The h Bar

Nihar Karve

02:14

You might wanna see this if you haven't already: physics.stackexchange.com/questions/83260/…

RewCie

03:48

Prime Minister Boris Jonson sent Diwali Wishes

His previous Dw wishes and this one makes him looks smart tho... How does he knows all these?

RewCie

03:59

Epic!

BioPhysicist

04:45

-1

Q: What is the mathematical foundation of the bra and ket formalism?

I answered this quite general question in several posts below, to make it easier to read. Context: Number of bras and kets My answer there and comments beneath it.

Is it allowed to break your answer up into many parts?

RewCie

@BioPhysicist What you mean by many parts?

Many paragraphs is okay

BioPhysicist

Sorry, I meant many posts. Although I thought the link I gave was enough clarification.

RewCie

NVM, the user did something for more upvotes is now getting more and more downvotes now.

BioPhysicist

They are getting some up votes. And since up votes give 5x as many points as down votes lose, the up votes are still beneficial even if it's net down

RewCie

ok, he won

BioPhysicist

04:53

Oddly enough the question and only 2 answers have up votes at the moment. The other two answers do not have up votes

RewCie

k....ok

Nihar Karve

06:19

Does an n-simplex have $\begin{pmatrix} r+1 \\ n \end{pmatrix}$ r-cells?

Qmechanic

07:08

Today, November 9th, 2020, is the 10 year anniversary of Phys.SE's public beta. Congratulation!

15

Nihar Karve

Congratulations, especially to the old-timers!

Emilio Pisanty

07:35

@FadedGiant This looks like it needs dedicated attention from a subject-matter expert

2

Q: Was Chernobyl Explosion a Small Nuclear Detonation?

The Chernobyl reactor was designed to produce a maximum of 3.2 megawatts of power. It is now estimated to have produced 3.2 gigawatts of power in the final second of its life. The original hypothesis is that the the explosions at Chernobyl were either steam/steam or steam/hydrogen explosions....

thermodynamics nuclear-physics nuclear-engineering explosions

The answers don't seem to address the cited paper. Is the paper wrong? Or is OP simply misinterpreting it?

Faded Giant

@EmilioPisanty ugh

> The Chernobyl reactor was designed to produce a maximum of 3.2 megawatts of power.

No. The design value was 3.2 GW of thermal power.

skullpatrol

07:50

:O

Massive!

::renews interest in Chernobyl HBO series::

Faded Giant

Well, other contemporary nuclear power plants were even larger.

But for an RBMK, the large size of the core creates its own problems.

@EmilioPisanty Okay, I will have to read that paper first.

Emilio Pisanty

09:35

@FadedGiant yeah, my initial reaction to the question was a similar, strong "ugh"

but it does seem to be based on a peer-reviewed paper that, at least from the abstract, seems to claim some form of explosive nuclear process

and I guess OP is indeed owed the benefit of addressing whatever is in that paper

Faded Giant

10:33

@EmilioPisanty Of course, the original energy source for the first explosion was a nuclear process. The question is, how this energy is transferred and released considering the various transfer and feedback mechanisms.

Clearly, it wasn’t like a small nuclear bomb exploding in air. Nevertheless, the fuel became prompt critical at one point. However, when looking from the outside, the final effective energy release during the first explosion was mostly via a steam explosion.

123

Hi All,

If two boxes of different masses placed on frictionless surface one is lighter and another is much heavier.

How do we know how much force is needed to slight move the object.

Actually they won't move. Just force at which slight more force we apply it move.

In other words. Is it possible to calculate inertia of an object at rest.

ACuriousMind

11:02

@123 arguably that's what scales do, no?

Emilio Pisanty

@ACuriousMind scales measure gravitational weight, not inertial mass

ACuriousMind

yeah, that's the point one can argue about - if you're trying to do this before you're convinced that inertial and gravitational mass are the same, it doesn't work

Emilio Pisanty

@123 If you're thinking that there is some threshold force $F_{thr}$ such that if you apply a force $0<F<F_{thr}$ the box won't move and for $F>F_{thr}$ it will start moving, then that is in direct contradiction with your assertion that the masses are placed on a frictionless surface

@123 in short: if you want to explicitly measure the inertial mass, then no, this is not possible.

The mechanism you described requires a surface with friction, which in turn depends on the weight of the object, i.e. its gravitational mass.

and as @ACuriousMind mentioned, if you want to measure the gravitational mass while the object is at rest, the simplest method is a set of scales.

@FadedGiant I guess I mostly still don't understand what mechanism is being proposed by De Gheer et al.

