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The K-Topology
Recall from the Topological Spaces page that a set $X$ an a collection $\tau$ of subsets of $X$ together denoted $(X, \tau)$ is called a topological space if:
$\emptyset \in \tau$ and $X \in \tau$, i.e., the empty set and the whole set are contained in $\tau$. If $U_i \in \tau$ for all $i \in I$ where $I$ is some index set then $\displaystyle{\bigcup_{i \in I} U_i \in \tau}$, i.e., for any arbitrary collection of subsets from $\tau$, their union is contained in $\tau$. If $U_1, U_2, ..., U_n \in \tau$ then $\displaystyle{\bigcap_{i=1}^{n} U_i \in \tau}$, i.e., for any finite collection of subsets from $\tau$, their intersection is contained in $\tau$.
We will now look at an important topology known as the $K$-topology.
Definition: Consider the set of real numbers $\mathbb{R}$ and let $K = \left \{ \frac{1}{n} : n \in \mathbb{N} \right \}$. The $K$-Topology of $\mathbb{R}$ is the collection of subsets $\tau$ such that: 1) $\emptyset \in \tau$. 2) All intervals of the form $(a, b)$ where $a < b$ are in $\tau$. 3) All sets of the form $(a, b) \setminus K$ are in $\tau$. 4) All unions of the sets from (1) - (3).
It's rather difficult to verify the $K$-topology at this point and time, so we will only state it to be a topology and verify it later. |
Characteristic Subgroups of a Group
Characteristic Subgroups of a Group
Definition: Let $G$ be a group. A subgroup $H$ of $G$ is said to be a Characteristic Subgroup of $G$ if for all automorphisms $\phi \in \mathrm{Aut}(G)$ we have that $\phi(H) \subseteq H$.
In other words, a subgroup $H$ of a group $G$ is a characteristic subgroup if every automorphism $\phi$ of $G$ maps $H$ to $H$.
Proposition 1: Let $G$ be a group and let $H$ be a subgroup of $G$. If $H$ is a characteristic subgroup of $G$ then $H$ is a normal subgroup of $G$. (1) Proof: Consider the subset $\mathrm{Inn}(G)$ of $\mathrm{Aut}(G)$, i.e., $\mathrm{Inn}(G)$ contains all inner automorphisms of $G$, i.e., all automorphisms of the form $i_a : G \to G$ where $a \in G$ and $i_a$ is defined for all $g \in G$ by:
\begin{align} \quad i_a(g) =aga^{-1} \end{align}
Since $H$ is a characteristic subgroup of $G$ we have that for all $\phi \in \mathrm{Aut}(G)$ that $\phi(H) \subseteq H$. In particular, for all $a \in G$ we have that $i_a(H) \subseteq H$, i.e., $aha^{-1} \in H$ for all $h \in H$ and for all $a \in G$. So $aHa^{-1} \subseteq H$ for all $a \in G$. So $H$ is a normal subgroup of $G$. $\blacksquare$
Proposition 2: Let $G$ be a group. Then $Z(G)$ is a characteristic subgroup of $G$.
Recall that if $G$ is a group then $Z(G)$ denotes the center of $G$, i.e., the set of all elements in $G$ that commute with every element of $G$. (2) Proof: Let $x \in Z(G)$. Then $xg = gx$ for all $g \in G$. Then for all $g \in G$ and for all $\phi \in \mathrm{Aut}(G)$ we have that:
\begin{align} \quad \phi(xg) &= \phi(gx) \\ \quad \phi(x)\phi(g) &= \phi(g)\phi(x) \end{align}
Since every $\phi \in \mathrm{Aut}(G)$ is a bijection, we equivalently have that $\phi(x)g = g \phi(x)$ for all $\phi \in \mathrm{Aut}(G)$. So $\phi(x) \in Z(G)$ for each $x \in Z(G)$ and for each $\phi \in \mathrm{Aut}(G)$. That is, $\phi(Z(G)) \subseteq Z(G)$ for all $\phi \in \mathrm{Aut}(G)$. So $Z(G)$ is a characteristic subgroup of $G$. $\blacksquare$
Proposition 3: Let $G$ be a group. Then $G'$ is a characteristic subgroup of $G$.
Recall that if $G$ is a group then $G'$ denotes the derived subgroup of $G$, i.e., the smallest subgroup of $G$ that contains all commutators of $G$. (3) Proof: Let $G'$ be the derived subgroup of $G$. Observe that:
\begin{align} \quad G' = \langle [g_1, g_2] : g_1, g_2 \in G \rangle = \langle g_1g_2g_1^{-1}g_2^{-1} : g_1, g_2 \in G \rangle \end{align}
(4) For all $g_1, g_2 \in G$ and for all $\phi \in \mathrm{Aut}(G)$ we have that:
\begin{align} \quad \phi([g_1, g_2]) &= \phi(g_1g_2g_1^{-1}g_2^{-1}) \\ &= \phi(g_1) \phi(g_2) \phi(g_1^{-1}) \phi(g_2^{-1}) \\ &= \phi(g_1) \phi(g_2) [\phi(g_1)]^{-1} [\phi(g_2)]^{-1} \in G' \end{align}
So for each $\phi \in \mathrm{Aut}(G)$, $\phi$ maps each generator of $G'$ to $G'$. So $\phi(G') \subseteq G'$. Since this holds for all $\phi \in \mathrm{Aut}(G)$ we conclude that $G'$ is a characteristic subgroup of $G$. $\blacksquare$ |
A well known representation of Dirac's delta-distribution is via the Fourier transform of distributions: \begin{equation} \delta[f]:=f(0)=\int_{\mathbb{R}}\int_{\mathbb{R}} e^{\mathrm{i}xk}f(k)\mathrm{d}k\mathrm{d}x. \end{equation} Can this be used to define the delta distribution composed with a function $\phi:\mathbb{R}\to\mathbb{R}$ via \begin{equation} (\delta\circ\phi)[f]:=\int_{\mathbb{R}}\int_{\mathbb{R}} e^{\mathrm{i}x\phi(k)}f(k)\mathrm{d}k\mathrm{d}x? \end{equation} Does this make sense? Heuristically and in a more physics-style notation, I would argue that \begin{equation} \int\int e^{\mathrm{i}x\phi(k)}f(k)\mathrm{d}k\mathrm{d}x "=" \int\left(\int e^{\mathrm{i}x\phi(k)}\mathrm{d}x\right)f(k)\mathrm{d}k "=" \int\delta(\phi(k))f(k)\mathrm{d}k. \end{equation} If it does not make sense, what can I say about the above expression for general functions $\phi$? E.g. I would expect the above integral to be positive if $f$ is point-wise positive.
