Mathematics - 2017-12-21 (page 5 of 5)

Clarinetist

23:03

This will be the last time I ask today:

do any of you guys know the calculus of variations?

(why have people been starring my posts here? Lol)

3

Ted Shifrin

you have gained in popularity, Clarinet.

Clarinetist

Ted Shifrin

I know some calculus of variations, yeah. But there's a lot of different aspects.

Secret

So, under this adjustment, the pdf of $X$ independent of $Y$ is given by $P(X) : \Bbb{R}^2 \to [0,1]$ and the pdf of Y independent of X is given by $P(Y) : \Bbb{R}^2 \to [0,1]$. The quesiton then becomes an optimisation problem of minimising the eucledian metric of points X and Y subjected to the constraint that the correlation of X and Y is maximised.

The dream's version has P(X) and P(Y) having some asymmetric lorentzian shape, but I think it isn't hurt to generalise from that into arbitrary continuous pdfs.

Clarinetist

This, in short, is the problem:

Eric Silva

23:06

Yo

Clarinetist

Let $p: \mathbb{R} \to \mathbb{R}_{\geq 0}$ be such that $\int_{\mathbb{R}}p = 1$. Let
$$H(p) = -\int_{\mathbb{R}}p\log p\text{.}$$
We define $p(x)\log[p(x)] = 0$ whenever $p(x) = 0$.
Consider the problem
> $$\max H(p) \text{ subject to }\int_{\mathbb{R}}p = 1\text{.}$$

Ted Shifrin

yoyo @Eric

Eric Silva

How's it going

Clarinetist

I wish I knew something about the calc of variations, but I know... nothing. All I keep reading about are these Euler-Lagrange equations

Ted Shifrin

All that is is Lagrange multipliers in function space.

Clarinetist

23:07

Yeah, I figured that out as much

When I saw that, I thought, hmm, Lagrange multipliers

Ted Shifrin

Just think about it in terms of basic calculus. Suppose you're at a maximum point $p$. What do you know about $H(p+tq)$ for all $q$ with $\int (p+tq)=1$?

Clarinetist

Hm

Balarka Sen

Note that you're moving in a particular direction now, as $t$ is the only parameter varying

@Eric Hey

Daminark

Hey @Eric, @Ted, @Balarka!

Balarka Sen

Hey

Clarinetist

23:10

Well, $H(p+tq) \leq H(p)$... and also, $H^{\prime}(p+tq) \neq 0$

Balarka Sen

Vsauce Michael here

Ted Shifrin

So the directional derivative of $H$ in every direction $q$ subject to the constraint must be $0$.

Balarka Sen

@Clarinetist That's true. So what if you take $f(t) = H(p + tq)$. What can you say about $f(t)$?

Oh, Ted is being less Socratic than me

Ted Shifrin

LOL, for a change, yes.

Balarka Sen

But yeah $f'(t) = 0$ as $t = 0$

Ted Shifrin

23:11

DogAteMy!

Akiva Weinberger

You got a hat!

Ted Shifrin

I have 6 hats :P

Clarinetist

Let me process that quick

(or not so quick)

Ted Shifrin

I was wearing my menorah hat, but I got tired of it (and isn't Chanukah over?).

Akiva Weinberger

Yup

Balarka Sen

23:13

I am super wearied out after struggling with $\chi^2$ tests

DarkRunner

How do you get these hats?

Clarinetist

@BalarkaSen $\chi^2$ tests are useless IMO

Akiva Weinberger

So, something exiting happened recently:

pbs.twimg.com/media/DRezf3BU8AASJRT.jpg:large

Ted Shifrin

Balarka and MikeM used to work very hard for hats in years past. I never did. They just show up when they show up.

