Mathematics - 2020-07-18 (page 1 of 2)

Sophie

12:00 AM

:54981574hmm that's true and I hadn't realized it so I just need $\int c_1 h = \int c_2 h = \int c_3 h = 0$

Mike Miller

Right, and that's just a "codimension 3" subspace of $L^2(\Bbb R)$

Vey infinite dimensional still

Sophie

what's $L$?

Thorgott

Why codimension $3$ in quotation? Isn't it literally that?

Mike Miller

What's codimension mean

Yeah sure that's fine actually whatever

Thorgott

dimension of quotient

Sophie

12:04 AM

@MikeMiller Mike, so if we require $h$ to be infinitely differentiable or we change to $\mathbb{C}$ and require it to be holomorphic does that make it finite dimensional?

Mike Miller

no

Unless the space of functions $f$ is finite dimensional to begin with, no

There's also no reasonable measure on most infinite-dimensional spaces, I should mention. That's physics tomfoolery

Thorgott

you need to change the setup for $\mathbb{C}$

Balarka Sen

@Fargle OK cool. I will tell you the h-principle someday

Thorgott

integrating holomorphic functions over the entire plane is not gonna happen

Sophie

hmm that's an unsatisfying outcome for my idea, but I'm glad it has a mathematical object corresponding to it

Fargle

12:05 AM

@BalarkaSen "not until you're older"

Balarka Sen

For now construct an immersion of RP^2 in R^3

Mike Miller

I don't really care how the setup gets changed, $h \mapsto \langle \int c_1 h, \int c_2 h, \int c_3 h\rangle$ is a linear map to a 3-dimensional space, and so its kernel is at most 3 dimensions less than the dimension of the space $h$ lives in

Thorgott

that's definitely true

Balarka Sen

@MikeMiller I know what you mean but what do you have in mind when you say this

Thorgott

I'm just saying you want a setup in which the integrals make sense

Ted Shifrin

12:07 AM

@Fargle You need to (re)read Oscar Wilde's The Importance of Being Earnest. One of my favorite lines: "Indeed, the immersion of adults is a quite canonical practice."

2

Balarka Sen

Like what kind of infinite dimensional spaces

Mike Miller

Hilbert spaces

Fargle

@BalarkaSen well, time to do like 30 preliminary things

Ted Shifrin

@Fargle: Oh, I got one word wrong. A perfectly canonical practice.

Fargle

I like that

Balarka Sen

12:12 AM

@MikeMiller Space of W^{1, 2} paths on a manifold is an important example of a Hilbet space for me and they do admit good measures, is my qualm

That's part of Kolmogorov's construction of a Brownian motion

I think by and large construction nice measures on Hilbert spaces are not actually a problem; you can do this with space of maps between manifolds or something

Thorgott

I think it's a matter of what properties you want your measure to have to make it reasonable

Balarka Sen

My point is I have heard that physicists do bullshit but I no longer know what the bullshit is

Thorgott

functional integration and what-not

look at this shit en.wikipedia.org/wiki/Functional_integration

Balarka Sen

The Wiener integral is actually meaningful

Physicists write it in the following nonsense way: $d\Bbb P_{physics}(\gamma) = e^{-1/2 \int_0^1 \|\gamma'\|^2} d\gamma$

Thorgott

is the stuff in the Examples section meaningful?

Balarka Sen

12:19 AM

The way it's written is most certainly not but I don't actually know what can't be formalized

Thorgott

why are they putting a $\square$ in the middle of two functinos

Balarka Sen

whatever I have seen being claimed to be bullshit is completely formal in Kolmogorov's language

Mike Miller

The space of paths on a manifold is not a Hilbert space

lmfao

But sure on $\Bbb R^n$

Balarka Sen

take R^n

It's a Hilbert manifold is what I meant

Ted Shifrin

What's the inner product of two paths? What does it mean for them to be orthogonal?

Mike Miller

12:21 AM

There's no issue in defining $W^{1,2}([0,1]; \Bbb R^n)$ as a Hilbert space

Balarka Sen

It's the L^2 inner product of the derivatives, Ted

Ted Shifrin

Integrated.

Not much geometry to that.

OK.

