you don't need a basis to write $V^*\otimes V\to{\rm End}(V)$. define $\phi\otimes v\mapsto \phi(-)v$ for pure tensors, check it makes sense to extend linearly. proving it's an isomorphism might require picking a basis though

hey @RandomVariable :)

@r9m Hello. I think you're the first person to say hello to me on here in like 6 months.

@RandomVariable :o c'mon I say hello to you whenever I drop by here and see you online :)

@r9m Maybe I'm exaggerating a little bit.

hi pal @RandomVariable

:P

@RandomVariable :) my courses are about to begin .. I havent been able to visit the chat room as frequently as I did before ..

@r9m Usually I'm very quiet unless I want to ask/annoy Daniel Fischer about something. And few people initiate conversations with me.

ping @Chris'ssis have you seen Tauraso's solution to AMM11821 (he updated the page it seems) :-)

@skullpatrol Hello. All of sudden I'm popular.

@RandomVariable I rarely see you joining in a conversation that has little to do with mathematics problems :P so I guess it's an expected consequence :P just jokin

@RandomVariable nyway .. it seems galactus is back at I&S :-) I hope the forum becomes active again :)

@skillpatrol yo pal !! :D

@r9m hi pal :-)

how are you?

@skillpatrol fine! and you?

@r9m fine thanks.

@anon -that makes sense, so, for example to check for injectivity, check LI of $(e_i)^{\ast}(-)e_j$?

@r9m what is I&S?

@r9m He said that Shobhit is not responding to messages and emails.

@skillpatrol integralsandseries.prophpbb.com :)

@r9m thanks

@DavidWheeler it takes the basis $\{e_i^*\otimes e_j\}$ to the matrix basis $\{e_{ij}\}$, yes

@RandomVariable ah! he must be busy ..

or $e_{ji}$, I don't feel like checking which one

@anon lol, I have the same problem-and I don't really care.

@r9m He hasn't posted anything on the site this year.

@DanielFischer I ended up convincing myself (perhaps wrongly) that the answer I posted the other day was just Parseval's theorem in disguise. So I deleted it since there is already an answer that uses Parseval's theorem.

Can someone help me out with number 3 here: math.arizona.edu/files/grad/past_quals/Anal-Aug2008.pdf

Specifically, the last part about describing all measurable f(x,y) with respect to M_F

Warning: pdf link

@RandomVariable Aha. Didn't look very Parsevalish to me, but your call.

morning folks

hi pal

@RandomVariable link please :)

@DanielF: How much do you need for an inverse function theorem? You can get one in Banach manifolds. How about Frechet?

@r9m Unless your reputation recently went over 10,000, you won't be able to see it.

@RandomVariable ah! but I don't know the question

@MikeMiller Not sure, but I think it should work.

My ignorance re: infinite-dimensional manifolds is showing lately. I have no idea what carries over, what doesn't. I guess it's time for a crash course.

It was one of Chris's sis the artist's recent questions. The one with $\text{artanh}(x)$.

@RandomVariable ah! I remember .. :)

If we use words to define ideas-then how do we define the idea of a word?

by meaning

@DanielFischer I noticed that I'm not being notified about reputation changes. Does it have anything to do with deleting that answer?

meaning...what?

the meaning of the word

@RandomVariable Probably. You lost some reputation by the deletion, and the notifications wait until your reputation rises above the level you had before the deletion. Could be that they come back before that if you close your browser window, but maybe not.

the meaning of the word "meaning", or the meaning of the word "word"?

@DanielFischer: Google tells me that apparently the naive statement in Frechet manifolds fails but that subtle modifications make it work.

@MikeMiller Aha. Does it tell you what the naive statement is and what modifications one needs?

The naive statement is that a smooth map with surjective differential has a local section. It doesn't say what the modifications are.

For today, this is mostly a curiosity, since my situation is Banach.

@MikeMiller Huh? For the inverse function theorem, I would expect that the differential is an isomorphism.

I always get the two statements mixed up. I meant implicit.

Well, I meant whatever statement is what I just said.

Hmm, for a local section, the kernel of the differential would need to be a complemented subspace, wouldn't it?

Oh, I guess you're right, the statement there can't possibly be true, can it? Aren't there surjections of Banach spaces that don't split?

@DavidWheeler we define the idea of a word as that which has meaning

I tried to warn you that I'm an ignoramus.

@MikeMiller Take any non-complemented closed subspace and look at the canonical projection $E \to E/F$. Doesn't split. And I don't think it has a (differentiable) [local] section.

