text:
# Disproof of Bell’s Theorem

Why is this evidence so feared by the dark forces?

In 1964 John Stewart Bell proved a mathematical theorem and claimed that “no physical theory which is local and realistic (in the sense espoused by Einstein) can reproduce all of the statistical predictions of quantum mechanics.” However, in a series of papers written between 2007 and 2014 I constructed just such a local and realistic framework for physics. Consequently, today Bell’s theorem no longer has the fundamental significance for physics it was thought to have.

In a nutshell, my disproof of Bell’s theorem goes as follows: In 2007 I noticed that there is a serious error in the very first equation of Bell’s famous paper [Physics 1, 195 (1964)]. Bell simply made an elementary but fatal mistake, and this mistake lies in the formulation of the very first equation of his famous paper. The correct form of his proposed functions to reproduce the quantum correlations local-realistically must be

$\displaystyle \pm\,1\,=\,{\cal A}({\bf a},\,\lambda):{\rm I\!R}^3\!\times\Lambda\longrightarrow S^3 \sim {\rm SU(2)} \ \ \ \ \ \ \ (1)$

$\displaystyle \pm\,1\,=\,{\cal A}({\bf a},\,\lambda):{\rm I\!R}^3\!\times\Lambda\longrightarrow S^0 \subset{\rm I\!R}, \ \ \ \ \ \ \ \ \ \ \ \ (2)$

for them to provide a complete local description of the physical reality demanded by Einstein. The latter form, with ${{S^0=\{-1,\,+1\}}}$ (or a subset of the real line) as the codomain of the functions ${{{\,\cal A}({\bf a},\,\lambda)}}$, is what Bell assumed in the first equation of his paper. But the complete description can be provided by such functions if and only if their codomain is an absolutely parallelized 3-sphere, ${{S^3}}$, which is a closed set of unit quaternions, homeomorphic to the fermionic rotation group SU(2):

$\displaystyle S^3:=\left\{\,{\bf q}(\psi,\,{\bf a},\,\lambda):= \,\lambda\,\cos\frac{\psi}{2}\,+\,\lambda\,{\boldsymbol\beta}({\bf a})\,\sin\frac{\psi}{2}\; \Bigg|\;||\,{\bf q}(\psi,\,{\bf a},\,\lambda)\,||^2=1\right\}\!. \ \ \ \ \ \ \ (3)$

Here the complete state $\lambda$ of the spin system is represented by the initial orientation of the 3-sphere, and the results ${\cal A}=-1$ and $+1$ are limiting cases of ${\bf q}(\psi,\,{\bf a},\,\lambda)$ for the rotation angles $\psi=2\pi$ and $4\pi$. Note also that a 3-sphere cannot be charted by a single coordinate system, but an anholonomic frame, made up of graded basis $\{\,1\,,\;{\boldsymbol\beta}_1\,,\;{\boldsymbol\beta}_2\,,\;{\boldsymbol\beta}_3\,\}$, can be defined on it, fixing each of its points uniquely. It is the impossibility of defining a global coordinate chart on ${{S^3}}$ but possibility of defining an anholonomic frame on it that is ultimately responsible for the existence of EPR-Bohm correlation.

It is very important to note that the image points of both forms (1) and (2) of the above functions, namely the measurement results ${{\pm\,1\,}}$ themselves, are the same:

$\displaystyle {\cal A}({\bf a},\,\lambda)\,=\,\;either\;\;+\!1\;\;\;\;or\;\;\;-\!1\,,\ \ \ \ \ \ \ \ \$

.

with

$\displaystyle {\cal E}({\bf a})\,=\,\lim_{\,n\,\gg\,1} \left[\frac{1}{n}\sum_{k\,=\,1}^{n}\,{\cal A}({\bf a},\,{\lambda}^k)\right] \,=\,0\,. \ \ \ \ \ \ \ (4)$

They are exactly what Bell assumed them to be in his first equation. Only the codomain of the two forms of the functions differ from each other. Failing to take the physically and mathematically correct codomain into account, Bell’s argument is thus indeed a non-starter. That is to say, his argument simply does not go through without the assumption of totally disconnected set $\cup\,S^0$ as the set of all possible measurement results (actual as well as counterfactual). What is more, once the correct form (1) of the measurement functions is recognized, the familiar singlet correlation predicted by quantum mechanics between the measurement results ${\cal A}$$({\bf a},\,\lambda)$ $=\pm\,1$ and ${\cal B}$$({\bf b},\,\lambda)$ $=\pm\,1$, namely

$\displaystyle {\cal E}({\bf a},\,{\bf b})\,=\,\lim_{\,n\,\gg\,1} \left[\frac{1}{n}\sum_{k\,=\,1}^{n}\,{\cal A}({\bf a},\,{\lambda}^k)\;{\cal B}({\bf b},\,{\lambda}^k)\right] \,=\,-\,{\bf a}\cdot{\bf b}\,, \ \ \ \ \ \ \ (5)$

is relatively easy to reproduce in a strictly local and realistic manner. In fact, once the topology of the codomain of the measurement functions ${{{\,\cal A}({\bf a},\,\lambda)}}$ is correctly taken into account in the above manner, the observed EPR-Bohm correlation predicted by quantum mechanics inevitably falls out as a classical, deterministic correlation among the points of an absolutely parallelized 3-sphere. This is a robust result. It is not difficult to prove that correlation between any two points ${\bf q}({\bf a})$ and ${\bf q}({\bf b})$ of $S^3$ is necessarily equal to $\,-\,{\bf a}\cdot{\bf b}\,$, with ${\cal A}$$({\bf a},\,\lambda)$ $=\pm\,1$ and ${\cal B}$$({\bf b},\,\lambda)$ $=\pm\,1$ being the limiting cases of ${\bf q}(\psi,\,{\bf a},\,\lambda)$ and ${\bf q}(\psi,\,{\bf a},\,\lambda)$, respectively. Thus all the perplexity generated by Bell’s impossibility claim is much ado about nothing. It stems from a topologically incorrect accounting of measurement results.

You can verify these facts in an explicit computer simulation of the model, as discussed in this paper. At least five explicit, event-by-event, numerical simulations of my local model for the EPR-Bohm correlation have been independently produced by five different authors, with codes written in Java, Python, Mathematica, Excel VB, and “R“. The above 2D-surface version of the simulation of my local model can be found here. I discuss two of these simulations in the appendix of this paper. A compact translation of one of these simulations (from Python to Mathematica) can be found here. Each simulation has given different statistical, geometrical, and topological insights into how my local-realistic framework works, and indeed how Nature herself works.

The most accurate simulation of my local model for the EPR-Bohm correlation can be found here.

From the above discussion Bell’s mistake is now very clear. Instead of investigating correlation among the points of a parallelized 3-sphere, Bell naïvely investigated correlation among the points of a real line to prove his celebrated theorem. But even an elementary topological consideration is sufficient to recognize the fact that correlation among the points of a real line cannot possibly be the same as those among the points of any topologically nontrivial space such as a parallelized 3-sphere.

But what about the famous Bell-CHSH inequality on which Bell’s supposedly iconic theorem is based? It turns out that it too can be violated local-realistically. In doing so, however, we gain a tremendous insight into the very origins of quantum correlations. This is discussed in greater detail on the next page, but first let us recall that for the set of four measurement directions, a, a$'$, b$'$, and b, the Bell-CHSH inequality, which claims an upper bound on local-realistic correlations, is given by

$\displaystyle |\,{\cal E}({\bf a},\,{\bf b})\,+\,{\cal E}({\bf a},\,{\bf b'}) \,+\,{\cal E}({\bf a'},\,{\bf b})\,-\,{\cal E}({\bf a'},\,{\bf b'})\,|\, \leqslant 2\,, \ \ \ \ \ \ \ (6)$

provided we assume the corresponding measurement functions to be of the incorrect form (2) assumed by Bell. Thus it is in this step—the crucial step in the derivation of Bell-CHSH inequality and the proof of Bell’s theorem—that the assumption of incorrect codomain gets smuggled-in. The four pairs of measurement results occurring in the above expression clearly cannot all occur at the same time. If, however, we conform to the usual assumption of counterfactual definiteness and pretend that they do all occur at the same time at least counterfactually, then we must specify the correct codomain for the functions ${\cal A}$$({\bf a},\,\lambda)$ $=\pm\,1$ and ${\cal B}$$({\bf b},\,\lambda)$ $=\pm\,1$ to represent them correctly. But for the measurement functions of the form (1) with the correct codomain $S^3$ the story turns out to be dramatically different from (6). Because then the measurement results $\pm\,1$ occur as points of an absolutely parallelized 3-sphere, which is the set $S^3$ of unit quaternions. As a result, we must specify the joint probability distribution for the occurrence of the four pairs of measurement results on such a 3-sphere, defined by the graded basis $\{\,1\,,\;{\boldsymbol\beta}_1\,,\;{\boldsymbol\beta}_2\,,\;{\boldsymbol\beta}_3\,\}$. Consequently, the upper bound on the correlations depends on the torsion ${\cal T}_{\,{\bf a\,a'}}$ at each point of the manifold $S^3$ as follows:

