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13:02
hello?
Anonymous
Sorry, went for a while
Anonymous
Yep
Anonymous
So how many numbers you need to specify over phone?
Anonymous
For the picture?
6? (x, y, z, rx, ry, rz)
Anonymous
13:04
Cool :)
Anonymous
You're getting it
Well, with practical examples it's so much easier to understand and imagine :)
Anonymous
Now if you want to hang N pictures in a tilted fashion in your room, how many numbers will you have to specify?
Anonymous
Yesh!
13:06
:thumbs up:
Anonymous
Now suppose any two pictures are connected by a rigid rod (but at the ends you are able to freely rotate the pictures...think of a ball and socket joint)
Anonymous
So specifying the (x,y,z) of one will automatically specify the (x',y',z') of another
Anonymous
But you'd still have to specify the (rx',ry',rz') for the other picture
Anonymous
Right?
Umm, if I set 0, 0, 0 to be the position of one of the pictures, the others will get a 0, 0, 0 position and overlap? Something is wrrong :D
Anonymous
13:08
@NovaliumCompany The rod has some finite length...
Anonymous
So both pictures can't be at (0,0,0)
Anonymous
One will be at (2,3,4) say
Anonymous
While the other is at (0,0,0)
So if I have a picture at 3, 0, 0 and another one at 5, 0, 0 then the length of the rod is 2?
Anonymous
Right
Anonymous
13:10
Now you can freely rotate the picture at (5,0,0), remember that
Anonymous
Also you can freely rotate the picture at (3,0,0)
Anonymous
But the length between them is constant
Anonymous
And what's the maximum number of rods that can connect N pictures?
I just use the formula?
Anonymous
13:11
Yes
Anonymous
That's what the formula is for
Anonymous
I will tell you the derivation later
Anonymous
Okay, so are done with the basics it seems
Anonymous
13:12
Let's start with the Lagrangian now
All points must be at an equal distance of each other, since the length of the rod is constant, isn't it?
Anonymous
@NovaliumCompany All points of what?
All pictures
Anonymous
One rod only connects two pictures
Anonymous
Not all pictures
13:13
Yes but you have many rods, are they with an equal length?
Anonymous
Finite length
Anonymous
May not be equal
what does finite mean?
Anonymous
Could be anything: 1 metre or 5000 miles
oks
o specifying the (x,y,z) of one will automatically specify the (x',y',z') of another
that sentence makes no sense to me sorry ;\
Anonymous
13:16
I have a picture and a rod connected to it using a ball and socket joint
Anonymous
The picture is at (1,2,3)
Anonymous
And the rod is rotated by an angle $(r_x,r_y,r_z)$
Anonymous
From that, wouldn't you be able to find the coordinate of the picture at the other end of the rod?
Anonymous
So that's what I meant
13:19
It would be on a circle with radius the length of the rod?
Anonymous
No
Anonymous
I have specified all the angles $(r_x,r_y,r_z)$
Anonymous
It would be a fixed point in space
You have specified the position of the first picture, thats it?
1, 2, 3
Anonymous
No, I also specified the way the rod is tilted with respect to the picture
13:22
You haven't?
where?
Anonymous
4 mins ago, by Blue
And the rod is rotated by an angle $(r_x,r_y,r_z)$
Ohhh.
I thought I meant with numbers.
With example numbers.
Ok I got it now.
Anonymous
Dude just take $(30^o,60^o,30^o)$
Anonymous
Clear now?
I got it, sorry
Im stupid sometimes...
Anonymous
13:25
Now, a rigid body can have 3 translation degrees of freedom and three rotational degrees of freedom
Anonymous
Right?
by degrees you mean parameters?
Anonymous
Yeah
Anonymous
And a system of N such bodies can have 6N such degrees of freedom
Anonymous
13:26
If you don't have things like rods connecting them
Anonymous
Now if you have rods connecting them then the degrees of freedom reduces
Anonymous
Since you have to specify lesser parameters
Anonymous
These rods are the constraints
Why? I mean, each picture would still have it's own pos and rot?
Anonymous
13:28
@NovaliumCompany Think of the previous example I gave
Anonymous
If you specify the $(x,y,z,r_x,r_y,r_z)$ of one, you automatically get the $(x',y',z')$ of the other
Anonymous
Right?
