I am not familiar with the book, but I imagine, since it is a functional analysis book, it would actually prove that. And yes it is right, even for infinite dimensional hilbert spaces

@PaulPlummer Thanks you. I was asking because later it says that every Banach space which is not isomorphic to a Hilbert space, has closed subspace that has no complement. And since every finite-dimensional Banach space has complement for every closed subspace, I was on doubt, so thanks!

Hello !

@user17629 yes that's right

in the finite dimensional case the result is trivial

or, better say, a result of linear algebra

closedness is automatic!

closedness is the hard part in the infinite dimensional case

Yup, in the book (until I have read it), only mentioined it, so I'm looking for some examples

um does anyone have information on solving for the integer solutions of a hyperbola

thanks!

I stumbled upon the fanciest function : $$f(x)=\frac{\sqrt{x}-\sqrt{k}}{x-\lfloor x \rfloor} \; ,\text{for}k \in \Bbb N$$

np

@user17629 Finite dimensional stuff is very nice. I think if every closed subspace of a banach space has a "complement" is actually a characterization for when the space is hilbert @user17629

It has no limit at any positive integer $k$

Is it ridiculous to consider forms like this ? $$f_2(x)=e^{\left(\operatorname{floor}\left(\tan \left(\frac{1}{\ln \left(\frac{1}{x}\right)}\right)\right)\right)}$$ @TedShifrin ?

I mean, could this come up as a way to solve a concrete problem ? If so, is it possible to study this ?

o.O

ugh @Mahmoud ... who cares!

@TedShifrin The fundamental nature of Analysis might.

I bet not.

@TedShifrin Today we were introduced to the notion of the almighty derivative.

Um, OK.

Hi @Mahmoud @Ted

Hi @Astyx

But it had something strange in the definition we saw @TedShifrin

Comment va ?

Hi @Ted

Our first definition dealt with it as a value, not a function.

Ça va, Astyx, et toi?

rehi @Zach

J'ai appris une subtilité du français que j'ignorais hier

@Mahmoud. You know a function if you know its values at every point of its domain, @Mahmoud. No big deal.

currently working on one of the problems... i think im gonna have to bring out a script

Eh bien, @Astyx?

ie que l'expression "au finale" était correcte à partir du moment où l'on mettait un e à la fin

@TedShifrin But we'll have to take the limit each time, rather than getting the general function.

@ZachHauk Not really sure, have you seen Pells equation? There is a nice book, by Silverman and Tate called rational points on elliptic curves, and he discusses how to find rational points on quadratic curves in the first chapter. If you have a single rational point you can find many/all by considering lines through that one point, and an auxillary rational line and study when the intersection s rational.

It is defined as a limit, yes, @Mahmoud. What's your point?

yeah i was just researching that

but i didn't find anything useful

Do you need to find all points on a specific hyperbola?

I suppose I can't talk much more about this, due to violation of the rules, except for just looking at resources

Alors que je croyais (en partie à juste titre) que "au final" était une erreur de français et qu'il était mieux d'utiliser "en fin de compte", "finalement", etc

I was gonna say, @Zach ...

Bref histoire passionnante

Oh nevermind

@TedShifrin if i do find a resource however, I can cite it

Hi @Ted

Heya @Alessandro

@TedShifrin Do $\frac{df}{dx}(x)=f'(x)$, for all $x$, rather than $\frac{df}{dx}(x_0)=f'(x_0)$, for each $x_0$ of the usually infinite domain.

@Astyx: Je ne l'ai jamais entendu, autant que je sache ... :)

But it's still defined as a limit, @Mahmoud. I don't get your point.

The first gives us a function, the second gives a special value of that function.

"Au final" ? Justement c'est une erreur de français qui se répand de plus en plus (une substantivation abusive de l'adjectif)

There is absolutely no difference. You define a function by its values at points in its domain.

@Astyx: Au bout?

@Mahmoud: Whether you call a point $x$ or $x_0$ makes no difference.

The domain of $f'$ is the set of all $x$ for which the limit defining $f'(x)$ exists.

@TedShifrin $x$ as a variable, $x_0$ as a uniform constant.

Nah.

C'est synonyme à "Au bout du compte", "En fin de compte", "Pour finir", "Finalement", "À la fin", "En définitive" et j'en passe

Mais la forme correcte "Au finale" fait référence au finale musical

Ah, OK.

I also learned that I lived in the same town as three Fields medalists

town? city?

What's the nuance ?

Paris is one of the big cities in the world. A town is small.

town then

Oh, you're not actually from Paris?

Not exactly no

Ah.

In a city of around 20k people just south of Paris

Next to L'École Polytechnique

Same with École Polytechnique in Palaiseau.