Clarinetist

Yes, I'm a stats grad student who said that

Akiva Weinberger

23:14

They finally duct-taped three space rockets together to get a bigger rocket

Balarka Sen

@Clarinetist O:

Akiva Weinberger

pbs.twimg.com/media/DRezf3EVAAAqm2s.jpg:large

It's gonna fly for the first time somewhere in January

Balarka Sen

I suppose the tape is made of something interesting

Akiva Weinberger

pbs.twimg.com/media/DRezf3DUEAAubA3.jpg:large

Balarka Sen

what is the point of SpaceX lol

Clarinetist

23:15

The thing that bugs me about hypothesis testing in general is that people who use it collect some data and do some $p$-hacking. VOILA, $p < 0.05$, PUBLISH!

Balarka Sen

they just do these weird things

Akiva Weinberger

They put up satellites

and people pay them for it

Clarinetist

Okay, here's the thing

Akiva Weinberger

They also resupply the ISS on occasion

Ted Shifrin

Clarinet: Back to calculus of variations before I leave.

Clarinetist

23:15

I understand that the directional derivative of $H$ in every direction $q$ subject to the constraint must be $0$

Now what?

Ted Shifrin

Well, $p+tq$ satisfies the constraint.

Akiva Weinberger

@BalarkaSen For its first flight, they don't want to put anything important on it in case it explodes, so it won't be carrying a satellite

Ted Shifrin

So what does that mean about $q$?

Balarka Sen

@AkivaW I see

Akiva Weinberger

So, instead, it will send "the silliest thing they can think of" to interplanetary space -

which, apparently, is the CEO's car.

So, assuming all goes according to plan, there will soon be a Tesla orbiting the sun

in an elliptical orbit tangent to Earth's and Mars's orbits

@BalarkaSen In all seriousness, it wasn't as simple as duct-taping them together. Apparently there's some engineering challenges that come with having 27 rocket engines firing next to each other at the same time. They've been planning it for a while, which is why I'm so excited at the fact that it's finally real.

(It's called the Falcon Heavy if you're curious)

Balarka Sen

23:20

I figured

Ted Shifrin

@Clarinet: So have you figured out that $\int q = 0$, i.e., $q\perp 1$?

Clarinetist

@TedShifrin Yeah, I was thinking that, but I couldn't justify to myself why

Ted Shifrin

Because $\int p = 1$ and $\int (p+tq) = 1$ for all (small) $t$.

Clarinetist

Ah, that's what I thought, but it seemed too simple

Akiva Weinberger

In other news, I was in my school play and my mother broke her leg

Secret

23:22

We don't need a floating tesla to orbit the sun to potentially collide to. However, it seems that SpaceX are finally launching, which is indeed exciting as these private companies will be supporting the space exploration stuff just in case the US government try to do something very bad to NASA that close it down

Balarka Sen

@Akiva Oh dear

Akiva Weinberger

(She was turning and braking too hard to avoid a car and the bike went out from under her)

Ted Shifrin

But now compute the directional derivative of $H$ in the direction of $q$. You should find that $q\perp$ some functional thing in $p$.

I'm glad nothing worse happened, DogAteMy.

Akiva Weinberger

@Secret "SpaceX are finally launching" They've launched 17 rockets this year

Ted Shifrin

I mean, my condolences to your mom, but it could have been scarier worse.

Clarinetist

23:23

hasn't thought of directional derivatives for at least 7 years ... I think I still remember the definition

Balarka Sen

I hope she recovers fast

Akiva Weinberger

Thank you. Me too

But it'll probably be several months

Ted Shifrin

Just take $d/dt\Big|_{t=0} H(p+tq)$, Clarinet.

Balarka Sen

Naturally. My broken finger took like a month to recover

Nowhere close to a broken leg.

Clarinetist

@TedShifrin Oh, easy enough. $\left. qH^{\prime}(p+tq)\right|_{t=0} = qH^{\prime}(p)$

Ted Shifrin

23:24

I had a sprained ankle that took something like 9 months to get back to normal. But I've been told that sprained ankles are worser than broken ankles.

