Mathematics

6:24 AM

hi @Ted

Must be morning for you, Balarka ☺️

yeah, it is :P

No school?

nope. exams finished 2 weeks ago.

wow, you're retired just

like me!

6:27 AM

heh. where are you now? i was surprised at first that you are awake, but then i recalled you were on your road trip

so time zones might have changed a bit.

ah.

Agh, can't stand doing this on my phone.

hmm. i'm trying to prove that f.g. projective modules over PID's a free. let $M$ be a proj. module. there is some $N$ such that $M \oplus N$ is free. free implies no torsion, so $Tor(M \oplus N) = Tor(M) \oplus Tor(N) = 0$. each factor is zero, so $M$ has no torsion either. invoke classification of f.g. moduels over PID to conclude $M$ is free.

is there no way other than classification theorem to do this?

G'night! ..: Be carful

'night. yeah, modules are subtle, so i know what you mean.

6:54 AM

@BalarkaSen Well, you can shorten it by noting that projective implies flat, and flat implies torsionfree over a domain (and then it is basically the classification in a different disguise)

right

@BalarkaSen Though I guess to make it completely the classification, you need to wrap back to get that torsionfree implies free (or just that it implies projective)

7:32 AM

Hello@Tobias @Balarka

@Rememberme Hi

@Tobias We have the characteristic polynomials for a linear operator on a finite dimensional vector space $V$. We can always associate with this polynomial a space $F[x]$ where F is a field. Right? What all can we say about $F[x]$ with just the linear operator given?

@Rememberme not sure what space you want to associate to it. The notation suggests just the space of polynomials in one variable

@Rememberme hi

My notation might be a bit misleading .

7:48 AM

how are you associating a polynomial ring given a certain polynomial? that doesn't seem a good thing to do.

What I actually want to say is that with the characteristic polynomial of a linear operator on a finite dimensional vector space $V$ (dim=n). We can always associate this polynomial with the space of polynomials in one variable(Can we actually do that? always). Now my question is with just the linear operator given we what can we say about this space of all polynomials in one variable

"We can always associate this polynomial with the space of polynomials in one variable" sure, then that'd be just boring.

for every linear transformation on an $F$-vector space, you are associating $F[x]$.

Yes.

@Rememberme what you are doing does not seem to be dependent on the polynomial

it seems a completely uninteresting thing to do, as it doesn't depend on the linear transformation you choose

7:52 AM

Okay.

why do you want to do anything like that anyway?

I want to think of a polynomial ring in terms of a linear operator from a vector space V to V whose characteristic polynomial lies in that ring.

why? (btw, you can associate to any polynomial ring F[x] an F-vector space with a linear transformation : it's just what an F[x]-module is, but that seems different from what you're trying to do)

Because characteristic polynomials seem very fascinating to me. And to think of a ring which can be thought of using this polynomial might be more interesting in my point of view. But as you say it is boring

i don't see why it is interesting, but whatever

8:01 AM

@Balarka What are the interesting things we can say if we think of a polynomial ring of the form $\Bbb{Q}[\sqrt{x}]$

it's not a polynomial ring.

the only way to make sense of what you wrote is as the quotient ring $\Bbb Q[x, y]/(y^2-x)$.

@BalarkaSen Not even that way

ah?

whoops, typo

More like $\mathbb{Q}[x,y]/(y - x^2)$

yep, typo

8:04 AM

well, now we certainly agree :)

If x is rational and not a perfect square in $\Bbb{Q}$

then it's not even a quotient ring of a poly. ring

it's a field

Ahh.. Okay

Thanks for that . My internet is a bit slow so taking time to write

Samuel Yusim

why can't it be both?

Which both @SamuelYusim

8:08 AM

polynomial ring over $R$ is a ring of the form $R[x_1, x_2, \cdots, x_n]$ where $x_i$ are $R$-algebraically independent transcendentals.

it's the definition of a polynomial ring.

@BalarkaSen Well, it does become a quotient of a polynomial ring

ok, fair enough

$\Bbb Q[x]/(x^2 - a)$

Samuel Yusim

a quotient of a polynomial ring $F[x]$ by a maximal ideal $M$ gives you a field extension of $F$

in the case of multiple variables it's the same

My question :I have been looking at the quotient notation for rings. How do we define quotients for rings. The same way as we did for groups?

