Mathematics - 2016-03-10 (page 1 of 2)

user19405892

01:42

hi

Axoren

04:03

How do you see which chat messages were starred for outspoken? I don't remember ever being starred before, let alone by 9 other people.

usukidoll

hiiiiiiiiii

Axoren

Yo.

Also, jeez. I just found all my past starred messages. From dark times, those.

usukidoll

do you know real analysis

Axoren

I know it exists, but I'm likely not very helpful in that regard.

usukidoll

T_T

Axoren

04:14

If you ask a question, I can be better sure of my uselessness.

And someone else might chime in.

Looking through the topics of real analysis, I seem to know more than I expected. I haven't taken the class at the university, yet, however.

I might be able to help.

usukidoll

o rly?
Suppose $f : D \rightarrow R$ has a limit at $x_{0}$. Prove that $\mid f \mid : D \rightarrow R$ has a limit at $x_{0}$ and that $\lim_{x \to x_{0}} \mid f(x) \mid = \mid \lim_{x \to x_{0}} f(x) \mid$ \\

Axoren

Is $R$ supposed to be $\mathbb R$?

usukidoll

yup

where R is the real numbers and D is the domain

I'm supposed to use this theorem

hold on

Theorem: Let $f : D \rightarrow R$ with $x_{0}$ an accumulation point of D. Then f has a limit at $x_{0}$ iff for each sequence $\{x_{n} \}$ n = 1 to infinity converging to $x_{0}$ with $x_{n} \in D$ and $x_{n} \neq x_{0} $ for all n, the sequence$ f(x_{n})$ n = 1 to infinity converges

so that would mean that we have a limit x_0 which is the accumulation point of D and as limit of x approaches that point the absolute value of f(x) is going to be equal to that lim as x approaches x_o f(x) . It's like I have to use that theorem twice

so that absolute value f has that sequence absolute value of x_n

?

Axoren

I'm still trying to interpret unfamiliar jargon. The use of sequences here is what's throwing me off.

usukidoll

oh yeah because the book intro to analysis by gaughan uses shxx notation for sequences

Axoren

04:30

Okay, I see what's going on.

usukidoll

$(x_n)_{n=1}^\infty$

that's better

Theorem: Let $f : D \rightarrow R$ with $x_{0}$ an accumulation point of D. Then f has a limit at $x_{0}$ iff for each sequence $(x_n)_{n=1}^\infty$ converging to $x_{0}$ with $x_{n} \in D$ and $x_{n} \neq x_{0}$ for all n, the sequence $(f(x_n))_{n=1}^\infty$ converges

Axoren

So, if $x_0$ is an accumulation point of $D$, then $f$ converges to as we approach $x_0$ from any direction in the neighborhood.

usukidoll

yeah

Axoren

So, if those sequences converge, so do their absolute value sequences.

usukidoll

but isn't the absolute value of f : D - > R also converge as well if the original f: D-> R converges

I was getting there... I'm kind of slow at this...but at least I got 3 out of 4 proofs done early by myself ^_^

Axoren

04:34

Could you repeat that? The phrasing's throwing me off.

Were you agreeing with what I said or finding fault in it?

usukidoll

if f: D -> R which is a function mapped from the DOmain to the set of all reals converges, then the absolute value of f : D-> R converges too.. I was agreeing with you

Axoren

Okay, sorry. It's pretty late and I've spent today doing nothing useful.

So the brain's not been primed.

usukidoll

but then how do we go beyond that point though

besides we know the statement is true x.x

Axoren

The absolute value of the sequences converge to $|f(x_0)|$ because the original converged to $f(x_0)$

And since the other sequences converged to $f(x_0)$, taking the absolute value of that limit gives you $|f(x_0)|$

usukidoll

so since the original f converged to $f(x_{0})$ then the $\mid f \mid$ converged to $ \mid f(x_{0}) \mid $

take absolute value of the limit? like a $\mid \mid f \mid \mid$ ?