Akiva Weinberger

@Secret Might be more than 17, actually, not sure. Another is scheduled for tomorrow, putting some Iridium NEXT satellites into space

Ted Shifrin

Um, no, @Clarinet.

Akiva Weinberger

The most recent launch was a successful ISS resupply mission

Ted Shifrin

You need to write things with the integral :P

Clarinetist

@TedShifrin Oh man. Hah.

Secret

23:25

@AkivaWeinberger I see, I am massively out of date in the space development, lol

Clarinetist

Let's see

Balarka Sen

Yeah sprains are bad. I was initially diagnosed with a sprain, but it turned out it's broken too

Clarinetist

$$H(p+tq) = \int_{\mathbb{R}}(p+tq)\log(p+tq)$$

so

assuming we can interchange differentiation and integration

Ted Shifrin

Of course ... This is applied stuff :P

Clarinetist

I should simplify first...

Ted Shifrin

23:27

No, you should just differentiate with product rule.

Clarinetist

Oh duh

$$H^{\prime}(p+tq) = \int_{\mathbb{R}}[(p+tq)\cdot\dfrac{q}{p+tq}+q\log(p+tq)]$$

Ted Shifrin

What's the derivative at $t=0$?

TheNotMe

@anon I got a similar question to the one I asked before. If you want a random subset of ${1,...,n}$ with probability proportional to $k(k-1)$ if the size of the subset is $k$, you must have the probability function $\frac{k(k-1)}{2^{n-2}n(n-1)}$.

Clarinetist

oops

$$H^{\prime}(p)=\int_{\mathbb{R}}\log(p)$$

Ted Shifrin

Your notation isn't quite right. As Balarka said, you're differentiating $f(t) = H(p+tq)$.

So no derivatives of $H$.

TheNotMe

23:29

this is kind of similar to picking a subcommittee of any size of a committee of length $n$ and then choosing 2 distinguished members out of this subcommittee

Ted Shifrin

That's not right, Clarinet. Come on.

Clarinetist

Ugh -_- I'm terrible at fast calculations. Let me give this another try

$$f(t) = H(p+tq)$$

Ted Shifrin

You need a $q$ in the answer under the integral, too, kiddo.

Clarinetist

so $f^{\prime}(t) = H^{\prime}(p+tq) \cdot q$

Ted Shifrin

Please don't write $H'$. You don't know what that is.

What you were writing above is basically right (with the integral).

Clarinetist

23:31

Oh... I completely get why $H^{\prime}$ is bad notation now -_-

$$f^{\prime}(t)=\int_{\mathbb{R}}\log(p)q$$

Ted Shifrin

NO ... First, it's $f'(0)$, not $f'(t)$ ... Second, you lost a term.

Clarinetist

Wow, why am I being so sloppy....

Ted Shifrin

shrugs

Lucas Henrique

hey guys.

Ted Shifrin

Serves me right for helping :P

Lucas Henrique

23:34

again

Ted Shifrin

hi again, @LucasH

Clarinetist

$$f^{\prime}(0) = \int_{\mathbb{R}}[q\log(p)+q]$$

Daminark

Hey @LucasHenrique!

Antonios-Alexandros Robotis

what's up

hi @LucasHenrique

Ted Shifrin

OK, so that has to be $0$ for all $q$ with $\int q = 0$, i.e., $q\perp 1$. What do you conclude, Clarinet?

I think I said hi to @Antonios already, but, if not ... hello.

Daminark

23:35

Hey @Antonios!

Clarinetist

So... we have $$\int_{\mathbb{R}}q\log(p)+0 = 0$$
hence $$\int_{\mathbb{R}}q\log(p) = 0$$

Antonios-Alexandros Robotis

Every year semester this happens... I get sick the day after my finals are done.

Daminark

So apparently argument principle is a thing for poles as well

Clarinetist

I know the above is an inner product (assuming $q$ and $\log p$ continuous)... but I'm not sure how I can make use of the $q \perp 1$ assumption

Akiva Weinberger

Isn't that one the one where you integrate the logarithmic derivative or something and the answer ends up being $N-P$?