@Rememberme Yes, though we need to quotient by ideals instead of normal subgroups

8:10 AM

@Rememberme read it from somewhere.

you need to know what an ideal is

@BalarkaSen That is only for alg. closed $F$

the idea is the same though. an ideal is precisely what kernel of a ring morphism is.

@Tobias i'm still sleepy, sorry!

Okay . Need to learn that

Samuel Yusim

I was gonna say

and regardless, $F$ is a field extension of $F$ in my book

it is, but that is just a trivial thing to say. i agree with you that qt. by a maximal idea is a field extension of $F$

this is just because qt. by maximal ideal is always a field, and that field contains $F$ by correspondence of ideals and qts.

8:12 AM

@BalarkaSen Well, because nothing but the $0$ from $F$ can be in the maximal ideal

so the composed map becomes injective

right

Robert Cardona

8:39 AM

at any algebraic topologists, category theorists, what are the main mailing lists for each one respectively? I did a preliminary search and found quite a few, but would like to just add the main one of each.

@RobertCardona Mailing lists for what purpose? conference announcements and such?

Hi

How to write a numbered equation in latex?

@Gigili try \tag{#}, for example x^2 \tag{2} gives $$x^2 \tag{2}$$

It's \gather in normal latex, but asking a question I dont know how to make the equation numbered so that I could refer to it later

@AntonioVargas Ah OK, thanks

Wait a minute

No, that's not what I need

$$||Y-DX||_F^2\\=||Y-\sum_{j=1}^K d_jx_T^j||_F^2\\=||\big(Y-\sum_{j \neq k} d_jx_T^j\big)-d_kx_T^k||_F^2\\=||E_k-d_kx_k^T||_F^2 \tag{1}$$

Ah, you want the whole thing numbered, rather than just one line?

8:50 AM

The other way around, I want every line of the above formula to be numbered

I would recommend using align then

just add \tag{#} at the end of each line

before the \\

also, use \| instead of ||

$\|$ vs $||$

@AntonioVargas Noted, thank you

I would probably format it something like $$\begin{align} &\|Y-DX\|_F^2 \tag{1} \\ &\qquad = \left\|Y-\sum_{j=1}^K d_jx_T^j\right\|_F^2 \tag{2} \\ &\qquad =\left\|\big(Y-\sum_{j \neq k} d_jx_T^j\big)-d_kx_T^k\right\|_F^2 \tag{3} \\ &\qquad = \left\|E_k-d_kx_k^T\right\|_F^2 \tag{4} \end{align}$$

@Gigili glad to help :)

Thank you very much

What difference does it make if you use \qquad instead of \quad?

@Gigili Bigger space

A: First order logic expression of "Each finite state automaton has an equivalent push-down automaton"?

9:04 AM

Ah, interesting

@AntonioVargas Your formatting looks much better than mine

$$\begin{align}
\|Y-DX\|_F^2 \tag{1}\\
=\|Y-\sum_{j=1}^K d_jx_T^j\|_F^2 \tag{2}\\
=\|\big(Y-\sum_{j \neq k} d_jx_T^j\big)-d_kx_T^k\|_F^2 \tag{3}\\
=\|E_k-d_kx_k^T\|_F^2 \tag{4}
\end{align}$$

Couldn't you do something so that the first equation will be center aligned?

hmm, never mind! Spent enough time on formatting, enough for asking a question on main

Thanks again

user4791206

10:25 AM

hi ! may I ask quest . here ?

0

In your second formula, x is not even bounded in the second part, so $$(∀x~ fsa(x))→(∃y~ pda(y)∧equivalent(x,y))$$ contains a free variable. Even considering its universal closure, it would not mean what you want.

I Want To Remain Anonymous

12:25 PM

@Gigili Here's the difference $\varepsilon\quad \delta$ & $\varepsilon\qquad \delta$, the difference is that /qquad gives more space

I Want To Remain Anonymous

3 hours ago, by Tobias Kildetoft

@Gigili Bigger space

1:12 PM

@Chris'ssistheartist

6

Q: How to calculate the limit that seems very complex..

Someone gives me a limit about trigonometric function and combinatorial number. $I=\displaystyle \lim_{n\to\infty}\left(\frac{\sin\frac{1}{n^2}+\binom{n}{1}\sin\frac{2}{n^2}+\binom{n}{2}\sin\frac{3}{n^2}\cdots\binom{n}{n}\sin\frac{n+1}{n^2}}{\cos\frac{1}{n^2}+\binom{n}{1}\cos\frac{2}{n^2}+\binom...

