« first day (3636 days earlier)      last day (1681 days later) » 
00:00 - 22:0022:00 - 00:00

Sophie
Sophie
00:00
: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
Mike Miller
Right, and that's just a "codimension 3" subspace of $L^2(\Bbb R)$
Vey infinite dimensional still
Sophie
Sophie
what's $L$?
Thorgott
Thorgott
Why codimension $3$ in quotation? Isn't it literally that?
Mike Miller
Mike Miller
What's codimension mean
Yeah sure that's fine actually whatever
Thorgott
Thorgott
dimension of quotient
Sophie
Sophie
00:04
@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
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
Thorgott
you need to change the setup for $\mathbb{C}$
Balarka Sen
Balarka Sen
@Fargle OK cool. I will tell you the h-principle someday
Thorgott
Thorgott
integrating holomorphic functions over the entire plane is not gonna happen
Sophie
Sophie
hmm that's an unsatisfying outcome for my idea, but I'm glad it has a mathematical object corresponding to it
Fargle
Fargle
00:05
@BalarkaSen "not until you're older"
Balarka Sen
Balarka Sen
For now construct an immersion of RP^2 in R^3
Mike Miller
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
Thorgott
that's definitely true
Balarka Sen
Balarka Sen
@MikeMiller I know what you mean but what do you have in mind when you say this
Thorgott
Thorgott
I'm just saying you want a setup in which the integrals make sense
Ted Shifrin
Ted Shifrin
00:07
@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
Balarka Sen
Like what kind of infinite dimensional spaces
Mike Miller
Mike Miller
Hilbert spaces
Fargle
Fargle
@BalarkaSen well, time to do like 30 preliminary things
Ted Shifrin
Ted Shifrin
@Fargle: Oh, I got one word wrong. A perfectly canonical practice.
Fargle
Fargle
I like that
Balarka Sen
Balarka Sen
00:12
@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
Thorgott
I think it's a matter of what properties you want your measure to have to make it reasonable
Balarka Sen
Balarka Sen
My point is I have heard that physicists do bullshit but I no longer know what the bullshit is
Thorgott
Thorgott
functional integration and what-not
look at this shit en.wikipedia.org/wiki/Functional_integration
Balarka Sen
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
Thorgott
is the stuff in the Examples section meaningful?
Balarka Sen
Balarka Sen
00:19
The way it's written is most certainly not but I don't actually know what can't be formalized
Thorgott
Thorgott
why are they putting a $\square$ in the middle of two functinos
Balarka Sen
Balarka Sen
whatever I have seen being claimed to be bullshit is completely formal in Kolmogorov's language
Mike Miller
Mike Miller
The space of paths on a manifold is not a Hilbert space
lmfao
But sure on $\Bbb R^n$
Balarka Sen
Balarka Sen
take R^n
It's a Hilbert manifold is what I meant
Ted Shifrin
Ted Shifrin
What's the inner product of two paths? What does it mean for them to be orthogonal?
Mike Miller
Mike Miller
00:21
There's no issue in defining $W^{1,2}([0,1]; \Bbb R^n)$ as a Hilbert space
Balarka Sen
Balarka Sen
It's the L^2 inner product of the derivatives, Ted
Ted Shifrin
Ted Shifrin
Integrated.
Not much geometry to that.
OK.
Balarka Sen
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
Balarka Sen
00:43
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
Mike Miller
@BalarkaSen Still, that's an infinite dimensional space. Still interesting.
user265855
user265855
01:40
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
Calvin Khor
02:14
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
user265855
02:28
would it be right to say that $f(n) = \mathcal{O}(n^{\epsilon})$ for any $\epsilon$
$\epsilon > 0$
Calvin Khor
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
Ted Shifrin
03:09
@user265855 No. For example, $\log n = \mathscr O(n^\epsilon)$ but it is not $\mathscr O(n^{1/n})$.
Calvin Khor
Calvin Khor
03:35
@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
 
1 hour later…
user265855
user265855
05:03
Ah, it needs to work for one $g(n) \to 0$ for $n \geq N$ but not all such $g(n)$
Ted Shifrin
Ted Shifrin
05:31
@user265855 It actually means $\lim\limits_{n\to\infty}\dfrac{\log f(n)}{\log n} = 0$.
 