Anonymous
So you no longer need to specify $(x',y',z')$
Yes, if you know the rod's length, the pos of the first pic and the rot of the first pic.
Anonymous
@NovaliumCompany Right
13:29
Ok, clear.
Anonymous
2 mins ago, by Blue
Now if you have rods connecting them then the degrees of freedom reduces
Anonymous
Agreed?
Anonymous
For such a system of N rigid bodies w.r.t. to a coordinate system I need to specify a large number of $(x^{1},y^{1},z^{1},r_x^{1},r_y^{1},r_z^{1},x^{2},y^{2},..)$
Anonymous
Where the rigid bodies are labelled 1,2,3,...
Anonymous
13:32
Yes?
Anonymous
"with respect to"
Anonymous
That's basically a list of numbers I need to specify for a system of N bodies
Anonymous
Including positions and rotations
Yes but you talked about reduced degrees of freedom?
Anonymous
13:34
Yes, I did.
did y'all hear about the data from the MiniBooNE experiment at Fermilab? Their data is pointing to a fourth flavor of neutrino (though obviously not enough data yet for conclusive proof)
Anonymous
@NovaliumCompany But is this okay?
Ok got it.
Anonymous
@NovaliumCompany That list of numbers is called the "generalized coordinates" of a system
Hey people. I have got a question in mind does more entropy always make the system of concern more disordered!
Anonymous
Can you say what would be the generalized velocities of the system? (Including rotational velocities)
Anonymous
@ParthaSarker "disorder" is a nonsense description of entropy
Anonymous
Think of it in terms of "number of configurations"
@Blue The velocity of each of the pictures?
Anonymous
13:37
@NovaliumCompany Yes, including rotational velocity
Anonymous
$dx^{1}/dt$ would be velocity of the particle labelled 1 in x-direction
Well, by just knowing the pos and rot of each of the pictures, I cannot know the velocity nor angular velocity of any of the pictures?
youtu.be/vSgPRj207uE I am particularly interested in the simulation done in this video. The system gets more ordered with more entropy.
I know that @Blue
Anonymous
@NovaliumCompany Right. You wouldn't know that. But suppose someone tells you how they change with time, then you would, right?
@Blue isn't the whole system of pictures have one velocity and one angular velocity?
Anonymous
13:40
@NovaliumCompany No, all the pictures can have different velocities
Anonymous
Why can't they?
They are connected by rods?
Anonymous
(Unless of course they're connected by a rod or something)
Anonymous
@NovaliumCompany In this case not all pictures are connected by rods
Anonymous
Let us assume
Anonymous
13:41
Maybe a few are connected, but not all
@Blue Hey.
We will have different results with different pictures being connected with rods?
Anonymous
@NovaliumCompany Sure
Ok, got it. What's next?
Anonymous
Given the generalized coordinates can you write down the generalized velocities now?
13:43
@Blue Have you watched the simulation in the video I shared the link of?
Anonymous
Given all the generalized coordinates are functions of time
@heather A 1eV sterile neutrino sounds interesting. I wonder what we can do with it...?
Anonymous
@ParthaSarker no
Just watch it. It's very interesting. The system gets more ordered (as in the common way of thinking) with more entropy.
@vzn see quantamagazine link above, paradigm shift $^{TM}$ may have a humble origin
Anonymous
13:44
@ParthaSarker Sure, that's a very well known phenomenon
@Blue I can find them, using V = d/t for each picture?
@Secret that's kind of the question, isn't it? It's not heavy enough to say it's dark matter. in the words of I.I. Ravi, I think: "Who ordered that?" but then again, probably not enough data to conclusively say they exist.
Anonymous
I'm not particularly interested in watching it now
here is the relevant paper for those who are interested.
Anonymous
@NovaliumCompany Yes. Given $(x^{1},y^{1},z^{1},r_x^{1},r_y^{1},r_z^{1},x^{2},y^{2},..)$ for the system the generalized velocities would simply be $d/dt(x^{1},y^{1},z^{1},r_x^{1},r_y^{1},r_z^{1},x^{2},y^{2},..)$
Anonymous
13:46
That is, take the derivative of each of the coordinates w.r.t. time
I forgot what was to 'take the derivative' but yes, I understand what you are saying :)
Anonymous
@NovaliumCompany Derivative of $x=t^2$ would be $dx/dt=2x$
Anonymous
For example
Anonymous
That is, $x$ coordinates changes with square of time passed
By putting d, you mean delta?
