i want to continue with statistics

cool

what do you study

measure theory and linear algebra

and you?

the courses iam taking this semester is: real analysis course, a course in programming (scheme,python) and numerical analysis

but numerical analysis is starting next month

measure theory sound intresting

what do you want to have for occupation?

mathematican

Hello people

charles

yes

hi Charlie

when are you finished with your bachelor

almost

Danny do you know what a Twink is?

@Danny how are you Danny?

twink , yes i know

iam fine charlie

@Charlie how are you

Good

nice to hear that!

why do you ask twink

curiosity..

No one needs to know anything, @twink . We have google.

ok xD

I put my hands up in the air sometimes

are you guys open-minded?

i dont mind what you are, so long as you are happy!

@Twink define it

define what?

you have google

@Twink what matters is your definition, now. Once it's you who asked.

Charlie

?

@Charlie ?

you didn't ask for my definition of twink when I asked if you knew what a twink is

@Twink you used a pre-existing definition. Not one you created.

@marcelolpjunior sim?

the definition of open-minded exists already

@Twink though each one has a concept over it.

Everyone has prejudice. It is a lie who says that has not prejudice. I'm not receptive to a new idea at first. Not until I fully comprehend it and understand.

hi guys

ok so you're not open minded

jack

I'm cautious

come an

hey guys

I was just asking if you were tolerant with the twinks

i have a question about the equation of a plane, but its not really question worthy, more like a chat question

anyone interested to help me ?

shoot

thanks

so i have 3 points and want to find the equation

@Twink then you should have asked it at first.

lets say we have (0,1,1), (0,1,0) and (-2,-1,-1).

sometimes when i read about plane equation, the do like p3 - p1 and p2-p3 to find the v1 and v2

or some other others

ok but it was evident that I was referring to that since I had just asked if you knew what a twink is

but can i always do like, v1 = p2-p1 and v2 = p3-p1 ?

@Twink you assume too much

Dave came up with a good answer

I'm not sure what you mean by "can I"

it means is it correct

you can always perform those subtractions

is it kay to do v1 = p2-p1 and v2 = p3-p1

ok good, and then i use it with cross product

however, they may not always result in answers that are useful to you

I have to go

Twink nobody cares we all like you

have a good day guys

bye

thanks Danny

bye

Bye

@marcelo algum problema?

Jack

@Dave what are you aiming to do with a cross product?

What is the correct way to find v1 and v2 with 3 points

to eventually get the equation

maplesoft.com/support/help/Maple/…

first of all, if you have 3 points, it's not even guaranteed that there'll be an equation for their plane

if the three points are on a line

from that website they do v1 = p2-p1 and v2 = p3-p1, but from other web sites they do some other substractions

lets assume that the 3 points are not linearly dependant

personally if I wanted the equation for the plane, I would just write down three equations in three variables a, b, and c, plug in the coordinates for the points as the coefficients, and solve for a b and c

I've never been comfortable with cross products to begin with

okay so the cross product of two vectors lying on a plane results in a vector normal to that plane

yeah

if you need two vectors lying in that plane, you just take two subtractions of points on the plane

my question is : what is the proper way to compute those 2 vectors

it doesn't matter if you do p2-p1, p1-p2, p3-p2, or any other combination

okay but, i can go and substract and 2 combination of 2 points

okay thanks!

as long as you do two different ones

good!

and obviously not two different ones that are just the same points swapped around

yeah

ie. using v1=p1-p2 and v2=p2-p1 isn't any good

thanks its clear now,

yea,

thanks, i kind of 'knew the answer' but wasnt sure , and the exam is tomorrow,

thank you jack

that's fine

erm, ive got a rather dumb question: is the space of bounded continuous functions uniformly bounded?

@Charlie Como eu acho o valor de $a$ aqui $$(116+17^{17})^{21}\equiv a\pmod8$$

?

@Ryan consider the constant functions

oh, pointwise bounded then? in general, when a set of functions is said to be "bounded", do they mean pointwise (and uniformly so only if explicited indicated)?