Axoren

04:39

$| \lim f |$ converges to $|f(x_0)|$

Yeah

usukidoll

so... when our original lim f converged to $f(x_{0})$ the $\mid lim f \mid$ converged to $ \mid f(x_{0}) \mid $

Axoren

Right.

usukidoll

ah..ok... now how do I clean this up lol XD

Axoren

There's two parts. First prove that $|f|$ has a limit.

usukidoll

O_O

Axoren

04:48

Then, prove that its limit is the same as $|\lim_{x \to x_0} f(x) |$

usukidoll

ragequits sorry I have a migraine .......... so if there's a limit then it's

Axoren

To clean up the proof, do these parts in order.

usukidoll

absolute value f has a limit...hmmm

Like ...
By the Theorem, if $x_{0}$ is an accumulation point of D, then f converges as we approach $x_{0}$ from any direction in the neighborhood. So, if $ f: D \rightarrow R$ converges then $ \mid f \mid : D \rightarrow R$ also converges. Since the original sequence converged to $f(x_{0})$, the absolute value of the sequence also converges to $ \mid f(x_{0}) \mid$. ????

Axoren

Remember, if we take the absolute value of the elements of all the sequences that converge at $x_0$ to $f(x_0)$, we have a new set of sequences that all converge at $x_0$ to $|f(x_0)|$.

Yeah.

user19405892

hi

can someone help me with a geometry question

Axoren

04:51

That means the limit at $x_0$ exists for $|f(x_0)|$.

usukidoll

that does that even sound like a proof or am I just goofing off? :/

Axoren

How you wrote it? It's fine.

user19405892

usukidoll

Like ...
By the Theorem, if $x_{0}$ is an accumulation point of D, then f converges as we approach $x_{0}$ from any direction in the neighborhood. So, if $ f: D \rightarrow R$ converges then $ \mid f \mid : D \rightarrow R$ also converges. Since the original sequence converged to $f(x_{0})$, the absolute value of the sequence also converges to $ \mid f(x_{0}) \mid$. Therefore, the limit at $x_{0}$ exists for $\mid f(x) \mid$.

Axoren

The dreaded missing \$

usukidoll

04:52

ikr

Axoren

Therefore, the limit at $x0$ exists for $|f(x)|$.*

usukidoll

XD. maybe it's because I'm still doubting myself at whether or not I'm at the right track T_T. I know proofs are hard but on some problems I can see right through them and know what to do while others takes time

Axoren

Some proofs are simply presentable by drawing, and it's great.

Makes graph theory tolerable.

usukidoll

so now we need to prove that the limit is the same as $|\lim_{x \to x_0} f(x) |$

Axoren

@user19405892 That's one hell of a question.

@usukidoll Correct.

usukidoll

04:54

whistles damnnnnnnn

errr... so we need to use the theorem again

Axoren

Do you? At that point it's just equality.

usukidoll

:O

Axoren

I'm not sure how streamlined you can get at this point. Again, I haven't taken the class yet.

usukidoll

so it's equal to each other...
$\lim_{x \to x_{0}} \mid f(x) \mid = \mid \lim_{x \to x_{0}} f(x) \mid$

but how?!

Axoren

But if $\lim_{x \to x_0} f(x) = y$, then all the sequences that converge at $x_0$ converge to $y$.

When you took those sequences and took their absolute value, you get $|y|$.

usukidoll

04:58

so for some reason when I took the absolute value of f(x) then I get the absolute value of whatever that thing was

Axoren

So, $\lim_{x \to x_0} |f(x)| = |y|$

And since $\lim_{x \to x_0} f(x) = y$, take absolute value of both sides.

