« first day (2939 days earlier)      last day (1994 days later) » 
00:00 - 21:0021:00 - 00:00

12:01 AM
You could always try going to the "other side", i.e. banks, where results are published still I guess (I suppose it's more mathematical, though, and the "framework" is generally and academically agreed upon)
 
I don't think I would actively apply for a banking position
I'd rather apply for positions like the one at openai
 
Did you apply to Google and/or DeepMind?
 
I applied to google
I applied to deepmind a while back
neither have ever gotten back to me lol
openai was cool in that they at least got back to me I guess...
 
 
2 hours later…
1:58 AM
@Semiclassical are you around?
If S is real symmetri, T is real anti-symmetric show that $det(T+iS-1)$ is non zero
Not quite sure on how to approach
 
2:12 AM
@JakeRose to be clear: $S,T$ are real matrices such that $T^\top =-T$ and $S^\top =S$?
 
yep
 
presumably that $1$ is actually the identity matrix as well
 
oh yes apologies
$det(T+iS-I)$
 
np. also, for future reference you can do \det in latex
 
$\det$
nice
 
2:13 AM
yeah
strangely, \tr doesn't give trace
anyways. There's a few different ideas which come to mind. one is to assume the opposite: What do we know about the case when $\det(T+iS-I)=0$?
or more generally when any matrix $A$ has zero determinant?
 
rows/columns are linearly dependent
 
sure, that's one. there's a few
 
erm
Theyre not coming to mind sadly
 
well, what do you call such a matrix?
there's a particular name
 
non-invertible?
 
2:18 AM
that'll do. I had singular in mind
but that's more to the point
and then you've got the invertible matrix theorem: en.wikipedia.org/wiki/…
So, that's a lot :)
 
very very tre
tru
true*
 
so we're basically trying to show that $T+iS-I$ is invertible
 
agreed
 
I'm not sure how they intend you to do that, and the following is probably not it. But here's one way to get some intuition for it
First, let $Z=T+iS$. Such a matrix with the stated symmetries is said to be Hermitian, but for the present case it's convenient
And second, we might as well look at $I-Z$ instead. So we're trying to see if $(I-Z)^{-1}$ exists.
 
Anti hermitian you mean?
 
2:23 AM
No. Hermitian means that you take complex conjugate and transpose
oh derp
yeah, $Z=S+iT$ would be hermitian. So what you've got is antihermitian
let me call that $A=T+iS$ instead to avoid making that mistake
aaaanyways
we're trying to see if $(I-A)^{-1}$ exists.
hmm. phrased like that, I'm a little surprised it's true in general.
 
It says deduce
So it should be pretty trivial
 
yeah
let's try a different angle. The symmetries suggest looking at $(I-A)^\dagger = I+A$
and if $I-A$ is to be invertible, so must this
which to my mind suggests looking at $(I-A)(I+A)=I-A^2$
hrm. No, this seems like the wrong thread to pull
 
Funnily enough thats what i was just trying
 
yeah. it does tell us that $I-A^2$ should be invertible
but I don't think it goes both ways
Maybe eigenvalues are the way to go. If $I+A$ were singular, it'd have a zero eigenvalue i.e. $(I+A)v=0$
or equivalently $Av=-v$, i.e. $A$ would have an eigenvalue of $-1$...ahah, I see it
What do you know about the eigenvalues of antihermitian matrices?
 
Theyre imaginary
 
2:32 AM
yep, or zero
so can $-1$ ever be an eigenvalue of an antihermitian $A$?
 
nope
oh
OH
ooooh
Thats an eigenvalue equation with eigenvalue of -1
 
I see that
 
This also tells us how to generalize the problem: If you had $A-\lambda I$ with $\lambda$ some nonzero real number, then this would also be invertible since $A-\lambda I$ can't possibly have any zero eigenvectors
 
But dont get how it implies the determinant isnt 0
oh
 
2:35 AM
Think back to the long list of invertible matrix statements.
 
if it was an eigenvalue it would = 0
 
and so if it isnt an eignevalue
it cant = 0
 
Right. To put that more formally: Since $A$ is antihermitian, it only has imaginary or zero eigenvalues. Hence there exists no nonzero vector such that $Av=\lambda v$ whenever $\lambda$ is some complex number off the imaginary axis.
 
yes
 
2:38 AM
Hence the kernel of $A-\lambda I$ is trivial and $A-\lambda I$ is invertible.
More broadly, this tells us that $A-\lambda I$ is singular if and only if $\lambda$ is an eigenvalue of $A$.
(I think the iff holds.)
in any case, that was pleasant :)
 
Whats a kernel?
Dont think we come across that until next term?
Definitely do it in quantum right?
 