13:47
If a sterile neutrino exists, it will be a window to some of the weakest fundemental forces (since a sterile neutrino does not even interact with the weak force), and may suggest we need to update our model to get a more accurate picture of our universe.

But the jury remains open until that follow up experiment is done
Hopefully experimental error won't kill this excitement like all the past hypes.
Anonymous
@NovaliumCompany I mean derivative w.r.t time. You should know it if you know little bit of calculus
I honestly get fed up with experimental error keep crushing interesting results, even though it is morally correct for that to happen because the alternative is to be trapped in delusions
Yes, but I have forgotten most of derivatives, or at least I didn't learn them right.
@heather "A standard two-neutrino model is assumed for the MiniBooNE oscillation fits." Um... Why?
Anonymous
It won't be possible for me to cover everything from grade 8 level to lagrangian mechanics
13:50
@NovaliumCompany if you want a good review, try 3blue1brown's videos, quite good and quick.
Anonymous
It's better you revise derivatives
Anonymous
And meet tomorrow
Ok then. Anything else I should know except derivatives?
@Mithrandir24601 i dunno what that even means or why it's a problem =P would you mind explaining?
@Mithrandir24601 I thought neutrino mixing need all 3 neutrinos? (but then I vaguely recall the $\theta_{13}$ mixing angle is very tiny)
Anonymous
13:51
@NovaliumCompany That'll do for now
Ok. @Blue Thanks so much for teaching me, I'll learn derivatives and be back :)
@heather I'll give it a look :)
See you @Blue :)
@heather There are 3 flavours of neutrinos (excluding this sterile one), not 2. On the other hand, if it's a relatively large mass, the oscillations between the others would be quite small, so maybe it's small enough to be ignored?
@Mithrandir24601 oh, wait, i think i know why: they were shooting a muon neutrino beam, and they were oscillating into electron neutrinos, so tau neutrinos weren't involved.
would that make sense?
Anonymous
@NovaliumCompany See you. If you continue on this track maybe you'll end up learning special relativity and quantum mechanics as well as the hamiltonian and lagrangian formulations by the end of the year.
13:54
@Secret Not at all - you can still get oscillations between 2 flavours, although I certainly wouldn't go around assuming that oscillations only ever occur between 2 flavours either, hence my question
@heather Ahhhh!! Yeah, that's it, thanks :)
Anonymous
But please don't get lost in pop science articles which claim particles are waves and waves are particles
@Mithrandir24601 so muon neutrinos can't oscillate into tau? or we can't detect tau neutrinos in the same way?
@heather Although depending, it's not impossible to get tau neutrinos involved as well, but "there were only 2 flavours in the experiment" is good enough for now, considering I'm about to head out soon
ah, well then. i actually listened to a presentation on neutrino research the other day (in an odd twist of fate) so this is rather timely.
our school's physics club took a trip to the nearest university's physics department =D
@Blue That is the dream :D
14:05
I wonder if the contribution mu->tau->electron is insignificant enough to assume the 2 flavour oscillation model, probably this one is a higher order process than a direct oscillation between two flavours
hmm, yeah. i might ask that question on the main site...
14:29
@Mithrandir24601 i asked a question on the subject of the 2-neutrino model, so hopefully that'll answer this problem.
vzn
vzn
@Secret nice article, thx for tip, there are other "small" )( discrepancies with the standard model, it would be nice to see a broad survey, have misc collected links myself & need to write them up at some pt. as for paradigm shifts aka scientific earthquakes, my feeling is the Big One is away from a particle-centric pov of the universe toward a soliton model, many (now mostly disconnected) researchers are now independently taking tentative-to-bold steps in that direction.
btw, what does ")(" mean?
vzn
vzn
its an emoticon variant for iota with other meanings :)
Anonymous
@Secret A squished face
vzn
vzn
some hardcore physicist believers will argue that standard model discrepancies are all ")(". a contrarian would argue some will turn out to be not so ")(". neutrinos are very ")(" but could turn out to have significance far greater than very ")(" as the article lays out o_O :P
14:36
I do not understand the above sentence as the ")(" seemed to have 4 different meanings at least
vzn
vzn
lol emoticons can be like that try not to get triggered (reminder, its a chat line) :) :P
(picture the human conversational gesture of two fingers held together closely but not touching) :)
oh hey, i just broke 6k rep on physics.
huh, i didn't realize i was that close. i need to hang around physics more often.