What you want to do is the pull-back of distributions. And there is a theorem (cf. Hörmander 1, Theorem 8.2.4) that if the set $\{(\phi(x),\eta) \colon \phi'(x) \eta = 0\}$ and $\operatorname{WF}(\delta) = \{ (0,\eta) \colon \eta \not = 0\}$ have empty intersection, then the pull-back is well-defined.
If you are only interested in the delta-Distribution, then there is also Theorem 6.1.5 saying that for any smooth function $\phi : X \to \mathbb{R}$ with $|\phi'| \not = 0$ on $\phi = 0$, one has that $\delta^* \phi = \frac{dS}{|\phi'|}$, where $dS$ is the Euclidean surface measure on $\{\phi = 0\}$. Even though I mentioned no integrals, this has quite strong flavour of oscillatory integrals (cf. Shubin, Chapter 1) and FIOs (Hörmander 4) to it.
Literature:
L. Hörmander - The Analysis of Linear Partial Differential Operators 1-4
M. Shubin - Pseudodifferential Operators and Spectral Theory
If $\phi(k)$ vanishes at $k=k_n$, $n=1,2,\ldots$, and $\phi'(k_n)\neq 0$ for all $n$, then \begin{equation} \int\int e^{\mathrm{i}x\phi(k)}f(k)\,\mathrm{d}k\mathrm{d}x = 2 \pi \int\delta(\phi(k))f(k)\,\mathrm{d}k=2\pi\sum_{n}\frac{f(k_n)}{|\phi'(k_n)|}, \end{equation} so yes, the integral is positive for point-wise positive $f$. |
"Is this really a proof?" is the exact question e-mailed to me today from an undergraduate mathematics student whom I know as a highly competent student. The one sentence question was accompanied with the following demo: I am looking for a down-to-earth, non-authoritative answer who one may gi...
the 2005 AMS article/survey on experimental mathematics[1] by Bailey/Borwein mentions many remarkable successes in the field including new formulas for $\pi$ that were discovered via the PSLQ algorithm as well as many other examples. however, it appears to glaringly leave out any description of t...
I'm reading the paper "the classification of algebras by dominant dimension" by Bruno J.Mueller, the link is here http://cms.math.ca/10.4153/CJM-1968-037-9. In the proof of lemma 3 on page 402, there is a place I can't understand. Who can tell me what $E_R \oplus * \cong \oplus X_R$ and $_AHom_...
I've been really trying to prove Ramanujan Partition theory, and different sources give me different explanations. Can someone please explain how Ramanujan (and Euler) found out the following equation for the number of partitions for a given integer? Any help is appreciated thank you so much! $...
I was wondering what role non-rigorous, heuristic type arguments play in rigorous math. Are there examples of rigorous, formal proofs in which a non-rigorous reasoning still plays a central part? Here is an example of what I am thinking of. You want to prove that some formula $f(n)$ holds, and y...
Perhaps the "proofs" of ABC conjecture or newly released weak version of twin prime conjecture or alike readily come to your mind. These are not the proofs I am looking for. Indeed my question was inspired by some other posts seeking for a hint to understand a certain more or less well-establised...
I do not know exactly how to characterize the class of proofs that interests me, so let me give some examples and say why I would be interested in more. Perhaps what the examples have in common is that a powerful and unexpected technique is introduced that comes to seem very natural once you are ...
Some conjectures are disproved by a single counter-example and garner little or no further interest or study, such as (to my knowledge) Euler's conjecture in number theory that at least $n$ $n^{th}$ powers are required to sum to an $n^{th}$ power, for $n>2$ (disproved by counter-example by L. J. ...
There is a tag called proofs. This tag has empty tag-info. Without any usage guidance it is quite likely to be used incorrectly. The fact that there are many deleted questions having this tag can be considered a supporting evidence of this fact. (According to SEDE there are 26 such questions - ...
I would like to know how you would rigorously introduce the trigonometric functions ($\sin(x)$ and relatives) to first year calculus students. Suppose they have a reasonable definition of $\mathbb{R}$ (as Cauchy closure of the rationals, or as Dedekind cuts, or whatever), but otherwise require as...
After having a solid year long undergraduate course in abstract algebra, I'm interested in learning algebra at a more advanced level, especially in the context of category theory. I've done some research, and from what I've read, it seems that using Lang as a main text and Hungerford as a supple...
Dear MO-community, I am not sure how mature my view on this is and I might say some things that are controversial. I welcome contradicting views. In any case, I find it important to clarify this in my head and hope that this community can help me doing that. So after this longish introduction, h...
Usually, during lectures Turing Machines are firstly introduced from an informal point of view (for example, in this way: http://en.wikipedia.org/wiki/Turing_machine#Informal_description) and then their definition is formalized (for example, in this way: http://en.wikipedia.org/wiki/Turing_machin...
« first day (1811 days earlier) ← previous day next day → last day (434 days later) » |
@JosephWright Well, we still need table notes etc. But just being able to selectably switch off parts of the parsing one does not need... For example, if a user specifies format 2.4, does the parser even need to look for e syntax, or ()'s?
@daleif What I am doing to speed things up is to store the data in a dedicated format rather than a property list. The latter makes sense for units (open ended) but not so much for numbers (rigid format).
@JosephWright I want to know about either the bibliography environment or \DeclareFieldFormat. From the documentation I see no reason not to treat these commands as usual, though they seem to behave in a slightly different way than I anticipated it. I have an example here which globally sets a box, which is typeset outside of the bibliography environment afterwards. This doesn't seem to typeset anything. :-( So I'm confused about the inner workings of biblatex (even though the source seems....
well, the source seems to reinforce my thought that biblatex simply doesn't do anything fancy). Judging from the source the package just has a lot of options, and that's about the only reason for the large amount of lines in biblatex1.sty...
Consider the following MWE to be previewed in the build in PDF previewer in Firefox\documentclass[handout]{beamer}\usepackage{pgfpages}\pgfpagesuselayout{8 on 1}[a4paper,border shrink=4mm]\begin{document}\begin{frame}\[\bigcup_n \sum_n\]\[\underbrace{aaaaaa}_{bbb}\]\end{frame}\end{d...
@Paulo Finally there's a good synth/keyboard that knows what organ stops are! youtube.com/watch?v=jv9JLTMsOCE Now I only need to see if I stay here or move elsewhere. If I move, I'll buy this there almost for sure.