The kernel of a mapping is the part of the domain which gets mapped to zero
So for linear transformations it's vectors such that $Av=0$
So the kernel being trivial means that $Av=0$ is only true when $v=0$
It's on that list of invertible matrix stuff as well
this should be distinguished from the notion of an integral kernel. that's a different thing
(I'll never understand why kernel has two such different meanings. I guess they both arise from the word 'kernel' as root/seed, but it's a bit annoying)
 
3:19 AM
Huh I was always confused about what a kernel was and never knew that there are multiple things with the name
 
3:34 AM
yeah
reproducing kernel is the integral kernel meaning i think
and similarly the kernel trick
 
 
4 hours later…
7:21 AM
anyone here?
 
@EmilioPisanty I think that's a better question than many that make the HNQ, and I think your answer is a wonderful example of straightforward common sense in physics and the sort of thing all physics students could learn from.
 
Hey John! Question: if every possible path is considered in the path integral formulation of quantum mechanics, and is equally valid, doesn't that violate the speed limit of nature, c? If so, how can unobserved particles violate this speed limit? Is it akin to saying that the particle is actually the entire universe?
 
@dm__ hi. You shouldn't take the path integral formulation too literally. It doesn't mean you have some point particle actually following some convoluted path faster than light.
When you get down to quantum field theory particles are very strange things. They are described by excitations in quantum fields and they look nothing like the intuitive notion of a little ball.
 
And aren't those fields manifest throughout the entire universe? If the particles are only concentrations of the fields, and those fields are everywhere, then can't we just as well say that the particle is everywhere?
 
7:36 AM
The way we usually approach QFT we describe the particles as Fock states, which are effectively infinite plane waves so the particle has the same probability of existing everywhere in the infinite universe.
But the particles we observe in e.g. the LHC are clearly not well described by infinite plane waves. Instead we describe them by superpositions of plane waves to construct a wave packet.
 
So if a single particle has the same probability of existing everywhere in the infinite universe, is that what gives equal footing to the path integral formulation?
Of course we cannot prove that it takes any path other than one, but we cannot disprove it either, right?
 
What lots of students fail to appreciate is that physical theories like QFT are mathematical models that we construct in the hope they will correctly describe the results of our experiments. When a model gives the correct results it's tempting to think that model must be physically true i.e. reflect whatever it is that is really happening.
But there is no guarantee if this - none at all. So to take a mathematical model and start asking what it tells us about the fundamental nature of reality is a risky business.
If the LHC creates a Higgs boson that decays a fraction of a second later there is obviously no useful sense in which that Higgs boson existed everywhere in the universe, regardless of what the mathematical model suggests.
 
I can see that, for sure. Apart from the usefulness of it though, and with that formulation leading to all of the same experimental results, I think that it deserves more weight than, say, string theory -- sounds beautiful, but like a trap. For the Higgs boson example, apart from the useful sense of it, isn't that still a possibility? That once it decays, it has then been 'absorbed' back into the field, and then resumes all possible locations?
Just seems like a fascinating interpretation of simultaneity
 
No, the Higgs boson is an excitation in quantum field. It appears when energy is transferred to the Higgs field from some other quantum field, and it disappears when the Higgs field transfers the energy away again into some other field. Once the Higgs boson has decayed it's gone.
The Higgs field still exists of course, but remember that a quantum field is a mathematical object. It is a field whose value at every point in spacetime is a mathematical function called an operator. What it means physically is unclear.
 
7:51 AM
So the Higgs field becomes excited into an actualized boson, and then that energy is transferred to other fields? It does not settle back into some lower state, only manifest in the spread of the field itself?
 
The quantum field isn't a physical object. Only the excitations of it are physical i.e. the particles.
 