@heather You sure do ;)
i'll probably have more time now that it's summer, so that's nice.
vzn
vzn
15:07
@Blue lol de broglie 1924, nobel prize en.wikipedia.org/wiki/Matter_wave
Anonymous
@vzn It's been nearly a 100 years since then
what's the best stackexchange for questions regarding programming finite element/finite difference methods?
Anonymous
@vzn Also that doesn't actually say particles are waves or vice versa
os
@georg You're probably looking for Computational Science
15:13
So I just discovered that Persian ice cream made with rice milk, vanilla, rose water and mango = food heaven :)
Anonymous
"rose water"?!?
@Blue Yeah - I thought you have that in India as well?
Anonymous
I don't think that adds to the taste in any way
Anonymous
Just some fragrance maybe
@Blue taste, maybe not, but I can smell it and smell does add to the flavour
Anonymous
15:16
Well, I don't particularly like the smell of rose :P
@Blue Now you're the one missing out :P
vzn
vzn
@Blue its still true 1 century later. + figured you would say that
Anonymous
@vzn As always I don't get your point :P
vzn
vzn
@Blue maybe study so-called "matter waves" more then :)
Anonymous
I love your strategies to escape debates ;)
vzn
vzn
15:19
lol, sarcasm? sometimes weary of certain so-called "debates" on here... and who has "strategies" / is "escaping"?
@Mithrandir24601 I'm currently eating mint choc chip icecream, and it's pretty good :-)
@JohnRennie Nah, it's not the same :P One advantage of not being able to have milk is that vegan ice creams tend to be made to a much higher quality than your average, run-of-the-mill non-vegan ice creams, so taste much nicer :P
That's what the icecream followed :-)
vzn
vzn
sig other demands coldstone creamery, like dairy queen softserve myself, slowly turning into relationship-killing incompatibility (among "other" things) o_O
i personally think some chocolate gelato rules above all.
15:27
@heather As usual, it's something or other to do with atmospheric neutrinos (atmospheric B meson mixing is also where the CP violation comes from if memory serves me correctly)
i also had some butterbeer flavored gelato once, which was pretty darn good.
@JohnRennie English sausages = meh
@Mithrandir24601 them darn atmospheric neutrinos ;) i'll look it up.
@heather Whaaaat?
@Mithrandir24601 Wild boar and apple!!
vzn
vzn
15:27
mmm, pistachio flavored icecream :)
@Mithrandir24601 Whaaaat what?
@JohnRennie OK, that does sound delicious
i mean like harry potter butterbeer, to be clear, not buttery alcoholic beverages
which sounds just plain weird, but anyway.
Anonymous
"harry potter butterbeer"? :)
@heather Yeah, but I've never even seen butterbeer, nevermind butterbeer flavoured this or that
15:29
@Mithrandir24601 really? i saw bottles of butterbeer in a specialty candy shop the other day, and bertie bott's every flavor beans, and chocolate frogs.
maybe the midwest is just better at these things =P
@heather ... That's the USA for you...
Anonymous
harry potter is an alcoholic beverage (?)
to be fair, the gelato tasted pretty much like butterscotch, but.
Anonymous
gosh
@heather Yep. Maybe I'll try some when I finally get to visit America in whenever that will be (within a couple of years, definitely)
Anonymous
15:30
the names
IIRC butterbeer is butterscotch flavoured
Anonymous
@JohnRennie Eeew
i didn't know what it was supposed to taste like.
@JohnRennie Isn't that cheating slightly?
Anonymous
Butterscotch is awful
15:31
@Blue !?!?
@Mithrandir24601 huh?
vzn
vzn
@heather the univ physics trip sounds cool which university? were they doing any experiments?