$|\lim_{x \to x_0} f(x)| = |y|$

Both of them are equal to $|y|$, so it's proved.

usukidoll

oh.. like

I was getting there -_-

$\lim_{x \to x_0} |f(x)| = |y|$

$\mid \lim_{x \to x_0} |f(x)| \mid = \mid |y| \mid $ like that?

errrr why y? htough

Axoren

Not 100%. I can't tell where all those bars are going, there's so many.

usukidoll

*though

Axoren

I just called it $y$

Just to separate it and show that they were the same value at the limit point $x_0$

idonutunderstand

05:45

Ya'll hear about AlphaGo's latest victory?

user174558

06:37

@mreyeglasses

Albas

Hmm this version of chat us pretty nice!!

usukidoll

ikr ... all silence crickets

Axoren

07:18

@idonutunderstand Yeah.

idonutunderstand

07:32

I think abstract mathematics will be the last bastion of homo-sapiens ability to be conquered by A.I. @Axoren what is your reaction to the news?

Axoren

@idonutunderstand I, for one, welcome our new computer overlords.

idonutunderstand

Haha you're a computer science graduate student so yeah.

Axoren

I also don't suspect abstract mathematics to be that hard.

For the A.I. of course.

I can already think of a way to model and train a Deep network to solve proofs.

In the same way that we taught it how to play Go, we can model proofs as a game

Where we simply stack well-formed formulas as "moves" and a victory state is the original claim.

idonutunderstand

You mean using rules of inference to transition from one to the next?

Axoren

Yeah.

idonutunderstand

07:40

Are you going to pursue this way of thinking and actually try to implement it?

Axoren

I'd rather not, actually.

In theory, I like the idea.

In practice, I hate the idea.

idonutunderstand

Let someone else do it...

Axoren

Okay, you do it.

idonutunderstand

Haha no not me either.

Axoren

I often tell people ways to improve their methods when I can't be bothered to write the paper myself.

It's a bad habit of mine.

I could have written so many papers.

idonutunderstand

07:45

How about music? You have a way to do that with deep neural nets?

Axoren

Oh yeah, already formalized that a while back.

idonutunderstand

Poor Lee Sedol looks so stressed out.

usukidoll

any links to rsa encryption proof?

mr eyeglasses

11:40

@Jasper

Danu

11:59

Hi guys

@Axoren You know you sound kind of arrogant saying that, right?

user174558

@Danu Not to me.

user174558

@Axoren Sometimes, when one tries to write it all down, he discovers that he made a mistake in his thinking. It's always good to write things down to make sure it really works.

user174558

@idonutunderstand I prefer chess to weiqi.

FISOCPP

12:33

So if we have an arithmetic progression with first element 4 and common difference 3. What's the sum of the first 20 elements?

I'm using this formula to find the 20 element:

\ a_n = a_1 + (n - 1)d

How do I easily format expressions?

Is there some online program?

For this site.

$\ a_n = a_1 + (n - 1)d,$

mr eyeglasses

@Jasper I like Go better than Chess

FISOCPP

Any help please - I'm noob here?

How do I format formulas here?

Or is it only possible on the questions?

and not in chat?

I solved my problem anyway. Thanks for the help

mr eyeglasses

13:11

@FISOCPP you need this for the chat math.ucla.edu/~robjohn/math/mathjax.html

user116211

1

Q: How proper is it to turn an ODE into a wave equation?

I have seen the following method used a few times for finding solutions of wave equations. Take an ODE with a known solution, of the form $y''(x) + g(y(x), x) = 0$ Switch it to a wave equation of the form $\Box y(x) + g(y(x), x) = 0$, where $x = (t, x, y, z, ...)$ Make the substitution in the k...

user116211

Can anyone give a reasonable answer to this unanswered question?

Huy

13:47

@s.harp I'm not quite sure what you mean by that. Can you explain or give a reference?

Evinda

Hi!!! I have a question...

I want to find the volume of the parallelepiped that has as vertices the (0,1,0), (1,1,1), (0,2,0), (3,1,2).

