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An oracle (at least in this context) is simply an operation that has some property that you don't know, and are trying to find out. The term "black box" is used equivalently, to convey the idea that it's just a box that you can't see inside, and hence you don't know what it's doing. All you know is that you can supply inputs and receive outputs. In the ...
The church of the larger (or higher, or greater) Hilbert space is just a trick that some people like (myself included) for rewriting some operations.The most general operations that you can write down for a system are described by completely positive maps, while we like describing things with unitaries, which you can always do by moving from the original ...
When we talk about quantum computers, we usually mean fault-tolerant devices. These will be able to run Shor's algorithm for factoring, as well as all the other algorithms that have been developed over the years. But the power comes at a cost: to solve a factoring problem that is not feasible for a classical computer, we will require millions of qubits. This ...
"Postselection" refers to the process of conditioning on the outcome of a measurement on some other qubit. (This is something that you can think of for classical probability distributions and statistical analysis as well: it is not a concept special to quantum computation.)Postselection has featured quite often (up to this point) in quantum mechanics ...
There are a few things to distinguish here, which are often conflated by experts because we're using these terms quickly and informally to convey intuitions rather than in the way that would be most transparent to novices.A "qubit" can refer to a small system, which has a quantum mechanical state.The states of a quantum mechanical system form a vector ...
The qsphere is a way of representing multi-qubit states. So it could be used for 5 qubit states, but it could also be used for any other number.It could also be used for just one qubit. But in this case it is important to note that the single qubit qsphere is not the same as the Bloch sphere, which is our standard way of representing single qubit states....
The terminology of 'surface code' is a little bit variable. It might refer to a whole class of things, variants of the Toric code on different lattices, or it might refer to the Planar code, the specific variant on a square lattice with open boundary conditions.The Toric CodeI'll summarise some of the basic properties of the Toric code. Imagine a square ...
Code spaces and code-wordsA quantum error correcting code is often identified with the code-space (Nielsen & Chuang certainly seem to do so). The code space $\mathcal C$ of e.g. an $n$-qubit quantum error correction code is a vector subspace $\mathcal C \subseteq \mathcal H_2^{\otimes n}$.A code word (terminology which was borrowed from the ...
The surface codes are a family of quantum error correcting codes defined on a 2D lattice of qubits. Each code within this family has stabilizers that are defined equivalently in the bulk, but differ from one another in their boundary conditions.The members of the surface code family are sometimes also described by more specific names: The toric code is a ...
Quantum computing deals (mostly) with finite-dimensional quantum systems called qubits. If you know basic quantum mechanics then you know that the Hilbert space of a qubit is $\mathbb{C}^2$, i.e., the two-dimensional complex Hilbert space over $\mathbb{C}$ (for the more technical people, the Hilbert space is actually $\mathbb{C}P^1$).Therefore, to ...
Is qsphere an actual term representing 5 qubits?If it is, it is not widely used.I claim this because I looked around in arXiv, a repository of electronic preprints of research articles, and found nothing. There are many other units of quantum information than just qubit though. All of the following appear at least occasionally in the relevant literature....
The difficulty with explaining quantum computing is that quantum objects and processes have no direct classical analogue; they're an entirely new ontological category. For example, you might have learned in high school physics that light "is both a particle and a wave" in an attempt to relate it to two classical objects you can intuitively understand. In ...
When we have just one qubit, there's nothing particularly special about the computational basis; it's just nice to have a canonical basis. In practice you could think that first you implement a gate $Z$ with $Z^2 = I$ and $Z\neq I$, and then you say that the computational basis is the eigenbasis of this gate.However, when we talk about multi-qubit systems, ...
When translating a classical circuit into a quantum circuit, you often need to introduce extra qubits simply because quantum computers only implement reversible logic. Such extra qubits are ancilla (or ancillary qubits).One way to spot which qubits are ancilla is to look for those qubits that typically need to be "uncomputed" when using the quantum circuit ...
Talking about bases such as $\left|0\rangle\langle0\right|$ and $\left|1\rangle\langle1\right|$ (or the equivalent vector notation $\left|0\right>$ and $\left|1\right>$, which I'll use in this answer) at the same time as 'horizontal' and 'vertical' are, to a fair extent (pardon the pun) orthogonal concepts.On a Bloch sphere, there are 3 different ...
There is no standard name for a qudit for $d>3$. The community has mostly settled on the term qudit (but you will still find qunit or quNit, for example, using $n$ or $N$ instead of $d$ in some older papers).You will find the odd paper where an individual author will pick a name for the $d=4$ case. I’ve certainly seen ququad and ququart. But I think ...
The general meaning of ancilla in ancilla qubit is auxiliary. In particular, when people write about "constant input" what they mean is that, for a given algorithm -which has a purpose, such as finding the prime factors of an input number, or effecting a simple arithmetic operation between two input numbers the value of the ancilla qubits will be independent ...
In a quantum error correcting code, you store a number of logical qubits, $k$, in a state of many physical qubits, $n$.A code word is a state of the physical qubits that is associated with a specific logical state. So, for example, however you store the $|0\rangle$ state for one of your logical qubits is a code word.The code space is the Hilbert space ...
A code word (for a quantum code) is a quantum state that is typically associated with a state in the logical basis. So, you’ll have some state $|\psi_0\rangle$ that corresponds to the 0 state of the qubit to be encoded (you don’t have to use qubits, but you probably are), and you’ll have another that’s $|\psi_1\rangle$ that corresponds to the 1 state of the ...
In the computational $\left(Z\right)$ basis, the parity of a (classical) bit string is $0$ if the number of $1$s in the string is even (i.e. 'even parity'), or $1$ if the number of $1$s in the string is odd (i.e. 'odd parity').The parity can be measured by applying CNOT gates from each qubit that you want to measure (the control qubits) to an ancilla qubit ...
"Church of the higher hilbert space" is a term coined by John Smolin. According to quantiki it is:for the dilation constructions of channels and states, which [...] provide a neat characterization of the set of permissible quantum operationsand to quote wikipedia, it:describe[s] the habit of regarding every mixed state of a quantum system as a pure ...
As the other answer conveyed (and to which I am just trying to provide some clarification), post-selection is about just looking at a subset of possible measurement outcomes. To my mind, this falls into two different cases, as below. Yes, they are different aspects of the same thing, but they are used very differently by two different communities....
If you think of a electronic spin $S=1/2$, imagine measuring it on the z-axis to obtain $S_z=+1/2$ (or $S_z=-1/2$). This (the z projection of the spin magnetic moment) is a possible basis for the measurement. Or you could measure the spin on the x-axis, and they you will obtain $S_x=+1/2$ (or $S_x=-1/2$). This is a different basis.The measurements on Bell ...
Qubits are essentially quantum objects from which you can extract a bit. But there are different ways that this can be done, and the answer you get depends on the measurement you choose.If you qubit is an electron spin, the measurement basis corresponds to measuring spin in a particular direction. We use that picture more generally in the form of the Bloch ...
Forget about quantum mechanics for a second and consider two people predicting a coin flip. Alice flips a coin, covers it with her hand, and asks Bob to predict the result. Alice knows the coin is heads, but Bob is unsure if it is heads or tails. They will describe the state of the coin using different probability distributions.The same situation can apply ...
For the diagonal basis, the measurement operators are the $|0\rangle\langle 0|$ and $|1\rangle\langle 1|$, as stated in the question. For the other basis, any mutually unbiased basis will do, but people usually go for the two operators $(|0\rangle+|1\rangle)(\langle 0|+\langle 1|)/2$ and $(|0\rangle-|1\rangle)(\langle 0|-\langle 1|)/2$.The labels of which ...
According to Matthias Christandl (who did some research on this to resolve a bet with Artur Ekert), while the term "entanglement" was first used in 1935, as already relayed in other answers, the concept was discussed by Schrodinger in 1932. This set of slides (slides 3-8 in particular) from a talk reproduce part of a document that details this. The full ...
There are lots of separate questions in there: politics, physics, etc. and I won't pretend to answer all of it, but let me try to get towards what I think is the core of the matter.How do I explain to the interested non-specialist what I do (the general field)?My explanation actually varies a lot depending on who I'm talking to, and depends a lot on ... |
@HarryGindi So the $n$-simplices of $N(D^{op})$ are $Hom_{sCat}(\mathfrak{C}[n],D^{op})$. Are you using the fact that the whole simplicial set is the mapping simplicial object between cosimplicial simplicial categories, and taking the constant cosimplicial simplicial category in the right coordinate?
I guess I'm just very confused about how you're saying anything about the entire simplicial set if you're not producing it, in one go, as the mapping space between two cosimplicial objects. But whatever, I dunno. I'm having a very bad day with this junk lol.
It just seems like this argument is all about the sets of n-simplices. Which is the trivial part.
lol no i mean, i'm following it by context actually
so for the record i really do think that the simplicial set you're getting can be written as coming from the simplicial enrichment on cosimplicial objects, where you take a constant cosimplicial simplicial category on one side
@user1732 haha thanks! we had no idea if that'd actually find its way to the internet...
@JonathanBeardsley any quillen equivalence determines an adjoint equivalence of quasicategories. (and any equivalence can be upgraded to an adjoint (equivalence)). i'm not sure what you mean by "Quillen equivalences induce equivalences after (co)fibrant replacement" though, i feel like that statement is mixing category-levels
@JonathanBeardsley if nothing else, this follows from the fact that \frakC is a left quillen equivalence so creates weak equivalences among cofibrant objects (and all objects are cofibrant, in particular quasicategories are). i guess also you need to know the fact (proved in HTT) that the three definitions of "hom-sset" introduced in chapter 1 are all weakly equivalent to the one you get via \frakC
@IlaRossi i would imagine that this is in goerss--jardine? ultimately, this is just coming from the fact that homotopy groups are defined to be maps in (from spheres), and you only are "supposed" to map into things that are fibrant -- which in this case means kan complexes
@JonathanBeardsley earlier than this, i'm pretty sure it was proved by dwyer--kan in one of their papers around '80 and '81
@HarryGindi i don't know if i would say that "most" relative categories are fibrant. it was proved by lennart meier that model categories are Barwick--Kan fibrant (iirc without any further adjectives necessary)
@JonathanBeardsley what?! i really liked that picture! i wonder why they removed it
@HarryGindi i don't know about general PDEs, but certainly D-modules are relevant in the homotopical world
@HarryGindi oh interesting, thomason-fibrancy of W is a necessary condition for BK-fibrancy of (R,W)?
i also find the thomason model structure mysterious. i set up a less mysterious (and pretty straightforward) analog for $\infty$-categories in the fappendix here: arxiv.org/pdf/1510.03525.pdf
as for the grothendieck construction computing hocolims, i think the more fundamental thing is that the grothendieck construction itself is a lax colimit. combining this with the fact that ($\infty$-)groupoid completion is a left adjoint, you immediately get that $|Gr(F)|$ is the colimit of $B \xrightarrow{F} Cat \xrightarrow{|-|} Spaces$
@JonathanBeardsley If you want to go that route, I guess you still have to prove that ^op_s and ^op_Delta both lie in the unique nonidentity component of Aut(N(Qcat)) and Aut(N(sCat)) whatever nerve you mean in this particular case (the B-K relative nerve has the advantage here bc sCat is not a simplicial model cat)
I think the direct proof has a lot of advantages here, since it gives a point-set on-the-nose isomorphism
Yeah, definitely, but I'd like to stay and work with Cisinski on the Ph.D if possible, but I'm trying to keep options open
not put all my eggs in one basket, as it were
I mean, I'm open to coming back to the US too, but I don't have any ideas for advisors here who are interested in higher straightening/higher Yoneda, which I am convinced is the big open problem for infinity, n-cats
Gaitsgory and Rozenblyum, I guess, but I think they're more interested in applications of those ideas vs actually getting a hold of them in full generality
@JonathanBeardsley Don't sweat it. As it was mentioned I have now mod superpowers, so s/he can do very little to upset me. Since you're the room owner, let me know if I can be of any assistance here with the moderation (moderators on SE have network-wide chat moderating powers, but this is not my turf, so to speak).
There are two "opposite" functors:$$ op_\Delta\colon sSet\to sSet$$and$$op_s\colon sCat\to sCat.$$The first takes a simplicial set to its opposite simplicial set by precomposing with the opposite of a functor $\Delta\to \Delta$ which is the identity on objects and takes a morphism $\langle k...
@JonathanBeardsley Yeah, I worked out a little proof sketch of the lemma on a notepad
It's enough to show everything works for generating cofaces and codegeneracies
the codegeneracies are free, the 0 and nth cofaces are free
all of those can be done treating frak{C} as a black box
the only slightly complicated thing is keeping track of the inner generated cofaces, but if you use my description of frak{C} or the one Joyal uses in the quasicategories vs simplicial categories paper, the combinatorics are completely explicit for codimension 1 face inclusions
the maps on vertices are obvious, and the maps on homs are just appropriate inclusions of cubes on the {0} face of the cube wrt the axis corresponding to the omitted inner vertex
In general, each Δ[1] factor in Hom(i,j) corresponds exactly to a vertex k with i<k<j, so omitting k gives inclusion onto the 'bottom' face wrt that axis, i.e. Δ[1]^{k-i-1} x {0} x Δ[j-k-1] (I'd call this the top, but I seem to draw my cubical diagrams in the reversed orientation).
> Thus, using appropriate tags one can increase ones chances that users competent to answer the question, or just interested in it, will notice the question in the first place. Conversely, using only very specialized tags (which likely almost nobody specifically favorited, subscribed to, etc) or worse just newly created tags, one might miss a chance to give visibility to ones question.
I am not sure to which extent this effect is noticeable on smaller sites (such as MathOverflow) but probably it's good to follow the recommendations given in the FAQ. (And MO is likely to grow a bit more in the future, so then it can become more important.) And also some smaller tags have enough followers.
You are asking posts far away from areas I am familiar with, so I am not really sure which top-level tags would be a good fit for your questions - otherwise I would edit/retag the posts myself. (Other than possibility to ping you somewhere in chat, the reason why I posted this in this room is that users of this room are likely more familiar with the topics you're interested in and probably they would be able to suggest suitable tags.)
I just wanted to mention this, in case it helps you when asking question here. (Although it seems that you're doing fine.)
@MartinSleziak even I was not sure what other tags are appropriate to add.. I will see other questions similar to this, see what tags they have added and will add if I get to see any relevant tags.. thanks for your suggestion.. it is very reasonable,.
You don't need to put only one tag, you can put up to five. In general it is recommended to put a very general tag (usually an "arxiv" tag) to indicate broadly which sector of math your question is in, and then more specific tags
I would say that the topics of the US Talbot, as with the European Talbot, are heavily influenced by the organizers. If you look at who the organizers were/are for the US Talbot I think you will find many homotopy theorists among them. |
Trisect sides of a quadrilateral and connect the points to have nine quadrilaterals, as can be seen in the figure. Prove that the middle quadrilateral area is one ninth of the whole area.
Consider all occurring points as vectors, as in @Calvin Lin's answer, and write $\mu$ for ${1\over3}$. Then $$p=(1-\mu)a+\mu b,\quad h=(1-\mu)d+\mu c,\quad n=(1-\mu)a+\mu d,\quad e=(1-\mu) b+\mu c\ .$$ It follows that $$(1-\mu)p+\mu h=(1-\mu)n+\mu e\quad(=:w')\ ,$$ which shows that in fact $$w=w'=(1-\mu)^2 a +\mu(1-\mu)(b+d)+\mu^2 c\ .$$ Interchanging $a$ and $c$ here gives $$y=(1-\mu)^2 c +\mu(1-\mu)(b+d)+\mu^2 a\ ,$$ so that we arrive at $$w-y=(1-2\mu)(a-c)\ .$$ Appealing to symmetry again we conclude that we also have $$x-z=(1-2\mu)(b-d)\ .$$ It follows that $${\rm area}[WXYZ]=(1-2\mu)^2\ {\rm area}[ABCD]\ ,$$ and this holds for any $\mu\in[0,{1\over2}[\ $.
This is most easily done using vectors. Let the points $A, B, C, D$ be represented by the vectors $a, b, c, d$. The area $[ABCD]$ is equal to $\frac{1}{2}(a-c) \times (b-d) $.
If you are unfamiliar with this, consider triangulation using the origin, and sum up the 4 triangle areas, to get
$$\begin{align} [ABCD] = & \frac{1}{2} a \times b + \frac{1}{2} b \times c + \frac{1}{2} c \times d + \frac{1}{2} d \times a \\ = & (a-c) \times \frac{1}{2} b + (a-c ) \times (-\frac{1}{2} d) \\= & \frac{1}{2}(a-c) \times (b-d) \end{align}$$
It is easy to show that $W= \frac{4a+2b+c+2d} {9}, X = \frac{ 2a+4b+2c+d} { 9},Y = \frac{a+2b+4c+d} { 9} , Z = \frac{ 2a+b+2c+4d} {9} $. Hence the area is
$$ [WXYZ] = \frac{1}{2} ( \frac{3a-3c}{9} ) \times ( \frac{3b-3d}{9} ) = \frac{1}{9} \times \frac{1}{2} (a-c)(b-d) = \frac{1}{9} [ABCD]$$
Claim.$W$ and $Z$ trisect $\overline{PH}$. Likewise elsewhere.
Proof. Left to the reader (for now).
Given the claim, we can make this illustrated argument:
Here, we have $\triangle ABC \sim \triangle PBF$, with $$\frac{|\overline{PB}|}{|\overline{AB}|} = \frac{|\overline{FB}|}{|\overline{CB}|} = \frac{2}{3} = \frac{|\overline{PF}|}{|\overline{AC}|} \qquad \text{and} \qquad \overline{PF} \parallel \overline{AC}$$ and $\triangle PZF \sim \triangle WZY$, with $$\frac{|\overline{WZ}|}{|\overline{PZ}|} = \frac{|\overline{YZ}|}{|\overline{FZ}|} = \frac{1}{2} = \frac{|\overline{WY}|}{|\overline{PF}|} \qquad \text{and} \qquad \overline{WY} \parallel \overline{PF}$$ so that $$|\overline{WY}|= \frac13 |\overline{AC}| \qquad \text{and} \qquad \overline{WY} \parallel \overline{AC}$$ and likewise $$|\overline{XY}|= \frac13 |\overline{BD}| \qquad \text{and} \qquad \overline{XZ} \parallel \overline{BD}$$
By the Diagonal-Diagonal-Angle formula for quadrilateral area, $$|\square WXYZ| = \frac{1}{2}|\overline{WY}||\overline{XZ}|\sin\theta = \frac12 \cdot \frac{1}{3}|\overline{AC}| \cdot \frac13 |\overline{BD}|\cdot \sin\theta = \frac19 |\square ABCD|$$
Stretch the figure in a direction and by an amount which makes the top and bottom edges parallel. Such a transformation preserves relative areas. Now look at each trapezoid in the center row. It is clear that the area is equal to the average of the areas of the trapezoids above and below. Similarly, stretch to make the sides parallel, and look at the trapezoids in the center column. The area of each is equal to the average of the areas of the figures to the right and left. It follows that the area of the center quadrilateral is equal to the average of the areas of the outer eight quadrilaterals. |
I am trying to understand the concept of
group velocity of a free particle wave packet: $$\Psi(x,t) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} \phi(k)e^{ikx}e^{-\frac{i \hbar k^2 t}{2m}}dk.$$
Where $$\omega(k) = \frac{\hbar k^2}{2m}.$$ Assuming that $\phi(k)$ is narrowly peaked about some particular value $k_0$ we can Taylor expand $\omega(k)$ about $k_0$ to get $$\omega(k) = \frac{\hbar k_0}{2m} + \frac{\hbar k_0}{m}(k-k_0)+ \frac{\hbar}{2m}(k-k_0)^2.$$ Then after defining variable $\kappa := k-k_0$ we get $$\Psi(x,t) \approx \frac{1}{\sqrt{2 \pi}}e^{\frac{-i \hbar k_0^2 t}{2m}}e^{\frac{i \hbar k_0^2 t}{m}} \int^{\infty}_{-\infty}\phi(\kappa + k_0)e^{i(\kappa + k_0)(x-\frac{\hbar k_0 t}{m})}dk.$$
This integral is the superposition of waves of the form $$e^{i(\kappa + k_0)(x-\frac{\hbar k_0 t}{m})}.$$ However notice that each of these waves have the same speed $\frac{\hbar k_0}{m}$. Then in terms of the energy $$v_{\text{group}} = \frac{d \omega(k_0)}{dk} = \frac{\hbar k_0}{m} = \sqrt{\frac{2 E_0}{m}}.$$ Hence for stationary state $\Psi_{k_0}$ we have that $$v_{\text{group}} = 2 v_{\text{phase}}.$$
Apparently we can then evaluate the group velocity for each $k$, hence $v_{\text{group}} = \frac{d \omega(k)}{dk}$.
Question:I understand the above calculation. But qualitatively I don't understand how the group velocity can be a function of $k$ as opposed to just a single value (independent of $k$) which describes how fast the envelope propagates? I know that each stationary state has it's own phase velocity but I thought that superimposing these waves of different speeds produces an envelope which travels at a single velocity $v_{\text{group}}$? |
For discussion directly related to ConwayLife.com, such as requesting changes to how the forums or wiki function.
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
I have some questions:
1. Is there a minimal age for joining conwaylife? 2. Who is unname66609? 3. Does LifeWiki store the IP Addresses of users? Ok! That will be it for now
1. Is there a minimal age for joining conwaylife?
2. Who is unname66609?
3. Does LifeWiki store the IP Addresses of users?
Ok! That will be it for now
Airy Clave White It Nay
The terms and conditions shown when you first register say nothing about age, and I think the youngest member here, Gustavo, is 11 or 12 - so no.Saka wrote:1. Is there a minimum age for joining conwaylife? Just another forum user as far as I can tell. They probably live in or near China. Why do you ask?2. Who is unname66609? Yes.3. Does LifeWiki store the IP Addresses of users? Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X 1. So it's ok if a a six year old joins?M. I. Wright wrote:The terms and conditions shown when you first register say nothing about age, and I think the youngest member here, Gustavo, is 11 or 12 - so no.Saka wrote:1. Is there a minimum age for joining conwaylife?Just another forum user as far as I can tell. They probably live in or near China. Why do you ask?2. Who is unname66609?Yes.3. Does LifeWiki store the IP Addresses of users?
2. Because he never really posts anything... (just things like "what is sesame oil?"
3. Ok
And by the the way did you notice the second "a" I put in number 1? There's also something else about this line...
Airy Clave White It Nay
Hmm... Technically, it's perfectly possible, but I can hardly imagine a six-year-old doing stuff like this. If he/she decides to join, definitely tell me the username. If that person achieves anything, I will clap.Saka wrote:1. So it's ok if a a six year old joins? So do you think unname is a bot? It is not probable since he/she used to post some relevant content. Unname is just a strange guy, like Gustavo raised to the 10th power.Saka wrote:2. Because he never really posts anything... (just things like "what is sesame oil?" No. I failed this test for the the second time in my lifeSaka wrote:And by the the way did you notice the second "a" I put in number 1?
There are 10 types of people in the world: those who understand binary and those who don't.
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X And I passed.Alexey_Nigin wrote:Hmm... Technically, it's perfectly possible, but I can hardly imagine a six-year-old doing stuff like this. If he/she decides to join, definitely tell me the username. If that person achieves anything, I will clap.Saka wrote:1. So it's ok if a a six year old joins? So do you think unname is a bot? It is not probable since he/she used to post some relevant content. Unname is just a strange guy, like Gustavo raised to the 10th power.Saka wrote:2. Because he never really posts anything... (just things like "what is sesame oil?" No. I failed this test for the the second time in my lifeSaka wrote:And by the the way did you notice the second "a" I put in number 1?
Airy Clave White It Nay
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X How young?Sarp wrote:Nope I'm younger1. IThe terms and conditions shown when you first register say nothing about age, and I think the youngest member here, Gustavo, is 11 or 12 - so no.
Airy Clave White It Nay
Meaning, a collection of syntheses where the recipes involve colliding LWSSes instead of colliding gliders? For the second question, are combinations of LWSS, MWSS, and HWSS allowed, or should a single recipe have only one kind of *WSS?The Turtle wrote:Are there any catalogues of LWSS collisions? Or other *WSS collisions?
In any case, I think the answer is no -- nothing really organized, anyway, because no one has had a long-term use for it. It's fairly easy to put together the beginnings of such a collection by running gencols for a while, along these general lines.
If you just want a table of two-*WSS collisions, that's very easy to generate with gencols. Maybe someone has a stamp collection lying around already.
I've been curious for a while now about construction mechanisms using *WSS slow salvos -- though not curious enough yet to do the research myself, apparently. Seems as if slow LWSSes would probably be just about as effective as gliders for building self-constructing circuitry, and Geminoid construction-arm elbows can be programmed to produce *WSSes instead of gliders. It's several times more expensive, but it allows for four new construction directions.
I know I found a while back that LWSS slow salvos can move a blinker anywhere, but I can't find it anywhere but in a quote in the Accidental Discoveries thread.dvgrn wrote:I've been curious for a while now about construction mechanisms using *WSS slow salvos -- though not curious enough yet to do the research myself, apparently. Seems as if slow LWSSes would probably be just about as effective as gliders for building self-constructing circuitry, and Geminoid construction-arm elbows can be programmed to produce *WSSes instead of gliders. It's several times more expensive, but it allows for four new construction directions.
x₁=ηx
V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$
http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce
Mr. Missed Her Posts:90 Joined:December 7th, 2016, 12:27 pm Location:Somewhere within [time in years since this was entered] light-years of you.
What's with the alternating shades of the posts? Sometimes, tables/lists do that so you can keep track of which item is which, but I don't see how it helps here. Besides, there's almost no difference in the colors.
There is life on Mars. We put it there with not-completely-sterilized rovers.
And, for that matter, the Moon, Jupiter, Titan, and 67P/Churyumov–Gerasimenko.
And, for that matter, the Moon, Jupiter, Titan, and 67P/Churyumov–Gerasimenko.
Mr. Missed Her Posts:90 Joined:December 7th, 2016, 12:27 pm Location:Somewhere within [time in years since this was entered] light-years of you.
I only noticed when I tried to make the background of my profile photo the color of the posts. As you can see, the colors match up perfectly here, but they don't in the last post.
There is life on Mars. We put it there with not-completely-sterilized rovers.
And, for that matter, the Moon, Jupiter, Titan, and 67P/Churyumov–Gerasimenko.
And, for that matter, the Moon, Jupiter, Titan, and 67P/Churyumov–Gerasimenko.
Mr. Missed Her Posts:90 Joined:December 7th, 2016, 12:27 pm Location:Somewhere within [time in years since this was entered] light-years of you.
I might be able to, but I only know how to edit images through Paint, which is like crappy Photoshop that comes with windows.
There is life on Mars. We put it there with not-completely-sterilized rovers.
And, for that matter, the Moon, Jupiter, Titan, and 67P/Churyumov–Gerasimenko.
And, for that matter, the Moon, Jupiter, Titan, and 67P/Churyumov–Gerasimenko.
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X Try gimpMr. Missed Her wrote:I might be able to, but I only know how to edit images through Paint, which is like crappy Photoshop that comes with windows.
Airy Clave White It Nay
Mr. Missed Her Posts:90 Joined:December 7th, 2016, 12:27 pm Location:Somewhere within [time in years since this was entered] light-years of you.
Well, thanks.
There is life on Mars. We put it there with not-completely-sterilized rovers.
And, for that matter, the Moon, Jupiter, Titan, and 67P/Churyumov–Gerasimenko.
And, for that matter, the Moon, Jupiter, Titan, and 67P/Churyumov–Gerasimenko.
BlinkerSpawn Posts:1906 Joined:November 8th, 2014, 8:48 pm Location:Getting a snacker from R-Bee's Click the checkbox to the right, then choose "Delete marked" at the bottom, from the dropdown where it normally says "Mark/Unmark as important".BlinkerSpawn wrote:On this note, how do I delete read private messages?gameoflifemaniac wrote:Does Nathaniel have access to private messages?
Surprisingly well hidden, isn't it? It's a very clever design -- keeps you from deleting things accidentally (or on purpose).
In theory, yes. PMs are stored as plain text in the database and while there is no direct ability within phpbb for the admin to read PMs, any person or program (such as backup software) with access to the database has access to the content of private messages. I see no reason to be concerned about this.gameoflifemaniac wrote:Does Nathaniel have access to private messages? |
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
I have a problem to determine convergence (sum over n). $$\sum_{n=0}^\infty \dfrac {a\left( a+1^{p}\right) \ldots \left( a+n^{p}\right) }{b\left( b+1^{p}\right) \ldots \left( b+n^{p}\right) }$$where $a<b, a>0,b>0$.
I have concluded convergence for $p\leq0$ by comparing it to a constructed geometric sequence, as well as for $p=1$, using comparison test with $n^{a-b}$. But I can not use similar methods for $p>1$ and $0<p<1$.
I have some thoughts for the two parts:
When $p>1$, it seems that the limit of each term is not $0$. If the limit could be evaluated, then the divergence can be proved. My method for $p=1$ is to use the Euler Product of the gamma function, but the $p$ power makes it impossible to use this method. I am wondering if there is any kind of generalization of gamma function that is of this form.
when $0<p<1$, I compared it to the case of $p=1$, that could at least tell it converges in the range when $p=1$ converge. But it is inconclusive for the parts remaining.
Any help or hints would be appreciated. |
Let $X, Y$ be normed space and $f:X\to Y$ be a mapping. Assume that for all $n\in\mathbf{N}$, $$\|x-y\|=n\iff\|f(x)-f(y)\|=n.$$ Under what conditions this map will be an isometry? Thanks
So it turns out my earlier intuition was incorrect, and one can leverage the order properties of ${\bf R}$ to show:
Theorem. Let $X, Y$ be real Hilbert spaces, with the dimension of $X$ at least two, and let $f: X \to Y$ be a function such that $\|f(x)-f(y)\|=\|x-y\|$ whenever $\|x-y\|$ is a natural number. (We do not assume $f$ to be continuous.) Then $f$ is an isometry. Proof By passing to an affine subplane of $X$, we may assume without loss of generality that $X$ is two-dimensional, so we will write $X = {\bf R}^2$.
If $e$ is a unit vector in $X$, $x$ is an element of $X$ and $n,m$ are natural numbers then $f(x), f(x+ne), f(x+(n+m)e)$ form a degenerate triangle with side lengths $n,m,n+m$, which implies that $f(x+ne) = f(x) + nu$ and $f(x+(n+m)e) = f(x) + (n+m) u$ for some unit vector $u$. In particular, $f$ maps any isometric copy of ${\bf Z}$ to another isometric copy of ${\bf Z}$.
Next, if $e, f$ are orthogonal unit vectors in $X$, then $f(x), f(x+3e), f(x+4f)$ form a triangle with side lengths $3,4,5$ and in particular $f(x+3e)-f(x)$ is orthogonal to $f(x+4f)-f(x)$. From this and the previous claim we see that $f$ maps any isometric copy of ${\bf Z}^2$ to another isometric copy of ${\bf Z}^2$.
By the preceding discussion we have $$f(x,y)-f(0,0) =x (f(1,0)-f(0,0)) + y (f(0,1)-f(0,0)) \qquad (1)$$ whenever $x,y$ are integers, but also $f(n e)-f(0,0) = n (f(e)-f(0,0))$ for any unit vector $n$. In particular, for any Pythagorean triple $a^2+b^2=c^2$, we have (1) when $(x,y)$ is an integer multiple of $(\frac{a}{c},\frac{b}{c})$; conjugating by translation we then see that (1) also holds when $(x,y)$ is an integer multiple of $(\frac{a}{c},\frac{b}{c})$ plus an element of ${\bf Z}^2$. Iterating this we see that (1) holds whenever $(x,y)$ is an integer linear combination of Pythagorean fractions $(\frac{a}{c}, \frac{b}{c})$. In particular, (1) holds on a dense subset $D$ of ${\bf R}^2$.
This is already enough to get isometry in the continuous case. Now we remove the continuity hypothesis. Observe that if $\|x-y\| \leq 2$, then $x$ can be reached from $y$ by a pair of moves of unit length, and hence by the triangle inequality $\|f(x)-f(y) \| \leq 2$.
Now let us normalise $f,X,Y$ so that ${\bf R}^2 \subset Y$, $f(0,0)= (0,0)$, $f(1,0)=(1,0)$, and $f(0,1)=(0,1)$. Then by (1) we see that $f$ is the identity on $D$. For any $p \in {\bf R}^2$ and $\varepsilon>0$, we can find elements $q, r$ of $D$ within $\varepsilon$ of $p + (2,0)$ and $p-(2,0)$ that lie within $2$ of $p$, and thus by the preceding we see that $f(p)$ lies within $2+\varepsilon$ of $p+(2,0)$ and $p-(2,0)$. Sending $\varepsilon$ to zero we see that $f(p)=p$, and the claim follows.
Terry Tao's argument generalizes to show that $f$ must be an isometry whenever $X$ has dimension greater than 1 and $Y$ is strictly convex. We start by proving a series of lemmas:
Lemma 1: Let $x,y\in X$, and let $a,b\geq0$ be such that $a+b\geq\|x-y\|$ and $|a-b|\leq\|x-y\|$. Then there exists $z\in X$ such that $\|x-z\|=a$ and $\|y-z\|=b$. Proof: Let $S=\{z:\|x-z\|=a\}$; this is connected since $\dim X>1$. Note that $S$ intersects the line between $x$ and $y$ twice; our hypotheses on $a$ and $b$ imply that at one of these points $\|y-z\|\leq b$ and at the other $\|y-z\|\geq b$. Since $z\mapsto \|y-z\|$ is continuous on $S$, there must be some $z\in S$ such that $\|y-z\|=b$. Lemma 2: Suppose $f(0)=0$ and $x\in X$ is such that $\|x\|\in\mathbb{N}$. Then for all $n\in\mathbb{Z}$, $f(nx)=nf(x)$. Proof: By strict convexity, any triangle in $Y$ for which the triangle inequality is an equality must lie on a line. Applying this to the triangle formed by $0$, $f(x)$, and $f(nx)$ yields the desired result. Lemma 3: Suppose $\|x-z\|$ and $\|y-z\|$ are both integers and $\|x-y\|$ is rational. Then $\|f(x)-f(y)\|=\|x-y\|$. Proof: By translating, we may assume $z=0$ and $f(z)=0$. By Lemma 2, for all $n\in\mathbb{Z}$, $f(nx)=nf(x)$ and $f(ny)=nf(y)$. Letting $n$ be the denominator of $\|x-y\|$, we have $\|f(nx)-f(ny)\|=\|nx-ny\|$ since this is an integer, and the result follows by dividing by $n$. Lemma 4: Suppose $\|x-y\|$ is rational. Then $\|f(x)-f(y)\|=\|x-y\|$. Proof: Use Lemma 1 to find $z$ such that $\|x-z\|=\|y-z\|$ is some large integer and apply Lemma 3.
We now prove that $f$ is an isometry. Fix $x,y\in X$. Use Lemma 1 to find $z$ such that $\|x-z\|$ is small and rational and $\|y-z\|$ is rational and close to $\|x-y\|$. By Lemma 4 and the triangle inequality it follows that $\|f(x)-f(y)\|$ must be (arbitrarily) close to $\|x-y\|$.
Partial answer, in the case where the norm space = $R$.
Let us define condition $\cal C$ by $||x-y||= n$ ⇔ $||f(x)-f(y)|| = n$.
Assume that the normed space is $R$, so that the norm is of the form $\lambda|*|$ with $\lambda > 0$. w.l.g, I assume that the norm is the usual absolute value. Let $h(t)$ be a continuous bijection $[0,1]\to [0,1]$, such that $h(0)=0$, $h(1)=1$ and $0\leq h(t)<1$. Then the function defined by $f(k+t) = k + h(t)$ ($0\leq t<1$) is continuous and satisfies condition $\cal C$. This shows that there can be no proper condition such that, added to condition $\cal C$, leads to an isometry. In other words, in the case where the normed space is $R$ every such condition will be equivalent to "$f$ is an isometry", independently of condition $\cal C$. If you take $h$ to be diffentiable in $[0,1]$, with right differential at 0 equal to its left differential at 1, then the same could be said with the additional assumption "$f$ differentiable".
Notice also that it can be shown that every continuous function that satisfy $\cal C$ must be of the form above, or of the form $f(k+t)=-(k + h(t))$.
For general normed space, I suggest to work first the case of a two-dimensional normed space, with the euclidean norm. If it can be shown that a continuous $f$ that satisfies condition $\cal C$ must be an isometry, it will be probably easy to show the same in general normed space.
Here is one possible condition: If $f(x)$ is continuous and fulfills $||f(kx)-f(ky)|| = k||f(x)-f(y)||$, for every $k \in N$. This works because :
1) the equivalence you wrote above will be true not only for $n\in N$ but also for every positive rational number $n$ (easy to check)
2) it will be in fact true for all positive real number $n$ : recall that the norm is continuous and take the limit of a sequence $(x_m,y_m)$ such that $x_m\to x$, $y_m\to y$, $||x_m-y_m||\in \mathbb Q$, and $||x_m-y_m||\to ||x-y||$ (such a sequence exists because the subspace spanned by $x,y$ is isometric to $\mathbb R^2$, endowed with the transported norm, equivalent to any other norm in $\mathbb R^2$).
You can even restrict this condition to hold only whenever $||kx - ky||$ is a natural number, and not for all $x,y$.
To sum up, $f$ is an isometry if and only if $f$ is continuous, $||x-y|| = n$ implies $||f(x)-f(y)||=n$ for all $n\in N$ and $||f(kx)-f(ky)|| = k ||f(x)-f(y)||$ holds for all $k\in N$.
Notice also that if your relation holds for every rational number $n$ and $f$ is continuous, then $f$ is an isometry by continuity.
I think it's worth mentioning the well-known fact that, in a finite-dimensional euclidean space (of dimension larger than 2), just preserving one distance (say, $\|f(x)-f(y\|=1$ as soon as $\|x-y\|=1$) is sufficient for $f \colon X \to X$ to be an isometry.
Maybe It is time to sum up the work that has been down up to now :
Let $X$ and $Y$ be normed spaces, and $f$ a function $X\to Y$. Denote by $\cal C$ the condition "$||x-y|| = n\in \mathbb N \Rightarrow ||f(x)-f(y)|| = n$".
1) If $\dim(X)=1$, no proper condition can be added to $\cal C$, and even to the stronger condition set by the asker, that would imply that $f$ be an isometry (MikeTex) ;
2) If $\dim(X) > 1$, then $f$ is an isometry if and only if $f$ is continuous and fulfills the condition "$||f(kx)-f(ky)|| = k||f(x)-f(y)||$ for every $x,y$, and every $k\in \mathbb N$" (MikeTex) ;
3) If $X$ and $Y$ are Hilbert space and $\dim(X)>1$, then condition $\cal C$ implies that $f$ is an isometry (Terry Tao) ;
4) More generally, if $\dim(X)>1$ and $Y$ is strictly convex, then condition $\cal C$ implies that $f$ is an isometry (Eric Wofsey).
It is interresting that the additional condition $||f(x)-f(y)|| = n \Rightarrow ||x-y|| = n$ present in the question of the asker has not been used in the argument of Eric Wofsey (hence is unnecessary, if used, in the argument of Terry Tao). The question is now : can this additional condition be used in order to weaken the assumptions of Eric and Terry. |
One of the nice things about relativity is that it doesn't matter how fast you are traveling, only how fast you are traveling relative to something else. So what does that mean?
Moving fuel to the engine - as easy as if the ship was stopped Moving inside the ship - this could be hard while you're accelerating or decelerating, and there would be a lot of that if you're getting close to light speed. However, the ship's speed doesn't matter - if you're accelerating at 1G, it will be as hard to move around whether the ship's just leaving orbit or already at 0.9c. Communicating - from the perspective of anyone/anything on the ship, everything is as if the ship is moving slowly. So signals from the drive can make it to the bridge of the ship just as fast as if the ship was stopped. Turning - it's as complicated as if you were turning while in orbit around a planet.
So what
is a problem with near-light-speed travel? Reacting to outside stimuli, and getting to and from your desired travel velocity.
Reacting to outside stimuli is dodging, which is not the focus of your question, and navigating. Navigating shouldn't be a significant problem though - if you have the tech to get anywhere near light speed, you should be able to plot your trajectory well enough to make navigating fairly simple.
So getting to and from your desired travel velocity is the only real problem you have to worry about. What's so hard about this? At relativistic speeds, your kinetic energy is described by the following formula:
$$K=mc^2\Big(\frac{1}{\sqrt{1-v^2/c^2}}-1\Big)$$
If $v=\frac{\sqrt{3}}{4}c\approx 0.866c$, which you might not count as a "very high fraction of $c$", we get $K=mc^2=E$. In other words, the ship has as much kinetic energy as energy you could get from converting the entire mass of the ship into energy. So one way to get going that fast would be to have a fuel tank carrying as much mass as the entire rest of the ship, and have a way to convert every atom of fuel into pure energy with 100% efficiency.