@JosephWright most likely that I'm for a full str module ... but I need a little more reading and backlog clearing first ... and have my last day at HP tomorrow so need to clean out a lot of stuff today .. and that does have a deadline now
@yo' that's not the issue. with the laptop I lose access to the company network and anythign I need from there during the next two months, such as email address of payroll etc etc needs to be 100% collected first
@yo' I'm sorry I explain too bad in english :) I mean, if the rule was use \tl_use:N to retrieve the content's of a token list (so it's not optional, which is actually seen in many places). And then we wouldn't have to \noexpand them in such contexts.
@JosephWright \foo:V \l_some_tl or \exp_args:NV \foo \l_some_tl isn't that confusing.
@Manuel As I say, you'd still have a difference between say \exp_after:wN \foo \dim_use:N \l_my_dim and \exp_after:wN \foo \tl_use:N \l_my_tl: only the first case would work
@Manuel I've wondered if one would use registers at all if you were starting today: with \numexpr, etc., you could do everything with macros and avoid any need for \<thing>_new:N (i.e. soft typing). There are then performance questions, termination issues and primitive cases to worry about, but I suspect in principle it's doable.
@Manuel Like I say, one can speculate for a long time on these things. @FrankMittelbach and @DavidCarlisle can I am sure tell you lots of other good/interesting ideas that have been explored/mentioned/imagined over time.
@Manuel The big issue for me is delivery: we have to make some decisions and go forward even if we therefore cut off interesting other things
@Manuel Perhaps I should knock up a set of data structures using just macros, for a bit of fun [and a set that are all protected :-)]
@JosephWright I'm just exploring things myself “for fun”. I don't mean as serious suggestions, and as you say you already thought of everything. It's just that I'm getting at those points myself so I ask for opinions :)
@Manuel I guess I'd favour (slightly) the current set up even if starting today as it's normally \exp_not:V that applies in an expansion context when using tl data. That would be true whether they are protected or not. Certainly there is no big technical reason either way in my mind: it's primarily historical (expl3 pre-dates LaTeX2e and so e-TeX!)
@JosephWright tex being a macro language means macros expand without being prefixed by \tl_use. \protected would affect expansion contexts but not use "in the wild" I don't see any way of having a macro that by default doesn't expand.
@JosephWright it has series of footnotes for different types of footnotey thing, quick eye over the code I think by default it has 10 of them but duplicates for minipages as latex footnotes do the mpfoot... ones don't need to be real inserts but it probably simplifies the code if they are. So that's 20 inserts and more if the user declares a new footnote series
@JosephWright I was thinking while writing the mail so not tried it yet that given that the new \newinsert takes from the float list I could define \reserveinserts to add that number of "classic" insert registers to the float list where later \newinsert will find them, would need a few checks but should only be a line or two of code.
@PauloCereda But what about the for loop from the command line? I guess that's more what I was asking about. Say that I wanted to call arara from inside of a for loop on the command line and pass the index of the for loop to arara as the jobname. Is there a way of doing that? |
Kodai Mathematical Journal Kodai Math. J. Volume 41, Number 3 (2018), 475-511. A new formula for the spherical growth series of an amalgamated free product of two infinite cyclic groups Abstract
We consider a group presented as $G(p,q) = \langle x, y|x^p = y^q\rangle$, with integers $p$ and $q$ satisfying $2 \leq p \leq q$. The group is an amalgamated free product of two infinite cyclic groups and is geometrically realized as the fundamental group of a Seifert fiber space over the 2-dimensional disk with two cone points whose associated cone angles are $\frac{2\pi}{p}$ and $\frac{2\pi}{q}$. We present a formula for the spherical growth series of the group $G(p,q)$ with respect to the generating set $\{x,y,x^{-1}, y^{-1}\}$, from which a rational function expression for the spherical growth series of $G(p,q)$ is derived concretely, once $p$ and $q$ are given. In fact, an elementary computer program constructed from the formula yields an explicit form of a single rational fraction expression for the spherical growth series of $G(p,q)$. Such expressions for several pairs $(p,q)$ appear in this paper. In 1999, C. P. Gill already provided a similar formula for the same group. The formula given here takes a different form from his formula, because the method we used here is independent of that introduced by him.
Article information Source Kodai Math. J., Volume 41, Number 3 (2018), 475-511. Dates First available in Project Euclid: 31 October 2018 Permanent link to this document https://projecteuclid.org/euclid.kmj/1540951250 Digital Object Identifier doi:10.2996/kmj/1540951250 Mathematical Reviews number (MathSciNet) MR3870700 Zentralblatt MATH identifier 07000580 Citation
Fujii, Michihiko. A new formula for the spherical growth series of an amalgamated free product of two infinite cyclic groups. Kodai Math. J. 41 (2018), no. 3, 475--511. doi:10.2996/kmj/1540951250. https://projecteuclid.org/euclid.kmj/1540951250 |
Table of Contents
Heredity of the Hausdorff Property on Topological Subspaces
Recall from the Hereditary Properties of Topological Spaces page that if $(X, \tau)$ is a topological space that a property of $X$ is said to be hereditary if for all subsets $A \subseteq X$ we have that the topological subspace $(A, \tau_A)$ also has that property (where $\tau_A$ is the subspace topology on $A$).
We will now show that second countability is hereditary.
Theorem 1: The Hausdorff property is hereditary, that is, if $(X, \tau)$ is a Hausdorff topological space and $A \subseteq X$ then $(A, \tau_A)$ is a Hausdorff topological space where $\tau_A = \{ A \cap U : U \in \tau \}$ is the subspace topology on $A$. Proof:Let $(X, \tau)$ be a Hausdorff topological space and let $x, y \in A$. Then $x, y \in X$ since $A \subseteq X$. So since $X$ is Hausdorff there exists open neighbourhoods $U$ of $x$ and $V$ of $Y$ such that: Since $U$ and $V$ contain $x$ and $y$, we have that $A \cap U$ and $A \cap V$ are open neighbourhoods of $x$ and $y$ in $A$. Moreover we see that: So for all $x, y \in A$ there exists open neighbourhoods $A \cap U$ and $A \cap V$ of $x$ and $y$ respectively such that $(A \cap U) \cap (A \cap V) = \emptyset$. So $(A, \tau_A)$ is Hausdorff. This shows that the Hausdorff property is hereditary. $\blacksquare$ |
@Secret et al hows this for a video game? OE Cake! fluid dynamics simulator! have been looking for something like this for yrs! just discovered it wanna try it out! anyone heard of it? anyone else wanna do some serious research on it? think it could be used to experiment with solitons=D
OE-Cake, OE-CAKE! or OE Cake is a 2D fluid physics sandbox which was used to demonstrate the Octave Engine fluid physics simulator created by Prometech Software Inc.. It was one of the first engines with the ability to realistically process water and other materials in real-time. In the program, which acts as a physics-based paint program, users can insert objects and see them interact under the laws of physics. It has advanced fluid simulation, and support for gases, rigid objects, elastic reactions, friction, weight, pressure, textured particles, copy-and-paste, transparency, foreground a...
