The h Bar

4:00 AM

@Eulb no, they were commercial doughnuts. Very nice though :-)

4:16 AM

onlinelibrary.wiley.com/doi/abs/10.1002/anie.201713108

Turns out core electron topologies explains why it is very hard to have Si=Si bonds

Cows

5:32 AM

Writing an ios app

Gotta get into hyper mode, and finish this within 2 hours. Listening to music that is extra hardcore even for me

ok . . . .

Got to make moola

A friend of mine has just started writing an IOS app (not his speciality but it's needed as part of a bigger project he's doing). He's using Swift.

6:04 AM

I'm working on an app...but my programmer partner is busy traveling so nothing is being done...T_T

6:24 AM

Bell Inequality and PR box correlations simplified

2 hours ago, 1 hour 59 minutes total – 200 messages, 7 users, 0 stars

Bookmarked 17 secs ago by Secret

Rishi Kakkar

I just read according to Einstein Indians were geographically inferior or stupid

6:41 AM

Link?

It doesn't seem very likely given Einstein's collaboration with Satyendra Nath Bose.

Avnish Kabaj

haaretz.com/world-news/europe/…

Cows

hehe yes I am using swift too :P

lolz so Einstein had some questionable views . . . hmm hehe

Avnish Kabaj

@Cows hahahaha

[Random]

How high we can push correlations:

0. No correlation whatsoever: Independent events

1. Classical correlations: Two coins being flipped, such that if one shows heads, we knew the other one must shows heads, and no correlation otherwise

@AvnishKabaj hmm, I normally regard the Guardian as a reputable paper and I'm inclined to believe what I read there. However I'd want to see the quotes from the diaries in context before I comment.

I note the Guardian is just quoting Rosenkranz, not commenting on the diaries themselves.

6:51 AM

2. Quantum discord: Correlated outcomes due to noncomomutativity of operators
3. Entanglement: Correlated outcomes between subsystems which exceed classical correlations as covariance calculations will require impossible outcomes
4. PR boxes: Inputs and outcomes are correlated

5. Can we go higher without signalling?

Wassup.

I need to read the PR box paper again later, I think they have proved 4 is the highest we can go without nonloca signallinh

yep, it is

so <s>5</s>

7:17 AM

Wait a sec... I should be able to signal this way:

Fix A at 1

B choose between 0 or 1 to report about the status in its immediate area

actually no

A can only ever see 1 or 0 poping up, so is B

They won't see the correlation unless they get together

dang...I was hoping for another answer on my cross validated question...but nobody gave a second answer :(

So to recap:

0. Independent events

1. Classical correlations: The correlation is already embedded in the system as some preexisting property

2. Quantum discord: Outcomes can be influenced by the ordering of the measurements of two conjugate observables, thus the observables are correlated due to the algebraic structure of quantum mechanics

3. Entanglement: Probabilities of of the outcomes between the subsystems are correlated

4. PR boxes: Inputs from all subsystems picks the pdf to be sliced on, and the probabilities of the outcomes from all subsystems are correlated such that the different pdf slices cannot be distinguished from each other

3 and 4 thus give us all the non signalling nonlocal correlations

So... if 4 is the highest possible, that means considering the following setup

Take two non signalling boxes such that they have a joint pdf of the form as shown above

except that e.g. the 4th column has an extra 2/3 chance to show up compared to the other 3 columns.

Then the outcomes that pops out locally will not allow us to distinguish which column we are at, and the overall joint pdf will be modulated by the above extra two factors

which means when all the inputs and outputs are all brought together, we should see a correlation between all sets of A,B,a,b, and that the 4th column is slightly bised more than the probability to actually have the A,B that can send us to the 4th column

The CHSH inequality of this should alo give 4 otherwise it will be signalling

Koolman

8:17 AM

@JohnRennie Which would be the best place(book,website) to learn and practice JAVA ?

Java Tutorial for Beginners - Original Step by Step

►

CSI Las Vegas Season 2 Intro/Opening/Theme Song

@Koolman I'm not sure to be honest. I learned Java when it was first released back in 199<something> and there wasn't much online back in those days.