Balarka Sen

@MikeMiller Oh maybe I'm just wrong. I have constructed here the details of a measure on $C([0, 1]; \Bbb R^n)$, but a.s. a random path wrt this measure is not $1/2$-Holder continuous. But then $W^{1,2}([0, 1]; \Bbb R^n)$ is measure zero wrt this

Because it Sobolev embeds in $C^{1, 1/2}([0, 1]; \Bbb R^n)$

You might be right. This is surprising

Balarka Sen

12:43 AM

I looked around and found this: en.wikipedia.org/wiki/Abstract_Wiener_space. This seems to suggest one doesn't usually easily get good measures on separable Hilbert spaces, so one embeds it in separable Banach spaces to upgrade the "canonical Gaussian cylinder events" to a measure.

$W^{1, 2}([0, 1]; \Bbb R^n) \hookrightarrow C([0, 1]; \Bbb R^n)$ seems to be an example of my setup. OK I understand now

Thanks

So path integrals are indeed nonsense

Mike Miller

@BalarkaSen Still, that's an infinite dimensional space. Still interesting.

user265855

1:40 AM

What does $ f(n) = \mathcal{O}(n^{o(1)})$

mean

i know the definition of the landau big oh and little oh notation but intuitively I don't know that that expression mean

Calvin Khor

2:14 AM

it means there is some $C,N>0$ and some $g(n)\to 0$ such that $|f(n)|\le C n^{g(n)}$ for all $n>N$

user265855

2:28 AM

would it be right to say that $f(n) = \mathcal{O}(n^{\epsilon})$ for any $\epsilon$

$\epsilon > 0$

Calvin Khor

@BalarkaSen I don't remember very much about Gaussian measures but I'm pretty sure you could build one on a Hilbert space instead of Banach

Ted Shifrin

3:09 AM

@user265855 No. For example, $\log n = \mathscr O(n^\epsilon)$ but it is not $\mathscr O(n^{1/n})$.

Calvin Khor

3:35 AM

@user265855 it would be correct that $O(n^{1/n})$ implies $O(n^\epsilon)$ for any $\epsilon$, but this is a strictly weaker statement as @TedShifrin's example shows

user265855

5:03 AM

Ah, it needs to work for one $g(n) \to 0$ for $n \geq N$ but not all such $g(n)$

Ted Shifrin

5:31 AM

@user265855 It actually means $\lim\limits_{n\to\infty}\dfrac{\log f(n)}{\log n} = 0$.

Edward Evans

7:15 AM

mornin'

NoName

Mornin'

Calvin Khor

mornin'

NoName

7:35 AM

mornin'

NoseBleed

morning¿

(removed)

Balarka Sen

@CalvinKhor I would be surprised because it's a fact that any reasonable random model of paths is a Brownian motion which as I proved are not W^{1,2}

Where reasonable means iid increments with finite mean and variance, and sample continuous (almost surely every random path in your model is continuous). This forces the increments to be Gaussian

Essentially intuitively because of central limit theorem but I think a rigorous proof factors through stochastic calculus. See here.

I dont mean iid I mean independent, and stationary increments.

So I am certain every measure on C[0, 1] which are standard in this sense on the finite-dimensional cylinder events (path hits x_0, ..., x_n at time t = t_0, ..., t_n resp) restricts to zero measure on W^{1,2}[0, 1]

So there is no such measure as you propose

Calvin Khor

yes i agree

well except the last bit lol

Balarka Sen

Maybe I misunderstood what you meant then

Calvin Khor

7:52 AM

i just wanted a measure with the right errr... projections to finite dimensional space

its been a while so im gonna say wrong things

but i think you can do that just by prescribing a mean and covariance operator....?

Balarka Sen

yeah i thought that's captured by the cylinder event conditions

Calvin Khor

ok, it probably screws up something about paths then

Balarka Sen

hmm i see

maybe whatever measure on W^{1,2}[0, 1] you have in mind is just not extendable to all of C[0, 1] but that also seems suspicious because since its defined in a Kolmogorov consistent way on the finite dim cylinder events like you said, it Kolmogorov extends to all of R^[0, 1], and then Kolmogorov continuity forces it back on C[0, 1]