@NeuroFuzzy awesome what have you done with it? how long have you been using it?
it definitely could support solitons easily (because all you really need is to have some time dependence and discretized diffusion, right?) but I don't know if it's possible in either OE-cake or that dust game
As far I recall, being a long term powder gamer myself, powder game does not really have a diffusion like algorithm written into it. The liquids in powder game are sort of dots that move back and forth and subjected to gravity
@Secret I mean more along the lines of the fluid dynamics in that kind of game
@Secret Like how in the dan-ball one air pressure looks continuous (I assume)
@Secret You really just need a timer for particle extinction, and something that effects adjacent cells. Like maybe a rule for a particle that says: particles of type A turn into type B after 10 steps, particles of type B turn into type A if they are adjacent to type A.
I would bet you get lots of cool reaction-diffusion-like patterns with that rule.
(Those that don't understand cricket, please ignore this context, I will get to the physics...)England are playing Pakistan at Lords and a decision has once again been overturned based on evidence from the 'snickometer'. (see over 1.4 ) It's always bothered me slightly that there seems to be a ...
Abstract: Analyzing the data from the last replace-the-homework-policy question was inconclusive. So back to the drawing board, or really back to this question: what do we really mean when we vote to close questions as homework-like?As some/many/most people are aware, we are in the midst of a...
Hi I am trying to understand the concept of dex and how to use it in calculations. The usual definition is that it is the order of magnitude, so $10^{0.1}$ is $0.1$ dex.I want to do a simple exercise of calculating the value of the RHS of Eqn 4 in this paper arxiv paper, the gammas are incompl...
@ACuriousMind Guten Tag! :-) Dark Sun has also a lot of frightening characters. For example, Borys, the 30th level dragon. Or different stages of the defiler/psionicist 20/20 -> dragon 30 transformation. It is only a tip, if you start to think on your next avatar :-)
What is the maximum distance for eavesdropping pure sound waves?And what kind of device i need to use for eavesdropping?Actually a microphone with a parabolic reflector or laser reflected listening devices available on the market but is there any other devices on the planet which should allow ...
and endless whiteboards get doodled with boxes, grids circled red markers and some scribbles
The documentary then showed one of the bird's eye view of the farmlands
(which pardon my sketchy drawing skills...)
Most of the farmland is tiled into grids
Here there are two distinct column and rows of tiled farmlands to the left and top of the main grid. They are the index arrays and they notate the range of inidex of the tensor array
In some tiles, there's a swirl of dirt mount, they represent components with nonzero curl
and in others grass grew
Two blue steel bars were visible laying across the grid, holding up a triangle pool of water
Next in an interview, they mentioned that experimentally the process is uite simple. The tall guy is seen using a large crowbar to pry away a screw that held a road sign under a skyway, i.e.
ocassionally, misshaps can happen, such as too much force applied and the sign snapped in the middle. The boys will then be forced to take the broken sign to the nearest roadworks workshop to mend it
At the end of the documentary, near a university lodge area
I walked towards the boys and expressed interest in joining their project. They then said that you will be spending quite a bit of time on the theoretical side and doddling on whitebaords. They also ask about my recent trip to London and Belgium. Dream ends
Reality check: I have been to London, but not Belgium
Idea extraction: The tensor array mentioned in the dream is a multiindex object where each component can be tensors of different order
Presumably one can formulate it (using an example of a 4th order tensor) as follows:
$$A^{\alpha}_{\beta}_{\gamma,\delta,\epsilon}$$
and then allow the index $\alpha,\beta$ to run from 0 to the size of the matrix representation of the whole array
while for the indices $\gamma,\delta,epsilon$ it can be taken from a subset which the $\alpha,\beta$ indices are. For example to encode a patch of nonzero curl vector field in this object, one might set $\gamma$ to be from the set $\{4,9\}$ and $\delta$ to be $\{2,3\}$
However even if taking indices to have certain values only, it is unsure if it is of any use since most tensor expressions have indices taken from a set of consecutive numbers rather than random integers
@DavidZ in the recent meta post about the homework policy there is the following statement:
> We want to make it sure because people want those questions closed. Evidence: people are closing them. If people are closing questions that have no valid reason for closure, we have bigger problems.
This is an interesting statement.
I wonder to what extent not having a homework close reason would simply force would-be close-voters to either edit the post, down-vote, or think more carefully whether there is another more specific reason for closure, e.g. "unclear what you're asking".
I'm not saying I think simply dropping the homework close reason and doing nothing else is a good idea.
I did suggest that previously in chat, and as I recall there were good objections (which are echoed in @ACuriousMind's meta answer's comments).
@DanielSank Mostly in a (probably vain) attempt to get @peterh to recognize that it's not a particularly helpful topic.
@peterh That said, he used to be fairly active on physicsoverflow, so if you really pine for the opportunity to communicate with him, you can go on ahead there. But seriously, bringing it up, particularly in that way, is not all that constructive.
@DanielSank No, the site mods could have caged him only in the PSE, and only for a year. That he got. After that his cage was extended to a 10 year long network-wide one, it couldn't be the result of the site mods. Only the CMs can do this, typically for network-wide bad deeds.
@EmilioPisanty Yes, but I had liked to talk to him here.
@DanielSank I am only curious, what he did. Maybe he attacked the whole network? Or he toke a site-level conflict to the IRL world? As I know, network-wide bans happen for such things.
@peterh That is pure fear-mongering. Unless you plan on going on extended campaigns to get yourself suspended, in which case I wish you speedy luck.
4
Seriously, suspensions are never handed out without warning, and you will not be ten-year-banned out of the blue. Ron had very clear choices and a very clear picture of the consequences of his choices, and he made his decision. There is nothing more to see here, and bringing it up again (and particularly in such a dewy-eyed manner) is far from helpful.
@EmilioPisanty Although it is already not about Ron Maimon, but I can't see here the meaning of "campaign" enough well-defined. And yes, it is a little bit of source of fear for me, that maybe my behavior can be also measured as if "I would campaign for my caging". |
Recall that $\binom n r$ is the number of ways of choosing $r$ objects out of $n$ objects and it can be shown that, $$^nC_r :=\binom n r=\dfrac {n!}{r!(n-r)!}$$
Firstly, prove that $$\binom n r + \binom n {r+1}=\binom {n+1} {r+1}$$
Now, what happens to your equation?
