014$s ?

@Chris'ssis Yay!

@Ted the pearson coeffecient

@G.T.R Ils disent que c'est évident ... mais pas pour moi

@Hippalectryon use your brain. $x^{4^q}=x$

@Chris'ssis did you get the alternating series?

@G.T.R brain ? what is that ? :P - no seriously ... what if i take $x^3=0$ ? how do I use $x^{4^q}$ here ? (or, is $q$ not in $\mathbb{N}$ ?)

@robjohn Yes, but I need to check everything to be sure all is fine.

@Chris'ssis Does it match numerically?

The strategy is to multiply by $x$ until you reach $x^{4^q}$

@G.T.R Oh ok ty

@robjohn Well, there is a bit of problem. Somewhere there is a small mistake though ...

@G.T.R see, I did learn something after all :)

Gotta catch'em all math.stackexchange.com/questions/831022/…

@G.T.R Awesome one :D

HI all

Consider some reduced (univariate) polynomial ring : R[X]/G(X).

Let f(z) be a Taylor series.

let g(z) be the functional inverse of f(z).

How do you know if every element of R[X]/G(x) has a value g(element) in the same ring ?

@mick What is R[X]/G(X)? You mean R[X]/(G(X))?

For instance I know that if a sqrt does not exist for all nonzero elements then neither does the log.

@BalarkaSen yes !

And how could the Covariance, such as in my case, be negative..

Pokemon, pokemon, gotta catch 'em all

@G.T.R Ah, I just realized you posted the same thing.

@mick Taylor series? There ain't no infinite series in a polynomial ring!

I know , I asked a very general question.

Ahhh, it actually makes perfect sense.

@mick And I can't make sense of it.

@BalarkaSen yes actually there are !

@mick No, there aren't.

You have formal power series ring R[[X]] for that

example : In the ring R[X]/(X^3-1) you can take the exp of X, by using the Taylor series !

@BalarkaSen

@mick exp(X) is not in R[X] anymore then.

It's not defined in that ring.

If you replace in your Taylor series X^3,X^4,X^5 etc according to mod (X^3-1) it works out fine !

@BalarkaSen

@mick That makes sense. I didn't note the idea, sorry.

But can you guarantee that an infinite power series is reduced to a finite polynomial y modding out by an ideal?

@BalarkaSen Sometimes.

@mick Think about R((x)), formal laurent series rings.

@mick Not always. When?

OK, gotta go.

@BalarkaSen then i cant explain

anyone else listening ?

Im talking to myself again :/

Is $R$ meant to be the reals?

Hi guys can someone help me with some permutations?

there are 10 people to arrange in a raw. A need to sit near B, C need not to sit near D

the answer is $2!⋅9!−8!⋅2!⋅2!$

but why 8!2!(2!)

@G.T.R @TedShifrin Il y a des propriétés connues des matrices dont l'inverse est leur transposée ?

@mick every Taylor series has an inverse?

@Hippalectryon elles sont orthogonales

@G.T.R Pour M une matrice, M* c'est quoi ?

@Hippalectryon la transposée complexe

@G.T.R Uh ça représente quoi ?

@G.T.R nvm wiki m'as éclairé

this guy -> @r9m is too silent today. Why? :-)

well no

@cc adjointe ou transconjugée

adjointe voila ;) transconjuguée makes more sense

@cc si j'ai bien compris, tout matrice adjointe unitaire est normale non ?

@Hippalectryon Une matrice A est normale si et seulement s'il existe une matrice unitaire U telle que U-1 A U soit diagonale.

@cc donc normale => unitaire non ?

@cc @G.T.R Pour démontrer la complétude de $\mathbb{R}$ wiki pose 'Soit une suite de Cauchy dans ℝ. Elle est donc bornée, si bien qu'on peut en extraire une sous-suite convergente. On conclut en utilisant que toute suite de Cauchy admettant une sous-suite convergente est elle-même convergente.' .... où est-ce ce que ce raisonnement est faux pour $\mathbb{Q}$ ?

@Hippalectryon not necessarily

@Hippalectryon une suite de Q convergeant vers une nombre irrationnel par exemple

@cc j'avais compris que unitaire <=> $A^*A=AA^*$ or normale <=> $A^*=A^{-1}$ donc pour moi $A^*A=AA^*$ par commutativité des matrices inverses non ?

@Hippalectryon unitaire pour moi signifie de norme 1 (mais peut être que je me trompe)

o, ok, I was wrong

@Hippalectryon donc oui, c'est vrai

@cc ok merci

Hi

Combinatorics someone?