since you have variance on the increments and whatnot

Calvin Khor

hmmmm

Calvin Khor

8:16 AM

Theorem 2.49 (Sazonov, 1958). Let $\mu$ be a centred Gaussian measure on a separable Hilbert space $\mathcal{H} .$ Then $C_{\mu}: \mathcal{H} \rightarrow \mathcal{H}$ is trace class and
\[
\operatorname{tr}\left(C_{\mu}\right)=\int_{\mathcal{H}}\|x\|^{2} \mathrm{d} \mu(x).
\]
Conversely, if $K: \mathcal{H} \rightarrow \mathcal{H}$ is positive, self-adjoint and of trace class, then there is a Gaussian measure $\mu$ on $\mathcal{H}$ such that $C_{\mu}=K$.
Sazonov's theorem is often stated in terms of the square root $C_{\mu}^{1 / 2}$ of $C_{\mu}:$ $C_{\mu}^{1 / 2}$ is Hilbert-Schmidt, i.e. ha…

Balarka Sen

What's $C_\mu$?

Calvin Khor

Definition 2.41. A Borel measure $\mu$ on a normed vector space $\mathcal{V}$ is said to be a (non-degenerate) Gaussian measure if, for every continuous linear functional $\ell: \mathcal{V} \rightarrow \mathbb{R},$ the push-forward measure $\ell_{*} \mu$ is a (non-degenerate) Gaussian measure on $\mathbb{R} .$ Equivalently, $\mu$ is Gaussian if, for every linear map $T: \mathcal{V} \rightarrow \mathbb{R}^{d}, T_{*} \mu=\mathcal{N}\left(m_{T}, C_{T}\right)$ for some $m_{T} \in \mathbb{R}^{d}$ and some symmetric

@BalarkaSen That would be its covariance operator

Balarka Sen

Ah OK

Calvin Khor

i found this theorem without proof in a book 'intro. to uncertainty quantification' by Tim Sullivan

Balarka Sen

Ah this is interesting. The identity map $W^{1,2}[0, 1] \to W^{1,2}[0, 1]$ is not trace class is the point, then?

Calvin Khor

8:19 AM

i think i found sazonov's paper

Balarka Sen

Because our cylinder event conditions seem to imply that's the covariance operator

Calvin Khor

i guess so

Balarka Sen

You can't have variance of increments proportional to the length of the increment intervals if you really do want a Gaussian measure

Got it

Nice theorem

Calvin Khor

proven over 60 years ago

Balarka Sen

Very cool. I'll look into these matters deeply and write some blog sometime soon.

Thanks

Calvin Khor

8:25 AM

np

Thorgott

11:16 AM

@Balarka Look at the $2$-form $dx\wedge dy$ on $\mathbb{R}^2$ with an intent gaze. Independent of our starting point $p$ and our vector fields $X,Y$, $\frac{1}{\operatorname{vol}(P_{\varepsilon})}\int_{P_{\varepsilon}}dx\wedge dy=1$ for all $\varepsilon>0$, so particularly in the limit. Meanwhile $(dx\wedge dy)(X,Y)(p)$ depends linearly on both $X,Y$. So how can they be equal?

In general, if our two-form is $fdx\wedge dy$, then $\frac{1}{\operatorname{vol}(P_{\varepsilon})}\int_{P_{\varepsilon}}fdx\wedge dy$ should tend to $f(p)$ as $\varepsilon\rightarrow0$. The difference between these terms is bounded above by $\sup_{x\in P_{\varepsilon}}|f(x)-f(p)|$, which should tend to $0$ as $\varepsilon\rightarrow0$ for all reasonably nice polygons at least.

I think the directions only come into play when you integrate over the boundary instead or something. For a smooth function $f$, $\frac{1}{t^2}\int_{[x_0,x_0+t]\times[y_0,y_0+t]}f$ tends to $f(x_0,y_0)$, but something like the line integral $\frac{1}{t}\int_{\gamma}f$ where $\gamma$ is the straight path of length $t$ from $(x_0,y_0)$ in the direction of $(x_1,y_1)$ tends to $\partial_{(x_1-x_0,y_1-y_0)}f(x_0,y_0)$.

user123456789

12:05 PM

Hi

anyone online ?

I want to ask somehting very basic

Calvin Khor

hi, sure @user123456789

user123456789

Why do we write ax^2 + bx + c = k (x - alpha)(x - beta) ?