@JohnRennie There's no point in inventing a new beverage, only to have it taste just like an already existing one
Anonymous
@heather Never liked the butterscotch flavour
ah, here we go:
> J.K. Rowling said in her interview to Bon Appétit magazine that she imagines it "to taste a little bit like less-sickly butterscotch"
according to wikipedia.
Anonymous
15:32
Too sweet for my taste
also:
> The earliest reference to buttered beer is from The Good Huswifes Handmaide for the Kitchin, published in London in 1588. It was made from beer, sugar, eggs, nutmeg, cloves and butter. Another old recipe for buttered beer, published by Robert May in 1664 from his recipe book The Accomplisht Cook, calls for liquorice root and aniseed to be added.
@heather That sounds like a different variety of butterscotch, not a new thing altogether
@Mithrandir24601 true, but it also has a "warming effect" whether served hot or cold, again according to wikipedia.
@vzn iowa state university; we got to see a lab where they were doing some laser spectroscopy experiments, so yes. the other two "talks" we saw were theoretical.
@heather Wouldn't that be achieved by adding a bit of alcohol? (or, if you feel so inclined, ginger)
@Mithrandir24601 well, in harry potter it is very mildly alcoholic, so there you go. not enough to get wizards tipsy, though, just house-elves.
vzn
vzn
15:35
@heather cool, so do you have some inclination for future directions in college? physics vs engr etc?
@vzn i'm leaning more physics currently, but experimental physics. electrical engineering and aerospace engineering are cool too.
@heather So... It's butterscotch liqueur, only with less alcohol?
vzn
vzn
@heather had some ideas/ inspiration for a photon experiment recently, itd be great if you pushed fwd on yours, was very excited to hear about it awhile back, those types of efforts are rare on here...
@vzn Oy! :P
@Mithrandir24601 a lot less alcohol, i'd think.
but i mean, yeah.
how are you supposed to commercialize a completely fictional drink? =P no, really, though, i don't think it's cheating to make it butterscotch-y.
vzn
vzn
15:37
@Mithrandir24601 you are a delightful exception. also, a "pro physicist" was thinking "amateur science".
@vzn OK, that's fair, I'll let you get away with that :P
(although I'm not overly sure how much of a 'pro physicist' I actually am)
vzn
vzn
lol, whew! otherwise "gets away" with next-to-nothing )( on here o_O
@Mithrandir24601 (lol oh geez you phd guys and your "pro" modesty... wasnt your group published in nature or science recently... saw that link you posted awhile back but just barely!)
@vzn not really amateur scientist at this point, seeing as i haven't done any actual science =P
Mar 10 at 16:43, by Mithrandir24601
@vzn Ah! Perfectly timed question! The group I'm in just published a paper in Science: http://science.sciencemag.org/content/early/2018/03/07/science.aar7053?rss=1
@JohnRennie hi.
vzn
vzn
@heather you did significant bkg research on an independent/ amateur tabletop photon experiment, found some great/ excellent refs, found equip suppliers etc, + bet youre doing stuff at school.
15:42
@vzn Oh yeah, the last time one of my friends/people from the group I'm in published in Nature, when was it? Three days ago now ;)
vzn
vzn
@Mithrandir24601 elite crowd! :) so what are you doing on SE anyway? :P
@user187604 Hi
@vzn Actually, now I think about it, you have the most ridiculously good sense of timing :P
@JohnRennie did you change the time of your attendance in se chat?
vzn
vzn
@Mithrandir24601 lol, blushing now! not so much )( positive feedback on here sometimes :'( :P
15:46
@JohnRennie I mean due to work. Did you change the time of your presence in se chat which was from 5am—1pm. In case if I have to ask you In other time.
@user187604 I don't have a set time I join the chat room, though I am generally around between 05:00 and 13:00. Today I've been busy attending to a server.
@JohnRennie whenever I touch your icon it shows last seen 56 seconds ago. Do you turn the net on all time on your laptop? Asking if I could know you are online or not?
Anonymous
@Mithrandir24601 Digressing a bit: Any idea what exactly they mean by "For example $|x^{n}\rangle$ indicates the $n$ th training sample which is prepared in the quantum state with the value of $x^{n}$" here? Afaik, for a classical feedforward neural network the training sample are like a picture with say 100 pixels and the corresponding output. Say the picture could be of the digit '8' in pixelated format and the corresponding output is 8.