@gbox have you tried @Chris'ssis 's awful limit with binomial $nCr(2014+k,2014)$ everywhere ? :c

@Hippalectryon what?

@gbox $\lim_{n\to\infty} \left(\frac{1}{\displaystyle \binom{2014}{2014}}+\frac{1}{\displaystyle \binom{2014+1}{2014}}+\cdots+\frac{1}{\displaystyle \binom{2014+n}{2014}}\right)$

Sorry did not any course in calculs yet

I am stupid I know

take

@gbox don't worry I'm stupid too :)

maybe can you help me

@gbox what is your question ?

@gbox What are you counting?

arrange 10 people in a row, when A need to sit near B, and C should not sit near D

the answer is 9!2!-8!2!2!

but I do not get why it is 8!2!2!

OK.

You have 10 people.

yes

Now, you can divide things into cases.

@Hippalectryon $$\frac{1}{ \displaystyle \binom{ p+k}{p}}- \frac{1}{ \displaystyle\binom{p+k+1}{p}} =\frac{ p}{p+1}\frac{ 1}{\displaystyle\binom{p-k-1}{p-1}}$$

@Chris'ssis Oh thank you

ok so the first is A near B it is 9!2!

Case I: A sits first or last. Then Bs position is uniquely determined.

This are two configurations. Now consider the simple task of having 8 people sit in a row, and two not seating next to each other. How many configurations are there?

8!-6!?

@gbox Give the reasoning.

there are 8! ways to oragnize them all

My reasoning would be: there are 8! configurations all in all. If they sit together, identify them with one person, this can be done in two ways, and then shuffle 7 persons, so 2! 7! in total.

So 8!-7!2

@PedroTamaroff ok

Well, I guess we can do something similar in more generally.

Since A and B sit together, identify them as one person.

@Chris'ssis I remember using that recently :-)

This can be done in two ways.

Now suffle to get 9!2!.

ok

@robjohn I know, you did that. :-)

Well, now the analogous reasoning identifying C,D with one person gives 2!2!8! configurations where they do sit together, and $A,B$ also sit together.

That's the answer in your book?

@Chris'ssis ah, I see the same (or similar) question here.

@robjohn That is my question.

Yes @PedroTamaroff

but why an extra 2!?

You're suffling AB,BA and also CD,DC.

@PedroTamaroff wow thanks

it is 1:32 AM, now I can do to sleep finally :)

you made my day :)

NP.

@PedroTamaroff You made my day too.

@MikeMiller ORLY.

I won't tell you how.

You suck.

No u

Do not.

Ah, the children ...

@TedShifrin He started!

hides

pedro's the child, not me

@Mike: You have interesting problems to do.

@MikeMiller When are you turning 21?

@PedroTamaroff After you.

Muahaha.

@TedShifrin I solved your first one. :P

You're the child.

Which one?

The graph of $|x|$ is not an immersed submanifold.

a connected immersed submanifold

Oh, first part of problem 0 in first problem set. Longgggggg way to go.

I know. My phone died before I could look at the next one.

(You wrote that it's not an immersed submanifold - do immersed submanifolds not have boundary to you?)

How convenient.

No boundary.

Oh, that makes it easier.

I can delete a whole step.

I hardly think it's germane.

?

Anyway, I'm going to take a nap instead of being responsible.

Night.

Aha.

I was just wondering if we have a different name for such functions now as I have not seen "Class A" functions in texts

You want a continuous function on the positive axis that decays fast enough for the transform to exist, that is all.

Right. I was just wondering if we had a different name for these functions now other than Class A but it seems like we don't need to group them. As long as they are of exponential order, the transform exists.

@Ted Did you have a position at another university before going to UGA?

@masfenix They're kind of like Schwartz functions plus guys that are at mos tpolynomially large

@KevinDriscoll Thanks. I think they are talking exactly about Schwartz, however they do mentioned sectionally continous, a term I havn't heard of.

Also another quick question. Consider a Laplace Transform $$f(s) = \int_0^\infty e^{-st} F(t) \. dt $$. The derivative is $$f'(s) = \int_0^\infty (-t)e^{-st} F(t) \. dt$$

@masfenix Ya technically Schwartz guys are supposed to be smooth, so it looks like they want to relax tha ta bit. Because Laplace doesn't care if the function is 'kinky'

However, what are the conditions on $F(t)$ so that taking the derivative is well defined.

@masfenix Well it seems like the only way you're going to run into trouble if you're using a generalized Laplace thansform where it doesn't exist for all s