$2\pi i(N-P)$. I was close.

Daminark

23:40

Yeah

Balarka Sen

yep

Daminark

So the way my complex prof did it was rather different

Akiva Weinberger

Right, 'cause you're essentially tracing out the value of the multifunction $\ln(f(z))$

Balarka Sen

Ya

Daminark

So let $f$ be an analytic function whose domain contains a disk $D$

Let $K = f(\partial D)$ and let $a\in \mathbb{C}-K$

Eric Silva

23:41

@Antonios-AlexandrosRobotis this happened to me too

Clarinetist

@TedShifrin I end up with $$\int_{\mathbb{R}}q\log(p) = 0$$ Is there something else I'm missing?

Eric Silva

it's like my body was hanging on and the floodgates opened right when i finally had time to relax

Akiva Weinberger

If $z$ goes around $0$ then $\ln(z)$ increases by $2\pi i$. In a sense, if $z$ goes around $\infty$ ("makes a big circle in the wrong direction") then $\ln(z)$ decreases by $2\pi i$.

Daminark

Then $f^{-1}(a)\cap D$ counted with multiplicity should be the winding number of $f(\partial D)$ around $a$

Akiva Weinberger

If $z$ goes around a root of $f(z)$, then $f(z)$ goes around $0$ and $\ln(f(z))$ increases by $2\pi i$

If $z$ goes around a pole of $f(z)$, then $f(z)$ "makes a big circle in the wrong direction" and $\ln(f(z))$ decreases by $2\pi i$

Daminark

23:44

Hmm, maybe you can formalize that using residue theorem?

Like, let's say $f$ has a pole of order $k$ at $a$

Then you should be able to write $f(z) = (z-a)^{-k}g(z)$ where $g$ is analytic

Akiva Weinberger

I imagine you take $\int f'/f\ dz$ and do some sort of change of variables

So like if $w=f(z)$ then $dw=f'dz$ and it's $\int 1/w\ dw$

So the only question is, what happens to the path of integration

And the answer should be something like what I outlined above

Daminark

Then $f'(z) = -k(z-a)^{-k-1}g(z) + (z-a)^{-k}g'(z)$

Secret

finally seeing something on chat that is not CW complex (or other incomprehensible) stuff

2

Akiva Weinberger

Wait, are there restrictions on how the change of variables works?

Clarinetist

^^ I starred that

Balarka Sen

23:46

What these two people are doing are topology

Daminark

So $\frac{f'(z)}{f(z)} = \frac{-k}{z-a} + \frac{g'(z)}{g(z)}$

Balarka Sen

Argument principle is 10/10 topology

Akiva Weinberger

Or is it only things like ${\rm blah}(w)=z$ ($w$ being the new variable) that you have to worry about

Clarinetist

@BalarkaSen It seemed like you were able to follow what Ted was telling me. Do you know what I'm missing?

Akiva Weinberger

(Cont'd) Probably

Secret

23:47

Topology is comprehensible, algebraic topology and algebraic groups is not so

Daminark

So that does it @Akiva

Secret

It will be uncountably long while later before I can make them comprehensible to me

Akiva Weinberger

@Daminark Why do we know that $g'(z)/g(z)$ have no poles?

@Secret The boundary of a ball is a sphere. The boundary of a sphere is the empty set. Thus, the $\rm boundary^2$ of a ball is the empty set.

Daminark

Well, when we write $f(z) = (z-a)^{-k}g(z)$, where $k$ is the order of the pole, we know $g(a) \ne 0$

Akiva Weinberger

Now you know the most important idea behind homology

@Daminark Ah right

So that does it

but you never used the fact that it's $(\ln(f))'$

Daminark

23:49

Yup

Akiva Weinberger

the logarithmic derivative

Daminark

I didn't touch the log at all, no, just computed it out really

Akiva Weinberger

which makes it seem like magic

The main reason why it works is that it's the logarithmic derivative

skullpatrol

math is the magic of logical ideas

Secret

It's magic until you understand it, otherwise, it is magical but understandable

Akiva Weinberger

23:51

Neither of those sentences make sense in this context I think

Daminark

Yeah that's fair, my proof disguised the topology going on

Ted Shifrin

@Clarinet: Sorry. My landlord dropped in.