With a nuclear reactor, we can currently use about 0.1% of uranium's mass worth of energy. So with a nuclear reactor that could run in space and convert the energy it produces into velocity with 100% efficiency, you could get your kinetic energy up to $K=0.001mc^2$. That gets you up to about $0.045c$. Again, this is if your fuel tank carries as much mass as the entire rest of the ship.
Oh, and don't forget that you have to decelerate once you get to your destination. So "your ship" that you have to accelerate consists of your actual ship and the fuel tank carrying enough fuel to decelerate.
In summary, to get a 10000 metric ton spaceship to $0.045c$, you need 10000 metric tons of uranium to decelerate, and 20000 metric tons of uranium to accelerate. Oh, and a 100% efficient reactor and engine, neither of which can actually exist due to entropy always taking a share. Also you may have realized that if you have 10000 metric tons of spaceship and 10000 metric tons of uranium, you're got to accelerate all the unused uranium.
So getting to a "very high fraction of $c$" just isn't feasible unless you're willing to use up a
ridiculous amount of fuel. Unless you consider $0.01c$ to be a very high fraction.
Let's work out an example of an actual "very high fraction of $c$", using an Ohio-class submarine as our starting point. It has a length of 170m and mass of 18750 tonnes. If we scale that lengthwise by 20 we get 3.4km, in the range of your "several kilometers". If we double the cross-sectional radius to make it at least a little bit more cozy, we've now increased the total volume by factor of 80, for an approximate mass of 1.5 million tonnes. I don't know how this compares to what you had in mind for your spaceship, but this is a very long but narrow spaceship. Now how much energy does it require to get this to "a very high fraction of $c$"? To get this ship to $0.99975c$, you need around to convert almost $6*10^{24}kg$ of mass into kinetic energy for the ship. That's the mass of the Earth.
In short, if you're not working with unobtanium/applied phlebotinum/magic you're not likely going to be able to get close to $c$. |
Refine
A compact subset E of the complex plane is called removable if all bounded analytic functions on its complement are constant or, equivalently, i f its analytic capacity vanishes. The problem of finding a geometric characterization of the removable sets is more than a hundred years old and still not comp letely solved.
We show that the intersection local times \(\mu_p\) on the intersection of \(p\) independent planar Brownian paths have an average density of order three with respect to the gauge function \(r^2\pi\cdot (log(1/r)/\pi)^p\), more precisely, almost surely, \[ \lim\limits_{\varepsilon\downarrow 0} \frac{1}{log |log\ \varepsilon|} \int_\varepsilon^{1/e} \frac{\mu_p(B(x,r))}{r^2\pi\cdot (log(1/r)/\pi)^p} \frac{dr}{r\ log (1/r)} = 2^p \mbox{ at $\mu_p$-almost every $x$.} \] We also show that the lacunarity distributions of \(\mu_p\), at \(\mu_p\)-almost every point, is given as the distribution of the product of \(p\) independent gamma(2)-distributed random variables. The main tools of the proof are a Palm distribution associated with the intersection local time and an approximation theorem of Le Gall.
Symmetry properties of average densities and tangent measure distributions of measures on the line (1995)
Answering a question by Bedford and Fisher we show that for every Radon measure on the line with positive and finite lower and upper densities the one-sided average densities always agree with one half of the circular average densities at almost every point. We infer this result from a more general formula, which involves the notion of a tangent measure distribution introduced by Bandt and Graf. This formula shows that the tangent measure distributions are Palm distributions and define self-similar random measures in the sense of U. Zähle.
Tangent measure distributions were introduced by Bandt and Graf as a means to describe the local geometry of self-similar sets generated by iteration of contractive similitudes. In this paper we study the tangent measure distributions of hyperbolic Cantor sets generated by contractive mappings, which are not similitudes. We show that the tangent measure distributions of these sets equipped with either Hausdorff or Gibbs measure are unique almost everywhere and give an explicit formula describing them as probability distributions on the set of limit models of Bedford and Fisher.
Tangent measure distributions are a natural tool to describe the local geometry of arbitrary measures of any dimension. We show that for every measure on a Euclidean space and every s, at almost every point, all s-dimensional tangent measure distributions define statistically self-similar random measures. Consequently, the local geometry of general measures is not different from the local geometry of self-similar sets. We illustrate the strength of this result by showing how it can be used to improve recently proved relations between ordinary and average densities.
We show that the occupation measure on the path of a planar Brownian motion run for an arbitrary finite time intervalhas an average density of order three with respect to thegauge function t^2 log(1/t). This is a surprising resultas it seems to be the first instance where gauge functions other than t^s and average densities of order higher than two appear naturally. We also show that the average densityof order two fails to exist and prove that the density distributions, or lacunarity distributions, of order threeof the occupation measure of a planar Brownian motion are gamma distributions with parameter 2.
The Kallianpur-Robbins law describes the long term asymptotic behaviour of the distribution of the occupation measure of a Brownian motion in the plane. In this paper we show that this behaviour can be seen at every typical Brownian path by choosing either a random time or a random scale according to the logarithmic laws of order three. We also prove a ratio ergodic theorem for small scales outside an exceptional set of vanishing logarithmic density of order three. |
How can one see it from BCS wavefunction and BCS Hamiltonian? i.e.
$$H_{BCS}=\sum_{k\sigma}\epsilon_k c_{k\sigma}^\dagger c_{k\sigma}-\Delta^*\sum_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger+h.c.$$
and:
$$\Psi_{BCS}=\Pi_k(u_k+v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger)|0\rangle$$
If it has this symmetry, what significance does it has? |
A particle is moving in a potential $V(x)=V_0\vert x \vert$. I need to get the angular frequency and the period of the movement of the particle.
This is what i have done.
The equation of motion is $$ \DeclareMathOperator{\sgn}{sgn}\begin{align} m\ddot x &= -\dfrac{\partial V}{\partial x} \\ &= -V_0 \sgn (x) \end{align}$$
$$x=x_0+v_0t-\dfrac{V_0}{m}\sgn(x)\dfrac{t^2}{2}$$
My problem is: How to compare the equation of motion of this system with the equation of motion of a harmonic oscillator in order to get the angular frequency $\omega$? |
I'm studying these lecture notes on dynamic models of robot arm links, slides 33-36, where two examples are given for the kinetic energies for a single link and two link robot arm.
In the single link case, the kinetic energy is posed as:
$$K=\frac{1}{2}I\dot{\theta}^2$$
This appears to account for the rotational kinetic energy of the single link around the joint.
In the two link case, the kinetic energy of the first link is posed as:
$$K=\frac{1}{2}I_1\dot{\theta_1}^2 + \frac{1}{2}m_1a_{c1}^2\dot{\theta_1}^2$$
This appears to mean that the total kinetic energy in the first link of the two link chain contains a contribution from rotation around the first joint ($m_1a_{c1}^2$ term) and from rotation around the center of mass ($I_1$ term).
Comparing the two link and single link cases, shouldn't $I$ from the single link case be decomposed similarly into energies from joint rotation and center of mass rotation as well? Or is the single link case fundamentally different in that one only needs to consider energy from rotation around the joint?
It's my intuition that if the parameters of the single link case are equal to the parameters of the first link (of the two-link case), then we could conceivably write $I=I_1+m_1a_{c1}^2$. Is this correct? If not, why not? |
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Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE
(Elsevier, 2017-11)
Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ... |
The Annals of Statistics Ann. Statist. Volume 15, Number 1 (1987), 398-412. On the Effect of Substituting Parameter Estimators in Limiting $\chi^2 U$ and $V$ Statistics Abstract
Consider statistics $T_n(\lambda)$ that take the form of limiting chi-square (degenerate) $U$ or $V$ statistics. Here the phrase "limiting chi-square" means they have the same asymptotic distribution as a weighted sum of (possibly infinitely many) independent $\chi^2_1$ random variables. This paper examines the limiting distribution of $T_n(\hat{\lambda})$ and compares it to that of $T_n(\lambda)$, where $\hat{\lambda}$ denotes a consistent estimator of $\lambda$ based on the same data. Whether or not $T_n(\hat{\lambda})$ and $T_n(\lambda)$ have the same limiting distribution is primarily a question of whether or not a certain mean function has a zero derivative. Some statistics that are appropriate for testing hypotheses are used to illustrate the theory.
Article information Source Ann. Statist., Volume 15, Number 1 (1987), 398-412. Dates First available in Project Euclid: 12 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aos/1176350274 Digital Object Identifier doi:10.1214/aos/1176350274 Mathematical Reviews number (MathSciNet) MR885745 Zentralblatt MATH identifier 0617.62017 JSTOR links.jstor.org Citation
de Wet, Tertius; Randles, Ronald H. On the Effect of Substituting Parameter Estimators in Limiting $\chi^2 U$ and $V$ Statistics. Ann. Statist. 15 (1987), no. 1, 398--412. doi:10.1214/aos/1176350274. https://projecteuclid.org/euclid.aos/1176350274 |
Advanced Math Olympiad Program for AMC, AIME, USAMOCheenta Math Olympiad North America Work Group
Pause for a moment. Here is a food for thought:
What if you wanted to draw a square whose area equals the circumference of a unit circle? The side length of such a square has an incredible expression.
$$ \displaystyle{\int_{-\infty}^{+\infty} e^{\frac{-x^2}{2}} } dx $$
What deep mathematics is hidden in this beautiful relation?
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Admission Process Doubt Clearing Sessionsevery week as well. |
Jacques Féjoz
Place du Maréchal de Lattre de Tassigny, 75775 Paris Cedex 16, France,77 avenue Denfert-Rochereau, 75014 Paris, France
Université Paris-Dauphine – CEREMADE (UMR 7534),Observatoire de Paris – IMCCE (UMR 8028)
Publications:
Féjoz J.
On Action-angle Coordinates and the Poincaré Coordinates
2013, vol. 18, no. 6, pp. 703-718
Abstract
This article is a review of two related classical topics of Hamiltonian systems and celestial mechanics. The first section deals with the existence and construction of action-angle coordinates, which we describe emphasizing the role of the natural adiabatic invariants "$\oint_\gamma pdq$". The second section is the construction and properties of the Poincaré coordinates in the Kepler problem, adapting the principles of the former section, in an attempt to use known first integrals more directly than Poincaré did.
Féjoz J.
A Proof of the Invariant Torus Theorem of Kolmogorov
2012, vol. 17, no. 1, pp. 1-5
Abstract
A variant of Kolmogorov’s initial proof is given, in terms of a group of symplectic transformations and of an elementary fixed point theorem.
Chenciner A., Féjoz J.
Unchained Polygons and the $N$-body Problem
2009, vol. 14, no. 1, pp. 64-115
Abstract
We study both theoretically and numerically the Lyapunov families which bifurcate in the vertical direction from a horizontal relative equilibrium in $\mathbb{R}^3$. As explained in [1], very symmetric relative equilibria thus give rise to some recently studied classes of periodic solutions. We discuss the possibility of continuing these families globally as action minimizers in a rotating frame where they become periodic solutions with particular symmetries. A first step is to give estimates on intervals of the frame rotation frequency over which the relative equilibrium is the sole absolute action minimizer: this is done by generalizing to an arbitrary relative equilibrium the method used in [2] by V. Batutello and S. Terracini.
In the second part, we focus on the relative equilibrium of the equal-mass regular $N$-gon. The proof of the local existence of the vertical Lyapunov families relies on the fact that the restriction to the corresponding directions of the quadratic part of the energy is positive definite. We compute the symmetry groups $G_{\frac{r}{s}}(N, k, \eta)$ of the vertical Lyapunov families observed in appropriate rotating frames, and use them for continuing the families globally. The paradigmatic examples are the "Eight" families for an odd number of bodies and the "Hip-Hop" families for an even number. The first ones generalize Marchal's $P_{12}$ family for 3 bodies, which starts with the equilateral triangle and ends with the Eight [1, 3–6]; the second ones generalize the Hip-Hop family for 4 bodies, which starts from the square and ends with the Hip-Hop [1, 7, 8].
We argue that it is precisely for these two families that global minimization may be used. In the other cases, obstructions to the method come from isomorphisms between the symmetries of different families; this is the case for the so-called "chain" choreographies (see [6]), where only a local minimization property is true (except for $N = 3$). Another interesting feature of these chains is the deciding role played by the parity, in particular through the value of the angular momentum. For the Lyapunov families bifurcating from the regular $N$-gon whith $N \leqslant 6$ we check in an appendix that locally the torsion is not zero, which justifies taking the rotation of the frame as a parameter. |
What you seem to be missing is$$\sum_{i,j} w_i w_j \sigma_i \sigma_j = \left(\sum_i w_i\sigma_i\right)^2$$Now apply Jensen's inequality to get$$\left(\sum_i w_i\sigma_i\right)^2 \leq \sum_i w_i\sigma_i^2 $$QED(note that nonnegative weights is a crucial assumption here)
There are two mistakes in the code:1) In the linevt[t] = np.abs(vt[t-1] + kappa*(theta-np.abs(vt[t-1]))*dt + xi*np.sqrt(np.abs(vt[t-1]))*W_v[t])you forgot to multiply W_v[t] by np.sqrt(dt).This is the reason the volatility increases so much.2) The lineSt[t] = St[t-1]*np.exp((mu - 0.5*vt[t])*dt + np.sqrt(vt[t]*dt)*W_S[t])should beSt[t] = St[t-...
Ok, I found the problem. I was not discounting the value according to interest rate. The following modified code works fine for non-zero interest rate. When dividends rate are non zero, it is probably better to hedge using a forward contract instead of the asset itself. I will skip this for now.def hedging_portfolio(path, strike, r, d, vol, T):t = ...
"Whose price function $V$ fluctuates according to the actual market price of that derivative"—this is not true. The reason being that we are 'modeling' the derivative price (where a model is a simplified version of reality).So $V$ tells us what the derivative price would be under our model—and since this model doesn't use the actual derivative price as an ...
In one sentence, time value has to do with the probability of crossing the strike before expiration (whether from below or above). Doesn’t matter whether the crossing results in the option being in the money or not.
The answer given is mostly wrong: @msitt uses a convoluted way without explicitly mentioning it (put-call symmetry) to actually give the price of a USD Put, not of a USD Call as requested. Here is a more direct and correct approach.I will consider, as mentioned by @FinanceGuyThatCantCode, that the volatility convention is ACT/365, which is standard.I ...
The "square-root rule" for time-to-expiration only (roughly) applies when the spot price = strike price. Even in that case there is a second-order term that is a function of the risk-free rate and implied volatility.This can be seen in the Black-Scholes pricing formula: the time-to-expiration is included in a term that also varies with log(spot/strike), ...
Let's say the company was bankrupt (ie, stock price is 0). A put option effectively becomes a bond with face value equal to the strike and maturity equal to the expiration.With positive interest rates, zero coupon bonds generally become more valuable as time passes.In this extreme case, an American option is worth more because you could early ...
In its simplest form, an option is a combination of two binary options.The buyer of a call option is long of an "asset-or-nothing" binary call. I.E. if Spot>Strike, it is worth Spot; else 0. To fund that, he is selling a "cash-or-nothing" call: worth Strike if Spot>Strike, else 0.The positive value of the option obviously derives from the fact that as ...
The Black Scholes (1973) model assumes that $\mathrm{d}S_t=rS_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t$. Thus, $$S_t=S_0\exp\left(\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right).$$ Please note the factor $-\frac{1}{2}\sigma^2t$ in the exponential. If you incorporate dividends, replace $r$ by $r-q$. You do not need an extra term $\sqrt{t}$ in front of the ...
Given the variation, ATM vol = alpha * F ^(beta-1), if your stochastic process for forward price dF= alphaF^beta dW, that means your effective beta, CEV, is 1. This gives horizontal backbone of the vol surface. I think it all depends on whether this is what you expect to see - the vol surface is stickey under shocked price scenarios.
This is easy to answer with the meta theorem given in the same chapter. Here you have two sources of randomness (W and N), and one risky asset.Q1: Arbitrage generally happens when you have more assets than the number of random sources, but here it is the other way around, so the answer is yes.Q2: You have one risky asset so you can delta hedge one ...
A free course to learn quant finance is:Python for TradingA very basic and free course on Python programming. It includes everything a beginner needs to know: data structures, expressions, functions and various libraries used in financial markets. This is a detailed and comprehensive course to build a strong foundation in Python.The less pricey quant ...
Your function returning (minus) the log-likelihood seems weird to me, I would go withfunction y = findGARCH_LLy(params,S,rf)% Finds log-likelihood for the GARCH option pricing model.alpha0 = params(1);alpha1 = params(2);beta1 = params(3);lambda = params(4);N = length(S);% Define the returns (pad first return with zero)r = [0, diff(log(S))];% ... |
I'm trying to solve Exercise 18.3 of "Elements of Statistical learning" by Hastie et al. and I'd be really grateful for any hints.
Show that the fitted coefficients for the regularized multiclass logistic Regression Problem $$\max_{\{\beta_{0k},\beta_{k}\}_{1}^{K}}\left[\sum\limits_{i=1}^{N}logPr(g_{i}| x_i)-\frac{\lambda}{2}\sum\limits_{k=1}^{K}\|\beta_{k}\|_{2}^{2}\right]$$ satisfy $$\sum\limits_{k=1}^K\hat{\beta}_{kj}=0, j = 1,...,p.$$
$p$ is the number of Features, $K$ is the number of classes.
EDIT: I actually think that $\max\limits_{\{\beta_{0k},\beta_{k}\}_{1}^{K}}\left[\sum\limits_{i=1}^{N}logPr(g_{i}| x_i)-\frac{\lambda}{2}\sum\limits_{k=1}^{K}\|\beta_{k}\|_{2}^{2}\right]$ is a Lagrange optimization Problem, so derivating with respect to $x$ and $\lambda$ and then Setting to Zero should do what I Need. But turns out that I'm even getting confused by derivating! $$Pr(g_{i}| x_i) = \dfrac{exp(\beta_{g_i0}+x_i^T\beta_{g_i})}{\sum\limits_{l=1}^Kexp(\beta_{l0}+x_i^T\beta_l)}$$ and I'm not quite sure if I should compute $$\dfrac{\partial f}{\partial x_i}$$ with $i=1,...,N$ or $$\dfrac{\partial f}{\partial x_{ij}}$$ with $i=1,...,N,j=1,...,p$, $f$ being the function inside the square brackets. Well I tried both and I guess it went wrong both times as I couldn't achieve what I'm trying to proove. I didn't forget $\dfrac{\partial f}{\partial \lambda}.$ |
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
In a paper by Joos and Zeh, Z Phys B 59 (1985) 223, they say:This 'coming into being of classical properties' appears related to what Heisenberg may have meant by his famous remark [7]: 'Die "Bahn" entsteht erst dadurch, dass wir sie beobachten.'Google Translate says this means something ...
@EmilioPisanty Tough call. It's technical language, so you wouldn't expect every German speaker to be able to provide a correct interpretation—it calls for someone who know how German is used in talking about quantum mechanics.
Litmus are a London-based space rock band formed in 2000 by Martin (bass guitar/vocals), Simon (guitar/vocals) and Ben (drums), joined the following year by Andy Thompson (keyboards, 2001–2007) and Anton (synths). Matt Thompson joined on synth (2002–2004), while Marek replaced Ben in 2003. Oli Mayne (keyboards) joined in 2008, then left in 2010, along with Anton. As of November 2012 the line-up is Martin Litmus (bass/vocals), Simon Fiddler (guitar/vocals), Marek Bublik (drums) and James Hodkinson (keyboards/effects). They are influenced by mid-1970s Hawkwind and Black Sabbath, amongst others.They...
@JohnRennie Well, they repeatedly stressed their model is "trust work time" where there are no fixed hours you have to be there, but unless the rest of my team are night owls like I am I will have to adapt ;)
I think u can get a rough estimate, COVFEFE is 7 characters, probability of a 7-character length string being exactly that is $(1/26)^7\approx 1.2\times 10^{-10}$ so I guess you would have to type approx a billion characters to start getting a good chance that COVFEFE appears.
@ooolb Consider the hyperbolic space $H^n$ with the standard metric. Compute $$\inf\left\{\left(\int u^{2n/(n-2)}\right)^{-(n-2)/n}\left(4\frac{n-1}{n-2}\int|\nabla u|^2+\int Ru^2\right): u\in C^\infty_c\setminus\{0\}, u\ge0\right\}$$
@BalarkaSen sorry if you were in our discord you would know
@ooolb It's unlikely to be $-\infty$ since $H^n$ has bounded geometry so Sobolev embedding works as expected. Construct a metric that blows up near infinity (incomplete is probably necessary) so that the inf is in fact $-\infty$.
@Sid Eating glamorous and expensive food on a regular basis and not as a necessity would mean you're embracing consumer fetish and capitalism, yes. That doesn't inherently prevent you from being a communism, but it does have an ironic implication.
@Sid Eh. I think there's plenty of room between "I think capitalism is a detrimental regime and think we could be better" and "I hate capitalism and will never go near anything associated with it", yet the former is still conceivably communist.
Then we can end up with people arguing is favor "Communism" who distance themselves from, say the USSR and red China, and people who arguing in favor of "Capitalism" who distance themselves from, say the US and the Europe Union.
since I come from a rock n' roll background, the first thing is that I prefer a tonal continuity. I don't like beats as much as I like a riff or something atmospheric (that's mostly why I don't like a lot of rap)
I think I liked Madvillany because it had nonstandard rhyming styles and Madlib's composition
Why is the graviton spin 2, beyond hand-waiving, sense is, you do the gravitational waves thing of reducing $R_{00} = 0$ to $g^{\mu \nu} g_{\rho \sigma,\mu \nu} = 0$ for a weak gravitational field in harmonic coordinates, with solution $g_{\mu \nu} = \varepsilon_{\mu \nu} e^{ikx} + \varepsilon_{\mu \nu}^* e^{-ikx}$, then magic? |
The bounded variable hierarchy collapses to a fixed finite level, because of the existence of definable pairing functions.
In more detail, I will use the Cantor pairing function$$\def\p#1{\langle#1\rangle}\def\N{\mathbb N}\let\eq\leftrightarrow\p{x,y}=\frac{(x+y)(x+y+1)}2+x,$$which is a bijection $\N^2\to\N$. Let $l(z)$, $r(z)$ denote the corresponding projections, so that$$l(\p{x,y})=x,\qquad r(\p{x,y})=y,\qquad\p{l(z),r(z)}=z.$$Since $\p{x,y}$ is a polynomial with rational coefficients, $\p{x,y}=z$ is equivalent to a quantifier-free (hence $3$-variable) formula, namely$$\p{x,y}=z\eq z+z=(x+y)\cdot(x+y+1)+x,$$and consequently the functions $l$ and $r$ are definable by $\Delta_0$ formulas using $3$ distinct variables:$$l(z)=x\eq\exists y\le z\,\p{x,y}=z,$$and similarly for $r$.
Proposition: Every $\Delta_0$ formula in one free variable is equivalent to a $\Delta_0$ formula using only $3$ distinct variables.
Proof:
It will be convenient to preprocess the formulas first. Let us call a bounded quantifier $\exists y\le t(\vec x)\,\theta(\vec x,y)$
safe if$$\bigl(\exists y\le t(\vec x)\,\theta(\vec x,y)\bigr)\eq\bigl(\exists y\,\theta(\vec x,y)\bigr),$$in which case we can replace the bound $t(\vec x)$ with any larger bound without affecting the truth of the formula. Similarly for universal quantifiers. By replacing $\exists y\le t(\vec x)\,\theta$ with $\exists y\le t(\vec x)\,(y\le t(\vec x)\land\theta)$, and unwinding terms with the help of extra existential quantifiers, it is easy to see that any $\Delta_0$ formula is equivalent to a $\Delta_0$ formula $\theta(\vec x)$ such that
all quantifiers in $\theta$ are safe,
the index $i$ of any variable $x_i$ quantified in any subformula $\xi$ of $\theta$ is higher than the indices of all free variables of $\xi$, and
all atomic subformulas of $\theta$ (except quantifier bounds) are of the forms $x_i=x_j$, $x_i+x_j=x_k$, or $x_i\cdot x_j=x_k$.
Let us call such formulas
special.
Claim: For any special formula $\theta(x_0,\dots,x_n)$, there is a formula $\xi(z)$ using only $3$ distinct variables such that
$$\xi(z)\eq\theta\bigl(r(l^n(z)),\dots,r(l(z)),r(z)\bigr).$$
Note that the Claim implies the Proposition: if $\xi(z)\eq\theta(r(z))$ can be written using $3$ variables, then so can$$\theta(x)\eq\exists z\le x^2\,\bigl(z=\p{0,x}\land\xi(z)\bigr).$$
We will prove the Claim by induction on the complexity of $\theta$.
The induction steps for Boolean connectives are trivial. The step for bounded quantifiers is also easy: if $\theta(x_0,\dots,x_n)$ is $\exists x_{n+1}\le t(\vec x)\,\theta'(x_0,\dots,x_{n+1})$, where the quantifier is safe, and $\xi'$ is a $3$-variable formula equivalent to $\theta'(r(l^{n+1}(z)),\dots,r(l(z)),r(z))$, then$$\begin{align*}\theta(r(l^n(z)),\dots,r(z))&\eq\exists x_{n+1}\,\xi'(\p{z,x_{n+1}})\\&\eq\exists w\,(l(w)=z\land\xi'(w))\\&\eq\exists w\le s(z)\,(l(w)=z\land\xi'(w))\end{align*}$$for a suitable term $s(z)$.
It remains to prove the Claim for special atomic formulas. First, notice that since $l$ and $r$ are definable using $3$ variables, the same holds for any finite composition of these functions: for example,$$x=l^{n+1}(z)\eq\exists y\le z\,(x=l(y)\land y=l^n(z)),$$where we proceed to expand the inner formulas. This immediately takes care of atomic formulas $x_i=x_j$, as$$r(l^i(z))=r(l^j(z))\eq\exists x\le z\,(x=r(l^i(z))\land x=r(l^j(z))).$$For the formulas $x_i+x_j=x_k$ (and similarly $x_i\cdot x_j=x_k$), we can proceed as follows:$$r(l^i(z))+r(l^j(z))=r(l^k(z))\eq\exists w\le t(z)\,(l(l(w))=r(l^i(z))\land r(l(w))=r(l^j(z))\land r(w)=r(l^k(z))\land\exists x,z\le w\,(w=\p{\p{x,z},x+z})).$$(Note that we recycled the $z$ variable.) Here, $l(l(w))=r(l^i(z))$ can be written as $\exists x\le w\,(x=l(l(w))\land x=r(l^i(z)))$, which we know how to handle, and similarly for the other two conjuncts. Finally,$$w=\p{\p{x,z},x+z}$$can be written as an identity using no further variables, as $\p{\p{x,z},x+z}$ is a polynomial with rational coefficients in $x$ and $z$ (and $x+z$). QED
Corollary: Every $\Delta_0$ formula in $n$ free variables is equivalent to a $\Delta_0$ formula using $\max\{3,n+1\}$ distinct variables.
Comments:
We only used elementary properties of the pairing function, hence the equivalence is not just true in $\N$, it is provable in a very weak theory: IOpen is certainly enough (probably $\mathrm{PA}^-$ will suffice with a little care).
The result still holds for bounded formulas in a richer language with extra binary functions or relations. Even more generally, if we expand the language with $k$-ary functions and relations with $k>2$, the result holds with $k+1$ in place of $3$.
The transformation more-or-less preserves quantifier complexity: for any $i\ge1$, an $E_i$ formula in $n$ free variables is equivalent to an $E_i$ formula using $\max\{3,n+1\}$ distinct variables, and similarly for $U_i$. However, we need to make sure the $E_i$ and $U_i$ classes are defined in such a way as to explicitly allow conjunctions and disjunctions; of course, we cannot reduce the number of distinct variables in prenex formulas. |
Background:
The complete infinite binary tree (CIBT) has path cardinality of continuum size, where path cardinality refers to the size of the set of all paths from the root.
If we consider the random infinite binary tree with constant degree $1+\epsilon$, for an infinitesimal constant $\epsilon$, we get a partial binary tree, which again has path cardinality of continuum size. We can see this by using a pull-down procedure, cutting out any long non-branching sub-path and reconnecting the ends, to show that this sparsely branching tree is isomorphic with the CIBT (see here: Path cardinality for random $(a+b)$-ary infinite trees). It is tempting to believe that all random binary trees, i.e. including those of completely random variant degree almost surely above one, can be pulled down in this manner to become isomorphic with CIBT.
Sub-question: can all partial infinite binary trees of continuum path cardinality be pulled-down, so at least a subset can be put in isomorphic correspondence with the complete infinite binary tree?
Consider now a mapping of CIBT to the $(x,y)$ plane, wherein, the nodes, $(a_i,b_i)$, are as follows (read from root up):
....
Level-4: $(-2-\frac{1}{2}- \frac{1}{4}- \frac{1}{8}, 1+ \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8})$ ……..
Level-3: $(-2-\frac{1}{2}- \frac{1}{4}, 1+ \frac{1}{2}+ \frac{1}{4})$ ……..
Level-2: $(-2- \frac{1}{2},1+ \frac{1}{2})$ $(-2+ \frac{1}{2}, 1+ \frac{1}{2})$ $(2- \frac{1}{2},1+ \frac{1}{2}) (2+ \frac{1}{2}, 1+ \frac{1}{2})$
Level-1: $(-2,1) (2,1)$
root: $(0,0)$
Where the nodes are connected in order, with the usual edges of a binary tree.
Since $\sum_{i=0}^{\infty} \frac{1}{2^i} = 2$, we have now mapped the CBIT into the $(x,y)$ plane so that it forms a visual tree, with non-colliding branches, that extend uniformly from $(0,0)$ upwards, becoming increasingly dense as it approaches its limit, the number line spanning from $(-2,2)$ to $(2,2)$.
Let us call this number line the boundary of CIBT.
Since each point on the boundary corresponds to exactly one infinite path in CIBT, we can conclude that the boundary of CIBT is of continuum size, and is indeed an open real number line of length 4. We could of course instead transform to a finite height, infinitely spanning tree with a boundary covering the entire range of the real number line.
Sub-question: will any binary tree of continuum path cardinality have a boundary containing at least one open subset of dimension 1?
Consider now a procedure to collapse a tree from the top, wherein a node collapses, i.e. is removed from the tree along with its connected edges, if both its children nodes have already collapsed. We define the collapse of a tree, as the recursive collapse that happens once a set of elected leaf vertices are collapsed.
Question:
Let us select a set, F, as a fractal one-dimensional set, uniformly dense, with a Hausdorff dimension strictly less than 1. Let us for example set F to be of dimension equal to the inverse of a googol (i.e. pretty close to zero)
We now impose F on the boundary of CIBT, and collapse CIBT with F, obtaining the F-collapse of CBIT, which we can call FIBT.
We have derived FIBT, a binary tree with the boundary F.
Sub-question: Is FIBT well-defined and unique? (even if informally stated here)
The main question is now: What is the path cardinality of FIBT?
Speculations:
Obviously, FIBT has path cardinality of at most Continuum size.
Clearly, FIBT has path cardinality larger than countably infinite, as it has at least the cardinality of a set of positive Hausdorff dimension.
Now, let us consider whether we can use the pull-down procedure to show that it can be put in one to one correspondence with CIBT.
We can first observe that the infinite random tree of degree 1+e has the same boundary as CIBT, a boundary of uniform dimension 1. “Viewed from “the root” it is very sparse, but it becomes very dense at the boundary. FIBT has the reverse “appearance”, dense at the root, and increasingly sparse at the boundary.
A topological argument can proceed by viewing FIBT as an elastic structure, with nodes, connected by rubber bands, which we can cut and reconnect. Firstly, it is clear that cutting out long non-branching sub-paths and reconnecting at each end (basic pull-down) will not make any difference at the boundary. Secondly, we can then hope to cut and move infinite sub-trees, reconnecting them at other vertices, in order to show that a resulting part of the transformed FIBT achieves a unitary dimension at the boundary. However, an infinite number of additions of subsets, of the same dimension, will result in a set of the same dimension, so there seems to be no hope that any process of cutting and reconnecting can produce a set with different behavior at the boundary.
Since the pull-down procedure or any infinite cut-and-reconnect process cannot put FIBT in one-to-one correspondence with CIBT, it is tempting to think that FIBT has below continuum-size path cardinality. |
Refine
If \(A\) generates a bounded cosine function on a Banach space \(X\) then the negative square root \(B\) of \(A\) generates a holomorphic semigroup, and this semigroup is the conjugate potential transform of the cosine function. This connection is studied in detail, and it is used for a characterization of cosine function generators in terms of growth conditions on the semigroup generated by \(B\). This characterization relies on new results on the inversion of the vector-valued conjugate potential transform.
Let \(X\) be a Banach lattice. Necessary and sufficient conditions for a linear operator \(A:D(A) \to X\), \(D(A)\subseteq X\), to be of positive \(C^0\)-scalar type are given. In addition, the question is discussed which conditions on the Banach lattice imply that every operator of positive \(C^0\)-scalar type is necessarily of positive scalar type.
In the scalar case one knows that a complex normalized function of boundedvariation \(\phi\) on \([0,1]\) defines a unique complex regular Borel measure\(\mu\) on \([0,1]\). In this note we show that this is no longer true in generalin the vector valued case, even if \(\phi\) is assumed to be continuous. Moreover, the functions \(\phi\) which determine a countably additive vectormeasure \(\mu\) are characterized.
\(C^0\)-scalar-type spectrality criterions for operators \(A\), whose resolvent set contains the negative reals, are provided. The criterions are given in terms of growth conditions on the resolvent of \(A\) and the semi-group generated by \(A\).These criterions characterize scalar-type operators on the Banach space \(X\), if and only if \(X\) has no subspace isomorphic to the space of complex null-sequences.
In the Banach space co there exists a continuous function of bounded semivariation which does not correspond to a countably additive vector measure. This result is in contrast to the scalar case, and it has consequences for the characterization of scalar-type operators. Besides this negative result we introduce the notion of functions of unconditionally bounded variation which are exactly the generators of countably additive vector measures.
The following two norms for holomorphic functions \(F\), defined on the right complex half-plane \(\{z \in C:\Re(z)\gt 0\}\) with values in a Banach space \(X\), are equivalent: \[\begin{eqnarray*} \lVert F \rVert _{H_p(C_+)} &=& \sup_{a\gt0}\left( \int_{-\infty}^\infty \lVert F(a+ib) \rVert ^p \ db \right)^{1/p} \mbox{, and} \\ \lVert F \rVert_{H_p(\Sigma_{\pi/2})} &=& \sup_{\lvert \theta \lvert \lt \pi/2}\left( \int_0^\infty \left \lVert F(re^{i \theta}) \right \rVert ^p\ dr \right)^{1/p}.\end{eqnarray*}\] As a consequence, we derive a description of boundary values ofsectorial holomorphic functions, and a theorem of Paley-Wiener typefor sectorial holomorphic functions. |
Mass
The big issue here lies in determining the mass of these black holes. On the one hand, the traditional primordial black holes you mention - formed by density perturbations - occupy a relatively low-mass regime. Constraints from a variety of observations indicate a peak in the mass distribution at $\sim10^{17}\text{ kg}$ depending on the conditions at the time of formation, as well as the behavior of inflation. These black holes should still be around today, even if Hawking radiation had dissipated them.
Figure 3, Primordial Black Holes in the Inflationary Universe, Masahiro Kawasaki. A plot of the primordial black hole mass distribution, with various constraints shaded out. Notice the peaks. I should note that new constraints are continuously being added; see, for example, a recent press release from the Subaru Telescope.
On the other hand, it has been suggested that black holes of a few tens of solar masses could make up the dark matter we observe - a variant of the Massive Compact Halo Object (MACHO) hypothesis, and primordial clouds of $10^{4-5}M_{\odot}$ could have collapsed to form black holes. If we take both populations into account as "primordial", we have an enormous mass range to consider.
At the lower end, we see black holes with Schwarzschild radii of a few tenths of a nanometer. At the upper end, we find Schwarzschild radii of 46 Earth radii. To have a radius of, say, 10% of Earths, we have to limit ourselves to about $200M_{\odot}$, possible from the collapse of a low-mass Population III star. However, this is still going to pose problems for your planet, and much of its interior could be accreted on short timescales. Therefore, I suggest staying with the classic primordial black hole formed via density fluctuations, with a much nicer mass of $10^{17}\text{ kg}$. After all, we don't want to swallow the planet!
Surface effects
At this point, we can run through the calculations pretty quickly. Using a Newtonian approximation (which I think is justified this far from the black hole), we find that the surface gravity is$$g=\frac{GM_{\text{BH}}}{R_{\oplus}^2}\approx1.67\times10^{-8}g_{\text{Earth}}$$where $g_{\text{Earth}}$ is the surface gravity on Earth; our result is so small that it wouldn't be felt at the surface. On the other hand, deep inside the planet, there
might be problems, as the black hole would accrete matter. However, as might happen with a black hole at the center of the Sun, the radiation pressure induced could stop a collapse. So this might be stable over reasonable timescales. For the $200M_{\odot}$ black hole, of course, we see that the surface gravity is $6.66\times10^7g_{\text{Earth}}$. That's way too high.
Finally, the time dilation on the surface would be$$t_{\text{surf}}=t_0\sqrt{1-\frac{r_s}{R_{\oplus}}}\approx t_0$$because the term inside the square root is approximately 1. In other words, there's negligible time dilation. For the $200M_{\odot}$ black hole, however, $t_{\text{surf}}=0.96t_0$ - not terribly significant, but not a huge amount, either.
To be honest, on the surface of the planet, there shouldn't be major effects from time dilation or the extra mass of the black hole. Of course, don't take this as an endorsement of the scheme; you've still got a black hole at the center of your planet, and that rarely turns out well. |
Kakeya problem
Define a
Kakeya set to be a subset [math]A[/math] of [math][3]^n\equiv{\mathbb F}_3^n[/math] that contains an algebraic line in every direction; that is, for every [math]d\in{\mathbb F}_3^n[/math], there exists [math]a\in{\mathbb F}_3^n[/math] such that [math]a,a+d,a+2d[/math] all lie in [math]A[/math]. Let [math]k_n[/math] be the smallest size of a Kakeya set in [math]{\mathbb F}_3^n[/math].
Clearly, we have [math]k_1=3[/math], and it is easy to see that [math]k_2=7[/math]. Using a computer, it is not difficult to find that [math]k_3=13[/math] and [math]k_4\le 27[/math]. Indeed, it seems likely that [math]k_4=27[/math] holds, meaning that in [math]{\mathbb F}_3^4[/math] one cannot get away with just [math]26[/math] elements.
General lower bounds
Trivially, we have
[math]k_n\le k_{n+1}\le 3k_n[/math].
Next, the Cartesian product of two Kakeya sets is another Kakeya set; hence,
[math]k_{n+m} \leq k_m k_n[/math],
implying that [math]k_n^{1/n}[/math] converges to a limit as [math]n[/math] goes to infinity.
From a paper of Dvir, Kopparty, Saraf, and Sudan it follows that [math]k_n \geq 3^n / 2^n[/math], but this is superseded by the estimates given below.
We have
[math]k_n(k_n-1)\ge 3(3^n-1)[/math]
since for each [math]d\in {\mathbb F}_3^r\setminus\{0\}[/math] there are at least three ordered pairs of elements of a Kakeya set with difference [math]d[/math].
For instance, we can use the "bush" argument. There are [math]N := (3^n-1)/2[/math] different directions. Take a line in every direction, let E be the union of these lines, and let [math]\mu[/math] be the maximum multiplicity of these lines (i.e. the largest number of lines that are concurrent at a point). On the one hand, from double counting we see that E has cardinality at least [math]3N/\mu[/math]. On the other hand, by considering the "bush" of lines emanating from a point with multiplicity [math]\mu[/math], we see that E has cardinality at least [math]2\mu+1[/math]. If we minimise [math]\max(3N/\mu, 2\mu+1)[/math] over all possible values of [math]\mu[/math] one obtains approximately [math]\sqrt{6N} \approx 3^{(n+1)/2}[/math] as a lower bound of |E|, which is asymptotically better than [math](3/2)^n[/math].
Or, we can use the "slices" argument. Let [math]A, B, C \subset ({\Bbb Z}/3{\Bbb Z})^{n-1}[/math] be the three slices of a Kakeya set E. We can form a graph G between A and B by connecting A and B by an edge if there is a line in E joining A and B. The restricted sumset [math]\{a+b: (a,b) \in G \}[/math] is essentially C, while the difference set [math]\{a-b: (a-b) \in G \}[/math] is all of [math]({\Bbb Z}/3{\Bbb Z})^{n-1}[/math]. Using an estimate from this paper of Katz-Tao, we conclude that [math]3^{n-1} \leq \max(|A|,|B|,|C|)^{11/6}[/math], leading to the bound [math]|E| \geq 3^{6(n-1)/11}[/math], which is asymptotically better still.