$\displaystyle |\,{\cal E}({\bf a},\,{\bf b})\,+\,{\cal E}({\bf a},\,{\bf b'}) \,+\,{\cal E}({\bf a'},\,{\bf b})\,-\,{\cal E}({\bf a'},\,{\bf b'})\,|\, \ \ \ \ \$

$\displaystyle \leqslant\sqrt{\,\lim_{\,n\,\gg\,1}\left[\,\frac{1}{n}\sum_{k\,=\,1}^{n}\, \big\{\,4\,+\,4\,{\cal T}_{\,{\bf a\,a'}}({\lambda}^k)\,{\cal T}_{\,{\bf b'\,b}}({\lambda}^k)\,\big\}\,\right]} \ \ \ \ \ \ \$

$\displaystyle \leqslant\,2\,\sqrt{\,1-({{\bf a}}\times{\bf a'})\cdot({\bf b'}\times{{\bf b}})\,}\, \ \ \ \ \ \ \$

$\displaystyle \leqslant\,2\,\sqrt{2}\,.\, \ \ \ \ \ \ \ \ \ (7)$

This upper bound has been derived in detail in the last appendix of this paper. This derivation has also been numerically verified in a GAviewer by Fred Diether and myself. It is of course exactly what is predicted by quantum mechanics. Thus, with the topologically correct choice (1) for the measurement functions, not only the strong correlations (5) but also the upper bound (7) on the correlations turns out be exactly what is predicted by quantum mechanics.

More importantly, the above derivation reveals a crucial role played by the non-vanishing torsion in our physical space on the strength of the observed correlations. If the physical space in which the measurement events ${\,{{\,\cal A}=\pm\,1\,}\,}$ and ${\,{{\,\cal B}=\pm\,1\,}\,}$ are occurring has vanishing torsion, then the correlation between these events cannot be as strong as those predicted by quantum mechanics. But vanishing torsion is precisely what all derivations of Bell-CHSH inequality unwittingly assume (see, for example, the discussion in this paper). Note also that parallelizability of spaces by torsion is a topological concept. A toy example of how topology can strengthen the observed correlation can be found in the first appendix of this paper.

Needless to say, the above discussion is merely a sketch of my disproof. In particular, I have glossed over the important issue of how errors propagate within a parallelized 3-sphere (a detailed statistical analysis of which can be found in this paper). Moreover, in my book I have shown that, not only the simple EPR-Bohm-type correlations, but ALL quantum correlations can be reproduced purely local-realistically in an analogous manner. For more general cases, however, the codomain of the measurement functions such as ${{{\cal A}({\bf a},\,\lambda)=\pm\,1}}$ must be extended to the absolutely parallelized 7-sphere. Thus quantum correlations in general turn out to be nothing but classical, local, and deterministic correlations among the points of an octonionic 7-sphere. On the next page I discuss how the parallelizing torsion ${\cal T}$ within such a 7-sphere is responsible for ALL quantum correlations.

The argument I have presented above is experimentally testable. I have proposed a macroscopic experiment for this purpose. Further discussion about this experiment can be found on this page.

————–

PS: A theoretical computer scientist, Paul Snively, has crystalized the essence of my argument in a logical sequence that I find quite interesting. According to Snively the logic behind my refutation of Bell’s theorem is:

algebra with operations lacking the closure property $\rightarrow$ mathematical singularities $\rightarrow$ partial functions $\rightarrow$ logical inconsistency.

A brief discussion of what he means by this sequence can be found on this page of his blog.

He summarizes the main point thus: “Dr. Bell used scalar algebra. Scalar algebra isn’t closed over 3D rotation. Algebras that aren’t closed have singularities. Non-closed algebras having singularities are isomorphic to partial functions. Partial functions yield logical inconsistency via the Curry-Howard Isomorphism. So you cannot use a non-closed algebra in a proof, which Dr. Bell unfortunately did. … This is a sufficient disproof of Bell’s theorem.”

Let me add to Paul’s observation that, not only Bell’s original 1964 proof, but *ALL* of the proofs of *ALL* of the Bell-type theorems, including the proofs of their variants and generalizations such as Hardy’s theorem or GHZ theorem, use only scalar algebra. Therefore all such no-go “proofs” are logically inconsistent (i.e., they are nonsense).

————————————————————————————————————

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### 23 Responses to Disproof of Bell’s Theorem

1. Fred Diether says:

Hi Joy,

I was just looking at this video on the Scientific American website about quantum entanglement presented on a blog by George Musser. I have to say that after understanding your model, I could immediately see where the argument for quantum entanglement breaks down in the video. I am wondering if anyone else reading this blog can point out those spots in the video?

I remember on the FQXi blogs that George Musser noted to you that he would like to have a better discussion with you about your model. I think it is only fair that a counter opinion about quantum entanglement should be offered to SciAM readers and you should take George up on his offer.

Best,

Fred

2. Hi Fred,

I watched the video. I must admit I am deeply disappointed and disturbed by how misleadingly the idea of entanglement is presented in the video. With all due respect to George Musser (who I think is a talented young journalist), what he is doing is no different from “spreading the good word” of entanglement. The video is misleading from the start. The two men set out to witness entanglement firsthand in a physics lab. But you can no more witness entanglement of a photon than you can witness phlogiston in a combustible body. All one can ever see in a lab are correlations between the supposedly entangled photons. No one has ever seen, or could ever see, entanglement directly. If we could, then all controversies over the interpretation of quantum mechanics would end instantly.

To be sure, we can prepare various states of the photons as shown in the video—say, a state in which the correlations exhibited by the photons is no better than that produced by coin tosses, or a state in which the correlations exhibited by the photons which can be explained by the concept of entanglement. But the latter is not the only explanation possible for the observed correlations, as we know from the results of this paper. I show in this paper how the observed correlations can be explained without the notion of phlogiston… I mean entanglement. They can be explained simply as classical, deterministic correlations among the points of a parallelized 3-sphere. So it is very disappointing indeed that Scientific American would continue to “spread the good word” of entanglement without even mentioning the much less mystical alternative. The day will come, however, when the idea of entanglement would be no more in vogue than the idea of phlogiston.

Best,

Joy

3. Hi Joy,

I appreciate your remarks and do hope to continue our discussions at some point. But I think you will admit that your views on entanglement are a minority opinion. Of course that doesn’t make them wrong! The history of science shows that minority opinions are often vindicated; all great truths are initially resisted. Nonetheless, Scientific American *has* to give higher weight to prevailing views — otherwise, the editors would be placing their own judgment above that of the majority of scientists, which is sometimes justified, but only in exceptional cases. To put it differently, once you convince your fellow physicists and philosophers of physics of your arguments, then you will have convinced us, too.

George

4. Hi George,

Thanks for dropping by and responding to my comments. I do agree that my views on entanglement are a minority opinion at the moment, and I do appreciate why Scientific American has to give higher weight to the prevailing views. But it seems to me that your video linked above gives a wrong impression of entanglement even from the prevailing point of view. It is understood by most experts (say, by Lucien Hardy, for example) that all we can see in the lab are correlations, not entanglement itself. The latter concept provides a good explanation for the observed correlations, but that is a different matter.

But again, I appreciate that you have a short time at your disposal in which you have to explain the essence of the idea to lay viewers, without getting into technicalities.

As for my own views on entanglement, I do hope that someday my arguments are better understood by my fellow physicists and philosophers of physics. In any event—leaving your journalist hat aside—please feel free to continue the discussion we started during the last FQXi conference.

Best,

Joy

5. I hope our world lines will intersect soon, so that we can continue our chat. Please let me know if you have plans to visit NYC.

George

6. T H Ray says:

Hi Joy,

I’d like to expand on the phlogiston example you used. Shorly before Lavoisier showed that combustion is caused by rapid oxidation, there were theoretically two kinds of phlogiston, deduced from the evidence of combustion — “positive” phlogiston in the form of rust, which apparently added matter to a substance such as iron, and “negative” phlogiston which apparently subtracted matter from a substance such as wood, which was reduced to ashes.

These conclusions are of course contradictory. The contradiction is of the same form [as certain mistaken claims about] your framework. Just as “phlogiston theorists” were mistaken about the true nature of combustion, so are these [claims] about the constructive nature of a continuous measurement process that obviates the identification of discrete outcomes with physically real, nonlocal results.

Best,
Tom

7. Hi Tom,

Thank you for your comments. I did not know that there were two kinds of phlogiston.

I am terribly sorry, however, because I had to edit out parts of your message. Names of certain individuals whom I cannot respect are not allowed on this blog. I hope you will understand. I have tried to preserve the integrity of your message as much as possible.

As for the mistaken claims of those individuals you mention, it is important to note that equations (1.22) to (1.26) on page 10 of my book, as well as similar set of equations in this paper, have been explicitly verified (in great detail) by Lucien Hardy, Manfried Faber, and several other high profile and exceptionally competent physicists and mathematicians around the world. In fact, any competent reader with only basic skills in mathematics should be able to reproduce equations (1.22) to (1.26) of my book rather effortlessly.