If we would have three linearly indepedendent vectors, the volume would be their triple product.
What do we do in this case? @DanielFischer

s.harp

14:19

@Huy I was thinking about this kind of thing en.wikipedia.org/wiki/Modular_curve#Compactified_modular_curves

Huy

@s.harp: It appears to me this requires me to add points on the real line.

s.harp

yeah, thats right

Huy

@s.harp: I'm looking for something like this, but compact.

Mark S.

14:50

I have a question related to moderation. Basically, what are the intended circumstances for posting an answer to meta.math.stackexchange.com/questions/19042/… ? There are a few things I might want to post there, but I really want to make sure I understand the purpose before inadvertently abusing it.

whoops, that should probably go in the mods office

Semiclassical

@evinda that's four vertices. a parallelpiped has 8. (it may be the case that four vertices are enough to determine a parallelpiped, but that's not obvious to me off the bat.)

(five definitely do, but i'm skeptical about 4)

Evinda

15:08

Ok, I will think about... If we are given a parallelogram , how can we find the point at which the diagonals intersect? @Semiclassical

Semiclassical

depends in what sense you're 'given' a parallelogram, i suppose

if you're given the vertices, then that point should be the midpoint from both pairs of opposite vertices

e.g. if you're given a parallelogram ABCD, then that point should be the midpoint both of AC and of BD

user19405892

hi

i need help wit this

here is the solution

what doesn't make sense to me is how 'no three elements from E can have distinct $x$-coordinates and distinct $y$-coordinates'

Semiclassical

15:26

what's the definition of the centroid?

user19405892

It is the barycenter so it would be $(\frac{x_a+x_b+x_c}{3}, \frac{y_a+y_b+y_c}{3})$

it is the intersection of the medians of the triangle

doesn't $(3,6),(4,7),(1,0)$ in their $E$ violate that?

user19405892

15:43

do they mean something else by that?

Evinda

@Semiclassical Could you give me a specific exercise?

Semiclassical

sure, that's simple enough. "Suppose that a parallelogram $ABCD$ has vertices $A=(0,0)$ and $B=(1,0)$. In terms of $C=(x,y)$, find the position of $D$ and where the two diagonals will intersect."

user19405892

anyone know about my question

Mike Miller

16:11

morning

mr eyeglasses

morning

Evinda

@Semiclassical So we will have something like that:

@Semiclassical How could we explain formally that C and D have to have the same y-coordinate?

Huy

parallel postulate & geodesics in $R^2$ are straight lines

@MikeMiller: any idea what I have to quotient $\mathbb{H}$ by to get the two-holed-torus? I think it should be a group generated only by hyperbolic isometries (not elliptic or parabolic), but which elements do I take to see best that the quotient is what I want?

Mike Miller

16:27

do elliptic and parabolic fix points

Huy

@MikeMiller: wat?

Mike Miller

do elliptic and parabolic isometries fix points on the plane

Huy

I understand all those words but not what you're saying

I read that if I have parabolic isometries, the quotient won't be compact

Mike Miller

I don't know what to clarify. Does a parabolic isometry have a fixed point?

That's what I'm asking. I do not know the answer. This is not a trick question.

Huy

on $\partial \mathbb{H}$, yes

I thought you were telling me how to proceed

Mike Miller

16:31

Eh, I meant on the interior.

Huy

no

Mike Miller

K.

Huy

only elliptic have on the interior

Mike Miller

So elliptic are garbage and parabolic are possibly ok

math.univ-toulouse.fr/~jaramayo/hypstr.pdf fig 3.1

since there should be a unique transformation that takes each circle to the appropriate circle, I think that's it

Huy

yeah I've encountered the construction with the octagon as an exercise in a different text but I was trying to come up with something directly in the upper half-plane

but I guess I'll just take that then

Semiclassical

16:34

@Evinda $A$ and $B$ both lie on the line $y=0$, and $CD$ must be parallel to $AB$.

