Mathematics - 2017-05-28 (page 2 of 4)

4:02 PM

@user379685 And what does that yield ?

$$\int \frac{P(x)e^x}{Q(x)}=\left(\int \frac{1}{Q(x)}\right)P(x)e^x-\int (P'(x)+1)e^x \left(\int \frac{1}{Q(x)}\right)$$

$(P'(x)+1)\left(\int \frac{1}{Q(x)}\right)=1$ when:

Semiclassical

Integration by parts is definitely the right thing to try with that integral.

@AbdullahUYU But it's best split it as $e^x\cdot \frac{x}{(x+1)^2}\,dx$

user379685

@Hippalectryon sketchtoy.com/68125999 that's the probability distribution for Z, when X>Y

?

Secret

$$Q(x)P''(x)\int \frac{1}{Q(x)}dx+P'(x)+1=0$$

Abdullah UYU

split and then?

I solved the problem with IBP btw

Semiclassical

4:17 PM

though I guess integrating that is indeed obvious once you write $\frac{x}{(x+1)^2}=\frac{1}{x+1}-\frac{1}{(1+x)^2}$

Yeah. I hadn't realized.

Hippalectryon

@user379685 Sorry, my connection dropped.

Secret

(sorry the above rambles are all wrong, retyping...)

user379685

It was fxy

Secret

(Nvm, that $Q(x)$ has to be quite specific to produce that desired result)

Hippalectryon

@user379685 What you wrote looks right, but it's more complicated than it needs to be

Secret

4:21 PM

guess that for this integral it is just a coincidence that $P,Q$ are in a form such that they cancel to 1 in the 2nd term of the IBP, thus allowing the $x$ that is stuck to the $e^x$ to be nuked

user379685

How would you do it

Can it be done without splitting it into two cases?

BAYMAX

IIRC means?

Semiclassical

@user379685 Probably the place to start is to draw contours of constant Z on that unit square.

Hippalectryon

@BAYMAX If I Recall

BAYMAX

thanks

Semiclassical

4:22 PM

@Secret Yeah, it's stuff like that which makes me dislike IBP examples. It's easy to make an example which looks artificially complicated just by starting with an obvious case and doing an integration by parts.

Hippalectryon

@user379685 Alright let me write it down :-)

Semiclassical

I thought it was "if i remember correctly" but I suppose that's not all that different

BAYMAX

yeah

Hippalectryon

@Semiclassical Oh, never heard of that one but it totally makes sense

Secret

@Semiclassical For that example, $e^x$ being an eigenfunction of the integral operator does contribute towards that result, but still, $P,Q$ has to be just right to allow the 2nd IBP term to cancel out the 1st, and it is quite unlikely to randomly select a $P,Q$ from the space $P(\Bbb{C})$ to make that work

since $Q$ has to satisfy two constraint equations which is quite specific to this particular integrand

Semiclassical

4:25 PM

@Lozansky The easiest way, if you're lucky, is to write down an $f$ such that $\mathbf{A}=\nabla f$. If you can do that, then the path independence is obvious.

Salech Alhasov

Just wondering: Soon I'm gonna teach 6 years old some mathematics, and I'm thinking from where to start?

Ted Shifrin

@Salech: Probably better to ask that on the Math Educators site.

Hippalectryon

@user379685 Alright here's how I'd do it. Do you agree that $P(Z>x)=\int_0^1P(X=t)[P(Y>t+x)+P(Y<t-x)]$ ?

Semiclassical

That's pretty open-ended, yeah.

Salech Alhasov

@Te

@TedShifrin: Thanks

Lozansky

4:27 PM

@Semiclassical Yeah but that's more work in most cases

Semiclassical

Eh, depends on the case.

Secret

The best I can say is that, the results of that integral relies on the following 2 properties:
1. $e^x$ is an eigenfunction of the integral operator
2. Polynomials are nilpotent under differential operator

After that, $P,Q$ has to satisfy two more equations to reproduce that result

Salech Alhasov

Probably I'll do some chess first

chess is fun

Semiclassical

There are some cases where you can see $f$ at a glance.