You have that $$\begin{align}\binom {17}{x+3}&=50\\\dfrac{17!}{(x+3)!(14-x)!}&=50\end{align}$$
Now we need to use number theoretic arguments to get the answer:
Note that as there is a factor of $25$ in RHS, there must be a factor of $25$ in LHS, which means the following:
$x+3<10 \implies x<7$ and $14-x<10 \implies x>4$. But, then that would mean that, $4<x<7$ and ass $\binom n r = \binom n {n-r}$ we'll have that $\binom {17} 9 = \binom {17} 8$, but none of which equal $50$ and hence no integral solutions.
However, there are solutions in real numbers using the generalized factorial (Gamma, $\Gamma$) functions! |
The spring 2019 Ritt Lectures, by
Professor Laure Saint Raymond, will take place on Wednesday, May 8 and Thursday, May 9, 2019 from 4:30 – 5:30pm in Rm 417. Professor Laure Saint Raymond (École Normale Supérieure de Lyon), will deliver a two talk series titled:
1. Disorder Increases Almost Surely
Consider a system of small hard spheres, which are initially (almost) independent and identically distributed. Then, in the low density limit, their empirical measure$\frac1N \sum_{i=1}^N \delta_{x_i(t), v_i(t)}$converges almost surely to a non reversible dynamics. Where is the missing information to go backwards?
2. The Structure of Correlations
Although the distribution of hard spheres remains essentially chaotic in this regime, collisions give birth to small correlations. The structure of these dynamical correlations is amazing, going through all scales. How can combinatorial techniques help analyze this departure from chaos?
Tea will be served at 4 pm in 508 Mathematics. Print this page |
Note: I’ve started putting together the material from these postings into a proper document, available here, which will be getting updated as time goes on. I’ll be making changes and additions to the text there, not on the blog postings. For most purposes, that will be what people interested in this subject will want to take a look at.
When a Lie group with Lie algebra [tex]\mathfrak g[/tex] acts on a manifold [tex]M[/tex], one gets two sorts of actions of [tex]\mathfrak g[/tex] on the differential forms [tex]\Omega^*(M[/tex]). For each [tex]X\in \mathfrak g [/tex] one has operators:
and
These operators satisfy the relation
[tex]di_X+i_Xd={\mathcal L}_X[/tex]
where [tex]d[/tex] is the de Rham differential [tex]d:\Omega^k(M)\rightarrow \Omega^{k+1}(M)[/tex], and the operators [tex]d, i_X, \mathcal L_X[/tex] are (super)-derivations. In general, an algebra carrying an action by operators satisfying the same relations satisfied by [tex]d, i_X, \mathcal L_X[/tex] will be called a [tex]\mathfrak g[/tex]-differential algebra. It will turn out that the Clifford algebra [tex]Cliff(\mathfrak g)[/tex] of a semi-simple Lie algebra [tex]\mathfrak g[/tex] carries not just the Clifford algebra structure, but the additional structure of a [tex]\mathfrak g[/tex]-differential algebra, in this case with [tex]\mathbf Z_2[/tex], not [tex]\mathbf Z[/tex] grading.
Note that in this section the commutator symbol will be the supercommutator in the Clifford algebra (commutator or anti-commutator, depending on the [tex]\mathbf Z_2[/tex] grading). When the Lie bracket is needed, it will be denoted [tex][\cdot,\cdot]_{\mathfrak g}[/tex].
To get a [tex]\mathfrak g[/tex]-differential algebra on [tex]Cliff(\mathfrak g)[/tex] we need to construct super-derivations [tex]i_X^{Cl}[/tex], [tex]{\mathcal L}_X^{Cl}[/tex], and [tex]d^{Cl}[/tex] satisfying the appropriate relations. For the first of these we don’t need the fact that this is the Clifford algebra of a Lie algebra, and can just define
[tex]i_X^{Cl}(\cdot)=[-\frac{1}{2}X,\cdot][/tex]
For [tex]{\mathcal L}_X^{Cl}[/tex], we need to use the fact that since the adjoint representation preserves the inner product, it gives a homomorphism
[tex]\widetilde{ad}:\mathfrak g \rightarrow \mathfrak{spin}(\mathfrak g)[/tex]
where [tex]\mathfrak{spin}(\mathfrak g)[/tex] is the Lie algebra of the group [tex]Spin(\mathfrak g)[/tex] (the spin group for the inner product space [tex]\mathfrak g[/tex]), which can be identified with quadratic elements of [tex]Cliff(\mathfrak g)[/tex], taking the commutator as Lie bracket. Explicitly, if [tex]X_a[/tex] is a basis of [tex]\mathfrak g[/tex], [tex]X_a^* [/tex] the dual basis, then
[tex]\widetilde{ad}(X)=\frac{1}{4}\sum_a X_a^*[X,X_a]_{\mathfrak g}[/tex]
and we get operators acting on [tex]Cliff(\mathfrak g)[/tex]
[tex]{\mathcal L}_X^{Cl}(\cdot)=[\widetilde{ad}(X),\cdot][/tex]
Remarkably, an appropriate [tex]d^{Cl}[/tex] can be constructed using a cubic element of [tex]Cliff(\mathfrak g)[/tex]. Let
[tex]\gamma= \frac{1}{24}\sum_{a,b}X^*_aX^*_b[X_a,X_b]_{\mathfrak g}[/tex]
then
[tex]d^{Cl}(\cdot)=[\gamma, \cdot][/tex]
[tex]d^{Cl}\circ d^{Cl}=0[/tex] since [tex]\gamma^2[/tex] is a scalar which can be computed to be [tex]-\frac{1}{48}tr\Omega_{\mathfrak g}[/tex], where [tex]\Omega_{\mathfrak g}[/tex] is the Casimir operator in the adjoint representation.
The above constructions give [tex]Cliff(\mathfrak g)[/tex] the structure of a filtered [tex]\mathfrak g[/tex]-differential algebra, with associated graded algebra [tex]\Lambda^*(\mathfrak g)[/tex]. This gives [tex]\Lambda^*(\mathfrak g)[/tex] the structure of a [tex]\mathfrak g[/tex]-differential algebra, with operators [tex]i_X, \mathcal L_X, d[/tex]. The cohomology of this differential algebra is just the Lie algebra cohomology [tex]H^*(\mathfrak g, \mathbf C)[/tex].
[tex]Cliff(\mathfrak g)[/tex] can be thought of as an algebra of operators corresponding to the quantization of an anti-commuting phase space [tex]\mathfrak g[/tex]. Classical observables are anti-commuting functions, elements of [tex]\Lambda^*(\mathfrak g^*)[/tex]. Corresponding to [tex]i_X, \mathcal L_X, d[/tex] one has both elements of [tex]\Lambda^*(\mathfrak g^*)[/tex] and their quantizations, the operators in [tex]Cliff(\mathfrak g)[/tex] constructed above.