It is also worth noting that all of the so-called arguments against my disproof to date are based on an elementary logical fallacy—the Straw-man Fallacy. What they do is replace my argument (or model), say X, with a grossly distorted misrepresentation of it, say Y, and then pretend—by refuting their own distortion Y (by resorting to deliberate dishonesty or out of sheer incompetence)—that they have undermined my actual argument (or model) X. Such a dishonest strategy defies reason at its very core (for more details, see this paper).

Unlike Bell himself, some of the followers of Bell are naïve, uninformed, and dishonest.

Best,

Joy

8. T H Ray says:

Hi Joy,

The presence of positive and negative phlogiston is implied in the link you gave. The author just failed to point out the contradiction between two *physical* results from a theory that can accommodate only a singular physical outcome. If phlogiston is assumed to be physically real, it must be causal, and if causal, attributable to only one result. It doesn’t do to say that in the one case phlogiston causes rust and in the other case, causes ashes. If rust and ashes have a common cause, it can’t be phlogiston.

By the same token, non-vanishing topological torsion analogous to the rapid oxidation that causes both rust and ashes, informs us that continuous functions imply classical orientation entanglement — (and not the entanglement of real physical quantities whether we call the phenomena phlogiston or quantum entanglement) — that guarantees a unitary cause.

In other words, if one assumes phlogiston, an experimental outcome of + 1 (rust) or – 1 (ashes) explains nothing. Both (contradictory) outcomes can’t result from the same physical cause. Your topological model, OTOH, has the same attributes as chemical oxidation in principle: a continuous process drives the measurement result from a deeper underlying physical cause of nondegenerate torsion.

Einstein faced the same problem in explaining how spacetime is physically real (” … independent in its properties, having a physical effect but not influenced by physical conditions…”) because one does not think of continuous phenomena as physically real. One intuitively wants to ascribe “reality” to the measurement apparatus rather than the measurement process.

Your critics are not subtle thinkers. They allow that 2 and 3 dimensional simulations are adequate for n-dimension phenomena. Thus, they reject *any* topological model out of hand; global results, as topology implies, have to be nonlocal by definition. I think that until one can grasp the analytical nature of your framework, it’s an uphill battle. Einstein had the advantage of large scale validation for continuous spacetime. Small scale classical experiments are tougher, less forgiving. Just as one has to accept a 4-dimension physical spacetime continuum for general relativity to make sense, one has to accept a parallelized physical 7-sphere for your framework to make sense. And just as Minkowski space-time requires a great deal of mathematical background to grasp, your framework requires that much more.

Best,

Tom

9. Hi Tom,

You make some very good points. Sadly, as you put it, my “critics are not subtle thinkers”, to say the least. Moreover, dogmas, even within science, are not easy to overcome, as we have witnessed over the past few years. But thank you for your thoughtful remarks, especially concerning the phlogiston analogy.

Best,

Joy

10. igael says:

Hello

I have on my website 2 formulae that work with good performance at the degree resolution level. There is an online and checkable javascript implementation for the ones that haven’t a tool to emulate probability functions and to compute correlations.

11. Hello Igael,

What you have is a nice demonstration of well known facts. It is well known (since 1970’s) that by reducing the efficiency of detection, to, say, 86%, one can reproduce the cosine correlation. Such a demonstration does not contradict Bell’s argument, and it does not have fundamental significance for physics.

Best,

Joy

12. Igael says:

Hi

Please note that I’m working on new variants with 89% pairs ( then 94% of arm efficiency ) and probably more. I use 80 servers , raising to ‘powers’ above 3.000.000.000 without Markov takes a lot of resources

But, you are true , even if I haden’t yet seen theses formulae on sites or books.
It is not so clear when you read QM curses or books on EPR to understand that the debate is not closed. How to start developping applications in such conditions ?

Please, let me know where I could find the formulae known since 70 ? I found only the Maudlin scheme B ( it seems to my algo 9 without Markov and has a limit around 66.6 % pairs and not 63.6 as I read in some articles ) ?

This is an amazing topic !
Thank you.
Igael

13. Igael,

I had this paper in mind: Hidden-Variable Example Based upon Data Rejection, by P. Pearle, Physical Review D 2, 1418–25 (1970).

Best,

Joy

14. Igael says:

Hi Joy

Many thanks.

15. Olav Helgeland (emeritus professor Materials Science) says:

Hi Joy.

BLOG: (revise or cut as you wish)

Eureca!

Checking ‘Internet’ it appears that you have been through some nasty ‘entanglements’ and have passed it in ‘local reality’, free and well, – with honour.
Good luck for your further work!

engaged Emeritus.no

PRIVATE:

If you check your e-mail box for 03.03.2014, Re: Bell’s theorem, with two enclosed items, you will find the background for my contact.
I would appreciate v.m. a reply and a comment to my paper. ( I can repeat the e-mail if you have lost it).

w.f.g. Olav

16. KL Rajpal says:

Linear Polarization http://vixra.org/pdf/1303.0174v5.pdf
Einstein was right when he did not agree with the EPR experiment conclusions and had said, “spooky action at a distance” cannot occur and that, “God does not play dice”.

Electron Spin
http://vixra.org/pdf/1306.0141v3.pdf

17. John R. Cox says:

Dr. Christian,
I would imagine Tom Ray might keep in touch with you, so you might know that the current FQXi essay contest has an entry by Edwin Eugene Klingman in which he challenges constraints inherent in Bell’s formulation, which produce the same distribution curve against the Bell straight-line eigenvector that your topological challenge results in. Not that I blame either you or Tom for not continuing to subject yourselves to the abuse of the blogosphere, I think it wise. I recall warning R. Gill that his posturing in Vienna made him a perfect fall guy for professional spooks. I had wondered if he ever made it back from his ‘sabatical’ to the Baltic, hope you actually collected the 10K Euros. Kudos, jrc

18. Brian Rosenblum says:

Dr Christian:

You are a diseased little turd swirling around the toilet drain en route to an obscurity even deeper than the one currently defining you.

Sincerely

19. Thank you, Mr. Rosenblum.

If your prophesy is right, then there is no reason for you to be concerned.

20. Martin says:

Dr. Christian,

As I understand (interpret) your theoretic conjectures, the problem arises from the interpretation of what a particle represents: QT versus your description – point vs 3D object (sphere), wave of probability with instantaneous communication between points in ST versus local ST with light speed limitation. I have always wondered how points in QT (Copenhagen Interpretation) can possess complex properties and give rise to complex interactions? QT begs for existence of hidden variables.
Do waves of probability also govern virtual particles? And how could we unite the Copenhagen interpretation with your view?
The answer could very well lay in an existence of a real De Broglie wave made of two sets of virtual photons (intersecting at opposite directions and at the speed of light) which express their circular polarization pattern only at a “point” at which the particle (the Higgs expression?) is located. Those photons would cancel to randomness in all other positions along their path of propagation, and shifting their frequency would lead to particle speed change. In this view matter would be a stable interference pattern. Particles would form as rain drops form in clouds and would exist intermittently pulsating to the beat frequency of the waves that build them.
So, does this expression fit yours?
Also you could have a look at the work of Dr. Yves Couder: https://www.youtube.com/watch?v=W9yWv5dqSKk

Cheers!

21. We have finally launched the long-awaited website for the Einstein Centre for Local-Realistic Physics: http://einstein-physics.org/.

22. I have posted a simplified version of my proposed macroscopic experiment to test Bell’s theorem here: http://libertesphilosophica.info/blog/wp-content/uploads/2016/05/PropExp1.pdf.


text:
Problem: Write the products for each of the following acid-base reactions.a) These reactions will be in equilibrium, and the equilibrium can favor one side or the other or be balanced (i.e. nearly equal). Indicate which side of the reaction is favored in the equilibrium either by use of arrows ⇌ or writing on one side of the reaction, “favored”.b) Draw the curved arrows to show the flow of e - for each of these reactions.

91% (39 ratings)
Problem Details

Write the products for each of the following acid-base reactions.

a) These reactions will be in equilibrium, and the equilibrium can favor one side or the other or be balanced (i.e. nearly equal). Indicate which side of the reaction is favored in the equilibrium either by use of arrows ⇌ or writing on one side of the reaction, “favored”.

b) Draw the curved arrows to show the flow of e for each of these reactions.

format: plain


text:
# Math Help - Relative Maximum/minimum help!

1. ## Relative Maximum/minimum help!

I know how to do A and B but can some one help me on C please.

2. Originally Posted by lilwayne
I know how to do A and B but can some one help me on C please.
what do the first and second derivative tests for extrema tell you?

3. Solve $f'(x) = 0, g'(x)=0, h'(x)=0$

4. Originally Posted by lilwayne
I know how to do A and B but can some one help me on C please.
Dear lilwayne,

Take the first derivative of f(x) and consider its behavior in the regions x<0 and x>0.

If $\frac{df(x)}{dx}<0~ when ~x<0 ~and ~\frac{df(x)}{dx}>0~ when~ x>0$ then there is a relative minimum at x=0.

