Mathematics

12:01 AM

Hi all. its been a while since I came in here. I have a quick question. I have a sequence of functions $f_n : [0, 1] \rightarrow [0, 1]$ such that $f_n(x) \geq f_{n+1}(x)$ for all $x$. I want to show that $g(x) = lim f_n(x)$ is well defined...

How should I go about showing this?

Daniel Fischer

@masfenix Use some theorem about bounded monotonic sequences.

Fernando Martin

Why are they bounded?

Oh, the codomain is $[0,1]$

masfenix

Sure. Thanks. Since [0, 1] is Lebesgue measurable, I can apply the Lebesgue's monotone convergence theorem

Mike Miller

A nice problem similar but different to that: A pointwise convergent sequence of monotonic functions converges uniformly

masfenix

@MikeMiller yes we did that in class (in measure theory). They converge uniformly except on a set with arbitrarily small measure.

Correct?

Mike Miller

12:07 AM

No, they literally converge uniformly. (I should say the domain is $[0,1]$.)

masfenix

@MikeMiller Ahh, I see. Cool!

I have another question (disclaimer from assignment - but I am not enrolled in the course)

Is there an example of such a sequence of functions that converge to g but g is not continuous?

Actually back to my original question: Wikipedia says that the Lebesgue's monotone convergence theorem is on increasing functions. ($f_n \leq f_{n+1}$) but this shouldnt' make a difference correct?

GBeau

is there a simple way to do this problem: "How many solutions will $x^2\equiv 0\mod 2^k$ have for integers $k\geq 1$ "

I think I've found the answer to be \[
n(k)=
\begin{cases}
2^{\frac{k}{2}} & : k\mod 2\equiv 0 \\
2^{\frac{k-1}{2}} & : k\mod 2\equiv 1
\end{cases}
\]

but the proof is quite long

Alizter

Seems like an interesting problem

Care Bear

12:30 AM

@Alizter Seems like Weyl's inequality should give some information there.

Alizter

@CareBear It is also interesting because it is old with not many veiws. I changed the title maybe that was the problem.

Care Bear

It was missing a number-theory tag, which is the actual area where the question belongs. Added now.

Ted Shifrin

Hi @FernandoMartin

Alizter

@CareBear Good.

GBeau

@TedShifrin Perhaps I didn't get off easy with regard to the counting ignorance, the next question was "How many solutions will $x^2\equiv 0\mod 2^k$ have for integers $k\geq 1$?"

(which I figured out)

Mike Miller

12:45 AM

@TedShifrin I object to "$X$ is parallel to $Y$" meaning $\nabla_X Y = 0$, since the first statement seems symmetric...

Ted Shifrin

No, that's not right @Mike. It means $Y$ is parallel along the integral curve of $X$.

Mike Miller

OK, fair. Then I'm happy.

Pedro Tamaroff

1:59 AM

@KarlKronenfeld

HELLO CORAL

Karl Kronenfeld

hey yo @Pedro

Pedro Tamaroff

@KarlKronenfeld What's up?

Karl Kronenfeld

now I'm sick

Pedro Tamaroff

Oh, noes.

What kind of sick?

Mike Miller

@PedroTamaroff Challenge; find a ring $A$ and two finitely generated modules $M, M'$ such that $M_{\mathfrak p} \cong M'_{\mathfrak p}$ for all primes $\mathfrak p \in \text{Spec}(A)$, but $M \not\cong M'$

Karl Kronenfeld

2:03 AM

a cold. my body had to deal with a cold not too long ago, so I suspect I'll get over it quickly. I feel tired mainly @pedro

Pedro Tamaroff

@KarlKronenfeld Oh, sucks.

@MikeMiller OK. I'll think about it.

@KarlKronenfeld Do you recall any proof of Zariski's lemma?

Karl Kronenfeld

@PedroTamaroff not really tbh

Pedro Tamaroff

@MikeMiller I bet I can find non f.g. modules first. =P

Mike Miller

I will accept that as partial credit.

Pedro Tamaroff

OK.

Karl Kronenfeld

2:09 AM

@PedroTamaroff oh, maybe i do.

Pedro Tamaroff

YAY!

@TedShifrin Hello.

Ted Shifrin

Hi @Pedro @Karl

Karl Kronenfeld

hey @Ted

Pedro Tamaroff

@TedShifrin What's up?

Ted Shifrin

So you aced your midterm, @Pedro?

Pedro Tamaroff

2:11 AM

@TedShifrin Not particularly. I answered 4 out of 5 questions, and one I answered incorrectly. =P

Mike Miller

I suppose I don't deserve hellos anymore.

Pedro Tamaroff

In my defense, it was 8 a.m.

Ted Shifrin

No, you don't ... We were just talking.

Kaj Hansen

Ugh. 8 AM exams.

Mike Miller

That's true, but I'm very needy, you see.

Ted Shifrin

2:12 AM

Crummy defense, @Pedro :D

Pedro Tamaroff

@TedShifrin The answer I decided not to do was about finding a Möbius transformation that sent two points to two others and a line to a circle. I was like NAH.

@MikeMiller hugs

Karl Kronenfeld

@PedroTamaroff lol that's the kind of thing I would love to do

Ted Shifrin

So you dictate where the two points and $\infty$ go ... Pretty routine.