Huy

@MikeMiller: btw I didn't understand your previous questions because you didn't write a question-mark so I mistook it for an answer

Mike Miller

Thanks for the clarification

Huy

I need a larger flat for a larger desk for more monitors to have more windows open simultaneously

Evinda

So both $C$ and $D$ must lie on the same horizontal line. Since C=(x,y) , D is of the form D=(x',y). Is the justification right? @Semiclassical

Semiclassical

sure, that works.

not going to be paying attention for a bit

Evinda

16:48

Are the edges of the parallelogram vectors? @Semiclassical

user174558

@evinda Aha!

Evinda

Hi @Jasper

The length of AB is 1. Since AB and CD have to be equal , it has to hold $|CD|=1 \Rightarrow |x'-x|=1 \Rightarrow x'=x \pm 1$.

Thus $D=(x+1,y)$ or $D=(x-1,y)$.

Right? @Semiclassical

Axoren

17:06

@Danu Arrogant telling people how to improve their methods? or arrogant saying that I tell people how to improve their methods?

Danu

@Axoren Never mind it---I was being grumpy :-)

Mike Miller

Hi @Danu.

Axoren

@Danu It's all good.

Danu

Hi Mike

Is it standard terminology to call all holomorphic maps between Riemann surfaces coverings, and to designate the ones I know from topology (i.p. the ones that are loc. homeo's) as "unbranched"?

Mike Miller

Whose terminology is this?

Danu

17:11

Forster - Lectures on Riemann Surfaces

Mike Miller

Not all holomorphic maps between Riemann surfaces are what you describe ("branched coverings").

They should have the word branched next to the word.

But a nonconstant holomorphic map is a branched covering.

Danu

He says all of them are coverings, some are branched and some are unbranched.

Mike Miller

That just means it's a covering map away from a finite set of points.

Evinda

Hey @DanielFischer
Did you see my question above?

Danu

@MikeMiller Right, that's true

Evinda

17:49

@AkivaWeinberger Hey!!! Could I ask you something?

I want to find $\phi=arc \tan{\left( \frac{1}{0}\right)}$. How can I find the $\phi$ ? @AkivaWeinberger

Huy

18:00

@MikeMiller: silly question, in the notes you linked, the proof of prop. 2.9, why is $d(z,z_0) < d(z, T^{-1}(z_0))$? why is equality impossible? I suspect it has something to do with $T$ being nontrivial but there must be an additional property necessary to get the inequality.

Akiva Weinberger

@Evinda I'll assume you mean $\arctan(\infty)$, which would be shorthand for:$$\lim_{x\to\infty}\arctan(x)$$

Evinda

And how do we find this value? @AkivaWeinberger

Akiva Weinberger

Do you know what the graph of the arctan looks like?

Also, note that $\frac10$ could mean $\pm\infty$, but $\arctan(\infty)\ne\arctan(-\infty)$ (using the shorthand again) so it doesn't make sense

Here's one way to do it:

\begin{align}\arctan(\infty)&=x\\\infty&=\tan x\\\infty&=\frac{\sin x}{\cos x}\end{align}

So it would make sense to look for values of $x$ where $\cos x=0$

On second thought, I take back what I said about it not making sense

Ted Shifrin

18:27

Hi, DogAteMy @Akiva

Akiva Weinberger

Hi!

Ted Shifrin

What're you learning in linear algebra these days?

Akiva Weinberger

I'm not sure I recall exactly what we did most recently, but we were doing orthogonal matrices and related things

Ted Shifrin

Cool. Well, I can always pass along some interesting exercises if you need some ... :)

Akiva Weinberger

At the moment I have quite a lot of things on my plate (including a fairly large history paper due Monday), but thanks

Ted Shifrin

18:33

LOL ... The offer has no expiration date. Good luck with the history paper :)

Akiva Weinberger

I'm learning about that time when millions of people died

Mike Miller

@Huy: I can think about it if you email that to me so I can star it or if you ask me again late tonight

Akiva Weinberger

Not that one, the other one

Ted Shifrin

Well, that was an explicit conversation with yourself, DogAteMy :D

good night @MikeM @Huy

Mike Miller

Morning

Huy

18:35

night @Ted

sent you a mail @Mike

Mike Miller

Thanks, starred it

Will respond when I get a chance

Akiva Weinberger

(It's on the Holodomor, kinda like the USSR's version of the Holocaust, @TedShifrin)

Ted Shifrin

Ah ... and we could go all the way back to the Crusades, too, but adjusting for population growth. ... Hope the elections don't lead us to massive world war ...