If $\frac{df(x)}{dx}>0 ~when~ x<0 ~and~ \frac{df(x)}{dx}<0 ~when ~x>0$ then there is a relative maximum at x=0.

If the derivative is positive or negative in both regions then x=0 is neither a ralative maximum nor a relative minimum.

Hope you understood.

5. Thanks a lot!


text:
102 views

A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at $25$ m intervals in this plot.

Which of the following is the steepest path leaving from $P$?

1. $\text{P to Q}$
2. $\text{P to R}$
3. $\text{P to S}$
4. $\text{P to T}$

edited | 102 views
Migrated from GO Electronics 5 months ago by Arjun

+1 vote

From the given contour plot of a geographical region, we get

• Height of point P > 575 m
• Height of point R < 425 m
• Height of point Q is in between 475 m to 500 m
• Height of point S is in between 500 m to 525 m
• Height of point T is in between 500 m to 525 m

among PQ, PR, PS and PT the one with largest height difference will be the steepest path.

since R is at lowest height , so path from P to R will be steepest path.

Correct Answer: B. $P$ to $R$

by Boss (15.1k points)
selected by
0

For the concept watch this video


text:
# For people with problems in my hidden links!

And that is just about it! Just click that sentence and it works!

Same for all the other problems as well...

Note by Frisk Dreemurr
3 months, 1 week ago

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- 3 months, 1 week ago

Well, it is a different type of link...

It is black, and it is in $\LaTeX$ as well @Páll Márton

- 3 months, 1 week ago

- 3 months, 1 week ago

Yes I can!, And it is quite cute!

- 3 months, 1 week ago

So you use a mobile. You can delete the en.m. section :)

- 3 months, 1 week ago

No I can't find it........is it under my bed...................Oh, its the text.............wait, no its a link...............to the wikipedia page on swiss cheese...........................Swiss Cheese?.........................I'm outta here........@Páll Márton.........................@Frisk Dreemurr...........................

- 2 months ago

@Percy Jackson, LOL

Though I changed my name, the gif. hasn't changed yet :)

- 1 week, 1 day ago


text:
### Institut de Recherche en Informatique Fondamentale

 L'IRIF est une unité mixe de recherche (UMR 8243) du CNRS et de l'Université Paris-Diderot, issue de la fusion des deux UMR LIAFA et PPS au 1er janvier 2016. Ses objectifs scientifiques se déclinent selon trois grandes thématiques au cœur de l'informatique : les fondements mathématiques de l’informatique ; les modèles de calcul et de preuves ; la modélisation, les algorithmes et la conception de systèmes.

#### Prochains séminaires

Lundi 29 août 2016 · 11h00 · Room 2002

Algorithmes et complexité · Sanjeev Khanna (University of Pennsylvania) · On the Single-Pass Streaming Complexity of the Set Cover Problem

In the set cover problem, we are given a collection of $m$ subsets over a universe of $n$ elements, and the goal is to find a sub-collection of sets whose union covers the universe. The set cover problem is a fundamental optimization problem with many applications in computer science and related disciplines. In this talk, we investigate the set cover problem in the streaming model of computation whereby the sets are presented one by one in a stream, and the goal is to solve the set cover problem using a space-efficient algorithm.

We show that to compute an $\alpha$-approximate set cover (for any $\alpha= o(\sqrt{n})$) via a single-pass streaming algorithm, $\Theta(mn/\alpha)$ space is both necessary and sufficient (up to an $O(\log{n})$ factor). We further study the problem of estimating the size of a minimum set cover (as opposed to finding the actual sets), and show that this turns out to be a distinctly easier problem. Specifically, we prove that $\Theta(mn/\alpha^2)$ space is both sufficient and necessary (up to logarithmic factors) for estimating the size of a minimum set cover to within a factor of $\alpha$. Our algorithm in fact works for the more general problem of estimating the optimal value of a covering integer program. These results provide a tight resolution of the space-approximation tradeoff for single-pass streaming algorithms for the set cover problem.

This is joint work with my students Sepehr Assadi and Yang Li.

#### Événements

Mercredi 7 – Vendredi 9 septembre 2016 · Institut Henri Poincaré (IHP)

Dimanche 15 - Vendredi 20 Janvier 2017 · Jussieu


text:
# Geometric perspective of the vector potential

1. Jun 9, 2012

### PerpStudent

I'm struggling with trying to visualize the vector potential as in the identity:

B = ∇⨯A

For starters, how does A relate to, say, a uniform magnetic field, which is quite easy to visualize. Then, how about the magnetic field around a bar magnet -- where is A?
Any help would be appreciated.

2. Jun 9, 2012

### Muphrid

The formula should look similiar to something you're already familiar with:

$$\mu_0 j = \nabla \times B$$

If you can imagine the magnetic field that goes with a specific current density, then the same picture applies to the vector potential that goes with the magnetic field.

3. Jun 10, 2012

### vanhees71

It's not easy to visualize $\vec{A}$ or give it a physical meaning since it is a gauge-dependent quantity. What's physical is the magnetic field, $\vec{B}$ which is given by

$$\vec{B}=\vec{\nabla} \times \vec{A}.$$

For a constant field, it's easy to get the vector potential. So let's consider

$$\vec{B}=B_0 \vec{e}_z.$$

You have quite some freedom to choose the vector potential. You can take always one constraint since it is only defined from $\vec{B}$ up to the gradient of a scalar field. Here, I'd choose the spatial axial gauge

$$A_z=0.$$

Then you have

$$\vec{B}=\begin{pmatrix} 0 \\ 0 \\ B_0 \end{pmatrix} = \vec{\nabla} \times \vec{A}=\begin{pmatrix} -\partial_z A_y \\ \partial_z A_x \\ \partial_x A_y-\partial_y A_x \end{pmatrix}.$$

Obviously our constraint doesn't fix the solutions completely, and you have some more freedom. You can, e.g., set $A_x=0$ and $A_y=B_0 x$. Then you have

$$\vec{A} = B_0 x \vec{e}_y.$$

Around a bar magnet you have to solve the magnetostatic Maxwell equations for a given magnetization of your bar. You find some calculations about this in Sommerfeld's Lectures on Theoretical Physics, vol. III.


text:
# Finding the Center of Percussion

1. Mar 3, 2013

### bd2015

1. The problem statement, all variables and given/known data
A rod of mass M and length L rests on a frictionless table and is pivoted on a frictionless nail at one end. A blob of putty of mass m approaches with velocity v from the left and strikes the rod a distance d rom the end as shown, sticking to the rod.

1. Find the angular velocity of the system about the nail after the collision (already done)
2. Is there a value of d for which linear momentum is conserved? If there were such a value, it would be called the center of percussion for the rod for this sort of collision.

2. Relevant equations

ω final = $\frac{mvd}{(1/3)ML^2 + md^2}$

3. The attempt at a solution
So, I know that linear momentum most be conserved. I don't know how to find the linear momentum of the rod

2. Mar 3, 2013

### TSny

The linear momentum of any object can be calculated as the total mass of the object multiplied by the linear velocity of the center of mass of the object.

3. Mar 3, 2013

### bd2015

So then do I just need to set P initial to P final, and solve for d, without having to worry about angular momentum?

4. Mar 3, 2013

### TSny

Essentially, yes. But you'll need to use your result from part 1 which came from conservation of angular momentum. Also, don't forget to include the final linear momentum of the putty.


text:
# 3d-diffusion equation in spherical coordinates (numerical), boundary problem

There is one boundary problem

$$\frac{\partial u}{\partial t}= \operatorname{div}\left(a^2 E \nabla u\left(r,\varphi,\psi \right) \right)$$ in a ball $$B_{1}(0)=\left\{x \in \mathbb{R^3}: \left\| x \right\| \le 1 \right\}$$

(boundary and initial values are now not principal). It's obvious that this equation can be expanded to: \begin{gather} \frac{\partial u}{\partial t}=a^2\left(\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^2 \frac{\partial u}{\partial r}\right)+\frac{1}{r\sin\varphi}\frac{\partial }{\partial \varphi}\left(\frac{\sin\varphi}{r} \frac{\partial u}{\partial \varphi} \right)+\frac{1}{r\sin\varphi}\frac{\partial}{\partial \psi}\left(\frac{1}{r\sin\varphi}\frac{\partial u}{\partial \psi} \right)\right).\end{gather} This boundary problem is solved numerically. So, we write an implicit finite-difference scheme

$$\frac{u^{n+1}_{i,j,k}-u^{n}_{i,j,k}}{\Delta t}=a^2 \left( \frac{2}{r_{1}+(i-1)\Delta r}\frac{u^{n+1}_{i+1,j,k}-u^{n+1}_{i-1,j,k}}{2\Delta r}+\frac{u^{n+1}_{i+1,j,k}-2u^{n+1}_{i,j,k}+u^{n+1}_{i-1,j,k}}{\Delta r^2}\right) +$$ $$+a^2\left(\frac{\cot \left(\varphi_{1}+(j-1)\Delta \varphi \right) ) }{\left(r_{1}+(i-1)\Delta r\right)^2}\frac{u^{n+1}_{i,j+1,k}-u^{n+1}_{i,j-1,k}}{2\Delta \varphi}+\frac{1}{\left(r_{1}+(i-1)\Delta r\right)^2}\frac{u^{n+1}_{i,j+1,k}-2u^{n+1}_{i,j,k}+u^{n+1}_{i,j-1,k}}{\Delta \varphi^{2}} \right) +$$ $$+a^2\left(\frac{1}{\left(r_{1}+(i-1)\Delta r\right)^2 \sin^2\left(\varphi_{1}+(j-1)\Delta \varphi \right)}\frac{u^{n+1}_{i,j,k+1}-2u^{n+1}_{i,j,k}+u^{n+1}_{i,j,k-1}}{\Delta \psi^2} \right).$$ The main problem appears in realization of this scheme. The values $r_{1},\varphi_{1}$ -- can be equal to zero in some moments of evaluation ,that's why we face with singularities during evaluation. Could you help me, what should I do with this scheme (perhaps, change it in the moments with singularities) in order to avoid singularities during evaluation?