For more details about the above, see the following references
Last Updated on |
In classification one usually computes $$ C = \operatorname*{argmax}_k p(C=k\mid X) $$ where $p(C=k\mid X)$ is the posterior distribution.
In a simple logistic regression setting with $C \in \{0, 1\}$ and $$ p(C=1\mid X)=\frac{\exp(\beta_0+\beta_1 x_i)}{1+\exp(\beta_0+\beta_1 x_i)} $$ and therefore $$ p(C=0\mid X)=\frac{1}{1+\exp(\beta_0+\beta_1 x_i)} $$ with $X=\{x_i\},\ i=1,\ldots,N$.
we estimate the parameters $\beta_0, \beta_1$ via the maximum likelihood estimation. To do so one has to compute the product of the likelihood function of all $N$ observations. So far, so normal. However, in all text books the authors plug in the posterior instead of the likelihood (e.g., Bishop, p. 206, Hastie, et al., p. 120): \begin{align} \ell(\beta) &= \log\left(\prod_{i=1}^N p(C_i=k\mid x_i, \beta)\right) \\[8pt] &= \log\left(\prod_{i=1}^N p(C_i=1\mid x_i, \beta)^{C_i}(1-p(C_i=1\mid x_i, \beta))^{1-C_i}\right) \end{align} And even though these probabilities are now conditioned on $\beta$ as well, they are still no likelihood to the posterior $p(C=k\mid X)$. So
How come we plug in the MLE just the posterior conditioned on the parameter $\beta$?
Why is $p(C=k\mid X)$ a posterior anyway? To me a posterior is a distribution of over a parameter given the observed data. But the class $C$ is to me not a parameter but a target just like the observations $y_i$ in a linear regression setting. |
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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 ... |
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Now showing items 1-10 of 27
Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider
(American Physical Society, 2016-02)
The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ...
Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(Elsevier, 2016-02)
Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < |\eta| < 4.0$) and associated particles in the central range ($|\eta| < 1.0$) are measured with the ALICE detector in ...
Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(Springer, 2016-08)
The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ...
Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2016-03)
The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ...
Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2016-03)
Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ...
Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV
(Elsevier, 2016-07)
The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ...
$^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2016-03)
The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ...
Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV
(Elsevier, 2016-09)
The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ...
Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV
(Elsevier, 2016-12)
We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ...
Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV
(Springer, 2016-05)
Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ... |
@Secret et al hows this for a video game? OE Cake! fluid dynamics simulator! have been looking for something like this for yrs! just discovered it wanna try it out! anyone heard of it? anyone else wanna do some serious research on it? think it could be used to experiment with solitons=D
OE-Cake, OE-CAKE! or OE Cake is a 2D fluid physics sandbox which was used to demonstrate the Octave Engine fluid physics simulator created by Prometech Software Inc.. It was one of the first engines with the ability to realistically process water and other materials in real-time. In the program, which acts as a physics-based paint program, users can insert objects and see them interact under the laws of physics. It has advanced fluid simulation, and support for gases, rigid objects, elastic reactions, friction, weight, pressure, textured particles, copy-and-paste, transparency, foreground a...
@NeuroFuzzy awesome what have you done with it? how long have you been using it?
it definitely could support solitons easily (because all you really need is to have some time dependence and discretized diffusion, right?) but I don't know if it's possible in either OE-cake or that dust game
As far I recall, being a long term powder gamer myself, powder game does not really have a diffusion like algorithm written into it. The liquids in powder game are sort of dots that move back and forth and subjected to gravity
@Secret I mean more along the lines of the fluid dynamics in that kind of game
@Secret Like how in the dan-ball one air pressure looks continuous (I assume)
@Secret You really just need a timer for particle extinction, and something that effects adjacent cells. Like maybe a rule for a particle that says: particles of type A turn into type B after 10 steps, particles of type B turn into type A if they are adjacent to type A.
I would bet you get lots of cool reaction-diffusion-like patterns with that rule.
(Those that don't understand cricket, please ignore this context, I will get to the physics...)England are playing Pakistan at Lords and a decision has once again been overturned based on evidence from the 'snickometer'. (see over 1.4 ) It's always bothered me slightly that there seems to be a ...
Abstract: Analyzing the data from the last replace-the-homework-policy question was inconclusive. So back to the drawing board, or really back to this question: what do we really mean when we vote to close questions as homework-like?As some/many/most people are aware, we are in the midst of a...
Hi I am trying to understand the concept of dex and how to use it in calculations. The usual definition is that it is the order of magnitude, so $10^{0.1}$ is $0.1$ dex.I want to do a simple exercise of calculating the value of the RHS of Eqn 4 in this paper arxiv paper, the gammas are incompl...
@ACuriousMind Guten Tag! :-) Dark Sun has also a lot of frightening characters. For example, Borys, the 30th level dragon. Or different stages of the defiler/psionicist 20/20 -> dragon 30 transformation. It is only a tip, if you start to think on your next avatar :-)
What is the maximum distance for eavesdropping pure sound waves?And what kind of device i need to use for eavesdropping?Actually a microphone with a parabolic reflector or laser reflected listening devices available on the market but is there any other devices on the planet which should allow ...
and endless whiteboards get doodled with boxes, grids circled red markers and some scribbles
The documentary then showed one of the bird's eye view of the farmlands
(which pardon my sketchy drawing skills...)
Most of the farmland is tiled into grids
Here there are two distinct column and rows of tiled farmlands to the left and top of the main grid. They are the index arrays and they notate the range of inidex of the tensor array
In some tiles, there's a swirl of dirt mount, they represent components with nonzero curl
and in others grass grew
Two blue steel bars were visible laying across the grid, holding up a triangle pool of water
Next in an interview, they mentioned that experimentally the process is uite simple. The tall guy is seen using a large crowbar to pry away a screw that held a road sign under a skyway, i.e.
ocassionally, misshaps can happen, such as too much force applied and the sign snapped in the middle. The boys will then be forced to take the broken sign to the nearest roadworks workshop to mend it
At the end of the documentary, near a university lodge area
I walked towards the boys and expressed interest in joining their project. They then said that you will be spending quite a bit of time on the theoretical side and doddling on whitebaords. They also ask about my recent trip to London and Belgium. Dream ends
Reality check: I have been to London, but not Belgium
Idea extraction: The tensor array mentioned in the dream is a multiindex object where each component can be tensors of different order
Presumably one can formulate it (using an example of a 4th order tensor) as follows:
$$A^{\alpha}_{\beta}_{\gamma,\delta,\epsilon}$$
and then allow the index $\alpha,\beta$ to run from 0 to the size of the matrix representation of the whole array
while for the indices $\gamma,\delta,epsilon$ it can be taken from a subset which the $\alpha,\beta$ indices are. For example to encode a patch of nonzero curl vector field in this object, one might set $\gamma$ to be from the set $\{4,9\}$ and $\delta$ to be $\{2,3\}$
However even if taking indices to have certain values only, it is unsure if it is of any use since most tensor expressions have indices taken from a set of consecutive numbers rather than random integers
@DavidZ in the recent meta post about the homework policy there is the following statement:
> We want to make it sure because people want those questions closed. Evidence: people are closing them. If people are closing questions that have no valid reason for closure, we have bigger problems.
