Mathematics

2:01 PM

well technically laws can be derived. But in doing so you are basically revisiting the whole cultural history that lead to the law (and why it is in that form), which I believe is a book longer than any existing proofs in mathematics

List of long mathematical proofs

This is a list of unusually long mathematical proofs. As of 2011, the longest mathematical proof, measured by number of published journal pages, is the classification of finite simple groups with well over 10000 pages. There are several proofs that would be far longer than this if the details of the computer calculations they depend on were published in full. == Long proofs == The length of unusually long proofs has increased with time. As a rough rule of thumb, 100 pages in 1900, or 200 pages in 1950, or 500 pages in 2000 is unusually long for a proof. 1799 The Abel–Ruffini theorem was nearly...

A typical history of an average nation goes over 10000 pages, right...?

Akiva Weinberger

> 1799 The Abel–Ruffini theorem was nearly proved by Paolo Ruffini, but his proof, spanning 500 pages, was mostly ignored and later, in 1824, Niels Henrik Abel published a proof that required just six pages

Um.

So what they're saying is that there's still hope for the FLT

Tobias Kildetoft

@AkivaWeinberger Depends on which part one considers "the proof"

Alessandro Codenotti

@BalarkaSen applied topology?

Balarka Sen

heh

so how's your numerical analysis going

Alessandro Codenotti

My exam was moved to Tuesday because there are too many people taking it

But I'm quite ready

Balarka Sen

2:22 PM

neat

DHMO

@Secret depends on how long that nation has been around

@Secret speaking of which, can you prove that exp(r ln x) is the extension of x^q?

Alessandro Codenotti

I would have preferred doing it tomorrow actually since I'm ready but a few more days won't hurt. I also aced the topology exam from Monday

Balarka Sen

Cool

Mary Star

How can we show that one vector space is the complement of an other vector space?

Secret

@DHMO I think I need hints for this one, as I felt like my current thoguth process is massively overthinking on the level of domains and codomains

Alessandro Codenotti

2:34 PM

Are you talking about 2 subspaces of a bigger space?@MaryStar

Mary Star

Let $V$ be the real vector space $\mathbb{R}[X]$ and $M \subset \mathbb{R}$ a set with $d$ elements. Let $$U_1 := \{ f \in \mathbb{R}[X] | \forall m \in M : f(m) = 0\}, \ \ U_2 := \{ f \in \mathbb{R}[X] \mid \deg(f) \leq d − 1\}$$ be two vector spaces of $V$. I want to show that $U_2$ is a complement of $U_1$. @AlessandroCodenotti

DHMO

@Secret well what are we trying to prove?

Mary Star

Do we have to show that the intersection is zero? @AlessandroCodenotti

Balarka Sen

@Alessandro Do you want to talk about differential forms now then?

I am more or less free for the next hour

Alessandro Codenotti

I don't know anything about infinite dimensional vector spaces @Marystar but I'd guess you also need their direct sum to be the whole of $\Bbb R[x]$

Sure @Balarka, thanks! So we defined what an alternating multilinear function is last time

Secret

2:41 PM

@DHMO that $x^q$ is the restriction of $exp(r \ln x)$?

DHMO

@Secret well... do we need to prove continuity as well?

Balarka Sen

@Alessandro Yeah, so it's just a multilinear map $T : V \times V \times \cdots \times V \to \Bbb R$ such that $T(\cdots, a_i, \cdots, a_j, \cdots) = -T(\cdots, a_j, \cdots, a_i, \cdots)$.

Examples include the determinant, as we mentioned - eating column vectors and spitting the det of the matrix.

Alessandro Codenotti

Every alternating multilinear function is $0$ if supplied the same argument twice, just like the determinant

Balarka Sen

Yeah.

Secret

@DHMO we do need to if $q$ is rational, and I am not good at continuity stuffs

DHMO

2:45 PM

alright, let's prove the weaker statement then

Simple Art

Does anyone know if this method for integrating has a name?

Mary Star

@AlessandroCodenotti So do we have to find a basis for each for them?

DHMO

@Secret Shall we start with the naturals?

Simple Art

$$\int te^t\ dt=f'(1),\quad f(x)=\int e^{tx}dt$$

DHMO

Hmm...

Balarka Sen

2:48 PM

@Alessandro So now I want to construct a vector space $\Lambda^k(V)^*$ of alternating multilinear maps $V^k \to \Bbb R$ (here $V$ is a vector space of dimension $n > k$); that it's a vector space is easy to check.

Because adding two alternating multilinear maps is alternating multilinear, multiplying by a scalar is obviously so.