@JackM thanks.

@Ryan you want to know if the set of continuous functions is pointwise bounded?

@JackM huh?

when you said "oh, pointwise bounded then?"

sorry, there should have been a question mark on that

@JackM i dont understand what you're saying. i was confirming your answer to my question (the space is pointwise bounded not uniformly)

I thought you were asking a second question

@JackM yes i was asking a new question. when a set of functions is said to be bounded, i should not assume the author means uniformly bounded, right? the set could either be continuous or not necessarily.

I don't know really

sets of bounded functions aren't something I've come across very often so far

@JackM Cool, thank you.

I guess it would depend on whether the author said "set of bounded functions" or "bounded set of functions"

@marcelolpjunior você esboçou algo?

Estou tentanto

:/ @Charlie

@JackM Hmm, im not so sure abt that. pointwise/uniform refers to the uniformity depending on the domain of the functions, not to the functions as points in the set

@marcelolpjunior tudo, eu nao acho isso muito facil. Tinha dificuldades também

Mas consegue me ajudar?

@marcelolpjunior hmm, não sei :/ vamos ver

@marcelolpjunior tem muitos exercícios similares

@JackM actually come to think of it, the space of bounded continuous functions isnt even pointwise bounded because the bound on the space would depend on the function (as per your suggestion to consider the constant functions)

@marcelolpjunior onde se fecha parênteses?

$\sum\limits_{n=1}^\infty \frac{\cos(x)}{x^2}$ is it possible to express this sum in rational numbers or known numbers such as $\pi$ or $e$

@Danny your latex isn't rendering

$$\sum\limits_{n=1}^{\infty} \frac{\cos x}{x^2}$$

you mean like that?

yes

i dont know why

I think it's the square bracket at the end

i cant edit

now

jack M any idea?

none

hehe

@Twink Open minded people really don't exist. There exist people who's open minded to a specific kind of stuff.

@Danny do you have ChatJax installed?

@GustavoBandeira but there are people who are receptive to new and different ideas or the opinions of others, that's being open minded

@robjohn are you open minded?

@Twink I think so, why?

Just asking...

I think algebraists have a right to exist :-)

lol xD

@robjohn why do you have an angry face in your profile picture?

that's casually contrary to my profile picture xD

@Twink You've asked about my avatar before, and we've discussed what is written in my profile. It is the mean square.

hmmm ok

@robjohn math.stackexchange.com/questions/507508/… can you take a look please?

Who do I have to blame for the fact that the Dedekind psi and digamma function both have the same signs? I bet it's Dedekind.

the one who used the signs later

The Greek alphabet was deficient

evening

GMT, yes

A nice problem: Show that for each $n$, it is possible to find integers $x$ and $y$ such that $\sqrt{x^2+nxy+y^2}$ is also an integer.

$y=0$, $x=1$?

did you mean for $x$ and $y$ to be non-zero?

@Daniel: Yes, actually I meant positive, but that word got lost somehow :)

Anyone available for a quick probability question?

@Codefun64: Just ask; don't ask to ask.

When I do that on most chats, I get no response ;)

en.wikipedia.org/wiki/… In regards to the "Four of a kind" probability, how would I change the function if I were drawing 7 cards instead of 5, for my hand?

@MårtenW it's incorrect to say that $ \text{the expression is integral} \Leftrightarrow x^2+nxy+y^2 \text{has repeated roots in x}$, isn't it?

I think it's only $\Leftarrow$

@Alyosha: I don't know what you mean with "repeated roots in $x$"..

I mean $n^2y^2-4y^2=0$, so that in the polynomial $(x-x_1)(x-x_2)=x^2+nxy+y^2$, $x_1=x_2$.

Actually, it's obvious, this is the case where $n=2$, the only case when it's $\Leftrightarrow$, not $\Leftarrow$.

Nice question

See? No love for statistics.

