Mathematics - 2018-02-24 (page 1 of 5)

@Semiclassical zero becoming both an additive and multiplicative absorber is as useless as a structure can get that is nontrivial

Xander Henderson

@AkivaWeinberger A question that you answered came across the review queue just now (it was a fake "Should this question be reopened" question). I like that example. +1

tbh, as pointed out by many algebra people here, the trivial object {0} is actually very useful due to it being a basic object inside many algebraic structures

but a (two sided) null semigroup?, nah, why would I want to map everything in the algebraic structure to one element no matter the operation anyway...

4:00 AM

@philmcole yes, by either counting the number of free variables in the row echelon form or using the rank nullity theorem, you can work out the nullity of the matrix which corresponds to the dimension of the basis of the null space, but to work out what that basis is, I don't quite remember but my vague memory said you cannot read it out from the column vectors like for the column space

@AkivaWeinberger Ok, I think I am still stuck. The issue is that while I can let some invariant $I$ that is additive and translational invariant in the polygonal lattice, I cannot just assume that $f(I)$ is also translationally invariant and additive where $f$ is the area function. Any hints to nudge me to the correct thinking process?

skullpatrol

\o @nitsua60

@skullpatrol o/

WDUK

4:15 AM

Can any one explain this in a simpler way i.imgur.com/wEUBOro.png the way its saying f(1,0) = (a,c) and f(0,1) = (b,d) is kinda hard to intepret

i dont "get it" from reading the statement

It means if you put in the x column vector, you get the vector (a,c) and if you put in the y column vector, you get the vector (b,d)

thus using the standard basis, the matrix representation of f is given as in the pics

WDUK

oh the column vector

it was not clear imo

skullpatrol

Have you seen this? @nitsua60

@skullpatrol No, I had not. Probably won't ever again, either.

skullpatrol

4:30 AM

:-)

4:42 AM

Tbh, what is the rigorous way to write the following?

$$\begin{pmatrix}\textbf{a}\cdot \textbf{a} & \textbf{a}\cdot \textbf{b} & \textbf{a}\cdot \textbf{c} \\
\textbf{b}\cdot \textbf{a} & \textbf{b}\cdot \textbf{b} & \textbf{b}\cdot \textbf{c} \\\textbf{c}\cdot \textbf{a} & \textbf{c}\cdot \textbf{b} & \textbf{c}\cdot \textbf{c}\end{pmatrix}$$

I want to wrote that as something like a matrix multiplication, but I cannot seem to ensure the elements dot together

Outer product

In linear algebra, an outer product is the tensor product of two coordinate vectors, a special case of the Kronecker product of matrices. The outer product of two coordinate vectors u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} } , denoted u ⊗ v {\displaystyle \mathbf {u} \otimes \mathbf {v} } , is a matrix w ...

something like this?

$$\begin{pmatrix}\textbf{a}^T \\ \textbf{b}^T \\ \textbf{c}^T\end{pmatrix} \otimes \begin{pmatrix}\textbf{a} & \textbf{b} & \textbf{c} \end{pmatrix}$$

Well, no, it'd just be $$\begin{bmatrix}\textbf{a} \\ \textbf{b} \\ \textbf{c}\end{bmatrix} \times \begin{bmatrix}\textbf{a} & \textbf{b} & \textbf{c} \end{bmatrix}$$

or, if A = [a b c], it's just $A \otimes A$. Or $A^T A$.

But I want the boldfaces (vectors) to become inner products after the outer product stuff, if we do the above, then e.g. I would end up with $\textbf{a}\textbf{a}$ and not $\textbf{a}\cdot \textbf{a}$. I am not sure how to adapt the definition of the outer product here because what I have here is basically a column vector of column vectors outer product with a row vector of column vectors, if that makes sense?

Oooooooh. I wasn't following that each of your a, b, c were vectors. Sorry.

4:56 AM

Or in other words, I am trying to do the usual vector outer product, except the issue I have here which I don't know how to interpret is that the components themselves are vectors and after the multiplication, the components will dot together inner product fashion

So you want to know what happens when you mix an outie with an innie...