This is an interesting statement.
I wonder to what extent not having a homework close reason would simply force would-be close-voters to either edit the post, down-vote, or think more carefully whether there is another more specific reason for closure, e.g. "unclear what you're asking".
I'm not saying I think simply dropping the homework close reason and doing nothing else is a good idea.
I did suggest that previously in chat, and as I recall there were good objections (which are echoed in @ACuriousMind's meta answer's comments).
@DanielSank Mostly in a (probably vain) attempt to get @peterh to recognize that it's not a particularly helpful topic.
@peterh That said, he used to be fairly active on physicsoverflow, so if you really pine for the opportunity to communicate with him, you can go on ahead there. But seriously, bringing it up, particularly in that way, is not all that constructive.
@DanielSank No, the site mods could have caged him only in the PSE, and only for a year. That he got. After that his cage was extended to a 10 year long network-wide one, it couldn't be the result of the site mods. Only the CMs can do this, typically for network-wide bad deeds.
@EmilioPisanty Yes, but I had liked to talk to him here.
@DanielSank I am only curious, what he did. Maybe he attacked the whole network? Or he toke a site-level conflict to the IRL world? As I know, network-wide bans happen for such things.
@peterh That is pure fear-mongering. Unless you plan on going on extended campaigns to get yourself suspended, in which case I wish you speedy luck.
4
Seriously, suspensions are never handed out without warning, and you will not be ten-year-banned out of the blue. Ron had very clear choices and a very clear picture of the consequences of his choices, and he made his decision. There is nothing more to see here, and bringing it up again (and particularly in such a dewy-eyed manner) is far from helpful.
@EmilioPisanty Although it is already not about Ron Maimon, but I can't see here the meaning of "campaign" enough well-defined. And yes, it is a little bit of source of fear for me, that maybe my behavior can be also measured as if "I would campaign for my caging". |
For discussion of specific patterns or specific families of patterns, both newly-discovered and well-known.
gmc_nxtman Posts: 1147 Joined: May 26th, 2015, 7:20 pm Kazyan wrote:
Component found in a CatForce result:
Code: Select all
x = 42, y = 67, rule = LifeHistory
A$.2A$2A5$7.A$8.2A$7.2A19$24.2A$24.2A2$39.A$37.A3.A$36.A$36.A4.A$36.
5A$14.2A.2D$13.A.AD.D$13.A$12.2A25$5.3A$7.A$6.A!
That can be done with 4 gliders, although it's still interesting that it was found accidentally:
Code: Select all
x = 21, y = 30, rule = B3/S23
10b2o$11bo$11bobo$12b2o14$10bo4bo$10b2ob2o$9bobo2b2o8$2o17bo$b2o15b2o$
o17bobo!
What were you looking for, exactly? A MWSS-to-herschel converter?
Kazyan Posts: 864 Joined: February 6th, 2014, 11:02 pm
gmc_nxtman wrote:What were you looking for, exactly? A MWSS-to-herschel converter?
I'd settle for any signal, but yes. The current Orthogonoids have geometry challenges that pad their size, and the limiting factor in their repeat time is the syringe. Repeat time is more important for single-channel operations than probably any other constructor design, so I'm trying to give that fire some better fuel.
Tanner Jacobi
mniemiec
Posts: 1055 Joined: June 1st, 2013, 12:00 am
gmc_nxtman wrote:4-glider trans-boat with tail edgeshoot: ...
Even though was already buildable from 4 gliders, this method improves syntheses of one still-life and 18 pseudo-objects.
Kazyan Posts: 864 Joined: February 6th, 2014, 11:02 pm
Potential component spotted in a failed eating reaction:
Code: Select all
x = 21, y = 17, rule = B3/S23
o$3o$3bo$2b2o2$6bo$5bobo2$5b3o$19bo$8bo9bo$18b3o3$15bo$14b2o$14bobo!
Tanner Jacobi
gmc_nxtman Posts: 1147 Joined: May 26th, 2015, 7:20 pm
Unusual still life in 8 gliders:
Code: Select all
x = 18, y = 26, rule = B3/S23
11bo$10bobo$10b2o2$10bo$9b2o$9bobo5$obo$b2o$bo2$3b2o$4b2o$3bo$9bo$9b2o
$8bobo2$15b3o$7b2o6bo$6bobo7bo$8bo!
EDIT:
This also gives 21.41458 in 9 gliders.
Kazyan Posts: 864 Joined: February 6th, 2014, 11:02 pm
Potentially grow out a BTS into a structure like a snorkel loop:
Code: Select all
x = 15, y = 25, rule = B3/S23
2b2obo$3bob3o$bobo4bo$ob2ob2obo$o4bobo$b3obo$3bob2o3$10b3o2$8bo5bo$8bo
5bo$8bo5bo2$10b3o6$11b2o$10bo2bo$11b2o$11bo!
I suspect that the drifter catalyst and its variants also have odd transformations, since both objects are robust.
Tanner Jacobi
gmc_nxtman Posts: 1147 Joined: May 26th, 2015, 7:20 pm
Haven't seen a component quite like this before:
Code: Select all
x = 27, y = 18, rule = B3/S23
20bobo$20b2o$21bo7$15bo$15bobo$15b2o2$3o9bobo$b3o9b2o$13bo10b3o$24bo$
25bo!
EDIT:
Better version:
Code: Select all
x = 15, y = 11, rule = B3/S23
13bo$12bo$12b3o3$7bo$6bobo$6bobo2b2o$7bo2b2o$3o9bo$b3o!
Gamedziner
Posts: 796 Joined: May 30th, 2016, 8:47 pm Location: Milky Way Galaxy: Planet Earth
p8 c/2 derived from blinker puffer 1
:
Code: Select all
2bo$o3bo$5bo$o4bo$b5o5$b2o2b2o$bob2ob2o$2b5o$3b3o$4bo$2bo3bo$7bo$2bo4bo$3b5o!