Is k here always equal to a ?

Calvin Khor

k=a

yes

not everyone will write it like that, its not so important that you use k or a to write it

user123456789

I have seen many time that in p(x) which I mentioned above, when a is equal to 1, we write the factors as (x-alpha)(x- beta) since when we multiply it coeff of x^2 is one

Yuvraj

hi guys

hi mathematics

Calvin Khor

12:14 PM

@Yuvraj hi

user123456789

But what about the cases like (1/2x- A)(2x - B)

Calvin Khor

that's correct, but i people also write e.g. $(x-x_0) (x-x_1)$, and so on, there are countless variations, what is important is the meaning, not the symbols

@user123456789 then you open the brackets and you can find out what k or a are

user123456789

Wait for a sec, I can't express what I want to ask, let me think how

Yuvraj

any help on this?

Calvin Khor

is namakeen really a word

here's my input, because I don't feel like doing combinatorics. If you type it, its easier to read and talk about.

15. If $p=2^{u} \cdot 3^{b} \cdot 5^{c}, a, b, c \in N,$ is the number of words which can be formed using all the letters of the word 'NAMAKEEN'so that no two alike vowels are together then :
(a) number of odd divisors of $p$ is 6
(b) number of even divisors of $p$ is 42 .
(c) number of zeroes at the end of $p$ is 2 .
(d) number of zeroes at the end of $p$ is 1 .

@user123456789 sure thing

Yuvraj

12:20 PM

38 secs ago, by Calvin Khor

15. If $p=2^{u} \cdot 3^{b} \cdot 5^{c}, a, b, c \in N,$ is the number of words which can be formed using all the letters of the word 'NAMAKEEN'so that no two alike vowels are together then :
(a) number of odd divisors of $p$ is 6
(b) number of even divisors of $p$ is 42 .
(c) number of zeroes at the end of $p$ is 2 .
(d) number of zeroes at the end of $p$ is 1 .

@CalvinKhor thanks

i want to solve it using the elimination of the possible outcomes.

user123456789

I understand!! Thanks @CalvinKhor

Calvin Khor

@user123456789 great! yw :)

Balarka Sen

12:45 PM

@Thorgott Ah yes I meant $1/\varepsilon^2 \int_{P_\varepsilon} f dx\wedge dy$. This is like $f(p)$ times $\vol(P_\varepsilon)/\varepsilon^2$, which in the limit goes to $f(p)$ times area of $X \wedge Y$, i.e., $f(p) \det(X, Y)$

Sorry about that

Approximate $X$ and $Y$ by their linearizations $X(p)$ and $Y(p)$ considered as vector fields on $\Bbb R^2$. Then $P_\varepsilon$ is approximated by the rectangle spanned by $\varepsilon X(p)$, $\varepsilon Y(p)$ on $\Bbb R^2$, taking the area suggests $\vol(P_\varepsilon) \approx \varepsilon^2 \det(X(p), Y(p))$.

Just have to verify this is correct upto $O(\varepsilon^3)$ terms

Calvin Khor

wow \varepsilon even in chat. maybe we should just silently redefine \epsilon every so often

Balarka Sen

$\renewcommand{\epsilon}{\varepsilon}$ $\newcommand{\vol}{\mathrm{vol}}$

$\vol(P_\epsilon)$

Nice

Calvin Khor

or we could star that message

i wonder if anyone would complain

Balarka Sen

nah

do it

Calvin Khor

kk should last for a while

Balarka Sen

12:56 PM

yup

Thorgott

ok, that sounds reasonable

Balarka Sen

I haven't verified though

Thorgott

I'll try

the potential difficulty is that I'm not sure how an error bound on approximating the vector field by its linearization gives a bound on approximating its flow by the linearization

Balarka Sen

yeah i feel like this is something i should know by heart but cant figure out when im half asleep

Suppose $X : \Bbb R^n \to \Bbb R^n$ is a vector field, solution to the IVP $\gamma'(t) = X(\gamma(t))$, $\gamma(0) = 0$ satisfies $\gamma(t) = \int_0^t X(\gamma(s)) ds$. Then $X(\gamma(s)) = X(0) + O(s)$, so $\gamma(t) = t X(0) + O(t^2)$, right?