2 hours later…
Edward Evans
Edward Evans
07:15
mornin'
NoName
NoName
Mornin'
Calvin Khor
Calvin Khor
mornin'
NoName
NoName
07:35
mornin'
NoseBleed
NoseBleed
morning¿
(removed)
Balarka Sen
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
Calvin Khor
yes i agree
well except the last bit lol
Balarka Sen
Balarka Sen
Maybe I misunderstood what you meant then
Calvin Khor
Calvin Khor
07:52
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
Balarka Sen
yeah i thought that's captured by the cylinder event conditions
Calvin Khor
Calvin Khor
ok, it probably screws up something about paths then
Balarka Sen
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
Calvin Khor
hmmmm
Calvin Khor
Calvin Khor
08:16
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
Balarka Sen
What's $C_\mu$?
Calvin Khor
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
Balarka Sen
Ah OK
Calvin Khor
Calvin Khor
i found this theorem without proof in a book 'intro. to uncertainty quantification' by Tim Sullivan
Balarka Sen
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
Calvin Khor
08:19
i think i found sazonov's paper
Balarka Sen
Balarka Sen
Because our cylinder event conditions seem to imply that's the covariance operator
Calvin Khor
Calvin Khor
i guess so
Balarka Sen
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
Calvin Khor
proven over 60 years ago
Balarka Sen
Balarka Sen
Very cool. I'll look into these matters deeply and write some blog sometime soon.
Thanks
Calvin Khor
Calvin Khor
08:25
np
 
3 hours later…
Thorgott
Thorgott
11:16
@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
user123456789
12:05
Hi
anyone online ?
I want to ask somehting very basic
Calvin Khor
Calvin Khor
hi, sure @user123456789
user123456789
user123456789
Why do we write ax^2 + bx + c = k (x - alpha)(x - beta) ?
Is k here always equal to a ?
Calvin Khor
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
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
Yuvraj
hi guys
hi mathematics
Calvin Khor
Calvin Khor
12:14
@Yuvraj hi
user123456789
user123456789
But what about the cases like (1/2x- A)(2x - B)
Calvin Khor
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
user123456789
Wait for a sec, I can't express what I want to ask, let me think how
Yuvraj
Yuvraj
user image
any help on this?
Calvin Khor
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
Yuvraj
12:20
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
user123456789
I understand!! Thanks @CalvinKhor
Calvin Khor
Calvin Khor
@user123456789 great! yw :)
Balarka Sen
Balarka Sen
12:45
@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
Calvin Khor
wow \varepsilon even in chat. maybe we should just silently redefine \epsilon every so often
Balarka Sen
Balarka Sen
$\renewcommand{\epsilon}{\varepsilon}$ $\newcommand{\vol}{\mathrm{vol}}$
$\vol(P_\epsilon)$
Nice
Calvin Khor
Calvin Khor
or we could star that message
i wonder if anyone would complain
Balarka Sen
Balarka Sen
nah
do it
Calvin Khor
Calvin Khor
kk should last for a while
Balarka Sen
Balarka Sen
12:56
yup
Thorgott
Thorgott
ok, that sounds reasonable
Balarka Sen
Balarka Sen
I haven't verified though
Thorgott
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
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
Thorgott
13:13
yeah, I have something similar
Balarka Sen
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
Thorgott
13:30
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
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
Thorgott
right, that's true
Balarka Sen
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
Thorgott
yeah, that's what I'm trying to do too
Balarka Sen
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
Thorgott
13:34
just getting that estimate on $\vol(P_{\epsilon})$ is kind of awkward
Balarka Sen
Balarka Sen
yeah, but it should be doable somehow
Thorgott
Thorgott
$P_{\varepsilon}$ is the already linearized pentagon to me
Balarka Sen
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
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
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
Thorgott
13:38
ah yeah, vertices
for some reason I always confuse vertices with edges
Balarka Sen
Balarka Sen
Gotcha, thats a good simplification
Ok now its some obvious Euclidean geometry lol
Thorgott
Thorgott
yeah, it should be, but how do I do Euclidean geometry
triangulate the pentagon
Edward Evans
Edward Evans
straight edge and compass
Thorgott
Thorgott
still ugly
Balarka Sen
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
Calvin Khor
14:01
@Thorgott did you notice that your \varepsilons and \epsilons are the same
Thorgott
Thorgott
14:12
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
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
Thorgott
14:31
$$p=p$$
$$\Phi_X^{\epsilon}(p)=p+\epsilon X_p+O(\epsilon^2)$$
$$\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p))=p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}+O(\epsilon^2)$$
$$\Phi_X^{-\epsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p)))=p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}-\epsilon X_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}}+O(\epsilon^2)$$
$$\Phi_Y^{-\epsilon}(\Phi_X^{-\epsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p))))=p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}-\epsilon X_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}}-\epsilon Y_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}-\epsilon X_{p+\epsilo…
Balarka Sen
Balarka Sen
Nice man
Thorgott
Thorgott
this is actually un-nice
cause I can't cancel simplify the vector field terms without losing an order of accuracy
let's plug this into the formula
Thorgott
Thorgott
14:45
$$\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
Edward Evans
ew
Thorgott
Thorgott
oh wait, I need the 5th term too for the formula
Balarka Sen
Balarka Sen
yeah but thats like order epsilon^2 anyway or something
Thorgott
Thorgott
no, it contributes a lot of summands, sec
ah, there's cancellation incoming
Balarka Sen
Balarka Sen
where are you getting O(eps^4) man
shouldnt all of those be eps^2 detblah + O(eps^4)
Thorgott
Thorgott
14:52
$$\det(\Phi_Y^{-\epsilon}(\Phi_X^{-\epsilon}(\Phi_Y^{\epsilon}(\Phi_X^{\epsilon}(p)))),p)=\epsilon\det(X_p,p)+\epsilon\det(Y_{p+\epsilon X_p},p)-\epsilon\det(X_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}},p)-\epsilon\det(Y_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}-\epsilon X_{p+\epsilon X_p+\epsilon Y_{p+\epsilon X_p}}},p)+O(\epsilon^2)$$
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
Balarka Sen
the [X, Y] edge is [Phi_X^eps^2, Phi_Y^eps^2]
not eps
Thorgott
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
Balarka Sen
ah ok gotchu
so this formula is useless lol
Thorgott
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
Balarka Sen
it makes logical sense yeah
Thorgott
Thorgott
15:04
or does it?
Balarka Sen
Balarka Sen
vsauce michael here
Thorgott
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
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
Thorgott
I agree it should be true
this really shouldn't be that bad
Balarka Sen
Balarka Sen
15:30
@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
Thorgott
16:27
that should be a proof upon triangulating, let's see
 