Semiclassical

18:50

yo

@MikeMiller well, I think i figured out a simple reason why the Hamiltonian flow problem we've talked about shouldn't have a direct physical interpretation

Mike Miller

:D

Semiclassical

namely: while there's no problem talking about how solutions behave on subsets of phase space, it's not very typical in physics to consider canonical transformations that leave boundaries fixed

for instance, one could pick the Hamiltonian $H=(x^2+p^2)/2$. this induces a rotation of the phase space, and this would map the disc $D^2$ to itself (along with its boundary)

Mike Miller

Really? The most natural automorphism groups of manifolds with boundary are those that fix the boundary.

I gotta go tho.

Semiclassical

but it doesn't leave the boundary fixed

mmkay. later

Jim

1

Q: Construction of Isomorphism from sets of Local Isomorphism(Graph Isomorphism)

Given two graphs $G, H$ (each has $n$ vertices). We, split $G$ into subgraphs $G_1, G_2... G_x$ (total $x$ vertex set). Similarly,assume $H$ has subgraphs $H_1, H_2... H_x$ (total $x$ vertex set). Consider,$\forall k \leq x$, a permutation $\pi_k \in \beta_k$ where $H_k^{\pi_k}=G_k$ such that ...

group-theory graph-theory proof-verification algebraic-graph-theory graph-isomorphism

Semiclassical

18:59

@mike for your later interest: What I think might still be 'physical' is to consider two different flows $\phi_1,\phi_2$ which both act as $f:\partial D^2\to \partial D^2$ but don't behave the same inside of $D^2$. In that case, $\phi_2^{-1} \phi_1$ should act as the identity on the boundary.

in that case i'd expect not that the relevant integral would vanish, but that it would give the same result for the two flows.

Abhishek Bhatia

19:56

Hi guys

I did p-test and t-test. My prof. told me do anova analysis. What does he mean?

Tobias Kildetoft

@AbhishekBhatia Ask him

Abhishek Bhatia

I have unbalanced data from 2 observation. Which additional test should I do?

Tobias Kildetoft

@AbhishekBhatia The one your prof wants you to

Abhishek Bhatia

What does someone mean why he says do anova?

Tobias Kildetoft

@AbhishekBhatia How many times do I have to tell you to ask him?

Abhishek Bhatia

20:07

haha sry.. thx

Vrouvrou

20:43

please how to find what the $f\in \mathcal{C}([0,1],\mathbb{R})$ which satisfy $$\int_0^1|f(x)-2|dx<1$$ ?

Tobias Kildetoft

@Vrouvrou What sort of description are you looking for?

Vrouvrou

i'm surching a ball

Tobias Kildetoft

What do you mean by a ball?

Vrouvrou

i have $E=\mathcal{C}([0,1],\mathbb{R})$ and $$d(f,g)=\int_0^1|f(x)-g(x)|dx$$

i want to find $B_d(2,1)$

Tobias Kildetoft

@Vrouvrou Again, what sort of a description are you hoping for?

Vrouvrou

20:47

i don't understand how description ?

Tobias Kildetoft

@Vrouvrou What form would a satisfactory description take? So far you have one description, in terms o that integral

You can rearrange stuff a bit, taking into account where the functions are smaller of greater than $2$, to make it look different, but I am not sure how much nicer it will look

Vrouvrou

i don't know i just want to find f such that the integral be less then 1

Tobias Kildetoft

@Vrouvrou There are obviously infinitely many such $f$. I still don't see why the current description is not fine

Vrouvrou

in the sotion of the book they say that the ball is rectangle

Tobias Kildetoft

@Vrouvrou How does that even make sense in something which is not just the euclidean plane?