Just using mean value inequality shit

Thorgott

1:13 PM

yeah, I have something similar

Balarka Sen

So the actual flowline of $X$ until time $t$ differs from the linearized version $tX(0)$ by $O(t^2)$

So if you have two things you expect area error to be $O(t^4)$

Which is good for us

$\vol(P_\epsilon) - \epsilon^2 \det(X(p), Y(p)) = O(\epsilon^4)$, so dividing by $\epsilon^2$ and taking limit gives the equality we want

I think some soft, dumb estimates like this are enough

Convinced?

Thorgott

1:30 PM

I think we need an estimate on their ratio, not their difference

Is there a non-headache way of writing down the area of an arbitrary pentagon

Balarka Sen

Well but if $\vol(P_\varepsilon) - \epsilon^2 \det(X(p), Y(p)) = O(\varepsilon^4)$ then $\lim_{\varepsilon \to 0} \vol(P_\varepsilon)/\varepsilon^2 = \det(X(p), Y(p))$

Thorgott

right, that's true

Balarka Sen

And then $1/\varepsilon^2 \int_{P_\varepsilon} \alpha = \vol(P_\varepsilon)/\varepsilon^2 \cdot 1/\vol(P_\varepsilon) \int_{P_\varepsilon} \alpha$, the first term goes to $\det(X(p), Y(p))$ in the limit and the second term goes to $f(p)$ in the limit where $\alpha = f dx \wedge dy$ by what you said earlier

So the whole thing goes to $\alpha_p(X, Y) = f(p) \det(X(p), Y(p))$

So this would settle everything I think

Thorgott

yeah, that's what I'm trying to do too

Balarka Sen

@Thorgott Yikes but I mean like sides of $P_\varepsilon$ are not straightlines. You really want to see that the edges are linear upto error $O(t^2)$ implies area is linear upto error $O(t^4)$... how to formalize this though?

Thorgott

1:34 PM

just getting that estimate on $\vol(P_{\epsilon})$ is kind of awkward

Balarka Sen

yeah, but it should be doable somehow

Thorgott

$P_{\varepsilon}$ is the already linearized pentagon to me

Balarka Sen

What is the linearized pentagon? Sides of $P_\varepsilon$ are flowlines of the nonlinear guys $X, Y, [X, Y]$

Once you linearized you get an actual rectangle because $[X(p), Y(p)] = 0$

Somehow want to compare the area of these guys

Thorgott

just a pentagon with edges $p,\Phi_X^{\epsilon}(p),\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p)),\Phi_X^{-\varepsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p))),\Phi_Y^{-\varepsilon}(\Phi_X^{-\varepsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p))))$

Balarka Sen

Ah ok like that

That works I guess yeah

As in not edges but vertices that?

Just draw a linear pentagon with the vertices of the nonlinear pentagon

Thorgott

1:38 PM

ah yeah, vertices

for some reason I always confuse vertices with edges

Balarka Sen

Gotcha, thats a good simplification

Ok now its some obvious Euclidean geometry lol

Thorgott

yeah, it should be, but how do I do Euclidean geometry

triangulate the pentagon

Edward Evans

straight edge and compass

Thorgott

still ugly

Balarka Sen

If you have a rectangle and a random polygon whose vertices are all $\delta$-close to vertices of the original rectangle then the area error should be $O(\delta^2)$

This feels like a Pick's theorem sorta deal

I dunno

Euclidean geometry is hard

i'll probably only verify this detail if someone puts a gun to my head

@Thorgott en.wikipedia.org/wiki/Shoelace_formula

That settles it right

Calvin Khor

2:01 PM

@Thorgott did you notice that your \varepsilons and \epsilons are the same

Thorgott

2:12 PM

ah, I didn't know that formula

I'll torture myself by writing this out in detail

@Calvin yeah, I did \epsilon, because I saw what you were doing earlier, but typing \varepsilon is so much in my habit that one slipped through

Balarka Sen

@Thorgott or just believe everything works out and use Stokes' theorem in this setup to get to the formula for $d\omega(X, Y)$

