Mathematics - 2017-07-05 (page 7 of 10)

Alessandro Codenotti

19:00

oh, while writing out my approach I also found the mistake

Semiclassical

Neat.

Alessandro Codenotti

that was the fastest help I ever received in chat

Semiclassical

The fastest help generally is the one you don't have to ask for :)

What was the approach, out of curiosity?

Krijn

The fastest help generally is to be very consistent and claim the opposite is true

Balarka Sen

@Krijn Well, no, not at all.

Typhon

19:04

@Semiclassical any word on that post I wrote?

Alessandro Codenotti

The CDF of $M_n$ is $F(t)=(1-e^{-t})^n$ (the max is below $t$ if all $n$ exponential distributions are below $t$), so $\Bbb P(M_n\le t+\log n)=(1-e^{-t-\log n})^n$ and that converges pointwise to the desired results for all $t$ for $n\to\infty$

Typhon

or were you only speaking hypothetically when you said it needed attention?

Semiclassical

haven't had a chance to look at it, no.

Typhon

fair enough

Semiclassical

@AlessandroCodenotti Neat.

Typhon

19:07

@SteamyRoot algebra ought to be illegal

Semiclassical

"If we outlaw algebra, only criminals will have algebra!!!"

Astyx

You'll see gangsters dealing algebra in front of schools

Typhon

but algebra is evil!

@Astyx that just sounds so wrong out of context.

Alessandro Codenotti

hey kids, wanna buy some integral domains?

Typhon

omfg

lmao

Alessandro Codenotti

19:10

that's good stuff, no zero divisors

Mike Miller

@BalarkaSen Fastest, not best

Semiclassical

Got any Euclidean domains? I only pay for premium, man

Mike Miller

@TobiasKildetoft In any case I should say I've thought a lot about this and various things and if you want to talk about it feel free to email me

Astyx

In the end, once you're addicted, they give you rings and low quality structures

Alessandro Codenotti

@Semiclassical I have some PIDs, is that good enough?

Tobias Kildetoft

19:11

@MikeMiller Thanks. I first need to read some more to have actual precise questions about it

Mike Miller

Fair enough

Though I (most?) find the equivariant literature hard to navigate

Tobias Kildetoft

The reason for my interest is that there have been some recent work finding uses for it in the representation theory of algebraic groups in positive characteristic

Sha Vuklia

Guys, my teacher says that
$$
O(x^5y^5)=O(\Vert(x,y)\Vert^{10}).
$$
We know that $\Vert(x,y)\Vert^{10}=(x^2+y^2)^5$. Was 10 really the smallest we could have chosen? I don't really see why he chose that number specifically.

Tobias Kildetoft

But what precisely it is they use is hard to figure out when you know as little geometry as I do

Semiclassical

Toss in a magma and I'll take it, my friend is into some weird s***

Alessandro Codenotti

19:13

lol

Sha Vuklia

he uses it here

Mike Miller

@TobiasKildetoft Gotcha. I'm especially busy this month but I'm glad to/interested in taking a look

Sha Vuklia

(It's just the Euclidean norm btw, to the power of 10)

Semiclassical

@ShaVuklia Hmm, seems like that should be equivalent to $O(xy)=O(x^2+y^2)$

Alessandro Codenotti

if you're a physicist $\sin x=x$, don't worry about that annoying $O()$ stuff

Semiclassical

19:15

which seems sensible enough

@AlessandroCodenotti don't go spreading lies and slander

Sha Vuklia

lol was that directed at me @Alessandro :P

@Semiclassical yea true, I'll try to show that then

Semiclassical

It is true, though, that physicists typically use O(x^n) as short-hand for "there's more terms in this series" rather than actually making a claim about big-O.

Sha Vuklia

true

Faust7

$\lim_{n \to \infty} \frac{a^{n}}{n!}$ is there some a that would make this a number?

other than 1 or 0

Semiclassical

Suppose you take the ratio of successive factors. What does that give you?

Faust7

19:24

1?

Sha Vuklia

@Faust do you mean you want $a\neq 1,0$ of you want the limit to be $\neq 0,1$?

Semiclassical

nah. I mean $(a^{n+1}/(n+1)!)/(a^n/n!)$

Alessandro Codenotti

yes. What's an $a$ such that this limit isn't a number?