If you can't find such a potential function, though, then taking the curl probably is the best route.

user379685

@Hippalectryon could you elaborate

Secret

4:29 PM

I wonder if there exists a function such that if I plug in any integrand, it spits out all the system of constraint equations the integrand has to satisfy when operating under the integral sign, but then I might be accidentally entering hodge conjecture territory from this overthinking...

Hippalectryon

@user379685 Alright, first of all do you agree that $P(Z>x)=\int_0^1P(X=t)P(Z>x|X=t)$ ?

Lozansky

I want to do $\oint z^4 dS$ with $S:|r|=R$. Since $z^4 = (R^2-x^2-y^2)^2$, I think I can say that $\int z^4 = \int (R^2-2z^2)^2 = \int R^4-4R^2z^2+4z^4 \leftrightarrow \int z^4 = -\dfrac{1}{3} \int R^4-4R^2z^2 = -\dfrac{1}{3} \int R^4-\dfrac{4}{3}R^2z^2$. Does this look correct?

BAYMAX

thanks @AlessandroCodenotti

Ted Shifrin

Huh? @Lozansky

Semiclassical

Wait. Do you mean $dS$ or $ds$?

Lozansky

4:32 PM

dS

Zophikel

1

Q: Proving that $\frac{1}{2 \pi i}\oint_{\gamma_{1}} \frac{d \zeta}{(\zeta - 1)\zeta + 1}$ is equal to itself?

In the text "Functions of a Complex Variable" I'm having trouble verifying if my proof of $(0.)$ $(0.)$ $$\frac{1}{2 \pi i}\oint_{\gamma_{1}} \frac{d \zeta}{(\zeta - 1)(\zeta + 1)} = \frac{1}{2 \pi i}\oint_{\gamma_{2}}\frac{d \zeta}{(\zeta - 1)(\zeta + 1)}$$ Remark: $\gamma_{1}$ is $\partial{d...

Lozansky

@TedShifrin I don't write out all the notation, but the integral is over the sphere

Ted Shifrin

The first step looks like garbage.

BAYMAX

oh yes @nbro I thought as I saw you in cs SE so I thought you may help about the default directory

Ted Shifrin

I know what you're doing in general.

Abdullah UYU

4:32 PM

How to find $$\lim_{x \to \pi/6} \frac{8\sin x -4}{\cos (x) (\frac{\pi}{6}-x)}$$

Lozansky

@TedShifrin Which step is garbage exactly?

Ted Shifrin

@AbdullahUYU: If you mean $(\pi/6-x)\cos x$ in the denominator, please write that.

Semiclassical

$(\cos x)(\pi/6-x)$ or $\cos(x(\pi/6-x))$?

Ted Shifrin

Where did $R^2-2z^2$ come from, @Lozansky?

Astyx

@AbdullahUYU Make the variable change $h = x - {\pi \over 6}$

Abdullah UYU

4:33 PM

first one

Lozansky

@TedShifrin I figured $-x^2-y^2 = -2z^2$ under integration

Astyx

Hi @Ted, @Semi, @everyoneelse

Semiclassical

Okay. First thing to note is that $\cos x\to \sqrt{3}/2$ as $x\to \pi/6$, so that's finite and can be taken out of the limit without issue.

Lozansky

But that was a shot in the dark

Ted Shifrin

You can't do symmetry arguments in the middle of algebraic computations. Integrals do not commute with algebra.

Lozansky

4:34 PM

Argh

Semiclassical

(supposing the rest of the limit exists, of course)

Ted Shifrin

That's why I said it was garbage.

Abdullah UYU

@Semiclassical OK

Lozansky

Okay then I guess I should just Gauss this thing

Semiclassical

Similarly, you can pull 8 out of the top.

Abdullah UYU

4:35 PM

yeah

Secret

One thing I don't really understand, despite $\lim_{x_i\to a_i}$ and $\int$ are linear operators, they are like the most nontrivial ones to evaluate. I suspect these operators in general don't have a matrix representation (even an infinite one)

Ted Shifrin

Remember that to "Gauss" it, you have to have the integral as the flux of a force field.

Semiclassical

So that gives $\frac{48}{\pi}\lim_{x\to\pi/6}\frac{\sin x-1/2}{x-\pi/6}$.