Anonymous
15:51
@Blue This link works, the other one is slow to load
Anonymous
Page 4
Anonymous
The link is paywalled, lol
Anonymous
Anyway, you could get it
@Blue Not having read it yet, I suspect that here, n is just a label
15:52
Most entertaining fictional news read so far
vzn
vzn
@heather fyi recently was looking into polarizing beamsplitters re tabletop bell experiment and found one for $27 on amazon, remarkably rare at that price, many easily over 10x cost. (Mithrand could likely attest to that) amazon.com/10x10mm-Prism-630nm-Polarizing-Splitter/dp/…
Anonymous
@Mithrandir24601 But even then, how are they considering training samples to be quantum states?
Anonymous
How's a pixelated image of the digit 8 a quantum state?
Anonymous
Or rather how are they converting it to one?
@JohnRennie likewise it's showing 30 seconds ago..
15:53
@vzn ah, nice.
@user187604 My computer is on all the time and I often leave Chrome running, so my last seen time doesn't mean I'm actually at the computer.
When Slereah is on next time, I am gonna to post that snapshot of EVE to create a contradiction in h bar
vzn
vzn
@heather are you too busy now to pursue your bell experiment ideas? fyi this was my latest wild + crazy idea, needs +1s physics.stackexchange.com/questions/402680/…
@JohnRennie oh so so far I have made a wonderful world record of pinging you 5 Times (I guess) :p
That will be soooooo interesting for our goals of The Plan
15:56
@vzn i am rather stuck currently, at least partly because BBO crystals are, well, rather expensive, and at least partly because i have no idea what i'm doing =P
@JohnRennie had a hard work today?
@vzn Not always 10x the cost but near enough, yeah
Oh I guess I'll beat my record :p
@Blue Hmm... I'm having a look now
vzn
vzn
@heather yep BBO very tricky & maybe the major hurdle but maybe a small one wouldnt be so expensive... have wondered about that myself... afaik there are other materials that are "similar"... think you may have mentioned another material or maybe saw one in a paper... trying to remember...
15:59
@vzn there are other crystals, but as far as I can tell just as specialized and expensive, unfortunately.
KDP crystals, for example, can be used.
vzn
vzn
@heather small ones work and maybe can be found from places other than "pro optical suppliers" (like the beamsplitter) which are killer $ typically. :| ... esp if they are used as electronics parts somehow...
@vzn yeah, i just haven't found the right spots yet, probably.
oh, yeah, i love electronics surplus stores.
@Blue I'm going to hazard a guess that, for a given sample $n$, $x_j^n$ represents the value of the $j^{th}$ neuron for that sample
vzn
vzn
@heather its a (very) tricky problem, almost nobody else in the world is attempting it, so was defn impressed with your effort/ ideas so far, even maybe cited it in a blog iirc
@JohnRennie all right professor take a great rest and food. I'm not gonna vex you anymore today. I'll come back tomorrow. Goodbye. Good night from India....
Anonymous
16:02
@Mithrandir24601 Is $x_{j}^{n} \in\{0,1\}$?
@user187604 I'm afraid my presence here is a bit intermittent right now ...
@Blue In the classical case, yeah
Or at least, that's what I'd assume
Anonymous
For real neurons (classical) the activation of each is a number in between 0 and 1
I don't see any reason that you couldn't have more than one bit per neuron...
@JohnRennie today or forever?
16:04
@Blue but in the quantum case, this happens naturally using a single qubit
Anonymous
@Mithrandir24601 Ah, they took 10 qubits per neuron in the input layer. That means each neuron in their case can take $2^{10}$ values between $0$ and $1$ (representing the activation)?
Anonymous
@Mithrandir24601 Wait wait. How does a qubit take the value (say) 0.3 ?
@Blue Yeah, this makes sense. I'm really not sure though
Anonymous
Activation of a neuron could be anything between 0 and 1
Anonymous
Not only 0 and 1
16:06
@Blue Yeah, multiple qubits per quron makes more sense
vzn
vzn
@heather bet just like your dad, inherited... my dad/ brother liked em lots too :)
@Blue Sup.