Vrouvrou

20:54

i don't know

What i can do when i have $\int_0^1 |f(x)-2| dx<2$

Tobias Kildetoft

@Vrouvrou So you decided that even after my questions, it was a fine formulation to just throw on the main site?

Vrouvrou

i really don't understand your question

i found this question in a book

i just want to find f

Tobias Kildetoft

@Vrouvrou No, you want to find all such $f$

and you have no idea what a proper answer would look like

Vrouvrou

yes

LeGrandDODOM

what is the chance that the guy who asks and the guy who answers have the same Gravatar picture? math.stackexchange.com/questions/1692085/…

Vrouvrou

21:09

@TobiasKildetoft see my edite please: math.stackexchange.com/questions/1692084/how-to-find-the-ball

mr eyeglasses

i thought gravatar was based on your e-mail

Mary Star

I am looking at the following exercise:

Let the finite group $G$ act transitively on the set $\Omega$. Then the action of $G$ and on $\Omega\times\Omega$ is defined as follows $(a,b)\cdot x=(a\cdot x, b\cdot x)$.
Let $a\in \Omega$.
Show that the number of orbits of $G$ on $\Omega\times\Omega$ is equal to the number of orbits of $G_a$ on $\Omega$.

I have done the following:

From the fact that the finite group $G$ acts transitively on the set $\Omega$, we have that there is just one orbit on $\Omega$.

Evinda

Hey @TobiasKildetoft
Could I ask you something?

Tobias Kildetoft

@Evinda Is it something I am usually interested in answering questions about?

Evinda

Hmm... I don't know... Is it known that diagonals in a parallelogram are divided in half by the intersection point or do we have to show it? @TobiasKildetoft

Tobias Kildetoft

21:18

@Evinda No idea. And no, basic geometry is not generally something I am interested in

user174558

I am interested in basic geometry, but it is something I know nothing about.

Huy

how you doing @Jasper?

user174558

@Huy So so. I am going to try learn some English, French, German, and LaTeX this year.

Huy

did you start on any of it yet?

user174558

Not yet. I have my first 6 holy math books planned, final list.

Huy

21:25

can I see?

Evinda

With use of algebra I want to prove the Lagrange property:
For any real numbers $x_1, \dots, x_n$ and $y_1, \dots, y_n$, $$\left( \sum_{i=1}^n x_i y_i\right)^2=\left(\sum_{i=1}^n x_i^2 \right)\left(\sum_{i=1}^n y_i^2 \right)- \sum_{i<j} (x_i y_j-x_j y_i)^2$$

Can you give me a hint what I could do?

user174558

@Huy Petersen's Linear Algebra; Jacobson's Basic Algebra I and II; Kaplan and Lewis's Calculus and Linear Algebra I and II; Protter and Morrey's A First Course in Real Analysis

Evinda

@DanielFischer Do you maybe have an idea?

Huy

is the PDF showing up as first google result an officially freely available version of Petersen's Linear Algebra?

user174558

@Huy It has been published as a book. Not the set of lecture notes. As usual, all can be obtained from Russian sites...

Huy

21:28

ah, I see

books of what topics are still missing for your holy list?

user174558

Too many to list here. The list will probably go up to 18.

Tobias Kildetoft

@MaryStar Hmm, if you take some the orbit of some $(b,c)$ and inside that take all elements of the form $(a,d)$ will these $d$'s not be an orbit of $G_a$?

user174558

I discovered that Springer sometimes sells cheaper than Amazon.

Huy

and most of content found on Springer is freely available for me

user174558

I watched Saw 1,2,3,4,5,6,7 and Wrong Turn 1,2,3,4,5,6. Extreme violence.