I don't think it really matters. Just find a web site that you're comfortable with.

Koolman

@Secret I also have newboston videos

Emilio Pisanty

8:43 AM

@JohnRennie so I imagine you'll not be pleased by the notion that for a substantial number of people, the main association of that song is "oh, yeah, that's the intro to CSI, no?"

►

@EmilioPisanty No I'm not fussed at all. If anyone enjoys music for any reason that's just fine with me (though I tend to grit my teeth when Nicki Minaj is on the radio :-)

Emilio Pisanty

@JohnRennie look at you and your Zen outlook on life

Oops, mis-spelling of Nicki - there goes my credibility with the Nicki Minaj fans

@EmilioPisanty I don't mean to be precious. It's just that I work hard at being chilled these days :-)

Emilio Pisanty

=P

9:01 AM

In relativity, what does the notation $\Lambda_0^{0'}$ mean?
Is it standing for the element (0,0) of $\Lambda$? Is it some form of summation convention? (I'm pretty sure you sum over the matching upper and lower indices's)

Just starting general relativity study for the first time (done SR before but it was very basic).

Typically it means $\Lambda$ is a mixed tensor i.e. with one upper and one lower index. Then $\Lambda_0{}^0$ is the top left element of the matrix.

Thank you

If $\Lambda$ is the boost then that is indeed the case.

Oh, yes, it is a boost in the content I'm reading it in (sorry about the vague wording.)
The book is Sean Carrolls spacetime and geometry, I'm on the first chapter.

9:42 AM

Over here, what does the d stand for? I mean, I know velocity is change in x over t, but what is the d there?

I know it has something to do with derivatives, but what?

There is a really good answer to that question here:
https://math.stackexchange.com/questions/340744/what-do-the-symbols-d-dx-and-dy-dx-mean

The $\frac{d}{dx}$ is all one thing. Its an operator that you can apply to a differentiable function to get a new function, which corresponds to rate of change of the first.

Did that make sense?

9:57 AM

I'll be back after school sorry.

11:58 AM

In Sean Carroll's Spacetime and Geometry book, on page 29, there is reference to a rotation matrix $\Lambda$. On the page it states:

$\eta_{\rho\sigma}=\Lambda_\rho^{\mu'}\eta_{\mu'\nu'}\Lambda_\sigma^{\nu'}$

Using the summation convention mentioned earlier in the book, I have interpreted this as:

\[
\begin{bmatrix}
-1&&&\\
&1&&\\
&&1&\\
&&&1
\end{bmatrix}=
\begin{bmatrix}
-{\Lambda_0^0}^2&&\\
&{\Lambda_1^1}^2&&\\
&&{\Lambda_2^2}^2&\\
&&&{\Lambda_3^3}^2
\end{bmatrix}
\]

This leads to the magnitude of $\Lambda_0^0$ been equal to one.

Also, if this is a good example of a question that could be asked on the main site (meets the standards and all) please let me know. I'm asking here partially because I can discuss an answer, but mostly because I am very rusty with asking things on S.E. and the chat is more appropriate for practice.

12:20 PM

If you take the $\rho=0, \sigma=0$ component of the master equation you get: $\eta_{00}=\eta_{00}(\Lambda^0{}_{0})^2+\delta_{ij}\Lambda^i{}_{0}\Lambda^j{}_{0‌}$, where I have split the RHS into time and space (latin indices) parts

Rearrange this to get $(\Lambda^0{}_{0})^2=1+\sum (\Lambda^i{}_{0})^2$

so that its modulus is greater or equal to 1

I dont think what you have written in matrix form is correct

Is $\eta_{00}$ the top left element of the metric $\eta$? In the textbook, its the first time double subscript notation has been used.

Yep, if you want to think of it in terms of matrices (which is sometimes dangerous) the first index that appears is always the row index whilst the second is the column index

12:37 PM

Why the split in notation? Its a subscript for $\eta$ and both a superscript and subscript for $\Lambda$.