Akiva Weinberger

@Daminark Like, you know how $fg$ has $N_f+N_g$ roots and $P_f+P_g$ poles

nbro

Hi

Akiva Weinberger

(where $f$ has $N_f$ roots etc)

nbro

23:52

I am curious, what's your background Akiva?

Ted Shifrin

You should have $\int q[\log p + 1] = 0$, @Clarinet.

For all $q\perp 1$. So what do you conclude?

Akiva Weinberger

So if you do the arg principle on $fg$ you should get $(N_f-P_f)+(N_g-P_g)$, the sum of doing it on the two thingies individually

Secret

Akiva: It's more generalised to all maths, not just this result. Recall all those first encounters when people taught us who to solve problems by doing nothing, how to find the best contour so we can blow the imaginary part to zero, as well other neat tricks

(and my response is mainly based on skullpatrol, as otherwise that post of mine will not exist)

Akiva Weinberger

which should scream "logarithms" probably

@nbro Last year of high school

skullpatrol

yeah, it was a play on the famous Einstein quote about math :P

Akiva Weinberger

23:55

(and very tired at the moment, which you might be able to tell)

nbro

Oh, it looks like you're already into advanced stuff and you very well trained

Ted Shifrin

as one of my best friends loves to say, DogAteMy, 'wait until you're my age!'

Clarinetist

@TedShifrin Hmm. I'm trying to figure out what happens with the $\log p$, because clearly $\int q = 0$. looks back at directional derivative stuff

Daminark

@Akiva yeah that's true, I guess I'm less comfortable with the whole "going around a loop" stuff because I'm not really too good with jumping back and forth between the picture and the formal argument

Ted Shifrin

Oh. Leave it as $\int q[1+\log p] = 0$ for all $q$ with $\int q = 0$. Think $L^2$ orthogonality, as I kept hinting.

Akiva Weinberger

23:56

I just did Poincare's lemma in Rudin (if $d\omega=0$ then $\omega$ is the derivative of something), and I felt like the proof was horribly written

Secret

nbro: He has background on very high level stuff such as model theory, as well various analysis and geometry stuff

Ted Shifrin

DogAteMy: Rudin's multivariable is horrendous. Use my book.

Secret

Daminark prefer picture free proofs

while I am opposite

Ted Shifrin

We're not talking about Demonark.

Daminark

Not exactly

Akiva Weinberger

23:57

Wait, does he?

Do you?

Daminark

It's more that I'm not good at generating the pictures, so while I'm happy to see the pictorial idea of something, I'm only truly comfortable when things have been written down

Clarinetist

@TedShifrin This can't be right... does that mean that $1+\log p = 0$?

Ted Shifrin

NOOO.

Clarinetist

I was going to say

Ted Shifrin

Think about orthogonal complements.

Secret

23:58

any questions : ) ?

Akiva Weinberger

@Daminark Oh, also

skullpatrol

How's the tooth professor @TedShifrin?

Daminark

Who's the "tooth professor"? :P

Ted Shifrin

LOL, thanks for asking, skull. Got a brand new spanking permanent crown at 8 AM today.

Akiva Weinberger

If you do it to $1/f$ you should get $P_f-N_f$, right?

The negative, 'cause poles and roots switch

Ted Shifrin

23:59

Demonark: I applaud your plea for correct comma usage. Now if only we could fix apostrophes.

skullpatrol

Cool :-)

Ted Shifrin

DogAteMy: What is $d(1/f)/(1/f)$?

Akiva Weinberger

@TedShifrin $-df/f$

I assume

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$Mathematics$ Mathematics