But that limit should look awfully familiar, moreso if you let $f(x)=\sin x$.

Ted Shifrin

Provided you remember what $\sin(\pi/6)$ is.

SBM

huh hello

Semiclassical

4:37 PM

True. (Though the limit would likely be trivial if it wasn't of that form)

Ted Shifrin

Of course. I was being explicitly helpful.

SBM

Gaussian integrals?

like

Lozansky

@TedShifrin What? Don't I just need $\mathbf{u}$ to be $\mathcal{C}^1$ in $|r|\leq R$?

Semiclassical

True.

Ted Shifrin

What is $\mathbf u$, @Lozansky?

user379685

4:38 PM

@Hippalectryon yes

SBM

$$\int_0^{\infty} e^{-x^2} \mathrm{d} x$$

Lozansky

@TedShifrin Oh sorry, I thought I had written that out. $\mathbf{u} = (x^2y^2,xy,z^3)$

Astyx

$\sqrt\pi \over 2$

Alessandro Codenotti

@BAYMAX was it correct? That's the default on linux and I guesses it should be more or less the same on window

Hippalectryon

@user379685 Great. Now, can you see why $P(Z>x|X=t)=P(Y>t+x)+P(Y<t-x)$ ?

Alessandro Codenotti

4:39 PM

Hi @Ted

Ted Shifrin

Oh, so your surface integral started with a flux integral. I had no idea.

Hi @Alessandro.

Of course, @Lozansky: Whenever possible you should use Stokes's Theorem and Gauss's Theorem to avoid line integrals and surface integrals and do easier integrals instead.

Lozansky

Yeah, the integral itself wasn't the problem. I wanted to do it using symmetry arguments but that failed

Abcd

Hello.

How do I find the cube root of a number with a decimal?

Lozansky

Well I see it as good practise to do some brute force computing every now and then

Abcd

(without a calculator)

Abdullah UYU

4:41 PM

@Semiclassical after pulling $\cos x$ and $8$, are you sure you write it true?

Ted Shifrin

How do you find cube root of a number without a decimal, like 24, @Abcd?

Abcd

@TedShifrin No idea :(

Ted Shifrin

Why are you trying to do this?

Abcd

@TedShifrin whom was that addressed to?

Ted Shifrin

You.

Abcd

4:43 PM

@TedShifrin I have a stoichiometry question that requires this

Ted Shifrin

Use a calculator, for goodness sake.

In my high school days, we would have used log tables.

Abcd

@TedShifrin That's not allowed in my exams :(

Ted Shifrin

Or slide rule. Now you use calculators.

SBM

We still do use at school

Ted Shifrin

Well, they do not expect you to do something ridiculous like that by hand.

Abcd

4:44 PM

That's not allowed here.

user379685

@Hippalectryon yes

Astyx

These values should be given, like $\sqrt 3$ etc

SBM

log tables

Abcd

@Astyx They aren't :(

None.

SBM

but remember trig values

Hippalectryon

4:45 PM

@user379685 Great, so combining those two together gives us my original formula, $P(Z>x)=\int_0^1P(X=t)[P(Y>t+x)+P(Y<t-x)]$. So far so good ?

Abcd

@SBM I do.

Astyx

What's the general question ?

As in, what's the context of these ?

Ted Shifrin

If you can't use a calculator and they give you a cube root of some non-obvious number, you'd better use log tables. You aren't expected to calculate Taylor series by hand in a chemistry test when there are so many other things to do.

user379685

@Hippalectryon yes

Ted Shifrin

Or maybe you're doing the problem wrong and there's really no cube root in the first place.

Hippalectryon

4:46 PM

@user379685 Great. Now, can you calculate each term in this integral ?

Abcd

@TedShifrin I am sure of my solution as there is a solved example given.

Astyx

Or approximate your number by cubes

Abcd

of the same question type.

BAYMAX

oh sorry no @AlessandroCodenotti

I searched it but could not get it

user379685

@Hippalectryon could you show me how?

Ted Shifrin

4:47 PM

Well, @Abcd, I've told you all I know.