Thorgott

2:31 PM

$$p=p$$
$$\Phi_X^{\epsilon}(p)=p+\epsilon X_p+O(\epsilon^2)$$
$$\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p))=p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}+O(\epsilon^2)$$
$$\Phi_X^{-\epsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p)))=p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}-\epsilon X_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}}+O(\epsilon^2)$$
$$\Phi_Y^{-\epsilon}(\Phi_X^{-\epsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p))))=p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}-\epsilon X_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}}-\epsilon Y_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}-\epsilon X_{p+\epsilo…

Balarka Sen

Nice man

Thorgott

this is actually un-nice

cause I can't cancel simplify the vector field terms without losing an order of accuracy

let's plug this into the formula

Thorgott

2:45 PM

$$\det(p,\Phi_X^{\epsilon}(p))=\epsilon\det(p,X_p)+O(\epsilon^2)$$
$$\det(\Phi_X^{\epsilon}(p),\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p)))=\epsilon\det(p,Y_{p+\epsilon X_p})+O(\epsilon^2)$$
$$\det(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p)),\Phi_X^{-\epsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p))))=-\epsilon\det(p,X_{p-\epsilon X_p+\epsilon Y_{p+\epsilon X_p}})+O(\epsilon^2)$$
$$\det(\Phi_X^{-\epsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p))),\Phi_Y^{-\epsilon}(\Phi_X^{-\epsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p)))))=-\epsilon\det(p,Y_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}-\epsilon X_{…

Edward Evans

ew

Thorgott

oh wait, I need the 5th term too for the formula

Balarka Sen

yeah but thats like order epsilon^2 anyway or something

Thorgott

no, it contributes a lot of summands, sec

ah, there's cancellation incoming

Balarka Sen

where are you getting O(eps^4) man

shouldnt all of those be eps^2 detblah + O(eps^4)

Thorgott

2:52 PM

$$\det(\Phi_Y^{-\epsilon}(\Phi_X^{-\epsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p)))),p)=\epsilon\det(X_p,p)+\epsilon\det(Y_{p+\epsilon X_p},p)-\epsilon\det(X_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}},p)-\epsilon\det(Y_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}-\epsilon X_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}}},p)+O(\epsilon^2)$$

wait, now everything cancels

something is off

this just shows the area is $O(\epsilon^2)$

which isn't tight enough

I'm not getting any $O(\epsilon^4)$ is the issue

well, I can

Balarka Sen

the [X, Y] edge is [Phi_X^eps^2, Phi_Y^eps^2]

not eps

Thorgott

that edge doesn't exist in this picture

just look at the first determinant, $\det(p,\Phi_X^{\epsilon}(p))=\det(p,p+\epsilon X_p+O(\epsilon^2))$, agree?

I can't approximate that any better than up to second order

Balarka Sen

ah ok gotchu

so this formula is useless lol

Thorgott

maybe it's just that I'm useless, probably both

ok, let's just try triangulating the pentagon, can't be worse than this

let's look at the first triangle with vertices $p$, $\Phi_X^{\epsilon}(p)$, $\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p))$, what's its area

urgh, this is horrible too

but it should be true that knowing the vertices up to $O(\epsilon^2)$ should give the area up to $O(\epsilon^4)$

Balarka Sen

it makes logical sense yeah

Thorgott

3:04 PM

or does it?

Balarka Sen

vsauce michael here

Thorgott

Consider a rectangle with fixed bottom left vertex and edges of length $x+O(\epsilon^2)$ and $y+O(\epsilon^2)$. The area is $xy+O(\epsilon^2)$ still, no? That's the same issue as happening in the above.

Because $O(\epsilon^2)O(\epsilon^2)=O(\epsilon^4)$, but $xO(\epsilon^2)=O(\epsilon^2)$

Even simpler, assume the height of the rectangle is fixed $y$ and the bottom is $x+O(\epsilon^2)$

the area is a linear function of the bottom length and approximated as $xy+O(\epsilon^2)$

Balarka Sen

im too sleepy to do this right now but it should be true that vol(P_eps)/eps^2 -> area(X, Y)

Thorgott

I agree it should be true

this really shouldn't be that bad

Balarka Sen

3:30 PM

@Thorgott The point is the "two adjacent edges" of P_eps are of length eps|X| + O(eps^2) and eps|Y| + O(eps^2) respectively with angle theta between them being the angle as X and Y on the tangent space at p. So the "area of P_eps" is (eps|X| + O(eps^2))*(eps|Y| + O(eps^2))sin(theta) = eps^2|X||Y|sin(theta) + O(eps^3) = eps^2*area(X, Y) + O(eps^3)