Anonymous
Also their next line: "Moreover $y_k^{n}$" is used to denote the output conditional probability $p(|y^{n}\rangle = |k\rangle |x^{n})$ sounds cryptic to me
Anonymous
Wtf is $|k\rangle$
Anonymous
Any natural number
16:09
@JohnRennie I'll ask you tomorrow good night from India....
@Blue Actually, no, they just use $x_j\in\lbrace 0, 1\rbrace$, at least for this bit
@Blue I think I've revised derivatives good enough. If you don't want now, we could continue later?
Anonymous
@Mithrandir24601 Umm, so each quron either has activation 1 or activation 0?
Anonymous
@NovaliumCompany Also learn the derivation of the $\binom{N}{2}$ thing
Anonymous
We will continue tomorrow
16:11
@Blue Considering that "binary values of a McCulloch–Pitts neuron" is apparently a thing, that seems possible
Ok @Blue. What should I google to see the derivation of the N/2 thing?
why not use a textbook?
@Blue From Fig. 2, I'm fairly sure that's the case, at least to begin with
Anonymous
@NovaliumCompany Permutations and combinations
Anonymous
@Mithrandir24601 Okay, but I still don't get why later in the paper they took 10 qubits per quron and what the significance of that is
Anonymous
16:14
From the first part of the paper it seems they are taking only 1 qubit per quron
Anonymous
And a state like $|000101010101000001011010101010101111\rangle$ is a training sample
So @Blue I should learn how to find the derivative of Permutations and combinations?
Anonymous
@NovaliumCompany Learn what permutations and combinations are
Anonymous
Just the basics will do
Anonymous
derivative of P&C doesn't make sense
16:16
Ok. So that's all for now? Just Permutations and combinations?
Anonymous
Yep
Anonymous
7 mins ago, by Blue
Also their next line: "Moreover $y_k^{n}$" is used to denote the output conditional probability $p(|y^{n}\rangle = |k\rangle |x^{n})$ sounds cryptic to me
Ok, see you guys.
@Blue I've just found the sentence "i.e., each quron is a single qubit", so that helps clarify that, at least. Later, more qubits per quron means that it looks more continuous than discrete (until you get to the $10^{-m}$ level anyway), so presumably you get more accuracy or something
cya @NovaliumCompany
Anonymous
16:18
@Mithrandir24601 "more continuous than discrete"...so are they actually trying to take into account activations other than 0 and 1, by introducing more qubits later on in the paper?
Anonymous
Also, what do you mean by "$10^{-m}$" level?
@Blue I believe so (bear in mind I don't actually do machine learning :P )
@Blue As in, if you've got $m$ qubits (maybe $m+1$?)
(or something similar to that) you can then measure down to... Whoops
That should be $2^{-m}$
I'm thinking in binary, sorry :P
2
Anonymous
Yeah. Breaking down the length between 0 and 1 into $2^{m}$ parts
Anonymous
I guess
@Blue yeah, that
Anonymous
16:22
@Mithrandir24601 Lol
Anonymous
@Mithrandir24601 It's okay, I too just started learning this stuff a couple of days back :P
@Blue Sometimes, the computer scientist in me pops out
@Blue machine learning's highly sought after, so no better time to start than the present
:)
Anonymous
Hehe, anyway, any idea about they're saying about the $|y\rangle$ thingy?
Anonymous
(BTW if you're busy or have some other work to do, let me know. Hope I'm not bothering much)
Anonymous
Couldn't really understand the term "output conditional probability"
16:28
@Blue ah, from equation 11, it's just the output value i.e. What is the probability of the output being the state $\left| k\right>$, given the input $x^n$?
Anonymous
@Mithrandir24601 Is it something like, given the input $|010000010100100101010101010101\rangle$ (representing activation of pixels due to the input image....say of a handwritten symbol), what is the probability of it producing the output $3$ (representing that the input was actually the handwritten symbol 3)?
Anonymous
They're just representing the output $3$ as $|3\rangle$ I suppose
@Blue Something along those lines, presumably?
@Blue yeah
Anonymous
I think that sounds more or less clear as of now. Proceeding to the next part :)

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