binary exponentiation algorithm ?

nvm I made my argument not reliant on it.

anyone would like to check my argument ?

I am proving the following.

o.o

Let R be a ring and consider new ring M = R[x]. Let $f = a_0 + ... a_n x^n \in M$ Then if $a_0,...,a_n$ are nilpotent then f is nilpotent.

here is my proof

um

Suppose $a_0,...,a_n$ are nilpotent, then there exists $i_1,...,i_n$ : $a_0^{i_1} = ... = a_n^{i_n} = 0$. Let $N := i_1 ... i_n$, then my claim is $f^N = 0$.

did you forget an "if" somewhere

eah

yeah

your $N$ is very large

if deg f = 0, then $f = a_0$, then $N = i_1$ so $f^N = a_0^{i_1} = 0$.

yes

Suppose the result is true for $deg f \leq n$, then we want to prove it is true for n + 1. $f = (a_0 + ... + a_nx^n) + a_{n + 1}x^n = u + s$

is R commutative ?

$(u + s)^{N} = \Sigma_{k = 0}^{k = N} (N k) u^{N - k} s^k$

yeah

(n k) represent n choose k

yes

suppose N - k and k are both less than N then N - k + k < N which contradiction.

So one of them must be bigger or equal to N.

what

sorry I adjusted the indices

N = max{i_1,...,i_{n + 1}} remember ?

what²

?

which squared ?

i am doubly confused now

why ?

because your N was the product of the ai

and because your "suppose N-k and k are blablabla" message makes no sense

my N is product of the $i_{j}$

._.

?

the i_j yes

still

product <> max

I am raising (u + s) to the power of big N

right and I am just using bionomial theorem ?

is big N the product of the ijs or the max of the ijs ?

big N is the products of ijs

I don't think max works

@Adeek so I should forget this ?

forget what ?

I am confused

when you said "N = max{i_1,...,i_{n + 1}} remember ?"

ohh

sorry my mistake

$N = i_1 ... i_{n + 1}$

alright so now I'm still stuck trying to understand why among N-k and k, one of them is greater than N

well by bionomial theorem their addition must add up to N right ?

yes

um

if both of them are less than N then we will have what ?

I wouldn't say it's because of the binomial theorem

why ?

a-b+b = a is like, a basic rule of rings

I hope you see there's something wrong ?

7-2<7 and 2<7, but maybe I don't understand

ohh

I am not sure

I think if N-k <= N and k <= N, then N-k+k <= 2N

yeah

I guess there is no contradiction here.

but I don't think that's very useful for your problem

yeah

n-k will always be smaller than n, for positive k

also

@mercio I want one of them to be bigger than N, because if that is the case then we will get 0 * bla or bla * 0

and n-k+k=n will always be smaller than 2n, for positive n

it looked like you were doing a proof by induction

yeah

so how come you've never said "suppose we have shown that (a0+... + a(n-1)x^(n-1))^M = 0 for some integer M"

I said suppose the result is true for $deg (f) \leq n$

that is my claim is true

hm ? ah maybe

but then

you are in a tricky spot where your N can mean 2 different things at once if you're not careful

Yeah

I need to indice the N

I guess I should indice the N according to the degree.

that is $N_n := i_1 ... i_n$

also you may want to do things one at a time

what do you mean ?

1. show that the set of nilpotent elements is stable by +

2. do the induction

or even better

2. show it that it is an ideal of R

okay

I guess if I think locally that is consider the ring M / (a_1,...,a_n)

if I show that this has no nilpotent element then I am done ?

your trouble came from having the same name N for two different things so doing things separately can help you

what no, why would it have no nilpotent element

oh wait no

and when i said an ideal of R I meant an ideal of the ring we're in, so of R[X] in your problem o.o

yeah

I see

I have to curl up for the night

okay thanks @mercio

I will go buy a sandwich I am so hungry

hi

hi

I want to find the quotient map of an abelian group to a quotient

so for example

if G = $ \mathbb{Z}^3 $

and H = $(15,10,6)$

I can see by normal form that $G/H$ is isomorphic to $\mathbb{Z}^2$

but if i want to find the matrix $P:G \to G/H$ I need to get a basis for $G/H$

right? it's been ages since i've worked with these groups

sorry $(15,10,6)$ generates $H$ I mean

oh @Ramanujan is here, you'll be able to help! haha

@MikeMiller sorry to bother you but i think i'm just being really stupid again - is the above an obvious thing?