Ask your teacher; don't ask us.

Abcd

@TedShifrin okay.

Hippalectryon

@user379685 Well, what's $P(Y<t-x)$ for instance ? You know that $Y$ follows a uniform distribution on $[0,1]$

user379685

t-x?

Hippalectryon

@user379685 Exactly (as long as $t-x>0$). Now, likewise, what's $P(Y>t+x)$ ?

Semiclassical

@AbdullahUYU should be 8/cos(pi/6)=8/(pi/2)=16/pi. so yeah, i was being silly

@Abcd If it's coming up in the context of gen-chem, they may want you to do a few successive approximations

user379685

4:51 PM

@Hippalectryon 1-t-x?

Semiclassical

(I have vague recollections of needing to do that in pH-problems)

Abcd

@Semiclassical Nope. Mole concept.

Semiclassical

:/

Abdullah UYU

cos(pi/6)=pi/2 ? you should take a rest

Semiclassical

...wow. nah, I just need to actually wake the f* up :P

Hippalectryon

4:52 PM

@user379685 Yep. So, what's the whole integral ?

SBM

$0.5\sqrt{3}$

user379685

isn't P(X=t)=0?

Hippalectryon

@user379685 Not exactly. $P(x=t)=dt$

@user379685 Basically it's $P(x\in[t,t+dt])$

Zophikel

Suppose U is an open subset of the complex plane C, f : U → C is a holomorphic function and the closed disk D = { z : | z − z0| ≤ r} is completely contained in U. Let γ be the circle forming the boundary of D. Then for every a in the interior of D:f(a)=$$\frac 1 {2 \pi i} \oint \frac{\phi (\zeta)}{\zeta - z} \, d \zeta.$$

      ^ Is this definition valid

Abdullah UYU

and why didn't we continue to calculate the limit? @Semiclassical

user379685

4:54 PM

@Hippalectryon =dt*(1-t-x+t-x)=dt*(1-2x)

Semiclassical

Because I was waiting for you to recognize it as something more familiar. But I may have missed your reply on that regard.

Namely, what kind of ratio is $\frac{\sin x-1/2}{x-\pi/6}$?

Abdullah UYU

and the denominator should be $\pi/6 -x$ btw

Hippalectryon

@user379685 Nearly; be careful with the integral's bounds, you need to make sure that $t+x<1,t-x>0$. So that'd give you $\int_0^{1-x} P(Y>t+x)dt +\int_x^1P(Y<1-x)dt$. Can you simplify that ?

SBM

Some limit

Semiclassical

Eh. I prefer my version :)

SBM

4:56 PM

What's it?

Semiclassical

To see where I'm going, let $f(x)=\sin x$.

What's $f(\pi/6)$?

Abdullah UYU

1/2

SBM

$$\lim_{x \to \frac{\pi}{6}} \frac{\sin x - 0.5}{\frac{\pi}{6} - x}$$ ?

Semiclassical

Right. So we can rewrite that ratio as $\frac{f(x)-f(\pi/6)}{x-\pi/6}$

SBM

hoh

Semiclassical

4:58 PM

And you're interested in the limit as $x\to\pi/6$. That should look familiar!

Abdullah UYU

sorry i am so dumb :)

Semiclassical

$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=?$

user379685

@Hippalectryon in the right integral where's t?

Hippalectryon

@user379685 oops. $\int_0^{1-x} P(Y>t+x)dt +\int_x^1P(Y<t-x)dt$

Salech Alhasov

f'(a) :)

Abdullah UYU

5:00 PM

l'hopital's rule?

because it is not trivial with this form, is it?

oh, derivative's definition :D @Semiclassical

user379685

@Hippalectryon x^2-2x+1?

Semiclassical

yuuup

Abdullah UYU

i can't believe i can't remember it quickly

Hippalectryon

@user379685 Exactly, $(1-x)^2$. Which makes a lot of sense since $P(Z>0)=1,P(Z>1)=0$. To finish, $P(Z<x)=1-P(Z>x)$

SBM

uh oh

Semiclassical

5:04 PM

Happens.