That seems like a blurry line. That something physical can come out of a non physical object
 
The field is just a function of position in spacetime, so it exists everywhere in the sense that for any point (t,x,y,z) you choose the field will have some value at that point.
@dm__ well the physical thing is the energy that's being transferred around and turned into different particles. The quantum field is telling us how this happens.
 
So the quantum field is viewed solely as a convenient mathematical tool, not at all something physical?
 
The quantum field obviously describes something physically real because, well, it works - it gives the correct predictions for experiments. But what exactly it represents is something the philosophers worry about. Physicists tend to get on with the job.
 
7:57 AM
How should I view the difference between electromagnetic fields and quantum fields?
Or the gravitational field
 
They are utterly different. All they have in common is the word field.
This is one of the things that makes QFT hard to understand for outsiders.
 
It seems that it would be very unlikely if something that yields the most detailed experimental results is solely a mathematical tool, that it does not have a physical basis in nature. Perhaps we blind ourselves by holding onto the intuitive notion of locality, which these fields, if interpreted physically, are inherently not.
 
You need to be a little cautious about the meaning of non-locality. Physicists have known that quantum mechanics is nonlocal from the very early days. That's exactly what the EPR thought experiment is about (and it's now been confirmed by experiment). But QM is still causal.
 
Does it violate causality to say that an unobserved particle, embedded into the path integral formulation, travels every possible path in the universe? It seems like these two things do not conflict. My professor said today "The paths considered in the path integral do not obey the speed limit of light. This is also true in a relativistic version." Which has led me down this entire train of thought
 
You're no doubt familiar with the idea that two entangled particles can be a large distance apart and yet can influence each other instantaneously (also experimentally proved). However this doesn't affect causality since you can't use this behaviour to generate any acausal behaviour.
The FTL travel in the path integral is sort of similar in the sense that it doesn't give any problems with causality.
 
8:13 AM
So in the path that travel outside of what would be permitted by the speed of light, can we understand that with entanglement? Is the particle just relaying its state, in some sense, to entangled particles, and then we measure the result of that entanglement? We cannot distinguish between the particles, so then it would be just as well that 'our' particle did the travel
 
The path integral does not actually involve a particle travelling along some weird path faster than light.
You are taking literally a mathematical model that is not intended to be taken literally.
 
I'm not too sure about that. Feynman seemed to take it quite literally, as well as my current professor. Perhaps cautioning that we cannot know it has or has not, but the experimental result is the same as if it had. Why discard the possibility?
 
Sid
If anyone's familiar with Graph theory for Network analysis, how do we represent two elements which are connected in parallel? With a single line Or with two separate lines?
 
@dm__ You can't measure anything travelling FTL
 
@dm__ I guess ultimately this comes down to personal opinion. All we know is that the maths works. We don't, and can't, know or certain how far this reflects what is actually happening. Personally I am agnostic.
 
8:24 AM
As you can show by computing the measurement of two fields at spacelike separations
 
@joh
 
You'll see that measuring $\phi(x) \phi(y)$ is the same as $\phi(y) \phi(x)$ if $x,y$ are spacelike separated
ie the measurement of one doesn't affect the other
 
@JohnRennie I suppose so. Thanks for the stimulating conversation! I must be going, take care
 
Ultimately my view is that only experimentally measurable quantities (measurable in principle if not in practice) are physically meaningful.
 
@JohnRennie And I think that is very grounding :)
 
8:25 AM
(I would use the example of a compact support function propagating, but turns out you can't have compact support solutions in QFT!)
 
@dm__ you're welcome. All physicists secretly love speculating about this sort of thing :-)
 
(Bloody Malament theorem)
 
@JohnRennie Haha it's the fun (bizarre) part! Talk to you later
 
Oh wait, you can have compact support function for KG
Nevermind
I think you can probably show that the spread of a compact support KG wavefunction travels slower than light
 
 
1 hour later…
10:29 AM
Hey it's written by Hegerfeldt!
Of the eponymous theorem
 
 
2 hours later…
12:43 PM
Hey look who's working at a nearby lab
It's Kleinert!
Of the path integral book fame!
 
1:10 PM
just wtf is this weird special character that is the cause of hunging up all my code:
 
2:06 PM
the nearby lab works on theoretical calculations of gravitational waves
 
3:03 PM
419
Q: What does ^M character mean in Vim?