So there is no technical problem in the naive reasoning as I see it

You're right you don't get O(eps^4). You do get O(eps^3)

It's stronger than saying the error in edges is O(eps^2), it's that while you shrink everything at the same time

But I dont have anything rigorous to say so I'll back off and go to sleep

Thorgott

4:27 PM

that should be a proof upon triangulating, let's see

Thorgott

5:37 PM

yeah, I'm absolutely convinced this works, will write down the details later

Actually, this tells us something more about the picture we would have intuitively expected either way. Triangulate the pentagon as follows: One triangle has vertices $p,\Phi_X^{\epsilon}(p),\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p))$, the next one has vertices $\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p)),\Phi_X^{-\epsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p))),p$ and the last one has vertices $\Phi_X^{-\epsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p))),\Phi_Y^{-\epsilon}(\Phi_X^{-\epsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p)))),p$.

sai-kartik

6:12 PM

So I recently came across the topic of Quantum Calculus on wiki. Where and why is it used?

Quantum calculus

Quantum calculus, sometimes called calculus without limits, is equivalent to traditional infinitesimal calculus without the notion of limits. It defines "q-calculus" and "h-calculus", where h ostensibly stands for Planck's constant while q stands for quantum. The two parameters are related by the formula q = e i h = e 2 π i ℏ {\displaystyle q=e^{ih}=e^{2\pi i\hbar...

Mike Miller

7:13 PM

In my uneducated guess: "Not often and for spurious reasons"

UmbQbify -Key20-

7:24 PM

Some people say that title should be informative and subjective, (put the formula in title) some say question should be in body and not in title (avoid formula in title) so which one is correct..?

Ted Shifrin

The latter

The title should be informative but NOT the question.

user76284

@sai-kartik en.wikipedia.org/wiki/…

UmbQbify -Key20-

8:07 PM

I found an answer, which had some useful facts.. and I closed the tab it because it very basic, now I need it but I can't find it (T_T)

Thorgott

check your browser history?

Alessandro Codenotti

For the future ctrl+shift+t opens the last tab you closed on firefox and I suppose other browsers have similar shortcuts

nbro

8:25 PM

Can someone explain to me what the author means by "variable" in section 1 of the paper arxiv.org/pdf/1601.00670.pdf (page 2)? Is he referring to a random variable or what?

I don't think he's talking about random variables here, because if z is a random variable or vector, then p(z) isn't meaningful or am I wrong? p seems to be a density

but if he's not talking about random variables, then how can we formulate the same thing by taking into account also the random variables

UmbQbify -Key20-

@Alessandro Codenotti, well it wasn't the immediate previous tab by that time

@Thorgott, well, I had already browsed a lot

Drathora

8:51 PM

@nbro why don't you think that the density of a random variable isn't meaningful?

These are exactly the kinds of variables that might have associated densities wouldn't you say?

nbro

9:10 PM

@Drathora I am not saying of a random variable, I am saying of a "variable"

variable != random variable (in general)

The author says "suppose we have a VARIABLE z with density p(z)"

but how come a variable have a density?

That's my question

I don't understand if the author us talking about variables or random variables

Drathora

Ah

Yes, it's a "latent" variable

nbro

omg, no

that's has nothing to do with my question

I am not talking about latent vs observable variables

I am talking about VARIABLE vs RANDOM VARIABLE

Drathora

Okay fine, I'll give you the short answer. Yes, z is a random variable

nbro

but if z is a random variable, then what's the meaning of p(z)?

Drathora

The density of the distribution that's associated with z

nbro

9:17 PM

but z in p(z) is the input to the density

it's like x in f(x), where f is some function like x^2

Drathora

Yes, because the "output" of z is the point at which we want to evaluate the density

So we evaluate p at the point z

It's awkward notation since the variable name is also being used to represent the output of the variable

nbro

so z represents both the random variable and its realization?

that's also the most likeliky thing that I thought he was doing and meaning

or, let's pretend that this was the case

Drathora

Yes. Some authors will distinguish and use Z as the variable name and z as the individual values

Something that might be useful is to forget that "variable != random variable" in general

And just view every variable as a random variable

With deterministic variables having an associated distribution that assigns all probability mass to a single point

nbro

so when someone says "consider variable z with density p(z)" he means that there's a random variable Z with realization z and then we use another variable with the same name z as the realization to index the density p?