So, let's review. We wanted to find $\lim_{x \to \pi/6} \frac{8\sin x -4}{\cos (x) (\frac{\pi}{6}-x)}$

Noting the linear factor on the bottom and the resemblance to a difference quotient, we rewrite that quotient as $\frac{8}{-\cos{x}}\cdot \frac{\sin x-1/2}{x-\pi/6}$

user379685

@Hippalectryon 1-(1-x)^2 and differentiate it to get the distribution?

Hippalectryon

@user379685 yes

user379685

@Hippalectryon thank you very much

Semiclassical

At this point, we've factored the quotient into two terms which each have a well-defined limit.

Hippalectryon

@user379685 :-)

SBM

5:07 PM

:)

Semiclassical

So we can use the product rule to say that the limit of the product is the product of the limits, and just compute those two limits.

yeah, exactly

So, do that :)

i am doing it now

mmkay.

it'll end up being rather simple.

Abdullah UYU

5:09 PM

$-8$. Thanks for patience

Semiclassical

Right.

SBM

I hope nobody tries $$\int \ln (\sin x) \mathrm{d} x$$ at least indefinitely to waste a few hours
Great

Semiclassical

Effectively the limit ends up being $\frac{1}{f'(x)}\frac{f(x)-f(a)}{x-a}$ as $x\to a$

Which, yay, is just 1.

SBM

absolutely

reciprocals

Semiclassical

(so long as $f'(a)\neq 0$ etc. etc.)

SBM

5:13 PM

true

Secret

Ordinal question: Is $\sup (\{{}^j\epsilon_{{}^k\omega}|j,k\in \Bbb{N}\})=\epsilon_{\epsilon_0}$ or $=\epsilon_{\epsilon_0+1}$?

Simply Beautiful Art

@Secret Don't quite understand what $k_\omega$ means

Ah, okay

Secret

${}^k\omega=\omega^{\omega^{\omega^{\ddots^{\omega}}}}$ k times

Simply Beautiful Art

It's supposed to read $~^k\omega$

It's equal to $\varepsilon_{\varepsilon_0}$

Daminark

Hey there chat!

Simply Beautiful Art

5:25 PM

Hey there! @Daminark

How's it going?

Good

\('-')/

Kirby praises the sun!

Daminark

Wait hold on @Semi this is actually the best description of that I've ever seen

Secret

5:28 PM

@SimplyBeautifulArt If I try to evaluate $k$ first I get ${}^j\epsilon_{\epsilon_0}$. If I then evaluate $j$, shouldn't I get the successor epsilon number $\epsilon_{\epsilon_0+1}$ as I have a tower of $\epsilon_{\epsilon_0}$, or that does not happen because since $\epsilon_{\epsilon_0}$ is already an infinite tower of the $\alpha < \epsilon_0$ many epsilon exponential tower thus that extra tower just end up bing absorbed by $\epsilon_{\epsilon_0}$?

Semiclassical

lol

Simply Beautiful Art

@Secret But that's not how it works

Semiclassical

Now I'm suddenly imagining Kirby doing a Dark Souls run

Simply Beautiful Art

You can't evaluate with respect to $k$ first

Daminark

As in, Kirby being a character in Dark Souls? Or Kirby at the controller?

Simply Beautiful Art

5:29 PM

What you can do is apply a squeeze

Semiclassical

Former.

Hippalectryon

Kirby: Nightmare in Dark Souls

►

Semiclassical

Which, given how much one inevitably dies during that, means that I also am envisioning a Hollowed Kirby.

Alessandro Codenotti

Ok so I'm doing this complex analysis exercise. Suppose $f(z)$ and $g(z)$ both have an essential singularity at $z_0$, must $f(z)g(z)$ have an essential singularity at $z_0$? My first thought was to look for a function $f$ such that both $f$ and $\frac1f$ have an essential singularity at $z_0$

Simply Beautiful Art

$$\varepsilon_{~^k\omega}<~^j\varepsilon_{~^k\omega}<\varepsilon_{~^{k+1}\omega}‌$$

SBM

5:30 PM

essential singularity ...