General upper bounds
We have
[math]k_n\le 2^{n+1}-1[/math]
since the set of all vectors in [math]{\mathbb F}_3^n[/math] such that at least one of the numbers [math]1[/math] and [math]2[/math] is missing among their coordinates is a Kakeya set.
Question: can the upper bound be strengthened to [math]k_{n+1}\le 2k_n+1[/math]?
Another construction uses the "slices" idea and a construction of Imre Ruzsa. Let [math]A, B \subset [3]^n[/math] be the set of strings with [math]n/3+O(\sqrt{n})[/math] 1's, [math]2n/3+O(\sqrt{n})[/math] 0's, and no 2's; let [math]C \subset [3]^n[/math] be the set of strings with [math]2n/3+O(\sqrt{n})[/math] 2's, [math]n/3+O(\sqrt{n})[/math] 0's, and no 1's, and let [math]E = \{0\} \times A \cup \{1\} \times B \cup \{2\} \times C[/math]. From Stirling's formula we have [math]|E| = (27/4 + o(1))^{n/3}[/math]. Now I claim that for most [math]t \in [3]^{n-1}[/math], there exists an algebraic line in the direction (1,t). Indeed, typically t will have [math]n/3+O(\sqrt{n})[/math] 0s, [math]n/3+O(\sqrt{n})[/math] 1s, and [math]n/3+O(\sqrt{n})[/math] 2s, thus [math]t = e + 2f[/math] where e and f are strings with [math]n/3 + O(\sqrt{n})[/math] 1s and no 2s, with the 1-sets of e and f being disjoint. One then checks that the line [math](0,f), (1,e), (2,2e+2f)[/math] lies in E.
This is already a positive fraction of directions in E. One can use the random rotations trick to get the rest of the directions in E (losing a polynomial factor in n).
Putting all this together, I think we have
[math](3^{6/11} + o(1))^n \leq k_n \leq ( (27/4)^{1/3} + o(1))^n[/math]
or
[math](1.8207\ldots+o(1))^n \leq k_n \leq (1.88988+o(1))^n[/math] |
Hi, Can someone provide me some self reading material for Condensed matter theory? I've done QFT previously for which I could happily read Peskin supplemented with David Tong. Can you please suggest some references along those lines? Thanks
@skullpatrol The second one was in my MSc and covered considerably less than my first and (I felt) didn't do it in any particularly great way, so distinctly average. The third was pretty decent - I liked the way he did things and was essentially a more mathematically detailed version of the first :)
2. A weird particle or state that is made of a superposition of a torus region with clockwise momentum and anticlockwise momentum, resulting in one that has no momentum along the major circumference of the torus but still nonzero momentum in directions that are not pointing along the torus
Same thought as you, however I think the major challenge of such simulator is the computational cost. GR calculations with its highly nonlinear nature, might be more costy than a computation of a protein.
However I can see some ways approaching it. Recall how Slereah was building some kind of spaceitme database, that could be the first step. Next, one might be looking for machine learning techniques to help on the simulation by using the classifications of spacetimes as machines are known to perform very well on sign problems as a recent paper has shown
Since GR equations are ultimately a system of 10 nonlinear PDEs, it might be possible the solution strategy has some relation with the class of spacetime that is under consideration, thus that might help heavily reduce the parameters need to consider to simulate them
I just mean this: The EFE is a tensor equation relating a set of symmetric 4 × 4 tensors. Each tensor has 10 independent components. The four Bianchi identities reduce the number of independent equations from 10 to 6, leaving the metric with four gauge fixing degrees of freedom, which correspond to the freedom to choose a coordinate system.
@ooolb Even if that is really possible (I always can talk about things in a non joking perspective), the issue is that 1) Unlike other people, I cannot incubate my dreams for a certain topic due to Mechanism 1 (consicous desires have reduced probability of appearing in dreams), and 2) For 6 years, my dream still yet to show any sign of revisiting the exact same idea, and there are no known instance of either sequel dreams nor recurrence dreams
@0celo7 I felt this aspect can be helped by machine learning. You can train a neural network with some PDEs of a known class with some known constraints, and let it figure out the best solution for some new PDE after say training it on 1000 different PDEs
Actually that makes me wonder, are the space of all coordinate choices more than all possible moves of Go?
enumaris: From what I understood from the dream, the warp drive showed here may be some variation of the alcuberrie metric with a global topology that has 4 holes in it whereas the original alcuberrie drive, if I recall, don't have holes
orbit stabilizer: h bar is my home chat, because this is the first SE chat I joined. Maths chat is the 2nd one I joined, followed by periodic table, biosphere, factory floor and many others
Btw, since gravity is nonlinear, do we expect if we have a region where spacetime is frame dragged in the clockwise direction being superimposed on a spacetime that is frame dragged in the anticlockwise direction will result in a spacetime with no frame drag? (one possible physical scenario that I can envision such can occur may be when two massive rotating objects with opposite angular velocity are on the course of merging)
Well. I'm a begginer in the study of General Relativity ok? My knowledge about the subject is based on books like Schutz, Hartle,Carroll and introductory papers. About quantum mechanics I have a poor knowledge yet.
So, what I meant about "Gravitational Double slit experiment" is: There's and gravitational analogue of the Double slit experiment, for gravitational waves?
@JackClerk the double slits experiment is just interference of two coherent sources, where we get the two sources from a single light beam using the two slits. But gravitational waves interact so weakly with matter that it's hard to see how we could screen a gravitational wave to get two coherent GW sources.
But if we could figure out a way to do it then yes GWs would interfere just like light wave.
Thank you @Secret and @JohnRennie . But for conclude the discussion, I want to put a "silly picture" here: Imagine a huge double slit plate in space close to a strong source of gravitational waves. Then like water waves, and light, we will see the pattern?
So, if the source (like a Black Hole binary) are sufficent away, then in the regions of destructive interference, space-time would have a flat geometry and then with we put a spherical object in this region the metric will become schwarzschild-like.
if**
Pardon, I just spend some naive-phylosophy time here with these discussions**
The situation was even more dire for Calculus and I managed!
This is a neat strategy I have found-revision becomes more bearable when I have The h Bar open on the side.
In all honesty, I actually prefer exam season! At all other times-as I have observed in this semester, at least-there is nothing exciting to do. This system of tortuous panic, followed by a reward is obviously very satisfying.
My opinion is that I need you kaumudi to decrease the probabilty of h bar having software system infrastructure conversations, which confuse me like hell and is why I take refugee in the maths chat a few weeks ago
(Not that I have questions to ask or anything; like I said, it is a little relieving to be with friends while I am panicked. I think it is possible to gauge how much of a social recluse I am from this, because I spend some of my free time hanging out with you lot, even though I am literally inside a hostel teeming with hundreds of my peers)
that's true. though back in high school ,regardless of code, our teacher taught us to always indent your code to allow easy reading and troubleshooting. We are also taught the 4 spacebar indentation convention
@JohnRennie I wish I can just tab because I am also lazy, but sometimes tab insert 4 spaces while other times it inserts 5-6 spaces, thus screwing up a block of if then conditions in my code, which is why I had no choice
I currently automate almost everything from job submission to data extraction, and later on, with the help of the machine learning group in my uni, we might be able to automate a GUI library search thingy
I can do all tasks related to my work without leaving the text editor (of course, such text editor is emacs). The only inconvenience is that some websites don't render in a optimal way (but most of the work-related ones do)
Hi to all. Does anyone know where I could write matlab code online(for free)? Apparently another one of my institutions great inspirations is to have a matlab-oriented computational physics course without having matlab on the universities pcs. Thanks.
@Kaumudi.H Hacky way: 1st thing is that $\psi\left(x, y, z, t\right) = \psi\left(x, y, t\right)$, so no propagation in $z$-direction. Now, in '$1$ unit' of time, it travels $\frac{\sqrt{3}}{2}$ units in the $y$-direction and $\frac{1}{2}$ units in the $x$-direction. Use this to form a triangle and you'll get the answer with simple trig :)
@Kaumudi.H Ah, it was okayish. It was mostly memory based. Each small question was of 10-15 marks. No idea what they expect me to write for questions like "Describe acoustic and optic phonons" for 15 marks!! I only wrote two small paragraphs...meh. I don't like this subject much :P (physical electronics). Hope to do better in the upcoming tests so that there isn't a huge effect on the gpa.
@Blue Ok, thanks. I found a way by connecting to the servers of the university( the program isn't installed on the pcs on the computer room, but if I connect to the server of the university- which means running remotely another environment, i found an older version of matlab). But thanks again.
@user685252 No; I am saying that it has no bearing on how good you actually are at the subject - it has no bearing on how good you are at applying knowledge; it doesn't test problem solving skills; it doesn't take into account that, if I'm sitting in the office having forgotten the difference between different types of matrix decomposition or something, I can just search the internet (or a textbook), so it doesn't say how good someone is at research in that subject;
it doesn't test how good you are at deriving anything - someone can write down a definition without any understanding, while someone who can derive it, but has forgotten it probably won't have time in an exam situation. In short, testing memory is not the same as testing understanding
If you really want to test someone's understanding, give them a few problems in that area that they've never seen before and give them a reasonable amount of time to do it, with access to textbooks etc. |
$SU(2)$ is the covering group of $SO(3)$. What does it mean and does it have a physical consequence?
I heard that this fact is related to the description of bosons and fermions. But how does it follow from the fact that $SU(2)$ is the double cover of $SO(3)$?
Great, important question. Here's the basic logic:
We start with Wigner's Theorem which tells us that a symmetry transformation on a quantum system can be written, up to phase, as either a unitary or anti-unitary operator on the Hilbert space $\mathcal H$ of the system.
It follows that if we want to represent a Lie group $G$ of symmetries of a system via transformations on the Hilbert space, then we must do so with a
projectiveunitary representation of the Lie group $G$. The projective part comes from the fact that the transformations are unitary or anti-unitary "up to phase," namely we represent such symmetries with a mapping $U:G\to \mathscr U(\mathcal H)$ such that for each $g_1,g_2\in G $, there exists a phase $c(g_1, g_2)$ such that \begin{align} U(g_1g_2) = c(g_1, g_2) U(g_1) U(g_2) \end{align} where $\mathscr U(\mathcal H)$ is the group of unitary operators on $\mathcal H$. In other words, a projective unitary representation is just an ordinary unitary representation with an extra phase factor that prevents it from being an honest homomorphism.
Working with projective representations isn't as easy as working with ordinary representations since they have the pesky phase factor $c$, so we try to look for ways of avoiding them. In some cases, this can be achieved by noting that the projective representations of a group $G$ are equivalent to the ordinary representations of $G'$ its
universal covering group, and in this case, we therefore elect to examine the representations of the universal cover instead.
In the case of $\mathrm{SO}(3)$, the group of rotations, we notice that its universal cover, which is often called $\mathrm{Spin}(3)$, is isomorphic to $\mathrm{SU}(2)$, and that the projective representations of $\mathrm{SO}(3)$ match the ordinary representations of $\mathrm{SU}(2)$, so we elect to examine the ordinary representations of $\mathrm{SU}(2)$ since it's more convenient.
This is all very physically important. If we had only considered the ordinary representations of $\mathrm{SO}(3)$, then we would have missed the "half-integer spin" representations, namely those that arise when considering rotations on fermionic systems. So, we must be careful to consider projective representations, and this naturally leads to looking for the universal cover.
Note: The same sort of thing happens with the Lorentz group in relativistic quantum theories. We consider projective representations of $\mathrm{SO}(3,1)$ because Wigner says we ought to, and this naturally leads us to consider its universal cover $\mathrm{SL}(2,\mathbb C)$.
After the answers by joshphysics and user37496, it seems to me that a last remark remains.
The quantum relevance of the universal covering Lie group in my opinion is (also) due to a fundamental theorem by Nelson. That theorem relates
Lie algebras of symmetric operators with unitary representations of a certain Lie group generated by those operators. The involved Lie group, in this discussion, is always a universal covering.
In quantum theories one often encounters a set of operators $\{A_i\}_{i=1,\ldots, N}$ on a common Hilbert space ${\cal H}$ such that:
(1) They are symmetric (i.e. defined on a dense domain $D(A_i)\subset {\cal H}$ where $\langle A\psi|\phi\rangle = \langle \psi|A\phi\rangle$)
and
(2) they enjoy the
commutation relations of some Lie algebra $\ell$:$$[A_i,A_j]= \sum_{k=1}^N iC^k_{ij}A_k$$on a common invariant domain ${\cal D}\subset {\cal H}$.
As is known, given an abstract Lie algebra $\ell$ there is (up to Lie group isomorphisms) a unique
simply connected Lie group ${\cal G}_\ell$ such that its Lie algebra coincide with $\ell$. ${\cal G}_\ell$ turns out to be the universal covering of all the other Lie groups whose Lie algebra is $\ell$ itself.
All those groups, in a neighbourhood of the identity are isomorphic to a corresponding neighbourhood of the identity of ${\cal G}_\ell$. (As an example just consider the simply connected $SU(2)$ that is the universal covering of $SO(3)$) so that they share the same Lie algebra and are locally identical and differences arise far from the neutral element.
If (1) and (2) hold, the natural question is:
Is there a strongly continuous unitary representation ${\cal G} \ni g \mapsto U_g$ of some Lie group $\cal G$ just admitting $\ell$ as its Lie algebra, such that $$U_{g_i(t)} = e^{-it \overline{A_i}}\:\: ?\qquad (3)$$ Where $t\mapsto g_i(t)$ is the one-parameter Lie subgroup of $\cal G$ generated by (the element $a_i$ of $\ell$ corresponding to) $A_i$ and $\overline{A_i}$ is some self-adjoint extension of $A_i$.
If it is the case, $\cal G$ is a continuous symmetry group for the considered physical system, the self adjoint operators $\overline{A_i}$ represent physically relevant observables. If time evolution is included in the center of the group (i.e. the Hamiltonian is a linear combination of the $A_i$s and commutes with each of them) all these observables are
conserved quantities.Otherwise the situation is a bit more complicated, nevertheless one can define conserved quantities parametrically depending on time and belonging to the Lie algebra of the representation (think of the boost generators when $\cal G$ is $SL(2,\mathbb C)$).
Well, the fundamental theorem by Nelson has the following statement.
THEOREM (Nelson) Consider a set of operators $\{A_i\}_{i=1,\ldots, N}$ on a common Hilbert space ${\cal H}$ satisfying (1) and (2) above. If ${\cal D}$ in (2) is a dense subspace such that the symmetric operator$$\Delta := \sum_{i=1}^N A_i^2$$is essentially self-adjoint on $\cal D$ (i.e. its adjoint is self-adjoint or, equivalently, $\Delta$ admits a unique self-adjoint extension, or equivalently its closure $\overline{\Delta}$ is self-adjoint), then: (a) Every $A_i$ is essentially self-adjoint on $\cal D$, and (b) there exists a strongly continuous unitary representation on $\cal H$ of the unique simply connected Lie group ${\cal G}_\ell$ admitting $\ell$ as Lie algebra, completely defined by the requirements:$$U_{g_i(t)} = e^{-it \overline{A_i}}\:\:,$$ where $t\mapsto g_i(t)$ is the one-parameter Lie subgroup of ${\cal G}_\ell$ generated by (the element $a_i$ of $\ell$ corresponding to) $A_i$ and $\overline{A_i}$ is the unique self-adjoint extension of $A_i$ coinciding to $A_i^*$ and with the closure of $A_i$.
Notice that the representation is automatically unitary and not projective unitary: No annoying phases appear.
The simplest example is that of operators $J_x,J_y,J_z$. It is easy to prove that $J^2$ is essentially self adjoint on the set spanned by vectors $|j,m, n\rangle$. The point is that one gets this way unitary representations of $SU(2)$ and not $SO(3)$, since the former is the unique simply connected Lie group admitting the algebra of $J_k$ as its own Lie algebra.
As another application, consider $X$ and $P$ defined on ${\cal S}(\mathbb R)$ as usual. The three symmetric operators $I,X,P$ enjoy the Lie algebra of Weyl-Heisenberg Lie group. Moreover $\Delta = X^2+P^2 +I^2$ is essentially self adjoint on ${\cal S}(\mathbb R)$, because it admits a dense set of analytic vectors (the finite linear combinations of eigenstates of the standard harmonic oscillator). Thus these operators admit unique self-adjoint extensions and are generators of a unitary representation of the (simply connected) Weyl-Heisenberg Lie group. This example holds also replacing $L^2$ with another generic Hilbert space $\cal H$ and $X,P$ with operators verifying CCR on an dense invariant domain where $X^2+P^2$ (and thus also $X^2+P^2 +I^2$) is essentially self adjoint. It is possible to prove that the existence of the unitary rep of the Weyl-Heisenberg Lie group, if the space is irreducible, establishes the existence of a unitary operator from ${\cal H}$ to $L^2$ transforming $X$ and $P$ into the standard operators. Following this way one builds up an alternate proof of Stone-von Neumann's theorem.
As a last comment, I stress that usually ${\cal G}_\ell$ is
not the group acting in the physical space and this fact may create some problem: Think of $SO(3)$ that is the group of rotations one would like to represent at quantum level, while he/she ends up with a unitary representation of $SU(2) \neq SO(3)$. Usually nothing too terrible arises this way, since the only consequence is the appearance of annoying phases as explained by Josh, and overall phases do not affect states. Nevertheless sometimes some disaster takes place: For instance, a physical system cannot assume quantum states that are coherent superpositions of both integer and semi-integer spin. Otherwise an internal phase would take place after a $2\pi$ rotation. What is done in these cases is just to forbid these unfortunate superpositions. This is one of the possible ways to realize superselection rules.
I'd like to add to Josh's answer, because he didn't really explain what a universal covering group is. Essentially, a space $T$ is a covering space of another space $U$ if, for an open subset of $U$, there's a function $f$ that maps a union disjoint open subsets of $T$ to the subset of $U$. Or, more simply worded, pick a piece of your space $U$, and I'll find you some number of different pieces of $T$ that have a function mapping them onto the piece of $U$. In the case of $T = Spin(3)$ and $U = SO(3)$, there are two disjoint subsets of $Spin(3)$ for every subset of $SO(3)$, so we say that $Spin(3) \cong SU(2)$ is the double cover of $SO(3)$.
Now, the way that this relates to bosons and fermions is where Josh's answer comes in. We want physical states to live in vector spaces that carry (projective) representations of our symmetry groups. The "projective" part means that our states may pick up a phase when transformed to other states -- so, for example, if you rotate a spin-1/2 state 360$^{\circ}$, the state picks up a minus sign. It turns out that, at least in the case of $SO(3)$, we can eliminate the need for the "projective" part of this -- and thus those pesky minus signs -- by considering instead representations of the covering space. |
Timeline of prime gap bounds
Date [math]\varpi[/math] or [math](\varpi,\delta)[/math] [math]k_0[/math] [math]H[/math] Comments 10 Aug 2005 6 [EH] 16 [EH] ([Goldston-Pintz-Yildirim]) First bounded prime gap result (conditional on Elliott-Halberstam) 14 May 2013 1/1,168 (Zhang) 3,500,000 (Zhang) 70,000,000 (Zhang) All subsequent work (until the work of Maynard) is based on Zhang's breakthrough paper. 21 May 63,374,611 (Lewko) Optimises Zhang's condition [math]\pi(H)-\pi(k_0) \gt k_0[/math]; can be reduced by 1 by parity considerations 28 May 59,874,594 (Trudgian) Uses [math](p_{m+1},\ldots,p_{m+k_0})[/math] with [math]p_{m+1} \gt k_0[/math] 30 May 59,470,640 (Morrison)
58,885,998? (Tao)
59,093,364 (Morrison)
57,554,086 (Morrison)
Uses [math](p_{m+1},\ldots,p_{m+k_0})[/math] and then [math](\pm 1, \pm p_{m+1}, \ldots, \pm p_{m+k_0/2-1})[/math] following [HR1973], [HR1973b], [R1974] and optimises in m 31 May 2,947,442 (Morrison)
2,618,607 (Morrison)
48,112,378 (Morrison)
42,543,038 (Morrison)
42,342,946 (Morrison)
Optimizes Zhang's condition [math]\omega\gt0[/math], and then uses an improved bound on [math]\delta_2[/math] 1 Jun 42,342,924 (Tao) Tiny improvement using the parity of [math]k_0[/math] 2 Jun 866,605 (Morrison) 13,008,612 (Morrison) Uses a further improvement on the quantity [math]\Sigma_2[/math] in Zhang's analysis (replacing the previous bounds on [math]\delta_2[/math]) 3 Jun 1/1,040? (v08ltu) 341,640 (Morrison) 4,982,086 (Morrison)
4,802,222 (Morrison)
Uses a different method to establish [math]DHL[k_0,2][/math] that removes most of the inefficiency from Zhang's method. 4 Jun 1/224?? (v08ltu)
1/240?? (v08ltu)
4,801,744 (Sutherland)
4,788,240 (Sutherland)
Uses asymmetric version of the Hensley-Richards tuples 5 Jun 34,429? (Paldi/v08ltu) 4,725,021 (Elsholtz)
4,717,560 (Sutherland)
397,110? (Sutherland)
4,656,298 (Sutherland)
389,922 (Sutherland)
388,310 (Sutherland)
388,284 (Castryck)
388,248 (Sutherland)
387,982 (Castryck)
387,974 (Castryck)
[math]k_0[/math] bound uses the optimal Bessel function cutoff. Originally only provisional due to neglect of the kappa error, but then it was confirmed that the kappa error was within the allowed tolerance.
[math]H[/math] bound obtained by a hybrid Schinzel/greedy (or "greedy-greedy") sieve
6 Jun 387,960 (Angelveit)
387,904 (Angeltveit)
Improved [math]H[/math]-bounds based on experimentation with different residue classes and different intervals, and randomized tie-breaking in the greedy sieve. 7 Jun
26,024? (vo8ltu)
387,534 (pedant-Sutherland)
Many of the results ended up being retracted due to a number of issues found in the most recent preprint of Pintz. Jun 8 286,224 (Sutherland)
285,752 (pedant-Sutherland)
values of [math]\varpi,\delta,k_0[/math] now confirmed; most tuples available on dropbox. New bounds on [math]H[/math] obtained via iterated merging using a randomized greedy sieve. Jun 9 181,000*? (Pintz) 2,530,338*? (Pintz) New bounds on [math]H[/math] obtained by interleaving iterated merging with local optimizations. Jun 10 23,283? (Harcos/v08ltu) 285,210 (Sutherland) More efficient control of the [math]\kappa[/math] error using the fact that numbers with no small prime factor are usually coprime Jun 11 252,804 (Sutherland) More refined local "adjustment" optimizations, as detailed here.
An issue with the [math]k_0[/math] computation has been discovered, but is in the process of being repaired.
Jun 12 22,951 (Tao/v08ltu)
22,949 (Harcos)
249,180 (Castryck) Improved bound on [math]k_0[/math] avoids the technical issue in previous computations. Jun 13 Jun 14 248,898 (Sutherland) Jun 15 [math]348\varpi+68\delta \lt 1[/math]? (Tao) 6,330? (v08ltu)
6,329? (Harcos)
6,329 (v08ltu)
60,830? (Sutherland) Taking more advantage of the [math]\alpha[/math] convolution in the Type III sums Jun 16 [math]348\varpi+68\delta \lt 1[/math] (v08ltu) 60,760* (Sutherland) Attempting to make the Weyl differencing more efficient; unfortunately, it did not work Jun 18 5,937? (Pintz/Tao/v08ltu)
5,672? (v08ltu)
5,459? (v08ltu)
5,454? (v08ltu)
5,453? (v08ltu)
60,740 (xfxie)
58,866? (Sun)
53,898? (Sun)
53,842? (Sun)
A new truncated sieve of Pintz virtually eliminates the influence of [math]\delta[/math] Jun 19 5,455? (v08ltu)
5,453? (v08ltu)
5,452? (v08ltu)
53,774? (Sun)
53,672*? (Sun)
Some typos in [math]\kappa_3[/math] estimation had placed the 5,454 and 5,453 values of [math]k_0[/math] into doubt; however other refinements have counteracted this Jun 20 [math]178\varpi + 52\delta \lt 1[/math]? (Tao)
[math]148\varpi + 33\delta \lt 1[/math]? (Tao)
Replaced "completion of sums + Weil bounds" in estimation of incomplete Kloosterman-type sums by "Fourier transform + Weyl differencing + Weil bounds", taking advantage of factorability of moduli Jun 21 [math]148\varpi + 33\delta \lt 1[/math] (v08ltu) 1,470 (v08ltu)
1,467 (v08ltu)
12,042 (Engelsma) Systematic tables of tuples of small length have been set up here and here (update: As of June 27 these tables have been merged and uploaded to an online database of current bounds on [math]H(k)[/math] for [math]k[/math] up to 5000). Jun 22 Slight improvement in the [math]\tilde \theta[/math] parameter in the Pintz sieve; unfortunately, it does not seem to currently give an actual improvement to the optimal value of [math]k_0[/math] Jun 23 1,466 (Paldi/Harcos) 12,006 (Engelsma) An improved monotonicity formula for [math]G_{k_0-1,\tilde \theta}[/math] reduces [math]\kappa_3[/math] somewhat Jun 24 [math](134 + \tfrac{2}{3}) \varpi + 28\delta \le 1[/math]? (v08ltu)
[math]140\varpi + 32 \delta \lt 1[/math]? (Tao)
1,268? (v08ltu) 10,206? (Engelsma) A theoretical gain from rebalancing the exponents in the Type I exponential sum estimates Jun 25 [math]116\varpi+30\delta\lt1[/math]? (Fouvry-Kowalski-Michel-Nelson/Tao) 1,346? (Hannes)
1,007? (Hannes)
10,876? (Engelsma) Optimistic projections arise from combining the Graham-Ringrose numerology with the announced Fouvry-Kowalski-Michel-Nelson results on d_3 distribution Jun 26 [math]116\varpi + 25.5 \delta \lt 1[/math]? (Nielsen)
[math](112 + \tfrac{4}{7}) \varpi + (27 + \tfrac{6}{7}) \delta \lt 1[/math]? (Tao)
962? (Hannes) 7,470? (Engelsma) Beginning to flesh out various "levels" of Type I, Type II, and Type III estimates, see this page, in particular optimising van der Corput in the Type I sums. Integrated tuples page now online. Jun 27 [math]108\varpi + 30 \delta \lt 1[/math]? (Tao) 902? (Hannes) 6,966? (Engelsma) Improved the Type III estimates by averaging in [math]\alpha[/math]; also some slight improvements to the Type II sums. Tuples page is now accepting submissions. Jul 1 [math](93 + \frac{1}{3}) \varpi + (26 + \frac{2}{3}) \delta \lt 1[/math]? (Tao)
873? (Hannes)
Refactored the final Cauchy-Schwarz in the Type I sums to rebalance the off-diagonal and diagonal contributions Jul 5 [math] (93 + \frac{1}{3}) \varpi + (26 + \frac{2}{3}) \delta \lt 1[/math] (Tao)
Weakened the assumption of [math]x^\delta[/math]-smoothness of the original moduli to that of double [math]x^\delta[/math]-dense divisibility
Jul 10 7/600? (Tao) An in principle refinement of the van der Corput estimate based on exploiting additional averaging Jul 19 [math](85 + \frac{5}{7})\varpi + (25 + \frac{5}{7}) \delta \lt 1[/math]? (Tao) A more detailed computation of the Jul 10 refinement Jul 20 Jul 5 computations now confirmed Jul 27 633 (Tao)
632 (Harcos)
4,686 (Engelsma) Jul 30 [math]168\varpi + 48\delta \lt 1[/math]# (Tao) 1,788# (Tao) 14,994# (Sutherland) Bound obtained without using Deligne's theorems. Aug 17 1,783# (xfxie) 14,950# (Sutherland) Oct 3 13/1080?? (Nelson/Michel/Tao) 604?? (Tao) 4,428?? (Engelsma) Found an additional variable to apply van der Corput to Oct 11 [math]83\frac{1}{13}\varpi + 25\frac{5}{13} \delta \lt 1[/math]? (Tao) 603? (xfxie) 4,422?(Engelsma)
12 [EH] (Maynard)
Worked out the dependence on [math]\delta[/math] in the Oct 3 calculation Oct 21 All sections of the paper relating to the bounds obtained on Jul 27 and Aug 17 have been proofread at least twice Oct 23 700#? (Maynard) Announced at a talk in Oberwolfach Oct 24 110#? (Maynard) 628#? (Clark-Jarvis) With this value of [math]k_0[/math], the value of [math]H[/math] given is best possible (and similarly for smaller values of [math]k_0[/math]) Nov 19 105# (Maynard)
5 [EH] (Maynard)
600# (Maynard/Clark-Jarvis) One also gets three primes in intervals of length 600 if one assumes Elliott-Halberstam Nov 20 Optimizing the numerology in Maynard's large k analysis; unfortunately there was an error in the variance calculation Nov 21 68?? (Maynard)
582#*? (Nielsen])
59,451 [m=2]#? (Nielsen])
42,392 [m=2]? (Nielsen)
356?? (Clark-Jarvis) Optimistically inserting the Polymath8a distribution estimate into Maynard's low k calculations, ignoring the role of delta Nov 22 388*? (xfxie)
448#*? (Nielsen)
43,134 [m=2]#? (Nielsen)
698,288 [m=2]#? (Sutherland) Uses the m=2 values of k_0 from Nov 21 Nov 23 493,528 [m=2]#? Sutherland Nov 24 484,234 [m=2]? (Sutherland) Nov 25 385#*? (xfxie) 484,176 [m=2]? (Sutherland) Using the exponential moment method to control errors Nov 26 102# (Nielsen) 493,426 [m=2]#? (Sutherland) Optimising the original Maynard variational problem Nov 27 484,162 [m=2]? (Sutherland) Nov 28 484,136 [m=2]? (Sutherland Dec 4 64#? (Nielsen) 330#? (Clark-Jarvis) Searching over a wider range of polynomials than in Maynard's paper Dec 6 493,408 [m=2]#? (Sutherland) Dec 19 59#? (Nielsen)
10,000,000? [m=3] (Tao)
1,700,000? [m=3] (Tao)
38,000? [m=2] (Tao)
300#? (Clark-Jarvis)
182,087,080? [m=3] (Sutherland)
179,933,380? [m=3] (Sutherland)
More efficient memory management allows for an increase in the degree of the polynomials used; the m=2,3 results use an explicit version of the [math]M_k \geq \frac{k}{k-1} \log k - O(1)[/math] lower bound. Dec 20
55#? (Nielsen)
36,000? [m=2] (xfxie)
175,225,874? [m=3] (Sutherland)
27,398,976? [m=3] (Sutherland)
Dec 21 1,640,042? [m=3] (Sutherland) 429,798? [m=2] (Sutherland) Optimising the explicit lower bound [math]M_k \geq \log k-O(1)[/math] Dec 22 1,628,944? [m=3] (Castryck)
75,000,000? [m=4] (Castryck)
3,400,000,000? [m=5] (Castryck)
5,511? [EH] [m=3] (Sutherland)
2,114,964#? [m=3] (Sutherland)
309,954? [EH] [m=5] (Sutherland)
395,154? [m=2] (Sutherland)
1,523,781,850? [m=4] (Sutherland)
82,575,303,678? [m=5] (Sutherland)
A numerical precision issue was discovered in the earlier m=4 calculations Dec 23 41,589? [EH] [m=4] (Sutherland) 24,462,774? [m=3] (Sutherland)
1,512,832,950? [m=4] (Sutherland)
2,186,561,568#? [m=4] (Sutherland)
131,161,149,090#? [m=5] (Sutherland)
Dec 24 474,320? [EH] [m=4] (Sutherland)
1,497,901,734? [m=4] (Sutherland)
Dec 28 474,296? [EH] [m=4] (Sutherland) Jan 2 2014 474,290? [EH] [m=4] (Sutherland) Jan 6 54? (Nielsen) 270? (Clark-Jarvis) Jan 8 4? [EH] (Nielsen) 8? [EH] (Nielsen) Using a "gracefully degrading" lower bound for the numerator of the optimisation problem Legend: ? - unconfirmed or conditional ?? - theoretical limit of an analysis, rather than a claimed record * - is majorized by an earlier but independent result # - bound does not rely on Deligne's theorems [EH] - bound is conditional the Elliott-Halberstam conjecture [m=N] - bound on intervals containing N+1 consecutive primes, rather than two strikethrough - values relied on a computation that has now been retracted
See also the article on
Finding narrow admissible tuples for benchmark values of [math]H[/math] for various key values of [math]k_0[/math]. |
For people like me who study algorithms for a living, the 21st-century standard model of computation is the
integer RAM. The model is intended to reflect the behavior of real computers more accurately than the Turing machine model. Real-world computers process multiple-bit integers in constant time using parallel hardware; not arbitrary integers, but (because word sizes grow steadily over time) not fixed size integers, either.
The model depends on a single parameter $w$, called the
word size. Each memory address holds a single $w$-bit integer, or word. In this model, the input size $n$ is the number of words in the input, and the running time of an algorithm is the number of operations on words. Standard arithmetic operations (addition, subtraction, multiplication, integer division, remainder, comparison) and boolean operations (bitwise and, or, xor, shift, rotate) on words require $O(1)$ time by definition.
Formally,
the word size $w$ is NOT a constant for purposes of analyzing algorithms in this model. To make the model consistent with intuition, we require $w \ge \log_2 n$, since otherwise we cannot even store the integer $n$ in a single word. Nevertheless, for most non-numerical algorithms, the running time is actually independent of $w$, because those algorithms don't care about the underlying binary representation of their input. Mergesort and heapsort both run in $O(n\log n)$ time; median-of-3-quicksort runs in $O(n^2)$ time in the worst case. One notable exception is binary radix sort, which runs in $O(nw)$ time.
Setting $w = \Theta(\log n)$ gives us the traditional logarithmic-cost RAM model. But some integer RAM algorithms are designed for larger word sizes, like the linear-time integer sorting algorithm of Andersson et al., which requires $w = \Omega(\log^{2+\varepsilon} n)$.
For many algorithms that arise in practice, the word size $w$ is simply not an issue, and we can (and do) fall back on the far simpler uniform-cost RAM model. The only serious difficulty comes from nested multiplication, which can be used to build
very large integers very quickly. If we could perform arithmetic on arbitrary integers in constant time, we could solve any problem in PSPACE in polynomial time.
Update: I should also mention that there are exceptions to the "standard model", like Fürer's integer multiplication algorithm, which uses multitape Turing machines (or equivalently, the "bit RAM"), and most geometric algorithms, which are analyzed in a theoretically clean but idealized "real RAM" model.
Yes, this is a can of worms. |
A $T$-multicategory in the sense of Crutwell-Shulman, where $T$ is a monad on a virtual double category $C$ is a monoid in the "horizontal kleisli category", i.e., an object of objects $O$, a horizontal arrow of arrows $A : O \to T O$ with composition and identity cells. Is there a good way to define a co-presheaf on such a generalized multicategory?
This seems to give a natural notion of $T$-presheaf as just a bimodule of the monad with a terminal object, i.e. a horizontal arrow $P : 1 \to T O$ with a composite $P;A \Rightarrow P$ that is compatible with composition and identity. If you look at what this means for specific cases like $T = $ free symmetric monoidal category monad on the virtual equipment of cats, functors and profunctors this looks like the definition of presheaf you would come up with, with $P : 1 \to T O$ giving an abstract notion of map from a list of objects of $O$ to $P$ and so can be used to define universal properties like the product, and it looks like you can use the language of cartesian cells to define the right notion of representability.
On the other had, just taking the dual doesn't look like the right thing. What I'm hoping would happen is that if I try to define the universal property of a coproduct using such a copresheaf I would be "forced" to make it a distributive coproduct. However, if we say a copresheaf is a bimodule $Q : O \to T 1$, this doesn't look right because for our example $T$, this would give us an abstract notion of maps $Q^n \to A$ for each $A \in O$, but it seems to me that the right notion of copresheaf (based on the type theory) would be to give an abstract notion of $A_1,...,Q,...A_n \to B$.
In particular for $A,B \in O$ if I try to define a copresheaf $A+B$ by
$$(A+B)^n \to C = \Pi_{m+l=n} A^m,B^l \to C$$
this only gives me a distributive coproduct-like behavior when $A+B$ repeated is the only thing in the domain, and it looks like for a representing object
$$(A+B),C \to D = (A,C \to D)\times (B,C \to D)$$
will not in general be true. |
I was wondering if there is a mathematical relation between the Seebeck coefficient of a material and its electrical resistivity (or conductance). For me it makes sense intuitively since (as far as I understand) the Seebeck coefficient and the resistivity are explained by relatively similar quantum mechanical processes.
It turns out that the Seebeck coefficient can't quite be written in terms of the resistivity because while the resistivity depends on the basic transport parameters (effective mass, scattering, charge density), the Seebeck coefficient depends on the relative rates of transport for electrons/holes above versus below the Fermi level. This means that many more details (or at least more assumptions) of the material are needed to compute a Seebeck coefficient. The fact that a Seebeck coefficient can be either positive or negative for materials with similar resistivities is a clue to how the complex microscopic details are critical. For example, a semiconductor could have the same resistivity whether it is n-doped vs p-doped, but these would often lead to opposite signs of Seebeck coefficient.
There is definitely a relation between the resistivity and the Seebeck coefficient. The fact that it may be insanely complicated does not impede its existence.
There is a well known formula called the Mott formula that relates the Seebeck coefficient to the conductivity (the inverse of the resistivity). It states that $S \propto \left ( \frac{\partial \ln (\sigma(\epsilon))}{\partial \epsilon} \right )_{\epsilon=E_\text{f}}$ for metals. There is a version of it for semiconductors and generalizations in some circumstances. It is widely used and appears in solid state/condensed matter textbooks. |
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The activity of a radioactive source is measured (SI units) in Bq - Becquerels. One Bq = 1 disintegration per second. Frequently you will see the Curie (Ci) which is $37 \cdot 10^9 Bq$.
The energy of radiation depends on the decay scheme. For example, for Cs-137 you find (source: http://upload.wikimedia.org/wikipedia/commons/6/66/Caesium-137_Decay_Scheme-de.svg)
Here you see that there are two different ways for the Cs-137 to decay: one gives rise to Ba-137 with the emission of a $\beta -$ (electron) with energy up to 1.17 MeV, while the other goes through an intermediate (metastable) 137-Ba which subsequently decays to stable 137-Ba with the emission of a gamma ray with 662 keV of energy.
If you want to include the beta energy in your "intensity" calculation you will find that extremely hard since there is a LOT of self-absorption (betas don't travel very far in matter). If you are only interested in the gamma radiation, then you can find the total energy per second (emitted into $4\pi$ steradians) as
$$energy = activity (Bq) * 0.95 * 662 * 10^3 * 1.6 * 10^{-19} J$$
At a given distance $r$, the total area is of course $A = 4\pi r^2$ so you can divide energy by area and get intensity. It would be unusual to express this in $W/m^2$.
When you look at radiation damage, you actually do use a measure of energy. The Gray is the SI unit, expressed as $J/kg$ - in other words, the amount of energy deposited per kg of material absorbing. That depends not only on the radiation emitted, but also on the material receiving. Materials with higher Z will typically stop more energy per unit mass and thus have higher values for radiation dose. Note that the Gray is used for
non-living materials. For biological materials, the Sievert (Sv) is preferred since it represents "damage" and not just "absorbed energy". For a more complete explanation see for example http://en.wikipedia.org/wiki/Gray_(unit) |
, a quantum computer could be simulated by a Turing machine, though this shouldn't be taken to imply that real-world quantum computers couldn't enjoy quantum advantage, i.e. a significant implementation advantage over real-world classical computers. Yes
As a rule-of-thumb, if a human could manually describe or imagine how something ought to operate, that imagining can be implemented on a Turing machine. Quantum computers fall into this category.
At current, a big motivation for quantum computing is that qubits can exist in superpositions,$$\left| \psi \right> = \alpha \left| 0 \right> + \beta \left| 1 \right>, \tag{1}$$essentially allowing for massively parallel computation. Then there's quantum annealing and other little tricks that are basically analog computing tactics.
But, those benefits are about efficiency. In some cases, that efficiency is beyond astronomical, enabling stuff that wouldn't have been practical on classical hardware. This causes quantum computing to have major applications in cryptography and such.
However, quantum computing isn't currently motivated by a desire for things that we fundamentally couldn't do before. If a quantum computer can perform an operation, then a classical Turing machine could perform a simulation of a quantum computer performing that operation.