1 hour later…
Thorgott
Thorgott
17:37
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
sai-kartik
18:12
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
Mike Miller
19:13
In my uneducated guess: "Not often and for spurious reasons"
UmbQbify -Key20-
UmbQbify -Key20-
19:24
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
Ted Shifrin
The latter
The title should be informative but NOT the question.
user76284
user76284
@sai-kartik en.wikipedia.org/wiki/…
UmbQbify -Key20-
UmbQbify -Key20-
20:07
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
Thorgott
check your browser history?
Alessandro Codenotti
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
nbro
20:25
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-
UmbQbify -Key20-
@Alessandro Codenotti, well it wasn't the immediate previous tab by that time
@Thorgott, well, I had already browsed a lot
Drathora
Drathora
20:51
@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
nbro
21:10
@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
Drathora
Ah
Yes, it's a "latent" variable
nbro
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
Drathora
Okay fine, I'll give you the short answer. Yes, z is a random variable
nbro
nbro
but if z is a random variable, then what's the meaning of p(z)?
Drathora
Drathora
The density of the distribution that's associated with z
nbro
nbro
21:17
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
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
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
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
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
Drathora
yup
nbro
nbro
21:26
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
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
nbro
but we write p(D|x)
What does it mean then that we are defining the likelihood of realizations?
Drathora
Drathora
So that's the probability of our dataset D given our observations x
nbro
nbro
no, sorry, I didn't mean x to mean observations
forget that notation
use this one p(D|theta)
Drathora
Drathora
Alright, and what are D and theta representing here?
nbro
nbro
21:28
D is your dataset
theta are tha parameters you want to find by maximizing the likelihood
Drathora
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
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
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
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
Drathora
yes, or rather it's a family of functions from the domain of D to likelihoods
nbro
nbro
21:37
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
Drathora
yes
nbro
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
Drathora
21:53
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
nbro
I want to assume that p is equal to the Gaussian density or Gaussian function
Drathora
Drathora
And then for specific values of y, p(y:x,y) will give you the conditional density at that point
nbro
nbro
right, so I should probably distinguish between y (NOT bold) and x (BOLD, which is an actual realization), right?
Drathora
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
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
user image
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
00:00 - 22:0022:00 - 00:00

« first day (3636 days earlier)      last day (1681 days later) »