I have to think of them as matrices at the moment because I haven't been introduced to tensors yet, (but I will be shortly introduced).

Linear transformations (which is what Lambda is) carry mixed indices, i.e. one index up and one down. The reason for the index placement will become clearer after you have learned about tensors

Ok then, I'll stave off that question until I get to them

In your explanation, am I correct that within,
$\eta_{00}=\eta_{00}\Lambda_0^{0}+\delta_{ij}\Lambda^i_0\Lambda^j_0$
the first term on the RHS is time, and the second all the space terms?

1:04 PM

When I said that I split it into time and space components I only meant that I was distinguishing between the $\mu=0$ values of the indices and $\mu=i={1,2,3}$ values of the indices. This doesn't always have an interpretation of space and time components so you should be careful about taking it in the literal sense

1:18 PM

@NormalsNotFar I'm confused as to where the $\delta_{ij}\Lambda_0^i\Lambda_0^j$ part comes from in:

$\eta_{00}=\eta_{00}(\Lambda^0_0)^2+\delta_{ij}\Lambda_0^i\Lambda_0^j$

If the syntax is consistent in $\eta_{\rho\sigma}=\Lambda_\rho^{\mu'}\eta_{\mu'\nu'}\Lambda_\sigma^{\nu'}$, then when $\mu'=\nu'=0$, and $\rho=\sigma=0$ it should read as:

\begin{align}
\eta_{00}&=\Lambda_0^{0}\eta_{00}\Lambda_0^{0}\\
&=\eta_{00}(\Lambda_0^{0})^2
\end{align}

Thanks for the help so far by the way

@user400188 we have, remembering that we need to sum over all indices which appear twice (once up, once down), $\eta_{00}=\Lambda^{\mu}{}_{0}\eta_{\mu\nu}\Lambda^{\nu}{}_{0}=\Lambda^{\mu}{}_{‌0}\eta_{\mu\nu}\Lambda^{\nu}{}_{0}+\Lambda^{0}{}_{0}\eta_{01}\Lambda^{1}{}_{0}+..‌.$

Since we are summing over all the values that $\mu$ and $\nu$ can take, there will be a total of 16 terms. But then we remember that only the diagonal elements of $\eta_{\mu\nu}$ are non-zero, so that only the four terms with $\eta_{00}, \eta_{11},\eta_{22}$ and $\eta_{33}$ are non-zero

so we only need to include those ones

and no problem :)

We still have $\eta_{00}$ on the left hand side, so the first element of the metric contains four terms.
Have I got that right?

I'm not entirely sure what you mean, but $\eta_{00}=-1$ ?

@NormalsNotFar Well, in this comment $\eta_{00}$ is on the left hand side of the equation, and earlier it was mentioned that $\eta_{00}$ corresponded to the top left element of $\eta$, which is negative one.

So does this mean that $-1$ is equal to the sum of those four terms?

1:33 PM

Yep that's right

Ok, I think I'm starting to get the notation now

It also seems to clarify just about all the questions I had. Thanks a bunch @NormalsNotFar :)

No problem, and good job, index notation will become second nature after a while :)

Just write it out carefully: From
$$x'^2 = x^2 \leftrightarrow x'^{\mu} x'_{\mu} = x^{\rho} x_{\rho} \leftrightarrow \Lambda^{\mu} \, _{\rho} x^{\rho} \Lambda_{\mu \sigma} x^{\sigma} = x^{\rho} \eta_{\rho \sigma} x^{\sigma} \leftrightarrow \Lambda^{\mu} \, _{\rho} \eta_{\mu \nu} \Lambda^{\nu} \, _{\sigma} x^{\rho} x^{\sigma} = \eta_{\rho \sigma} x^{\rho} x^{\sigma}$$
we have
$$\Lambda^{\mu} \, _{\rho} \eta_{\mu \nu} \Lambda^{\nu} \, _{\sigma} = \eta_{\rho \sigma} $$
and from this we can analyze the $\Lambda^0 \, _0$ component of $\Lambda$ by first considering

Thanks for the detailed explanation @bolbteppa. I just went through it myself

If I had asked on the main site, I'd be awarding you points now (although I'm still very grateful of @NormalsNotFar for sticking around for 2 hours to discuss the notation.)