(and how different it is from the consensus view of what happened)

JingleBells

12:20

When I have a linear model of $wx + b$ where $w$ is the weight and $b$ is the bias, it's a simple linear regression model used in machine learning. So a way to fit nonlinear data to capture more complex relationships would be to use polynomial features, where I add new features such as $x^2$, $x^3$... that in a way stretches the data allowing for the linear model to be fit. And when I plot it, I just plot the x with the trained model and it looks as if the linear model is curved,

but it's just a linear model fitted to stretched data, right?

Since it has multiple weights for each polynomial feature ($x^2, x^3$...) the model becomes $w_1x + w_2x^2 + w_3x^3 + b$, but that's no longer a linear model, right? It's a polynomial regression model (curved)?

Or is it just a linear model where we fit a straight hyperplane to stretched data ($x^2, x^3...$)?

I don't know which way to view the model. Is it a linear model that fits stretched data, or is it a curved polynomial model that fits the original unstretched data?

I think it's the former, pliz help

Emilio Pisanty

12:40

@JingleBells that's a linear model, because the fit parameters $(w_0,w_1,w_2,\ldots)$ appear as the coefficients of a linear combination of functions of the data.

JingleBells

@EmilioPisanty Got it, makes sense! So the model fits a linear hyperplane to stretched data?

If the weights were $w_1^2, w_2^3...$ something like this, then it would be considered a nonlinear model, right? But in the above case, the model is linear and it's fitted to the $x^2, x^3...$

So I should view the scenario above as a linear hyperplane that's fitted to stretched exponential data ($x, x^2, x^3$...), right?

And we're introducing those polynomial exponents of x for the sake of actually being able to fit a straight hyperplane, which would not be possible otherwise.

RewCie

13:11

Today's challenge... Complete 100 page of a 1200 page x86/IBM-6000OS DEV Book!

#DeveloperSE #CodingLife

Linear Algebra is Linear because of $\mathbf A \mathbf x = \mathbf b$

not powers

klol

Mainly study of straight functions... $f(x+y) = f(x) + f(y)$ and $f(ax) = a.f(x)$

Emilio Pisanty

13:33

@JingleBells "stretched data" is not a particularly useful concept

Linear regression

In statistics, linear regression is a linear approach to modelling the relationship between a scalar response (or dependent variable) and one or more explanatory variables (or independent variables). The case of one explanatory variable is called simple linear regression. For more than one explanatory variable, the process is called multiple linear regression. This term is distinct from multivariate linear regression, where multiple correlated dependent variables are predicted, rather than a single scalar variable.In linear regression, the relationships are modeled using linear predictor functions...

the Wikipedia section on Assumptions > Linearity has a good discussion

RewCie

Interestingly polynomial regression doesn't need matrices, vectors...

Emilio Pisanty

> Linearity. This means that the mean of the response variable is a linear combination of the parameters (regression coefficients) and the predictor variables. Note that this assumption is much less restrictive than it may at first seem. Because the predictor variables are treated as fixed values (see above), linearity is really only a restriction on the parameters.

RewCie

It can be done with hand only.

Emilio Pisanty

> The predictor variables themselves can be arbitrarily transformed, and in fact multiple copies of the same underlying predictor variable can be added, each one transformed differently. This technique is used, for example, in polynomial regression, which uses linear regression to fit the response variable as an arbitrary polynomial function (up to a given rank) of a predictor variable.

> With this much flexibility, models such as polynomial regression often have "too much power", in that they tend to overfit the data.

@RewCie That is incredibly misleading. @JingleBells, I would advise you to disregard this completely.

Polynomial regression "needs" matrices and vectors as much as any other type of linear regression

you can do it "with hand only" as much as any other linear regression.

RewCie

@EmilioPisanty It doesn't need Matrices, Vectors because it can be done with hand

What's wrong here?

Emilio Pisanty

13:40

@RewCie are you talking about polynomial regression specifically, or linear regression in general?

If the former, then the exclusion of the latter is dead wrong

RewCie

@EmilioPisanty Polynomial regression only!

Emilio Pisanty

@RewCie then it's wrong, period.

RewCie

No, it isn't!

Emilio Pisanty

There are no "by-hand-only" methods for polynomial regression which discard the explicit concepts of matrices and vectors, and which wouldn't also work for any other type of linear regression.

RewCie

If you are to find, coefficients $a_0, a_1, a_2, \dots, a_n$ such that $\hat y = a_0 + \sum a_i x^i$ fits, it, you just need basic calculus for it. I don't see any usage of Matrices here.

Emilio Pisanty

13:42

If you want to dispute that, then you need to provide an explicit example (and somehow prove that it does not work for general linear regression)

RewCie

and you have, dataset as $(x, y)$

then, you have to pick the best estimate of $a_i \, \forall i \in \mathbb N$ such that the $\hat y$ fits it properly.

Emilio Pisanty

I'm uninterested in discussing the specifics. This is not a controversial point. If you're confused about it then the Cross Validated SE chat is the place for it.