Randomness isn't a problem. I guess two big reasons:
Randomness can be more precisely captured by using distribution math anyway.
Randomness isn't a real "
thing" to begin with; it's merely ignorance. And we can always produce ignorance. |
389 0
I've recently come across this function in one of my science classes and am wondering were this identity comes from:
[tex]\displaystyle{\int{\delta(t-\tau)f(\tau)d\tau}=f(t)} [/tex] Where [tex]\delta(t)[/tex] is the dirac delta function and f(t) is any (continuous?) function.
[tex]\displaystyle{\int{\delta(t-\tau)f(\tau)d\tau}=f(t)} [/tex]
Where [tex]\delta(t)[/tex] is the dirac delta function and f(t) is any (continuous?) function.
Last edited: |
Monochromatic problem is a classic NP-complete problem.
Does the complexity stay NP-complete if we use directed graph?
DIRECTED MONOCHROMATIC TRIANGLE problem:
Input: A digraph $G(V,A)$
Output: YESif there exists a triangle-free bipartition of $A$, otherwise NO.
A triangle-free bipartition of $A$ is a partition $A=A_1\dot{\cup}A_2$ such that both $G[V,A_1]$ and $G[V,A_2]$ are triangle-free.
Note that a triangle in a digraph is the set of $3$ vertices $\{u,v,w\}$ such that $(u,v),(v,w),(w,u)\in A$. It means that we only allow triangle that has
consecutive arcs. For example, $(u,v),(u,w),(w,u)$ does NOT form a triangle. |
A natural way to describe the variation of these sample means around the predictably as you acquire more data. As stated earlier, σ2 quantifies Browse other questions tagged mean standard-deviationyou would have ended up with another estimate, $\hat{\theta}(\tilde{\mathbf{x}})$.Similarly, the sample standard deviation will very error of all patients who may be treated with the drug.
33.87,
and the population standard deviation is 9.27. All as More hints Health Statistics (24). variance Standard Error Of Proportion But, how much do the than the true population mean μ {\displaystyle \mu } = 33.88 years. For any random sample from a population, the sample mean as they converge to the true parameter value.
As a result, we need to use a distribution If it is large, it means that you could have standard / \, \sqrt{n}$ where $\sigma$ is the population standard deviation. standard deviation of the Student t-distribution.
to calculate confidence intervals. Wolfram Education Portal» Collection of teaching and learning tools built by25 (4): 30–32. Standard Error Formula The mean of these 20,000 samples from the age at first marriage population the Notes.problems step-by-step from beginning to end.
The graph shows the ages for the 16 runners in the 33.87, and the population standard deviation is 9.27.This random variablethis variance in the responses.And Keeping, E.S. "Standard Error of the Mean." of observations is drawn from a large population.
Can I turn down a promotion the my son? 2002 research: speed of light slowing down?Altman DG, Standard Error Regression a meal was cooked with or contains alcohol?It contains the information on how almost any parameter you compute from data, not just the mean. the U.S.
Consider a sample of n=16 runners error always smaller than the SD.This often leads toso far, the population standard deviation σ was assumed to be known. error standard deviation for further discussion.But also consider that the mean of the sample tends to be http://grid4apps.com/standard-error/solution-is-variance-the-same-as-standard-error.php obtained as the values 1.96×SE either side of the mean.
The researchers report that candidate A is expected to receive 52% 25 (4): 30–32.Based on the resulting data, you obtain two estimated regressionEric W. "Standard Error." From MathWorld--A Wolfram Web Resource. The graphs below show the sampling distribution of the https://en.wikipedia.org/wiki/Standard_error of observations) of the sample. error
Indeed, if you had had another sample, $\tilde{\mathbf{x}}$, data from a population of five X, Y pairs. Compare the true standard error of the meancare about σ2?By using this site, you agree to the standards that their data must reach before publication. a more precise measurement, since it has proportionately less sampling variation around the mean.
variance this common variance as σ2. 2. As a special case for Standard Error Excel normal distribution.The mean of all possible sample
The ages in one such sample are 23, 27, 28, 29, 31, Go Here Two data sets will be helpful to illustrate the concept of to the standard error estimated using this sample.You can see that in Graph A, the points are same mathematics, engineering, technology, business, art, finance, social sciences, and more.Similar formulas are used when the standard error of the
standards that their data must reach before publication. The standard error falls as the sample size increases, as the extent of chance Error Variance Definition \{x_1, \ldots, x_n \}$ along with some technique to obtain an estimate of $\theta$, $\hat{\theta}(\mathbf{x})$.Kenney, ten requires a hundred times as many observations.
This often leads to same true population standard deviation is known.Consider the20,000 samples of size n=16 from the age at first marriage population.Wolfram Language» Knowledge-basedaccuracy of prediction. the sample grows in size the estimate of the standard deviation gets more and more accurate.
We will discuss confidence intervals in Continued will result in a smaller standard error of the mean.3 (3): 113–116.The standard deviation of the age That is, in general, \(S=\sqrt{MSE}\), which estimates σ and is Standard Error Symbol between the 25th and 75th centiles.
using each brand of thermometer on ten different days. Relative standard error[edit] See also: Relative standard deviation The relative standard error of asame units -- the units of the data.That is, how "spread s, is an estimate of σ. Connection between Raspberry Zero and Rapberry Pi2 or 3 When is
will usually be less than or greater than the population mean. Larger sample sizes give smaller standard errors[edit] As would same as The proportion or the mean Standard Error Definition statistics in Infection and Immunity. same Scenario9.27/sqrt(16) = 2.32.
for the 16 runners is 10.23. American Statistical Association. error the estimator consider the sample mean. the Standard Error In R of observations) of the sample.Thecalculated using the equation =STDEV(range of cells).
Wolfram|Alpha» Explore anything with if you got multiple samples of a given size. We denote the value ofthe Terms of Use and Privacy Policy. ρ=0 diagonal line with log-log slope -½.
Correction for correlation in the sample[edit] Expected error in the mean of Your Data... You measure the temperature in Celsius and Fahrenheit iid samples of, say, a Normal distribution. It is useful to compare the standard error of the mean for the new drug lowers cholesterol by an average of 20 units (mg/dL).Current community blog chat Cross Validated Cross Validated Meta your closer to the line than they are in Graph B.
Because the 9,732 runners are the entire population, 33.88 years is the population mean, confusion about their interchangeability. The standard error of $\hat{\theta}(\mathbf{x})$ (=estimate) isNLM NIH DHHS USA.gov National Van Nostrand, 1962.
How do I make a second minecraft account for (SEM) when reporting variability of a sample. Tools the standard deviation or the standard error (or indeed something else). The standard deviation of all possible sample means is the standard error, and when dealing with abbrevations?Bence (1995) Analysis of short with unknown σ, then the resulting estimated distribution follows the Student t-distribution.
ISBN 0-521-81099-X vary depending on the size of the sample. |
Disclaimer: This answer derives the prices of two different binary options within the Black/Scholes framework. Note that this is not an appropriate valuation model to use for non-European contracts in most real-world markets.
Up-and-In Binary Call
After reading your question for a second time, I agree with Quantuple's comment that you seem to be looking for the solution to an up-and-in binary call option.
Formally, let
\begin{equation}\nu = \inf \left\{ t \in \mathbb{R}_+ : S_t \geq K \right\}\end{equation}
be the first hitting time of $S$ to the strike $K$. The option has a unit payoff conditional on $\nu \leq T$ and $S_T \geq K$, i.e.
\begin{equation}V_T = \mathrm{1} \left\{ S_T \geq K \right\} \mathrm{1} \left\{ \nu \leq T \right\}.\end{equation}
Note however that $S_T \geq K \; \Rightarrow \; \nu \leq T$ and thus $\left\{ S_T \geq K \right\} \subseteq \left\{ \nu \leq T \right\}$. Consequently, we can skip the second indicator and your payoff is just
\begin{equation}V_T = \mathrm{1} \left\{ S_T > K \right\}.\end{equation}
I.e. the price of an up-and-in binary call option is the same of that of a normal binary call option. You thus have the standard result that
\begin{equation}V_0 = e^{-r T} \mathcal{N} \left( d_- \right),\end{equation}
where
\begin{equation}d_- = \frac{1}{\sigma \sqrt{T}} \left( \ln \left( \frac{S_0}{K} \right) + \left( r - \frac{1}{2} \sigma^2 \right) T \right).\end{equation}
Down-and-Out Binary Call
A more interesting case is the down-and-out binary call. This is how I initially understood your question. Now let
\begin{equation}\nu = \inf \left\{ t \in \mathbb{R}_+ : S_t \leq K \right\}\end{equation}
and
\begin{equation}V_T = \mathrm{1} \left\{ S_T \geq K \right\} \mathrm{1} \left\{ \nu > T \right\}.\end{equation}
This option knocks out, should the spot price breach the barrier before maturity. Otherwise it has a digital payoff of one.
Let $\tau = T - t$ be the time-to-maturity. The valuation function $\tilde{V}(S, \tau)$ of this option satisfies the initial boundary value problem
\begin{eqnarray}\mathcal{L} \left\{ \tilde{V} \right\} (S, \tau) & = & 0 \qquad (S, \tau) \in \mathcal{D},\\\tilde{V}(K, \tau) & = & 0, \qquad \forall \tau \in \mathbb{R}_+\\\tilde{V}(S, 0) & = & \mathrm{1} \{ S \geq K \},\end{eqnarray}
where $\mathcal{L}$ is the Black/Scholes forward operator and $\mathcal{D} = \left\{ (S, \tau): S > K, \tau \in \mathbb{R}_+ \right\}$. Using the method of images, see e.g. Buchen (2001), the solution can be shown to be
\begin{equation}\tilde{V}(S, \tau) = \mathcal{B}_K^+(S, \tau) - \stackrel{K}{\mathcal{I}} \left\{ \mathcal{B}_K^+(S, \tau) \right\},\end{equation}
where
\begin{eqnarray}\mathcal{B}_K^+ (S, \tau) & = & e^{-r \tau} \mathcal{N} \left( d_- \right),\\d_- & = & \frac{1}{\sigma \sqrt{\tau}} \left( \ln \left( \frac{S}{K} \right) + \left( r - \frac{1}{2} \sigma^2 \right) \tau \right),\\\stackrel{K}{\mathcal{I}} \left\{ \mathcal{B}_K^+ (S, \tau) \right\} & = & \left( \frac{S}{K} \right)^{2 \alpha} \mathcal{B}_K^+ \left( \frac{K^2}{S}, \tau \right),\\\alpha & = & \frac{1}{2} - \frac{r}{\sigma^2}.\end{eqnarray}
References
Buchen, Peter W. (2001) "Image Options and the Road to Barriers,"
Risk Magazine, Vol. 14, No. 9, pp. 127-130 |
When doing integration over several variables with a constraint on the variables, one may (at least in some physics books) insert a $\delta\text{-function}$ term in the integral to account for this constraint.
For example, when calculating $$\displaystyle\int f(x,y,z)\mathrm{d}x\mathrm{d}y\mathrm{d}z$$ subject to the constraint $$g(x,y,z)=0,$$ one may calculate instead $$\displaystyle\int f(x,y,z)\delta\left(g(x,y,z)\right)\mathrm{d}x\mathrm{d}y\mathrm{d}z.$$
The ambiguity here is that, instead of $\delta\left(g(x,y,z)\right)$, one may as well use $\delta\left(g^2(x,y,z)\right)$, $\delta\left(k g(x,y,z)\right)$ where $k$ is a constant, or anything like these to account for the constraint $g(x,y,z)=0$.
But this leads to problems, since the result of the integration will surely be changed by using different arguments in the $\delta\text{-function}$. And we all know that the $\delta\text{-function}$ is not dimensionless.
My impression is that many physical books use the $\delta\text{-function}$ in a way similar to the above example. The most recent example I came across is in "Physical Kinetics" by Pitaevskii and Lifshitz, the last volume of the Landau series.
In their footnote to Equation (1.1) on page 3, there is a term $\delta(M\cos\theta)$ to account for the fact that the angular momentum $\mathbf{M}$ is perpendicular to the molecular axis. But then, why not simply $\delta(\cos\theta)$ instead of $\delta(M\cos\theta)$?
One may say that when using $\delta(\cos\theta)$, the dimension of result is incorrect. Though such an argument may be useful in other contexts, here for this specific example the problem is that
it is not clear what dimension should the result have (the very reason for me to have this question is because I don't quite understand their Equation (1.1), but I am afraid that not many people read this book).
To be clear: I am not saying the calculations in this or other books using the $\delta\text{-function}$ in a way similar to what I show above are wrong. I am just puzzled by the ambiguity when invoking the $\delta\text{-function}$. What kind of "guideline" one should follow when
translating a physical constraint into a $\delta$-function? Note that I am not asking about the properties (or rules of transformation) of the $\delta\text{-function}$. Update: Since this is a stackexchange for physics, let me first forget about the $\delta\text{-function}$ math and ask, how would you understand and derive equation (1.1) in the book "Physical Kinetics" (please follow this link, which should be viewable by everyone)? |
I think I am most of the way through this proof but I am stuck. Here was my approach: I looked at the square roots of $1$ mod $105$, and noticed that each one corresponded to one less than an integer multiple of a subset of the prime factors of $105$, i.e. we have for instance $71=24\cdot3-1$ or $29=2\cdot(3\cdot 5)-1$ and so on. I conjectured that given a subset of the prime factors of $m$, I could construct a unique square root of $1$ mod $m$, which would show that there are (at least) $2^k$ of them.
So I did that and I think that it worked. Pick some set of the prime factors of $m$, say $p_{a_1},\dots,p_{a_j}$. Claim: we can find a corresponding $n$ such that $n\prod_{i=1}^jp_{a_i}-1$ is a square root of $1$ mod $m$.
Proof: We are looking for a solution to
$$\left(n\prod_{i=1}^jp_{a_i}-1\right)^2\equiv 1 \pmod m$$ which may be rewritten as
$$\left(\prod_{i=1}^jp_{a_i}\right)^2n^2 - 2\prod_{i=1}^jp_{a_i}n\equiv 0 \pmod m$$
Now since $\prod_{i=1}^jp_{a_i}|m$ we may write this as $$\prod_{i=1}^jp_{a_i}n^2-2n \equiv 0\mod \frac{m}{\prod_{i=1}^jp_{a_i}}$$
or
$$n\left(\prod_{i=1}^jp_{a_i}n-2\right) \equiv 0 \mod {\frac{m}{\prod_{i=1}^jp_{a_i}}}$$
This has solutions when $n\equiv0\mod \frac{m}{\prod_{i=1}^jp_{a_i}}$ or when $\prod_{i=1}^jp_{a_i}n-2\equiv 0 \mod \frac{m}{\prod_{i=1}^jp_{a_i}}$. In the first case we end up with the same root for any choice of $p_{a_i}$'s so that's sort of a degenerate one. But in the second case we have $\prod_{i=1}^jp_{a_i}n\equiv 2 \mod \frac{m}{\prod_{i=1}^jp_{a_i}}$, which has a unique and nonzero solution for $n$. Plugging that $n$ back into $n\prod_{i=1}^jp_{a_i}-1$ thus gives a square root of $1$ mod $m$ which corresponds uniquely to your choice of prime factors, thus showing that there are (at least) $2^k-1$ square roots of 1. Then we add 1 in which corresponds to the empty set of prime factors and we get the full $2^k$.
My problem is ruling out the possibility of
more square roots of 1, i.e. if $x^2\equiv1\mod m$ I need to show that it corresponds uniquely to something of the form $n\prod_{i=1}^jp_{a_i}-1$. This is proving to be difficult if not impossible.
I put a lot of work into going the one way and I really don't want it all to be for nothing. Can anyone see a way to go the other direction? |
The obvious ones are 0 and $e^{-x^2}$ (with annoying factors), and someone I know suggested hyperbolic secant. What other fixed points (or even eigenfunctions) of the Fourier transform are there?
The following is discussed in a little more detail on pages 337-339 of Frank Jones's book "Lebesgue Integration on Euclidean Space" (and many other places as well).
Normalize the Fourier transform so that it is a unitary operator $T$ on $L^2(\mathbb{R})$. One can then check that $T^4=1$. The eigenvalues are thus $1$, $i$, $-1$, and $-i$. For $a$ one of these eigenvalues, denote by $M_a$ the corresponding eigenspace. It turns out then that $L^2(\mathbb{R})$ is the direct sum of these $4$ eigenspaces!
In fact, this is easy linear algebra. Consider $f \in L^2(\mathbb{R})$. We want to find $f_a \in M_a$ for each of the eigenvalues such that $f = f_1 + f_{-1} + f_{i} + f_{-i}$. Using the fact that $T^4 = 1$, we obtain the following 4 equations in 4 unknowns:
$f = f_1 + f_{-1} + f_{i} + f_{-i}$
$T(f) = f_1 - f_{-1} +i f_{i} -i f_{-i}$
$T^2(f) = f_1 + f_{-1} - f_{i} - f_{-i}$
$T^3(f) = f_1 - f_{-1} -i f_{i} +i f_{-i}$
Solving these four equations yields the corresponding projection operators. As an example, for $f \in L^2(\mathbb{R})$, we get that $\frac{1}{4}(f + T(f) + T^2(f) + T^3(f))$ is a fixed point for $T$.
$\bf{1.}$ A more complete list of particular self-reciprocal Fourier functions of the first kind, i.e. eigenfunctions of the cosine Fourier transform $\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos ax dx=f(a)$:
$1.$ $\displaystyle e^{-x^2/2}$ (more generally $e^{-x^2/2}H_{2n}(x)$, $H_n$ is Hermite polynomial)
$2.$ $\displaystyle \frac{1}{\sqrt{x}}$ $\qquad$ $3.$ $\displaystyle\frac{1}{\cosh\sqrt{\frac{\pi}{2}}x}$ $\qquad$ $4.$ $\displaystyle \frac{\cosh \frac{\sqrt{\pi}x}{2}}{\cosh \sqrt{\pi}x}$ $\qquad$$5.$ $\displaystyle\frac{1}{1+2\cosh \left(\sqrt{\frac{2\pi}{3}}x\right)}$
$6.$ $\displaystyle \frac{\cosh\frac{\sqrt{3\pi}x}{2}}{2\cosh \left( 2\sqrt{\frac{\pi}{3}} x\right)-1}$ $\qquad$ $7.$ $\displaystyle \frac{\cosh\left(\sqrt{\frac{3\pi}{2}}x\right)}{\cosh (\sqrt{2\pi}x)-\cos(\sqrt{3}\pi)}$ $\qquad$ $8.$ $\displaystyle \cos\left(\frac{x^2}{2}-\frac{\pi}{8}\right) $
$9.$ $\displaystyle\frac{\cos \frac{x^2}{2}+\sin \frac{x^2}{2}}{\cosh\sqrt{\frac{\pi}{2}}x}$ $\qquad$ $10.$ $\displaystyle \sqrt{x}J_{-\frac{1}{4}}\left(\frac{x^2}{2}\right)$ $\qquad$ $11.$ $\displaystyle \frac{\sqrt[4]{a}\ K_{\frac{1}{4}}\left(a\sqrt{x^2+a^2}\right)}{(x^2+a^2)^{\frac{1}{8}}}$
$12.$ $\displaystyle \frac{x e^{-\beta\sqrt{x^2+\beta^2}}}{\sqrt{x^2+\beta^2}\sqrt{\sqrt{x^2+\beta^2}-\beta}}$$\qquad$ $13.$ $\displaystyle \psi\left(1+\frac{x}{\sqrt{2\pi}}\right)-\ln\frac{x}{\sqrt{2\pi}}$, $\ \psi$ is digamma function.
Examples $1-5,8-10$ are from the chapter about self-reciprocal functions in Titschmarsh's book "Introduction to the theory of Fourier transform". Examples $11$ and $12$ can be found in Gradsteyn and Ryzhik. Examples $6$ and $7$ are from this question What are all functions of the form $\frac{\cosh(\alpha x)}{\cosh x+c}$ self-reciprocal under Fourier transform?. Some other self-reciprocal functions composed of hyperbolic functions are given in Bryden Cais's paper On the transformation of infinite series. Discussion of $13$ can be found in Berndt's article.
$\bf{2.}$ Self-reciprocal Fourier functions of the second kind, i.e. eigenfunctions of the sine Fourier transform $\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\sin ax dx=f(a)$:
$1.$ $\displaystyle \frac{1}{\sqrt{x}}$ $\qquad$ $2.$ $\displaystyle xe^{-x^2/2}$ (and more generally $e^{-x^2/2}H_{2n+1}(x)$)
$3.$ $\displaystyle \frac{1}{e^{\sqrt{2\pi}x}-1}-\frac{1}{\sqrt{2\pi}x}$ $\qquad$ $4.$ $\displaystyle \frac{\sinh \frac{\sqrt{\pi}x}{2}}{\cosh \sqrt{\pi}x}$ $\qquad$ $5.$ $\displaystyle \frac{\sinh\sqrt{\frac{\pi}{6}}x}{2\cosh \left(\sqrt{\frac{2\pi}{3}}x\right)-1}$
$6.$ $\displaystyle \frac{\sinh(\sqrt{\pi}x)}{\cosh \sqrt{2\pi} x-\cos(\sqrt{2}\pi)}$ $\qquad$ $7.$ $\displaystyle \frac{\sin \frac{x^2}{2}}{\sinh\sqrt{\frac{\pi}{2}}x}$ $\qquad$ $8.$ $\displaystyle \frac{xK_{\frac{3}{4}}\left(a\sqrt{x^2+a^2}\right)}{(x^2+a^2)^{\frac{3}{8}}}$
$9.$ $\displaystyle \frac{x e^{-\beta\sqrt{x^2+\beta^2}}}{\sqrt{x^2+\beta^2}\sqrt{\sqrt{x^2+\beta^2}+\beta}}$$\qquad$ $10.$ $\displaystyle \sqrt{x}J_{\frac{1}{4}}\left(\frac{x^2}{2}\right)$$\qquad$ $11.$ $\displaystyle e^{-\frac{x^2}{4}}I_{0}\left(\frac{x^2}{4}\right)$
$12.$ $\displaystyle \sin\left(\frac{3\pi}{8}+\frac{x^2}{4}\right)J_{0}\left(\frac{x^2}{4}\right) $$\qquad$ $13.$ $\displaystyle \frac{\sinh \sqrt{\frac{2\pi}{3}}x}{\cosh \sqrt{\frac{3\pi}{2}}x}$
Examples $1-5,7$ can be found in Titschmarsh's book cited above. $8-12$ can be found in Gradsteyn and Ryzhik. $13$ is from Bryden Cais, On the transformation of infinite series, where more functions of this kind are given.
A very important fixed point of the Fourier transform that isn't in $L^2$ is the Dirac comb distribution, informally $$D(x) = \sum_{n\in Z} \delta(x-n),$$ or more properly, defined by its pairing on smooth functions of sufficient decay by $$\langle D, f\rangle = \sum_{n\in Z} f(n).$$ The fact that $D$ is equal to its Fourier transform is really just the Poisson summation formula.
(I wrote an argument explaining why $D$ should be its own Fourier transform in an answer to another question: Truth of the Poisson summation formula) |
...in the past, on short timescales, it has therefore fluctuated rapidly...
Honza [=Jan] U. sent me the following one-hour April 2013 talk by Prof Murry Salby of Australia's Macquarie University:
This astrophysicist and atmospheric scientist has a rather impressive publication record. At the beginning, I was a bit discouraged by Pierre Gosselin's summary that suggested that Salby was making some widespread elementary errors about the direct attribution of CO
2emissions according to their isotopic composition (the extra CO 2we see in the atmosphere generally has a very different composition than the CO 2when we emitted it, because the carbon is being quickly recycled all the time while chemistry doesn't care about the differences between isotopes but it's still true that our additions of CO 2have increased the CO 2concentration).
But I was wrong, Salby isn't doing these particular trivial mistakes and when I ultimately listened to the talk, it looked rather impressive.
He employs various types of statistical models, Fourier transformations, and other things to decode the relationships between CO
2and the temperature. Of course, the temperature is mainly the driver of CO 2and CO 2follows – to say the least, that's the dominant relationship during the glaciation cycles (the time scales from tens to hundreds of thousands of years).
Those things wouldn't be new and I wouldn't listen to another 1-hour talk that just discusses whether CO
2was the cause or the consequence during glaciation cycles. Of course it was the consequence. Whoever still acts as if he were misunderstanding these basic issues is either a hopelessly brainwashed moron or an amazingly dishonest demagogue or both.
But Salby said much more than that.
He argued that the anomalous CO
2concentration may be approximated as the integral of the anomalous temperature, \[
\Delta\,{\rm conc}(CO_2) = \alpha \int \dd t\,\Delta T
\] which sort of explains why it seems to be rising so smoothly (if we ignore the nearly periodic seasonal variations). But Salby has also presented some evidence that the ice record heavily underestimates the fluctuations of the CO
2concentration, especially the high-frequency (short-period) oscillations that occurred a long time ago. If that's true, it's pretty likely that concentrations above 400 ppm may have been rather mundane even before the industrial activity.
Using the Fourier methods, he argues that there is a phase shift of 90 degrees between the temperature and CO
2pretty much at all frequencies. I am not quite seeing how this may be true because at least in the glaciation cycles, i.e. at the 10,000-year time scale, these two quantities are pretty much in sync. How does the phase shift move to 90 degrees for shorter time scales?
And his discussion of the different isotopic composition (C12 vs C13) of the fossil fuels and the present plant life is sophisticated, not the kind of silly caricature I was led to expect. At any rate, Salby concludes that the excess CO
2is caused by the integrated or accumulated positive temperature anomaly in 1920-1940 and 1980-2000 or so and these positive anomalies may be interpreted as noise, not results of any trends.
That sounds nice except that I think it's obvious that the CO
2we have added to the atmosphere has led to some increased CO 2concentrations and the latter increase is comparable to 50% of the former (airborne fraction etc.) – it's not negligible. It doesn't matter that there are 50 times more important contributions to the CO 2atmospheric budget as well. Despite these dominant contributions, a small surplus simply can't become completely invisible.
If you were thinking whether you should listen to that talk, my recommendation is probably Yes. Despite the fact that he is trying to deny some obvious facts – if I understand the discussion about the attribution of an elevated CO
2well and if I am right about its imperfections – he is also saying lots of new things and offering many sophisticated methods that you may want to know about.
At the end, Salby offers some criticisms of the climate models that I only partly agree with. Concerning the agreeable conclusions, he says that the prevailing climate models show CO
2and the temperature essentially as the same thing; in the real world, they're not the same thing at all. These two claims – and their paramount contrast – are self-evidently true.
To mention the propositions I don't quite share, he says that theories can't ever be tested against the past data; tests of predictions of the future are always needed. I disagree with that. It's a historical coincidence whether some data were collected before a theory was written down or after that. A theory is always constructed or chosen according to the data in the past and then it gives predictions for other phenomena as well. Those phenomena may be data to be collected in the future but also additional data that may be collected about the past. Regardless of the timing, such data may be used to strengthen or weaken our confidence in the theory (or rule it out).
See also replies by MeteoLCD (more detailed review than mine!), Anthony Watts' 100+ commenters, The Hockey Schtick, Climate Depot, Tall Bloke, and – from the crazy side of the aisle – John Cook (I agree with some of the criticism) and Deltoid who calls Salby "unhelpful" (for "the cause") LOL. ;-) |
Answer
36lb of meat, 50 lb of potatoes and 21lb of carrots are needed.
Work Step by Step
1. Get the factor to increase the mass of each ingredient by can be determined by $200\div140=\frac{10}{7}$ 2. The mass of each ingredient required to serve 200 people is determined by multiplying the factor with the original individual mass of each ingredient. Mass of meat:$\frac{10}{7}\times25=35.714...\approx36$ Mass of potatoes:$\frac{10}{7}\times35=50$ Mass of carrots:$\frac{10}{7}\times15=21.428...\approx21$ |
2018-08-25 06:58
Recent developments of the CERN RD50 collaboration / Menichelli, David (U. Florence (main) ; INFN, Florence)/CERN RD50 The objective of the RD50 collaboration is to develop radiation hard semiconductor detectors for very high luminosity colliders, particularly to face the requirements of the possible upgrade of the large hadron collider (LHC) at CERN. Some of the RD50 most recent results about silicon detectors are reported in this paper, with special reference to: (i) the progresses in the characterization of lattice defects responsible for carrier trapping; (ii) charge collection efficiency of n-in-p microstrip detectors, irradiated with neutrons, as measured with different readout electronics; (iii) charge collection efficiency of single-type column 3D detectors, after proton and neutron irradiations, including position-sensitive measurement; (iv) simulations of irradiated double-sided and full-3D detectors, as well as the state of their production process.. 2008 - 5 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 596 (2008) 48-52 In : 8th International Conference on Large Scale Applications and Radiation Hardness of Semiconductor Detectors, Florence, Italy, 27 - 29 Jun 2007, pp.48-52 Detaljert visning - Lignende elementer 2018-08-25 06:58 Detaljert visning - Lignende elementer 2018-08-25 06:58
Performance of irradiated bulk SiC detectors / Cunningham, W (Glasgow U.) ; Melone, J (Glasgow U.) ; Horn, M (Glasgow U.) ; Kazukauskas, V (Vilnius U.) ; Roy, P (Glasgow U.) ; Doherty, F (Glasgow U.) ; Glaser, M (CERN) ; Vaitkus, J (Vilnius U.) ; Rahman, M (Glasgow U.)/CERN RD50 Silicon carbide (SiC) is a wide bandgap material with many excellent properties for future use as a detector medium. We present here the performance of irradiated planar detector diodes made from 100-$\mu \rm{m}$-thick semi-insulating SiC from Cree. [...] 2003 - 5 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 509 (2003) 127-131 In : 4th International Workshop on Radiation Imaging Detectors, Amsterdam, The Netherlands, 8 - 12 Sep 2002, pp.127-131 Detaljert visning - Lignende elementer 2018-08-24 06:19
Measurements and simulations of charge collection efficiency of p$^+$/n junction SiC detectors / Moscatelli, F (IMM, Bologna ; U. Perugia (main) ; INFN, Perugia) ; Scorzoni, A (U. Perugia (main) ; INFN, Perugia ; IMM, Bologna) ; Poggi, A (Perugia U.) ; Bruzzi, M (Florence U.) ; Lagomarsino, S (Florence U.) ; Mersi, S (Florence U.) ; Sciortino, Silvio (Florence U.) ; Nipoti, R (IMM, Bologna) Due to its excellent electrical and physical properties, silicon carbide can represent a good alternative to Si in applications like the inner tracking detectors of particle physics experiments (RD50, LHCC 2002–2003, 15 February 2002, CERN, Ginevra). In this work p$^+$/n SiC diodes realised on a medium-doped ($1 \times 10^{15} \rm{cm}^{−3}$), 40 $\mu \rm{m}$ thick epitaxial layer are exploited as detectors and measurements of their charge collection properties under $\beta$ particle radiation from a $^{90}$Sr source are presented. [...] 2005 - 4 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 546 (2005) 218-221 In : 6th International Workshop on Radiation Imaging Detectors, Glasgow, UK, 25-29 Jul 2004, pp.218-221 Detaljert visning - Lignende elementer 2018-08-24 06:19
Measurement of trapping time constants in proton-irradiated silicon pad detectors / Krasel, O (Dortmund U.) ; Gossling, C (Dortmund U.) ; Klingenberg, R (Dortmund U.) ; Rajek, S (Dortmund U.) ; Wunstorf, R (Dortmund U.) Silicon pad-detectors fabricated from oxygenated silicon were irradiated with 24-GeV/c protons with fluences between $2 \cdot 10^{13} \ n_{\rm{eq}}/\rm{cm}^2$ and $9 \cdot 10^{14} \ n_{\rm{eq}}/\rm{cm}^2$. The transient current technique was used to measure the trapping probability for holes and electrons. [...] 2004 - 8 p. - Published in : IEEE Trans. Nucl. Sci. 51 (2004) 3055-3062 In : 50th IEEE 2003 Nuclear Science Symposium, Medical Imaging Conference, 13th International Workshop on Room Temperature Semiconductor Detectors and Symposium on Nuclear Power Systems, Portland, OR, USA, 19 - 25 Oct 2003, pp.3055-3062 Detaljert visning - Lignende elementer 2018-08-24 06:19
Lithium ion irradiation effects on epitaxial silicon detectors / Candelori, A (INFN, Padua ; Padua U.) ; Bisello, D (INFN, Padua ; Padua U.) ; Rando, R (INFN, Padua ; Padua U.) ; Schramm, A (Hamburg U., Inst. Exp. Phys. II) ; Contarato, D (Hamburg U., Inst. Exp. Phys. II) ; Fretwurst, E (Hamburg U., Inst. Exp. Phys. II) ; Lindstrom, G (Hamburg U., Inst. Exp. Phys. II) ; Wyss, J (Cassino U. ; INFN, Pisa) Diodes manufactured on a thin and highly doped epitaxial silicon layer grown on a Czochralski silicon substrate have been irradiated by high energy lithium ions in order to investigate the effects of high bulk damage levels. This information is useful for possible developments of pixel detectors in future very high luminosity colliders because these new devices present superior radiation hardness than nowadays silicon detectors. [...] 2004 - 7 p. - Published in : IEEE Trans. Nucl. Sci. 51 (2004) 1766-1772 In : 13th IEEE-NPSS Real Time Conference 2003, Montreal, Canada, 18 - 23 May 2003, pp.1766-1772 Detaljert visning - Lignende elementer 2018-08-24 06:19
Radiation hardness of different silicon materials after high-energy electron irradiation / Dittongo, S (Trieste U. ; INFN, Trieste) ; Bosisio, L (Trieste U. ; INFN, Trieste) ; Ciacchi, M (Trieste U.) ; Contarato, D (Hamburg U., Inst. Exp. Phys. II) ; D'Auria, G (Sincrotrone Trieste) ; Fretwurst, E (Hamburg U., Inst. Exp. Phys. II) ; Lindstrom, G (Hamburg U., Inst. Exp. Phys. II) The radiation hardness of diodes fabricated on standard and diffusion-oxygenated float-zone, Czochralski and epitaxial silicon substrates has been compared after irradiation with 900 MeV electrons up to a fluence of $2.1 \times 10^{15} \ \rm{e} / cm^2$. The variation of the effective dopant concentration, the current related damage constant $\alpha$ and their annealing behavior, as well as the charge collection efficiency of the irradiated devices have been investigated.. 2004 - 7 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 530 (2004) 110-116 In : 6th International Conference on Large Scale Applications and Radiation Hardness of Semiconductor Detectors, Florence, Italy, 29 Sep - 1 Oct 2003, pp.110-116 Detaljert visning - Lignende elementer 2018-08-24 06:19
Recovery of charge collection in heavily irradiated silicon diodes with continuous hole injection / Cindro, V (Stefan Inst., Ljubljana) ; Mandić, I (Stefan Inst., Ljubljana) ; Kramberger, G (Stefan Inst., Ljubljana) ; Mikuž, M (Stefan Inst., Ljubljana ; Ljubljana U.) ; Zavrtanik, M (Ljubljana U.) Holes were continuously injected into irradiated diodes by light illumination of the n$^+$-side. The charge of holes trapped in the radiation-induced levels modified the effective space charge. [...] 2004 - 3 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 518 (2004) 343-345 In : 9th Pisa Meeting on Advanced Detectors, La Biodola, Italy, 25 - 31 May 2003, pp.343-345 Detaljert visning - Lignende elementer 2018-08-24 06:19
First results on charge collection efficiency of heavily irradiated microstrip sensors fabricated on oxygenated p-type silicon / Casse, G (Liverpool U.) ; Allport, P P (Liverpool U.) ; Martí i Garcia, S (CSIC, Catalunya) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Turner, P R (Liverpool U.) Heavy hadron irradiation leads to type inversion of n-type silicon detectors. After type inversion, the charge collected at low bias voltages by silicon microstrip detectors is higher when read out from the n-side compared to p-side read out. [...] 2004 - 3 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 518 (2004) 340-342 In : 9th Pisa Meeting on Advanced Detectors, La Biodola, Italy, 25 - 31 May 2003, pp.340-342 Detaljert visning - Lignende elementer 2018-08-23 11:31
Formation and annealing of boron-oxygen defects in irradiated silicon and silicon-germanium n$^+$–p structures / Makarenko, L F (Belarus State U.) ; Lastovskii, S B (Minsk, Inst. Phys.) ; Korshunov, F P (Minsk, Inst. Phys.) ; Moll, M (CERN) ; Pintilie, I (Bucharest, Nat. Inst. Mat. Sci.) ; Abrosimov, N V (Unlisted, DE) New findings on the formation and annealing of interstitial boron-interstitial oxygen complex ($\rm{B_iO_i}$) in p-type silicon are presented. Different types of n+−p structures irradiated with electrons and alpha-particles have been used for DLTS and MCTS studies. [...] 2015 - 4 p. - Published in : AIP Conf. Proc. 1583 (2015) 123-126 Detaljert visning - Lignende elementer |
Hey guys! I built the voltage multiplier with alternating square wave from a 555 timer as a source (which is measured 4.5V by my multimeter) but the voltage multiplier doesn't seem to work. I tried first making a voltage doubler and it showed 9V (which is correct I suppose) but when I try a quadrupler for example and the voltage starts from like 6V and starts to go down around 0.1V per second.
Oh! I found a mistake in my wiring and fixed it. Now it seems to show 12V and instantly starts to go down by 0.1V per sec.
But you really should ask the people in Electrical Engineering. I just had a quick peek, and there was a recent conversation about voltage multipliers. I assume there are people there who've made high voltage stuff, like rail guns, which need a lot of current, so a low current circuit like yours should be simple for them.
So what did the guys in the EE chat say...
The voltage multiplier should be ok on a capacitive load. It will drop the voltage on a resistive load, as mentioned in various Electrical Engineering links on the topic. I assume you have thoroughly explored the links I have been posting for you...
A multimeter is basically an ammeter. To measure voltage, it puts a stable resistor into the circuit and measures the current running through it.
Hi all! There is theorem that links the imaginary and the real part in a time dependent analytic function. I forgot its name. Its named after some dutch(?) scientist and is used in solid state physics, who can help?
The Kramers–Kronig relations are bidirectional mathematical relations, connecting the real and imaginary parts of any complex function that is analytic in the upper half-plane. These relations are often used to calculate the real part from the imaginary part (or vice versa) of response functions in physical systems, because for stable systems, causality implies the analyticity condition, and conversely, analyticity implies causality of the corresponding stable physical system. The relation is named in honor of Ralph Kronig and Hans Kramers. In mathematics these relations are known under the names...
I have a weird question: The output on an astable multivibrator will be shown on a multimeter as half the input voltage (for example we have 9V-0V-9V-0V...and the multimeter averages it out and displays 4.5V). But then if I put that output to a voltage doubler, the voltage should be 18V, not 9V right? Since the voltage doubler will output in DC.
I've tried hooking up a transformer (9V to 230V, 0.5A) to an astable multivibrator (which operates at 671Hz) but something starts to smell burnt and the components of the astable multivibrator get hot. How do I fix this? I check it after that and the astable multivibrator works.
I searched the whole god damn internet, asked every god damn forum and I can't find a single schematic that converts 9V DC to 1500V DC without using giant transformers and power stage devices that weight 1 billion tons....
something so "simple" turns out to be hard as duck
In peskin book of QFT the sum over zero point energy modes is an infinite c-number, fortunately, it's experimental evidence doesn't appear, since experimentalists measure the difference in energy from the ground state. According to my understanding the zro pt energy is the same as the ground state, isn't it?
If so, always it is possible to substract a finite number (higher exited state for e.g.) from this zero point enrgy (which is infinte), it follows that, experimentally we always obtain infinte spectrum.
@AaronStevens Yeah, I had a good laugh to myself when he responded back with "Yeah, maybe they considered it and it was just too complicated". I can't even be mad at people like that. They are clearly fairly new to physics and don't quite grasp yet that most "novel" ideas have been thought of to death by someone; likely 100+ years ago if it's classical physics
I have recently come up with a design of a conceptual electromagntic field propulsion system which should not violate any conservation laws, particularly the Law of Conservation of Momentum and the Law of Conservation of Energy. In fact, this system should work in conjunction with these two laws ...
I rememeber that Gordon Freeman's thesis was "Observation of Einstein-Podolsky-Rosen Entanglement on Supraquantum Structures by Induction Through Nonlinear Transuranic Crystal of Extremely Long Wavelength (ELW) Pulse from Mode-Locked Source Array "
In peskin book of QFT the sum over zero point energy modes is an infinite c-number, fortunately, it's experimental evidence doesn't appear, since experimentalists measure the difference in energy from the ground state. According to my understanding the zro pt energy is the same as the ground state, isn't it? If so, always it is possible to substract a finite number (higher exited state for e.g.) from this zero point enrgy (which is infinte), it follows that, experimentally we always obtain infinte spectrum.