Although with lattices if I have an inclusion $A: H \to G$ can I just pick any $P$ such that $PA=0$ and $P$ has rank equal to $G/H$ ?

by dimensions etc

must be the quotient with respect to SOME basis, am i right?

yes i think i'm right, up to an automorphism of $G/H$

which is basically choosing a basis in this case

so in that example above we just pick a matrix with rank $2$ such that it sends $(15,10,6)$ to 0

would love any comments or corrections from anyone :)

Can I think of a limit like this?

$$\lim_{x\to a}f(x)=\mathbb R(f(a\pm\epsilon))$$

where $\epsilon=1/\omega$ is a hyperreal (en.wikipedia.org/wiki/Hyperreal_number) and $\mathbb R(\cdot)$ truncates the result to the nearest real number?

well, goodniight

Can someone give me a hint to find $f'$

Does anyone know a function that is infinitely differentiable with a horizontal asymptote in both directions?

(other than 0 as that is trivial)

x exp(-x^2)

@arctictern is that for me?

thanks.

@SimplyBeautifulArt You need to specify that the real closest to $f(a+\epsilon)$ is the same no matter which infinitesimal $\epsilon$ we choose (not just $1/\omega$ and $-1/\omega$). But that happens iff the limit exists, and if it does, that is indeed the equation for it. (Cont'd)

Two notes: Normally the closest-real function is written ${\rm st}(\cdot)$ (from "standard part"). Also, you should have $f^*$ instead of $f$ on the right-hand side of that equation: $f$ has domain $\Bbb R$, and $f^*$ is its hypperreal extension with domain $\Bbb R^*$ (the hyperreals). The fact that this extension is always defined is part of the reason that hyperreals are so useful.

@arctictern around ?

I just want to check a argument with you.

maybe

Let K be a field. Prove that R = k[ [x] ] is a local ring. I proved that $f(x) \in k[ [x] ]$ is a unit iff $a_0$ constant term is non-zero.

Let $M = \{f \in k[ [x] ] : a_0 = 0\}$ Then for all $x \notin M$ x is a unit.

therefore M is a unique maximal ideal thus we are done.

good ?

@SimplyBeautifulArt Also note that the existence of the hyperreals relies on the axiom of choice — or, at least, the ultrafilter lemma (a weaker version of the axiom of choice but still independent from ZF).

for all $x\not\in M$? aren't you already using the letter $x$?

but anyway yes good

oh I guess $s \notin M$

@arctictern Now I have to prove the following result when I go home. Let A be a ring and R = A[x]. Then $f = a_0 + ... + a_nx^n \in R$ is a unit iff $a_0$ is a unit in A.

Brb I will go home.

@SimplyBeautifulArt By the way, true or false: A function $f$ with domain $\Bbb R$ is continuous iff, for every $x$ and $y$ where $x$ is infinitely close to $y$, we have that $f^*(x)$ is infinitely close to $f^*(y)$.

I'll spoil it for you: That's false. That's the definition of uniform continuity, not continuity!

For example, take $f(x)=x^2$. Let $N$ be an infinite hyperreal. Then $N$ and $N+\frac1N$ are infinitely close to each other, but $N^2$ and $(N+\frac1N)^2=N^2+2+\frac1{N^2}$ are not (since their difference, $2+\frac1{N^2}$, is not infinitesimal).

Hey folks

Zup,

$\lim{\sqrt[n]{\frac{\arctan{n}}{1+e^n}}}$

Could someone shine some light on solving this?

@BernardoMeurer Break it into $\lim\frac{\sqrt[n]{\arctan n}}{\sqrt[n]{1+e^n}}$

and remember that $\arctan n\to\frac\pi2$

@AkivaWeinberger In regards to my question, is it a no-go to write such a statement in a thesis?

@AkivaWeinberger Yeah, but then the root of that will go to 0, no?

I'm not the person to ask, sorry.

(of arctan that is)

@BernardoMeurer No, it will go to $1$.

What's $\sqrt[100]{2}$, for example?