• Depending on the boundary conditions, you may be able to identify $i = 0$ with $i = N-1$. What are the boundary conditions? Also, can you explain to me where the $i-1$ terms in the denominator come from in the discretization? – Eric Kightley Jul 23 '13 at 10:28
• The boundary conditions are Dirichlet conditions: $u(\partial D)=0$. They are defined on $r_{n}$'s layer -- on the surface of $B_{1}(0)$. $i−1$ comes from the next fact: there is a summand 1r...∂∂.. in the equation, in the process of discretization we replace r by the value $r_1+(i−1)\Delta r$, where i -- index of the position in a ball (coordinate). When $i=1$, $r=r_{1}=0$ - here we are in the center of the ball. In other cases the same situation occurs with $φ_{1}=0$ – cool Jul 29 '13 at 18:57


text:
# How to setup a MCP79410 real time clock?

I was trying to set up a MCP79410 real time clock (RTC) and had to search quite a bit until I found a way to integrate the RTC's time signal into my Raspbian Jessie:

Apparently, even there, some things changed over the last updates of Raspbian and there does not seem to be a huge crowd using this particular chip (as opposed to, e.g. the DS1307).

I specifically wanted to connect a MCP79410 already hooked up to an ABS07 quartz crystal and a HU2032-LF battery to my Raspbbery Pi Model 3B running Raspbian Jessie 4.4.38-V7+ #938. The implementation should be as easy as possible and simply provide me with a correct system time even if the RPi is disconnected from the internet.

In order to set up a MCP79410 real time clock (RTC, see manual), one needs to start by connecting it to one's RPi and a quartz crystal. The following is supposed to be a rough sketch on how to connect the pins to point one in the right direction - in practice, one might need capacitors to smooth signals or resitors to pull-up the SCL- and SDA-interface, depending for example on the battery and quartz crystal used.

``````X1: Connect to external 32.768 kHz quartz crystal
X2: Connect to external 32.768 kHz quartz crystal
VBAT: Connect to external battery
VSS: Connect to GND (e.g. RPi-Pin 9)
VCC: Connect to RPi-Pin 1 (3.3V)
MFP: -
SCL: Connect to RPi-Pin 5 (GPIO3_SCL1)
SDA: Connect to RPi-Pin 3 (GPIO3_SDA1)
``````

After connecting the hardware components, one needs to set up a communication inferace between the MCP79410 and the RPi via I2C (can be activated via sudo raspi-config and therein "7 Advanced Options" -> "A7 I2C" -> "Yes" ). To do so, one should check their availability first:

`i2cdetect -y 1` or, for older RPis, `i2cdetect -y 0` should return

``````     0  1  2  3  4  5  6  7  8  9  a  b  c  d  e  f
00:          -- -- -- -- -- -- -- -- -- -- -- -- --
10: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
20: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
30: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
40: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
50: -- -- -- -- -- -- -- 57 -- -- -- -- -- -- -- --
60: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- 6f
70: -- -- -- -- -- -- -- --
``````

This means: At 0x57 and 0x6f, the MCP79410 is listening for I2C communication. 0x57 leads to the EEPROM, a status register and a unique ID register whereas 0x6f is the address that leads to the RTC itself.

In order to establish a connection, the line `rtc-mcp7941x` should be added to `/etc/modules` and the line `dtoverlay=i2c-rtc,mcp7941x` to `/boot/config.txt`.

After a reboot, the `i2cdetect`-command from above should return

``````     0  1  2  3  4  5  6  7  8  9  a  b  c  d  e  f
00:          -- -- -- -- -- -- -- -- -- -- -- -- --
10: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
20: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
30: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
40: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
50: -- -- -- -- -- -- -- 57 -- -- -- -- -- -- -- --
60: -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- UU
70: -- -- -- -- -- -- -- --
``````

In order to use the newly established connection to provide one's RPi with a time signal, one needs to comment out the lines

``````if [ -e /run/systemd/system ] ; then
exit 0
fi
``````

from `/lib/udev/hwclock-set` .

Before the next step, one should make sure, one's system time is accurate (e.g. by comparing it to a radio controlled clock nearby).

Now, one sends one's current system time to the RTC using the command

`````` hwclock -w
``````

From now on, the RTC will keep the time and resynchronize the RPi's system time automatically on startup or manually by entering

``````hwclock -r
``````

I see these instructions and the lack of seeing your wiring nor the output of your i2cdetect query. This means that you have the correct instructions to make it work. Answer: Simply follow them.

Let me point out that this clock is to be set by the Pi in the first place. That implies that the Pi is connected to the Internet and gets a NTP-based time and date setting periodically.

Then you must do extra work to keep this clock synchronized to the Pi.

So I don't see any use-case for using this module in the first place.

My suggestion is that you just let the Pi keep its own time.

• Well, yes, myinstructions work - that's why I put them here. To make it easier for others ;-) – Fantilein1990 May 19 '17 at 15:38
• Regarding the purpose of the clock: My Pi had a very unreliable internet connection. It was rebooted every now and then and I needed it to keep the time very accurately, even if it didn't have an internet connection upon reboot. In this case, a Pi can run into trouble keeping the time, so I added a RTC and a small battery feeding the RTC. Problem solved. Is this an acceptable use-case? :-) – Fantilein1990 May 19 '17 at 15:43


text:
MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

7 added 149 characters in body

I think that the best way to capture the idea beyond the proof of the fixed point theorem is to mirror it in an ordinary language formulation and then translate it back to the first order language of arithmetic (cf. J.N. Findlay, Goedelian Sentences: A Non-numerical Approach}, Mind, Vol. 51, 1942, pp. 259-65.). Clearly, what we seek is a sentence asserting that it has a given property, that is, a sentence that says "I have the property p". But, in order for it to be formalizable, our sentence should consist of components with easily identifiable formal first-order counterparts. Therefore we cannot use such indexicals as I'.
In order to circumvent the need for indexicals, we reformulate Grelling's paradox applying it to open sentences instead of adjectives:

(1) "x is heterological" is heterological,

where an open sentence is called autological if the property it attributes to x possessed by the sentence itself, otherwise it is called heterological. For example, "x consists of five words", "x is English", are autological, while "x is long", "x is German", are heterological. On the other hand, both in formal languages and in informal ones, the fact that an object has a property is expressed by a substitution of the name of the object into the open sentence expressing that property. Consequently,

(2) x is heterological just in case the sentence obtained by substituting the name of x for the variable in it is false.

Now (using the convention that he name of linguistic objects are the object itself between quotation marks), if we replace "being false" by "having property p", (1) and (2) together yield:

(3) the sentence obtained by substituting the name of "the sentence obtained by substituting the name of x for the variable in it has property p" for the variable in it has property p.

It can directly be checked that this

This is the sentence we need. On the one hand, it does not use indexicals, on the other, it indeed says of itself that it has property p (and says nothing else), since it is built up in such a way that if we perform the substitution described in it, then we obtain the sentence itself, which is stated to have property p.

Now, let s denote the open sentence between the quotation marks in (3), that is, let s be:

(4) the sentence obtained by substituting the name of x for the variable in it has property p.

Then, clearly, the whole sentence (3) is s("s"). That isIn order to obtain the fix point lemma, we should translate it into the language of formal arithmetic. Clearly, the formalization process should consist consists of two main steps. In the first step, we have to find the formal version $\eta$ of s, and then the second step is obvious: the desired sentence $\lambda$ will simply be $\eta(g(\eta))$ (where $g(\varphi)$ is the G\"odel number of $\varphi$ and plays, of course, the formal counterpart of name of $\varphi$, and, for simplicity, I leave out of consideration the difference between numbers and their formal counterparts in the language).

Now, that is all. That is the essence of the proof. What remains to do is simply translate the ordinary language argument into the formal language of arithmetic. That is a completely mechanical task.