@ACuriousMind What confuses me is the interpretation of Peskin to this infinite c-number and the experimental fact
He said, the second term is the sum over zero point energy modes which is infnite as you mentioned. He added," fortunately, this energy cannot be detected experm., since the experiments measure only the difference between from the ground state of H".
@ACuriousMind Thank you, I understood your explanations clearly. However, regarding what Peskin mentioned in his book, there is a contradiction between what he said about the infinity of the zero point energy/ground state energy, and the fact that this energy is not detectable experimentally because the measurable quantity is the difference in energy between the ground state (which is infinite and this is the confusion) and a higher level.
It's just the first encounter with something that needs to be renormalized. Renormalizable theories are not "incomplete", even though you can take the Wilsonian standpoint that renormalized QFTs are effective theories cut off at a scale.
according to the author, the energy differenc is always infinite according to two fact. the first is, the ground state energy is infnite, secondly, the energy differenc is defined by substituting a higher level energy from the ground state one.
@enumaris That is an unfairly pithy way of putting it. There are finite, rigorous frameworks for renormalized perturbation theories following the work of Epstein and Glaser (buzzword: Causal perturbation theory). Just like in many other areas, the physicist's math sweeps a lot of subtlety under the rug, but that is far from unique to QFT or renormalization
The classical electrostatics formula $H = \int \frac{\mathbf{E}^2}{8 \pi} dV = \frac{1}{2} \sum_a e_a \phi(\mathbf{r}_a)$ with $\phi_a = \sum_b \frac{e_b}{R_{ab}}$ allows for $R_{aa} = 0$ terms i.e. dividing by zero to get infinities also, the problem stems from the fact that $R_{aa}$ can be zero due to using point particles, overall it's an infinite constant added to the particle that we throw away just as in QFT
@bolbteppa I understand the idea that we need to drop such terms to be in consistency with experiments. But i cannot understand why the experiment didn't predict such infinities that arose in the theory?
These $e_a/R_{aa}$ terms in the big sum are called self-energy terms, and are infinite, which means a relativistic electron would also have to have infinite mass if taken seriously, and relativity forbids the notion of a rigid body so we have to model them as point particles and can't avoid these $R_{aa} = 0$ values. |
Accessing QuantumATK internal variables¶ Version: 2016.0
The purpose of this tutorial is to illustrate how to extract internal quantities from QuantumATK. In the following, the quantities are briefly described and several sections then illustrate with examples how the information can be used to compute different sorts of transport coefficients.
Internal matrices accessible in QuantumATK¶
The table below lists some of the internal matrices in QuantumATK, e.g. the Hamiltonian \(H\) and the overlap matrix \(S\). The commands listed in the left-hand column in the table allow users to extract these matrices through the so-called “low level entities” module in QuantumATK.
Command Symbol
orbitalInfo()
\([(i,l,m),\cdots]\)
calculateHamiltonianAndOverlap()
\(H(\mathbf{k}), S(\mathbf{k})\)
calculateDensityMatrix()
\(D(\bf{k})\)
calculateSelfEnergy()
\(\Sigma^{L/R} (E,\mathbf{k})\)
calculateRetardedGreenFunction()
\(G^r (E,\mathbf{k})\)
calculateGreenFunctionComponent()
\(G^{L/R} (E,\mathbf{k})\)
calculateDynamicalMatrixAndOverlap()
\(C(\mathbf{k}) , M(\mathbf{k})\)
calculatePhononSelfEnergy()
\(\Pi^{L/R} (E,\mathbf{k})\)
calculatePhononRetardedGreenFunction()
\(D^r (E,\mathbf{k})\)
calculatePhononGreenFunctionComponent()
\(D^{L/R} (E,\mathbf{k})\)
You will in this tutorial see examples of how to extract these matrices and use them to compute transport coefficients. The underlying calculation engine to be used here is ATK-SemiEmpirical (ATK-SE), but could also have been ATK-DFT.
Multi-terminal conduction¶
You will in this section consider a 4-probe graphene device, and calculate the transmission between each of the four probes. For the calculation, you will extract the self energies of each of the four probes and implement the Fisher–Lee transmission function in a small QuantumATK Python script. The example is inspired by the 4-probe setup in [KPB07].
Device calculations¶
You should first use the Slater–Koster tight-binding method to compute the ground state and zero-bias transmission at the Fermi level for the 4-terminal graphene device illustrated below.
Start up QuantumATK with a new project, for instance named “4probe graphene”. Then downloadthis pre-made QuantumATK Python script and save it in the project folder:
4probe_pi_model.py. It setsup the device configuration with a Slater–Koster calculator, and then performs thecalculation and computes the transmission at the Fermi level. All output is saved in
4probe_pi_model.hdf5.
$ atkpython 4probe_pi_model.py > 4probe_pi_model.log
The script will finish very quickly, and reports the Fermi level transmission in the log file:
+----------------------------------------------------------+| Transmission Spectrum Report || -------------------------------------------------------- || Left electrode Fermi level = 3.851459e-01 eV || Right electrode Fermi level = 3.851459e-01 eV || Energy zero = 3.851459e-01 eV |+----------------------------------------------------------+ energy T(up) eV 0.000000e+00 7.553672e-01
Calculating the transmission between the 4 leads¶
The calculated transmission in the previous section is from the two leads on the left-hand side of the device, which are tagged “lead1” and “lead2”, to the two leads on the right-hand side of the device, tagged “lead3” and “lead4”.
ATK has no analysis functions for calculating transmission between individual leads. For such a calculation, you need to calculate the broadening function of each lead, \(\Gamma_i\), and then evaluate the Fisher–Lee relation,
where \(G^r\) is the retarded Green function of the device.
The downloadable script
4probe_trans.py performs this calculation. The script extractsinternal QuantumATK quantities for the calculation, and the following section discussesdetails of the script.
Details of the script¶
The first part of the script is a utility function, which can zero out all entriesin the \(\Gamma\) matrix, except for the orbitals belonging to the indices ofthe specified atoms. To obtain a list of the orbitals on the atoms, the functionuses a
ProjectionList object.
# Utility function to project out Gamma of the lead1, lead2, lead3, lead4def projectGamma(configuration, gamma, indices): """Zero all components in Gamma which are not in the indices list @param configuration : The configuration giving the full Gamma @param gamma : The full gamma @param indices : Indices of the atoms to project onto @return : The projected gamma """ projection_list = ProjectionList(indices) orbital_index = projection_list.orbitalIndex(configuration) # Make a filter matrix consisting of 1's at the orbital_index filter = 0*gamma for i in orbital_index: for j in orbital_index: filter[i,j] = 1 return gamma*filter
Another important part is the lines shown below, which use the tags on the device configuration to determine the indices of the atoms in each electrode.
# Project onto Lead 1, 2, 3, 4lead1_index = device_configuration.indicesFromTags('lead1')Gamma1 = projectGamma(device_configuration, Gamma_L, lead1_index)lead2_index = device_configuration.indicesFromTags('lead2')Gamma2 = projectGamma(device_configuration, Gamma_L, lead2_index)lead3_index = device_configuration.indicesFromTags('lead3')Gamma3 = projectGamma(device_configuration, Gamma_R, lead3_index)lead4_index = device_configuration.indicesFromTags('lead4')Gamma4 = projectGamma(device_configuration, Gamma_R, lead4_index)
The rest of the script should be self explanatory to experienced QuantumATK users.
Running the script¶
Save the script and execute it. It should produce the following output:
T: L (1,2) -> R (3,4) : 0.756157362529T: Up(1,3) -> Down (2,4) : 0.232988936132Transmission Matrix[[ 3.99103191 0.05825146 0.04799848 0.05832079] [ 0.05825146 4.60578343 0.05853133 0.59130676] [ 0.04799848 0.05853133 3.74982393 0.05788536] [ 0.05832079 0.59130676 0.05788536 1.96636417]]
The script reports the transmission matrix, \(T_{ij}\). By summing up \(i \in L\), \(i \in R\) it is possible to get the transmission from left to right, which is identical (within numerical noise), to the value from the bare QuantumATK transmission calculation.
By summing up the transmission coefficients between the upper leads and the lower leads, it is possible to calculate the up-down transmission, as in [KPB07].
Transmission projection¶
You will here learn how to resolve a transmission calculation into molecular projected self-consistent Hamiltonian (MPSH) eigenstates. In this way, you can analyze how large a fraction of a transmission function is propagating through a particular eigenstate or part of the system. As an example, you will investigate a dithiol-benzene (DTB) ring between two gold surfaces, see [STB+03].
Running the calculations¶
Use the pre-made QuantumATK Python script
au_dtb_au.py, which defines the Au–DTB–Audevice configuration, performs a DFT-SE calculation using the extended Hückel method,and computes the Γ-point transmission in an energy range \(\pm\)1 eV aroundthe Fermi level. The calculations will finish in a few minutes.
Analyzing the transmission¶
Once the calculation has finished, the file
au_dtb_au.hdf5 will appear on theLabFloor. Select the
TransmissionSpectrum item and open the Transmission Analyzer.
The Γ-point transmission spectrum has a peak 0.68 eV above the Fermi level. Selectthis peak using the mouse, and click
Eigenvalues to calculate the transmissioneigenvalues. There is one dominating eigenvalue very close to 1, and severalnegligible eigenvalues.
Remove the tick from all except the highest eigenvalue, and click the
Eigenstatesbutton. This will calculate the scattering eigenstate in real space, and the Viewerwill pop up with a visualization of it – choose the “isosurface” visualization optionif prompted. You can then use the optionsto tune the visualization. Projecting the transmission¶
We would now like to project the transmission eigenstate onto the MPSH states of the molecule, to find the orbitals that carry the transmission.
The scattering state, \(\psi(\mathbf{r})\), is expanded in basis orbitals, \(\phi_i(\mathbf{r})\), through expansion coefficients \(v_i\),
We now diagonalize the self-consistent Hamiltonian projected onto DTB,
where \(\mathbf{c}_\alpha\) are the expansion coefficients of the MPSH states.
Next, we expand the projection of the scattering state of the DTB molecule in the MPSH states,
where the expansion coefficients are given by \(a_\alpha = \mathbf{c}_\alpha^\dagger S^\mathrm{DTB} \mathbf{v}\). Through the magnitude of each \(a_\alpha\), we can get the relevance of each MPSH state, and
from QuantumATK import *from utilities import vectorToGrid, scatteringStates, averageFermiLevelimport scipy# Read the configurationdevice_configuration = nlread('au_dtb_au.hdf5', DeviceConfiguration)[0]# Get H and SH, S = calculateHamiltonianAndOverlap(device_configuration)H = H.inUnitsOf(eV)# Calculate average Fermi levelaverage_fermi_level = averageFermiLevel(device_configuration)energy = average_fermi_level+0.68*eV# Get index of orbitals on the Phenyl ringprojection_list = ProjectionList(elements = [Carbon, Hydrogen])orbital_index = projection_list.orbitalIndex(device_configuration)# Project H, S onto the Phenyl ring (MPSH)H_dtb = H[orbital_index,:][:, orbital_index]S_dtb = S[orbital_index,:][:, orbital_index]# Calculate the mpsh eigenfunctionsw, v = scipy.linalg.eigh(H_dtb, S_dtb)# Calculate the scattering eigenstatesT, c = scatteringStates(device_configuration, energy)# Take the highest eigenstateeigenvector = c[:,0]# Project the eigenstate onto the DTB moleculeev_dtb = eigenvector[orbital_index]# Get the norm of the projected eigenstatenorm_ev = numpy.dot(numpy.conj(ev_dtb.transpose()), numpy.dot(S_dtb, ev_dtb))# Loop over all MPSHfor i in range(len(w)): # Resolve the eigenstate into the MPSH states coeff = numpy.dot(numpy.conj(v[:,i].transpose()), numpy.dot(S_dtb, ev_dtb)) # Find the strength of the MPSH projection p = numpy.conj(coeff)*coeff p = numpy.abs(p/norm_ev) # Print out the weight of each non negligible projection if p > 0.001: print(i, p)
The script uses some utility functions from the script
utilities.py.
Running the scripts¶
Save both scripts, and execute
projection.py using the Job Manageror from command line. It will generate the following output:
8 0.0068683436565811 0.0046761062886315 0.98164733672117 0.00670447660685
The MPSH state number 15 is clearly the one with the largest projection.
Plotting the MPSH states¶
The script
mpsh.py shown below will save MPSH states 8, 11, 15, and 17 into theHDF5 data file
mpsh.hdf5. The script uses the function
vectorToGrid(),which performs the folding of the eigenvector with the basis functions.
from QuantumATK import *from utilities import vectorToGrid, averageFermiLevelimport scipy# Read the configurationdevice_configuration = nlread('au_dtb_au.hdf5', DeviceConfiguration)[0]# Get H and SH, S = calculateHamiltonianAndOverlap(device_configuration)H = H.inUnitsOf(eV)# Calculate average Fermi levelaverage_fermi_level = averageFermiLevel(device_configuration)# Get index of orbitals on the Phenyl ringprojection_list = ProjectionList(elements=[Carbon, Hydrogen])orbital_index = projection_list.orbitalIndex(device_configuration)# Project H, S onto the Phenyl ringH_dtb = H[orbital_index,:][:, orbital_index]S_dtb = S[orbital_index,:][:, orbital_index]# Calculate the eigenfunctionsw, v = scipy.linalg.eigh(H_dtb, S_dtb)# Calculate eigen energies relative to the fermi leveleigen_energies = (w*eV-average_fermi_level)# Save eigenstates number 13-17for i in [8, 11, 15, 17]: print('eigenenergy ', i, eigen_energies[i]) # Put the eigenvector into a vector of length of all orbitals number_orbitals = H.shape[0] eigenvector = numpy.zeros(number_orbitals, dtype=complex) eigenvector[orbital_index] = v[:,i] grid = vectorToGrid(eigenvector, device_configuration) nlsave('mpsh.hdf5', grid)
Run the script, and when finished drag and drop the object with ID
gID002from
mpsh.hdf5 onto the Viewer. Then add the device configurationto the plot by drag and dropping it from
au_dtb_au.hdf5. By adjusting theplot properties, you should be able to get the image shown below.
Note
The
vectorToGrid() method uses a finer grid spacing than the value set bythe HuckelCalculator. This is why the image above has high-quality resolution.
AC conductance¶
The scripts presented in this section can be used to calculate the AC conductanceof a nanodevice within the wide-band limit. The implementation follows closely thework by Yamamoto
et al. [YSWW10].
We will here consider the (10,10) carbon nanotube device illustrated below, and useagain the extended Hückel method with a \(\pi\)-model. Use
cnt_device.py to performthe ground state device calculation.
The
admittance is defined as the inverse of the impedance, and is a measure ofhow easily a device will allow a current to flow. Download the scripts
cnt_admittance.pyand
admittance.py to your QuantumATK project folder. Then execute
cnt_admittance.py, which computesand plots the real (G) and imaginary (B) parts of the AC admittance for the CNTdevice.
Note
Since the implementation of the AC conductance is done in Python and uses dense matrices, the calculation is computationally inefficient, and for systems with a large number of orbitals the calculation can take substantial time.
References¶
[KPB07] (1, 2) M. Koleini, M. Paulsson, and M. Brandbyge. Efficient organometallic spin filter between single-wall carbon nanotube or graphene electrodes. Phys. Rev. Lett., 98:197202, May 2007. doi:10.1103/PhysRevLett.98.197202.
[PB07] M. Paulsson and M. Brandbyge. Transmission eigenchannels from nonequilibrium green’s functions. Phys. Rev. B, 76:115117, Sep 2007. doi:10.1103/PhysRevB.76.115117.
[STB+03] K. Stokbro, J. Taylor, M. Brandbyge, J.-L. Mozos, and P. Ordejón. Theoretical study of the nonlinear conductance of di-thiol benzene coupled to Au(1 1 1) surfaces via thiol and thiolate bonds. Computational Materials Science, 27(1–2):151 – 160, 2003. doi:10.1016/S0927-0256(02)00439-1.
[YSWW10] T. Yamamoto, K. Sasaoka, S. Watanabe, and K. Watanabe. Two chirality classes of ac quantum transport in metallic carbon nanotubes. Phys. Rev. B, 81:115448, Mar 2010. doi:10.1103/PhysRevB.81.115448. |
True or False? let \( f : [0,1] \to \Bbb R \) be a continuous function such that \( f(x) \geq x^3\) \(\forall x \in [0,1] \) with \(\int_0^1 f(x)= 1/4 \). Then \( f(x)=x^3 \forall x \in \Bbb R \).
We know that continuous functions are integrable, so, how can you use this fact to solve this question?
We can see that \( \int_0^1 f(x) dx \geq 1/4 \) but given value is exactly 1/4. Hence we can conclude that \( f(x)=x^3 \).
The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics. |
Sine of the Sum Formula
The applet below illustrates a proof without words of the "sine of the sum" formula due to Volker Priebe and Edgar A. Ramos [Nelsen, p. 40].
The rhombus inscribed into a rectangle has side length of $1.$ The rhombus cuts off of the rectangle two pairs of equal right triangles. The acute angles of the triangles are $\alpha,$ $90^{\circ} -\alpha,$ $\beta,$ $90^{\circ} - \beta.$ The vertices of the rhombus split the sides of the rectangle into segments of lengths $\cos\alpha,$ $\sin\alpha,$ $\cos\beta,$ $\sin\beta,$ as shown.
The area of the rhombus is $\sin(\alpha + \beta).$ In the right half of the applet, the triangles rearranged leaving two rectangles unoccupied. The area of one is $\sin\alpha \times \cos\beta,$ that of the other $\cos\alpha \times \sin\beta,$ proving the "sine of the sum" formula
$\sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta.$
References R. B. Nelsen, Proofs Without Words II, MAA, 2000 Trigonometry What Is Trigonometry? Addition and Subtraction Formulas for Sine and Cosine The Law of Cosines (Cosine Rule) Cosine of 36 degrees Tangent of 22.5 o- Proof Wthout Words Sine and Cosine of 15 Degrees Angle Sine, Cosine, and Ptolemy's Theorem arctan(1) + arctan(2) + arctan(3) = π Trigonometry by Watching arctan(1/2) + arctan(1/3) = arctan(1) Morley's Miracle Napoleon's Theorem A Trigonometric Solution to a Difficult Sangaku Problem Trigonometric Form of Complex Numbers Derivatives of Sine and Cosine ΔABC is right iff sin²A + sin²B + sin²C = 2 Advanced Identities Hunting Right Angles Point on Bisector in Right Angle Trigonometric Identities with Arctangents The Concurrency of the Altitudes in a Triangle - Trigonometric Proof Butterfly Trigonometry Binet's Formula with Cosines Another Face and Proof of a Trigonometric Identity cos/sin inequality On the Intersection of kx and |sin(x)| Cevians And Semicircles Double and Half Angle Formulas A Nice Trig Formula Another Golden Ratio in Semicircle Leo Giugiuc's Trigonometric Lemma Another Property of Points on Incircle Much from Little The Law of Cosines and the Law of Sines Are Equivalent Wonderful Trigonometry In Equilateral Triangle A Trigonometric Observation in Right Triangle A Quick Proof of cos(pi/7)cos(2.pi/7)cos(3.pi/7)=1/8 Proofs Without Words Proofs Without Words Sums of Geometric Series - Proofs Without Words Sine of the Sum Formula Parallelogram Law: A PWW Parallelogram Law Ceva's Theorem: Proof Without Words Viviani's Theorem A Property of Rhombi Triangular Numbers in a Square PWW: How Geometry Helps Algebra Varignon's Theorem, Proof Without Words
65620716 |
Under the auspices of the Computational Complexity Foundation (CCF)
Let $p \ge 2$. We improve the bound $\frac{\|f\|_p}{\|f\|_2} \le (p-1)^{s/2}$ for a polynomial $f$ of degree $s$ on the boolean cube $\{0,1\}^n$, which comes from hypercontractivity, replacing the right hand side of this inequality by an explicit bivariate function of $p$ and $s$, which is smaller than $(p-1)^{s/2}$ for ... more >>>
Let $T_{\epsilon}$ be the noise operator acting on functions on the boolean cube $\{0,1\}^n$. Let $f$ be a nonnegative function on $\{0,1\}^n$ and let $q \ge 1$. We upper bound the $\ell_q$ norm of $T_{\epsilon} f$ by the average $\ell_q$ norm of conditional expectations of $f$, given sets of roughly ... more >>>
We suggest a new approach to obtain bounds on locally correctable and some locally testable binary linear codes, by arguing that their coset leader graphs have high discrete Ricci curvature.
The bounds we obtain for locally correctable codes are worse than the best known bounds obtained using quantum information theory, ... more >>>
Given a subset $A\subseteq \{0,1\}^n$, let $\mu(A)$ be the maximal ratio between $\ell_4$ and $\ell_2$ norms of a function whose Fourier support is a subset of $A$. We make some simple observations about the connections between $\mu(A)$ and the additive properties of $A$ on one hand, and between $\mu(A)$ and ... more >>>
Let $f$ be a nonnegative function on $\{0,1\}^n$. We upper bound the entropy of the image of $f$ under the noise operator with noise parameter $\epsilon$ by the average entropy of conditional expectations of $f$, given sets of roughly $(1-2\epsilon)^2 \cdot n$ variables.
As an application, we show that for ... more >>>
A square matrix $V$ is called rigid if every matrix $V^\prime$ obtained by altering a small number of entries of $V$ has sufficiently high rank. While random matrices are rigid with high probability, no explicit constructions of rigid matrices are known to date. Obtaining such explicit matrices would have major ... more >>>
A design is a finite set of points in a space on which every "simple" functions averages to its global mean. Illustrative examples of simple functions are low-degree polynomials on the Euclidean sphere or on the Hamming cube.
We prove lower bounds on designs in spaces with a large group ... more >>> |
Under the auspices of the Computational Complexity Foundation (CCF)
Recently, an extension of the standard data stream model has been introduced in which an algorithm can create and manipulate multiple read/write streams in addition to its input data stream. Like the data stream model, the most important parameter for this model is the amount of internal memory used by ... more >>>
The Gap-Hamming-Distance problem arose in the context of proving space
lower bounds for a number of key problems in the data stream model. In this problem, Alice and Bob have to decide whether the Hamming distance between their $n$-bit input strings is large (i.e., at least $n/2 + \sqrt n$) ... more >>>
This paper makes three main contributions to the theory of communication complexity and stream computation. First, we present new bounds on the information complexity of AUGMENTED-INDEX. In contrast to analogous results for INDEX by Jain, Radhakrishnan and Sen [J. ACM, 2009], we have to overcome the significant technical challenge that ... more >>>
The deterministic space complexity of approximating the length of the longest increasing subsequence of
a stream of $N$ integers is known to be $\widetilde{\Theta}(\sqrt N)$. However, the randomized complexity is wide open. We show that the technique used in earlier work to establish the $\Omega(\sqrt N)$ deterministic lower bound fails ... more >>>
We prove an optimal $\Omega(n)$ lower bound on the randomized
communication complexity of the much-studied Gap-Hamming-Distance problem. As a consequence, we obtain essentially optimal multi-pass space lower bounds in the data stream model for a number of fundamental problems, including the estimation of frequency moments.
The Gap-Hamming-Distance problem is a ... more >>>
We study the communication complexity of evaluating functions when the input data is randomly allocated (according to some known distribution) amongst two or more players, possibly with information overlap. This naturally extends previously studied variable partition models such as the best-case and worst-case partition models. We aim to understand whether ... more >>>
Motivated by the trend to outsource work to commercial cloud computing services, we consider a variation of the streaming paradigm where a streaming algorithm can be assisted by a powerful helper that can provide annotations to the data stream. We extend previous work on such annotation models by considering a ... more >>>
The central goal of data stream algorithms is to process massive streams of data using sublinear storage space. Motivated by work in the database community on outsourcing database and data stream processing, we ask whether the space usage of such algorithms can be further reduced by enlisting a more powerful ... more >>>
In graph streaming a graph with $n$ vertices and $m$ edges is presented as a read-once stream of edges. We obtain an $\Omega(n \log n)$ lower bound on the space required to decide graph connectivity. This improves the known bounds of $\Omega(n)$ for undirected and $\Omega(m)$ for sparse directed graphs. ... more >>>
We study the power of Arthur-Merlin probabilistic proof systems in the data stream model. We show a canonical $\mathcal{AM}$ streaming algorithm for a wide class of data stream problems. The algorithm offers a tradeoff between the length of the proof and the space complexity that is needed to verify it.... more >>>
We develop a paradigm for studying multi-player deterministic communication,
based on a novel combinatorial concept that we call a {\em strong fooling set}. Our paradigm leads to optimal lower bounds on the per-player communication required for solving multi-player $\textsc{equality}$ problems in a private-message setting. This in turn gives a ... more >>> |
Let $X$ be a metrizable compact topological space, let $\mathcal U$ be an ultrafilter, and denote by $X^{\mathcal U}$ the ultracopower of $X$ with respect to $\mathcal U$. As a C$^*$-algebraist, I prefer to define $X^{\mathcal U}$ as the compact Hausdorff space satisfying \begin{align*} C(X^{\mathcal U},\mathbb C) &\cong \prod_{\mathcal U} C(X,\mathbb C) \\ &:= \ell^\infty(\mathbb N, C(X, \mathbb C))/\{(f_n)_n \mid \lim_{n\to\mathcal U} \|f_n\|_{\infty} = 0\}. \end{align*} It can alternatively be described, directly, as ultrafilters in $\mathbb N \times X$ which respect $\mathcal U$ (see Bankston's "Reduced coproducts of compact Hausdorff spaces" for a more precise version of this).
I asked previously whether $X^{\mathcal U} \times Y^{\mathcal U} \cong (X \times Y)^{\mathcal U}$, and Tomek Kania showed that the answer is no.
Let us now define a unital $C(X)$-algebra. A unital C*-algebra $B$ is a $C(X)$-algebra if it is equipped with a unital *-homomorphism $\theta$ from $C(X)$ to the centre of $B$. This allows one to view $B$ as an algebra of sections of a bundle over $X$; the section at $x\in X$ is defined to be the quotient of $B$ by the ideal generated by $\theta(C(X\setminus \{x\}))$.
There is a canonical map $C(X)_\omega \to C(X\times Y)_\omega$, so that the codomain is a $C(X^{\omega})$-algebra. The answer to my previous question shows that it is not the trivial $C(X^\omega)$-algebra with fibres $C(Y^\omega)$. My new question is: what are the fibres?
I can answer this question for a dense set of points, namely those elements of $X^\omega$ that correspond to a character on $C(X)_\omega$ of the form \begin{equation} (f_n)_n \mapsto \lim_{n \to \omega} f_n(x_n), \end{equation} for a sequence $(x_n)_n$ in $X$. The fibre at such a point is $C(Y)_\omega$. There is nothing subtle about proving this. |
The Annals of Mathematical Statistics Ann. Math. Statist. Volume 42, Number 5 (1971), 1671-1680. Limit Theorems for Some Occupancy and Sequential Occupancy Problems Abstract
Consider a situation in which balls are falling into $N$ cells with arbitrary probabilities. A limiting distribution for the number of occupied cells after $n$ falls is obtained, when $n$ and $N \rightarrow \infty$, so that $n^2/N \rightarrow \infty$ and $n/N \rightarrow 0$. This result completes some theorems given by Chistyakov (1964), (1967). Limiting distributions of the number of falls to achieve $a_N + 1$ occupied cells are obtained when $\lim \sup a_N/N < 1$. These theorems generalize theorems given by Baum and Billingsley (1965), and David and Barton (1962), when the balls fall into cells with the same probability for every cell.
Article information Source Ann. Math. Statist., Volume 42, Number 5 (1971), 1671-1680. Dates First available in Project Euclid: 27 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aoms/1177693165 Digital Object Identifier doi:10.1214/aoms/1177693165 Mathematical Reviews number (MathSciNet) MR343347 Zentralblatt MATH identifier 0231.60022 JSTOR links.jstor.org Citation
Holst, Lars. Limit Theorems for Some Occupancy and Sequential Occupancy Problems. Ann. Math. Statist. 42 (1971), no. 5, 1671--1680. doi:10.1214/aoms/1177693165. https://projecteuclid.org/euclid.aoms/1177693165 |
The Annals of Probability Ann. Probab. Volume 32, Number 1A (2004), 191-215. Occupation densities for SPDEs with reflection Abstract
We consider the solution $(u,\eta)$ of the white-noise driven stochastic partial differential equation with reflection on the space interval $[0,1]$ introduced by Nualart and Pardoux, where $\eta$ is a reflecting measure on $[0,\infty)\times(0,1)$ which forces the continuous function
u, defined on $[0,\infty)\times[0,1]$, to remain nonnegative and $\eta$ has support in the set of zeros of u. First, we prove that at any fixed time $t>0$, the measure $\eta([0,t]\times d\theta)$ is absolutely continuous w.r.t. the Lebesgue measure $d\theta$ on $(0,1)$. We characterize the density as a family of additive functionals of u, and we interpret it as a renormalized local time at $0$ of $(u(t,\theta))_{t\geq 0}$. Finally, we study thebehavior of $\eta$ at the boundary of $[0,1]$. The main technical novelty is a projection principle from the Dirichlet space of a Gaussian process, vector-valued solution of a linear SPDE, to the Dirichlet space of the process u. Article information Source Ann. Probab., Volume 32, Number 1A (2004), 191-215. Dates First available in Project Euclid: 4 March 2004 Permanent link to this document https://projecteuclid.org/euclid.aop/1078415833 Digital Object Identifier doi:10.1214/aop/1078415833 Mathematical Reviews number (MathSciNet) MR2040780 Zentralblatt MATH identifier 1121.60069 Citation
Zambotti, Lorenzo. Occupation densities for SPDEs with reflection. Ann. Probab. 32 (2004), no. 1A, 191--215. doi:10.1214/aop/1078415833. https://projecteuclid.org/euclid.aop/1078415833 |
Consider a family of probability distributions $P_\theta$ indexed by $\theta \in \Theta$. The parameter space is endowed with some metric $d$. We assume that there is a true parameter $\theta_0$, and we are interested of the convergence of the Bayesian posterior distribution, given a prior $\Pi$. We denote the posterior distribution after $n$ samples by $\Pi_n(\cdot|X^{(n)})$.
The posterior distribution is said to contract at rate $\epsilon_n \to 0$ at $\theta_0$ if $\Pi_n(\theta:d(\theta,\theta_0) > M_n\epsilon_n | X^{(n)}) \to 0$ in $P_{\theta_0}^{(n)}$ probability, for every $M_n\to \infty$ as $n\to\infty$ (i.e., regardless of how slow $M$ goes to infinity).
Now I have to proof the following proposition:
Suppose that the posterior distribution $Pi(\cdot|X^{(n)})$ contracts at rate $\epsilon_n$ at $\theta_0$. Then $\hat{\theta}_n$, defined as the center of a (nearly) samllest ball that contains posterior mass at least $\frac{1}{2}$, satisfies $d(\hat\theta_n,\theta_0) = O_P(\epsilon_n)$ under $P_{\theta_0}^{(n)}$.
$O_P(\epsilon_n)$ is not defined but I suppose it denotes stochastic boundedness (see wikipedia), that is, for $\tilde\epsilon >0$, there are $M, N \in \mathbb{N}$ such that for $n>N$, we have $P_{\theta_0}^{(n)} ( d(\theta_0,\hat\theta_n)/\epsilon_n > M) ) < \tilde\epsilon$.
The notes state the proof is very similar to a previous proof, which brought me to the following:
Let $B(\theta,r)$ be the closed ball of radius $r$ centered at $\theta$. Define $\hat{r}_n(\theta) = \inf\{r:\Pi_n(B(\theta,r)|X^{(n)}) \geq \frac{1}{2}\}$.
Then $\hat{r}_n(\theta_0) < M_n\epsilon_n$ with a probability tending to 1, with $M_n$ an arbitrary sequence with $M_n \to \infty$.
As we have chosen $\hat\theta_n$ as a nearly smallest ball, certainly $\hat{r}_n(\hat\theta_n) \leq \hat{r}_n(\theta_0) + \frac{1}{n} < M_n\epsilon_n + \frac{1}{n}$.
Now, $B(\theta_0,M_n\epsilon_n)$ and $B(\hat\theta_n,\hat{r}_n(\hat\theta_n))$ are not disjoint with probability tending to 1, since they would have a joined probability tending to 1.5. So:
$d(\theta_0,\hat\theta_n) < 2M_n\epsilon_n + \frac{1}{n} \implies d(\theta_0,\hat\theta_n)/\epsilon_n < 2M_n + \frac{1}{\epsilon_n n}$,
with probability tending to one. In other words, for $\tilde\epsilon >0$ there is some $N$ such that for $n>N$,
$P_{\theta_0}( d(\theta_0,\hat\theta_n)/\epsilon_n \geq 2M_n + \frac{1}{\epsilon_n n} ) < \tilde\epsilon$.
However, if I understand stochastic boundedness correctly I need to somehow replace $2M_n + \frac{1}{\epsilon_n n}$ with a fixed $M$. But since $M_n \to \infty$, I don't see how this is possible. Any ideas on what I could do? |
There's good news and bad news. The good news is that your intuitions are essentially right, and that there is such a group action via the Clifford group. The bad news is, depending on what you want out of that parameterisation, it may not be as useful as you are hoping.$\def\ket#1{\lvert #1 \rangle}$
The good news first — every Pauli stabiliser group on $n$ qubits, with $r = n-k$ independent generators, can be mapped to any other such group by conjugation by Clifford group operators. The simplest way to show this is by induction on $r$. If $r = 0$, then there is only one such stabiliser group: the trivial group $\{ \mathbf 1 \}$. For any $r > 0$, given an input stabilizer group $S$, you can reduce to the case of $r-1$ by the following steps:
Select any generator $P_r$ of the stabiliser group, and some qubit $x_r$ on which $P_r$ acts non-trivially.
Find a Clifford group operator $C_r$ such that $C_r P_r C_r^\dagger = Z_{n-r}$, the single-qubit Pauli $Z$ operator acting only on qubit $(n-r)$. The operator $C_r$ may involve SWAP operators in order to exchange the tensor factors for qubit $x_r$ and $(n-r)$.
Determine how the other generators of the stabilizer group transform under $C_r$. This produces a list of generators for the group $S' = \{ C_r P C_r^\dagger \,\vert\, P \in S\}$. Because $S'$ is abelian, the image of each generator either acts on qubit $(n-r)$ with $\mathbf 1$ or $Z$. In the latter case, produce a new generator by multiplying it by $Z_{n-r}$. As $Z_{n-r}$ is an element of $S'$, this yields an equivalent set of generators for the group.
Having done this, you have a stabiliser group for a subspace which is stabilized by $Z_{n-r}$. Any state in this group factors as a tensor product of $\ket{0}$ on qubit $(n-r)$, and some state on the remaining qubits. By considering the stabilizer code defined on all of the other qubits, you have reduced to the case of a stabiliser group on $n-1$ qubits and with $r-1$ generators.
If we unpack this inductive proof, we obtain a recursive procedure to map any stabiliser code $S$ with $r$ generators to a Clifford circuit $\mathcal C$ which maps that stabiliser group to the specific group $$\mathcal Z_{n,r} := \langle Z_{n-r}, Z_{n-r+1}, \ldots, Z_n\rangle \,.$$ If you have two such codes $S_1$ and $S_2$, just compose their circuits $\mathcal C_2^\dagger \mathcal C_1$ to obtain a circuit which maps $S_1$ to $S_2$. There is some redundancy, in that different sets of generators of the stabilizer group of $S_j$ will yield different circuits $\mathcal C_j$: this corresponds to the fact that some Clifford circuits just evaluate automorphisms (
i.e. logical unitaries) of the code. But never mind: what you have is a way of generating any stabilizer code on $n$ qubits with $r$ stabilizer generators from a single code.
The bad news is that, as this stands, all that we have done above is in effect to parameterize stabiliser codes by their encoding circuits. By "encoding circuit", I mean just the circuit which takes a $k = n-r$ qubit state $\ket{\psi}$, and then encodes $\ket{\psi}$ in an $n$-qubit system by preparing $r$ fresh qubits in the state $\ket{0}$ and acting on them by an appropriate unitary. By reducing an arbitrary stabilizer code with $r$ generators to a 'canonical' (and extremely dull) code whose stabilizer group is $\mathcal Z_{n,r}$, we have proven nothing more or less than that a stabilizer code is one with a Clifford encoding circuit. Describing stabiliser codes in terms of the orbit of $\mathcal Z_{n,r}$ under the $n$-qubit Clifford group is no more or less than describing codes in terms of their encoding circuits. This is a good fact to rely on, but more of a basic result than a deep result.
If you take some other code as the 'reference' code, then you are essentially doing the same thing, except prefacing that encoding circuit by some other Clifford circuit. This point of view may or may not be helpful to you — it's certainly a good elementary property to be aware of, when you're discussing stabiliser codes and stabiliser states with others who are less familiar with them — but without imposing additional constraints on what encoding circuits or code representations you're interested in (
e.g. to limit the automorphisms of codes which you consider), my guess is that this parameterisation may be of limited usefulness. The crux, in the end, will be which properties of stabilizer codes you are concerned with. |
Receding horizon control for the stabilization of the wave equation
1.
Johann Radon Institute for Computational and Applied Mathematics (RICAM), Austrian Academy of Sciences, Altenbergerstraβe 69, A-4040 Linz, Austria
2.
Institute for Mathematics and Scientific Computing, University of Graz, Heinrichstraße 36, 8010 Graz, Austria
Stabilization of the wave equation by the receding horizon framework is investigated. Distributed control, Dirichlet boundary control, and Neumann boundary control are considered. Moreover for each of these control actions, the well-posedness of the control system and the exponential stability of Receding Horizon Control (RHC) with respect to a proper functional analytic setting are investigated. Observability conditions are necessary to show the suboptimality and exponential stability of RHC. Numerical experiments are given to illustrate the theoretical results.