Mohammad Areeb Siddiqui

2:12 PM

if $\Psi = Ae^{ikx} + Be^{-ikx}$ where $A, B \in \mathbb{C}$ and $k \in \mathbb{Z}$ and I have to find the probability current:

I started by calculating $(\Psi)^* = Ae^{-ikx} + Be^{ikx}$

Then $\dfrac{d\Psi}{dx} = Akxe^{ikx} - Bkxe^{-ikx}$

i should just multiply the two

and take ou the imainary part right?

but i cant seem to know what will be the imaginary part

except instead for terms like $|A|^2kx$

Eulb

2:37 PM

@JohnRennie you'd eat all that commercial stuff? ;'(

@Eulb yes, they were very nice. And making doughnuts yourself is exceedingly messy - I speak from experience.

@MohammadAreebSiddiqui the probability current is the operator version of the classical velocity density $J = \rho v = \rho \frac{p}{m}$ for $\rho = \psi^* \psi$ which means you symmetrize it $\mathbf{j} = \frac{1}{2m}(\psi^* \hat{\mathbf{p}} \psi + \psi \hat{\vec{p}}^* \psi^*)$ so just calculate that for $\psi = A e^{ikx} + B e^{-ikx}$

Mohammad Areeb Siddiqui

2:55 PM

and for 3 dimensions $\dfrac{\partial \rho}{\partial t} = -\div J$

@MohammadAreebSiddiqui you can formally derive the expression for $\mathbf{j}$ by working out $\frac{\partial }{\partial t} \rho = \frac{\partial }{\partial t} \psi^* \psi$ and then using the Schrodinger equation

Mohammad Areeb Siddiqui

I see. But is all this working worth it for 3 points?

marks*

nvm it doesn't matter

philmcole

4:57 PM

I have a question regarding capacitors. The capacitance is $C=Q/V$ where $Q$ is the charge on the positive plate (and $-Q$ on the negative plate) and $V$ the potential difference between the plates. Must the charge always be equal and opposite on the plates? The capacitor is often explained by "picking a positive charge from the negative plate and moving it to the positive plate". What if I took the positive charge from somewhere else than the negative plate?

5:15 PM

The charge must indeed be equal and opposite. This is charge conservation at work.

iiScotchtape

5:25 PM

0-0 So much complicated physics talk here, I can't wait till I can actually learn all this =)

5:39 PM

hmmm...I feel like this other answer is wrong: stats.stackexchange.com/questions/351596/…

My answer contradicts his...anybody here understand neural nets well enough to back me up on this one. If I'm wrong, I want to delete my post before it spreads false knowledge, but I'm 99% certain I'm right on this one.

oh, the other guy edited his answer...alrighty then...

Anonymous

6:09 PM

@enumaris Your answer looks fine for feed-forward networks. Intuitively speaking, if a certain neuron in the input layer was completely shut off during training, it would never learn how much a nudge in the value of that neuron would affect the cost function. Thus, the weights which are connected to that neuron would never be updated to take into account the effect when that neuron's value changes.

Anonymous

There might be unexpected results in case the weights connected to that neuron were high initially.

Anonymous

@enumaris But, can you say the same for recurrent networks?

hmmm...