Semiclassical

Probably wasn't the clearest way to put it.

Faust7

what

no i want an a so that the limit isnt infinity

but a cannot be 1 or 0

Sha Vuklia

19:25

the limit is never infinity

Semiclassical

in other words, by what factor does the function change as you go from n to n+1?

Sha Vuklia

i think it's zero always, in fact, which you can prove using semi's hint

Alessandro Codenotti

what's an $a$ such that this limit is $+\infty$?

hi @dami

Faust7

somehow n! grows faster than exponential eventually? for small numbers it gets its ass kick

Semiclassical

@AlessandroCodenotti Not really sure what you've got in mind.

I mean, one can easily enough make the limit fail to exist.

(say, a<0)

But that's not the same as limit -> infinity.

Tobias Kildetoft

19:28

@Semiclassical why would that make it fail to exist?

Faust7

cause the lim sup and lim inf wouldnt be equal

Semiclassical

Because it's oscillatory with n.

Alessandro Codenotti

The same thing you've got in mind (I think), but I wanted to know why he thinks that some $a$ will give limit infinity

Semiclassical

One subsequence goes to -infinity, another to +infinity.

Faust7

anyway semi is right ratio test shows it always goes to 0 for positive a

Alessandro Codenotti

19:29

@Semiclassical it's still going to 0

Semiclassical

deeerp

Sha Vuklia

yea @Semi by your exact theorem

Semiclassical

I'll just be over here, extracting my foot from my mouth.

Sha Vuklia

@Faust $\lim\left\vert\dfrac{a_{n+1}}{a_n}\right\vert<1\implies\lim a_n=0$. Now just write out this ratio and you're done. also negative $a$

Semiclassical

So yeah, goes to zero for all a.

@sha absolute value of that, but yes

Alessandro Codenotti

19:31

lol, we all have brainfarts (actually I had a worse one with that probability exercise earlier, but realized it while typing)

Sha Vuklia

whoa

Faust7

yeah i got it awhile ago >.>

thanks though

Semiclassical

ah, latex issues.

Faust7

was just a random thought

Semiclassical

how much they bedevil us

Sha Vuklia

19:31

lol the struggles

Faust7

i had anthor wierd question

Alessandro Codenotti

we love weird questions

Semiclassical

Another way to to be sure that it goes to zero: the Taylor series for e^a converges everywhere in the complex plane.

And that wouldn't work if a^n/n! didn't converge to zero.

(That's not a proof, that's just a consistency check)

Typhon

heh

@Semiclassical Hey kid, want to buy some algebraic extensions?

Semiclassical

man, I could go for a whole chain of them right now

been craving some of that galois action

Typhon

19:38

well too bad

its illegal

;-)

Faust7

if something is eventually monotonic and bounded can i say it converges?

Typhon

@Faust7 convergent at what point?

Faust7

say its bounded above and eventually monotonic increasing sequence

can i say it converges to the upper bound?

or do i need to use cauchy?

Typhon

@Semiclassical yup. $Z[\sqrt{3}]$ is for you man!

Faust7

@Typhon lol sorry had to catch an airplane yesterday

Typhon

19:40

hmm?

Faust7

our generators discussion that got cut short

that i was incredibly poorly explaining

:P

Typhon

ah

why arent sqrt{3} and 1 generators?

all extended integers are of the form a + b\sqrt{3}

Faust7

because the dont generate the entire group

they*

for example 1

can produce all of Z

but nothing with $ a+b \sqrt3 $ where b is non zero

Typhon

$1 + 1 +1 +1 + 1 ... + \sqrt{3} + \sqrt{3} + \sqrt{3} + \sqrt{3} + \sqrt{3}$?

Alessandro Codenotti

Consider the set of sequences of natural numbers $\{a_n\}$ with "polynomial growth", in the sense that there is a polynomial $g(n)\in\Bbb N[x]$ such that $a_{n+1}-a_n\le g(n)$ for all $n\in\Bbb N$. What's the cardinality of this set?

Faust7

19:43

so 1 is not a generator of the exstended ring

Typhon

but 1 is a unit and you said units were generators...

Alessandro Codenotti

@Balarka this polynomials question is related to the discussion from this morning

Faust7

you said that u found a list of all the units

Typhon

ok?