@DanielFischer: If it had a differentiable section, that should provide a section on the level of tangent spaces; since the map is linear the differential should be exactly the same as the original map.

Yes, that's what I think.

I don't know what the definitions of tangent space and so are in this context, but I can't imagine a correct definition that doesn't imply the above.

Thanks for your help.

Does anyone know if the following identity holds: Suppose $F$ is the CDF of some RV. So $F(x) = \Pr(X < x)$. I am reading a paper, where they come across the inverse of the survival function $[1 - F(x)]^{-1}$ and reduce it to $F^{-1}(1 - x)$

Does such an identity hold and if so, can someone point me to a quick proof of it

if you compose the functions $1-F(x)$ and $F^{-1}(1-x)$ what do you get?

if you write $G(x)=1-x$, this is the same as saying $(G\circ F)^{-1}=F^{-1}\circ G^{-1}$, which is a general identity (note $G^{-1}=G$ here)

@BalarkaSen May I ask who that was written by?

grothendieck

Thanks, @MikeMiller. :-)

@DanielFischer: In my case, the map has Fredholm derivative, so it's automatically complemented. I wonder if the correct statement is just that the "kernel of the derivative can be given a continuously chosen complement" implies that there's a smooth section.

Harro.

Horra.

Holy.

Daniel, hello.

Is it possible to parametrize the ellipsoid x^2 + y^2/4 + (z+1)^2 = 1 in terms of one parameter t?

I have a question for you.

It can be shown every $\sigma$-finite measure is equivalent to a finite measure. Is the converse true?

@PedroTamaroff What does "equivalent" mean here?

Same sets of measure zero.

And the converse would be "If $\mu$ is equivalent to a finite measure then $\mu$ is $\sigma$-finite"?

Yes.

Hmmmmmmmmmmmmmmmmmmm.

Hehehehe.

I was told it was true by someone that was convinced it wasn't.

I guess the trivial counterexample doesn't count?

Oh?

Wait.

$X = \{x\}$, $\mu(\{x\}) = \infty$, $\nu(\{x\}) = 21$.

Oh, no nontrivial sets of finite measure. OK.

PLOP.

Is that onomatopoetic or an acronym?

It's from a comic, a chilean comic.

By the way, @anon @MikeMiller @DanielFischer check this out.

I'm trying to prove that for every $\varepsilon>0$ there exists $N>0$ such that $\left|\arctan(x)-\dfrac{\pi}{2}\right|<\varepsilon$ whenever $x>N$. I thought we could let $N=\tan\left(\dfrac{\pi}{2}-\varepsilon\right)$ but it's not defined for every $\varepsilon>0$. What $N$ should I use?

split into two cases: $\epsilon$ too big, and $\epsilon$ not too big

$$\Huge\varepsilon$$ and $\scriptscriptstyle\varepsilon$

When is $\varepsilon$ too big in this case?

$\pi/2$

Hmm. Okay, let me think...

If $\varepsilon\leq \pi/2$ then we use $N=\tan(\pi/2-\varepsilon)$ ?

@anon @DanielFischer Is what I said correct?

Yup.

Well, if you insist on $N > 0$, then take $\varepsilon < \pi/2$, or add something to $\tan(\pi/2-\varepsilon)$.

@DanielFischer Oh, good. And if $\varepsilon\geq\pi/2$ then we use $N=1$ ?

For example.

@DanielFischer Great :) Thanks to both of you

Not sure if it's my browser that's crazy or me... does $f_-'$ render a lot uglier than $f_+'$ ? I mean the position of subscript.

Yes.

It's horrible.

Weird.

Okay, I'll open an issue on GitHub. This is something new; I noticed it in my old post which (I think) rendered okay at the time of writing.

Thanks for confirming.

... turns out it was already reported and will be fixed in a future release: github.com/mathjax/MathJax/issues/1187

Hey guys

I was wondering as grad students are you allowed to take courses from other department

like 1 or 2 ?

Yes.

nice

@KarimMansour Depends on the educational system and the school policies. At my university: sure, why not, just pay $3000 per course.

I.e., the tuition waiver covers some courses but not others.

I see yeah makes sense

@NormalHuman Yikes.

@NormalHuman: You're talkative today.

Hey, can I get a little help in Algebra?

Hi @Pedro, @MikeM ....

Morning.

Hello.

What help, @Sahar?

@TedShifrin What's up?

75% moved in.

200% in need of chiropractor for my back ...

Is NormalHuman the seventeenth alias?

There is no evidence whatsoever that chiropractic does anything for your health that a masseuse doesn't. Of course, they do plenty of harm.