As far as this chat is concerned, I will ask you not to provide beginners with misleading explanations.

But that's as much as I will say about it.

RewCie

It's not disputed but works very well in most of the cases.

Okay, I'm off the conversation.

Emilio Pisanty

@JingleBells You can, yes. In effect, when you're doing polynomial regression, you are stating that you have reasons (external to the data itself, and to do with additional understanding of the process that produced it) to believe that if you expand the dimensionality of the data from $(1+1)$-dimensional $(x_i,y_i)$ to $(N+1)$-dimensional $(x_i,x_i^2,\ldots, x_i^N, y_i)$, then it will lie reasonably flat in an $N$-dimensional hyperplane, and you do linear regression there.

There is nothing specific to polynomials in that. You could do the same with e.g. $(\sin(x_i), \sin(2x_i), \ldots, \sin(Nx_i),y_i)$. As far as the linear algebra is concerned, it's exactly the same.

Polynomial regression

In statistics, polynomial regression is a form of regression analysis in which the relationship between the independent variable x and the dependent variable y is modelled as an nth degree polynomial in x. Polynomial regression fits a nonlinear relationship between the value of x and the corresponding conditional mean of y, denoted E(y |x). Although polynomial regression fits a nonlinear model to the data, as a statistical estimation problem it is linear, in the sense that the regression function E(y | x) is linear in the unknown parameters that are estimated from the data. For this reason...

and the multiple linear regression section

are good resources

and, as always with polynomial regression, it's crucial to keep in mind that the degree of the polynomial must be kept low, or you're overfitting.

Any set of $n$ points can be exactly fit by a polynomial of degree $n-1$ -- but that fit does not tell you anything, and it will be completely unpredictive for any new data points

geometrically, if you have any $n$ points and you try to fit them with a polynomial of degree $N=n-1$, then you're trying to find an $N$-hyperplane in $(N+1)$-space that will pass through $N+1$ points, and this is always possible

it's easiest to see with trying to fit a quadratic to $n=3$ points. That means you're unfolding to the 3D space $(x_i,x_i^2, y_i)$, and you're trying to find a plane that passes through three points. which is always possible.

schn

16:42

This is maybe a question for Cross Validated SE, but does the biased sample variance converge to the unbiased one for a very large sample?

Prateek Mourya

Ah i know there is rule that we don't ask before asking question

But still can anyone tell if i can still ask question or later as right now important discussion is going on

I will not bug people to answer my question just some conceptual doubts

Emilio Pisanty

> Don't ask about asking, just ask

"I know that there are good reasons why people don't like it when folks do what I'm about to do, but I'm going to do it anyway"?

If you're worried about interrupting anything -- the room's been quiet for three hours

from the guidelines:

> Don't interrupt ongoing conversations. One of the great things about Stack Exchange chat is that the reply system allows multiple conversations to take place simultaneously. But that only works if you're discreet, and you don't post walls of text

in other words, just ask.

RewCie

17:20

Aww! As soon as the Vaccines were announced the stock market peaked for them! A sudden 90 degree sharp straight surge.... Never seen something like that before.

Also, stock prices of Ubisoft and so fell sharply (no idea why)

Why such strange and sharp change after vaccine announcement?

The P/E ratio of the Pfizer is more than 15, showing overvalued stock price, whereas, if I calculate correctly, the Ubisoft stocks seem have fallen below intrinsic value.

Let's see more few more days... I'm curious about it.

Disney up by 10%, zoom down by 15%

Prateek Mourya

Nihar Karve

@RewCie pure speculation, but maybe it's because people will be out and about again, so less videoconferencing and video games?

Prateek Mourya

The container is filled with fluid

And i am trying to find the shape of fluid w.r.t

Container

Nihar Karve

17:36

@schn at any rate, large n means that Bessel's correction to the biased sample variance approaches 1

peterh - Reinstate Monica

17:46

@JohnRennie I grabbed and processed the wikipedia isotope tables just for this. It was a big work.

@JohnRennie I found some nuclei pairs with the same nucleon number but more than 106MeV mass difference. Some of them, like tin-139 and antimony-139, has a mass difference close to a proton mass. Unfortunately, all of them are very unstable nuclei. All the stable nucleus mass difference are below 2-3 MeV or yet more less.

But it had been so beautiful :-)

@RewCie Possibly they announced that they are working on a game where a modified covid virus makes zombie from the humanity, and the user can shot all of them.

JMac

18:21

@RewCie Only thing that immediately comes to mind for me is that Assassin's Creed: Valhalla comes out tomorrow and reviews came out today. They don't seem particularly negative, but it's a major release for them so it seems like it could make sense that it's tied to that.

vzn

19:02

in theory salon, 3 mins ago, by vzn

Blueprint for a Scalable Photonic Fault-Tolerant Quantum Computer / Bourassa et al

Transcript for