Let us recall that what we should show is that, for any arithmetical formula $\varphi$ with at most one free variable (this fact will be denoted by $\varphi=\varphi(x)$), there is a sentence $\lambda$ such that

$Q\vdash \lambda \longleftrightarrow \varphi(g(\lambda)),$

where $Q$ is Robinson arithmetic (essentially Peano arithmetic without induction).

Now, let the formula corresponding to the property p be $\varphi=\varphi(x)$. Then, obviously, the formal version of s is $\varphi(g[x(g(x)])$. In order to continue the formalization process, we should find a formula that can play the role of $\varphi(g[x(g(x))])$, that is, a formula $\eta=\eta(x)$ such that $\eta(g(\psi))$ is provably equivalent to $\varphi(g[\psi(g(\psi))])$ for every $\psi=\psi(x)$, or equivalently (denoting the inverse of $g$ by $g^{-1}$), for any $n \in N$,

$Q\vdash \eta(n)\longleftrightarrow\varphi(g[g^{-1}(n)(n)]).$

In order to find the appropriate formula $\eta$, let us consider the expression substituted into the formula $\varphi$, and define the function $f:\omega\longrightarrow \omega$ accordingly:

$f(n)=g[g^{-1}(n)(n)]$ if $n \in N$ and $f(n)=0$ otherwise.

Since this function is obviously recursive and hence representable, and, up to provable equivalence, the result of substituting a representable function into a formula can also be expressed by a formula, there is a formula $\eta$ such that, for any $n\in N$,

(5) $Q\vdash\eta(n)\longleftrightarrow\varphi(f(n))$

Thus we have obtained what we need, we have shown that there exists an $\eta$ that can be considered to be the formal version of s. Now, all that remains to do is straightforward: it follows from (5) that, for every $\psi$,

$Q\vdash \eta(g(\psi))\longleftrightarrow \varphi\big(g[\psi(g(\psi))])$,

which, in turn, choosing $\psi$ to be $\eta$, yields

$Q\vdash \eta(g(\eta))\longleftrightarrow \varphi(g[\eta(g(\eta))])$,

showing that the sentence $\lambda =\eta(g(\eta))$ indeed has the desired property.

6 deleted 13 characters in body

I think that the best way to capture the idea beyond the proof of the fixed point theorem is to mirror it in an ordinary language formulation and then translate it back to the first order language of arithmetic (cf. J.N. Findlay, Goedelian Sentences: A Non-numerical Approach}, Mind, Vol. 51, 1942, pp. 259-65.). Clearly, what we seek is a sentence asserting that it has a given property, that is, a sentence that says "I have the property p". But, in order for it to be formalizable, our sentence should consist of components with easily identifiable formal first-order counterparts. Therefore we cannot use such indexicals as I'.
In order to circumvent the need for indexicals, we reformulate Grelling's paradox applying it to open sentences instead of adjectives:

(1) "x is heterological" is heterological,

where an open sentence is called autological if the property it attributes to x possessed by the sentence itself, otherwise it is called heterological. For example, "x consists of five words", "x is English", are autological, while "x is long", "x is German", are heterological. On the other hand, both in formal languages and in informal ones, the fact that an object has a property is expressed by a substitution of the name of the object into the open sentence expressing that property. Consequently,

(2) x is heterological just in case the sentence obtained by substituting the name of x for the variable in it is false.

Now (using the convention that he name of linguistic objects are the object itself between quotation marks), if we replace "being false" by "having property p", (1) and (2) together yield:

(3) the sentence obtained by substituting the name of "the sentence obtained by substituting the name of x for the variable in it has property p" for the variable in it has property p.

It can directly be checked that this sentence indeed says of itself that it has property p (and says nothing else), since it is built up in such a way that if we perform the substitution described in it, then we obtain the sentence itself, which is stated to have property p. Now, let s denote the open sentence between the quotation marks in (3), that is, let s be:

(4) the sentence obtained by substituting the name of x for the variable in it has property p.

Then, clearly, the whole sentence (3) is s("s"). That is, the formalization process should consist of two main steps. In the first step, we have to find the formal version $\eta$ of s, and then the second step is obvious: the desired sentence $\lambda$ will simply be $\eta(g(\eta))$ (where $g(\varphi)$ is the G\"odel number of $\varphi$ and plays, of course, the formal counterpart of name of $\varphi$, and, for simplicity, I leave out of consideration the difference between numbers and their formal counterparts in the language).

Now, that is all. That is the essence of the proof. What remains to do is simply translate , step by step, the ordinary language argument into the formal language of arithmetic. That is a completely mechanical task.

Let us recall that what we should show is that, for any arithmetical formula $\varphi$ with at most one free variable (this fact will be denoted by $\varphi=\varphi(x)$), there is a sentence $\lambda$ such that

$Q\vdash \lambda \longleftrightarrow \varphi(g(\lambda)),$

where $Q$ is Robinson arithmetic (essentially Peano arithmetic without induction).

Now, let the formula corresponding to the property p be $\varphi=\varphi(x)$. Then, obviously, the formal version of s is $\varphi(g[x(g(x)])$. In order to continue the formalization process, we should find a formula that can play the role of $\varphi(g[x(g(x))])$, that is, a formula $\eta=\eta(x)$ such that $\eta(g(\psi))$ is provably equivalent to $\varphi(g[\psi(g(\psi))])$ for every $\psi=\psi(x)$, or equivalently (denoting the inverse of $g$ by $g^{-1}$), for any $n \in N$,

$Q\vdash \eta(n)\longleftrightarrow\varphi(g[g^{-1}(n)(n)]).$

In order to find the appropriate formula $\eta$, let us consider the expression substituted into the formula $\varphi$, and define the function $f:\omega\longrightarrow \omega$ accordingly:

$f(n)=g[g^{-1}(n)(n)]$ if $n \in N$ and $f(n)=0$ otherwise.

Since this function is obviously recursive and hence representable, and, up to provable equivalence, the result of substituting a representable function into a formula can also be expressed by a formula, there is a formula $\eta$ such that, for any $n\in N$,

(5) $Q\vdash\eta(n)\longleftrightarrow\varphi(f(n))$

Thus we have obtained what we need, we have shown that there exists an $\eta$ that can be considered to be the formal version of s. Now, all that remains to do is straightforward: it follows from (5) that, for every $\psi$,

$Q\vdash \eta(g(\psi))\longleftrightarrow \varphi\big(g[\psi(g(\psi))])$,

which, in turn, choosing $\psi$ to be $\eta$, yields

$Q\vdash \eta(g(\eta))\longleftrightarrow \varphi(g[\eta(g(\eta))])$,

showing that the sentence $\lambda =\eta(g(\eta))$ indeed has the desired property.

5 added 8 characters in body

I think that the best way to capture the idea beyond the proof of the fixed point theorem is to mirror it in an ordinary language formulation and then translate it back to the first order language of arithmetic (cf. J.N. Findlay, Goedelian Sentences: A Non-numerical Approach}, Mind, Vol. 51, 1942, pp. 259-65.). Clearly, what we seek is a sentence asserting that it has a given property, that is, a sentence that says "I have the property p". But, in order for it to be formalizable, our sentence should consist of components with easily identifiable formal first-order counterparts. Therefore we cannot use such indexicals as `I'.
In order to circumvent the need for indexicals, we reformulate Grelling's paradox applying it to open sentences instead of adjectives:

(1) "x is heterological" is heterological,

where an open sentence is called autological if the property it attributes to x possessed by the sentence itself, otherwise it is called heterological. For example, "x consists of five words", "x is English", are autological, while "x is long", "x is German", are heterological. On the other hand, both in formal languages and in informal ones, the fact that an object has a property is expressed by a substitution of the name of the object into the open sentence expressing that property. Consequently,

(2) x is heterological just in case the sentence obtained by substituting the name of x for the variable in it is false.

Now (using the convention that he name of linguistic objects are the object itself between quotation marks), if we replace "being false" by "having property p", (1) and (2) together yield:

(3) the sentence obtained by substituting the name of "the sentence obtained by substituting the name of x for the variable in it has property p" for the variable in it has property p.

It can directly be checked that this sentence indeed says of itself that it has property p (and says nothing else), since it is built up in such a way that if we perform the substitution described in it, then we obtain the sentence itself, which is stated to have property p. Now, let s denote the open sentence between the quotation marks in (3), that is, let s be:

(4) the sentence obtained by substituting the name of x for the variable in it has property p.

Then, clearly, the whole sentence (3) is s("s"). That is, the formalization process should consist of two steps. In the first step we have to find the formal version $\eta$ of s, and then the second step is obvious: the desired sentence $\lambda$ will simply be $\eta(g(\eta))$ (where $g(\varphi)$ is the G\"odel number of $\varphi$ and plays, of course, the formal counterpart of name of $\varphi$, and, for simplicity, I leave out of consideration the difference between numbers and their formal counterparts in the language).

Now, that is all. That is the essence of the proof. What remains to do is simply translate, step by step, the ordinary language argument into the formal language of arithmetic. That is a completely mechanical task.