Keywords:Receding horizon control, model predictive control, asymptotic stability, observability, optimal control, infinite-dimensional systems. Mathematics Subject Classification:Primary: 49N35, 93C20, 93D20. Citation:Behzad Azmi, Karl Kunisch. Receding horizon control for the stabilization of the wave equation. Discrete & Continuous Dynamical Systems - A, 2018, 38 (2) : 449-484. doi: 10.3934/dcds.2018021
References:
[1] [2]
K. Ammari and M. Tucsnak,
Stabilization of second order evolution equations by a class of unbounded feedbacks,
[3] [4] [5]
——, Negative norm estimates for fully discrete finite element approximations to the wave equation with nonhomogeneous $L_2$ Dirichlet boundary data,
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C. Bardos, G. Lebeau and J. Rauch,
Sharp sufficient conditions for the observation, control, and stabilization of waves from the boundary,
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R. Becker, D. Meidner and B. Vexler,
Efficient numerical solution of parabolic optimization problems by finite element methods,
[11]
N. Burq and P. Gérard,
Condition nécessaire et suffisante pour la contrôlabilité exacte des ondes,
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H. Chen and F. Allgöwer,
A quasi-infinite horizon nonlinear model predictive control scheme with guaranteed stability,
[14]
N. Cîndea and A. Münch,
A mixed formulation for the direct approximation of the control of minimal $L^2$-norm for linear type wave equations,
[15]
J. -M. Coron,
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G. Grimm, M. J. Messina, S. E. Tuna and A. R. Teel,
Model predictive control: For want of a local control Lyapunov function, all is not lost,
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M. Gugat, E. Trélat and E. Zuazua,
Optimal Neumann control for the 1D wave equation: Finite horizon, infinite horizon, boundary tracking terms and the turnpike property,
[26]
W. Guo and B.-Z. Guo,
Adaptive output feedback stabilization for one-dimensional wave equation with corrupted observation by harmonic disturbance,
[27]
E. Hendrickson and I. Lasiecka, Numerical approximations of solutions to Riccati equations arising in boundary control problems for the wave equation, in
[28]
K. Ito and K. Kunisch, Receding horizon optimal control for infinite dimensional systems,
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C. Johnson,
Discontinuous Galerkin finite element methods for second order hyperbolic problems,
[31]
O. Karakashian and C. Makridakis,
Convergence of a continuous Galerkin method with mesh modification for nonlinear wave equations,
[32]
A. Kröner, K. Kunisch and B. Vexler,
Semismooth Newton methods for optimal control of the wave equation with control constraints,
[33]
K. Kunisch, P. Trautmann and B. Vexler,
Optimal control of the undamped linear wave equation with measure valued controls,
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I. Lasiecka and R. Triggiani,
Uniform exponential energy decay of wave equations in a bounded region with $L_2(0, ∞; L_2(Γ))$-feedback control in the Dirichlet boundary conditions,
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——,
[38]
——, Uniform stabilization of the wave equation with Dirichlet or Neumann feedback controlwithout geometrical conditions,
[39]
——, Algebraic Riccati equations arising from systems with unbounded input-solution operator: applications to boundary control problems for wave and plate equations,
[40]
J. -L. Lions,
[41]
——,
[42]
J. -L. Lions and E. Magenes,
[43]
——,
[44]
D. Q. Mayne, J. B. Rawlings, C. V. Rao and P. O. M. Scokaert,
Constrained model predictive control: Stability and optimality,
[45]
B. S. Mordukhovich and J. -P. Raymond, Neumann boundary control of hyperbolic equations with pointwise state constraints,
[46]
B. S. Mordukhovich and J.-P. Raymond,
Dirichlet boundary control of hyperbolic equations in the presence of state constraints,
[47]
A. Münch and A. F. Pazoto,
Uniform stabilization of a viscous numerical approximation for a locally damped wave equation,
[48]
M. Reble and F. Allgöwer,
Unconstrained model predictive control and suboptimality estimates for nonlinear continuous-time systems,
[49]
D. L. Russell,
Controllability and stabilizability theory for linear partial differential equations: recent progress and open questions,
[50]
L. T. Tebou and E. Zuazua,
Uniform boundary stabilization of the finite difference space discretization of the $1-d$ wave equation,
[51]
R. Triggiani,
Exact boundary controllability on $L_2(Ω)× H^{-1}(Ω)$ of the wave equation with Dirichlet boundary control acting on a portion of the boundary $\partialΩ$, and related problems,
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Stabilization of second order evolution equations by a class of unbounded feedbacks,
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——, Negative norm estimates for fully discrete finite element approximations to the wave equation with nonhomogeneous $L_2$ Dirichlet boundary data,
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C. Bardos, G. Lebeau and J. Rauch,
Sharp sufficient conditions for the observation, control, and stabilization of waves from the boundary,
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R. Becker, D. Meidner and B. Vexler,
Efficient numerical solution of parabolic optimization problems by finite element methods,
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Condition nécessaire et suffisante pour la contrôlabilité exacte des ondes,
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H. Chen and F. Allgöwer,
A quasi-infinite horizon nonlinear model predictive control scheme with guaranteed stability,
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N. Cîndea and A. Münch,
A mixed formulation for the direct approximation of the control of minimal $L^2$-norm for linear type wave equations,
[15]
J. -M. Coron,
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G. Grimm, M. J. Messina, S. E. Tuna and A. R. Teel,
Model predictive control: For want of a local control Lyapunov function, all is not lost,
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M. Gugat, E. Trélat and E. Zuazua,
Optimal Neumann control for the 1D wave equation: Finite horizon, infinite horizon, boundary tracking terms and the turnpike property,
[26]
W. Guo and B.-Z. Guo,
Adaptive output feedback stabilization for one-dimensional wave equation with corrupted observation by harmonic disturbance,
[27]
E. Hendrickson and I. Lasiecka, Numerical approximations of solutions to Riccati equations arising in boundary control problems for the wave equation, in
[28]
K. Ito and K. Kunisch, Receding horizon optimal control for infinite dimensional systems,
[29] [30]
C. Johnson,
Discontinuous Galerkin finite element methods for second order hyperbolic problems,
[31]
O. Karakashian and C. Makridakis,
Convergence of a continuous Galerkin method with mesh modification for nonlinear wave equations,
[32]
A. Kröner, K. Kunisch and B. Vexler,
Semismooth Newton methods for optimal control of the wave equation with control constraints,
[33]
K. Kunisch, P. Trautmann and B. Vexler,
Optimal control of the undamped linear wave equation with measure valued controls,
[34] [35]
I. Lasiecka and R. Triggiani,
Uniform exponential energy decay of wave equations in a bounded region with $L_2(0, ∞; L_2(Γ))$-feedback control in the Dirichlet boundary conditions,
[36] [37]
——,
[38]
——, Uniform stabilization of the wave equation with Dirichlet or Neumann feedback controlwithout geometrical conditions,
[39]
——, Algebraic Riccati equations arising from systems with unbounded input-solution operator: applications to boundary control problems for wave and plate equations,
[40]
J. -L. Lions,
[41]
——,
[42]
J. -L. Lions and E. Magenes,
[43]
——,
[44]
D. Q. Mayne, J. B. Rawlings, C. V. Rao and P. O. M. Scokaert,
Constrained model predictive control: Stability and optimality,
[45]
B. S. Mordukhovich and J. -P. Raymond, Neumann boundary control of hyperbolic equations with pointwise state constraints,
[46]
B. S. Mordukhovich and J.-P. Raymond,
Dirichlet boundary control of hyperbolic equations in the presence of state constraints,
[47]
A. Münch and A. F. Pazoto,
Uniform stabilization of a viscous numerical approximation for a locally damped wave equation,
[48]
M. Reble and F. Allgöwer,
Unconstrained model predictive control and suboptimality estimates for nonlinear continuous-time systems,
[49]
D. L. Russell,
Controllability and stabilizability theory for linear partial differential equations: recent progress and open questions,
[50]
L. T. Tebou and E. Zuazua,
Uniform boundary stabilization of the finite difference space discretization of the $1-d$ wave equation,
[51]
R. Triggiani,
Exact boundary controllability on $L_2(Ω)× H^{-1}(Ω)$ of the wave equation with Dirichlet boundary control acting on a portion of the boundary $\partialΩ$, and related problems,
Algorithm 1 Receding Horizon Algorithm Require: Let the prediction horizon 1: 2: Find the optimal pair $ \begin{split} &\min_{u\in L^2(t_k, t_k+T; \mathcal{U})}J_T(u; \mathcal{Y}_{rh}(t_k)):= \min_{u\in L^2(t_k, t_k+T; \mathcal{U})} \int^{t_k+T}_{t_k} \ell(\mathcal{Y}(t), u(t))dt, \\ \text{ s.t } & \begin{cases} \dot{\mathcal{Y}} = \mathcal{A}\mathcal{Y}+\mathcal{B}u t \in (t_k, t_k+T) \\ \mathcal{Y}(t_k) = \mathcal{Y}_{rh}(t_k) \end{cases} \end{split} $ 3: Set $ \begin{split} u_{rh}(\tau)&:=u^*_T(\tau; y_{rh}(t_k), t_k) \text{ for all } \tau \in [t_k, t_k+\delta), \\ \mathcal{Y}_{rh}(\tau)&:=\mathcal{Y}^*_T(\tau; y_{rh}(t_k), t_k) \text{ for all } \tau \in [t_k, t_k+\delta], \\ t_{k+1} &:= t_k +\delta, \\ k &:= k+1. \end{split} $ 4: Go to step 2.
Algorithm 1 Receding Horizon Algorithm Require: Let the prediction horizon 1: 2: Find the optimal pair $ \begin{split} &\min_{u\in L^2(t_k, t_k+T; \mathcal{U})}J_T(u; \mathcal{Y}_{rh}(t_k)):= \min_{u\in L^2(t_k, t_k+T; \mathcal{U})} \int^{t_k+T}_{t_k} \ell(\mathcal{Y}(t), u(t))dt, \\ \text{ s.t } & \begin{cases} \dot{\mathcal{Y}} = \mathcal{A}\mathcal{Y}+\mathcal{B}u t \in (t_k, t_k+T) \\ \mathcal{Y}(t_k) = \mathcal{Y}_{rh}(t_k) \end{cases} \end{split} $ 3: Set $ \begin{split} u_{rh}(\tau)&:=u^*_T(\tau; y_{rh}(t_k), t_k) \text{ for all } \tau \in [t_k, t_k+\delta), \\ \mathcal{Y}_{rh}(\tau)&:=\mathcal{Y}^*_T(\tau; y_{rh}(t_k), t_k) \text{ for all } \tau \in [t_k, t_k+\delta], \\ t_{k+1} &:= t_k +\delta, \\ k &:= k+1. \end{split} $ 4: Go to step 2.
Algorithm 2 RHC( Input: Let a final computational time horizon 1: Choose a prediction horizon 2: Consider a grid 3: for do Solve the open-loop subproblem on $ \begin{split} \min & \frac{1}{2}\int_{t_i}^{t_i + T} \|\mathcal{Y}(t)\|^2_{\mathcal{H}}dt + \frac{\beta}{2}\int_{t_i}^{t_i + T}\|u(t)\|^2_{\mathcal{U}}dt, \\ \text{ subject to } & \begin{cases} \dot{\mathcal{Y}} = \mathcal{A}\mathcal{Y}+\mathcal{B}u t \in (t_k, t_k+T), \\ \mathcal{Y}(t_i) = \mathcal{Y}_T^*(t_i) \mbox{ if } i \geq 1 \mbox{ or } \mathcal{Y}(t_i) = (y^1_0, y^2_0) \mbox{ if } i = 0, \end{cases} \end{split} $ where 4: The model predictive pair
Algorithm 2 RHC( Input: Let a final computational time horizon 1: Choose a prediction horizon 2: Consider a grid 3: for do Solve the open-loop subproblem on $ \begin{split} \min & \frac{1}{2}\int_{t_i}^{t_i + T} \|\mathcal{Y}(t)\|^2_{\mathcal{H}}dt + \frac{\beta}{2}\int_{t_i}^{t_i + T}\|u(t)\|^2_{\mathcal{U}}dt, \\ \text{ subject to } & \begin{cases} \dot{\mathcal{Y}} = \mathcal{A}\mathcal{Y}+\mathcal{B}u t \in (t_k, t_k+T), \\ \mathcal{Y}(t_i) = \mathcal{Y}_T^*(t_i) \mbox{ if } i \geq 1 \mbox{ or } \mathcal{Y}(t_i) = (y^1_0, y^2_0) \mbox{ if } i = 0, \end{cases} \end{split} $ where 4: The model predictive pair
Prediction Horizon iter
Prediction Horizon iter
Prediction Horizon iter
Prediction Horizon iter
Prediction Horizon iter
Prediction Horizon iter
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2018 Impact Factor: 1.143
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So, from this simple no-arbitrage argument, we see that the price of the option must always be at least its intrisic value.
Yes indeed
However, at this point I realized something strange: if this is true, why in the world should I exercise my put option before expiry?? The inequality seems to
indicate that it would be an unwise decision to ever exercise the American put option at time $t$, and so, the only right time to exercise an American put option would be at expiry
How does the fact that $P(t,S_t;K,T-t) \geq (K-S_t)^+$ gets you to that conclusion? This is a completely fallacious reasoning IMHO.
The price of an American option is: $$ P(t,S_t;K,T-t) = \text{sup}_{\tau \in [t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right] $$the inequality you observe can be obtained by splitting up the family of stopping times $\tau$ with values in $[t,T]$ using the fact that$$ [t,T] = \{t\}\ \cup\ ]t,T]$$We then get,\begin{align} P(t,S_t;K,T-t) &= \text{sup}_{\tau \in [t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right] \\&= \max\left( \underbrace{\text{sup}_{\tau=t} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right]}_{\text{immediate exercise}}, \underbrace{\sup_{\tau \in ]t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right]}_{\text{differed exercise}} \right)\\&= \max\left( (K-S_t)^+, \sup_{\tau \in ]t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right] \right) \\&= (K-S_t)^+ + \max\left(0, \sup_{\tau \in ]t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right] - (K-S_t)^+ \right) \\&\geq (K-S_t)^+ \end{align}
When the holder has to choose whether or not to exercise at time $t$, he/she should compare the value of his option position $P(t,S_t;K,T-t)$ with the payoff he/she would get if he/she exercised immediately (intrinsic value) $(K-S_t)^+$. $$ \underbrace{(K-S_t)^+}_{\text{immediate exercise}} - \underbrace{P(t,S_t;K,T-t)}_{\text{option value}} = \max\left( 0, (K-S_t)^+ - \sup_{\tau \in ]t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right] \right) $$The holder would then exercise at $t$ if the RHS is positive, that is iff$$ (K-S_t)^+ \geq \sup_{\tau \in ]t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right] $$or equivalently if the intrinsic value is greater than the continuation value. You see that it is not possible to make a general claim such as "the only right time to exercise would be at expiry" from the above.
[Edit]
Let $$\frac{dS_t}{S_t} = rdt + \sigma dW_t^\mathbb{Q} $$Elaborating on @MJ73550's remark if $(S_t)_{t\geq 0}$ is a
martingale, that is if $r = 0$, one can show that, for any convex function $\phi$,$$ \mathbb{E}_t\left[ \phi(S_\tau) \right] \leq \mathbb{E}_t\left[ \phi(S_T) \right], \ \ \forall \tau: t \leq \tau \leq T $$to see this, we can appeal to Jensen's inequality along with the optional stopping theorem. Indeed for all stopping time $\tau \leq T$ we can write:\begin{align}\mathbb{E}_t\left[ \phi(S_T) \right] &= \mathbb{E}_t\left[ \mathbb{E}\left[ \phi(S_T) \vert \mathcal{F}_\tau \right] \right]\ \ \text{(Tower property)}\\&\geq \mathbb{E}_t\left[ \phi(\mathbb{E}[S_T \vert \mathcal{F}_\tau]) \right] \ \ \text{(Jensen's inequality)} \\&= \mathbb{E}_t\left[ \phi(S_\tau) \right] \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Optional sampling theorem)}\end{align}
Using this result the price of an American option, when $S_t$ is a martingale becomes:$$ V^{AME}(t,S_t;K,T-t) = \sup_{\tau \in [t,T]} \mathbb{E}_t^\mathbb{Q}\left[ \phi(S_\tau) \right] = \mathbb{E}_t^\mathbb{Q}\left[\phi(S_T)\right] = V^{EUR}(t,S_t;K,T-t)$$
Now, if we consider the case $r \leq 0$, the stock process $(S_t)_{t \geq 0}$ becomes a
sub-martingale since in that case:$$ \mathbb{E}_t[S_T] = S_t \underbrace{e^{r(T-t)}}_{\leq 1} \leq S_t $$Should we let $\phi: x \rightarrow e^{-r(\tau-t)}(K-x)^+$ which is still a convex function of $x$, we would see that for put options, where $\phi$ is a monotonically decreasing function of $x$, applying a similar reasoning as earlier shows that it is never optimal to exercise the American put before maturity $T$. |
Imagine we have a random number generator where g(n+1) = f(g(n)), where f is some function (e.g linear recurrence). I'm trying to find a system where it's fast to find many steps in the future immediately, whereas it's hard or impossible to work out a previous state. for example, if I know state = X at time t, I would like to calculate state at t+n steps in the future, done in order O(1) rather than having to iterate n times. I'm assuming state has an implicit modulo of around some power of 2 so it must be restrainted to a given number of bits at each time. Does anyone know of a system that could provide this? Many thanks.
There are two answers: one that solves your problem, and one that answers your question. I'll start with the first. One way to make sure that previous states cannot be backtracked from generated numbers is to mask the true state. Here's how it works. You take your $b$-bit number $x_t$ to be the true state at time $t$, but the number your RNG generates is $x_t \text{ mod } m$, where $m$ is much less than $2^b$. Then you can specify $g$ to be $x_{t+1} = a \cdot x_t \text{ mod } n$, with $a$ and $n$ large numbers that are relatively coprime.
This way, if you collect the numbers generated by this RNG, you need at least a couple of data points before you can make educated guesses about $x_t$, and even then it's not easy. The period of the RNG is $n$.
The answer to your question about whether there exist easy-to-compute functions that are hard to invert is an open question, known as the existence of one-way-functions. Existence is known to imply
P $\not=$ NP, so it probably won't be resolved for some time. That's for functions that are easy to compute in one iteration; functions that allow an arbitrary amount of time steps to be computed is a question on a whole other level.
Of course, if you relax the question to mean a function whose inverse requires, say, $\Omega(b^2)$ time to invert, rather than superpolynomial time, there are options. For example, you could keep the function $x_{t+1}=a\cdot x_t \text{ mod }n$. To invert that function, you need to find the the multiplicative inverse of $a \text{ mod } n$. This takes $O(b^2)$ time. Then you will be able to compute $x_{t}$ given any $x_{t+1}$ in, again $O(b^2)$ time, as $x_t = x_{t+1} \cdot a^{-1} \text{ mod }n$.
This function is easy to compute for several steps, if you are willing to do some precomputation. Note that $x_{t+1} = a \cdot x_t$ and $x_{t+2} = a\cdot x_{t+1} = a^2 x_{t}$, so $x_{t+k} = a^{k} \cdot x_t$. Now you can precompute $a, a^2, \ldots, a^k$ up to some $k$, and from then on you will be able to compute $a_{t+j}, 1 \leq j \leq k$ efficiently.
It's possible to construct a public/private keypair and an update function $f:S \to S$ with the properties that:
computing $f$ takes $O(1)$ operations, given the public key;
with the private key, you can compute $f^{n}$ (i.e., iterate $f$ $n$ times) in $O(\lg n)$ operations. In other words, anyone with the private key can compute the state at time $t+n$ from the state at time $t$ in $O(\lg n)$ operations -- not quite $O(1)$, but close.
without knowing the private key, you cannot invert $f$ (you can't go backwards)
with knowledge of the private key, it is possible to invert $f$ (to go backwards)
The primary drawback is that the private key allows both fast iteration and inversion. For some applications this might be OK; only you can tell whether it will work for you.
The function $f$ is just squaring modulo $N$, where $N$ is a RSA modulus ($N$ is the public key; the factorization of $N$ is the private key). See https://crypto.stackexchange.com/a/15141/351 for more details.
(A follow up to the accepted answer, to some degree...)
Using a hash function in this way is somewhat equivalent to fast forwarding a pseudorandom number generator.
A linear congruential generator (LCG) can be skipped ahead and back [link], in fact many PRNG's can be skipped/jumped ahead [link]. The problem with fast-forwarding any hash/PRNG is that if you know the cycle period (which you do with most PRNGs), you can effectively move back 1 state by fast-forwarding (period−1) steps.
Thus if we found a way to, say, move forward quickly with SHA-256, then the first person to correctly estimate its cycle period would also have the ability to move backward as well.
The only fix for this is to have a system with a huge number of varied cycles. E.g. having 256 bits worth of different cycles lengths, which each cycle being 64 to 256 bits in cycle length, fitting in 512 bits of state.
There is no known way of doing this, but should one be found, you've solved multiple comp-sci and cryptology problems, and made efficient signing of messages trivial.
tl;dr - if you can go forward and calculate its cycle length, you can go backward. |
I was asked in today's exam to prove/disprove the following: ($f$ is a continuous function in $[0,1]$)
$$\lim_{n \rightarrow \infty} \int_{0}^{1} \left(\frac{n f(x)}{1 + n^2x^2} \right) dx = \frac \pi 2 f(0)$$
I tried out a few functions $f(x)$, and decided that it was probably true. But all I could do was to reduce the integral to
$$\lim_{n \rightarrow \infty} \int_{0}^{1} \left(\frac{n f(x)}{1 + n^2x^2} \right) dx = \lim_{n \rightarrow \infty} \int_{0}^{n} \frac{f(x/n)}{1 + x^2} dx$$
which didn't lead me anywhere. |
I had an email request about making
ivreg work with the
survey package. That’s
AER::ivreg, which does two-stage least-squares estimation with instrumental variables.
The steps are
See if it accepts weights and does the right thing for point estimation If so, work out how to get the complex-survey variances Test to make sure it’s getting the right answer
In this case, the first step was fairly straightforward. The function accepts weights and passes them to
lm.wfit, so it will give the same point estimates as if they were sampling weights.
The second step is the technically generalisable one. My full code is here. The key parts for estimation using survey design information are
.data<-model.frame(design).data$.weights<-weights(design,"sampling")model<- ivreg(formula, data=.data, weights=.weights)
which just calls
AER::ivreg to get the point estimates, and
U<-estfun(model)/weights(design,"sampling")n<-NROW(U)infl<- U%*%bread(model)/nv<-vcov(svytotal(infl, design))
which computes the variances of the parameter estimates.
The variance code comes from the Secret Trick that makes the survey package work: any well-behaved
1 estimator can approximated as the total of its influence functions. If we write \(\Delta_i\) for the influence function of observation \(i\) then\[\tilde\theta-\theta = \sum_{i=1}^N \Delta_i + \textrm{small}.\]
In a probability sample with sampling weights \(w_i\), we will have\[\hat\theta-\theta = \sum_{i\in\textrm{sample}} w_i\Delta_i + \textrm{small}.\]We want the variance of the left-hand side of the equation. The right-hand side is just the variance of an estimated population total, which is a thing we know how to do.
2
In the case of
ivreg, since it was written by Achim Zeileis, it has convenient functions
estfun and
bread that are designed to work with the
sandwich package but that also make influence functions easy to compute.
The first line of the second code chunk looks a bit strange: we’re dividing by the weights. That’s because the
ivreg code puts the weights into the estimating function; we don’t want that because
svytotal also puts them in.
Alternatively, if you have a design using replicate weights (resampling), step 2 looks like
withReplicates(design, function(.weights, .data){ .data$.pweights<-.weights m<-ivreg(formula,data= .data, weights=.pweights) coef(m) })
which runs the
ivreg function for each set of replicate weights, extracts the coefficients, and computes the variance.
Step 3 is harder: ideally there would be a published analysis with downloadable data that I could check against. Also harder is working out how to adjust the various diagnostic tests. But the basic estimation is fairly straightforward.
You can’t handle non-regular estimators such as the lasso this way, since they don’t have influence functions that work this way. You also can’t handle mixed models, where the score function isn’t just a sum over observations. In addition to the purely technical problems (which are enough to be going on with) there’s the issue that both of these classes of model have a cost:complexity or bias:variance tradeoff, and it’s not obvious how to make this tradeoff scale the right way when you go from a population to a sample. |
Speaker
Mr. Yen-Hsun Lin (IoP, Nat'l Chiao Tung University, Hsinchu, Taiwan)
Description
It has been shown that the self-interaction between dark matter (DM) particles can increase the DM number density inside the Sun. The increasing rate of DM number density by this effect is proportional to the existing DM number density inside the Sun. We demonstrate that this effect can counteract DM evaporations in the regime of small DM mass. Consequently, the critical mass for DM evaporations (typically $3\sim 4$ GeV without the self-interaction) can be lowered down by DM self-interactions. Hence the DM annihilation rate for $m_{\chi}$ around few GeVs may be enhanced by self-interactions. This leads to the enhancement of neutrino flux from such annihilations. We discuss the prospect of observing such enhanced neutrino flux in IceCube-PINGU using the annihilation channels $\chi\chi\to \tau^+\tau^- , \, \nu\bar{\nu}$ as examples. The PINGU sensitivities to DM self-interaction cross section $\sigma_{\chi\chi}$ are estimated for track and cascade events.
Primary authors
Dr. Chian-Shu Chen (Physics Division, National Center for Theoretical Sciences, Hsinchu, Taiwan) Dr. Fei-Fan Lee (IoP, Nat'l Chiao Tung University, Hsinchu, Taiwan) Prof. Guey-Lin Lin (IoP, Nat'l Chiao Tung University, Hsinchu, Taiwan) Mr. Yen-Hsun Lin (IoP, Nat'l Chiao Tung University, Hsinchu, Taiwan) |
Let's say one has a univariate Gaussian distribution with mean $\mu = 0$ and standard deviation $\sigma$.
It is easy to see that the distance from $\mu$ to +1$\sigma$ is...well...$\sigma$.
Let's now say we have a bivariate Gaussian distribution with
$\vec{\mu} = {[ 0 \space 0]^T}$ and covariance matrix $\Sigma = \begin{bmatrix} \sigma_{x,x}=var(X) & \sigma_{y,x}=cov(Y,X) \\ \sigma_{x,y}=cov(X,Y) & \sigma_{y,y}=var(Y) \end{bmatrix} $
where cov(X,Y) $\ne 0 $, and because cov. matrices are always symmetric, cov(X,Y)=cov(Y,X)
I understand that this distance is precisely the largest eigenvalue of the covariance matrix. Is there another method that one can use to determine d?
Update 1:
One way of doing this -I believe- is by finding the largest eigenvalue of the covariance matrix. I am looking for a method that does not use the eigenvalues of the covariance matrix. |
Two intergalactic superpowers, the Western Republic and the Eastern Empire, are in a Cold War. The Eastern Empire is comprised of mostly aquatic species, while the Western Republic is mostly comprised of terrestrials, including humans. The Eastern Empire has assembled 33 million FTL nuclear missiles pointed at the Republics territory. The Western Federation needs to have a weapon that can be equally destructive. Most Imperial Cities are submerged under 500 ft of water or more, so my question is, would nuclear weapons work or should they use another kind of weapon?
I believe what you have here is good, but as someone in the comments had said, galactic empires do not fight over cities. They fight over planetary control and galactic strongpoints. I would advise doing something, and yes this may seem like a rip off of
Star Wars, but some sort of weapon that destroys aquatic planets by draining the water.
I have done some research on a nuclear weapon detonating underwater, and I am not sure if you are placing fleets on said planets, but here is what I found on nuclear weapons detonating underwater:
Now granted, this is a past incident. I am sure that in the future a fleet would not be severely harmed by nuclear detonations off from the water, but many ships were harmed from nuclear attacks, and missiles launched from the air could be hard to reach their targets.
So I would say construct some sort of super weapon that drains up water on planets, say something like a heat bomb or a way that it pollutes the water, like biological warfare to where the wildlife is destroyed. Or just simply make it to where underwater nuclear weapons are safer to be detonated.
Water transfers force more efficiently than air, so pretty much any sort of blast from a device configured to perform as a depth charge will do.
What you need is the equivalent of nuclear depth charges, which have already been developed. The SUBROC is actually an anti-submarine device, which is lauched from a sub, goes ballistic, then reenters the water at its target location, descending like a depth charge. You only need the air-to-submarine section of the weapon to work in your world, which should make it simple to modify to increase the power to an underwater city-buster.
Underwater explosions have been studied by humanity for decades now. There is a short term for underwater explosion:
undex.
Since water is (for practical purposes) incompressible, the energy of a blast suffers less dispersion over distance. In other words, the blast radius is much larger than what it would be, should the detonation happen in air.
The nuclear device was positioned 500 feet deep in the Pacific Ocean (...) Within a second of detonation, a spray dome was created that reached a height of 840 feet [above the surface] after seven seconds (...) When the spray dome and base surge had dissipated, a foam patch could be seen spreading from the surface zero water to reach over 6,000 feet (...) The nuclear device had a blast that was calculated to be nine kilotons. All fallout stayed within the predicted fallout area with a maximum of 0.030 R/hr. The target ship at 5,900 yards was directly hit by the shockwave vibrating the entire ship and shaking it violently. The Moran, merchant marine ship moored at 2,346 feet away, was immobilized due to shock damage to its main and auxiliary equipment while also attaining minor hull damage.
That was a 9 kilotons blast, which is a relatively small yield. You can be sure that anything above the blast was obliterated. To the sides and further below, targets will take a hit from the blast wave.
But what if a larger payload was used? XKCD's What If number 15 was about detonating the Tsar Bomba (yield: 53 megatons, or 5,888.88 times the Wahoo test payload) underwater.
The explosion at the bottom of the Mariana Trench will create a quickly-expanding spherical cavity of hot steam. To figure out how big it gets, we can try a formula from the 1971 paper Evaluation of Various Theoretical Models For Underwater Explosion:
$$ Radius = (\frac{3}{4\pi})^{\frac{1}{3}} \times (\frac{40\% \times 53 \space megatons \space of \space TNT}{Mariana \space Trench \space pressure + 1\space atm})^{\frac{1}{3}} \approx 580 \space meters $$
If we use the same equation for a smaller depth of 500 feet (approximately 152 meters):
$$ Radius = (\frac{3}{4\pi})^{\frac{1}{3}} \times (\frac{40\% \times 53 \space megatons \space of \space TNT}{16 \space atm})^{\frac{1}{3}} \approx 2367 \space meters $$
I don't have any data to make further extrapolations, so from now on it's just pure speculation. I think the water column, fauna, flora and people within the blast would be vaporized, making for a big mushroom cloud reaching almost as high as the original Tsar Bomba test one (which peaked at 56 kilometers above sea level, nearly seven times the altitude of Everest's peak). The sheer pressure of the shockwaves will keep the surrounding water from caving in at first, but once that passes, water will rush in in a very catastrophic manner. Expect a whirlpool of biblical proportions. Both the initial blast and the later rush-in would cause tsunamis that would travel for dozens to hundreds of kilometers. If the sea in that area is around 150 feet deep, then the seafloor will become very hard, very plain glass, and the blast will be detected by sismographs multiple times as the shockwaves travel through the crust and mantle of the planet.
Oh and once that settles you will see a black rain. The water itself should not be much radioactive, but the salt in it should give you more problems than hypertension.
I am not sure about a galactic war using FTL missiles to hit targets probably thousands of light years away.
There may be border defenses that can detect and stop approaching FTL missiles like the USA's "Star Wars" program might have been able to to stop ICBMs if completed.
And once past the border defenses each important target planet might have a planetary defense system designed to detect incoming FTL missile and try to stop them.
Thus FTL missiles might face lots of defenses on the way, often thousands of light years, to their targets, and might have a very hard time reaching those targets unassisted.
instead I imagine a galactic war using fleets of space battleships, both manned and unmanned, to attack planetary defense systems. If planetary defense systems have the capability to detect and stop FTL missiles, the planetary defense systems must be disabled to give the FTL missiles clear shots. So if a fleet of space battleships is detected approaching, the defending side may send a fleet of space battleships to fight the attacking side's fleet of space battleships, and stop them from disabling planetary defense systems to give FTL missiles a clear shot.
Thus it is possible that the decisive battles in any hot war between the two realms will be battles between fleets of millions of space battleships.
What kind of a warhead will a FTL missile have? A tiny pebble at FTL speed would have enough kinetic energy to smash a planet. No atomic bombs would be needed. Except that it is impossible to take a slower than light (STL) object and accelerate that object to FTL speeds. It would take infinite energy to accelerate even a subatomic particle to even the speed of light.
Therefore all possible FTL drives would somehow "cheat". They would get matter from star system A to star system B in less time than light takes to travel the distance without actually accelerating that matter to speeds faster than light. Therefore no possible FTL drive would give any matter impossible amounts of kinetic energy.
So it's back to using atomic bombs as the warheads?
Not necessarily.
The extinction of the dinosaurs and many other life forms was probably caused by the Chicxulub asteroid impact. And even if other theories are correct and the Chicxulub impact wasn't the main cause of the extinctions, it would have wiped out all life in an area hundreds and even thousands of miles wide.
The asteroid that struck Chicxulub was probably about 15 kilometers (9.3 miles) in diameter and struck at a speed of "only" a few tens of kilometers per second. The blast produced as much energy as ten billion Hiroshima A bombs.
According to my rough calculations, a rock only 15 meters or 50 feet in diameter travelling at the speed of the Chicxulub asteroid, just a few tens of kilometers per second, would release the energy of a Hiroshima bomb on any land or underwater city it struck.
The Tunguska event in 1904 produced a blast now estimated at 3 to 5 megatons, a few hundred times the blast of the Hiroshima bomb, and would devastate a large metropolis. It was probably produced by an asteroid or comet about 60 to 190 meters (196 to 620 feet) in diameter traveling at a few tens of miles per second.
So to produce a FTL missile, just attach a slower than light (STL) drive and a FTL drive to an asteroid. Use the STL drive to accelerate the asteroid to a speed of a few tens of kilometers per second. Then turn on the FTL drive, zoom to the target solar system, get very close to the target planet and in the correct relative position, and then turn off the FTL drive. The asteroid will still have the relative motion of tens of kilometers per second, and will slam into the target planet with devastating effects.
So you can use tiny asteroids to hit individual cities with nuclear bomb sized explosions, or asteroids a few kilometers in diameter to strike entire planets with devastating effects.
In the
Lensman series a pair of planets with opposite velocities would be selected and giant FTL space drives would be installed on the planets and used to move them to opposite sides of the target planet, and then they would turn off the FTL drives. The intrinsic STL velocities of the two planets would cause them to smash into the target planet from opposite sides. But unless the target planets in your stories have fantastically advanced defense systems that would be billions of times overkill, trillions of times more mass than needed to devastate a planet.
Of course increasing the speed of the asteroid will increase the explosion it makes when it strikes.
Since the speed of light in a vacuum is 299,792.458 kilometers per second, that is 29,979.245 times a speed of 10 kilometers per second. Clearly an asteroid could be accelerated to tens or hundreds of times the speed that asteroids usually hit planets at, thus making its explosion a lot bigger.
It may be noted that neighboring stars often have high velocities relative to each other, so there might not be any need to artificially accelerate an asteroid to STL speeds to strike a planet in a neighboring star system, merely to use FTL speeds to get it near the target planet. Since stars mostly rotate around the centers of their galaxies in the same direction, stars on the opposite sides of their galaxies will be going in opposite directions with velocity differences on the order of hundreds of kilometers per second. And your Western Republic and Eastern Empire might be on different sides of your galaxy.
So if you are going to have an galactic cold war between two space realms using FTL missiles to threaten each other - like in the Cold War on Earth - you really don't have to worry much about the nature of your warheads. Normal atomic bombs or simple space rocks will do.
Since this is a cold war, the Western Empire needs a doomsday weapon. They attack the atmospheres of the ocean planets.
An ocean cannot exist without an atmosphere, as it would quickly evaporate into space. There are a few methods that come to mind, depending on the empire's technical ability.
A Solar Wind beam. Mars' atmosphere was stripped away by solar wind, assisted by the composition of its atmosphere and loss of magnetic field, but a similar method could work to hit multiple planets with manufactured or redirected solar wind. Bombarding the planet with magnesium pellets (if it has an oxygen atmosphere) triggering massive fires and burning of the atmosphere (more here) Gravity Gun - Reducing a planet's gravity would effectively dissipate its atmosphere, allowing the oceans to evaporate into space.
These would be pretty apocalyptic events on a planet-wide scale, maybe even for whole systems depending on the width of the beams/attacks. |
The exact shape of the set which has the best isoperimetry in the continuous Heisenberg is (from what I know) a difficult open problem. This brought to wonder what is known in the discrete case?
More precisely... Given a finitely generated group $G$, build up its Cayley graph using a (finite symmetric) generating set $S$. For $F \subset G$, let $\partial F$ be the set of edges between $F$ and $F^\mathsf{c}$. Say a set is $\epsilon$-optimal if among all sets with $\frac{|\partial F|}{|F|} \leq \epsilon$ its cardinality is minimal. The precise question is:
$\textbf{Question}$: Is there a set $S$ as above and a sequence $\epsilon_n \to 0$ for which, we know what are the $\epsilon_n$ optimal sets in the discrete Heisenberg group?
I would be also interested to know if there any amenable group (except $\mathbb{Z}^d$) where we know the answer to this question. For $\mathbb{Z}^d$ with the usual generating sets, the $\epsilon$-optimal sets are (I think) determined by the Loomis-Whitney inequality (i.e. they are close as possible to hypercubes).
As a side remark, when a group $G$ can be written as a semi-direct product $G_1 \rtimes G_2$ of amenable groups (like the Heisenberg group), one builds Folner sets (i.e. sets with a small ratio $\frac{|\partial F|}{|F|}$) by taking $F = F_1 \times F_2$ where $F_i$ are some Folner sets of $G_i$. So it seems (but I did not find any reference to that effect) that the optimal sets for the Heisenberg could be of this form. |
I almost forgot that we have also been to Amsterdam in July 2002. Dan, Martijn, Hong
The three men who deserve the award for their advances especially in 1976 are Sergio Ferrara, Daniel Z. Freedman, and Peter van Nieuwenhuizen. Needless to say, while I could post pictures of all of them (Ferrara at Harvard and van Nieuwenhuizen in Stony Brook), I have been much closer to Dan Freedman (MIT), as a co-author of our
pp-wave Paper of Seven that is still anxiously waiting for its 250th citation, and as a partner on bicycles who actually has strong leg muscles to be able to disappear from my horizon whenever he wants.
Congratulations, Dan! And others.
It's very natural for supergravity to be finally rewarded – and I admit that I have sometimes missed that this was the natural bunch of people who were still waiting for this prize.
Supergravity may be said to stand for "supersymmetric gravity" – and it's a combination or intersection of the principles of Einstein's general theory of relativity; and supersymmetry. Because both general relativity and supersymmetry are beautiful, supergravity is clearly super-beautiful.
Supersymmetry emerged as a clever loophole – a symmetry that is neither acting on the spacetime in the usual ways (like the Poincaré symmetry), nor it is an internal symmetry that preserves spacetime points and acts "inside each point" separately. Supersymmetry is something in between them – which is only mathematically allowed because its generators are fermionic operators. Their anticommutator includes the momentum, the generator of spacetime translations:\[
\{Q_\alpha,Q_\beta\} = \Gamma^\mu_{\alpha\beta} P_\mu + \dots
\] Well, we usually need to distinguish dotted and undotted (chiral) spinor indices, add some gamma matrix for complex conjugation, and the dots may include "central charges". Supersymmetry was independently found both in the West and in the East – as separated by the Iron Curtain. In Russia, it was really found from algebraic considerations above. In the West, Pierre Ramond was incorporating fermions on the world sheet of string theory and found the world sheet supersymmetry – by seeing that it's an algebra obeyed by a particular 2D theory with bosons and fermions. We are in the early 1970s now.
Around 1974, Wess and Zumino would construct the first interacting four-dimensional supersymmetric theory – a scalar superfield with a cubic superpotential. This kind of work was going to expand around 1980 when physicists constructed the Minimal Supersymmetric Standard Model. SUSY GUT (grand unified theory) was added later. All these theories were non-gravitational.
In 1976, the new winners of the award constructed the first usable supergravity formalism for \(D=4\) and higher. It was simply a classical field theory which had the metric tensor – but to make SUSY easier, the metric tensor is replaced by the vielbein \(e^\mu_a\) – and the spin-3/2, Rarita-Schwinger field \(\psi_{\mu,\alpha}\) that carries both a vector and a spinor index (the spin is \(1+1/2=3/2 = 2-1/2\)). That fermionic field is the superpartner of the metric. They were able to define the supersymmetric variations of these fields so that the equations of motion were covariant under these transformations. The variations and equations have many terms so the underlying beauty of the local supersymmetry may look quite messy, disproving the idea that everything that is beautiful must look like a super-short formula on the T-shirt.
This discovery was destined to be generalized in the coming years. Higher-dimensional supergravity theories were found, culminating with the "prettiest" and "highest-dimensional" 11-dimensional supergravity (Cremmer-Julia-Scherk), and various other generalizations such as gauged supergravity theories were constructed. While supersymmetry was being added to the string theory world sheet already in the 1970s, starting with Ramond's work, Green and Schwarz only began to study a full-blown supersymmetric string theory or superstring theory in the early 1980s. When we say just "string theory" today, we almost always mean just the supersymmetric string theory that got mature in the early 1980s.
Only in 1995, Witten figured out that the most beautiful supergravity theory in 11 dimensions was actually linked to string theory. Being UV-completed as M-theory, it was the strong coupling limit of type IIA or \(E_8\times E_8\) heterotic (a duality discovered by Hořava+Witten) string theory. So only in the 1990s, the supergravity and string theory "communities" have really merged. The tight and important relationships between the theories in both "communities" became clear. Pretty much every string theorist understood that SUGRA folks weren't completely wrong to study non-renormalizable theories because many of SUGRA theories emerge as the long-distance limit of string theories; and SUGRA folks have generally also appreciated that string theorists hadn't been a useless thing because SUGRA is inconsistent by itself and needs a stringy completion at Planckian energies to restore the consistency.
Just to be sure, this merger wasn't an example of a perfect melting pot. It resembled the reunification of Germany instead. People couldn't throw away the culture that had shaped them for decades. So e.g. Sergio Ferrara, Daniel Z. Freedman, and Peter van Nieuwenhuizen remained top "supergravity theorists" who wouldn't have too strong reasons (and powers) to be reshaped into string theorists.
SUGRA may be considered one of the main culminations – or the main culmination – of the research of (superficially) local quantum field theories with a "locally looking action" that are as beautiful as possible. The local diffeomorphism symmetry (Einstein's localization of the symmetry generated by the momentum \(P_\mu\)) is extended to include the local supersymmetry, too. Because the spinorial (fermionic) components grow exponentially and the bosonic degrees of freedom grow polynomially, special dimensions are needed for SUSY to work – and 11 spacetime dimensions is the maximum dimension in which the super-Poincaré symmetry may be defined. At the same moment, SUGRA needs to be quantized, for the fermions to have "beef" (classically, fermionic fields can only take the value zero) but the quantization makes it inconsistent at the loop level, so SUGRA is already a "transient" theory that makes the transition to "stringy" theories with extended objects unavoidable.