Vrouvrou

hello

Alessandro Codenotti

And I think $\frac{e^{1/z}}{z\sin(1/z)}$ could work

Simply Beautiful Art

@Secret So it tends to $\varepsilon_{\varepsilon_0}$

Daminark

Essential is when there is no limit, right?

Secret

Ah ok

Semiclassical

5:31 PM

@Hippalectryon ...that is horrifying.

Daminark

I... guess I'm not surprised that someone has thought to do that but dear god why? @Hippalectryon

Alessandro Codenotti

@Daminark Essential is when the Laurent series has infinitely many nonzero coefficents with a negative index. There also isn't the limit (actually if $f(z)$ has an essential singularity at $z_0$ and $U$ is a punctured nbhd of $z_0$ then $f(U)$ is dense in $\Bbb C$)

Hippalectryon

@Daminark Why not :-)

Daminark

You make a convincing point (and I'm not even being sarcastic) @Hippa

Simply Beautiful Art

@Secret Have you made it to the ordinal collapsing functions yet?

Secret

5:33 PM

@SimplyBeautifulArt I do, that's how that question pop up

Daminark

@Alessandro Ah, right, and pole is finite

Secret

Incoming ordinal collapsing function workflow

$$C_0(\alpha)_0=\{0,1,...,\omega,\omega_1\}$$

\begin{align}
C_\beta(\alpha)_0 & =\{\gamma,\omega_{\beta+1}|\gamma\le\omega_\beta\}\\
C_\beta(\alpha)_{n+1} & =C_\beta(\alpha)_n\cup\{\gamma+\delta,\gamma\delta,\gamma^\delta,\omega_\gamma,\psi_\Gamma(\eta)|\gamma,\delta,\Gamma,\eta\in C_\beta(\alpha)_n,\eta<\alpha\}\\
C_\beta(\alpha) & =\bigcup\limits_{n<\omega}C_\beta(\alpha)_n\\
\Psi_\beta(\alpha) & =\min\{\gamma|\gamma\notin C_\beta(\alpha)\}\\
i=\{1,2,3\}
\end{align}

\begin{align}
C_0(0)_1 & =\{j,\omega+j,\omega j,\omega^j,\omega^{\omega},...|j\in \Bbb{N}\}\\

Ah, okay

@Daminark yup

Oh, okay

5:34 PM

@SimplyBeautifulArt Is it correct so far? (note I omitted many many terms that are not interesting, and place them all beyond the dot dot dot)

Vrouvrou

can we prove that $\int_0^1 |f(t)| dt\leq \alpha \sqrt{\int_0^1 |f(t)|^2 dt}$ ?

Daminark

OK, well, instinct says that since it's analytic nearby, you should be able to do formal products of power series

Though I'm very suspicious that this is true

Simply Beautiful Art

@Secret Give me a few minutes x.x

Daminark

(I also don't know complex analysis)

Hippalectryon

@Vrouvrou Cauchy–Schwarz ?

Daminark

5:36 PM

:P

This isn't quite Cauchy-Schwarz, it's saying that $\|f\|_1 \le \alpha \|f\|_2$

Hippalectryon

@Daminark But doesn't it follow from CS ? Doesn't it just say that $<f,1>\le \sqrt{<f,f>}$ ?

Vrouvrou

we must say that $||f||^2_2$ is a scalar pruduct @Hippalectryon

Secret

Guys, I need to head to sleep. I will respond any replies tomorrow

Hippalectryon

Oh right, my bad @Vrouvrou

Simply Beautiful Art

@Secret g'night!

Hippalectryon

5:40 PM

@Vrouvrou Wait, what about $\int|f|=<f,sgn(f)>\le\sqrt{<f,f><sgn(f),sgn(f)>}=\sqrt{<f,f>}$ ?

Simply Beautiful Art

@Secret Doing quite look, it looks right.

Balarka Sen

@AlessandroCodenotti Er, why can't $\exp(1/z)$ and $\exp(-1/z)$ work for a counterexample?

Nevermind, the latter does not have an essential singularity.

Yeah, so you need an $f$ such that Laurent expansion has infinitely many terms of the form $z^n$ both towards the right and left.