Depends on your implementation, but would $\sum_i \begin{bmatrix}a_i & b_i & c_i\end{bmatrix}^T \begin{bmatrix}a_i & b_i & c_i\end{bmatrix}$ work?

hmm... I think we might need i,j,k indices, the above will end up summing the 9 vector entries when seen in the formal manipulation, but what I only need is each entries only summing up their components

that is, each entry will be of the form $\sum_{ijk} x_{ij}y_{jk}$ where $x,y, \in [\textbf{a},\textbf{b},\textbf{c}]$. Let me grab a piece of paper to check...

I'm not sure I'm following you there ^^ but I'm pretty sure that sum gives you the same matrix quantity you originally specified. You get a 3x3 matrix foreach i, then sum on i. It's just a change in order of operations; in your formulation you get a 3x3 matrix of sums on i.

Now, depending on why you're doing this that might be important, it might not be. In any case, I'm past-due to hit the sack. Good luck!

I thought your original a, b, c were vectors? Then why do x, y in your last line need two indices?

If the whole thing is viewed as a 3x3 matrix of column vector entries, then the first index notate the position of the entry, and the 2nd index notate the x,y,z components of each vector entry where the inner product is carried out

(yeah typo, I think I actually need three indices and somehow contract one of them)

Let me see what your suggestion expands to. It might be actually end up what I am looking for. I am not very good at dealing with rank 3 tensor like objects rigorously, so it is time to train myself up

Semiclassical's map, I think when written in full, is actually $F: \Bbb{R}^3 \otimes \Bbb{R}^3 \otimes \Bbb{R}^3 \mapsto \Bbb{R}^3$ thus $F$ is necessary rank 3

13 hours ago, by Semiclassical

I tend to think of the mapping as $[a,b,c]\mapsto [a,b,c]^T [a,b,c]$ and then project to the three off-diagonal components.

so we are close

yeah, I have checked now with paper, yours give exactly what I want:
$$\begin{pmatrix}\textbf{a}\cdot \textbf{a} & \textbf{a}\cdot \textbf{b} & \textbf{a}\cdot \textbf{c} \\ \textbf{b}\cdot \textbf{a} & \textbf{b}\cdot \textbf{b} & \textbf{b}\cdot \textbf{c} \\\textbf{c}\cdot \textbf{a} & \textbf{c}\cdot \textbf{b} & \textbf{c}\cdot \textbf{c}\end{pmatrix}$$
thanks

so that means:

$$\sum_i \begin{bmatrix}a_i & b_i & c_i\end{bmatrix}^T \begin{bmatrix}a_i & b_i & c_i\end{bmatrix} = \begin{pmatrix}\textbf{a}\cdot \textbf{a} & \textbf{a}\cdot \textbf{b} & \textbf{a}\cdot \textbf{c} \\ \textbf{b}\cdot \textbf{a} & \textbf{b}\cdot \textbf{b} & \textbf{b}\cdot \textbf{c} \\\textbf{c}\cdot \textbf{a} & \textbf{c}\cdot \textbf{b} & \textbf{c}\cdot \textbf{c}\end{pmatrix}$$

This complete the description of $[a,b,c]\mapsto [a,b,c]^T [a,b,c]$

Now I just need to figure out:

$F(\textbf{a},\textbf{b},\textbf{c}) \mapsto (u,v,w)$ then I will be able to find the condition such that I specify any triplet of unit vectors, $(u,v,w) \in [-1,1]^3$

13 hours ago, by Semiclassical

My personal suggestion is to write out what $\det([a,b,c]^T [a,b,c])$ would have to be in the (1,1,-1) case

now to compute the det...

5:37 AM

@Secret What's up my man !

I am trying to compute something from semiclassical's

okay

But my poor explanation skills end up winding this thing too long

@Secret Tell me when you're free

@Secret I'm sure the person you're explaining it to must be grateful to you for giving him your precious time

well ,nitsua helped me to understand it better and help me to get one of the required objects to be formalised as above, so right now I am working out the rest using semiclassical's suggestion

5:42 AM

okay cool

@Secret I've got something to ask , ping me when you're free.

