Mathematics - 2017-08-02 (page 4 of 4)

SteamyRoot

20:00

Spoiler alert: $df$ is the Jacobian :^)

skullpatrol

:D

Astyx

Ssshhh

Kirill

For $\Omega \in \mathbb{R}^n, f: \Omega \to \mathbb{R}^m, x \in \Omega$. We say $f$ is differentiable in $x$, if there exists a linear map $T: \mathbb{R}^n \to \mathbb{R}^m$ with $$\lim_{h \to 0, h \ne 0} \frac{\Vert f(x+h) - f(x) - T(h)\Vert}{\Vert h\Vert} = 0.$$

"Quotient of norms"

$T$ is called the total derivative of $f$ in $x$.

So, $T=df_a$?

$T \sim df_a$?

Astyx

That's completely equivalent since the norm is continuous

$T = df_a$

Kirill

...$=\Delta x \frac{\mathrm{d}f}{\mathrm{d}\text{something}}$?

so, in other words, $f$ is differentiable if you can approximate it with some lines?

Astyx

20:09

What ?

Kirill

if differential is linear part of $\Delta f(x)$, then it is a gradient times $\Delta x$, isnt it?

or perhaps a Jacobian matrix times $\Delta x$.

Astyx

Ugh, perhaps, but that's kinda doing things backwards

Kirill

@Astyx ok, perfect. So I rarely understand these definitions, especially the one with the differential, as I have solved 0 exercises using it.

Why these two are similar? They look pretty different.

Pedro Tamaroff

@anon

anon

yo

Pedro Tamaroff

20:15

I was looking at this post of yours.

I think you can also interpret you claim combinatorially. If $\lambda$ is a partition, write $e_\lambda = e_{\lambda_1}e_{\lambda_2}\cdots$ where $e_r$ is the elementary symmetric polynomial on the variables $x_i$ (be there infinitely many or not).

It is known that if $\lambda'$ denotes the conjugate permutation to $\lambda$, then
$$ e_{\lambda'} = m_\lambda + \sum_{\mu<\lambda} a_{\lambda\mu}m_\mu$$
where $m_{\lambda}$ is the symmetric polynomial obtained as the sum of permutations of $x^\lambda$. Because the $m_\lambda$ are obviously a basis for the symmetric polynomials, the formula above shows the same is true for the $e_\lambda$.

Astyx

Which definition are you talking about ? @Kirill

Pedro Tamaroff

This you can read in MacDonald's book on Symmetric Functions.

On the other hand, if $\sigma$ is a permutation written in one line notation, let $a_i$ be the number of entries $j$ to the left of $i$ such that $j>i$. The tuple $(a_1,a_2,\ldots)$ built in this way, which satisfies $a_i\leqslant n-i$ for each $i$, is called the inversion table of $\sigma$ and it is known this assignment is a bijection between permutations and inversion tables.

What you are looking then is at monomials indexed by inversion tables.

I would guess someone who knows more about this than me (anyone, basically) can join the dots and give a nice combinatorial explanation of why that indexed basis by inversion tables (=permutations) is a basis of Z[polinomials] over Z[symmetric polinomials]

anon

interesting

I hadn't heard of inversion tables

I'll have to think about that some time

skullpatrol

Welcome back Pedro.

Pedro Tamaroff

I will think about it, @anon, but I confess I am busy.

@skullpatrol But I never left. ;)

skullpatrol

20:20

:)

Kirill

@Astyx lets take this one again.

why is $r$ defined this way?

you said you understand this

Astyx

What do you mean by why ?

Do you agree with the proof ?

Kirill

@Astyx let me hyperbolize. Why $r(h)$ is not $\sqrt{2}$?

Astyx

Because then the inequalities wouldn't be true

Kirill

@Astyx no, I am not. The proof is a manipulation for me at the moment. I understand only the triangle unequailty there.

Astyx

20:24

Don't you agree with the first equality ?

Kirill

if you define $r(h)$ as such, then the equality is true, so the triangle unequality is also true.

I still do not see, why we need $r(h)$, what is $r(h)$ among that that that is a "remainder" of some approximation. I thought we want to show that $f$ is continuous and not approximate something.

Astyx

Once you have the triangle inequality, do you understand why there exists $C$ such that $\Vert T(h)\Vert \le C\Vert h\Vert$ ?

Kirill

I have 4 versions: 1) somehow $T$ is bounded as linear functions seem to be continuous. 2) T(h)=O(h). 3) Something about the Lipschitz-continuity, as $T$ seems to be Lipschitz continous. 4) something else?

Astyx

$T$ is Lipschitz indeed

Kirill

That implies that $C$ is a constant that is independent from $h$, so from $x$, right?

Astyx

20:32

Only $h$ here

But that's sufficient

Pedro Tamaroff

@Kirill What is your problem?

Astyx

Since you need to prove continuity for a fixed $x$

Kirill

@PedroTamaroff mathematics

Pedro Tamaroff

Yes, that I can figure out.

Kirill

@PedroTamaroff or, more precisly, some steps in the proof above.

Astyx

20:33

Mathematics is not the problem, it's the solution

3

Pedro Tamaroff

I cannot read German, though. What do you want to show?

Astyx

That a differentiable function is continuous

Unless I'm mistaken

Pedro Tamaroff

OK. I can help you with that.