MaxI keep getting ^M character in my vimrc and it breaks my configuration.

^M, breaking linux code since 2000
 
@Secret ^M theory
 
3:58 PM
control-M is a carriage return. probably this file was created in DOS.
 
excel cut and paste fail
 
@enumaris I took a look at pytorch last night and it seemed pretty awesome. Automatic gradient calculation is certainly interesting
Seems a lot easier than the plain tf I've messed with
 
4:25 PM
Also is weight initialization something people often tinker with? I've always seen Gaussian initialization, but looks like Ng recommends He (square root of 2 / neurons in previous layer) for ReLU layers
 
4:47 PM
I think Xavier-Glorot initialization is pretty standard now
and yeah, I think generally speaking Pytorch is "better" than TF from what I can tell
but again...no keras for pytorch
Pytorch code looks more like python code though because you're actually executing operations on a Variable rather than setting up a graph and then initiating a session to run stuff through that graph
I don't have anything to do today
ain't nobody here today
but I gotta be here or else I don't get paid
gotta find a paper to read or something...
 
5:19 PM
any suggestions for a good paper to read...
 
ah, c'mon, Robert, why'd you take away the pleasure of closing this?
 
404
 
@enumaris no kidding
 
0
Q: Extract relations from a text using spacy(python)

Mohammed Rasfai'm trying to extract relations between Persons (Father of, Mother of .. ) to construct my ontology. Here is my code: ny_bb = url_to_string('https://www.biography.com/people/brad-pitt-9441989') article = nlp(ny_bb) len(article.ents) labels = [x.label_ for x in article.ents] Counter(labels) p...

such a broad question...
well maybe not broad...but like...open ended
there's no 1 solution to that problem
 
5:41 PM
Maybe you'll find a paper there. Supposedly it gives papers, but I only saw GitHub stuff on mobile
 
5:58 PM
looks like there is some work done on sparse attention
probably should read those papers...
 
6:35 PM
Do you ever feel like you've hit a limit of knowledge intake?
 
hmmm don't think so
unless by limit you mean like limit of how much stuff I can study in a day or something like that
then obviously yes
there's only so many hours in a day lol
 
Yeah like at the end of the day you're reading and it just feels like you aren't able to take in any more
What do you do?
 
stop
 
Also Mendeley managed to somehow get a materials science title and NLP abstract for a GAN paper o.o
 
o.o
 
6:41 PM
It's been getting some stuff messed up the past couple days, but usually the titles and abstracts math at least
arxiv search didn't come up with the right paper either
 
what paper are you looking for?
google scholar is a really good resource for papers as well
 
arxiv.org/abs/1406.2661 I got it by google -> arxiv -> mendeley arxiv id
I have no idea why it was trying to recognize it as a different paper or arxiv search wasn't working
Or maybe I just don't know how to use arxiv's search. I figured title + author would do it
 
ah, is that the original paper on GANs?
 
The paper does have "Generative Adversarial Nets" while arxiv has "Generative Adversarial Networks"
yeah I think that's the original
 
cool
I'm reading the paper on sparsemax atm
 
6:50 PM
> In this paper, we propose the sparsemax transformation. Sparsemax has the distinctive feature that it can return sparse posterior distributions, that is, it may assign exactly zero probability to some of its output variables
Like dropout but not random or something?
I should probably look more into softmax though. I thought it was a smooth version of ReLU rather than what I'd think of as a transformation
 
o.O
it ain't like a relu
 
Ahh I was thinking softplus (...I think)
That would explain why a lot of softmax things I've seen didn't make sense
 
ugh, I don't feel like dealing with all this math atm
$softmax_i(z) = e^{z_i}/\sum_j e^{z_j}$
 
Yeah I've seen that...I just didn't connect the dots
I'm not a smart person at times
 
I guess the $i$ should be outside since the softmax "function" need the whole z
the definition of sparsemax is not nearly as intuitive
 
7:07 PM
Speaking of which, I recently felt the level of fear that I think some people have when they see any math. It was caused by Penrose's tensor diagrams
 
lol
I never feel fear...just laziness
math is time consuming
 
Do you work out all the math when going through those papers?
 
generally no
I can generally tell at a high level what's being done so I don't go through the details
I will go through the math if I ever have to implement a model myself from scratch -.-
it's tedious and boring
I'm not seeing how the "projection" is generally sparse in the sparsemax
"In words, sparsemax returns the Euclidean projection of the
input vector z onto the probability simplex. This projection
is likely to hit the boundary of the simplex, in which case
sparsemax(z) becomes sparse."
I think I'm trying to visualize this geometrically and I'm failing since I can't visualize >3D lol
 
You lost me at "probability simplex"...though it seems to me that the boundary of such a thing may be where one class has probability 1 and the rest 0?
 