Anonymous

@enumaris For a recurrent network, I think the chances of unexpected results would be much lesser as the network would evolve to settle down in a "minimum energy" state which most closely matches the input at inference time

hmmm

well the feature vector is concatenated with the previous state

it's not added

but depending on the cell, it could go through a bunch of different transformations

a bit too complicated to figure out a priori what would happen imo lol

but my inclination is, since there's a concatenation, there's still a weight vector which won't ever get updated

one row of the weight matrix

and adding the feature back in would simply re-activation this random vector in the weight matrix

so it should still produce some random noise

The random noise might be negligible

Anonymous

6:20 PM

The updating of the weights depends on the learning rule we use for the recurrent network

but all learning rules have a grad_weight(J)

and that will always be 0 for the set of weights that couple to a 0 input feature

even if you add momentum and the like, it should still always be 0

Anonymous

If you consider something like say $w_{ij}=\frac{1}{n}\sum x_i x_j$

unless maybe if you had a non-gradient-descent based method

Anonymous

Then all the weights connected to that shut off neuron will be set at 0

Anonymous

So switching that on wouldn't really "fire" any other neuron

Anonymous

6:22 PM

And thus, no unintended consequences

I've never seen a weight get set to such a term...o.O

Anonymous

@enumaris Check this

I'm talking about in the case where weights are learned by some Gradient descent method

Hopfield networks I think are out of scope for the question...

I admit I am not expert enough to be able to say what the answer would be for a hopfield network...but does such a network even have a well defined notion of a "input node"?

all the units interact with each other right...so...how would you distinguish input vs output. I would think that "input" would be some initialization of the neurons and then "output" would be the steady state of those neurons?

Anonymous

@enumaris Yes

hmmm...

I don't think we have to get into those type of details for this question T_T

Anonymous

6:29 PM

I'm not really concerned about that question in particular :)

A: Can somebody better explain the Tenacious badge?

but I suppose for those networks a neuron that always got shut off as its initialization might activate due to interaction with other neurons and then the steady state could be unaffected by that initialization...

This would certainly be true if Hopfield nets are ergodic...but...I don't believ ethey are?

cus if they are then the intializations don't matter at all lol

hmmm I just earned the Tenacious badge on stack overflow...

but I don't see any of my answers on SO being accepted recently...

how did that happen o.O

Anonymous

@enumaris It would get activated during the inference stage only if some of the training samples had it activated. If all of the training samples had it inactive, then during inference all the steady states would have it inactive, too

Anonymous

32

Basically, you have a ton of answers that have a score of zero and are also accepted. In particular, you have more than 5 answers that have a score of zero and are accepted. And, those zero-score-accepted answers make up at least 20% of your total accepted answers.

But the neuron that gets initialized to 0 could still activate (get a value other than 0) simply from interaction with other neurons can't it? I.e. during time evolution for one input case.

Anonymous

@enumaris Yes, that's possible (but iff some of the training samples had it activated)

Anonymous

6:35 PM

Really depends on your training samples

@Blue right, I figured that's what the badge meant, but how can I gain that badge at a time when I didn't get any answers accepted.

hmmmm

well maybe at some point I'll learn a bit more about hopfield nets

Anonymous

@enumaris You have 6 zero score accepted answers

Anonymous

I think it is awarded periodically

Anonymous

Not immediately

oh

lol that's confusing then :P

Anonymous

6:39 PM

"And, those zero-score-accepted answers make up at least 20% of your total accepted answers."

Anonymous

Makes sense

Anonymous

@enumaris Hopfield nets and Boltzmann machines are very interesting :)

The last time I got a 0 score accepted answer is early June tho so...

I guess they just waited 10 days

I've studied restricted boltzmann machines

but not full on boltzmann machines

my machine learning related knowledge is more about practical use cases than theoretical interest

Anonymous

This is a very good book in case you're interested (huge PDF alert!)

Anonymous

@enumaris In case you want a practical application of Hopfield nets and Boltzmann machines: "they are damage tolerant"

6:42 PM

meaning if you remove a neuron it still works?

Anonymous

Yes. The results don't vary too much if you damage 10 out of 1000 neurons

that should be the case for feed forward neural nets trained with drop out as well...as long as you rescale the other neuron's activations accordingly

Anonymous

Which isn't true for feedforward

Sup ma fellas.