Let us recall that what we should show is that, for any arithmetical formula $\varphi$ with at most one free variable (this fact will be denoted by $\varphi=\varphi(x)$), there is a sentence $\lambda$ such that

$Q\vdash \lambda \longleftrightarrow \varphi(g(\lambda)).$

Let varphi(g(\lambda)),$where$Q$be is Robinson arithmetic (essentially Peano arithmetic without induction) and induction). Now, let the formula corresponding to the property p be$\varphi=\varphi(x)$. Then, obviously, the formal version of s is$\varphi(g[x(g(x)])$. In order to continue the formalization process, we should find a formula that can play the role of$\varphi(g[x(g(x))])$, that is, a formula$\eta=\eta(x)$such that$\eta(g(\psi))$is provably equivalent to$\varphi(g[\psi(g(\psi))])$for every$\psi=\psi(x)$, or equivalently (denoting the inverse of$g$by$g^{-1}$), for any$n \in N$,$Q\vdash \eta(n)\longleftrightarrow\varphi(g[g^{-1}(n)(n)]). $In order to find the appropriate formula$\eta$, let us consider the expression substituted into the formula$\varphi$, and define the function$f:\omega\longrightarrow \omega$accordingly:$f(n)=g[g^{-1}(n)(n)]$if$n \in N$and$f(n)=0$otherwise. Since this function is obviously recursive and hence representable, and, up to provable equivalence, the result of substituting a representable function into a formula can also be expressed by a formula, there is a formula$\eta$such that, for any$n\in N$, (5)$Q\vdash\eta(n)\longleftrightarrow\varphi(f(n))$Thus we have obtained what we need, we have shown that there exists an$\eta$that can be considered to be the formal version of s. Now, all that remains to do is straightforward: it follows from (5) that, for every$\psi$,$Q\vdash \eta(g(\psi))\longleftrightarrow \varphi\big(g[\psi(g(\psi))])$, which, in turn, choosing$\psi$to be$\eta$, yields$Q\vdash \eta(g(\eta))\longleftrightarrow \varphi(g[\eta(g(\eta))])$, showing that the sentence$\lambda =\eta(g(\eta))\$ indeed has the desired property.

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text:
# Opening tab-delimited file in Calc on Windows

Windows 7 and LO 6.0.1.1 64 bit

I have a number of tab-delimited files, with a file extension of .txt or .ods or .csv or .xls

When I open Calc, then use File/Open to open a tab-delimited file, the file always opens in Writer. I don't get the dialog allowing me to specify the column separator, etc

I have various versions of LO4 and LO5 and they show the same problem. Also the same with Windows 10

However, various versions of OpenOffice work fine, as does LibreOffice on Linux...... I get the dialog allowing me to specify the column separator, and the data opens in Calc

Any ideas?

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@David: Would you mind to mark the parts of the question edited by you?
BTW There is also a user DAVID written in upper case and I couldn't insert the usual @David ("David") here at the start of the comment without getting it changed to upper case.

( 2018-03-08 13:05:27 +0100 )edit

Hi @Lupp - I only added the "windows-7" tag, and the "...on Windows" part in the question title, since this was explicitly delimited in OP's text. Hope that helps! || It's odd that title revisions are not part of the general revision history; you can get that (normally) by clicking the revision date in the upper right of the question, or by manually changing the URL to: .../questions/123456/revisions/ (note the "s" on the end of "questions"). FWIW!

( 2018-03-08 16:22:58 +0100 )edit

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If it has file name extension .txt it will open in Writer, if it is .csv it will open in Calc. To force a .txt file to open in Calc you'll have to select the Text - txt - csv (StarCalc) filter when opening the file, or open it when having already another spreadsheet open as the current document (can be a new empty document).

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The files will not open in Calc regardless of extension. When I open a .txt file in Calc, I am given no opportunity to select the Text-txt-csv filter. The file just immediately opens in Writer. I already have Calc open when I try to open a tab-delimited file. From the Calc menu, I use FILE/OPEN and select the .txt file and it just opens in Writer

( 2018-03-09 13:51:15 +0100 )edit

If a .ods file actually is a table in plain text (delimited by tab or otherwise) the extension is lying. Eliminate this nonsense first.

LibO will assume Writer the applicable component for text files. If you want to order it otherwise permanently you need to change the assignments to extensions for each concerned one for your OS giving scalc.exe instead of soffice.exe as the application. (Not recommended.)
You can open a plain-text file inside Calc when inserting a sheet 'From file'.
Also possible: Use a command interpreter ('Execute' tool probably) and enter something like
"C:\Program Files\LibreOffice\program\soffice.exe" -calc "C:\Users\MyUserName\documents\MyFurtherFolderPath\MyFileName.csv"

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The file extension should only matter if I am double-clicking on the file to open it in Windows using the default application. When opening from within Calc using FILE/OPEN, the extension should not matter. It certainly does not matter when using OpenOffice, or LibreOffice on Linux. That works fine. As does Excel on Windows. And using such a command in a command line interpreter is total overkill for such a problem. I tried SHEET/INSERT SHEET FROM FILE and the resultant dialog has OK greyed out

( 2018-03-09 13:58:31 +0100 )edit

And I should emphasise that using OpenOffice on Windows or LibreOffice on Linux, I can open a tab-delimited file in Calc using FILE/OPEN or SHEET/INSERT SHEET FROM FILE. Both provide the dialog to select the column separator, etc. And I have uninstalled LO and downloaded a fresh copy. Still have the problem

( 2018-03-09 14:02:50 +0100 )edit

Well, of course I don't actually use the commanmd line for the purpose. The hnt was for cpmpleteness.
On my Win-10-system .csv is asociated with soffice.exe (not specifically with scalc.exe). LibO opens the files automatically in Calc.if they actually contain plain text.
A plain text file wrongly given the extension .ods is opened in Writer. I don't worry about if this is clever. I don't use lying extensions. Opening it from a running Calc the same file opens in Calc.

( 2018-03-09 15:48:08 +0100 )edit

If you have actually wrong behaviour insofar beyond the probability of lying extensions on Win, you may check if your user profile is corrupted. Plain text files with extension .txt are openedin Calc from running Calc.
I keep also a facultative 'Open With' scalc.exe on my system for .txt files. It works as expected,; the dialog for csv import is opened.

( 2018-03-09 15:52:55 +0100 )edit


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## LaTeX forum ⇒ Math & Science ⇒ Improving a Matrix

Information and discussion about LaTeX's math and science related features (e.g. formulas, graphs).
Lucasqcgroup
Posts: 1
Joined: Fri Jan 04, 2013 4:32 pm

### Improving a Matrix

Dear all,

I'm trying to type the following matrix:
matrixf.png (3.24 KiB) Viewed 1802 times

But I could only get this:
mymatrix.png (10.32 KiB) Viewed 1802 times

I used the following code:
\renewcommand{\arraystretch}{2}$g :=\left(\begin{array}{c|c}\mathbf{2^D} & \textbf{1} \\ \hline\textbf{1} & \begin{array}{cc}\boldsymbol{1^\alpha} & \textbf{0} \\\textbf{0} & \boldsymbol{1^\beta}\end{array} \\ \end{array}\right)$

There is too much space between the entries of the matrix and the vertical and horizontal lines.

How could I improve my matrix [typeset it correctly] ?

Any help would be greatly appreciated.

Tags:

percusse
Posts: 3
Joined: Mon Nov 12, 2012 11:24 pm
You might better avoid the nesting of arrays as much as you can in order to avoid such spacing and padding problems.

Here is something that comes to close to the given example. You can still modify the array stretch to 1.3 or higher, alternatively add some \mathstruts etc. I recommend reading Hendrik Vogt's nice answer on TeX.sx.

\documentclass{article}\usepackage{amsmath,array,multirow} \begin{document}     $g :=\left( \begin{array}{c|cc} 2^D & \multicolumn{2}{c}{1} \\ \hline \multirow{2}{*}{1} & 1^\alpha & 0 \\ & 0 & 1^\beta \end{array} \right)$   \end{document}
Attachments
texse.png (7.14 KiB) Viewed 1804 times


text:
# Tag Info

3

If $A$ is any unital C*-algebra, the hypotheses are satisfied with $C=A$ and $B=\mathbb C1$.

2

No. For example, $S_3$ contains a characteristic, abelian subgroup not contained in the centre of the group. Both of the automorphisms of this subgroup extend to (inner) automorphisms $G$.

2

1: On a quick look, I don't immediately see how to show that $\psi$ is contractive in general. But note that when $A$ is unital so is $\psi$, which together with positivity makes it contractive. I think this idea can be extended to the non-unital case. 2: When you prove that $\psi$ is positive, you don't need to use that $A$ is $A$, just that it is a ...

1

We can say that $\sigma(p) \subseteq \{0,1\}$. $p - p^2 = 0$, and hence, by the spectral theorem $$\{0\} = \sigma(0) = \{\lambda - \lambda^2 \mid \lambda \in \sigma(p) \}$$

1

Martini's answer is the canonical one. But in this case one can check things explicitly: We can assume that $A$ is unital, as we can always work on the unitization (we actually need it to define the spectrum). If $p=I$ of $p=0$, it is straightforward to check that the spectra are respectively $\{1\}$ and $\{0\}$. For non-trivial $p$: If $\lambda=0$, $p$ ...

