If every set contains the empty set, why is the first axiom of a topology/$\sigma$-algebra that it contains the empty set not a redundant statement?

@LeakyNun: Any hint here please?

@Koro It is most certainly not. Try looking at some concrete examples.

Hi @TedShifrin

I'm thinking about it.

@Koro look at the clock :)

Right now I feel like I'm in a car whose tyres are moving but not the car.

@Charlie, you are confusing the operators $\in$, and $\subseteq$

Sure: I'll think about it @TedShifrin @LeakyNun

Oh is the empty set a subset of every set, not a member

:-)

@Charlie correct

Got it, thanks :D

I have an $f\in \mathbb{F}_2[x]$, with $f = x^4 + x + 1$ an irreducible polynomial. What is $\mathbb{F}_2[x]/ \langle f \rangle$? Specifically $\langle f \rangle$?

@TedShifrin according to you , which book is good for 3D geometry ?

I like Pedoe’s Geometry: A Comprehensive Course. I have no idea what you mean by 3D geometry.

Geometry in 3D, like equation of plane

That's all? Not very interesting.

Basic linear algebra is all that is.

Hmm, Ok and about regular tetrahedron

Ok , so we can just work that all by linear algebra and some visualisation ..

Any hints for above question ?

Have you tried several examples to guess an answer?

Yes, I get 0 .

I want to do without taking a example.

Well, I didn't suggest that examples gave a proof. I would suggest thinking about the areas of the triangles $\triangle OAB$, $\triangle OBC$, $\triangle OCA$. This might be helpful.

@Tangoed Your first question makes good sense. Your "specifically" does not. When $f$ is irreducible, the quotient ring is a field. How many elements does it have?

I just suddenly wondered that if can we give the velocity of the growth of function. I mean using big O or small o, we can compare the growth of the function. But I wonder if I can give some real value to the growth of a function like 20m/s.

(Sorry for stupid question. I can't delete the chat)

@barista growth of a function is its first derivative, so you're talking about second derivative?

@Tangoed it's the field with 2^4 elements, what more do you want

I mean the how fast the function diverges like $\log x << x << e^x$

I wonder if there is a quantitative definition of such thing like a 'velocity' in physics

often the point of asymptotic analysis is to take a step away from point values (like 20 m/s) while still retaining some hint of being quantitative. you can get more precise by adding more stuff inside the o( ) or Omega( ). the thing inside there functions a bit like a generalized 'number' that can be compared against others. sort of.

Other than comparing derivatives as you go to infinity?

@barista, yes, the first derivative

@leslietownes So you're saying we can see $O(n)$ or $O(log n)$ as a generalized number

@barista you would observe that $\frac{\log x}{x} \rightarrow 0$ as $x \rightarrow \infty$

Are big O or small o things are algebraic object?

@TedShifrin $$ar($\triangle OAB$)=$\frac{1}{2}R^2sin2C$$,$$ar( $\triangle OBC$)=$\frac{1}{2}R^2sin2A$$$ and $$ ar($\triangle OCA$)=$\frac{1}{2}R^2sin2B$$$

it's landau notation. great way of keeping track of the magnitude of your "infinitesimals" or "infinities"

@Rover: one more step to go.

$$ar(\triangle OAB)=\frac{1}{2}R^2sin2C$$,$$ar( \triangle OBC)=\frac{1}{2}R^2sin2A$$ and $$ ar(\triangle OCA)=\frac{1}{2}R^2sin2B$$

barista yes although it is an analogy. it forgets some detail of the original function (as you do when you replace a function with its value at a point) while retaining enough info to admit comparison against other functions

@leslietownes Yes I know that but we can't quantitatively compare how fast $e^x$ is than $x$.

Rover, yes, you said that. Now see why the sum of areas times vectors gives $0$.

@TedShifrin I am not getting how ? Will the area times vector will become side of triangle ?

@Thorgott I'm trying to understand how this field is built

you just gave an explicit construction

@Thorgott Oh boy, $\mathbb{F}_2/\langle f \rangle$ is the same as saying $\mathbb{F}_2/( x^4+x+1 )$

No, Rover. Weighted average of the vectors will be $0$.

What got me confused was the $\langle \cdot \rangle$, I thought this meant something else, like some operation on $f$

Ideal!

oh, ok

hey ted

how are you

Hi Shmo.

how are things down by you

Still chugging.

good to hear. is the math still adding up?

Nah. Math is broken.

its maths, of course

So, maths are broken?

math has a split personality complex?

@TedShifrin how ?

A good vector exercise.

Okay

Are you considering are as vector? I think no right ?

whats the exercise?

No, but you can express area in terms of vectors.

oh, euclidean geometry exercise?

fun

@TedShifrin Ah yes

I come here when I want to procrastinate an exercise I don't want to start doing

So, I get $R^2(sin2A+sin2B+sin2C)=\vec{OA

Huh?

So, I get $R^2(sin2A+sin2B+sin2C)=\vec{OA.OB}sin2C+\vec{OA.OC}sin2B+\vec{OB.OC}sin2A$

I don't understand, and doesn’t seem relevant.

I found area by cross product of vectors

And equated both the areas

I don't see cross products. But the areas are the sins of the angles?

Yes, like ar($\triangle OAB=\frac{\vec{OA}×\vec{OB}}{2}$)

I understand that. We want the sins as coefficients of appropriate vectors? Review what you want to prove.

=($OA.OBsin2C$)/2

Don't write dot for cross. And the cross is a vector.

smacks rover on the head

how am I doing Ted?

Ok, I want to write that cross product of two vectors is magnitude of two vectors multiplied by sine of angle between them

No you don't.

You want a linear combination of $\overrightarrow{OA}$, etc.