It has to be more than $1$, because the hundredth power of something less than one is even smaller.

Derp

Of course!

(Apparently $\sqrt[100]{2}\approx1.007$, according to my calculator)

and the bottom goes to $e$ as is clear

Sweet

got it :)

Thanks!

For the denominator, to be rigorous, I think you could say something like, $e^n\le1+e^n\le2e^n$ for $n\ge0$

so $e\le\sqrt[n]{1+e^n}\le e\sqrt[n]2$

and the left and right go to $e$, so by the squeeze theorem, the middle also goes.

Oh, another one I was having trouble with earlier

But, yeah, like you said, it's kind of intuitively clear that $\sqrt[n]{1+e^n}\approx\sqrt[n]{e^n}=e$ and that it goes to $e$.

$$\lim{\frac{2n! + 3^n}{n^{50}+n!}}$$

I can see why it's $2/1$

Yeah, that'd be my guess also

But I don't know how to calculate it

I can argue that the factorial will grow much faster than the exponent and the n^50, so those will become insignificant as we get to infinity

Try dividing numerator and denominator by $n!$ (since it's of the fastest-growing order)?

Lemme see

$\lim\frac{2+3^n/n!}{1+n^{50}/n!}$

Yep

Now it's clear

And then if you can prove that $\frac{3^n}{n!}$ and $\frac{n^{50}}{n!}$ go to $0$, you're good.

Since it would become $\lim\frac{2+0}{1+0}$.

@AkivaWeinberger $1+0$ you mean?

Yes. That. Sorry.

:)

Nice, got it

In general with fractions like that it's a good idea to divide by what you think is the fastest-growing function in the fraction.

@AkivaWeinberger Why is $\lim{e^{1-x^2}} = 0$?

Ah

derp

So subrings of Noetherian rings aren't usually Noetherian

But I guess summands are? ($R\subset S$, $S$ noetherian is a summand if there is a $R$ module homomorphism $\pi:S\to R$ such that every point of $R$ is fixed).

Why is this? Is the kernel an $S$-module somehow?

@FTem because if S->R is onto then any ascending chain in R lattice corresponds to an ascending chain in S.

@arctictern But we don't know that $S$ is Noetherian as an $R$-module, right?

Wouldn't the correspondence theorem only work for $R$-modules? We only know that $S$ is Noetherian as a ring so as an $S$-module, right?

hey @arctictern

I am having troubles for the following proof.

Suppose f is nilpotent iff $a_0,...,a_n$ are nilpotent.

I proved the forward direction

I am having troubles for the reverse direction

I got it I think

@arctictern would you like to hear the argument ?

If an editor edited a question a second time, changing for a second time (after it got rerolled) the original meaning and intents of the question. What should be done?

Because clearly it was not enough to reroll the edit.

why is the maximum-column-sum norm larger than the spectral norm of a matrix?

i figured it out. gershgorin circle theorem gives you this

@Lepidopterist can you explain me the gershgorin circle theorem?

hi

Hello world!

@KasmirKhaan hi

@TedShifrin can you explain to me how Riemans sphere works in locating points?

hi @Socrates

I worked hard vs I hardly worked

Hello, i was doing a question on real analysis and saw this notation {x-[x] : x is a real number} , what does [x] means? It does not look like the floor or ceiling of x to me.

@yh05 I think it denotes the equivalence class of $x$

@Perturbative hmm the question didn't mention a particular equivalence relation on real numbers

Either floor or nearest integer

oh god

hi balarka sen

I have a question about intersection of a plane with riemans sphere

can you tell me when we get a circle and when a line ?

You always get a circle and you never get a line

Well, sometimes a point

hmm that cant be right

because the two options i have on my question is line or circle

Intersection of a plane with the unit sphere in R^3 can't be a line, because it has to be compact. A line is not compact.

Either that or you didn't quote the question right

hmm am doing complex analysis

its a Riemans sphere

But that makes no sense though. A Riemann sphere is P^1. You are intersecting it with a plane... where? What's the ambient space?

maybe a plane through the north pole

that doesn't answer my question at all

I did not understand it am new on this topic

Well, unless you have a sensible question I can't answer it

Quote the exact question you are given

well the image will be on the complex plane

ok ill post it its from staff snider

@Alessandro How did your exam go?