So, there are 3 different entities:

1. The random variable Z

2. The realization of the random variable z

3. The variable of the density function z

So, we use the same notation to refer to both 2 and 3

and sometimes we use the same notation to refer to 1 and 2, so it may happen that we refer to both 1, 2, 3 by the same letter, although they are different

Drathora

yup

nbro

9:26 PM

but there's still something I don't understand

sometimes we want to know the likelihood of a dataset. In that case, we want to know the likelihood of realizations of random variables

Drathora

In my own work I'd usually phrase it as "consider a random variable z with associated density p". But I guess this author did it differently

nbro

but we write p(D|x)

What does it mean then that we are defining the likelihood of realizations?

Drathora

So that's the probability of our dataset D given our observations x

nbro

no, sorry, I didn't mean x to mean observations

forget that notation

use this one p(D|theta)

Drathora

Alright, and what are D and theta representing here?

nbro

9:28 PM

D is your dataset

theta are tha parameters you want to find by maximizing the likelihood

Drathora

Alright. So yeah so p(D:theta) is the conditional density of the random variable D given the parameters theta

So when evaluating at a point, you'd of course have a realization of D

nbro

so, we can indeed evaluate a density at a realization

and so x in p(x) is not always the dummy variable but it can also be a realization of X

So, we can talk about "likelihood of dataset"

because are indeed fixing where we are evaluating the likelihood

Drathora

Yeah, you'll either be fixing a point D and evaluating there

Or you'll have it as a variable D and p(D:theta) will be written as a function of D and theta

nbro

So, I am confused for a good reason

@Drathora right, but in that case it won't be a likelihood, right, but a family of conditional probabilities, one for each configuration of theta, right?

Drathora

yes, or rather it's a family of functions from the domain of D to likelihoods

nbro

9:37 PM

I am still confused because I wanted to show that maximization of the log-likelihood is equivalent to the minimization of MSE

I showed it and the math works

but I am still confused even though I did it

I mean, I don't really and fully understand what I did

or what I said

This is because I assumed I am given a dataset D = {(x_i, y_i)}

So, these are realizations of Z_i = (X_i, Y_i)

So far so good, right?

Drathora

yes

nbro

Then I said that Z_i follows the joint p(x, y)

So for so good

right?

Then I assume that I am given an arbitrary x (in bold)

and I say I define $p(\mathbf{y} | \mathbf{y}, \mathbf{w})$

is a normal distribution

Sorry, I meant $p(\mathbf{y} | \mathbf{x}, \mathbf{w})$

Now, what the heck is $\mathbf{y}$ and $\mathbf{x}$ here?

$\mathbf{y}$ should be a dummy variable (I think)

but $\mathbf{x}$ is actually the given arbitrary input

maybe I should use $p(y | \mathbf{x}, \mathbf{w})$?

Drathora

9:53 PM

Sorry, is p a probability here or a density?

Either way, yes, if you want it to be a distribution or density function, y should be a variable

nbro

I want to assume that p is equal to the Gaussian density or Gaussian function

Drathora

And then for specific values of y, p(y:x,y) will give you the conditional density at that point

nbro

right, so I should probably distinguish between y (NOT bold) and x (BOLD, which is an actual realization), right?

Drathora

All three of these can be variables though, and then you end up with a complicated conditional function

But yeah, I would distinguish in some way between variables and fixed realizations of those variables

It helps a lot with clarity

nbro

right, that's what I want to do: achieve clarify and remove ambiguity

but there's still a problem

I will be finding a point estimate of the Gaussian by maximizing it with respect to w

right

and to show that max of log-likelihood is equivalence to minimization of MSE, then, at some point, I will need to have the actual CORRECT label

let me show you what I did

you see the problem there?

in step 3.11, $\mathbf{y}$ seems to be fixed (i.e. the correct label), but in step 3.5 it's a variable

I think that in all cases $\mathbf{y}$ is fixed

Because we are assuming that we know $\mathbf{y}$

I think that's the explanation

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