@TedShifrin yes

Maybe just use $\vec A$, $\vec B$, $\vec C$.

smacks JoeShmo for practice

Yes, and how do derivatives disappear in the right-hand side?

....exactly

@TedShifrin okay, I have to think on that tomorrow now good night.

Good night, @Rover. Cool question.

and then $(x \times \nabla_x) \cdot (x \times \nabla_x)$ would formally be the standard dot product, sure

and call that ^ $R_x^2$. But then for some reason the claim is that the oeprator $R_x^2 = -x - (I - xx^T)^2$ ??

oh and since we're working on the unit sphere, $(I - xx^T)^2 = I - xx^T$

So I assumed. Let's use the formula for dot product of two cross products.

it's a projection

Hmm, I get a Laplacian term, and $\nabla\cdot x = 3$, $x\cdot\nabla = x\partial_x+\dots$.

for the formal norm on the lhs?

Yes, using formula for dot product, or just work it out.

yeah, sure. one sec

I get Laplacian minus the product of those last two things.

Interesting Jacobi identity (relevant to Rover's problem): $$(A\times Bj\times C + (B\times C)\times A + (C\times A)\times B=0.$$

typo

In mine?

I think so, no?

I fixed it?

idk

there's an unclosed parentheses

Hi

and Bj?

Typing on ipad. Too late.

j is parenthesis

ok

I should refuse to chat on ipad.

I am getting some trailing terms too, but more than you

laplacian plus some other stuff

Whatever, the statement needs serious clarification.

Right now doing some financial coding :)

I already wrote a 500 line python webdriver that can make binary option orders in Nadex.com.

I think I have some probabilities based upon a very complicated string repetition algorithm

Anyway, just wanted to spill the beans on what I'm up to... ^_^

Hi, @geocalc33

Hi, @JackOhara

high

@geocalc33 what are you studying on?

@StudySmarterNotHarder hi

high there

High, how are you doing?

If you smoke grass, you can be a real grassmanian.

Misspelled Grassmannian :)

Ah

Grassmannian matrix method

he smoked grass, that's why he misspelled the name

Um …

I guess.

Well, I bet Grassmann had the dankest hearb in his pantry

Oh, I see it's the space of all $1$-dimensional vector spaces (or just lines) through the origin of a $k$-dimensional vector space.

No, general dimension subspaces. What you said is just projective space.

I see

So each is guaranteed to be a quotient space of something?

Or in otherwords itself forms a vector space?

No to the last.

A beautiful manifold, though.

Why not the last?

If it's a quotient space, (I meant an algebraic one, not necc. topological)

@StudySmarterNotHarder I want to generalize $(x,y)\mapsto (ax,y/a)$ to the third dimension

Compact, for starters, and no additive structure or scalar mult.

I see

it is a topological quotient, though

very nice quotient by a Lie group action

@geocalc33 Notice that you have $a, b = 1/a$ and that $ab = 1$. The obvious generalization is $abc = 1$!

So you'd have $f(x,y,z) := (ax, by, cz), \\ a,b,c \in \Bbb{C}, abc = 1$.

That's very interesting looking to me

@StudySmarterNotHarder okay, another thing is that it should preserve oriented volume, as it does preserve oriented area in the plane

Let $g$ be another such mapping $\Bbb{R}^3 \to \Bbb{R}^3$. Then you can componentwise multiply the maps: $f\cdot g (x,y,z) := (aa'x, bb'y, cc' z)$ so under that, you immediately have a group law.

@geocalc33 what are those?

You mean a rectangle with boundaries oriented or, a rectangle with a normal vector?

Say you represent the group of maps as $G = \{ (a,b,c) : a, b, c \in \Bbb{C}, \ abc = 1\}$.

Then $S_3$ acts on this space by permuting the indices.

If $\sigma \in S_3$ is a permutation and $g, h \in G$. Then $\sigma(f\cdot g) = \sigma(f) \cdot \sigma(g)$ i.e. permutation of indices respects multiplication. So each permutation of indices lies in $\text{Aut}(G)$.

@geocalc33 does that help?

@StudySmarterNotHarder yeah that's one generalization, interesting

It's an abelian group, so write $+$ instead of $\cdot$.

Then define $F: G \to G, F(f) = \sum_{\sigma \in S_3} \sigma f$

Then the result should always equal $(1,1,1)$ I think.

so if I know that $\int{\frac1{x}}dx = \ln|x| + C$. Does this still hold if I replace $x$ with some function of $x$, $f(x)$? ie do I still get: $\int{\frac1{f(x)}}dx = \ln|f(x)| + C$? This feels wrong to me.

for reference I'm trying to follow this explaination I kindly recieved a while back.

You can answer your own question. Take the derivative.

@AndrewMicallef No, but $\int f'(x)/f(x)\, \mathrm dx = \ln|f(x)|+C$, which is what's used at the end of the accepted answer if that's what you're confused by.

Why is it that $f(x^q) = f(x)^q$ for each $f\in \mathbb{F}_q[x]$?

ahhh....! gonna swing from chandler

I may be mistaken, but i think this can be proved with induction on the degree of $f$ and the binomial theorem

I think the first part is unnecessary

polynomials commute with ring homomorphisms, so all you need to show is that $x\mapsto x^q$ is a ring homomorphism (which is where the binomial theorem comes in)

Hmmm ok!! Thanks!

yeah that's also a nicer way to think about it!

i just realized that one can export geogebra as tikz. this would have literally saved me HOURS over the past year. smh

As long it's hours and not days!

i hate to think too hard about it but it is likely into the days category at this point. i feel like a fool ha

i was doing a lot of stuff with planar polygons, and they would take forever to get right in tikz. I would actually put them in geogebra then use that to physically read off the coordinates

such a waste

I always exported Mathematica to Adobe Illustrator. In the olden days.