Tobias Kildetoft

@MaryStar You just need to show that the spaces intersect trivially and that any polynomial can be written as a sum of an element from each

Balarka Sen

*$n \geq k$. Do you see why it's not interesting if $n < k$?

Mary Star

When $f\in U_2$ it holds that $\deg (f)\leq d-1$. Since $M$ is a set of $d$ elements, it can be that $f\in U_1$ only if $f=0$, since it must have $m$ roots. From that we get that the intersection is trivial, right?
@TobiasKildetoft

Tobias Kildetoft

@MaryStar right

Mary Star

But how can we show that any polynomial can be written as a sum of an element from each?
Suppose that $f\in U_1$ and $g\in U_2$.
If $\deg h\leq d-1$ then $h=0+g$, or not? @TobiasKildetoft

Tobias Kildetoft

2:54 PM

sure

Alessandro Codenotti

@BalarkaSen in that case the only multilinear map is the 0 one I think?

Because I can write one of its arguments as a linear combination of the others

Mary Star

@TobiasKildetoft And if $\deg h>d-1$, what happens then?

Balarka Sen

@Alessandro Bingo. Right.

Tobias Kildetoft

@MaryStar You want to subtract something of degree at most $d-1$ such that you end up with something that vanishes on all the given elements

Balarka Sen

So, ok, suppose $v_1, \cdots, v_n$ is a basis of $V$. Let's try to write down a basis for $\Lambda^k(V)^*$ too.

Later we'll do the wedge product construction, which is possible to do without fiddling with basis, but I find the basis-wise construction cleaner.

Alessandro Codenotti

3:00 PM

Ok, let me think about it

Balarka Sen

Try for $k = n$ first.

Mary Star

So, if we have a polynomial of degree >d-1 then we write as f-g, so that the terms that don't exist vanished? @TobiasKildetoft

Or did I misunderstand it? @TobiasKildetoft

Secret

@DHMO Ok I am somewhat stuck. If both $x \in \mathbb{N}$ then there is no difference between the two functions as both their domain and images have the same set of points

DHMO

@Secret both?

Tobias Kildetoft

@MaryStar I am not sure what you mean

DHMO

3:12 PM

@Secret I don't understand. Domain and image having the same set of points doesn't prove anything...

Secret

No I mean the images of exp(r ln x) and x^q will be the same if $x \in \mathbb{N}$

Balarka Sen

@Alessandro Here's the thing. If $T : V^n \to \Bbb R$ is an alternating $n$-multilinear form then for random vectors $w_1, \cdots, w_n$, $T(w_1, \cdots, w_n)$ is stuff times $T(v_1, \cdots, v_n)$, by expressing $w_i$'s in terms of the basis $\{v_1, \cdots, v_n\}$. What's that stuff?

DHMO

@Secret ah, I meant when r in N.

I'm not sure how you'd prove it if x in N instead.

Mary Star

@TobiasKildetoft I haven't understood why a polynomial of degree >d-1 can be written as a sum of an element of $U_1$ und one of $U_2$. Could you explain it to me?

Tobias Kildetoft

@MaryStar Try to do the case where $d=1$ first

Balarka Sen

3:20 PM

Also you should just set $V = \Bbb R^n$ and the basis to be the standard basis $\{e_1, \cdots, e_n\}$ (ie $e_i$ = vector with 1 at $i$-th slot and zero everywhere else)

Mary Star

So, we have a set of constant polynomials and a set of polynomials that have one root, right? Suppose we have a polynomial of degree n. Then we have to write it as an polynomial of the set $U_1$, since the other one are constants, or not? @TobiasKildetoft

Tobias Kildetoft

@MaryStar no

Mary Star

But how can we write it as a sum? I haven't understood it yet. @TobiasKildetoft

Alessandro Codenotti

So I have $T(w_1,\cdots,w_n)=T(\sum\alpha_iv_i,\cdots,\sum\gamma_iv_i)=\sum...\sum\alpha_i‌...\gamma_jT(v_i,...,v_j)$ if I opened it correctly (that sure is uncomfortable to type on mobile)

Balarka Sen

@Alessandro There are a bunch of signs involved though. What's $T(a_1 e_1 + a_2 e_2, b_1 e_1 + b_2 e_2)$?

Note that $T(e_1, e_2) = -T(e_2, e_1)$, not $T(e_2, e_1)$.

Alessandro Codenotti

3:31 PM

Ah, right, what I wrotr above was for a multilinear map

@BalarkaSen $(a_1b_2-a_2b_1)T(e_1,e_2)$

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$Mathematics$ Mathematics