Pedro Tamaroff

@KarlKronenfeld Really? It's just computations. And I don't know how to do it.

Mike Miller

That was probably the easiest one on your test

Karl Kronenfeld

2:14 AM

@PedroTamaroff easy points

Ted Shifrin

Let's all ridicule @Pedro :)

Pedro Tamaroff

@MikeMiller I wouldn't know. I had to find the domain of convergence of $$\sum w^n/n$$ where $w=i\frac{1-z}{1+z}$

Mike Miller

That one's the second easiest.

Pedro Tamaroff

I also had to find all entire functions for which $f(x+iy)=f(x)+if(y)$.

Mike Miller

Third easiest.

Ted Shifrin

2:15 AM

Or maybe easier.

Mike Miller

Sounds like you shoulda aced it, @Pedro

Pedro Tamaroff

And then find all branches of the logarithm in $\Bbb C\setminus \Bbb R^{\leqslant 0}$ such that $$\int_{|z-i|=1/2}\frac{z^{4/3}}{(z^2+1)^2}dz=-\frac\pi 6$$

@MikeMiller No, I got one wrong.

Mike Miller

Didn't say you did ace it. :P

Pedro Tamaroff

I had to prove that an entire function with strictly positive imaginary part was constant.

I tried to Liouville $1/f(z)$ but I should have taken $(f(z)-i)/(f(z)+i)$ which is holomorphic and sits inside $B(0,1)$.

Ted Shifrin

Damn conformal maps everywhere ...

Mike Miller

2:20 AM

I'm the king of those.

Fernando Martin

Hey @MikeMiller

If $\varphi_p:M_p\rightarrow M'_p$ is injective (surjective) for each prime ideal $p$, then $\varphi$ is injective (surjective)

But maybe the isomorphisms are not induced by the same morphism

Mike Miller

yes, but I never said the isomorphisms were the restriction of an is

yeah

Fernando Martin

yes

Karl Kronenfeld

wtf

Fernando Martin

Well, it's true for $\mathbb{Z}$ at least

Mike Miller

2:24 AM

yes

there are a few classes of rings I can prove it can't happen for

Fernando Martin

I'm thinking about $R=k[x_1,x_2,\dots]/\langle x_1^2, x_2^2,\dots\rangle$

Mike Miller

yikes

Pedro Tamaroff

@FernandoMartin IDEMPUHTETNZ

Mike Miller

use your words, pedro

you're an adult now

Fernando Martin

@MikeMiller: that ring is somewhat "nice" for this problem, since it has only one prime ideal

Mike Miller

2:28 AM

@FernandoMartin it's not possible for any local ring

:)

Fernando Martin

$\langle x_1,\dots\rangle$

dammit

why not?

Mike Miller

because $A-\mathfrak m$ consists of units, $M_{\mathfrak m} = M$

Because of reasons.

right

that's the reason

2:30 AM

Well, that excludes this ring

but not all local rings

wait, is it possible for them to have non-maximal prime ideals?

Karl Kronenfeld

yep

Fernando Martin

no

really?

Karl Kronenfeld

for a local ring to have non-maximal prime ideals: yes. localize $\mathbb Z$ at a prime.

Mike Miller

$\Bbb Z_p$

Fernando Martin

weren't all ideals $\langle x^k\rangle$?

for $x$ some uniformizing parameter

Mike Miller

2:32 AM

any local ring that's an integral domain has non-maximal primes

how does that not exclude all local rings, fernando?

if $M_{\mathfrak m} \cong M'_{\mathfrak m}$ then $M \cong M'$, which we didn't want

Fernando Martin

I'm stupid

Hmmm, but local rings don't have any non-zero non-maximal primes, right?

Karl Kronenfeld

lol

Mike Miller

I can make the spectrum of a local ring as complicated as I want

Pedro Tamaroff

You're so cool Mike.

Mike Miller

while that's true, I detect an air of sarcasm, which will not do

Pedro Tamaroff

2:45 AM

@MikeMiller The modules cannot be of length $1$.

Fernando Martin

Ok, I was confusing local rings with DVRs

meh

Pedro Tamaroff

All is better now @FernandoMartin

Clarinetist

3:53 AM

I've already surpassed my limit on questions, and I have a polar->Cartesian question:

Consider $r^2 \cos(2\theta) = 1$ in polar coordinates. Solving for $r$, we have $r = \sqrt{\dfrac{1}{\cos(2\theta)}}$.
Then
$$\begin{align}
&x = r\cos(\theta) = \sqrt{\dfrac{\cos^{2}(\theta)}{2\cos^{2}(\theta)-1}} \\
&y = r\sin(\theta) = \sqrt{\dfrac{\sin^{2}(\theta)}{1-2\sin^{2}(\theta)}}
\end{align}$$
after making use of the identity $\cos(2\theta) = 2\cos^{2}(\theta)-1 = 1-2\sin^{2}(\theta)$. In their current forms, $x$ and $y$ didn't seem useful to me, so I considered the reciprocals:
$$\begin{align}

Pedro Tamaroff

Do you even lemniscate?

Clarinetist

Nope.