Interesting one.

$$I=\displaystyle \lim_{n\to\infty}\left(\frac{\sin\frac{1}{n^2}+\binom{n}{1}\sin\frac{2}{n^2}+ \binom{n}{2}\sin\frac{3}{n^2}\cdots\binom{n}{n}\sin \frac{n+1}{n^2}} {\cos\frac{1}{n^2}+\binom{n}{1}\cos\frac{2}{n^2}+\binom{n}{2}\cos\frac{3} {n^2}\cdots\binom{n}{n}\cos\frac{n+1}{n^2}}+1\right)^n$$

@IWantToRemainAnonymous it's limit of $e$ at the big fraction multiplied by n. Euler formula should be useful. All reduces to boring calculations.

OK, the limit is $e^{1/2}$.

1:31 PM

$$\lim_{n\to \infty } \, \exp \left(\frac{n \Im\left(e^{\frac{i}{n^2}} \left(1+e^{\frac{i}{n^2}}\right)^n\right)}{\Re\left(e^{\frac{i}{n^2}} \left(1+e^{\frac{i}{n^2}}\right)^n\right)}\right)=e^{1/2}$$

As lab bhattacharjee, Moivre gives immediately the way to go, and to get rid of that ugly power in $n$ use that $1+\cos(x)=2\cos^2(x/2)$ and then place that $n$ in power in the arguments of sine and cosine in brackets.

That's all.

Well, not immediately since you have squared cosine, it doesn't work the foregoing operation.

With twice the argument it works as lab did it $1+e^{2iy}=2\cos y(\cos y+i\sin y)$. to have it nicely arranged.

or otherwise write that $\sin(x)= 2\sin(x/2) \cos(x/2)$ to factorize $\cos(x/2)$ such that you can place the power in argument (then you can keep the form $1+\cos(x)=2\cos^2(x/2)$)

There is nothing amazing in these limits, but only boring stuff, that is a classical limit of the type $1^{\infty}$ that reduces to $e^{\lim \text{something}}$. Play with Euler and Moivre formulas and you're done.

1:51 PM

Boring?

I've never heard you use that word with limits :P

@Rigor I did tons of them. :-)

I would like to see an unusual approach, something like slaying the limit from one single shot.

Like David meets Goliath?

@Rigor Yeah :-))))))))

:-)

@Rigor I see such a way now.

1:57 PM

Cool.

A: How to calculate the limit that seems very complex..

My way

0

With rough approaximations, we can avoid the use of trigonometric stuff, that is $$\lim_{n\to \infty } \, \left(1+\frac{2^{n-1} (n+2)}{n^2 2^n}\right)^n=\lim_{n\to \infty } \, \exp\left(\frac{2^{n-1} (n+2)}{n 2^n}\right)=\sqrt{e}$$ where $\displaystyle \sin\left(\frac{k}{n^2}\right)$ behaves lik...

Initially I let things unsimplified in the fraction such that my way is better understood. Now I see another answer using the idea in my answer.

To make it rigorous, use Taylor with error term and bla bla. It's not for me this kind of job now.

Soham Chowdhury

2:23 PM

@Balarka?

hi

Soham Chowdhury

come over to the room

My way allows you to calculate the limit WITHOUT PEN AND PAPER, but this might not be useful for MSE users.

Back to my research.

2:41 PM

nothing beats pen and paper

what if i told you, the hardest problems i have solved using a pen and paper away from my laptop screen ?

@Agawa001 If you wanna do the hardest stuff in integrals, series and limits you need to push your mind extremely hard.

one cant focus with all these screen radiations targetting his faceskin

@Agawa001 Maybe. :-)

@Chris'ssistheartist especially with a cup of coffee and fresh air

@Agawa001 :-) yeah, especially fresh air :-)

2:51 PM

especially the sea breeze :)))

ohhhhh, don't say it ... :-)

lol why ?

@Agawa001 I miss it, I didn't see the sea for some years ... :-(

@Chris'ssistheartist thats not same, i was born in the coast, i cant imagine a nature without sea :D

@Agawa001 nnniiiiiiiiiccccccccceeeeeeeeeeeee :-)

2:57 PM

i m looking up romania at the map

@Agawa001 and the Black sea. :-)

you have still a little opening to the sea :)

@Agawa001 I live in the other side of the country. :-)

@Chris'ssistheartist it looks much like a basin than a sea , well i guess streams and whirls are hardly found there isnt it ?