I'm also not seeing how their definition reduces to a Euclidean projection onto the simplex atm
a probability simplex in $\mathbb{R}^n$ is a $n-1$ dimensional surface where basically the vectors are all valid probability distributions
so all the components of the vectors are positive and add up to 1
by "boundary" they mean where the simplex hits some axis
since you are confined only to the "all positive" sectors
 
7:20 PM
what kind of things is the simplex used to describe?
 
so you can have boundaries that lie on 1 axis, or 2 axes, or up to n-2 axes
 
the simplex is simply the space of all acceptable probability distributions
 
I need more learning
My probability/stats class assumed we had no idea what a surface/derivative/anything was
 
woah, cool
 
7:21 PM
yo in 2-D the simplex is simply the part of the line y=1-x that lies in the x>0 and y>0 quadrant
 
Is it like a triangle plot?
 
in 3-D it's the plane z=1-x-y that lies in the x>0,y>0,z>0 sector and so on
the "boundaries" would be where that plane hits the axes, the x-z, y-z, or x-y planes
 
so these are always 'flat' in some sense to get the projections always adding to 1?
 
the projection is onto this subspace
 
So the projection is just to obtain a normalized probability distribution?
 
7:24 PM
so you have some general vector, e.g. (3,6,9) in 3-D space, and then you Euclidean project it onto the plane z=1-x-y
yep
then that's your sparsemax
what I'm not seeing is why in general the projection would hit some boundary where the resulting probability distribution would be sparse
this statement: "This projection
is likely to hit the boundary of the simplex, in which case
sparsemax(z) becomes sparse. "
is not obvious to me
I guess it's just known to be true...
 
"likely to hit the boundary" means the vectors in the higher space have to have a certain distribution, right?
 
don't think so
as far as I can tell, they are not applying any restrictions on the vectors that you project onto the simplex
specifically, they have not said that the vectors must all lie in the positive space either...they haven't said you MUST put the vector through a ReLU or some such.
 
so is there anything underlying the distribution of vectors, or they are completely randomized to begin with?
 
they can be the output of some generic neural network
so I don't think there's any restrictions on their distribution
 
does a generic neural network have some tendency to have these vectors be near boundaries?
 
7:30 PM
no
 
well, their projection
 
apparently the projection has some tendency to be on some boundaries
and in order for the "sparse" part to be really true
it's gotta be actually on some boundary points that have "sparse" representations
in fact
 
yeah, I can't make sense out of that. seems there would be no reason for that to be the case
 
I'm uncertain how a generic Euclidean projection is done onto a surface with boundary
like
imagine 2-D space
how would you generically project onto the line segment y=1-x where x>0 and y>0?
projecting onto y=1-x is no problem and is well known to me
but you have to truncate that line to x>0 and y>0
 
wait, how is that projection done? min distance?
 
7:34 PM
yeah I think that's probably where I'm having trouble
when I imagine project, I'm thinking of right angle projections
 
if its a min distance projection then I can definitely see why there would be a tendency for the boundaries
 
I could see how it ends up on a boundary if outside projections get moved to the boundary, but I don't imagine that's a reasonable thing to just do
 
and I can't do that if there's some truncation going on since there's no right angles you can make for a lot of points in that case
it's a min distance projection
ok that makes sense now
 
:D
 
it's all clear
I was thinking right angle projections cus that's what I'm used to lol
 