1

I think it is simply this fact: take $x\in\ker\pi$. From the relation in the first paragraph of the proof, $$x=\lim_n\Psi_n(\pi(x))=\lim_n\Psi(0)=0.$$ So $\pi$ is an isometric epimorphism, and so an isomorphism.

1

This looks like an "associativity" property. Both algebras live in $B(H_B\otimes H_A\otimes \ell^2(\Gamma))$, so the topology is the same. At the pre-closure level, you should convince yourself that $C_c(\Gamma,B\odot A)$ and $B\odot C_c(\Gamma,A)$ are equal, and that their closures give the two algebras you want to consider.

Only top voted, non community-wiki answers of a minimum length are eligible


text:
## Algebra: A Combined Approach (4th Edition)

$x = (3\sqrt 5, -3\sqrt 5)$
$x^{2}$ = 45 $x = ± \sqrt45$ (Use the square root property) $x = ±5\sqrt 2$ (Simplify the radical) Check- 1. Let $x = 3\sqrt 5$ $x^{2} = 45$ $(3\sqrt 5)^{2}$ could be $45$ $9\times5$ could be $45$ $45 = 45$ Hence, true. 2. Let $x = -3\sqrt 5$ $x^{2} = 45$ $(-3\sqrt 5)^{2}$ could be $45$ $9\times5$ could be $45$ $45 = 45$ Hence, true. Therefore, the solution set is $x = (3\sqrt 5, -3\sqrt 5)$


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# Learn more about

The database contains extension fields of $\Q_p$ of degree $n$ up to isomorphism including all extensions for primes $p<200$ and $1\leq n \leq 15$.

## Browse local number fields

The table gives for each $p$ and $n$ shown, the number of degree $n$ extension fields of $\Q_p$ up to isomorphism.

$p$ \ $n$ 2 3 4 5 6 7 8 9 10 11 12 13 14 15
2 7 2 59 2 47 2 1823 3 158 2 5493 2 590 4
3 3 10 5 2 75 2 8 795 6 2 785 2 6 1172
5 3 2 7 26 7 2 11 3 258 2 17 2 6 1012
7 3 4 5 2 12 50 8 7 6 2 20 2 654 8
11 3 2 5 6 7 2 8 3 18 122 13 2 6 12
13 3 4 7 2 12 2 11 7 6 2 28 170 9 8
17 3 2 7 2 7 2 15 3 6 2 17 2 6 4
19 3 4 5 2 12 2 8 13 8 2 20 2 6 8
23 3 2 5 2 7 2 8 3 6 12 13 2 6 4

A random local number field from the database.

## Find a specific local number field

 Search by label a field label, e.g. 2.4.6.7


text:
Address 2401 Dawson Rd Ste N, Albany, GA 31707 (229) 234-8033 http://dotcomputer.net

# error in the slope of a linear regression Sasser, Georgia

From the t Distribution Calculator, we find that the critical value is 2.63. I remember when I learnt statistics, an estimator was framed as a transformation/function on Random Variables( i.e $\hat{\beta} = g(x_1,x_2,\cdots))$. The confidence interval for 0 takes the form b0 + t*sb0, and the confidence interval for 1 is given by b1 + t*sb1. In the example above, a 95% confidence interval And in a regression we assume $Y = \beta X + \epsilon$ where $\epsilon \sim N(0,\sigma^2 I)$.

See that the estimator $\widehat{b}$ of the slope $b$ is just the 2nd component of $\widehat{\beta}$ --- i.e $\widehat{b} = \widehat{\beta}_2$ . If you need to calculate the standard error of the slope (SE) by hand, use the following formula: SE = sb1 = sqrt [ Σ(yi - ŷi)2 / (n - 2) Well, what if need to find the uncertainty in the slope? The adjective simple refers to the fact that the outcome variable is related to a single predictor.

This is not as good as the slope because the slope essentially uses all the data points at once. In this example, the standard error is referred to as "SE Coeff". As an exercise, I leave you to perform the minimisation to derive $\widehat{\sigma}^2 = ||Y - X\widehat{\beta}||^2$. The value t* is the upper (1 - C)/2 critical value for the t(n - 2) distribution.

T Score vs. Here the dependent variable (GDP growth) is presumed to be in a linear relationship with the changes in the unemployment rate. The MINITAB output provides a great deal of information. Stat Trek Teach yourself statistics Skip to main content Home Tutorials AP Statistics Stat Tables Stat Tools Calculators Books Help   Overview AP statistics Statistics and probability Matrix algebra Test preparation

However, you can use the output to find it with a simple division. In many labs, you will collect data, make a graph, find the slope of a function that fits that data and use it for something. In statistics, simple linear regression is the least squares estimator of a linear regression model with a single explanatory variable. Andale Post authorApril 2, 2016 at 11:31 am You're right!

The dependent variable Y has a linear relationship to the independent variable X. s actually represents the standard error of the residuals, not the standard error of the slope. A scatterplot of the two variables indicates a linear relationship: Using the MINITAB "REGRESS" command with "sugar" as an explanatory variable and "rating" as the dependent variable gives the following result: A confidence interval for the mean response is calculated to be y + t*s, where the fitted value y is the estimate of the mean response.

The original inches can be recovered by Round(x/0.0254) and then re-converted to metric: if this is done, the results become β ^ = 61.6746 , α ^ = − 39.7468. {\displaystyle Suppose I measure the diameter and the circumference of several roundish objects. The formulas all work out the same whether you treat x as fixed or random (the fixed is just a little easier to show). In the least-squares model, the best-fitting line for the observed data is calculated by minimizing the sum of the squares of the vertical deviations from each data point to the line

Note how all the regression lines pass close to the centroid of the data. thanks! –aha Dec 11 '15 at 4:05 @aha, The x values in regression can be considered fixed or random depending on how the data was collected and how you Elsewhere on this site, we show how to compute the margin of error. In particular, when one wants to do regression by eye, one usually tends to draw a slightly steeper line, closer to the one produced by the total least squares method.

Correlation Coefficient Formula 6. Step 7: Divide b by t. A plot of the residuals y - on the vertical axis with the corresponding explanatory values on the horizontal axis is shown to the left. However, it is important enough that I talk about it.

The standard error of regression slope for this example is 0.027. The table below shows hypothetical output for the following regression equation: y = 76 + 35x . This value follows a t(n-2) distribution. If those answers do not fully address your question, please ask a new question. 1 see stats.stackexchange.com/questions/88461/… –TooTone Mar 28 '14 at 23:19 It's reasonably straightforward if you

Normality assumption Under the first assumption above, that of the normality of the error terms, the estimator of the slope coefficient will itself be normally distributed with mean Î² and variance You may need to scroll down with the arrow keys to see the result. The uncertainty in the intercept is also calculated in terms of the standard error of the regression as the standard error (or deviation) of the intercept, sa: The corresponding confidence interval How to Find the Confidence Interval for the Slope of a Regression Line Previously, we described how to construct confidence intervals.

copy file to current directory What would be a good approach to make sure my advisor goes through all the report? Either way, I would get something like this (I did this in Logger Pro): This gives a slope of 3.28 (compare to pi = 3.14). WikipediaÂ® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. How do you do that?

asked 2 years ago viewed 6230 times active 2 years ago Blog Stack Overflow Podcast # 90 - Developer Stories, Charger Butts, and Joel's… Bye Bye, Bullets: The Stack Overflow Developer Under the equation for the regression line, the output provides the least-squares estimate for the constant b0 and the slope b1. In the example above, the slope parameter estimate is -2.4008 with standard deviation 0.2373. Standard Error of Regression Slope was last modified: July 6th, 2016 by Andale By Andale | November 11, 2013 | Linear Regression / Regression Analysis | 3 Comments | ← Regression

If you don't know how to enter data into a list, see:TI-83 Scatter Plot.) Step 2: Press STAT, scroll right to TESTS and then select E:LinRegTTest Step 3: Type in the Find critical value. The system returned: (22) Invalid argument The remote host or network may be down. Step 4: Select the sign from your alternate hypothesis.

Here is some sample data. Back to the top Back to uncertainty of the regression Back to uncertainty of the slope Back to uncertainty of the intercept Skip to Using Excel’s functions Using Excel’s Functions: So The Y values are roughly normally distributed (i.e., symmetric and unimodal). The MINITAB "BRIEF 3" command expands the output provided by the "REGRESS" command to include the observed values of x and y, the fitted values y, the standard deviation of the

The least-squares regression line y = b0 + b1x is an estimate of the true population regression line, y = 0 + 1x. The heights were originally given in inches, and have been converted to the nearest centimetre. Note that $\widehat{\beta}$ is now expressed as some constant matrix multiplied by the random $Y$, and he uses a multivariate normal distribution result (see his 2nd sentence) to give you the Regression equation: Annual bill = 0.55 * Home size + 15 Predictor Coef SE Coef T P Constant 15 3 5.0 0.00 Home size 0.55 0.24 2.29 0.01 What is the