>:|

@Codefun64: I'm typing an answer.. Be patient :)

@Codefun64: The first two factors will remain the same, as those represent the four-of-a-kind. Then the question is how many variants we have for the remaining three cards. If you have chosen four cards, there should be 48 left, from which you should choose three. Therefore, I think you would get $\binom{13}{1}\binom{4}{4}\binom{48}{3}$.

But wouldn't that mean if I drew a five card hand, the last term would be binom(48,3)? The article says to use binom(12,1) * binom(4,1), not binom(48,1).

Unless, there's some sort of identity or rule I don't know?

@Codefun64: As far as I know, 12*4=48..

Well there's the rule I didn't know - you can multiple the binomial coefficient numbers together to condense it into one binomial coefficient instead of 2. I thought that binom(48,1) wasn't equal to binom(12,1) * binom(4,1).

@Codefun64: You are familiar with binomial coefficients? Well, $\binom{n}{1}=n$, which gives that $\binom{12}{1}\binom{4}{1}=12\cdot 4=48=\binom{48}{1}$.

@Codefun64: It is not possible in general! Only when the "lower" number is zero or one (or $n-1$ or $n$).

is numerical analysis useful?

@DonLarynx: yes.. why?

Is it true that there is a field with $2^n$ elements?

@leo: yes

do you have to be really good at combinatorics to be a good mathematician

combinatorics is so fugding hard

you can summon people to chat by typing their name, right?

@leo Yeah.

IIRC, there is a field $F$ with $|F|=p^n$ for each pair $(n,p)\in\Bbb N\times \{\rm primes\}$

(And one can prove every field is of this kind of cardinality, @leo)

@PedroTamaroff Yeah, all this follows from the Klein-Smith theorem.

@KarlKronenfeld Hmmm....

@KarlKronenfeld You silly. I have a question.

More seriously, you can induct on $n$, recalling that the group of nonzero elements is cyclic.

Justo estaba leyendo ese teorema. Pregunté porque la demostración es considerar $x^{p^n}-x$ sobre $\Bbb Z_p$, extender el cuerpo a uno donde este polinomio escinda, ver que este pol no tiene raíces repetidas y entonces el cuerpo será el conjunto de ceros del polinomio en esa extensión. No entiendo como esos ceros forman un cuerpo en el caso en que $p=2$. Por ejemplo, si $-a$ es un cero puede que $a$ no. Debe ser por otros métodos entonces...

Which is the Klein-Smith theorem? Google gives nothing

Hi all

@leo He was being funny.

@leo Pero $-a=a$, o no?

O por lo menos $-1=1$ en tal cuerpo.

What do you guys use to take notes? Pen and paper or a computer? If a computer, what software? I would really like to use a computer because I am not organized enough with paper notes (yes, I know that's my own fault), but I haven't really found a tool which I'm satisfied with. LaTeX notation is just too much work to type and it distracts me from the problem I'm thinking about.

spiral bound french paper large notebook, cheap black ballpoint

I buy the ballpoints in bulk

well, in boxes of 20

Just to clarify, I do not mean lecture notes. I mean notes on the topic you're working on.

I could never use anything else than pen and paper for lecture notes.

I occasionally type up thoughts on a computer

especially in the final stages of working on a proof

Heya @Pedro

@TedShifrin Hey there.

Something interesting came up in my linear algebra class today.

Ah?

@PedroTamaroff Oh, I didn't notice this ping, sorry.

@Szabolcs: LyX can be fast once you learn/configure enough keyboard shortcuts

but nothing really matches handwriting for note-taking

@TedShifrin Well, our assistance prof. said that if we obtain an equation satisfied by invertible matrices, then this equation is also true for every matrix.

maybe consider a tablet PC?

@KarlKronenfeld Let me tell this to Ted first =)

if this doesn't work I'm gonna feel so dumb

@Ryan right, that sequence isn't bounded because there's no way to simultaneously bound every element of it

@pedro: maybe if "equation" means something specific involving continuity ;)

As I guessed, my prof. told me this is a consequence of ${\rm GL}\;(n,K)$ being dense in $K^{n\times n}$.