Faust7

there will be 2 of those in that form that are able to generate your whole group

Typhon

19:44

why 2?

Faust7

because Z has only 2 generators

and exactly 2

Typhon

and why can I not just pick the pair 1 and \sqrt{3}?

you're picking a pair

all extended integers are an integer linear combination of 1 and \sqrt{3}

Faust7

yes

its very well possible that

Typhon

so... 1 and square root of 3 together form a generator?

Faust7

$1+ \sqrt 3$ and $-1 - \sqrt 3 $ are your two generators

could*

but u need to prove that

Typhon

19:47

uuuh

those aren't even units

Faust7

well youll need to find a unit that looks like them

that will be your generator

Typhon

you're being unclear

I can add 1 and the square root of 3 together to get your "generator"

Faust7

ok ill write it up gimmie 10 minutes and ill find your generators n prove it but i need a cup of coffee

Typhon

so doesn't that make them a generator

are you saying the generator is always just one element?

Faust7

no

w89

yes

one element applied over and over again with a binary operation

the resuly must yield the entire group though

Semiclassical

19:49

I forget, what kind of object is Z[sqrt(3)] here?

Typhon

so you cant put 1 and sqrt{3} together as a pair?

Faust7

its a ring

Typhon

@Semiclassical it has a division theorem

Eric Silva

one element does not generate $\mathbb{Z}[\sqrt{3}]$

Faust7

adjioned with root 3

Typhon

19:51

@Faust7 the units are all numbers of the form $(2 + \sqrt{3})^n$ and their conjugates and negations

n is any integer

note:

n can be an arbitrarily small negative number

@Faust7 generator is the element closest to 0, no?

Semiclassical

negative number, or specifically integer?

Eric Silva

usually you define a generating set $S$ of a ring $R$ as a set with the property that no proper subring of $R$ contains all the elements of $S$.

Typhon

@Semiclassical n is any integer

which includes negatives

Eric Silva

under this defn $S = \{1, \sqrt{3}\}$ is a generating set for $\mathbb{Z}[\sqrt{3}]$.

Typhon

in fact, you cannot take the successor of an extended integer

Semiclassical

19:52

right.

Typhon

that is why norms are used for ordering

generator cannot be one lone element

as that would imply it is the element succeeding 0 in the positive or negative direction

Faust7

Generating set of a group

In abstract algebra, a generating set of a group is a subset such that every element of the group can be expressed as the combination (under the group operation) of finitely many elements of the subset and their inverses. In other words, if S is a subset of a group G, then 〈S〉, the subgroup generated by S, is the smallest subgroup of G containing every element of S, meaning the intersection over all subgroups containing the elements of S; equivalently, 〈S〉 is the subgroup of all elements of G that can be expressed as the finite product of elements in S and their inverses. If G = 〈S〉, then we say...

Semiclassical

well, you do have $\sqrt{3}^2-\sqrt{3}^0-\sqrt{3}^0=1$ here. But somehow I doubt that counts.

Typhon

@Faust7 and I said $(1,\sqrt{3})$ is a generating set.

Semiclassical

Generating set of a ring, not a group.

Eric Stucky

19:55

Eric: what is a proper subring of $\Bbb Z[\sqrt 3]$ that contains $\sqrt 3$? Do you not assume rings have identity?

Faust7

ok gimmie 10 minutes ill type it up then try n explaining this is getting us nowhere ^^

Semiclassical

oh lawd, rings vs rngs

Typhon

@Semiclassical rings are groups?

and I fail to see the relevance with this in factoring stuff

Semiclassical

Eh, my point is that bringing up the wiki article for the generating set of a group rather than the one for a ring seems a bit strange

Faust7

he doesnt know what a ring is

or do you?

Eric Stucky

19:58

LOL you're serious

Faust7

no not getting dragged back into the vortex lol

Eric Stucky

Typhon, you really don't want to define the generating set of a ring as a generating set of its underlying group. Otherwise $\Bbb Z[x]$, for instance, is not finitely generated.

Eric Silva

@EricStucky, oh I usually don't assume it has an identity

Eric Stucky

Okay, figured that was the issue :)

Eric Silva

i guess you can say the $R$-submodule generated by $S$ is $R$ if you do assume it

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$Mathematics$ Mathematics