Mathematics

12:20 AM

Can anyone point me to the paper that Andrew Granville generalized Lucas's Theorem to prime powers and if it's paywalled send me the pdf in a pm?

Secret

12:33 AM

[Random]

dx is an infintesimal

$dx^{j}$ is a tangent vector

$dx^{j}$ can capture an infinitesimal change in a function value, thus bypasses the non uniqueness problem of infintesimals as traditionally outlined by Lebniz

Now applying explosive generalisation, we can consider the following mathematical object:

$M(dx)$ is a mathematical object with the property being infinitesimal or capture an infinitesimal property, meaning:

$M(dx) = \{x \in M | \phi(x)\}$

where $\phi$ reads "is related to or has the property of being infintesimal"

Now recall the definition of an infinitesimal as an element $\epsilon$ in a nonarchimedian ring $R$ such that for any $x \in R$, $x\epsilon < 1$

hmm... need to figure out how to formalise $M(dx)$... perhaps the literature already have some examples

ÍgjøgnumMeg

10:05 AM

@Adam web.archive.org/web/20170202003812/http://www.dms.umontreal.ca/… google is your friend

Manolis Lyviakis

11:28 AM

hey

im studying dfferential geometry

and we construct and study the tangent space of a surface

but i dont see the reaso why we do that

Tangent space

In mathematics, the tangent space of a manifold facilitates the generalization of vectors from affine spaces to general manifolds, since in the latter case one cannot simply subtract two points to obtain a vector that gives the displacement of the one point from the other. == Informal description == In differential geometry, one can attach to every point x {\displaystyle x} of a differentiable manifold a tangent space—a real vector space that intuitively contains the possible directions in which one can tangentially pass through ...

how does that help us study the manifold itself?

the generalization of vectors from affine spaces to general manifolds what does that mean?

"the tangent space of a manifold facilitates the generalization of vectors from affine spaces to general manifolds"

Daminark

So, a priori a manifold need not live in $\mathbb{R}^n$. It might just be some topological space kinda floating around and minding its own business

Manolis Lyviakis

ok

Daminark

The point is, you're trying to learn about manifolds via calculus

Manolis Lyviakis

the tangent space is usefull to define

Daminark

But calculus only makes any sense when there's a vector space structure to work with, so the tangent space is there to make sense of that

Manolis Lyviakis

11:34 AM

the fundamental forms

ok

hm..

is it like studying

the derivative of the surface?

how that plane behaves from point to point it tells you how much the sphere curves? and so on.

Daminark

There's a sense in which that's the right way to think about it. For example, let's say you have some manifold which is just the level set of a smooth function. Then the tangent space of the manifold is just the kernel of the derivative

The way I tend to think about it is that if you have a smooth function between manifolds, then the derivative of that map at that point is going to be a linear map on tangent spaces

Manolis Lyviakis

yes

ohh

on tangent spaces

ohh nice didnt thought of that

Daminark

As for what you said about the tangent space telling you about curvature, that's something I know little about, I know some general facts here and there about manifolds and some topology but geometry is mostly outside of what I've been thinking about

Manolis Lyviakis

oh

Daminark

But yeah that sorta captures the idea of how this tangent space structure lets you do calculus, the derivative literally lives in these spaces. If you know anything about differential forms, say in $\mathbb{R}^n$, then they are defined out of tangent spaces, and that's the mechanism by which you want to talk about integration on manifolds

Manolis Lyviakis

11:42 AM

oh ok . it helped me thanks

so they come up naturally

from the derivatives

(the tangent spaces)

Daminark

Yup, and no problem! :)

Manolis Lyviakis

link to see that

derivative of a manifold?(should i google that?)

derivative of multyvariable function?

Daminark

I think any book on manifolds will probably include that soon enough

You could Google something to the effect of "tangent spaces and smooth functions" to see what you get?

Manolis Lyviakis

ohh so i have the total derivative

of a function

$R^n$

now for a function

between manifolds

the derivative is called the differential

$df:Tdf(p)\colon T_{p}M\to T_{f(p)}N.$

$df(p)\colon T_{p}M\to T_{f(p)}N.$

which is a generalization of the total derivative

im startng to get it

Daminark

Fantastic!

Well, anyway, I think I'm gonna have to go now

Alessandro Codenotti

11:52 AM

Right, another important thing is that, skipping details, you can give a manifold structure to the set of all tangent spaces to a manifold $X$, which is called the tangent bundle of $X$, denoted by $TX$. Now if $f:X\to Y$ is a smooth function between manifolds its differential is a smooth function $TX\to TY$!