Consider the intersection of the plane 6x+3y+7z =6 with the Riemann sphere. Does this curve correspond to a line (L) , or a circle (C) in the complex plane?

In the complex plane. That's the crucial part you did not mention.

So after you intersect the plane with the Riemann sphere (thought as the sphere in R^3) and you get a circle on the Riemann sphere. Now you project through the north pole to the equatorial plane.

That is a line if and only if the circle on the Riemann sphere passes through the north pole.

@BalarkaSen I'm waiting my turn to do it

I still have a couple of people in front of me

Ah. Wish you a good luck!

How do I know if it passes or not ?

and thanks for the help :)

Thanks

@AlessandroCodenotti Good luck Alex

@Jacksoja You compute it. The unit sphere in R^3 is x^2 + y^2 + z^2 = 1 and the north pole is (0, 0, 1). Does the plane 6x + 3y + 7z = 6 pass through the north pole?

If the plane passes through the north pole so does the circle on the sphere. If it doesn't then neither does the circle.

i think it creates a circle if I understood correct

Why?

if that 7z were 6z then it would be a line right?

That's right :) The plane as it is does not pass through the north pole.

If it was 6z then it would indeed be a line.

okay I think I get :D thanks

I need more practice tho because i cant visualise this yet

@TobiasKildetoft we were discussing whether the existence of a highest trajectory places a ceiling on the growth of the Collatz Conjecture. Every path is an ordering of the odd integers of which the highest trajectory is the maximal such ordering: math.stackexchange.com/questions/2109281/…

@TobiasKildetoft If every maximal ordering reaches 2^n then every number beneath it has a finite number of steps before it would have to cross that ordering, which it cannot do, since every upwards step of any trajectory is exactly parallel to the maximal ordering.

@TobiasKildetoft therefore it must move forwards with respect to the ordering towards 2^n. It only remains to prove that 2^n is orthogonal to the ordering, which it clearly is, away from the vicinity of the number 1

@RobertFrost The answer you got there already states that this thing being an ordering is equivalent to Collatz

@TobiasKildetoft No, that states it's an ordering if and only if there is no non-trivial loop.

@TobiasKildetoft there are two alternatives to Collatz. One is that the exists a non-trivial loop, the 2nd is that there is some path ascending to infinity.

@TobiasKildetoft In this chat I'm only arguing no path may ascend to infinity. The ordering was the part of my argument I was missing; I wanted to get it clear and check it first.

@TobiasKildetoft p.s. you're underestimating me.

I still have no idea what you mean by a number crossing the ordering. Clearly some of the possible trajectories pass each other.

@TobiasKildetoft Arrange the ordering created by 3x+1/2^n going from right to left, and the ordering given by the number of factors of 2 in any number from bottom to top, so odd numbers are at the bottom and even numbers having high powers of 2 at the top. The function (3x+1)/2 runs from right to left and slightly upwards on this map. Place the line 2^n on the left axis and it should become clear that any application of the Collatz function moves unambiguously to the left; towards 2^n.

@TobiasKildetoft they can only pass each other if they move unambiguously to the left in the ordering created by f(x)=(3x+1)/2

@RobertFrost What do you mean by arranging from right to left? this is not going to be a total order, is it?

If numbers are connected then it is a total order

What do you mean by connected?

@TobiasKildetoft Comparable

Comparable by means of the ordering

@RobertFrost Sure, but why would they be?

@TobiasKildetoft because they may only not be, if there is a non-trivial loop or a path which ascends to infinity. In testing whether a path ascends to infinity we may assume that there is no path which ascends to infinity.

@RobertFrost I don't see how that is related to any two number being comparable.

@TobiasKildetoft Because if the conjecture is true all numbers are comparable

@RobertFrost Still, I don't see why that would be. If $x$ and $y$ are distinct odd number such that applying the function to either results in a power of two, how are those comparable?

@TobiasKildetoft because 3x+1/2 and 3x+1/4 are equivalent in this ordering. I mentioned above, 2^n is orthogonal to it.

Sorry (3x+1)/2 and (3x+1)/4

I think I have misunderstood the ordering then. So you are not ordering the numbers but some equivalence classes of them?