@Agawa001 Yeap.

3:01 PM

that makes it safer than other seas

@Agawa001 you live in US?

lol no

norrth africa

@Agawa001 Ah, OK. Hot places there. :-)

i live exactly here google.dz/maps/place/Roumanie/@36.6044211,2.3762222,13z/…

no not hot, just in summer

columbus8myhw

Algeria has a nice anthem

3:06 PM

@Agawa001 Nice anyway. :-)

@Agawa001 Have you ever visited our country?

@Chris'ssistheartist no, i just visited italy and france

my sojourn there was limited

I see. I lived in Italy for some time.

3:23 PM

@DanielFischer Would I be right in stating that for a unital Banach algebra the spectrum $\sigma(0) = 0$?

That's correct.

@MikeMiller Thanks. Could I ask for a hint regarding something. I want to prove that for idempotent members $a^{2} = a$ in a unital Banach algebra, it follows $\sigma(a) = \{0,1\}$. So in order to this I will use the Spectral Mapping Theorem. Hence I will consider the analytic function $f(x) = x^{2}-x$ on some neighbourhood of the spectrum of $a$. Then the Spectral Mapping Theorem yields $\sigma(a^{2}-a) = \sigma(a)^{2}-\sigma(a) = \sigma(a) -\sigma(a) = 0$. Is this the correct approach so far?

3:41 PM

Yeah, that's good. The point being that an element of
the spectrum would have to be idempotent, and there are only two idempotents in $\Bbb C$.

So you've proved that the apectrum is a subset of $\{0,1\}$.

@MikeMiller Yeah that's what I proved. Do you have any small hint on how to proceed?

@MikeMiller Why can't it be the full interval $[0,1]$? Since you would have $[0,1] = [0,1]^{2}$

Q: Calculating in closed form $\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$

3:58 PM

Upvote (and provide with an amazing solution)

0

It's not hard to see that for powers like $1,2$, we have a nice closed form. What can be said about the cubic version, that is $$\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$$ What are your ideas on it? Differentiation under the integral sign? Other ways?

calculus real-analysis integration definite-integrals

@Moses: 1) For a nontrivial idempotent you can prove those are both in the spectrum by the definition, basically.

I think the holomorphic function calculus is point wise. That is, when you say $f(\sigma(a))$, you mean $\{f(s) \mid s \in \sigma(a)\}$.

@Moses: So in particular this means that if $s \in \sigma(a)$ then $s^2 = s$. This is must stronger than saying that the set $\sigma(a)$ satisfies $\sigma(a)^2 = \sigma(a)$.

Shay Ben Moshe

4:19 PM

Hi there!
Given two rings $A,B$, homomorphism $f$ between them, and an $A$-module $M$,
how do we give $B\otimes_A M$ the structure of a $B$-module?

I think I got it, thanks anyway...

St3114

4:39 PM

hello can someone shortly write or give me a link about why is if S is transition matrix from base $V=\left[ \left[ v_1\right] ,\left[ v_{2}\right] \right] $ to
base $W=\left[ \left[ w_1\right] ,\left[ w_{2}\right] \right] $ then $S=W^{-1}.V$

@MikeMiller I'm not following you reasoning regarding the conclusion of your statement:
"That is, when you say $f(\sigma(a))$, you mean $\{f(s) \mid s \in \sigma(a)\}$. So in particular this means that if $s \in \sigma(a)$ then $s^2 = s$."

Do you agree that that is the definition of $f(\sigma(a))$?

@MikeMiller Yes that is just the image of $f$ on $\sigma(a)$. Are you saying that because we have $\sigma(a)^{2} = sigma(a)$ we have that $s^{2} = s$ or that we always have $s^{2} = s$ for any $s \in \sigma(a)$?

Well, we know $f(\sigma(a)) = 0$, like you pointed out. So, by the definition of $f(\sigma(a))$, we have for $s \in \sigma(a)$, $f(s) = 0$. Well, $f(s) = s^2-s$.

It is not true that $C^2 = C$ implies that $C^2 - C = 0$. You can't subtract sets wildly, unfortunately.

We have the latter statement. It's stronger.

(The point being that $C - C$ is not actually zero if $C$ is not a singleton. If $C = [0,1]$, then $C - C = [-1,1]$.)

Hi @Ted.

Goodnight @Mike ... Inconsistent wifi here ...

4:53 PM

RIP.