Mathematics - 2014-11-10 (page 4 of 4)

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11:00 PM

Today somebody wrote $1\cdot 2^5 = 1\cdot32=\frac{33}{25}$.

I like my math class.

Best part about algebra for me was getting to the finite fields chapter and being able to write $(a+b)^p = a^p + b^p$. It feels so....dirty :)

Alizter

@KajHansen Was it mod 6 or something?

well mod 2 is trivial

Kaj Hansen

That holds in fields of characteristic $p$ @Alizter

Alizter

@KajHansen same p?

well y naught

@KajHansen the dream has come true!

Ted Shifrin

11:02 PM

true for any $p$-power, @Kaj, as you know.

Kaj Hansen

Indeed @Alizter

Frobenius endomorphism

In commutative algebra and field theory, the Frobenius endomorphism (after Ferdinand Georg Frobenius) is a special endomorphism of commutative rings with prime characteristic p, an important class which includes finite fields. The endomorphism maps every element to its p-th power. In certain contexts it is an automorphism, but this is not true in general. == Definition == Let R be a commutative ring with prime characteristic p (an integral domain of positive characteristic always has prime characteristic, for example). The Frobenius endomorphism F is defined by for all r in R. Clearly th...

Alizter

thats like a multiplyy thing with characteristics then like fermats little theorem then wow.

Hmm my thoughts are messier than I thought.

Kaj Hansen

Oh lord, I didn't want that massive thing to go with my link, but okay.

Alizter

@KajHansen Don't take your wik out in public!

@TedShifrin says it is wrong.

Kaj Hansen

hahaha

Ted Shifrin

11:04 PM

smacks @Alizter ... I didn't say it was universally wrong, dammit.

Alizter

@KajHansen endomorphism is a made up word.

It is not cited.

expects slaps/smacks

Kaj Hansen

All words are made up words, are they not?

Alexander Gruber

does $\left( C_4\rtimes C_3 \rtimes C_2 \right) \times \left( C_8 \rtimes C_7 \rtimes C_2 \right)$ have an element of order $2\cdot 3 \cdot 7$?

it doesn't right?

Alizter

@AlexanderGruber draw the cayley graph :P

Studentmath

Oh gotta start doing toplogy

Américo Tavares

11:06 PM

The above meta question, which was not meant to publicize one of my answers, got a downvote.

Alizter

@Studentmath Do you get to apply it to molecules and stuff?

Alexander Gruber

we'd need $C_3 \times C_7$ and then some part of the 2

Studentmath

@Alizter haven't had the chance yet - I did have some graph applications on molecules, but I really didn't deal with chemistry lately

don't would be better grammar wise

Alizter

@Studentmath I remember someone (I think Mr. Yuan) had a book about Lies and Hydrogen atom

Lie algebras

not Wikipedia lies

ducks

Alexander Gruber

no it doesn't. because the centralizer of the $3$ involves only the $2$'s on the right, and they don't commute with the $7$.

Studentmath

11:08 PM

I always trusted wikipedia

It seems realiable

@Alexander we're always glad to be of service

Alexander Gruber

Yup.

Kaj Hansen

@ZachSaucier, one way of doing it is to show that your vector field is the gradient of some function.

Alexander Gruber

okay new question

how do i make a group so that $\pi(|G|)=\{p,q,r,s\}$ so that there are elements of order $pqr$, $qrs$, $prs$, and $pqs$, but no element of order $pqrs$

Studentmath

I will get my useless self back to studying and solving Ted's questions.. g'night!

Zach Saucier

@KajHansen Ya, thanks. I just remembered we covered it in class today

Kaj Hansen

11:15 PM

The super-duper useful thing is that line integrals over any path connecting point $a$ to point $b$ will come out the same.

In conservative vector fields that is.

Alexander Gruber

does this work? $(C_p \rtimes C_s)\times (C_q\rtimes C_s) \times (C_r \rtimes C_s)$

Fargle

@AlexanderGruber What does $\pi(|G|)$ denote?

Alexander Gruber

@Fargle prime divisors of $|G|$

Fargle

@AlexanderGruber ohhhh, okay, haha

Alexander Gruber

I am pretty sure the above example works.

Hippalectryon

11:31 PM

Is some dude in the chatroom downvoting half the new questions ? or is it someone else ?

Ashley Davies

Hey guys, I've got a query that doesn't justify a question really - I'm reading a textbook on integration and during explaining integration by substitution it states something along the lines of du/dx not being a fraction per sé but that multiplying both sides by dx is okay in that scenario, stating that there's a branch of mathematics that in part shows that this is a valid operation but saying that it's more complicated than the textbook is going to go into

what branch of mathematics would that be? :P

Fargle

non-standard analysis

Mike Miller

@Hippa Why would you assume it's someone in chat...?

Hippalectryon

@MikeMiller I didn't. I just thought that if it was, it would have been easier :D

Ashley Davies

well, it says "The last example included the statement 'du = dx.' Some mathematicians are reluctant to write such statements on the grounds that du and dx may only be used in the form du/dx as a gradient. This is not true, and there is a well defined branch of mathematics that justifies this but it is well beyond the scope of this book"

Ooh, thanks! :)

Mike Miller

11:42 PM

@Ashley Thwre have been quite a few discussions about the "meaning of" dx on main. You mightjust search dx.

Ashley Davies

Ah cool :) I've always struggled to understand how it works algebraically and I assumed it was down to bad teaching that it wasn't clear - I guess not then

Alexander Gruber

Okay number theory question

masfenix

Hey all, was wondering if anyone could just explain what this question is asking me to do. (i already have solutions, but its impossible to understand the solution to a question i dont fully grasp). Anyways, here goes: Prove that there exists continous $f : \rightarrow [0, \infty)$ such that $\int f = 1$ but $limsup_x f(x) = 1$.

Alexander Gruber

let $s$ be a prime number. There are infinitely many primes $r$ such that $s\mid r-1$.

are there infinitely many primes $r$ such that $s \mid r-1$, and also, $r \mid s^n-1$ for some $n$?

Ashley Davies

Ah so even though dy/dx can behave similarly to a fraction like with the reciprocal and chain rule it isn't actually one and these are just special cases and it's still just a notation?

Mike Miller

11:49 PM

@Ashley One can turn it into a fraction in the field of nonstandard analysis. One doesn't really gain anything by doing so, though...

You should not think of it as a fraction except possibly for the sake of intuition.

Ashley Davies

Awesome, feel a bit more confident using it now :) I find it useful to imagine them as fractions for using the chain rule which I guess comes under intuition - It eases the mental effort just imagining it to be a number of fractions that I'm trying to cancel down to the result I guess

Mike Miller

It's good intuition for now @Ashley but when you do multi variable calculus, the chain rule will look much weirder!

Ashley Davies

Maths was so much easier back when we just had to learn the multiples of 2 in a week :')

Mike Miller

But so much less interesting, too!

Ashley Davies

True! I think I have to teach myself the basics of multi-variable calculus for a course I'm hoping to do later in the year, but I don't think it goes into much depth so I hope it's not too difficult.