@AnthonyCarapetis Yes, LyX was one of the things I was thinking of. It has the advantage that it's easy to transfer pieces to LaTeX later. I only looked at it brifely. I tried working with TeXmacs for a while, which is much more beautiful than LyX, but I found it quite buggy and the project seems to be half dead. Also transferring to LaTeX turned out to be much ore difficult than I expected.

Assuming the equation is given by something continuous, yes.

@AnthonyCarapetis Well, more formally if $f(A)=0$ where $f$ continuous, yes.

* glares @ Anthony *

sniped

:D

Yup @Pedro.

@TedShifrin My question is, what is the topology?

Standard topology.

My boss uses a tablet computer of about A4 size, with one of those magnetic pens. It's very nice, and combines the advantages of digital and paper, but also a very expensive thing.

The subspace topology in $K^{n\times n}$ I think

@TedShifrin Induced by the usual matrix norm?

@Szabolcs: Yeah, unfortunately it's hard to find them for less than ~$1200

He also told me one can prove $\overline{{\rm GL}\;(n,K)}=K^{n\times n}$ using the Jordan form.

Or just $n\times n$ matrices as $K^{n^2}$.

Yes, although you don't need Jordan form.

Hmmm

@TedShifrin Explainz?

That holds for arbirtary fields, right?

Need alg closed for Jordan, but we want that.

What does?@Fernando

@TedShifrin He told me one can "disturb" the eigenvalues slightly to make the matrix invertible but keep it near out original matrix.

$\overline{GL(n,K)} = K^{n\times n}$

I think one needs $n>1$

You can do that with triangularization. Nah @Fernando

You're right Ted. I was thinking it wrong.

@TedShifrin How?

If $K$ is a field of finite char, topology gets interesting.

@TedShifrin Oh?

Good exercise @Pedro (in chap 9 of my book) ... If eigenvalues are in $K$, you can change basis to make the matrix triangular. Way easier than Jordan.

@TedShifrin Yes, I know that.

Oh, wait.

You mean even if the matrix is not diagonalizable?

Then you can wiggle $0$ eigenvalues.

Yes, of course, not diagonalizable.

can somebody answer my question please? it has a 50+ bounty math.stackexchange.com/questions/507508/…

@KarlKronenfeld My question.

I have read in MO that it is fallacious to say that $\chi_A(A)=0$ by "evaluating" at $A$.

@TedShifrin

@TedShifrin What do you mean?

(Oh, $\lVert A\rVert \geqslant \lambda$ for any eigenvalue, yes? @Ted)

Bleh, @KarlKronenfeld is gone now.

@FernandoMartin Seems we're alone.

@PedroTamaroff: They showed that to us in my Linear Algebra course

(Hamilton-Cayley's fake proof)

@FernandoMartin Right, I wanna know why it is fake.

Can you link me the thread? I don't remember much about it

@FernandoMartin Here.

Well, I think the mistake is in plugging $A$ in place of $t$; $t$ is a scalar, not a matrix.

$p(A)$ is an $n\times n$ matrix, while the determinant of some matrix is always a scalar

I'll be back in a while

@anon

yeah?

@anon See the above =)

the char poly thing?

@anon Yeah.

what fernando says, you don't automatically know det(T-xA)=p(T) for matrices T

in a sense, it's obviously false, since the left is a scalar and the right is a matrix

What is $p$?

the char poly

@anon But the char poly is p(X)=det(XI-A) yes?

for scalars X

@anon Aha.

in order for the equality p(x)=det(xI-A) to make sense in the first place x has to be a scalar

if X is a matrix of unknowns for example you get a very different multivariable polynomial

@anon What do you mean by "very different"?

Oh, right.

Because $xI$ and $X$ are markedly different when $X$ is a matrix.

well, it's multivariable first of all, and linear in each variable secondly

@anon Aha,.

@anon I have another question, too.

We can define a map $\det:K^{n\times n}\to K$ as usual.

And we can define a map $\det :K[X]^{n\times n}\to K[X]$ too.

By the same $\sum_{\sigma \in S_n}\cdots$ thingy but with polynomial coefficients.