Daminark

But good luck!

Manolis Lyviakis

bye!

let me think of it abit

i can make the set of all tangent spaces manifold-like stucture

and now $df$ is a smooth map between the bundles

ok im starting to get it

now the hard part is to do the actual math

Alessandro Codenotti

@ManolisLyviakis as always :P

Manolis Lyviakis

XD

Late347

12:20 PM

is anybody out there?

Alessandro Codenotti

12:36 PM

Well, only got an hour of daylight left. Better get started.

rschwieb

1:50 PM

This is a body out there

quallenjäger

2:25 PM

Suppose I have two infinite dimensional vector space $V$ and $W$ with hamel base $v_j$ and $w_i$. For the tensor product $V \otimes W$, is it true that I can write any element in $V \otimes W$ as finite linear combination of $v_j \otimes w_i$?

Leaky Nun

@quallenjäger yes

quallenjäger

How can I see that this is true?

Leaky Nun

Any element can be written as a finite combination of $v \otimes w$ where $v \in V$ and $w \in W$

any $v$ (resp. $w$) can be written as a finite linear combination of $v_j$ (resp. $w_i$)

small tensor distributes over addition

qed

quallenjäger

Oh yes, thanks

user193319

If $G$ is a topological group, $H$ a closed subgroup, and $G/H$ is given the topology with respect to which the canonical projection $\pi : G \to G/H$ is quotient map, is $\pi$ a closed map?

Nick

3:05 PM

Is polynomial regression typically linear or non-linear?

user76284

3:31 PM

@Nick Polynomial regression is considered to be a special case of multiple linear regression.

Is there a name for this kind of loss for predicting the value $x$ of a random variable $X$: $\frac{-\log \mathrm{P}(X = \hat{x})}{\mathrm{E}(X)}$. It's the ratio of the suprisal of this prediction under the true distribution to the entropy of the true distribution. Basically I'm comparing point-predictions across different distributions but don't want to unfairly penalize predictions for distributions which have inherently higher spread.

Adam

4:06 PM

depends if the order of the polynomial is 1 or not

Is there particular notation for declareing $q_k \in \mathbb Q$ to have infinite digits in a particular number base system and corresponding notation to indicate those elements of $\mathbb Q$ that only have a finite number of non zero terms in their digit sequence?

user76284

4:36 PM

"Although polynomial regression fits a nonlinear model to the data, as a statistical estimation problem it is linear, in the sense that the regression function E(y | x) is linear in the unknown parameters that are estimated from the data." I think this is what Nick is referring to.

@Adam en.wikipedia.org/wiki/…

For integers p and q with gcd(p, q) = 1, the fraction p/q has a finite representation in base b if and only if each prime factor of q is also a prime factor of b.

Adam

4:59 PM

@user76284 that's very impressive buddy did you come up with that one?

anon

you can also compute what the repeating segment (for any rational) is using modular arithmetic and geometric series

Semiclassical

they're quoting from the wikipedia article they linked

Adam

@anon well isn't that the definitive property that makes a number rational?

anon

a number is rational if it's a ratio of integers, having an eventually repeating digital expansion is a property that you prove (not part of the definition) by coming up with an actual algorithm

Semiclassical

not really. the defining property of a rational number $x$ is that there's a positive integer $q$ such that $xq$ is an integer. that'll be true regardless of the base you're in

Adam

5:05 PM

correct that is the definition but not always helpful in a rationality proof

anon

wat

Semiclassical

it's true that "if I can write it as a repeating base-b expansion, then it's rational" is a useful tool

Adam

ok then I think my point has been effectively made

Semiclassical

but it derives that efficacy from the definition of a rational number as a ratio of integers, not vice versa

(and it's not terribly useful in proving that a given number is not rational)

Adam

right, but my point is there are situations in which that is the shortest direct proof

you can conclude that it has a repeating p - adic expansion much easier than establishing there to be an integer denominator and integer numerator of the expression you are working with in some cases

robjohn

5:59 PM

Note that for any $n$ there are only $n$ residue classes mod $n$. Thus, by the pigeonhole principle, for any base $b$, in the set $\left\{b^k\right\}_{k=0}^n$, there must be two distinct elements that have the same residue class. That means there are $j\lt k$ so that $b^k-b^j\equiv0\pmod{n}$. This means that $n\mid\left(b^k-b^j\right)$

This means that $\frac mn=\frac{p}{b^k-b^j}$

Leaky Nun

6:30 PM

@user193319 Projections need not be closed however. Consider for instance the projection $\Bbb R^2 \to \Bbb R$ on the first component; then the set $A = \{(x,1/x) \mid x \ne 0 \}$ is closed in $\Bbb R^2$, but $p_1(A) = \Bbb R - \{0\}$ is not closed in $\Bbb R$.

user193319

6:46 PM

Thanks, Leaky!