Anonymous

@enumaris But for feed-forward nets some of the neurons very strongly affect the cost-function

Anonymous

6:43 PM

If you damage those, the output might completely change

drop out learning is actually killing off a bunch of neurons at each pass tho right. But I guess the rescaling of the activations might be an added inconvenience.

Anonymous

Drop out is for killing those neurons which don't affect the output much. But, what if you damage those neurons which affect the output strongly?

theoretically yes, if you trained a neural network without regularization. But the entire purpose of drop out regularization is to mitigate that factor - i.e. reduce the output's dependence on any one neuron

no, drop out regularization drops neurons at random

it is not affected by activation of the neurons

Anonymous

@enumaris I'm not talking about activation. Rather the weights associated with a certain neuron

yeah drop out is agnostic to those as well

Anonymous

6:46 PM

Some of them might be too high

A: Capacitor with different charges on each plate

drop out removes neurons at random

Like, the whole purpose of drop out regularization is to turn your single neural net into an ensemble classifier/regressor. It makes sure the output is not too highly dependent on any one set of neurons.

That's how, it's often possible to have a training loss with drop out that is higher than the test loss.

Because drop out is basically making you train one (random) classifier at a time, while at inference time you remove drop out and now you have an ensemble of classifiers.

But with drop out you do have to rescale the other neuron's activations. If you kill off 50% of neurons in a layer, you will have to multiply activations in other neurons by 2.

If you don't need to do such a rescaling with hopfield nets, that could still be an advantage

I'm sure there's a lot of other potential advantages to having a fully general set of connections as well.

rather than feed forward

but current use cases heavily favor the feed forward or simple recurrent architecture :P

Mithrandir24601

@JohnRennie Have you ever had issues with trying to write a column vector in a tag wiki and it only displaying as a row vector?

ahhh, interesting - I had to write 3 \s to get a new row instead of the usual 2

philmcole

7:15 PM

@enumaris Thanks, but can you explain why?

What would go wrong if I just put two plates together and only one of them had a charge on it? There is still a potential difference, isn't there..? Why would those not make a capacitor?

Anonymous

@enumaris Interesting! Drop out regularization is not what I was thinking it is. Was confusing it with dimension-reduction. However, it seems to be a very new technique introduced only after 2010, while Hopfield nets were introduced in the 80's

Anonymous

And it seems the Drop out technique can even be applied to Hopfield nets and Boltzmann machines to further improve them

Anonymous

7:32 PM

3

Systems of plates are not typically considered capacitors unless they are globally neutral. Nevertheless, capacitance is a geometric property that is to do with the system more than the actual voltages and charges you apply to it, so that your question still makes sense: the capacitance is the sa...

philmcole

What I mainly get from the answer is that this situation is rarely encountered because it is much more complicated.

Oh My God, @Blue you have the coolest icon I have ever seen.

Anonymous

@philmcole If you have two parallel plates with different charges $Q_1$ and $Q_2$ the charges will simply get re-arranged so that the inner surfaces have equal and opposite charges while the outer surfaces will have the same charge. You can justify using the Gauss law.

Anonymous

@NovaliumCompany Thanks ;)

@Blue Guess who finishes school year with A's in Physics, Math, Chemistry and Biology. Ha, this guy <--

Anonymous

7:43 PM

@NovaliumCompany Great! Congrats

Trigonometry finally makes sense :D

Now time to dig up these derivatives (and integrals) because they have been annoying me for a while.

Over here, what does the d stand for? I mean, I know velocity is change in x over t, but what is the d there?
I know it has something to do with derivatives, but what?

@NovaliumCompany Leibniz' notation for the derivative.

@ACuriousMind Thanks, also in our debate about whether teleportation exists or not, black holes?

The $\mathrm{d}$ itself stands for nothing. For any function $f(x)$, $\frac{\mathrm{d}f}{\mathrm{d}x}$ is a single expression denoting its derivative that cannot be decomposed into its constituent symbols.

@NovaliumCompany Hm? Please form a full sentence ;)

I mean, you said that teleportation isn't possible. Well, can't you teleport with black holes? Or they just speed you up to get you to another point in space?