@TobiasKildetoft I was hoping to stay away from equivalence classes because it becomes massively more complicated then and I am getting into language I don't fully know

But you just said two of the elements were equivalent

@TobiasKildetoft perhaps better to say they are equal in this ordering

but they cannot be equal unless they are the same number since this was an ordering on the numbers

@TobiasKildetoft can we not say they are equal under the ordering of (3x+1)/2^n

no, because the ordering does not redefine equality

Ok then we may need to use the language of equivalence classes which is foreign to me

So for an explicit example: $3$ is not comparable to $7$ in this ordering since $3$ is smaller than $5$ and $1$ and $7$ is smaller than $11,17,13,5,1$

How do we go about comparing them and preserving the measure in number of steps, how far greater or less they are?

So for example 21 is equal to 5

This is needed because it is a problem of discrete steps and a part of the argument needed is that paths may only cross when they take a step.

why not?

look it up :-)

I know what it is

The exam went well, now I finally have time for differential topology!

(Ignoring the fact I shoukd study Lagrangian mechanics too)

good

Heya @AlessandroCodenotti

Was it an oral exam?

Yes, I did the written exam last week and the oral one today

How many examiners?

Hi,

Had a linear algebra question

does the row reduced echelon form mean no rows are linear combination of earlier rows ?

does it ALSO mean that no cols are linear combination of earlier columns ?

@analyst How could it when all matrices have a reduced row echelon form, but not all matrices have that other property

so it just means that no rows are linear combinations of earlier rows

it does not apply to cols ?

no, it does not mean no rows are linear combinations of earlier rows unless it has a pivot on each row

same for columns

right I meant the ones without a pivot are free rows

but if it is the same for columns, then why is it not called column reduced echelon form ?

why would you call it that?

because the rref is both for col and rows

if it was for rows only, then you called it rref

but because it is for cols too why not the alternate name ? column ref

and hence the word "row reduced"

Because the transpose need not be a reduced echelon form

what does transpose have anything to do with it ?

that is what you do to go from rows to columns

if you change the definitions involved to involve columns instead of rows, the result will also not necessarily satisfy this

just consider the matrix $[1\ 1]$

sorry, I donot get that but maybe because I have to study some more

I can look into it

have another general question

Do you see that that matrix in on rref?

I am using strangs linear algebra book

however sometimes the statements are not explained in the section and I have to think a lot and finally figure it out

should I just ignore them till they become clear

OR

wait to figure them out

what is the right way to do math ?

or should I say learn math ?

definitely spend the time thinking about things until they make sense

I wish there was a study group online where I could just ask questions

:-(

@Perturbative did you see my comment?

@analyst no one can answer the questions for you. There is a difference between knowing the answer and working out the answer. However, you might have a good chance discovering the tools to build the answers. The most important one might be your own mind.

@AlessandroCodenotti Yep I saw your comment, thanks! If you want to make it into an answer I'd be happy to upvote and accept

Does someone of you have an idea about my question: math.stackexchange.com/questions/2079912/… ?

@AkivaWeinberger ah, ok. And I suppose you spoiled that uniform stuff, but the way you said it made it half obvious.

Nah, I see you already got a good answer

see scoliosis jones

@MikeMiller So once upon a long time ago you asked me whether every foliation by circles is a circle bundle.

I think there is a foliation on the Klein bottle which does not come from the fibers of a fiber bundle: think of the foliation by "longitudinal circles" instead of the "meridianal" ones (the latter being the standard nonorientable S^1-bundle on S^1); the quotient map to the leaf space is badly not locally trivial near eg the center circle of one of the two Mobius bands covering the bottle.

But this does not constitute a counterexample to what you asked anyway

Good!

That is not a circle bundle.

But weren't you asking for manifolds admitting circle foliations which is not total space of any circle bundle? I guess I misinterpreted

No, I was looking for the foliations themselves. That's exactly the answer I wanted.

Ah, great then.

Theorem: Every foliation of a surface by circles is either locally trivial or locally the Möbius foliation. I wonder if you can prove this when you get further in Candel-Conlon.

That's a wonderful fact.

Have you made any pushes on that book yet?

hi everyone

Hey

Hi

A bounty and I just would like some attention.