6:13 AM

The determinant:
$$\det([\textbf{a},\textbf{b},\textbf{c}]^T[\textbf{a},\textbf{b},\textbf{c}])=(\textbf{a}\cdot \textbf{a}) ((\textbf{b}\cdot \textbf{b})(\textbf{c}\cdot \textbf{c})-(\textbf{b}\cdot \textbf{c})(\textbf{c}\cdot \textbf{b}))-(\textbf{a}\cdot \textbf{b}) ((\textbf{b}\cdot \textbf{a})(\textbf{c}\cdot \textbf{c})-(\textbf{b}\cdot \textbf{c})(\textbf{c}\cdot \textbf{a}))+(\textbf{a}\cdot \textbf{c}) ((\textbf{b}\cdot \textbf{a})(\textbf{c}\cdot \textbf{b})-(\textbf{b}\cdot \textbf{b})(\textbf{c}\cdot \textbf{a}))$$

Simplified the original determinant further, I get:

$\det ([\textbf{a},\textbf{b},\textbf{c}]^T[\textbf{a},\textbf{b},\textbf{c}]) = 1 - (\textbf{b} \cdot \textbf{c})^2 - (\textbf{a} \cdot \textbf{b})^2 - (\textbf{a}\cdot \textbf{c})^2 + 2 (\textbf{a} \cdot \textbf{b})(\textbf{b} \cdot \textbf{c})(\textbf{c} \cdot \textbf{a})$

It kinda reminds me of the cosine law, I am not sure

and so for the output (1,-1,-1), the determinant is negative, while for (1,1,-1), the determinant is zero

6:29 AM

@MatheinBoulomenos you didn't, don't worry, I understood what you meant

In fact, for any coordinates (u,v,w), if it is any permutation of $\pm$(1,0,0), the determinant will be zero, if it is any permutation of $\pm$ (1,1,0), the determinant will be -1. Those cases where the determinant is negative will then correspond to impossible (u,v,w)

Since specifying any two inner products will specify the 3rd in 3D, it follows that the image of the map $[\textbf{a},\textbf{b},\textbf{c}]\mapsto (u,v,w)$ has to be two dimensional as there are only two degrees of freedom

so that means, the relation that express one of the inner product in terms of the other two will give the corresponding requirement for the spherical triangle inequality

\o

I am still looking for that relation

o/

Hah

Truly the peak of machine learning

o/

That doesn't trigger it? Hm

\o

@Pseudohuman \o

6:46 AM

Rather rudimentary algorithms I see

Balarka: Based on your experience in the past years, what type of maths do you think I actually know and not half baked? I would like to know from your perspective since analysing oneself is not going to be very effective due to bias and so on

@Secret Bro , you available ?

not yet, still thinking Akiva's problem

alright

6:57 AM

@Secret Knowledge is overrated. I think you are genuinely interested, which is what matters

7:10 AM

Partial progress: Ok, I think of a way to illustrate that I still got stuck in Akiva's question in the form of some commutative diagram, I will discuss with him on that later cause I am currently not on my desktop

but my doubt is that if the invariant in question is not a real number, it cannot be a multiple of the area

@Tanuj link to your room please. Also please show some effort

@Secret haha I've greyed my hair now in the effort to solve it

@Secret its h bar

ah johnrennie, that should keep h bar stable enough for my visit

sec..

What was Akiva's problem

7:31 AM

17 hours ago, by Akiva Weinberger

First, prove that anything translation invariant and additive on the set of lattice polygons is necessarily a constant multiple of the area

He is sharing about an interesting thing which only holds in 2D but not 3D, but he left the story hanging with this problem

probably because semiclassical and I are too busy discussing our spherical triangle inequality problem

Cool problem.

I think one should think about the determinant but do not quote me on that

@Secret $(2x^3+3x^2+x-3) + K(x-1)(x+2) = Ax(x-1)(x+2) - B(x+2) - C(x-1)$
Now plug in x=1,-2 for each case and B,C should pop out, after that you should be able to get A , how do i get A ?

and where I got stuck is the while I can just let some arbitrary invariant $I$ that depends on the polygonal lattice with the required properties, I have yet to find the link on how I can guarantee the invariant has to even be a real number because in general, what we have here is that $I = f(polygonal lattice) = f(T(polygonal lattice))$ where $T$ is the translation map, and f can potentially map to any structure as long it fulfils this equation, thus the invariant may not even be a real number