7:36 PM
yep, same here
 
whoop, that's 2 conundrums solved in 1 swoop
the procedure is then
given some z find the point in the simplex closest to z
that point is generally going to be on some boundary
and so you get a sparse representation out
I can also see how that could be differentiable in the sense that relu's are differentiable
it's only not differentiable at some boundary points when you move from the boundary to out of the boundary
otherwise if you stay along the same boundary the differentiation results in 0
thar we go
thx for the help :D
 
this builds a nice intuitive picture of it
 
yep
 
interesting chat!
gotta run now, take care :)
 
see ya
 
7:44 PM
Wait so is a min projection something completely different than a right angle projection?
Or is it the same as when you have oblique axes?
 
the right angle projection is basically a way to get to the min projection for cases where the right angle projection works
for
"flat" surfaces
you're projecting to
if your surface is generic, you could see that you might get multiple "right angle projections" or no right angle projections
so I guess min distance is probably the better way to think about it
 
Is this the same projection as dot product onto basis vector?
ahh nevermind I think I found a MSE question
 
so a projection is really generic and specifies any way you can map a higher dim space to a lower dim space I think
generally you want a "projection" to have some nice properties. The one we are familiar with in Linear Algebra for example would be problems like PCA where you're projecting points onto a linear subspace
in that case, one could generally think of things in terms of making right angles
but if the subspace is non-linear or with boundary or something then you either can't always make right angles or there are multiple ways you could make right angles
so I suppose in those cases it'd be nice to use "min distance"
min distance may not be unique either though
I think if you had some non-linear subsurface...it's probably just generically hard to find a "nice" projection absent some other restrictions
I dunno
I haven't had to think too much about projections lol
 
Yeah I suppose it's just like finding a nice projection of a sphere. I didn't know it could be more general though
 
I never thought about it much to try to figure out what "more general" would be lol
 
7:57 PM
Yeah about all I've done with projections is dot a vector onto orthonormal bases
 
yep...
so this sparsemax thing seems very useful...
its derivatives should be very easy to compute in general as well
and it's got a constant gradient like the ReLu
but...you might run into dead neurons like the ReLU as well maybe...
since it's got parts where the gradient is 0
 
8:09 PM
wooo and interviewer actually called me exactly on time for once
 
nice nice
 
I thought such a thing was impossible!
And it seems facebook has released a camera I can put in my house...I think I'll pass
Also "dead neurons" would be what Ng calls "vanishing gradient"?
 
for a ReLU unit it's kinda worse than the "vanishing gradient" problem
the vanishing gradient is when you have a bunch of gradients that are all <1
and so when you multiply them together through a very deep neural network you eventually get very small gradients
and things are hard to train
for a ReLU, if you have input x<0 you have no gradient for that unit, period.
so that neuron stays dead
 
ahh so it would basically be the same as what happens for zero initialization?
 
forever
 
8:16 PM
except with just a single neuron
 
just for those single neurons yeah
that's why people introduced leaky relu
 
man I'm behind the times
 
76
Q: What is the "dying ReLU" problem in neural networks?

tejaskhotReferring to the Stanford course notes on Convolutional Neural Networks for Visual Recognition, a paragraph says: "Unfortunately, ReLU units can be fragile during training and can "die". For example, a large gradient flowing through a ReLU neuron could cause the weights to update in such ...

 
I had the idea of trying to train the dropout probability and it seems people use entire networks for that
 
like turning drop out rate from a hyperparameter to a parameter?
seems non trivial
 
8:19 PM
Well it would be covered more if it was trivial!
 
this sparsemax paper doesn't actually say if sparsemax saved them computation time over non sparse attention mechanisms though...
since at each step still the entire input space has to be considered I'm guessing that the savings in computation time is slim
be careful turning hyperparameters into parameters though
you may run into over training on dev set
lol
over cross-validation!
 
Good thing I picked out some dev2 data!
 
XD
 
This is what I found in that direction, but it looks like I'm going to have to do some reading on Boltzmann machines and belief networks to understand it
 
the sparse max really is more of a softmax analogue...the sparse attention I was thinking of was more along the lines of not even considering some inputs
thereby saving time
 
8:23 PM
So like dropout on the inputs, but systematic?
say like something that may happen if you trained dropout
Actually the more that I think about that, the more non-trivial it seems. A couple of ways seem like they would either overfit or be the equivalent of training the weights like normal
 
00:00 - 21:0021:00 - 00:00

« first day (2939 days earlier)      last day (1994 days later) »