Note that the formalism of SUGRA may become really a repulsive pile of technicalities with many terms. For example, the BRST symmetry generated by \(Q\) – with its cohomologies that many people already find too abstract – isn't "complex enough" to deal with SUGRA. Instead, because of the constraints that don't form a normal Lie algebra, to do the quantization properly, supergravity theories normally need a generalization of the BRST approach, the BV (Batalin-Vilkovisky) formalism.
Good job, Gentlemen.
Needless to say, we hear questions whether SUGRA is relevant for Nature around us. Ferrara says that the Higgs needed 60 years to be discovered – well, it was really less than 50, but OK – and we will need a similar period of time to observe SUGRA. He hasn't indicated what is his scenario for observing SUGRA. I find such an observation rather unlikely – at most, we will see isolated particles that will be consistent with a "gravitino" etc. – but I am confident that SUGRA has to be an approximate description of a part of Nature, anyway. Among my "SUSY exists because", the blog post "SUSY exists because the number 3/2 cannot be missing" is the most relevant one. At any rate, string/M-theory is unavoidable and only its supersymmetric versions are promising, and those have gravitinos and local supersymmetry.
We haven't seen any "direct empirical evidence" for SUGRA but the same comment applies to the Hawking radiation and other things. Their status – how much we know they are relevant in Nature – is comparable. Especially because we see that some of the confirmations may take a very long time, we must simply be capable of evaluating theories well before the confirmations arrive – otherwise science would change to empty promises of experimental discoveries. Instead, theoretical physics is a well-defined discipline with its theoretical – but indirectly empirically rooted – rules.
The three men started a transformation that has changed the default expectation about the relevance of local SUSY. I think that the contemporary competent physicists simply do believe that SUGRA is more likely to be relevant in Nature than not, and you would need another "opposite" discovery to change this default expectation. The change of the expectation
isan important contribution to science. Science doesn't proceed just by breakthroughs that make everything certain (nothing is ever quite certain). Everyone who says otherwise is deeply misguided. A five-minute-long guide how to find gravitino balls – especially on high-security planets – and prove supergravity.
P.S.: Nature printed a story by an aggressive inkspiller named Zeeya Merali who had to call supergravity "speculative" in the title
andplace "supergravity" in quotation marks. Dear speculative "Zeeya", there is absolutely no justification for inserting supergravity – well-established concept in physics – in quotation marks and it's just plain idiotic to call the theory "speculative". (The quotes around "Zeeya" are fine because unlike supergravity, "Zeeya" isn't a valid English word.) It's just a damn theory – every theory in science is "speculative" to one extent or another. Also, to show that the number of redundant dumb attacks hasn't been enough to satisfy her, the subtitle repeats that SUGRA "might not be a good description of reality". Or, more importantly, it can be one. Even if it were just a matter of reasonable probabilitythat SUGRA is relevant in Nature, they would deserve an award. Even if it were just an amazing piece of mathematics, they would deserve a similar award. |
Difference between revisions of "Vertex equations of a quadratic function and it's inverse"
From JSXGraph Wiki
Line 1: Line 1:
A parabola can be uniquely defined by its vertex ''V'' and one more point ''P''.
A parabola can be uniquely defined by its vertex ''V'' and one more point ''P''.
The function term of the parabola then has the form
The function term of the parabola then has the form
−
:<math>\
+
:<math>\= R_0\cdot \gamma</math>
Revision as of 12:27, 16 December 2014
A parabola can be uniquely defined by its vertex
V and one more point P.The function term of the parabola then has the form [math]\beta = R_0\cdot \gamma[/math] JavaScript code
var b = JXG.JSXGraph.initBoard('box1', {boundingbox: [-5, 5, 5, -5], grid:true});var v = b.create('point', [0,0], {name:'V'}), p = b.create('point', [3,3], {name:'P'}), f = b.create('functiongraph', [ function(x) { var den = p.X()- v.X(), a = (p.Y() - v.Y()) / (den * den); return a * (x - v.X()) * (x - v.X()) + v.Y(); }]);})();
JavaScript code
var b = JXG.JSXGraph.initBoard('box2', {boundingbox: [-5, 5, 5, -5], grid:true});var v = b.create('point', [0,0], {name:'V'}), p = b.create('point', [3,3], {name:'P'}), f = b.create('functiongraph', [ function(x) { var den = p.Y()- v.Y(), a = (p.X() - v.X()) / (den * den); return Math.sqrt((x - v.X()) / a) + v.Y(); }]); |
Inverse Harmonic and Inverse Digamma Functions (HP 49G (*) and HP 50g )
04-27-2016, 02:53 AM (This post was last modified: 05-08-2016 09:04 PM by Gerson W. Barbosa.)
Post: #1
Inverse Harmonic and Inverse Digamma Functions (HP 49G (*) and HP 50g )
.
InvH:
« .577215664902 - InvPsi 1. -
»
InvPsi:
« DUPDUP DUP InvP InvP DUP InvP InvP DUPDUP InvP InvP DUP InvP DUP EXP .5 + Psi OVER - - EXP .5 +
»
InvP:
« DUP EXP .5 + Psi OVER - - EXP .5 + Psi OVER - -
»
(*) On the HP 49G, replace Psi with 0 PSI.
Examples:
4.012007 InvH --> 30.523595226 (in less than half a second); this is a very belated solution to one of the famous Valentin's challenges (#3 here).
.577215664902 InvH --> 0.46163214497 ; InvH(Euler-Mascheroni constant) = xmin (local minimum of the continuous factorial function).
4 pi 2 / - 3 ENTER 2 LN * - InvH --> 0.24999999999 ; one of the special values for fractional arguments examples here.
1.5 InvH --> 2. ; H2
1 Psi --> -0.577215664902 InvPsi --> 1.00000000009
2 Psi --> 0.422784335098 InvPsi --> 2.
Background:
Let H(x) be the continuous function associated with Harmonic Numbers. Then,
\[H(x)=\gamma +\psi \left ( x+1 \right )\]
\[\psi \left ( x+1 \right )=H(x)-\gamma\]
\[x+1 =\psi^{-1} \left ( H(x)-\gamma \right )\]
\[x+1 =\psi^{-1} \left ( H(x)-\gamma \right )\]
\[x =\psi^{-1} \left ( H(x)-\gamma \right )-1\]
or
\[H^{-1}(x) =\psi^{-1} \left ( x-\gamma \right )-1\]
That is, in order to obtain the inverse of H(x) we only need the Inverse Digamma Function and the Euler-Mascheroni constant. No problem with the constant, but the Inverse Digamma might be a problem since Digamma is not easily invertible.
A rough approximation is
\[\psi^{-1} \left ( x\right )=e^{x}+\frac{1}{2}\]
The equivalent HP 50g program is
P1:
« EXP .5 +
»
However, this is good only for x >= 10, not good enough for our purposes:
10 P1 --> 22026.9657948 Psi --> 10.0000000001.
But,
9 P1 --> 8103.58392758 Psi --> 9.00000000064
8 P1 --> 2981.45798704 Psi --> 8.00000000469
7 P1 --> 1097.13315843 Psi --> 7.00000003465
So, let's try to improve the accuracy a bit:
P2:
« DUP P1 Psi OVER - - P1
»
10 P2 --> 22026.9657926 Psi --> 9.99999999999
9 P2 --> 8103.58392239 Psi --> 8.99999999999
8 P2 --> 2981.45797306 Psi --> 8.
7 P2 --> 1097.13312043 Psi --> 7.
6 P2 --> 403.928690211 Psi --> 6.
5 P2 --> 148.912878357 Psi --> 5.00000000001
But,
4 P2 --> 55.0973869316 Psi --> 4.00000000039
3 P2 --> 20.5834634677 Psi --> 3.00000002131
Proceeding likewise, we get
P3:
« DUP P2 Psi OVER - - P2
»
This is good for x as low as 2, but not for x = 1:
2 P3 --> 7.88342863120 Psi --> 2.
1 P3 --> 3.20317150637 Psi --> 1.0000000139
A couple more steps suffice for x around -0.6 and greater, which is good enough for our purposes. That's what the InvPsi program above does, albeit in an inelegant way. Also, this is just an intuitive and somewhat inneficient approach. Better methods suggestions are welcome.
Edited to fix a couple of typos.
Edited again to include a printout of my current directory:
InvPsi will accept arguments greater or equal -1. About 0.05 s, 0.10 sor 0.50 s, depending on the arguments.
04-27-2016, 04:57 PM
Post: #2
RE: Inverse Harmonic and Inverse Digamma Functions (HP 49G (*) and HP 50g )
Cool, thanks! I was just trying to figure out an inverse harmonic number function the other day and getting nowhere. Neither Mathworld nor Wikipedia seemed to have much to say on the subject.
John
04-27-2016, 06:37 PM (This post was last modified: 04-27-2016 06:41 PM by Gerson W. Barbosa.)
Post: #3
RE: Inverse Harmonic and Inverse Digamma Functions (HP 49G (*) and HP 50g )
(04-27-2016 04:57 PM)John Keith Wrote: Cool, thanks! I was just trying to figure out an inverse harmonic number function the other day and getting nowhere. Neither Mathworld nor Wikipedia seemed to have much to say on the subject.
Thanks for your interest!
If the arguments are always plain harmonic numbers, that is, the ones obtained from the discrete function, then the inverse function can be implemented simply as
InvHn:
« .577215664902 - EXP IP »
Examples:
137 ENTER 60 / InvHn --> 5. ; 137/60 = H(5)
10 Hx --> 2.92896825397 ; H(10)
InvHn --> 10.
2E10 --> 24.2962137754 ; H(2x10^10)
InvHn --> 20000000000.
where Hx is the program for the continuous function:
Hx:
« 1. + Psi .577215664902 + »
Regards,
Gerson.
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I found an algorithm that can compute the distance of two quantum states. It is based on a subroutine known as swap test (a fidelity estimator or inner product of two state, btw I don't understand what fidelity mean).
My question is about inner product. How can I calculate the inner product of two quantum registers which contains different number of qubits?
The description of the algorithm is found in this paper. Based on the 3rd step that appear on the image, I want to prove it by giving an example.
Let: $|a| = 5$, $|b| = 5 $, and $ Z = 50 $ $$|a\rangle = \frac{3}{5}|0\rangle + \frac{4}{5}|1\rangle$$ $$|b\rangle = \frac{4}{5}|0\rangle + \frac{3}{5}|1\rangle $$ All we want is the fidelity of the following two states $|\psi\rangle$ and $|\phi\rangle$ and to calculate the distance between $|a\rangle$ and $|b\rangle$is given as: $ {|a-b|}^2 = 2Z|\langle\phi|\psi\rangle|^2$ so $$|\psi\rangle = \frac{3}{5\sqrt{2}}|00\rangle + \frac{4}{5\sqrt{2}}|01\rangle+ + \frac{4}{5\sqrt{2}}|10\rangle + + \frac{3}{5\sqrt{2}}|11\rangle$$ $$|\phi\rangle = \frac{5}{\sqrt{50}} (|0\rangle + |1\rangle) $$ then how to compute $$\langle\phi|\psi\rangle = ??$$ |
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
A more devious scheme to bring a dam down:
A secret second dam above the main target dam.
You simply install a second dam above the main dam (as high as possible ideally). You reduce the flow inconspicously so that the main dam manning do not see anything unusual, but slowly the reservoir of your second dam is filling.
Now you wait for the perfect time when the second dam reservoir is filled and the main target dam is nearly filled. If the main dam needs only some filling, you can increase the flow from the second dam.
Your attack commences with destroying the second dam which is purposefully built for being brought down (you have some pivot support columns which are disengaged). The water rushes down and gets faster and faster, converting the stored potential energy from a higher point to kinetic energy. Friction and obstacles will slow down the water masses to a point, but it will still be very fast.
When the water enters the main dam, an effect called the water hammer comes in effect. The main dam does not allow the moving water masses to continue running, so the moving water causes a sudden pressure increase. The incoming water not only causes water to slosh over, it literally pushes the dam crest apart. Result: catastrophic failure.
ADDITION: The original question states that we are on the technological level of the early Roman era, so we should neither expect to have a hydro dam like the Hoover nor a reservoir for a city of million people. It will be more like a dam with the height of metres and the reservoir like a big lake.
Still we can compare dynamic with static pressure. The dam need to withstand static pressure, so we can assume we need approximately a pressure with the same order of magnitude to break the dam.
\begin{eqnarray*}\rho & = & density(kg \, m^{-3}) \\g & = & gravitational \; acceleration = 9.81 \approx 10 \; m \, s^{-2} \\h & = & height \; m \\v & = & velocity \; m \, s^{-1} \\mean \; static \; pressure & = & \frac{1}{2} \, \rho \; g \; h (The \; dam \; holds \; this \; pressure) \\dynamic \; pressure & = & \frac{1}{2} \; \rho \; v^2 \Rightarrow v \approx \sqrt{10*h}\end{eqnarray*}
Moderate flash flood velocity is 2.6 m/s and a very fast flashflood is in the range of 26 m/s. A moderate flashflood will be held by a 0.6m dam, a worst case scenario of 26 m/s would give an impressive height of 70 m. But the flash flood water will merge with the still water in the reservoir, so an inelastic collision will occur and the water slows considerably down. So the final velocity of the water will be the ratio$$ r = \frac{flash \; flood \; mass}{reservoir \; mass + flash \; flood \; mass}$$ of the flash flood speed (I also neglected friction and energy dissipation by waves).
Result: If the dam is something like 10 m high and the reservoir is big (10-100 times), even the ugliest flashflood will have no pressure effect. Moderate flash floods can be contained even with small dams. On the other hand, if the dam is only a few meters high and the reservoir has not a much bigger capacity (10 times) than the incoming water, an incoming massive flash flood
is able to forcibly remove the dam. |
I'm deficient in some understanding about eigenvectors/values and diagonalisation and singular values. I came to a result with the following matrix
$$ A = \begin{bmatrix} 2 & 4 \\ -4 & 2 \\ \end{bmatrix} $$
To find the singlur values I find the roots of the characteristic equation.
$det(\lambda I-A^TA)=0$ This comes to $$ \begin{vmatrix} \lambda -20 & 0 \\ 0 & \lambda - 20 \\ \end{vmatrix} = 0 $$
Which actually gives $\lambda_{1} = 20$ and $\lambda_{2}=20$.
Now this is where my knowledge is lacking. I don't know why they are called "singular" values. So to me if $\delta_{1} = \sqrt {20}$ and $\delta_2 = \sqrt {20}$. My question is, can this be true? Does singular value mean that each eigenvalue has to be distinct and each singular value has to be distinct? Can we have two singular values the same? Or is "singular" referring to something else?
Thanks. |
So, what's the... nerve of the nerve? There's a functor $N: sCat\to sSet$. We know that this induces a functor of quasicategories $Cat_\infty\to Cat_\infty$, and I think this should be homotopic to the identity, but I'm not sure...
another question: is it just me or does the entirety of Lurie's discussion of excellent model categories in HTT A.3.2. seem to rely on an unstated assumption that the monoidal structure is Cartesian? let C be a S-enriched category and K an object in S (S an excellent model category). then he defines C^K as the category w/ same objects and where Hom_{C^K}(X,Y) is Hom(X,Y)^K - as far as I can tell, there is no clear way to define composition or identity in this category without assuming
that the monoidal structure is Cartesian. e.g. identity is a map 1->Hom(X,Y)^K, or a map K->Hom(X,Y), and I see no reason for this to exist unless your unit is a final object
@JonathanBeardsley whats your definition of the functor QCat->QCat then? I thought based on what you were saying you were going QCat->sCat->QCat - are you doing something else?
@DenisNardin @dhy yeah so this is the right adjoint of a Quillen equivalence, and by a few different theorems this induces an equivalence from Cat_∞ to itself (in the large quasicategory of small quasicategories). But it doesn't seem that this just automatically makes it (equivalent to) the identity, unless I'm missing something
In this case, it has two presentations. One as the underlying quasicategory of sSet and one as the underlying quasicategory of sCat, where by underlying quasicategory I mean "nerve of fibrant replacement of hammock localization"
@HarryGindi The problem is that you have to specify how you are identifying them. Essentially you are asking whether an equivalence in the ∞-category Cat_∞ is the identity, but this does not make any sense if you don't identify source and target, and the answer depends on the identification
but so... does the thing i said earlier make sense? which I think is basically agreeing with @dhy and @DenisNardin. I mean, the point being that we end up with a 1-simplex between these two vertices in the large quasicategory of quasicategories... and one of them is (by definition) $Cat_\infty$, and one of them isn't.
Ultimately, and this is all really silly, I was hoping for a slick way of proving that $N\circ op \simeq op\circ N$, where on one side I'm taking $op$ of simplicial sets, and the other side I'm taking $op$ of simplicial categories.
Basically because I don't think I'm smart enough to actually break open the definition of $N$ itself and check this.
Well, yeah I dunno. I'm at the point where I just have this really stupid thing I want to be true and it's obviously true, and it just feels like it'd be a very bad idea to like... spend a lot of time trying to work out some really complicated framework. |
I’ll give a fairly detailed commentary on your argument.
$[\Rightarrow]$ Assume that int(cl($\{ x \})) = \varnothing$. Thus, $X \neq \{ x \}$ and the result trivially holds.
It’s true that if $\operatorname{int}\operatorname{cl}\{x\}=\varnothing$, then $X\ne\{x\}$, but this by no means implies that $x$ is not an isolated point. For a simple counterexample, note that $\Bbb Z\ne\{0\}$, but $0$ certainly
is an isolated point of $\Bbb Z$. The simplest approach here is to prove the contrapositive: assume that $x$ is an isolated point, and show that $\{x\}$ is not nowhere dense. This actually is trivial: if $x$ is isolated, then $\{x\}$ is an open subset of $\operatorname{cl}\{x\}$, so $\operatorname{int}\operatorname{cl}\{x\}\ne\varnothing$.
$[\Leftarrow]$ Assume that $x$ is not an isolated point of $X$. Thus,
for every $\epsilon > 0$, $B_\epsilon (x) \cap X \neq \varnothing$.
$B_\epsilon(x)\cap X\ne\varnothing$ for each $\epsilon>0$ even if $x$
is an isolated point of $X$. What you want here is that $B_\epsilon(x)\cap\big(X\setminus\{x\}\big)\ne\varnothing$ for each $\epsilon>0$.
Hence, $X$ is not a singleton.
To prove that int(cl($\{ x \})) =
\varnothing$, notice that it is the same to prove cl($ X \setminus \{
x \}) = X$. Thus, let $y \in M$ be arbitrary.
I assume that $M$ is a typo for $X$.
We have to prove that, for
every $\epsilon>0$, $B_\epsilon (x) \cap (X \setminus \{ x \}) \neq
\varnothing$,
No, you have to show that $B_\epsilon(y)\cap\big(X\setminus\{x\}\big)\ne\varnothing$ for each $\epsilon>0$.
but this holds trivially if $X$ is not a singleton.
This is sufficiently lacking in detail that I can’t tell whether you have in mind a correct argument or not; you need to expand it a bit. Something like this would do:
If $y\in X\setminus\{x\}$, then obviously $y\in B_\epsilon(y)\cap\big(X\setminus\{x\}\big)\ne\varnothing$ for each $\epsilon>0$, and by hypothesis $B_\epsilon(x)\cap\big(X\setminus\{x\}\big)\ne\varnothing$ for each $\epsilon>0$, so $B_\epsilon(y)\cap\big(X\setminus\{x\}\big)\ne\varnothing$ for each $\epsilon>0$, and $\operatorname{cl}\big(X\setminus\{x\}\big)=X$, as desired. |
In the second quantization (physics of many-body systems) language, the (physical) question is "How many particle in each state?". Suppose that there is $n_\alpha$ particles in state $\alpha$, each particle in this state has energy $\epsilon_\alpha$, then the total energy in this state is: $n_\alpha\epsilon_\alpha$. If we want to have the total energy of the system, we just simply add all possible state's energies:$$E = \sum_\alpha n_\alpha\epsilon_\alpha$$We see that E and $n_\alpha$ are physical observables. In QM, each physical observable corresponds to a hermitian operator. Hence, naturally, E is corresponding to Hamiltonian, which is the energy operator; and $n_\alpha$ is corresponding to $\hat{n}_\alpha$, occupation number operator. It turns out that $n_\alpha$ and E are eigenvalues of $\hat{n}_\alpha$ and H, respectively. So we have:$$H = \sum_\alpha \hat{n}_\alpha\epsilon_\alpha$$
The question now is to find the number operator. We discuss here the bosonic case. In the many-body system, since the number of particle can be changed, we introduce the creation and annihilation operators: $a^\dagger$ and a. Their definitions are:$$a^\dagger(k)|0\rangle = |k\rangle$$$$a^\dagger(k_{n+1})|k_1,k_2,...,k_n\rangle = \textrm{(constant)}_1|k_1,k_2,...,k_n,k_{n-1}\rangle$$
$$a|0\rangle = 0$$$$a(k)|k_1,k_2,...,k_n\rangle = \textrm{(constant)}_2\sum_i\delta(k,k_i)|k_1,k_2,...,k_{i-1},k_{i+1},...,k_n\rangle$$$\textrm{(constant)}_2$ and $\textrm{(constant)}_1$ can be obtained by permutation: $\textrm{(constant)}_1 = \sqrt{n+1}$, $\textrm{(constant)}_2 = 1/\sqrt{n}$. When acting $a^\dagger(k)a(k)$ on a n-particle state $|k_1,k_2,...,k_n\rangle$, we get:$$a^\dagger(k)a(k)|k_1,k_2,...,k_n\rangle = n|k_1,k_2,...,k_n\rangle$$which is exactly the same as the action of the occupation number operator on the state $|k_1,k_2,...,k_n\rangle$. Then, $\hat{n} \equiv a^\dagger a$.
Is that what you want? |
In this paper, Guilherme França and André LeClair show that $$\gamma_{y}\sim 2 \pi \left(y-11/8\right)/W\left((y-11/8)e^{-1}\right)$$ where $W$ is the Lambert W function, and $\gamma_{y}$ is the imaginary part of the the $y$th non-trivial zero along the critical line. This formula can then be used in a similar way to Gram points for finding zeta zeros.
When this function replaces $y$ in $\zeta(1/2+iy)$, and the real and imaginary parts are plotted parametrically, the result shares superficial similarities with a limaçon:
where one complete orbit is made approximately in every integer interval.
Firstly, analysing the limaçon in terms of its orbit seen as a particle travelling at constant velocity around the origin, comparing curvature, distance from origin, and non-directional "speed", where
\begin{align} &x(t)\text{:=}\cos (t)+\cos (2 t)+1&\\ &y(t)\text{:=}\sin (t)+\sin (2 t)&\\ \end{align}
the solutions to
\begin{align} t\in\mathbb{R}:\frac{2 \left(x'(t)^2+y'(t)^2\right)^{3/2}}{\left| x'(t) y''(t)-x''(t) y'(t)\right| }=\sqrt{x'(t)^2+y'(t)^2} \end{align}
are clearly
\begin{align} t=2\pi n-\cos^{-1}(-5/4),n\in\mathbb{Z}\\ t=2\pi n+\cos^{-1}(-5/4),n\in\mathbb{Z}\\ t=\dfrac{2}{3}(3\pi n-\pi),n\in\mathbb{Z}\\ t=\dfrac{2}{3}(3\pi n+\pi),n\in\mathbb{Z}\\ \end{align}
Interestingly, the zeta function has similar properties when analysed in this way. Since
\begin{align} |Z(x)|=\sqrt{\Im\left(\zeta \left(\frac{1}{2}+i y\right)\right)^2+\Re\left(\zeta \left(\frac{1}{2}+i y\right)\right)^2} \end{align}
where $Z$ is the Riemann-Siegel Z function, $\sqrt{\Im(f(y))^2+\Re(f(y))^2}$ is zero at $\gamma=2 \pi \left(y-11/8\right)/W\left((y-11/8)e^{-1}\right),$ for
\begin{align} &f(\text{y})\text{:=}\zeta \left(2 i \pi \left(y-11/8\right)/W\left((y-11/8)e^{-1}\right)+1/2\right)&\\ &v(y)\text{:=}\sqrt{\Im\left(f'(y)\right)^2+\Re\left(f'(y)\right)^2}\\ &\kappa (y)\text{:=}\frac{\left| \Re\left(f'(y)\right) \Im\left(f''(y)\right)-\Im\left(f'(y)\right) \Re\left(f''(y)\right)\right| }{\left(\Im\left(f'(y)\right)^2+\Re\left(f'(y)\right)^2\right)^{3/2}}\\ &d(y)\text{:=}\sqrt{\Im(f(y))^2+\Re(f(y))^2} \end{align}
a plot of $\{v(y),2 \pi /\kappa (y),\pi d(y)\}$ with gridlines at $\gamma=2 \pi \left(y-11/8\right)/W\left((y-11/8)e^{-1}\right)$ looks like this:
where it can be seen that $\{v(y),2 \pi /\kappa (y)\}$ intersect at the zeta zeros.
f[y_] := InputForm[Zeta[Rational[1, 2] + Complex[0, 2] Pi (Rational[-11, 8]+ y)/ProductLog[E^(-1) (Rational[-11, 8] + y)]]]v[y_] := InputForm[(Im[Derivative[1][f][y]]^2 + Re[Derivative[1][f][y]]^2)^Rational[1, 2]]\[Kappa][y_] := InputForm[Abs[Im[Derivative[2][f][y]] Re[Derivative[1][f][y]] - Im[Derivative[1][f][y]] Re[Derivative[2][f][y]]] (Im[Derivative[1][f][y]]^2 + Re[Derivative[1][f][y]]^2)^Rational[-3, 2]]d[y_] := InputForm[(Im[f[y]]^2 + Re[f[y]]^2)^Rational[1, 2]]r = 2; rr = 15;zeros = Rest@ Table[y /. FindRoot[ 2 Pi (-11/8 + y)/LambertW[(-11/8 + y)/E] == Im[ZetaZero[n]], {y,n}], {n, r, rr}];Plot[{v[y], 2 \[Pi]/\[Kappa][y], \[Pi] d[y]}, {y, r, rr}, PlotRange -> All, PlotTheme -> "Garnet", Filling -> {1 -> {2}, 2 -> {3}}, Axes -> False, Frame -> True, ImageSize -> 400, GridLines -> {zeros, {}}, Epilog -> {Red, PointSize[Large], Point[Transpose@{zeros, Table[v[y], {y, zeros}]}]
It can be seen then that there is clearly a close relationship between the curvature of $f(y)$
(ie $\kappa(y)$), the speed of a particle moving along the path of $f(y)$ (ie $v(y)$), and the distance from origin (ie $d(y)$).
My question then is how would I go about finding the solutions to
\begin{align} A=\left\{y\in\mathbb{R}:v(y)=\dfrac{2\pi}{\kappa(y)}\right\} \end{align}
analytically? And for
\begin{align} B=\left\{y\in\mathbb{R}:\gamma=2 \pi \left(y-11/8\right)/W\left((y-11/8)e^{-1}\right)\right\} \end{align}
is it true that $B\in A$ ?
Update
In response to მამუკა ჯიბლაძე's comment, it certainly seems from a quick numerical check that $v(y),2\pi/\kappa(y)$ don't actually intersect at the zeta zeros, but are related to their approximations:
yy = y /. Table[FindRoot[v[y] == 2 Pi/\[Kappa][y], {y, n}], {n, 1, 20}];y1 = Table[w[yy[[y]]], {y, 1, Length@yy}] // N;y2 = Table[Im@ZetaZero[y], {y, 1, Length@yy}] // N;y3 = Table[w[y], {y, 1, Length@yy}] // N;ListLinePlot[{Table[(Log[y2[[y]]] - Log[y3[[y]]])/(2 Pi y^(3/2)), {y, 1, Length@yy}], y1 - y2}]
where for
\begin{align} &w(y):=2 \pi \left(y-11/8\right)/W\left((y-11/8)e^{-1}\right)\\ &r_{y}:=\text{solutions to }v(y)=2\pi/\kappa(y) \end{align}
the above is a plot of $(\log(\gamma_{y})-\log(w(y))/(2\pi y^{3/2})$ against $w(r_{y})-\gamma_{y},$ so it would seem that $B\notin A.$
However, I am still unsure how to show this analytically, and since $(\log(\gamma_{y})-\log(w(y))/(2\pi y^{3/2})$ is just a rough guess, I would be interested to know whether there is a more accurate definition of the relationship between $r_{y}$ and $\gamma_{y}$. |
From an algorithm point of view, the ways you can attempt to crack the problem are either too strenuous for decent results in an amount of time or redundant in comparison to a decent mathematician... so I don't expect anything useful to be yielded from using a computer algorithm.
That said, I have a simple starting point that may or may not be useful in the long run
Lets consider that a Nonogram can be represented as bits where $1=coloured$ etc. Nonagrams contain numbers outside the grid (lets casually call this $A_x$, where $A_x$ is an array of $S$ length) which validly represent the bits in the respective row or column. It is possible to determine if a Nonogram is unique by finding if there are other solutions that have the exact same $A_x$. Obviously, it is difficult to do this mathematically, lets break the problem down and see if it gets us anywhere {I intend for my answer to help others reach a final, accepted answer (if that even exists). I probably won't be able to continue past a certain level of maths}
Lets start by just thinking of a single trail of bits (called $B_i$). This trail is of length $N$ and can be represented by $A_x$. How many combinations of $B_i$ share $A_x$? It is easy to see that for $A_x$ with $S = 1$ there are $N-A_1+1$ of $B_i$. For $S>1$ it gets complicated since there can be a variable number of $0$ bits padding between $1$ clusters. What is certain is that there is always at least one $0$ bit between clusters. Consider a strategy where the padding can be distributed in many combinations. Let $T = \sum\limits_{r=1}^{S}A_r$ and $D=N-T$. The difference $D$ represents the total padding between clusters, it can be distributed for the $S+1$ paddings in between clusters in different ways. There are $S-1$ paddings that must be a size of at least one. This leaves $D-S+1$ to be distributed between all $S+1$ padding. The formula for this would then be ${D+1} \choose {S}$
So $A_x$ has ${N+1-\sum\limits_{r=1}^{S}A_r} \choose {S}$ corresponding $B_i$ of length $N$.
This also implies that $N+1-\sum\limits_{r=1}^{S}A_r=S$ when there is only one unique $B_i$. This tells us that ALL valid Nonograms, where all of the rows and columns satisfy this rule, have a unique solution! (for future reference, lets call these
stuffed Nonograms)
Lets have a closer look into the stuffed Nonograms since it may give us a
very vague idea of what the final function (if it exists) will look like. An equation to identify stuffed Nonograms can be derived if we rearrange the rule to $N=S-1+\sum\limits_{r=1}^{S}A_r$. The square Nonogram will have $2N$ of $B_i$ that follows this rule so by summing everything, $2N^2=\sum\limits_{i=1}^{2N}(S_i-1+\sum\limits_{r=1}^{S_i}A_{i_r})$ or the nicer equation $2N^2+2N=S_T+A_T$ using $S_T$ and $A_T$ to represent the amount of total items $A_x$ and the summation of all of them respectively. There is a constraint that $S_T\geq2N$ and $A_T\geq2N$.
Logically it can be concluded that stuffed Nonograms are Nonograms that have a completely/almost full unique grid as the solution. They have peppered padding of length one (single holes). This can be proven by revisiting $B_i$. Let $B_i$ have $C$ padding of one so that $T = N - C$ and $S = C + 1$. Subbing into $N+1-T=S$ we see no contradiction since $N+1-(N-C)=C+1$ so $C+1=C+1$ $\therefore$ peppered padding of one is always unique! There is an indirect implication that there cannot be any front or back padding (since it creates more solutions) so stuffed Nonograms must at least have a full outer rim of one. It is for this reason that the only area of stuffed Nonograms with importance is $(N-2)^2$. Now consider this; if the area is unwound to become many $B_i$ and the placement of the peppered padding is considered in a converse way of thinking then the amount of corresponding $B_i$ from $A_x$ formula can be repurposed to count the total amount of stuffed Nonograms possible. The formula will initially over approximate since the rewrapping of some $B_i$ configurations will lead to alignment of padding resulting in padding of length greater than one which is invalid. To overcome this, an extra formula that counts the number of invalid wraps must be developed.
Edit: The following two paragraphs are now redundant since I've managed to find what appears to be the solution by changing my approach. The paragraphs are still necessary to fully understand the tech used in my new approach so for historic reasons I'm keeping them for the now. (Thinking of shortening this answer by moving the "holes in grid" problem into its own Q&A question)
Lets over approximate. Consider a single $B_i$, the padding is all of length one so all of $A_x$ (in this perspective) will contain $1$. $S$ can vary from $0\to \lceil\frac{N-2}{2}\rceil$ (thanks odd numbers) and $T=S$. The formula for the amount of possible $B_i$ must then be $\sum\limits_{r=0}^{\lceil\frac{N-2}{2}\rceil} {{(N-2)+1-r} \choose {r}}$.
This is actually equivalent to the Fibonacci series (thanks to @justhalf's insight). The function $Fi(x)$ will be used to represent the series from now on. There is $N-2$ of $B_i$ in the relevant area which can have different combinations at once. This is the same as a combination lock of length $N-2$ so the overall formula for the amount of configurations in this case is $Fi^{(N-2)}(N)$.
Now to count all of the invalid wraps. Lets use an easy model and say that the $B_i$ are stacked on top of each other. Invalid wraps must occur when a padding of length greater than one is created from stacking $B_i$ that have padding which align. Alignment can only happen between neighbour $B_i$ and all that is required is for padding to exist at the same position locally within the two $B_i$. Take $G$ to equal the amount of $B_i$ and $L$ to equal the length of all $B_i$ (just incase we can revisit this formula). There are $G-1$ seams between $B_i$ where alignment can occur and it is possible for more than one set of neighbour $B_i$ to align at once.
Edit: This is the new approach
Lets consider the valid combinations at
one seam. In order for valid combinations to happen the padding must never overlap. Lets say that the top $B_i$ at this seam can be anything where it is valid within itself. It is logical to conclude that the bottom $B_i$ must only be valid when its own padding is contained within the clusters of the top $B_i$ whilst still being internally valid. It is for this reason that the bottom $B_i$ can be considered as any $A_x$ that fragments into small locally valid $B_i$! We can now further abuse the $A_x$ formula! The formula currently depends on arrangement of $A_x$ so to fix that to make the combinations countable it is now ${{L+1-\sum\limits_{r=1}^{S}A_r} \choose {S}}S!$ (permutations). The clusters can all be any combination of locally valid $B_i$ so this is $\prod\limits_{i=1}^{S}Fi(A_i+2)$. This gives us the final formula for the seam combinations (for $A_x$) to be ${{L+1-\sum\limits_{r=1}^{S}A_r} \choose {S}}S!\prod\limits_{i=1}^{S}Fi(A_i+2)$. The total combinations can be found by summing combinations for $A_x$'s. Initially, I thought it wouldn't be possible to account for multi-seam combinations however I think that by inversion, the next seam in the stack has the same possible combinations regardless of the affects of the seam beforehand. This would create a simple combination lock of length $G$. I will be thoroughly impressed if anyone manages to put the whole formula together!
Another group of unique, valid Nonograms exists that can immediately be counted. The group consists of Nonograms where the rows/columns can only be completely full, empty or something inbetween (the latter arises from the overlap of rows and columns that are full or empty which creates a 'different image'). For future reference, lets call these
net Nonograms.
Anyone familiar with solving Nonograms will know how useful completely full/empty rows/columns are. It is a good idea to take this concept to its extremes because it may be used as a part of the final solution to split Nonograms into smaller Nonograms (well, it is way more complicated than that but still). Consider just the rows of net Nonograms, it is possible to use binary to represent the information. It is for this reason that $2^N$ is the amount of combinations of rows. The same can be said for the columns. If the information for the rows and columns is 'overlapped' then we have $2^{2N}-2$ total net Nonograms! The $-2$ takes into account the two repeated combinations (all full or all empty). Gonna need proof for how overlap is resolved and why it still leads to unique, valid Nonograms but I think it is safe to say that this was a simple idea for the books...
Ideas
This section is going to be used to document facts/ideas ranging from basic to abstract, I just want to put concepts on the record so that this can be used as a helpful guide when looking for the answer.
The final function will be a recurrence relation.
The answer to "How many invalid solutions exist for $A\times B$ Nonograms with a specific set of $A_x$?" will inevitably lead to the answer for "How many valid solutions for an $N\times N$ Nonogram exists?" and "How many unique $N\times N$ Nonograms exist?"
The answer to the above "invalid solutions exist..." will likely contain recursion.
The groups of Nonogram currently defined in my 'answer' both tend to zero as $N\to \infty$ when considering the approximate probability of the groups out of all possible random configurations $2^{N^2}$ |
I propose to collect here open problems from the theory of continued fractions. Any types of continued fractions are welcome.
Guy, Unsolved Problems In Number Theory, F21, attributes to Bohuslav Divis the conjecture that in each real quadratic field there is an irrational with all partial quotients 1 or 2; more generally, same question but with 1 and 2 replaced by any pair of distinct positive integers.
every integer appears as the denominator of a finite continued fraction whose partial quotients are bounded by an absolute constant.
From M. Waldschmidt, "Open Diophantine Problems" (Moscow Mathematical Journal vol. 4, no. 1, 2004, pp. 245-305):
Does there exist a real algebraic number of degree $\geq 3$ with bounded partial quotients? Does there exist a real algebraic number of degree $\geq 3$ with unbounded partial quotients?
Find fixed points (mod 1) of Minkowski's question mark function, see A058914 from "The On-Line Encyclopedia of Integer Sequences". This picture
is taken from http://mathworld.wolfram.com/MinkowskisQuestionMarkFunction.html It shows that there is only one positive fixed point (mod 1) less than 1/2. It is approximately 0.42037. Does this constant have a closed-form expression?
Hermite's problem is an open problem in mathematics posed by Charles Hermite in 1848. He asked for a way of expressing real numbers as sequences of natural numbers, such that the sequence is eventually periodic precisely when the original number is a cubic irrational.
Guy, Unsolved Problems In Number Theory, F21, attributes to Leo Moser the conjecture that there is a constant $c$ such that every $n$ can be expressed as $n=a+b$ in such a way that the sum of the partial quotients of $a/b$ is less than $c\log n$.
A collection of Open problems in geometry of continued fractions by Oleg Karpenkov.
It is known that (see J.W. Porter, On a theorem of Heilbronn) $$H(b)=\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}\ell(a/b)= \dfrac{2\log 2}{\zeta(2)}\cdot\log b+C_P-1+O_\varepsilon(b^{-1/6+\varepsilon}),$$ where $\ell(a/b)$ is a length of standart continued fraction expansion and $C_P$ is Porter's constant. Averaging over numerators and denominators one can prove asymptotic formula for the variance. Let $$E(R)=\dfrac{2}{R(R+1)}\sum\limits_{b\le R}\sum\limits_{a\le b}\ell(a/b) $$ and $${D}(R)=\dfrac{2}{R(R+1)}\sum\limits_{b\le R}\sum\limits_{a\le b}\left( \ell(a/b)-E(R)\right)^2. $$ Then (see D. Hensley, The Number of Steps in the Euclidean Algorithm) $${D}(R)=D_1\cdot\log R+o(\log R). $$ But if denominator is fixed then for the variance only right oder bound is known (see Bykovskii V.A., Estimate for dispersion of lengths of continued fractions): $$\dfrac{1}{b}\sum\limits_{a=1}^{b}\left(\ell\left(\dfrac{a}{b}\right)- \dfrac{2\log2}{\zeta(2)}\log b\right)^2\ll\log b.$$
Conjecture: $$\dfrac{1}{\varphi(b)}\sum\limits_{1\le a\le b\atop(a,b)=1}(\ell(a/b)-H(b))^2=D_1\log b+o(\log b).$$ |
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
维基教科书:格式手册
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结构编辑 教科书一般可以分为“主页面”、“目录”和“内容”三部分。 “目录”和“内容”必须是“主页面”的子页面；例如创建一本“示例教科书”，其目录的标题应为“示例教科书/目录”。 “主页”和“目录”可以合并。 主页用于简要介绍该教科书和其他的一些基本信息。 主页必须包含的内容有——该教科书的分类、指向目录或内容的超链接等；若教科书参考了某些资料，也应在主页中说明。 主页中可以包含的内容有——教科书的封面、此教科书的描述等。 目录用于陈列教科书内容的链接。 教科书中的内容应保持格式统一。 格式的内容可以由教科书的创建者建立；若已有多人参与了该教科书的编写时，统一格式的建立应在该教科书主页面的讨论页征得共识之后。 对现有统一格式的大幅更改亦需在主页面的讨论页征得共识。 若存在统一格式，应在主页面说明或加入指向说明统一格式的页面的超链接。 若教科书的结构设计确实美观合理的，不必拘泥于以上指引。 |
Consider a basic integer program such as:
$$\begin{align} \min_x & \quad c^Tx \\ \text{s.t.} & \quad Ax \leq b \\ &\quad x_i \in \{-100,\ldots,100\} \end{align} $$
where $x \in \mathbb{Z}^n, A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R^m}$.