Perhaps I am overthinking on what f should look like

on the right hand side i have $A(x^3+x^2-2x)$,

Wait lattice polygon just means polygons with vertices having integer coordinates, I think

amirite @Akiva

7:37 AM

@Tanuj well, we are lucky in this problem that there's only one $x^3$ term in the LHS and RHS, thus A has to be 2

@Secret No but if we see on the right hand side , the only terms involving $x^2$ is $Ax^2$

so comparing with the LHS , isn't that a contradiction , as A should also be 3

nope, that's where the K came in:

The equality for $x^2$ is

$3 + K = A$

oh , i kept missing the K , thanks @Secret

that's why you must always add Const. when you indefinite integrate!

else you missed a family of solutions

wait , if i just do partial fractions of just the part inside the d/dx without integrating other side , i would not add K right ?

7:41 AM

@Alessandro Are you still interested in this?

@Tanuj yup, and that is expected, since d/dx constants = 0. thus the K has to be there. It may be fixed by some initial conditions

okay

and as Blue noted, there are 4 linear equations in 4 unknowns, thus K in this case is fixed

@BalarkaSen I am, I'm going to have the commutative algebra exam next Tuesday (the 27th) but I'm free after that

Mmkay

7:46 AM

yea @Secret

What's the book you were thinking about by the way?

It's a few pages in "Foliations I" by Candel-Conlon

[Random note]
If the zero product property is gone, actually I will argue almost everything in polynomial rings will break down. For example, one consequence is the above partial fraction question will become a lot harder to solve

As I will be discussing with Blue later, there is actually something interesting geometrically going on at those roots

Daminark

Foliations are for nerds

8:17 AM

how to differentiate $x|x^3|$

Silent

Let $\{\alpha_n\},\{\beta_n\}$ be real valued sequences, such that $\alpha_n<\beta_n$, $\alpha_{n-1}<\beta_n$ and $\beta_1>0$, such that $\alpha_n\to\alpha$, $\beta_n\to \beta$ where $\alpha\le\beta$. How to be sure that such $\{\alpha_n\},\{\beta_n\}$ exist for each $\alpha\le\beta$?

normally you cannot differentiate absolute values

8:38 AM

You might want to check, but I suspect the two sequence are only required to converge to some value and that the $\beta$ sequence will be larger than the $\alpha$ sequence, but otherwise it can oscillate thus not strictly decreasing

e.g. something like adding a zig zag to an exponential such that the zig zag slowly dies away

Aditya Pratap Singh

In the given figure, ABCD and AEFG are 2 squares. Prove that ∆ACF ~ ∆ADG

photos.app.goo.gl/h9TGW1E2hWuHxBSJ2

9:01 AM

To be investigated: A finite non Archimedean field

@Secret there is none

Because there aren't even just ordered finite fields

Actually there aren't ordered fields of finite characteristic

@AlessandroCodenotti Ah right, as finite fields cannot be linear ordered. I however don't know if they must be cyclically ordered like the Z/primes

Aditya Pratap Singh

Anyone who can help me?

as what I currently have in mind is e.g. {-1,1,a,b,c,d,e} and there's a cyclic order a<b<-1<1<c<d<e<a such that there is no products nor sums that can give e (thus e behaves like an unreachable element in the structure), while a,b,c,d can have their additive and multiplicative inverses as usual

typo: No product nor sums that does not have e in it

Alex

9:20 AM

@Secret What does this mean?

I am thinking about something like this:

First we have the following underlying set:

{0,a,b,c,d,1,-1,e}

Next we have the following pairs of additive inverses:

0+0=0
a+d=0
b+c=0
1-1=0
e+e=0

and multiplicative inverses:

1*1=1
a*c=1
b*d=1
-1*-1=1
e*e=1 (this one I am not sure, will figure out later as we go on)

We also have the cyclic ordering:

a<b<-1<0<1<c<d<e<a

jublikon

Hi, little question about matrix decomposition here :)

Let $P,D,L,R$ be matrices and matrix $A\in \mathbb{R^{3 \times 3 }}$ that it holds $PDA = LR$.

My task is now to solve $Ax=b$ using that LR decomposition above that is computed with pivotisation.

Now the solution says that: $Ax=b \Leftrightarrow D^{-1}P^{-1}LRx = b$ How can I get there?