So you can take $1/f$, which would also have essential singularity, and look at $f \cdot 1/f$

Alessandro Codenotti

why doesn't $\exp(-1/z)$ have an essential singularity? Isn't its Laurent series $\sum\limits_{k=0}^\infty \frac{(-1)^k}{k!x^k}$?

Balarka Sen

Er, yeah. Sorry, you just plug in $-1/z$ in the Taylor expansion, and that gives infinitely many $z^{-k}$ terms.

So that's an example?

user193319

I have another question. I am trying to show that the subset (0,1](0,1] is not compact in RR with the usual topology. Note that ⋃n∈N(1n,n)⋃n∈N(1n,n) covers (0,1](0,1]. Thus, if (0,1](0,1] were compact, then there would exist {n1,...,nk}⊆N{n1,...,nk}⊆N such that ⋃ki=1(1ni,ni)⋃i=1k(1ni,ni). WLOG, take 1n1≤1ni1n1≤1ni for every ii. Hence, there exists an m∈BbbNm∈BbbN such that 1m<1n11m<1n1 and...
therefore an element in (0,1](0,1] but not in not in ⋃ki=1(1ni,ni)⋃i=1k(1ni,ni), a contradiction.
Does this sound right?

Alessandro Codenotti

5:52 PM

@BalarkaSen I think so

Abdullah UYU

Alessandro Codenotti

I think the one I wrote earlier works too, but it's definitely much uglier

Balarka Sen

Ok. Sorry for being dumb.

Simply Beautiful Art

@AlessandroCodenotti Pretty sure it has an essential singularity.

Balarka Sen

Yes, we just decided that.

Daminark

5:54 PM

Yeah that's essentially correct

Alessandro Codenotti

...

Abdullah UYU

@Semiclassical picture above is the original question of the limit i wrote

Hippalectryon

@user193319 Can you write that in LaTeX ? Some of your symbols are not very clear (e.g: (1n,n))

Semiclassical

Seems legit.

Lozansky

Does anyone have a simple proof for $\nabla \cdot (\mathbf{A} \times \mathbf{B}) = (\nabla \times \mathbf{A}) \cdot \mathbf{B} - (\nabla \times \mathbf{B}) \cdot \mathbf{A}$ ?

Semiclassical

5:56 PM

Does index notation count as simple?

Lozansky

Yes

Hippalectryon

(sorry, gtg. Bbl)

user193319

I am trying to show that the subset $(0,1]$ is not compact in $\Bbb{R}$ with the usual topology. Note that $\bigcup_{n \in \Bbb{N}} (\frac{1}{n},n)$ covers $(0,1]$. Thus, if $(0,1]$ were compact, then there would exist $\{n_1,...,n_k\} \subseteq \Bbb{N}$ such that $\bigcup_{i=1}^k (\frac{1}{n_i},n_i)$. WLOG, take $\frac{1}{n_1} \le \frac{1}{n_i}$ for every $i$. Hence, there exists an $m \in Bbb{N}$ such that $\frac{1}{m} < \frac{1}{n_1}$ and...

therefore an element in $(0,1]$ but not in not in $\bigcup_{i=1}^k (\frac{1}{n_i},n_i)$, a contradiction.
Does this sound right?

Lozansky

$\partial_i \epsilon_{ijk} A_j B_k = \epsilon_{ijk} (A_{j,i}B_k+A_jB_{k,i})$ which I don't know how to complete

Maybe write out the implied sum?

Semiclassical

Well, start by writing out the first term as $\epsilon_{ijk}A_{j,i}B_k.$

Lozansky

5:58 PM

We should only get 6 terms (2 for every i = 1,2,3)

Balarka Sen

Fun fact about essential singularity: $f(z)$ has an essential singularity at $z = 0$ iff the various "directional limits" $\lim_{z \to 0} f(z)$ while approaching along various lines through origin in $\Bbb C = \Bbb R^2$, are lots and lots of stuff

To be rigorous, $f$ takes every possible complex value, with at most one exception, infinitely often in any puncture neighborhood of $0$.

Semiclassical

For that to match, it should be the case that $\epsilon_{ijk}A_{j,i}=(\nabla\times \mathbf{A})_k$.

Transcript for

$Mathematics$ Mathematics