Huy

21:33

Only Saw 1 and 2 are watch-worthy from those movies.

user174558

I think I am going to watch Scream 1,2,3,4 and Final Desination 1,2,3,4,5.

user174558

I also watched Hostel 1,2,3.

Huy

normally I'd ask if you're mentally ill but

=(

on a more serious note, I'm really liking Marvel's Agents of SHIELD atm.

user174558

Don't worry. I am not a killer.

MickLH

@Jasper lol

user174558

21:35

@MickLH Hi!!!

MickLH

@Jasper Hey :) How's things?

user174558

The usual, which means bad.

MickLH

Shit!

user174558

But as usual, I am trying to make some progress with my OCD themes.

Mary Star

Why will these $d$'s be an orbit of $G_a$ ? @TobiasKildetoft

MickLH

21:38

@Jasper I know it's cliche, but I started on psych meds recently and I've been... well I don't know how to explain the trade-off but it's better for me

Tobias Kildetoft

@MaryStar Given such two, they have the form $(x.g,y.g)$ and $(x.h,y.h)$ where $x.g = x.h = a$. This should tell you how to find an element in $G_a$ that sends $y.g$ to $y.h$

user174558

@MickLH I have been taking them the past few years as well.

user174558

This is an announcement that the fifth edition of More Math into LaTeX has been published. Buy your copy today!

MickLH

What will go wrong if I were to build a group by removing the number $0$ from $\Bbb{R}$ and replacing with a symbol $0 = \frac{1}{\infty}$ ?

Tobias Kildetoft

@MickLH A group under addition or multiplication?

MickLH

21:46

I think I mean multiplication

Tobias Kildetoft

@MickLH what would the be the inverse of you new element?

MickLH

The symbol $\infty$

Tobias Kildetoft

@MickLH But you didn't add that

MickLH

sorry, I meant to also add that

Tobias Kildetoft

@MickLH Ok, and how do you define all the rest of the multiplications?

MickLH

21:48

@TobiasKildetoft This one is most surprising and I think conveys the theme I'm after: $0\cdot 0 = 0^2$

Tobias Kildetoft

@MickLH You had removed $0$

MickLH

But I meant the symbol $0$ not the number. So what I mean is that $\infty^{-1}\cdot \infty^{-1}=\infty^{-2}$

Tobias Kildetoft

@MickLH Yes, multiplying an element with itself squares it, that is the definition of square

Huy

so you get one new infinite cyclic group

MickLH

I am curious why I do not see this convention used, is there something fundamentally wrong with it?

Huy

21:51

why do you suspect it to be useful?

Tobias Kildetoft

@MickLH What convention? So far you have not actually defined anything

Huy

yeah, what happens if you multiply $\infty$ with some other real number?

and what happens if you multiply $1/\infty$ with one?

MickLH

I treat it somewhat like an imaginary component

Tobias Kildetoft

@MickLH That does not answer what it should be

MickLH

I mean that for real $x$ I simply define $x\cdot\infty = x\infty$

Tobias Kildetoft

21:54

@MickLH But then you need to further add these symbols

MickLH

It leads to what (to me) is an intuitive interpretation of infinitesimals

Tobias Kildetoft

So it seems that you have really just taken the direct product of $\mathbb{R}$ with the infinite cyclic group (generated by $\infty$)

Huy

yeah this "R with one more element" is getting quite a few new elements

Tobias Kildetoft

@MickLH infinitesimals are only interesting if you can also add and subtract stuff (and more importantly, compare stuff)

MickLH

I'm not sure what to make of this, I am getting the impression that it's not necessarily broken yet, but not considered a useful path to explore?

Tobias Kildetoft

21:58

@MickLH There are formal ways to add infinitely large and infinitesimal elements to the reals. For example the surreals do this. But it will quickly become apparent, that no element should be named $\infty$ as there is no good candidate for this

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