7:58 PM

@NovaliumCompany I'm not sure how you imagine a black hole "teleporting" anything

Ok, then I need to stop watching movies :D

Note that, for an outside observer, objects take infinite time even reaching the event horizon!

Plus there's the whole "spaghettification" part

Yep, I've heard about it.

@Semiclassical Well, teleporting spaghetti would still be teleporting, I guess :P

7:59 PM

Cosmic Pasta Delivery Service

Ok, I'm making this a real company for the future.

Come to think of it, this may be another sign of the divinity of the Flying Spaghetti Monster

Guys, multivariable functions cannot be graphed on a 2D graph right?

@Semiclassical In another universe, some alien is very confused why their white hole is just producing pasta.

@NovaliumCompany Depends on what you want to graph :P

Well, example: f(x, y) = ...

8:03 PM

But sure, to graph a function of $n$ variables faithfully, you need an $n+1$ dimensional graph

You have 3 parameters here.

@ACuriousMind loool

Question: If I have 3D graph with f(x, z), then that means that the up part will depend on the x and z variables?

@NovaliumCompany People sometimes graph such functions "in 2D" by using color, e.g. see the pictures at en.wikipedia.org/wiki/Riemann_zeta_function

@NovaliumCompany If you're graphing $y$ in the "up/down" direction, then sure

And y is basically f(x, z) right?

8:06 PM

Yes

Yay.

In Mathematica your options range from a 3D plot (i.e. z=f(x,y)) to a contour plot (level sets of f(x,y)) to a density plot (different colors = different values of f(x,y))

(there's also a 3D contour plot but I don't remember what that's for)

Just like a geographical map can represent height?

sure. that's how a contour plot works

Omg I am actually smart :D

Just plotted $y = f(x, z) = x^2 + z^2$ on an online 3D grapher. Math is beautiful.

8:10 PM

try doing sqrt(x^2+y^2)

and see if you recognize the figure

Ahh, my old friend cone :D

yup

of course, that is not going to be the same as the graph of $y^2=x^2+z^2$

@Blue was just in a meeting lol. But yeah Dropout is quite a new technique. And it's goal is explicitly to make the output not be too dependent on any one set of neurons.

Anonymous

I'd like to learn how to do the 3Blue1Brown type animations sometime. Those tend to be excellent for visualisation. Maybe the graphing packages are open-source. Gotta search

with that one, you get negative as well as positive values of y

Anonymous

8:12 PM

@enumaris Right. I didn't know that one :)

whereas with y=sqrt(x^2+z^2) you won't get any negative values

:D

Anonymous

@enumaris Is the Goodfellow book good? That is, do you recommend it? :P

@Semiclassical $y^2=x^2+z^2$ wut?

sure

8:13 PM

What do you mean? I mean $f(x, z)^2$?

What goodfellow book? I don't think I mentioned a goodfellow book before? o.O

I mean the set of (x,y,z) values such that y^2=x^2+z^2

Do you mean Barto and Sutton's Reinforcement Learning?

Anonymous

@enumaris This. I guess it is quite popular for those who're into deep learning (?)

@Semiclassical Ok then :D

Anonymous

8:14 PM

@enumaris I don't know about that either. Is it good?

I have not read Goodfellow

Anonymous

I'm looking for some good machine learning books which are more focused on theory and mathematical justifications rather than the coding part

Barto and Sutton is quite good

That book is also very practical because they don't just give you the theory and algorithm, they give you pseudocode on how to implement the algorithm

Ah

Then maybe not lol

Anonymous

I don't really care much about ready-made pseudocode :P I can get those online too

But it definitely has a lot of theory and mathematical justifications...but it's not quite rigorous at the level of a mathematics textbook.