I have done that:

$\begin{align}
Ax=b &\Leftrightarrow LRx = PDAb \\
&\Leftrightarrow D^{-1}P^{-1}LRx = Ab \\
\end{align}$

I would be very glad if someone could help me :)

and e is an "unreachable" element which has the property that there exists no elements $x,y \neq e$ such that $xy=e$ or $x+y=e$

ok... I think I saw something wrong, cause ee=1 and e+e=0 will mean ee+ee=1+1=0, contradicting 1-1=0. But that's an easy fix. Let's see if the modified set is actually consistent...

{0,a,b,c,d,1,e}

Alex

What is your ternary relation?

I am aiming something that resembles a field, so I don't see why I will have a ternary relation instead of two binary relations

9:35 AM

@Alex Changed your name I see

who is he before?

Narcissusjewel

I see

Hm

o/

\o

9:37 AM

Aha

~$wikilookup Whitney_topology

en.wikipedia.org/wiki/Whitney topology

en.wikipedia.org/wiki/"Whitney topology"

Whitney topologies

In mathematics, and especially differential topology, functional analysis and singularity theory, the Whitney topologies are a countably infinite family of topologies defined on the set of smooth mappings between two smooth manifolds. They are named after the American mathematician Hassler Whitney. == Construction == Let M and N be two real, smooth manifolds. Furthermore, let C∞(M,N) denote the space of smooth mappings between M and N. The notation C∞ means that the mappings are infinitely differentiable, i.e. partial derivatives of all orders exist and are continuous. === Whitney Ck-topology... ===

@jublikon $Ax=b \implies PDAx=PDb \implies LRx = PDb \implies D^{-1}P^{-1}LRx =b$. You have an extra A in your orignal workings

@Pseudohuman Thanks

@BalarkaSen You're welcome!

~$poem

9:43 AM

Roses are red, Violets are blue, Haven't you got anything better to do?

Indeed so

~$poem2

(I might need to lookup all the commands)

~$help

Available commands:
Builtin: asm, coinflip, cointoss, eval, fibonacci, help, joinroom, learn, leaveroom, list, listcommands, rand, rolldice, room, unlearn, wotd
Learned: /o_o, /why, addition, calloakbot, cat, conv, downgoat, galactic_transform, greet, gypsytime, iwonderwhatthisdoes, lenny, look_of_disapproval, matrix, mithrandy, o_o, poem, salute, scary, se, tif, wikilookup, xyz

@Secret There you go

9:44 AM

~$iwonderwhatthisdoes

I don't understand.

ok

~$iwonderwhatthisdoes

I don't understand.

What a useless command

~$unlearn iwonderwhatthisdoes

9:46 AM

@BalarkaSen Forgot command: iwonderwhatthisdoes

Good riddance

"I don't understand" is literally the catch all whenever pseudohuman does not register a command in her database

but having a command that maps to that is weird

Yeah I deleted that

Hm how do I teach her a specific command

~$help

Available commands:
Builtin: asm, coinflip, cointoss, eval, fibonacci, help, joinroom, learn, leaveroom, list, listcommands, rand, rolldice, room, unlearn, wotd
Learned: /o_o, /why, addition, calloakbot, cat, conv, downgoat, galactic_transform, greet, gypsytime, lenny, look_of_disapproval, matrix, mithrandy, o_o, poem, salute, scary, se, tif, wikilookup, xyz

9:47 AM

~$learn

jublikon

Thank you !! @Secret

~$listcommands

Available commands:
Builtin: asm, coinflip, cointoss, eval, fibonacci, help, joinroom, learn, leaveroom, list, listcommands, rand, rolldice, room, unlearn, wotd
Learned: /o_o, /why, addition, calloakbot, cat, conv, downgoat, galactic_transform, greet, gypsytime, lenny, look_of_disapproval, matrix, mithrandy, o_o, poem, salute, scary, se, tif, wikilookup, xyz

That's just the same thing as help

protip, no matter what someone do, do not type unlearn <very important keyword>

otherwise you will do irreversible damage that only the devs of pseudohuman can fix

(assuming they have not put flags on those commands)

9:51 AM

~$learn command newcommand

@BalarkaSen Learned command: command