Mathematics

4:06 PM

In psychology research literature, the term child prodigy is defined as person under the age of ten who produces meaningful output in some domain to the level of an adult expert performer. Child prodigies are rare, and in some domains, there are no child prodigies at all. Prodigiousness in childhood does not always predict adult eminence. The persons listed here have come to the haphazard attention of history or current news and probably do not represent the typical experience of a child prodigy. == Mathematics and science == === Mathematics === ==== Born 1600–1699 ==== Juan Caramuel ...

Interesting read.

Norbert Wiener (1894–1964) began graduate studies at age 14 at Harvard and was awarded PhD at 18 for a dissertation on mathematical logic

@JasperLoy this means to be a genius ^^^

Carl Friedrich Gauss (1777–1855) made his first ground-breaking mathematical discoveries while still a teenager. Also at the age of 3 watched his father add up his accounts and corrected him

N3buchadnezzar

Heya

user 1618033

William James Sidis (1898–1944) set a record in 1909 by becoming the youngest person to enroll at Harvard College, at the age of 11 years. He entered Harvard at age eleven and, as an adult, was claimed to be conversant in over forty languages and dialects.

Charles Fefferman (born April 18, 1949) Entered college at age eleven, later becoming the youngest full professor in the United States. He has won many major awards in mathematics, including the Fields medal.

He received his B.S. from University of Maryland in mathematics, which is part of the University of Maryland College of Computer, Mathematical, and Natural Sciences at age 17. Then received his PhD in mathematics at 20 from Princeton University under Elias Stein, Fefferman achieved a full professorship at the University of Chicago at the age of 22. This made him the youngest full professor ever appointed in the United States.

I think it's enugh to have a picture of how a genius should look like. @JasperLoy

Balarka Sen

Hi @Krijn

Krijn

4:21 PM

Hey @Balarka

I saw your mention but I thought you must have mistyped someone else

Balarka Sen

No, I meant to ping you. Weren't you an arithmetic geometer?

Krijn

@user1618033 Sidis also holds the record for highest IQ, although that is a controversial subject of course. He ended up as a shoe maker or something, didn't he?

I took a course on Algebraic Geometry, that hardly qualifies, does it?

Balarka Sen

I think it does, but IDK. Soham's asking about the analog of branched covers for maps between affine schemes.

0celo7

@BalarkaSen Much easier than the covering argument.

user 1618033

@Krijn It seems a very controversial figure, some thinking that around him there were made many exaggerations. I also saw a documentary in the past that mentioned him, it was about the smartest people that ever lived on earth.

Balarka Sen

4:24 PM

But you looked it up. Does not qualify as being easier.

0celo7

@BalarkaSen I still do not understand the post in that thread

This is a much easier variant

Krijn

Anyway, I'm gonna cook. I'll be back here tonight I think

Balarka Sen

You filled in the details, but that hardly qualifies as original. I prefer the covering argument in any case.

0celo7

According to the discussion on that thread, you need a graph theory argument to construct the chain of Lipschitz neighborhoods

Balarka Sen

Yes.

0celo7

4:26 PM

that's hardly trivial

Balarka Sen

I didn't say it was. Just that geometrically very apparent.

I have used this kind of argument before in covering space theory, say.

0celo7

apparent to you, I still don't get it. Something with shrinking balls, but that's hardly nice

Balarka Sen

You cover your compact set by balls over which $f$ becomes Lipschitz. That can be done.

0celo7

you have to pick all of them to be the same size

Balarka Sen

Sure, why not.

0celo7

4:28 PM

that's nontrivial too

Balarka Sen

No.

0celo7

how does the argument go?

Balarka Sen

I am not going to do it for you.

0celo7

ok, I guess I'll be left to wonder

Balarka Sen

If you want, sure.

0celo7

4:30 PM

I like this argument far more, I'm not going to save the geometric proof

Balarka Sen

That's fine.

@0celo7 Ok, this is the Lebesgue number lemma.

0celo7

@BalarkaSen Ah, right.

I don't have that at my disposal.

Balarka Sen

The Lipschitz open sets cannot be arbitrary small inside your compact set, is the point.

@0celo7 Fair enough. This argument is much less ad-hoc however: it's worth working it out. But if you don't want to, sure.

0celo7

@BalarkaSen By "me" I mean "it hasn't been talked about in class"

Balarka Sen

Yes, I get you.

Not particularly an analysis fact, Lebesgue.

N3buchadnezzar

4:34 PM

anyone here with some knowledge of analysis? :p

Balarka Sen

@N3buchadnezzar I don't know much analysis, but you should post it.

N3buchadnezzar

@BalarkaSen Trying to understand the abstract of a paper

Just for a start, is the torus $\mathbb{T}$ defined as $\mathbb{C}/(\mathbb{Z} + \tau \mathbb{Z})$?

Balarka Sen

I think $\Bbb T$ means the one dimensional torus in the first snippet, aka, the circle.

N3buchadnezzar

Right, that makes sense

And in the first snippet, what do they mean by the orthogonal projection

Balarka Sen

If you have a closed subspace of a Hilbert space, you can always orthogonally project to that subspace with respect to the inner product on the space you have.

It just means "perpendicular projection" in the standard sense. If $W$ is a Hilbert space, $V$ a closed subspace, there is a canonical decomposition $W \cong V \oplus V^\perp$, just as in the finite dimensional case. Then you project to the first factor.

N3buchadnezzar

4:45 PM

Right, since it is closed and bounded.

Balarka Sen

Um, what is bounded?

Linear subspaces aren't bounded.

N3buchadnezzar

Well we are working on the circle?

Balarka Sen

We are working on the space of $L^2$-functions on the circle.

Transcript for

$Mathematics$ Mathematics