Well if Bures is Paris, I am in Paris :p

(Even though I'm not in Bures)

Hello world!

Hi @Kasmir

I think most of the math world thinks of IHES as being in Paris, as opposed to a 20-minute train ride from Paris.

Hi @Kasmir

:)

Les Paul lived in my town for quite a while

Who dat?

not that I was alive then, or anything...

not anyone to do with math

Imagine that ... a world pre-Zach. :D

a better world /s

A world pre-Zack should not have existed in first place

well, that could be due to other people

A world plunged in darkness and ignorance

The center of universe is Zack hauk

I'm afraid the darkness and ignorance is now

Who is zack? >:)

I only know a Zach, and he's me.

There's still light :)

Zack is the beginning and the end

Ok enough that ><

@KasmirKhaan that's somewhat true, regarding my existence anyways

If you belive that you belive anything @ZachHauk

Ted!

Anyways, umm @Ted. a couple people said i should ask you for a recommendation for Mathcamp. Would you be willing to do that for me?

Sure, Zach, although I obviously don't know your math as well as I know Akiva's or Balarka's. Maybe you should do a few more projective geometry problems when all the dust clears :)

@TedShifrin if I can represent a function as a power serie does that mean its analytic or what is the relation ?

Alright, thanks. I appreciate it :]

and I really dont get how we just copy the structure of power series from real case to complex case without anywork whatsoever @TedShifrin

That's the definition, yes, @Kasmir..

It really belonged in the complex setting in the first place, @Kasmir. The fact that the series for $1/(1+x^2)$ blows up in the reals makes no sense.

@Mystic If I'm not mistaken you can make this assumption without loss of generality

Hmm I feel like am being rubbed on many informations on this course :D

Anyway all I can do is trying to get some intuitive feeling now

@Mystic: Christian is very smart. I don't know why he didn't put one of the points at $1$ to simplify things. But you can certainly rotate to put one of the points anywhere you want. Let me ponder a bit more.

To understand complex analysis, you really need to know uniform convergence decently well, @Kasmir.

Because one of the angles made by two points of a triangle and the center of the associated circumscribed circle is necessarily smaller than ${2\pi\over3}$

OR equal to.

But the angle between $e^{i\alpha}$ and $-e^{-i\alpha}$ will be at least $2\pi/3$.

Meh. French terminology

@TedShifrin you think its better if I reread that topic ? what I know from uniform convergence is that we can interchange the order of limits and integration, and the defintion is that SUP for x E intervall , f(x) behaves like its limit function f_k

that was not very clear when i read it but

All that Weierstrass M-test stuff we discussed works just fine with complex functions, @Kasmir.

Okay thanks =p

The proof that you keep asking me about (how to prove holomorphic implies analytic) relies on changing integral with infinite sum.

Oh, Christian had two negatives. I missed one of 'em.

well our teacher told us that complex analysis is a big course and we only gonna focus on certain topics such as conformal mapping

so if for complex functions $f: \Bbb R^2 \to \Bbb R^2$ holomorphic implies analytic, then what about functions from $\Bbb R^3 \to \Bbb R^3$?

last 4 lectures are on that , and a small intro to multivarble complex analysis

That fixes it. So then the angle is at most $2\pi/3$. Good.

Holomorphic only makes sense with $\Bbb C$ in there, @Zach. You can do multivariable complex analysis, but you need $\Bbb C^n$.

well

a notion of differentiability in $\Bbb R^3$?

Yes, triangle inequality, @Mystic.

Differentiable (even infinitely differentiable) does not give you real analytic (convergent power series), @Zach.

is that special to $\Bbb C^n$?

It's special to complex differentiability, yes.

@Mystic. It isn't, but if you specifically show the maximum is attained, you are OK. :)

@Mystic No that's why he shows an example when it's attained

(last sentence)

math.stackexchange.com/questions/2146901/… could this problem use inversion ?

i.e. with respect to the origin

@Zach What is your idea ?

invert around a circle whose center is the origin

then it would become a bunch of lines

Have you read the answers ? :)

oh someone already said that lol

whatever

wasn't going to post an answer anyways

@Mystic You can always turn the circle so that the x-axis is bissects the angle

It is !

Glad to help

The answers proposed are quite beautiful by the way @Zach

At least to my uneducated eyes :p

I'm gone now, have a nice day everyone !

I find that solving a problem after a long bit of work is very rewarding

So, I'm attempting to read an article on the use of TDA on remote sensing, and I came across the following: line: "Descriptors are topological invariants that are weakly dependent on the metric distortion." I'm pretty sure the metric distortion they are referring to is the geometric distortion inherent in remote sensing. However, I'm not sure what they mean by weakly dependent in this context....any clues?