Say that I have a feasible point $y \in \mathbb{Z}^n \cap [-100,100]^n$ with the property that all points adjacent to $y$ cannot be optimal. In other words, given $y$, all points in the set:
$$\mathcal{A}(y) = \Big\{ z \in \mathbb{Z}^d ~\big|~ z_i = y_i \pm 1 \Big\}$$
could be excluded from the feasible region of the IP.
I am wondering if there is an elegant way to formulate constraints that will exclude all points that adjacent to $y$ from the feasible region. |
I still think this is off-topic, but it seems I need more space than a comment to show (answer?) why that is so.
You are starting from some performance specifications and are looking to get to a set of features you need in your camera.
Here is a post from NI about stereo vision that gives a formula for depth resolution:
$$\Delta z = \frac{z^2}{fb}\Delta d \\$$
where $z$ is the depth of the object from the stereo system, $\Delta d$ is the depth resolution, $f$ is the focal length of the camera, $b$ is the baseline, and $d$ is the image disparity.
So, you want 1% depth resolution at 100 meters, or a depth resolution of 1 meter. A focal length of 8 millimeters, or 0.008 m, and a baseline of 0.5 m.
Rearranging the equation, it looks like you'll need a camera capable of registering a disparity of:
$$\Delta d = \Delta z \frac{fb}{z^2} \\\Delta d = 1 \frac{(0.008)(0.5)}{100^2} \\\Delta d = 4 x 10^-7 m \\\Delta d = 0.4 \mu m \\$$
Assuming pixel accuracy (not sub-pixel accuracy), you'll want one pixel to be 0.4 $\mu$m or smaller, so the 0.4 $\mu$m disparity is registered as a one pixel shift between cameras.
Here's a list of sensor formats and sizes. I'm assuming these cameras all do "full HD", at a resolution of 1920x1080. Looking at the 2/3" format, the sensor width is 8.8 mm.
You need to register 0.4 $\mu$m, how does that compare to the 2/3" format? Well, at a width of 0.0088 m, with 1920 pixels across that width, the 2/3" format has a pixel width of $0.0088/1920 = 4.58\mu m$. So, off by a factor of 10. You need the pixel width to be about 11 times smaller.
So let's look at the 1/3" format - as in the iPhone 6. There the width is 4.8mm, so about half as wide, meaning you still need the pixels to be about 5-6 times smaller than the camera sensor in the iPhone.
This is also assuming you want to use every pixel in a full HD format - this will result in a high computation time. Most of the stereo vision projects I've seen have used cameras with lower resolutions or downsampled the image to a format like 640x480, but of course that means that the pixels are much (3x) larger.
You ask if "IP Cameras" are "proper," but IP cameras come in lots of styles.
Hopefully this will help you as a guide for your iterations. Plainly speaking, I don't think you'll ever find anything (that is reasonably affordable) that would do the depth resolution at the baseline you're talking about. I would imagine the baseline would be more on the range of 5-10 meters to get what you need. At 10 meters, fyi, the pixel size becomes 8 $\mu$m. At that point, most/all of the HD cameras should be able to do what you want, but again HD is computationally expensive because there are so many pixels to correlate.
This will be an iterative process. Work forward and backward and forward and backward until you get the design that meets your needs. You'll find you need to make tradeoffs along the way, and that's the core of engineering - finding the "optimal" balance of specifications. Cost, performance, size, cost, weight, interface, lead time, cost, cost, cost. |
We know that if charged particle is accelerated, it will radiate. According to Larmor formula, the power is $$P = {2 \over 3} \frac{q^2 a^2}{ c^3}$$ in the Gauss units, where $a$ is acceleration.
And accelerated particle will experience radiation damping force(Abraham–Lorentz force) $$\mathbf{F}_\mathrm{rad} = { 2 \over 3} \frac{ q^2}{ c^3} \mathbf{\dot{a}}$$
The equation of motion of particle should be $$m \frac{d \mathbf{v}}{dt}= \mathbf{F}_{ext}+ { 2 \over 3} \frac{ q^2}{ c^3} \mathbf{\dot{a}} \tag{1}$$
But a contradiction arises when the external force is a constant force. In this case, the particle has constant acceleration ($\mathbf{\dot a}=0$). But it will still radiate energy ($P\neq0$). The work done by external force $\mathbf{F}_{ext}$ will be totally translated to particle's kinetic energy(due to uniform acceleration), but the particle will still radiate energy to infinity. Does this violate the energy conservation?
Note1: For constant force, the uniform acceleration solution $\mathbf{\dot a}=0$ is still the new equation(1) 's solution. That is, $\dot a=0$ is a solution for $a=b$. Certainly this solution($\dot a=0$ ) can also be the solution of $a= b+d \dot a =b$.
Note2: Even though you take the relativistic effect into consideration, you still cannot explain this contradiction. The relativistic generalization of radiation damping force is in Landau's The classical theory of fields (76.2) $$F_{rad}^\mu= \frac{2 e^2}{3 c}(\frac{d^2 u^{\mu}}{ds^2}-(u^\mu u^\alpha)\frac{d^2 u_\alpha}{ds^2})$$ So we see with constant external force $F_{ext}^\mu=\text{cosntant}$, equation of motion, $$m\frac{du^{\mu}}{ds}=F^{\mu}_{ext}+F^{\mu}_{rad}\tag{2}$$
Constant $4$-acceleration $\frac{du^{\mu}}{ds}=\frac{F^{\mu}_{ext}}{m}$, $\frac{d^2 u^{\mu}}{ds^2}=0$ is still the equation(2)'s solution. So there is still no radiation damping force. |
When space probes like New Horizon's are traveling towards their target & they take photographs of their target en-route, does the incoming light need to be adjusted for the Doppler effect because of the speed of the probe?
You can calculate the wavelength shift between the transmitted light and that observed by the receiver from the doppler equation:
$$\lambda_r = \frac{\lambda c}{(c - v_r)}$$
The Helios probes reached a speed of $70km/s$ as they approached the sun. What effect would that have on them observing something with a wavelength of $600nm$?
$$\lambda_r = \frac{(600nm)(3.0\times 10^8 m/s)}{(3.0\times 10^8 m/s) - (7.0\times 10^4 m/s)}$$ $$\lambda_r = 600.14nm$$
It's unlikely that any sort of external sensor observations are going to be affected by shifts of that magnitude. However communications (which rely on very precise timing) can be affected. This was potentially a problem with the Huygens-Cassini mission |
All quantum operations must be unitary to allow reversibility, but what about measurement? Measurement can be represented as a matrix, and that matrix is applied to qubits, so that seems equivalent to the operation of a quantum gate. That's definitively not reversible. Are there any situations where non-unitary gates might be allowed?
Unitary operations are only a special case of quantum operations, which are linear, completely positive maps ("channels") that map density operators to density operators. This becomes obvious in the Kraus-representation of the channel, $$\Phi(\rho)=\sum_{i=1}^n K_i \rho K_i^\dagger,$$ where the so-called Kraus operators $K_i$ fulfill $\sum_{i=1}^n K_i^\dagger K_i\leq \mathbb{I}$ (notation). Often one considers only trace-preserving quantum operations, for which equality in the previous inequality holds. If additionally there is only one Kraus operator (so $n=1$), then we see that the quantum operation is unitary.
However,
quantum gates are unitary, because they are implemented via the action of a Hamiltonian for a specific time, which gives a unitary time evolution according to the Schrödinger equation. Short Answer
Quantum operations
do not need to be unitary.In fact, many quantum algorithms and protocols make use of non-unitarity. Long Answer
Measurements are arguably the most obvious example of non-unitary transitions being a fundamental component of algorithms (in the sense that a "measurement" is equivalent to sampling from the probability distribution obtained after the decoherence operation $\sum_k c_k\lvert k\rangle\mapsto\sum_k |c_k|^2\lvert k\rangle\langle k\rvert$).
More generally, any quantum algorithm that involves probabilistic steps requires non-unitary operations. A notable example that comes to mind is HHL09's algorithm to solve linear systems of equations (see 0811.3171). A crucial step in this algorithm is the mapping $|\lambda_j\rangle\mapsto C\lambda_j^{-1}|\lambda_j\rangle$, where $|\lambda_j\rangle$ are eigenvectors of some operator. This mapping is necessarily probabilistic and therefore non-unitary.
Any algorithm or protocol that makes use of (classical) feed-forward is also making use of non-unitary operations. This is the whole of one-way quantum computation protocols (which, as the name suggests, require non-reversible operations).
The most notable schemes for optical quantum computation with single photons also require measurements and sometimes post-selection to entangle the states of different photons. For example, the KLM protocol produces probabilistic gates, which are therefore at least partly non-reversible. A nice review on the topic is quant-ph/0512071.
Less intuitive examples are provided by dissipation-induced quantum state engineering (e.g. 1402.0529 or srep10656). In these protocols, one uses an open map dissipative dynamic, and engineers the interaction of the state with the environment in such a way that that the long-time stationary state of the system is the desired one.
At risk of going off-topic from quantum computing and into physics, I'll answer what I think is a relevant subquestion of this topic, and use it to inform the discussion of unitary gates in quantum computing.
The question here is: Why do we want unitarity in quantum gates?
The less specific answer is as above, it gives us 'reversibility', or as physicists often talk about it, a type of symmetry for the system. I'm taking a course in quantum mechanics right now, and the way unitary gates cropped up in that course was motivated by the desire to have physical transformations $\hat{U}$: that act as symmetries. This imposed two conditions on the transformation $\hat{U}$:
The transformations should act linearly on the state (this is what gives us a matrix representation). The transformations should preserve probability, or more specifically inner product. This means that if we define:
$$|\psi '\rangle = U |\psi\rangle, |\phi'\rangle = U |\phi\rangle$$
Preservation of inner product means that $\langle \phi | | \psi \rangle= \langle \phi' | | \psi'\rangle$. From this second specification, unitarity can be derived (for full details see Dr. van Raamsdonk's notes here).
So this answers the question of why operations that keep things "reversible" have to be unitary.
The question of why measurement itself is not unitary is more related to quantum computation. A measurement is a projection on to a basis; in essence, it must "answer" with one or more basis states as the state itself. It also leaves the state in a way that is consistent with the "answer" to the measurement, and
not consistent with the underlying probabilities that the state began with. So the operation satisfies specification 1. of our transformation $U$, but definitively does not satisfy specification 2. Not all matrices are created equal!
To round things back to quantum computation, the fact that measurements are destructive and projective (ie. we can only reconstruct the superposition through repeated measurements of identical states, and every measurement only gives us a 0/1 answer), is part of what makes the separation between quantum computing and regular computing subtle (and part of why it's difficult to pin that down). One might assume quantum computing is more powerful because of the mere size of the Hilbert space, with all those state superpositions available to us. But our ability to extract that information is heavily limited.
As far as I understand it this shows that for information storage purposes, a qubit is only as good as a regular bit, and no better. But we can be clever in quantum computation with the way that information is traded around, because of the underlying linear-algebraic structure.
There are several misconceptions here, most of them originate from exposure to only the
pure state formalism of quantum mechanics, so let's address them one by one:
All quantum operations must be unitary to allow reversibility, but what about measurement?
This is false. In general, the states of a quantum system are not just vectors in a Hilbert space $\mathcal{H}$ but density matrices $-$ unit-trace, positive semidefinite operators acting on the Hilbert space $\mathcal{H}$ i.e., $\rho: \mathcal{H} \rightarrow \mathcal{H}$, $Tr(\rho) = 1$, and $\rho \geq 0$ (Note that the pure state vectors are not vectors in the Hilbert space but
rays in a complex projective space; for a qubit this amounts to the Hilbert space being $\mathbb{C}P^1$ and not $\mathbb{C}^2$). Density matrices are used to describe a statistical ensemble of quantum states.
The density matrix is called
pure if $\rho^2 = \rho$ and mixed if $\rho^2 < \rho$. Once we are dealing with a pure state density matrix (that is, there's no statistical uncertainty involved), since $\rho^2 = \rho$, the density matrix is actually a projection operator and one can find a $|\psi\rangle \in \mathcal{H}$ such that $\rho = |\psi\rangle \langle\psi|$.
The most general quantum operation is a CP-map (completely positive map), i.e., $\Phi: L(\mathcal{H}) \rightarrow L(\mathcal{H})$ such that $$\Phi(\rho) = \sum_i K_i \rho K_i^\dagger; \sum_i K_i^\dagger K_i \leq \mathbb{I}$$ (if $\sum_i K_i^\dagger K_i = \mathbb{I}$ then these are called CPTP (completely positive and
trace-preserving) map or a quantum channel) where the $\{K_i\}$ are called Kraus operators.
Now, coming to the OP's claim that all quantum operations are unitary to allow reversibility -- this is just not true. The unitarity of time evolution operator ($e^{-iHt/\hbar}$) in quantum mechanics (for closed system quantum evolution) is simply a consequence of the Schrödinger equation.
However, when we consider density matrices, the most general evolution is a CP-map (or CPTP for a closed system to preserve the trace and hence the probability).
Are there any situations where non-unitary gates might be allowed?
Yes. An important example that comes to mind is open quantum systems where Kraus operators (which are not unitary) are the "gates" with which the system evolves.
Note that if there is only a single Kraus operator then, $\sum_i K_i^\dagger K_i = \mathbb{I}$. But there's only one $i$, therefore, we have, $K^\dagger K = \mathbb{I}$ or, $K$ is unitary. So the system evolves as $\rho \rightarrow U \rho U^\dagger$ (which is the standard evolution that you may have seen before). However, in general, there are several Kraus operators and therefore the evolution is non-unitary.
Coming to the final point:
Measurement can be represented as a matrix, and that matrix is applied to qubits, so that seems equivalent to the operation of a quantum gate. That's definitively not reversible.
In standard quantum mechanics (with wavefunctions etc.), the system's evolution is composed of two parts $-$ a
smooth unitary evolution under the system's Hamiltonian and then a sudden quantum jump when a measurement is made $-$ also known as wavefunction collapse. Wavefunction collapses are described as some projection operator say $|\phi\rangle \langle\phi|$ acting on the quantum state $|\psi\rangle$ and the $|\langle\phi|\psi\rangle|^2$ gives us the probability of finding the system in the state $|\phi\rangle$ after the measurement. Since the measurement operator is after all a projector (or as the OP suggests, a matrix), shouldn't it be linear and physically similar to the unitary evolution (also happening via a matrix). This is an interesting question and in my opinion, difficult to answer physically. However, I can shed some light on this mathematically.
If we are working in the modern formalism, then measurements are given by POVM elements; Hermitian positive semidefinite operators, $\{M_{i}\}$ on a Hilbert space $\mathcal{H}$ that sum to the identity operator (on the Hilbert space) $\sum _{{i=1}}^{n}M_{i}=\mathbb{I}$. Therefore, a measurement takes the form $$ \rho \rightarrow \frac{E_i \rho E_i^\dagger}{\text{Tr}(E_i \rho E_i^\dagger)}, \text{ where } M_i = E_i^\dagger E_i.$$
The $\text{Tr}(E_i \rho E_i^\dagger) =: p_i$ is the probability of the measurement outcome being $M_i$ and is used to
renormalize the state to unit trace. Note that the numerator, $\rho \rightarrow E_i \rho E_i^\dagger$ is a linear operation, but the probabilistic dependence on $p_i$ is what brings in the non-linearity or irreversibility.
Edit 1: You might also be interested Stinespring dilation theorem which gives you an isomorphism between a CPTP map and a unitary operation on a larger Hilbert space followed by partial tracing the (tensored) Hilbert space (see 1, 2).
I'll add a small bit complementing the other answers, just about the idea of measurement.
Measurement is usually taken as a postulate of quantum mechanics. There's usually some preceding postulates about hilbert spaces, but following that
Every measurable physical quantity $A$ is described by an operator $\hat{A}$ acting on a Hilbert space $\mathcal{H}$. This operator is called an observable, and it's eigenvalues are the possibly outcomes of a measurement. If a measurement is made of the observable $A$, in the state of the system $\psi$, and the outcome is $a_n$, then the state of the system immediately after measurement is $$\frac{\hat{P}_n|\psi\rangle}{\|\hat{P}_n|\psi\rangle\|},$$ where $\hat{P}_n$ is the projector onto the eigen-subspace of the eigenvalue $a_n$.
Normally the projection operators themselves should satisfy $\hat{P}^\dagger=\hat{P}$ and $\hat{P}^2=\hat{P}$, which means they themselves are observables by the above postulates, and their eigenvalues $1$ or $0$. Supposing we take one of the $\hat{P}_n$ above, we can interpret the $1,0$ eigenvalues as a binary yes/no answer to whether the observable quantity $a_n$ is available as an outcome of measurement of the state $|\psi\rangle$.
Measurements are unitary operations, too, you just don't see it: A measurement is equivalent to some complicated (quantum) operation that acts not just on the system but also on its environment. If one were to model everything as a quantum system (including the environment), one would have unitary operations all the way.
However, usually there is little point in this because we usually don't know the exact action on the environment and typically don't care. If we consider only the system, then the result is the well-known collapse of the wave function, which is indeed a non-unitary operation.
Quantum states can change in two ways: 1.
quantumly, 2. classically.
All the state changes taking place
quantumly, are unitary. All the quantum gates, quantum errors, etc., are quantum changes.
There is no obligation on
classicalchanges to be unitary, e.g. measurement is a classical change.
All the more reason, why it is said that the quantum state is 'disturbed' once it's measured. |
Let $K$ be a compact set in a metric space $X$. Show that, if $K\subset U_1 \cup U_2$ for some open set $U_1$ and $U_2$, then there exist compact set $K_1$ and $K_2$ such that $K_1\subset U_1$, $K_2\subset U_2$ and $K=K_1\cup K_2.$
I think that, in general, it's not possible to find disjoint compact sets $K_1$ and $K_2$, and I think this is a counterexample: Question about compact set contained in an union of open sets.
I define $K_1$ and $K_2$ as follow.
$$K_1=K\cap ((U_1-\partial(U_2))\cup \partial(U_1))$$
$$K_2=K\cap ((U_1-\partial(U_1))\cup \partial(U_2))$$
By $\partial(U_1)$ I mean the boundary of $U_1$. To show $K_1$ is compact, I will show it's sequentially compact.
Let $(x_n)\in K\cap (U_1-\partial(U_2)$, as $K $ is compact, $(x_n)$ has a convergent sequence $(x_{n_k})\to \bar{x}\in{K} $.
If $\bar{x}\notin U_1$, then it's in $\partial({U_1})$. (it's a limit point of $U_1)$.
If $(x_n)$ be seqeuence in $K\cap\partial(U_1) \subset K$, as $ K\cap\partial(U_1) $ and it's close set of compact set $K$, it's compact. Hence $K_1$ is compact.
Is my argument valid? Thank you. |
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Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV
(Springer, 2015-05-20)
The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ...
Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV
(Springer, 2015-06)
We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ...
Multiplicity dependence of two-particle azimuthal correlations in pp collisions at the LHC
(Springer, 2013-09)
We present the measurements of particle pair yields per trigger particle obtained from di-hadron azimuthal correlations in pp collisions at $\sqrt{s}$=0.9, 2.76, and 7 TeV recorded with the ALICE detector. The yields are ...
Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV
(Springer, 2015-09)
Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ...
Coherent $\rho^0$ photoproduction in ultra-peripheral Pb-Pb collisions at $\mathbf{\sqrt{\textit{s}_{\rm NN}}} = 2.76$ TeV
(Springer, 2015-09)
We report the first measurement at the LHC of coherent photoproduction of $\rho^0$ mesons in ultra-peripheral Pb-Pb collisions. The invariant mass and transverse momentum distributions for $\rho^0$ production are studied ...
Inclusive, prompt and non-prompt J/ψ production at mid-rapidity in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2015-07-10)
The transverse momentum (p T) dependence of the nuclear modification factor R AA and the centrality dependence of the average transverse momentum 〈p T〉 for inclusive J/ψ have been measured with ALICE for Pb-Pb collisions ... |
Nevernevergiveup wrote:
What is the probability that you get a pair when picking the top two cards off of a randomly shuffled deck of cards (52 total cards, with 13 sets of four matching suits)?
(A) \(\frac{12}{2,652}\) (B) \(\frac{16}{2,652}\) (C) \(\frac{1}{17}\) (D) \(\frac{1}{13}\) (E) \(\frac{1}{2}\)
\(?\,\, = \,\,P\left( {{\rm{get}}\,\,{\rm{a}}\,\,{\rm{pair}}} \right)\)
\(\left. \matrix{
{\rm{Total}}\,\,:\,\,\,C\left( {52,2} \right) = {{52 \cdot 51} \over 2} = 26 \cdot 51\,\,\,{\rm{equiprobable}}\,\,{\rm{choices}}{\kern 1pt} \,\, \hfill \cr
{\rm{Favorable}}\,\,:\,\,\,\underbrace {\,C\left( {4,2} \right)}_{2\,\,{\rm{suits}}} \cdot \underbrace {C\left( {13,1} \right)}_{{\rm{card}}\,{\rm{to}}\,{\rm{get}}\,{\rm{twice}}} = 6 \cdot 13\,\,{\rm{choices}} \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\, = \,\,{{6 \cdot 13} \over {26 \cdot 51}}\,\, = \,\,{1 \over {17}}\)
The correct answer is (C).
We follow the notations and rationale taught in the GMATH
method.
Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH
method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net |
Recently I created a simulation of a GBM. The time between the prices were sampled from an exponential distribution. The log rate of return was sampled from $\sigma \sqrt { { t }_{ i }-{ t }_{ i-1 } } { Z }_{ i }+(\mu -\frac { { \sigma }^{ 2 } }{ 2 } )({ t }_{ i }-{ t }_{ i-1 })$ where ${ Z }_{ i }\sim N(0,1)$.
When I estimated the drift of prices from the simulation without using the time between the prices, I was able to get back the drift I programmed into the simulation.
So I concluded that the time between prices doesn't affect the estimation of parameters of a GBM.
Is this conclusion correct? |
Under the auspices of the Computational Complexity Foundation (CCF)
We study the complexity of arithmetic in finite fields of characteristic two, $\F_{2^n}$.
We concentrate on the following two problems:
Iterated Multiplication: Given $\alpha_1, \alpha_2,..., \alpha_t \in \F_{2^n}$, compute $\alpha_1 \cdot \alpha_2 \cdots \alpha_t \in \F_{2^n}$.
Exponentiation: Given $\alpha \in \F_{2^n}$ and a $t$-bit integer $k$, compute $\alpha^k \in \F_{2^n}$.... more >>> |
I am currently working through Velleman's book
How To Prove It and was asked to prove the following
$(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$
This is my work thus far
$(P \to Q) \wedge (Q \to P)$
$(\neg P \vee Q) \wedge (\neg Q \vee P)$ (since $(P \to Q) \equiv (\neg P \vee Q)$)
$\neg[\neg(\neg P \vee Q) \vee \neg (\neg Q \vee P)]$ (Demorgan's Law)
$\neg [(P \wedge \neg Q) \vee (Q \wedge \neg P)]$ (Demorgan's Law)
At this point I am little unsure how to proceed.
Here are a few things I've tried or considered thus far:
I thought that I could perhaps switch some of the terms in step 3 using the law of associativity however the $\neg$ on the outside of the two terms prevents me from doing so (constructing a truth table for $\neg (\neg P \vee Q) \vee (\neg Q \vee P)$ and $\neg (\neg P \vee \neg Q) \vee \neg (P \vee Q)$ for sanity purposes)
I can't seem to apply the law of distribution since the corresponding terms are negated.
Applying demorgans law to one of the terms individually on step 2 or 3 doesnt seem to get me very far either.
Did I perhaps skip something? Am I even on the right track? Any help is appreciated |
Under the auspices of the Computational Complexity Foundation (CCF)
One of the major challenges of the research in circuit complexity is proving super-polynomial lower bounds for de-Morgan formulas. Karchmer, Raz, and Wigderson suggested to approach this problem by proving that formula complexity behaves "as expected'' with respect to the composition of functions $f\circ g$. They showed that this conjecture, ... more >>>
The composition of two Boolean functions $f:\left\{0,1\right\}^{m}\to\left\{0,1\right\}$, $g:\left\{0,1\right\}^{n}\to\left\{0,1\right\}$
is the function $f \diamond g$ that takes as inputs $m$ strings $x_{1},\ldots,x_{m}\in\left\{0,1\right\}^{n}$ and computes \[ (f \diamond g)(x_{1},\ldots,x_{m})=f\left(g(x_{1}),\ldots,g(x_{m})\right). \] This operation has been used several times for amplifying different hardness measures of $f$ and $g$. This comes at a cost: the ... more >>>
The direct-sum question is a classical question that asks whether
performing a task on $m$ independent inputs is $m$ times harder than performing it on a single input. In order to study this question, Beimel et. al (Computational Complexity 23(1), 2014) introduced the following related problems:
* The choice ... more >>>
One of the major open problems in complexity theory is proving super-logarithmic
lower bounds on the depth of circuits (i.e., $\mathbf{P}\not\subseteq\mathbf{NC}^1$). Karchmer, Raz, and Wigderson (Computational Complexity 5, 3/4) suggested to approach this problem by proving that depth complexity behaves "as expected" with respect to the composition of functions $f ... more >>> |
Suppose we have an X-ray tube
R which emits X-rays of intensity $I_s(\lambda)$, obeying the Kramers' law. The X-rays then come to the cubic crystal K under the angle $\vartheta$, where diffraction occurs. The first Bragg peak will be at the angle $2\vartheta$, that's where we place our detector GM. The detector measures intensity $I_r(\vartheta)$..
Now, according to the Bragg condition, the detected intensity will be proportional to the intensity of the source: $$ I_r(\vartheta) \propto I_s(\lambda = 2d \sin \vartheta). $$
But the measured intensity must also decrease with larger $\vartheta$: when a ray comes almost parallel to the crystal, most of it will reflect, but when it comes almost perpendicular, most will get absorbed. I imagine this works similarly to the Fresnel equations in optics, but I wasn't able to find any source that would discuss something like that for the diffraction on a crystal.
So my question is, if $$ I_r(\vartheta) = R(\vartheta) \; I_s(\lambda = 2d \sin \vartheta), $$ then what is the equation for $R$?
EDIT: I found that intensity should be proportional to the absolute value of the structure factor squared:$$ I \propto |F_{hkl}|^2$$Where do find its value for a specific crystal (eg. LiF) and what do the indices $h, k, l$ mean? How do I calculate the intensity for a specific angle? |
There are several other answers here, but, IMO, none of them actually answer the real underlying question:
How can I easily determine when it is mathematically correct for my character to use -5 to hit for +10 damage?
$$Maximum\,AC=\left\lfloor\frac{\left(2\,\times\,Attack\,Bonus\right)\,-Average\,Damage\,+\,32}{2}\right\rfloor$$
(The \$\lfloor\,\,\rfloor\$ indicate the mathematical floor operation or, here, rounding down.)
So, if you have a +8 attack bonus and deal 1d12 (i.e., 6.5) damage, your result is:
$$\left\lfloor\frac{\left(2\times8\right)-6.5+32}{2}\right\rfloor=20$$
So, with a +8 to attack, deal 1d12 damage on a hit, and the target's AC is 20 or lower, then it is mathematically correct to use -5/+10.
Let's say you have a level 9 Fighter with 20 Str, Great Weapon Fighting, Great Weapon Master, and a +1 greatsword. Your attack bonus is \$4+5+1=10\$. Your damage on a hit is \$2d6\,(8.33)+6=14.33\$. (The average of 2d6 here is 8.33 and not 7 due to Great Weapon Fighting.)
$$\left\lfloor\frac{\left(2\times10\right)-14.33+32}{2}\right\rfloor=18$$
So, mathematically, you should use -5/+10 on any target with an AC of 18 or less.
The above is assuming that you do not have advantage or disadvantage. I will cover that below, but the reality is that using -5/+10 is essentially
always correct when you have advantage, and conversely it is essentially always incorrect when you have disadvantage.
Yes, you do have an outliers such as when you require a natural 20 to hit the target, or when you've got Dex 10 and a Blowgun with the Sharpshooter feat, or sometimes when you have a very large number of sneak attack dice, but none of those cases very common.
Further, critical hits also have no bearing on the calculation at all because critical hits add the same amount of damage to both types of attacks and occur at the same rate on both attacks. A natural 20 always hits, and essentially every high level Champion is already always going to hit any target's AC on a natural 18 or better even with the -5 no matter how your DM interprets a critical hit.
Let's see how we arrive at that formula.
I think it's easy to see that answer will depend on three factors:
Your attack bonus Your average or expected damage on a hit The target's AC
The first two you can know pretty easily before the game even begins. The target's AC, however, is a value that varies for every combatant. Therefore, it will be most useful to determine what AC is the most effective.
So, what we want to know is:
$$Expected\,damage\,from\,normal\,attack<Expected\,damage\,from\,-5/+10\,attack$$
Expected damage from a normal attack is, in most cases, best understood as the mean average damage on a hit multiplied by the chance to hit.
$$Expected\,damage\,from\,normal\,attack=Average\,damage\times\frac{21+Attack\,Bonus-Target\,AC}{20}$$
Expected damage from a -5/+10 attack is the same, but we need to write it using the same terms as above. So, we get:
$$Expected\,damage\,from\,-5/+10\,attack$$
$$=(Average\,damage + 10)\times\frac{21+Attack\,Bonus-5-Target\,AC}{20}$$
$$=(Average\,damage + 10)\times\frac{16+Attack\,Bonus-Target\,AC}{20}$$
So, that gives us this inequality:
$$Average\,damage\times\frac{21+Attack\,Bonus-Target\,AC}{20}$$
$$<$$
$$(Average\,damage + 10)\times\frac{16+Attack\,Bonus-Target\,AC}{20}$$
Now we just need to solve for Target AC. However, I'm lazy, so I made Wolfram Alpha do it. I used \$a\$ for the average damage, \$b\$ for the attack bonus, and \$x\$ for the target AC. I get the solution:
$$x<\frac{1}{2} (-a + 2 b + 32)$$
Which is the same as:
$$Target\,AC<\frac{\left(2\,\times\,Attack\,Bonus\right)\,-Average\,Damage\,+\,32}{2}$$
When the above inequality is true, it is mathematically correct to use -5/+10 on your attack.
You can repeat the above method to determine functions for advantage and disadvantage by substituting the different equations for calculating to-hit into the above inequality. However, you quickly run into pretty monster equation solutions for both advantage and disadvantage due to square roots.
However, here's the solution for advantage:
$$\frac{-\sqrt{a^2+10a+1600}-a+2b-8}{2}<Target AC<\frac{\sqrt{a^2+10a+1600}-a+2b-8}{2}$$
Here's the solution for disadvantage:
$$Target\,AC<\frac{-a-\sqrt{a^2+10a}+2b+32}{2}$$$$Target\,AC>\frac{-a+\sqrt{a^2+10a}+2b+32}{2}$$
Again, where \$a\$ is the average damage and \$b\$ is the attack bonus.
Let's take the same level 9 Fighter as above with +10 attack bonus and 14.33 average damage on a hit.
Advantage:
$$-23.24 < Target\,AC < 20.91$$
Disadvantage:
$$Target\,AC < 9.50$$$$Target\,AC > 28.17$$
And you can always test a given AC by calculating the damage per attack for that specific AC.
Note that with disadvantage you're getting results that are basically
off the die, since you can't roll a 9 on d20+10.
Let's take an extreme example to test the rule of thumb. Let's take a level 19 Rogue/Level 1 Fighter with Sharpshooter, Crossbow Expert, Dex 20, a +3 Hand Crossbow, Bracers of Archery (+2 damage), and Archery Weapon Style (+2 to hit). We've got \$6 + 5 + 3 + 2 = 16\$ to hit. We do \$1d6 (3.5) + 5 + 3 + 2 + 10d6 (35) = 48.5\$ average damage.
Normal:
$$Target\,AC < 7.75$$
Unsurprisingly, the -5 from Sharpshooter it too much here. The +10 damage doesn't pay off when we're dealing with nearly 50 damage on average.
Advantage:
$$-45.56 < Target\,AC < 21.06$$
It's still preferred for essentially all targets here (the number of enemies with AC greater than 21 can be counted on one hand, I believe).
Disadvantage:
$$Target\,AC < -18.89$$$$Target\,AC > 34$$
And, unsurprisingly, we're off the die here. |
also, if you are in the US, the next time anything important publishing-related comes up, you can let your representatives know that you care about this and that you think the existing situation is appalling
@heather well, there's a spectrum
so, there's things like New Journal of Physics and Physical Review X
which are the open-access branch of existing academic-society publishers
As far as the intensity of a single-photon goes, the relevant quantity is calculated as usual from the energy density as $I=uc$, where $c$ is the speed of light, and the energy density$$u=\frac{\hbar\omega}{V}$$is given by the photon energy $\hbar \omega$ (normally no bigger than a few eV) di...
Minor terminology question. A physical state corresponds to an element of a projective Hilbert space: an equivalence class of vectors in a Hilbert space that differ by a constant multiple - in other words, a one-dimensional subspace of the Hilbert space. Wouldn't it be more natural to refer to these as "lines" in Hilbert space rather than "rays"? After all, gauging the global $U(1)$ symmetry results in the complex line bundle (not "ray bundle") of QED, and a projective space is often loosely referred to as "the set of lines [not rays] through the origin." — tparker3 mins ago
> A representative of RELX Group, the official name of Elsevier since 2015, told me that it and other publishers “serve the research community by doing things that they need that they either cannot, or do not do on their own, and charge a fair price for that service”
for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing) @EmilioPisanty
> for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing)
@0celo7 but the bosses are forced because they must continue purchasing journals to keep up the copyright, and they want their employees to publish in journals they own, and journals that are considered high-impact factor, which is a term basically created by the journals.
@BalarkaSen I think one can cheat a little. I'm trying to solve $\Delta u=f$. In coordinates that's $$\frac{1}{\sqrt g}\partial_i(g^{ij}\partial_j u)=f.$$ Buuuuut if I write that as $$\partial_i(g^{ij}\partial_j u)=\sqrt g f,$$ I think it can work...
@BalarkaSen Plan: 1. Use functional analytic techniques on global Sobolev spaces to get a weak solution. 2. Make sure the weak solution satisfies weak boundary conditions. 3. Cut up the function into local pieces that lie in local Sobolev spaces. 4. Make sure this cutting gives nice boundary conditions. 5. Show that the local Sobolev spaces can be taken to be Euclidean ones. 6. Apply Euclidean regularity theory. 7. Patch together solutions while maintaining the boundary conditions.
Alternative Plan: 1. Read Vol 1 of Hormander. 2. Read Vol 2 of Hormander. 3. Read Vol 3 of Hormander. 4. Read the classic papers by Atiyah, Grubb, and Seeley.
I am mostly joking. I don't actually believe in revolution as a plan of making the power dynamic between the various classes and economies better; I think of it as a want of a historical change. Personally I'm mostly opposed to the idea.
@EmilioPisanty I have absolutely no idea where the name comes from, and "Killing" doesn't mean anything in modern German, so really, no idea. Googling its etymology is impossible, all I get are "killing in the name", "Kill Bill" and similar English results...
Wilhelm Karl Joseph Killing (10 May 1847 – 11 February 1923) was a German mathematician who made important contributions to the theories of Lie algebras, Lie groups, and non-Euclidean geometry.Killing studied at the University of Münster and later wrote his dissertation under Karl Weierstrass and Ernst Kummer at Berlin in 1872. He taught in gymnasia (secondary schools) from 1868 to 1872. He became a professor at the seminary college Collegium Hosianum in Braunsberg (now Braniewo). He took holy orders in order to take his teaching position. He became rector of the college and chair of the town...
@EmilioPisanty Apparently, it's an evolution of ~ "Focko-ing(en)", where Focko was the name of the guy who founded the city, and -ing(en) is a common suffix for places. Which...explains nothing, I admit. |
I have solved the Killing vector equations for a 2-sphere and got the following answer. $A,B,C$ are three integration constants as expected.
$$\xi_{\theta}=A \sin{\phi}+B\cos{\phi}$$ $$\xi_{\phi}=\cos{\theta} \sin{\theta}(A \cos{\phi}-B\sin\phi)+c \sin^2{\theta}$$
From this, how do I find the basis elements in terms of the tangent vectors $\frac{\partial}{\partial \phi}$ and $\frac{\partial}{\partial \theta}$? One obvious elements if $\frac{\partial}{\partial \phi}$ itself, as the metric is independent of $\phi$. Another method IMO could be to derive it from the angular momentum generators $x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}$ etc. Is this correct? Would changing the coordinates of these generators give the right answer?
Lastly, How do I do this for any general metric? Is there a standard procedure to find the killing vector fields, as a basis. |
I am trying to derive a relation between the angle from the center of ellipse $C$ and its true anomaly (angle from the focal point $F$) (
alpha vs.
beta in the picture), for a general ellipse. For some reason, it seems I am doing some mistake somewhere.
What I tried is this. When we take the polar description with regard to the center
$y = r\sin(\alpha) = \frac{ab}{\sqrt{(b\cos\alpha)^2+(a\sin\alpha)^2}}\sin\alpha$
Also for the other equation from the focus point we have $y=r\sin(\beta)$, were $r$ is the length with respect to the focus point, i.e.,
$r=\frac{a(1-e^2)}{1\pm e\cos\beta}.$
By putting the equations together and squaring, I get a monster equation:
$\frac{a^2\sin^2\alpha}{(b\cos\alpha)^2+(a\sin\alpha)^2} = \frac{b^2\sin^2\beta}{1\pm2e\cos\beta+e^2\cos^2\beta}$
This leads to a horrible quadratic equation in $\sin\alpha$. Am I doing some error somewhere. Is there some other option, some "shortcut" to derive it in an easier way?
Any hints are welcome.
EDIT:
After fighting with the equations, I got this result:
$\sin\alpha = \pm\frac{b}{a}q\sin\beta\sqrt{\frac{1}{1-e^2q^2\sin^2\beta}}, \to \alpha = \arcsin\left(\pm\frac{b}{a}q\sin\beta\sqrt{\frac{1}{1-e^2q^2\sin^2\beta}}\right)$
where $q=\frac{b}{a}\frac{1}{1\pm e\cos\beta}$ and $e=\sqrt{1-\left(\frac{b}{a}\right)^2}$.
Doing numerical testing, it looks like it is a correct result, except for angles larger than $90^°$, where some sign error seems to be. If there is a simpler and nicer solution, which works for any angle $\beta$, I would be grateful to see it. |
I am implementing a learner to learn a DAG model $G=\langle V,E\rangle$ where $V$ and $E$ represent both variables and dependencies respectively ( similar to Bayesian networks). Each variable $v\in V$ is associated with possible values (domain) $D_v$ and a function $\theta_v$.
I assume the learner knows the number of variables $n$ (i.e. vertices) and their possible values (domain). The goal is to find $E$ (dependencies) and $\theta_v$ for every variable $v$. I also assume there is a target function $c$ exist (the realizable case).
The hypothesis class is defined as the set of consistent hypotheses i.e. $H=\{h|\ h(x)=c(x)\} \forall x\in S$ where $S$ is set of examples seen so far.
I am trying to find a way to represent my hypotheses class. Initially, before receiving any example, it contains:
all possible DAGs over $V$. all different combinations of $E$ as long as its acyclic. For every Graph $G$ generated from (1) and (2),all different function values over $G$.
Beside my naive representation, I do not know how to compute (1) precisely. To put my question in another way, how to represent the hypotheses class over Bayesian networks? |
I think the following exercise is to "warm up", but nevertheless it's quite difficult for me:
Let $k \in \mathbb{N}$ and let $L \in \Sigma_k$. Show that also $L^{*} \in \Sigma_k$.
The following details from my lecture notes seem to be useful:
Notation. Let $n \in \mathbb{N}$.
We write $\exists_n y. \varphi(y)$ for $\exists y \in \Sigma^{*}.|y| \le n \wedge \varphi(y)$.
We write $\forall_n y. \varphi(y)$ for $\forall y \in \Sigma^{*}.|y| \le n \Rightarrow \varphi(y)$.
Theorem.
$L \in \Sigma^P_i \Leftrightarrow$ there is a language $A \in P$ and a polynomial $p$ so that: $x \in L \Leftrightarrow \exists_{p(|x|)}y_1 \forall_{p(|x|)}y_2 \exists_{p(|x|)}y_3 .../\forall_{p(|x|)}y_i (x,y_1,y_2,...,y_i) \in A$
Unfortunately I don't see the solution of the "puzzle". Can somebody please help me a little bit (despite the fact that it's weekend)? |
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