Secret

7:26 PM

hmm...

$pe^{e} =q$; $p,q \in \Bbb{N}$

$\ln p + e = \ln q$

ÍgjøgnumMeg

ha pee

Leaky Nun

lol

Secret

suppose $pe^e = q$ is true. Then suppose $e^e$ is transcendental, then $pe^e$ is transcendental, thus contradiction since $q$ is algebraic

Now suppose $e^e$ is algebraic. Then $pe^e$ is algebraic thus *dies

rschwieb

7:54 PM

Looks like the transcript of a lecturer keeling over mid-lecture?

Leaky Nun

hi @rschwieb

rschwieb

@LeakyNun hi

Leaky Nun

what can I do for your site lol

rschwieb

Or should I be... Brews Chi

Your suggestions have been great so far

anything holding you back?

besides having a life?

Leaky Nun

I don't know what to do next lol

rschwieb

7:57 PM

That's funny, my site rep is 99064, but it says I'm over 101k here. I wonder if it doesn't count downvotes or something.

My next targeted goal is the FAQ page with conventions and helpful advice

Then after that I'll see if I can link the properties like you suggested

Leaky Nun

the larger number is the total of the reps across all SE site

rschwieb

That part needs a little work

ahh, really, interesting

Ribs Chew is pretty good too

@LeakyNun That 'inspiration list' isn't getting much shorter :) There have been a handful of suggestions, but nothing near the number of things that are requested in there

But probably don't work on "right PCI \neq left PCI" (because that's an open question)

Leaky Nun

@rschwieb ringtheory.herokuapp.com/search/results/?H=151&L=152

I thought of things to say but you would already have thought of what I want to say

rschwieb

@LeakyNun And when you run out of ideas, you could start trying to drag your friends into the fray

Not sure what you mean here... you mean about filtering out searches which are inconsistent and can't offer results?

I never thought of a good way to do that in general

but it might be simple enough to do for direct implications like that one.

The most general method woudl be to make a throwaway ring with the search properties, then process it with all the logic until it throws an exception because it hit a contradiction.

Leaky Nun

8:13 PM

no

rschwieb

oh sory

this search is the other way around

Leaky Nun

the thing is that $\Bbb Z/(3)$ is periodic and not boolean

but $\Bbb Z/(2)$ is boolean

rschwieb

oh yeah, I plan to add a finite field as soon as the competition si over

Leaky Nun

but this should do

rschwieb

I can't right now because I woudl be destroying the 100th ring prize :)

Leaky Nun

8:14 PM

because $2$ is not a prime

rschwieb

haha, good observation

Fargle

2 is prime in $\Bbb Z/2\Bbb Z$, though, because it generates a maximal ideal

rschwieb

I know the 2 not being prime remark is facetious, but I'm not quite sure how you're applying it here... $\mathbb Z/(n)$ is periodic anyway, right?

even if 2 divides n?

Leaky Nun

@rschwieb so it can clear that inspiration

it's periodic and not boolean

rschwieb

rtight

I'll do that

I overlooked taht Z/n would take care of it.. i was intending to make Z/(3) a concrete example

or to make Z/(2) an example and specify that $p>2$ in the Z/N example

that latter one may still happen

there ya go: ringtheory.herokuapp.com/search/results/?H=151&L=152

Leaky Nun

8:20 PM

it's still in the list lol

rschwieb

Yes: remember that it only gets refreshed when the instance powers off and then back on

I don't have any way of triggering a new spin-up where I am now

But at least the search now shows there is an example

good catch

there are just so many little things that can be fixed, but I can't see them all

and the site would come to a standstill if I did nothing but look for fillings for tiny gaps

So I really appreciate the community help

I will be really happy when all of those Inspiration items are wiped out

It looks like, if there were one new ring per item, it woud take about 50 rings to accomplish that

Sebastiano

Good evening to all users into this chat. Could someone help me on a math/physical issue?

The question is

0

Q: Alternative proof of $\mathbf{g}(\mathbf{r})=-{\mathbf \nabla}\psi(\mathbf r)$: I not find an error of signus $-$

Consider the following figure where $R=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}=|\mathbf{r}-\mathbf{r}'|$ is the module of the $\mathbf{R}$ vector depends not only on the location of the $P$ point but also on the location $P'$ where the $dV'$ volume is located (fixed once located in the volume $\math...