But it does have like "here's the convergence proof"

and stuff like that

Anonymous

8:17 PM

@enumaris Ah, that's the problem ;_;

but when things get too complicated they might only prove a subset and leave the rest to a reference

Anonymous

I don't feel satisfied with people skipping such proofs

Anonymous

Anyway, I'll give it a try ;)

heh

sometimes for practical results you can skimp on convergence proofs

because deep-Q learning is actually unstable theoretically

and yet people use it

and deepmind used it (with modifications) to play atari games

Lol I tried to graph a 4D function on a 3D grapher and nothing happened (obviously) :D

8:19 PM

@Blue I learned deep learning from a series of courses by Andrew Ng on Coursera

Anonymous

@enumaris I did see some of those. Andrew's voice makes me fall asleep :P Also he skips a lot of the mathematical justifications in the Coursera series. (However, they say that the Stanford version is much more rigorous)

well he does show you mathematical justification for a lot of it

I just watched one of the greatest football matches of all time

but often he will say that those parts are "optional"

Anonymous

@Sid The free kick!

Anonymous

8:21 PM

Once in a lifetime event ;)

But it's intended for a wider audience so it won't be like a math text

@Blue I know,right? We were screaming at half-past 1 in our room

Anonymous

@enumaris That's true, yes

Am I nowhere safe from the discussion of football? :P

Anonymous

They haven't yet released the Stanford version for the public

8:22 PM

There will invariably be a disconnect between things presented to academics and for academic research and things being used in a practical setting

@ACuriousMind are you worried your team won't win this year?

Anonymous

Lol ^

@enumaris I literally could not care less about that

Anonymous

Why the hate for football? :P

It's not "my team". It's a bunch of dudes living in the same country as me

Anonymous

It's cool stuff

8:25 PM

@Blue I don't hate football as such. What I hate is almost everyone going around assuming as a default that I should care about it.

Not here, so far. I guess I'm just venting.

I care that you used "could not care less" rather than "could care less"

kudos

Like, I get into the elevator at work, and the first question I get from someone I don't even know is "Nice game yesterday, eh?"

Ugh.

Germany didn't even play yesterday

what u talkin bout philis

The thing is, that never changes out in the real world

and then there is the gossip, office politics

etc

@enumaris Oddly enough, Germans seem to care a lot about matches Germany doesn't play in, too

8:29 PM

is it phyllis...

phylis...hmmm

Anonymous

I'm pretty sure that's the same feeling mathematicians have to face when they're told that CERN discovered a new particle :P

Yesterday's game was 5-0 RUS-RSA

@bolbteppa "the real world"?

how is that a nice game lol

Away from the fiction that is high energy physics :p

8:30 PM

@bolbteppa I'm not in high energy physics!

@enumaris People like to see Saudi Arabia lose?

(anymore)

lol

ACM is a COBOL engineer now

@enumaris He might have said "funny" instead of "nice", I'm not sure :P

Like from the movie Inception

8:34 PM

There was a COBOL programmer in inception?

Clearly I need to rewatch that movie!

Anonymous

imdb.com/title/tt5295894

Hmmmm

quora.com/…

Cobol engineering is the company that chased Cobb lol

that's why I said "COBOL engineer" rather than COBOL programmer :P

@enumaris What a neat...technicality ;)

:D

lılostafa

8:44 PM

@ACuriousMind Iran is currently 1st in the group of death

rob

@lılostafa Without knowing that's a World Cup link, there are lots of ways to misinterpret it.

lılostafa

Group of death

A group of death in a multi-stage tournament is a group which is unusually competitive, because the number of strong competitors in the group is greater than the number of qualifying places available for the next phase of the tournament. Thus, in the group phase, one or more strong competitors in the "group of death" will necessarily be eliminated, who would otherwise have been expected to progress further in the tournament. The informal term was first used for groups in the FIFA World Cup finals. It is now also used in other association football tournaments and other sports. After the draw for...

@lılostafa Eh, Group of Death is probably Group D or Group F.

lılostafa

Spain + C. Ronaldo (Portugal) > All other teams and players
That clearly suffices for Group B to qualify as the group of death :)

who else is in group B?

it's not a group of death if there's only 2 top teams

8:55 PM

@enumaris Iran and Morocco