Kirill

@PedroTamaroff or, more precisly, what is $r(h)$ among that tha that is a remainder of some approximation that I do not see. Why do we need it? Why we define $r(h)$ that way besides that we do tricks in order to get the right unequality?

Pedro Tamaroff

So $f$ is a function from Euclidean spaces, and let us assume it is differentiable at $x_0$.

Kirill

20:36

I mean there should be a logic of what we doing. I do not see any in the proof.

@PedroTamaroff agree.

Pedro Tamaroff

This means there is $T =Df(x_0)$ that satisfies what you wrote above, namely that the difference $r(h) = f(x_0+h) - f(x_0) - T(h)$ satisfies that $\|r(h)\|/\|h\|$ goes to zero as $h\to 0$ right?

anon

$f(x)$ is a $0$th order approximation of $f(x+h)$ (it is constant with respect to $h$)
$f(x)+Th$ is a $1$st order approximation of $f(x+h)$ (where the matrix $T$ depends on $x$)

$r$ is the difference between this approximation and $f(x+h)$

Astyx

You are approximating as best as possible $f$ with an affine map $f(a) + T$

Kirill

@PedroTamaroff yesno, there exists a linear map. Namely that the difference... I do not understand.

Pedro Tamaroff

Well, the definition of differentiability at a point is...?

Astyx

20:40

$r$ is the remainder of that approximation, ie the difference between the actual function and the approximation

Pedro Tamaroff

...that $\|f(x_0+h) - f(x_0) - T(h)\| /\| h\|\to 0$ as $h\to 0$.

Kirill

@PedroTamaroff "there exists a linear map, so that the limit of the quotient of some terms is 0". I do not understand the numerator.

Pedro Tamaroff

Consider the 1-dimensional case.

A linear transformation is multiplication by a scalar $T(h) = \lambda h$.

Astyx

$r(h) = o(\Vert h \Vert)$ means that $r$ is small enough around $0$

ie that the approximation is good enough around $a$

Kirill

@anon $r$ is a function that shows the difference?

Pedro Tamaroff

20:42

Then quotient is $\frac{|f(x+h) - f(x) -\lambda h|}{|h|} = \left| \dfrac{f(x+h)-f(x)}h - \lambda\right|$

Astyx

$r$ is the difference : $r(h) = f(x+h) - (f(x) + T(h))$

anon

"Hey, I wonder how closely this affine map $f(x)+Th$ approximates $f(x+h)$? Well, let's call the difference $r$ and see..."

Pedro Tamaroff

@Kirill Now convince yourself $\lambda = f'(x)$ in this case.

Astyx

Right, I'll leave you to it then :)

Pedro Tamaroff

The multidimensional case generalizes this: you want a linear map that makes $f(x+h)-f(x)-T(h)$ go to zero faster than $|h|$ as $h\to 0$.

Kirill

20:44

@PedroTamaroff why do I want to? :)

Pedro Tamaroff

"you want to define a derivative as"

Danu

o/

skullpatrol

\o

Astyx

o/

Kirill

@PedroTamaroff, why are you sure in the first case, that the $T$ you look for, is $T(h)=\lambda h$ and not $\lambda h + $some constant? Why did you mention a linear transformation?

Semiclassical

20:52

oh man, RobertIsrael's comment here: math.stackexchange.com/q/2380498/137524

Astyx

Lol

Akiva Weinberger

\_o_

Semiclassical

Also, I just tried a Mathematic calculation on a campus desktop, one which gave my laptop a heart attack

And it seems to give the campus computer a nervous breakdown as well :)

(It's trying to find, by brute force, the longest simple path on a 30-node graph with lots of cycles )

Kirill

@anon why do you multiply $T$ with $h$ in the first order of approximation? Are you multiplying $T(x)$ with $h$, or you set $T$ as a matrix corresponding to the linear map $T$?

Semiclassical

(Brute force meaning: for each of the 30 choose 2 = 435 pairs of vertices on the graph, compute all simple paths between them. That's a lot of paths :P )

Kirill

21:11

so, have I got it right? $r$ is a function that shows me the difference between the approximated value and the given value of the function. And, somehow, the approximation is much better, if I am closer to $x$.

So, using this fact, I compute the norm of $f(x+h)-f(x)$ and find an upper bound for it. And this one goes to zero as $h\to 0$. That means the last equation in the proof, which is equivalent to $\lim_{x \to a}=f(a)$ saying me that $f$ is continuous. Right?

@Astyx ?

at least I hope so, and thank you for help

Astyx

Sorry, I was away

Yes, right @Kirill

Kirill

@Astyx analysis seems to be so simple for me, but ONLY when I understood it

before that moment that is a pain

Astyx

It's the same for all of us I think :p

Kirill

so, I got $1/2$ of a one proof among 70 other proofs that are improtant. Do not know, if I shoulb be happy or sad about that...

:)

skullpatrol

21:26

...be happy because these things build on each other.

Astyx

It's only the first ones that are hard

Be happy because it's best to be happy too

Kirill

@skullpatrol I also think so. If you build a house, you are in dirt the first year building the foundation.

@Astyx btw. "math is not a problem, but a solution" was a really nice one!

Astyx

22:21

Hi @Daminark

Daminark

22:58

Hey @Astyx!

Simply Beautiful Art

23:10

@Kirill Is the cup half empty or half full? Be optimistic about it :-)

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$Mathematics$ Mathematics