Code: Select all
x = 81, y = 96, rule = LifeHistory
58.2A$58.2A3$59.2A17.2A$59.2A17.2A3$79.2A$79.2A2$57.A$56.A$56.3A4$27.
A$27.A.A$27.2A21$3.2A$3.2A2.2A$7.2A18$7.2A$7.2A2.2A$11.2A11$2A$2A2.2A
$4.2A18$4.2A$4.2A2.2A$8.2A!
mniemiec
Posts: 1055 Joined: June 1st, 2013, 12:00 am
This is known. It can be easily synthesized from 10 gliders:
Code: Select all
x = 88, y = 26, rule = B3/S23
34bobo$35boo$35bo3$45bo$bo44boo$bbo42boo$3o$20boo18boo6bo$bbo17bobo17b
obo4bo$boo18bo19bo5b3o$bobo$$43boo$44boo7b3o22bo4b3o$31b3o9bo9bobbo20b
3o3bobbo$33bo19bo16b3o3boobo3bo$32bo20bo3bo12bobbobb3o4bo3bo$53bo16bo
6boo4bo$54bobo13bo3bo3bo5bobo$70bo$71bobo$77bo$78bo$77bo!
gameoflifemaniac Posts: 774 Joined: January 22nd, 2017, 11:17 am Location: There too
Code: Select all
x = 17, y = 17, rule = B3/S23
8bo$7bobo$4bo2bobo2bo$3bobobobobobo$2bo2bobobobo2bo$3b2o2bobo2b2o$7bob
o$b6o3b6o$o15bo$b6o3b6o$7bobo$3b2o2bobo2b2o$2bo2bobobobo2bo$3bobobobob
obo$4bo2bobo2bo$7bobo$8bo!
Code: Select all
x = 17, y = 17, rule = B3/S23
8bo$7bobo$4bo2bobo2bo$3bobobobobobo$2bo2bobobobo2bo$3b2o2bobo2b2o$7bob
o$b6o3b6o$o7bo7bo$b6o3b6o$7bobo$3b2o2bobo2b2o$2bo2bobobobo2bo$3bobobob
obobo$4bo2bobo2bo$7bobo$8bo!
dvgrn Moderator Posts: 5878 Joined: May 17th, 2009, 11:00 pm Location: Madison, WI Contact:
While incompetently welding a tremi-Snark this evening...
Code: Select all
x = 23, y = 31, rule = LifeHistory
$3.4B$4.4B$5.4B5.2A$6.4B4.2A$7.9B$8.6B$8.4BA3B$6.7BA2B$6.5B3A2B$6.11B
$4.2AB.10B$4.2AB3.B2A4B$9.2B2A5B$10.8B$10.6B$11.5B$5.2A5.3B$5.A.A3.5B
$7.A2.B2AB2A$7.2A2.2A.AB2.2A$10.B3.A.A2.A$5.10A.2A$5.A$6.12A$17.A$8.
7A$8.A5.A$11.A$10.A.A$11.A!
... I ended up with a p3 that I didn't really want.
Doesn't seem worth keeping it around until people are synthesizing all the 58-bit p3's, but it seemed mildly entertaining anyway.
A for awesome Posts: 1901 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 Contact: dvgrn wrote:
Code: Select all
x = 23, y = 31, rule = LifeHistory
$3.4B$4.4B$5.4B5.2A$6.4B4.2A$7.9B$8.6B$8.4BA3B$6.7BA2B$6.5B3A2B$6.11B
$4.2AB.10B$4.2AB3.B2A4B$9.2B2A5B$10.8B$10.6B$11.5B$5.2A5.3B$5.A.A3.5B
$7.A2.B2AB2A$7.2A2.2A.AB2.2A$10.B3.A.A2.A$5.10A.2A$5.A$6.12A$17.A$8.
7A$8.A5.A$11.A$10.A.A$11.A!
Pointless reduction:
Code: Select all
x = 17, y = 25, rule = LifeHistory
4B$.4B$2.4B5.2A$3.4B4.2A$4.9B$5.6B$5.4BA3B$3.7BA2B$3.5B3A2B$3.11B$.2A
B.10B$.2AB3.B2A4B$6.2B2A5B$7.8B$7.6B$8.5B$9.3B$8.5B$7.B2AB2A$4.2A2.2A
.AB2.2A$4.A2.B3.A.A2.A$5.7A.3A2$7.2A.4A$7.2A.A2.A!
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
gmc_nxtman Posts: 1147 Joined: May 26th, 2015, 7:20 pm
Can someone salvage this? (Look at T≈20)
Code: Select all
x = 16, y = 12, rule = B3/S23
bo$2bo$3o7$3b2o5b2o2b2o$4b2o3bobob2o$3bo7bo3bo!
BlinkerSpawn Posts: 1906 Joined: November 8th, 2014, 8:48 pm Location: Getting a snacker from R-Bee's gmc_nxtman wrote:
Can someone salvage this? (Look at T≈20)
Code: Select all
x = 16, y = 12, rule = B3/S23
bo$2bo$3o7$3b2o5b2o2b2o$4b2o3bobob2o$3bo7bo3bo!
The red pattern inserted at gen 16 would do it:
Code: Select all
x = 17, y = 14, rule = LifeHistory
13.D$11.2D$.A14.D$2.A8.5D$3A7$3.2A5.2A2.2A$4.2A3.A.A.2A$3.A7.A3.A!
AbhpzTa
Posts: 475 Joined: April 13th, 2016, 9:40 am Location: Ishikawa Prefecture, Japan gmc_nxtman wrote:
Can someone salvage this? (Look at T≈20)
Code: Select all
x = 16, y = 12, rule = B3/S23
bo$2bo$3o7$3b2o5b2o2b2o$4b2o3bobob2o$3bo7bo3bo!
Code: Select all
x = 27, y = 19, rule = B3/S23
16bo$4bo9b2o$5bo9b2o$3b3o$22bo$20b2o$21b2o$bo$2bo$3o2$25b2o$24b2o$26bo
3$3b2o5b2o2b2o$4b2o3bobob2o$3bo7bo3bo!
Iteration of sigma(n)+tau(n)-n [sigma(n)+tau(n)-n : OEIS A163163] (e.g. 16,20,28,34,24,44,46,30,50,49,11,3,3, ...) :
965808 is period 336 (max = 207085118608). gmc_nxtman Posts: 1147 Joined: May 26th, 2015, 7:20 pm
Reduced an old synthesis from eleven (I think) down to eight gliders:
Code: Select all
x = 34, y = 34, rule = B3/S23
10bo$bobo7bo19bobo$2b2o5b3o19b2o$2bo29bo3$32bo$30b2o$31b2o3$12bo$12bob
o$12b2o11$14b2o$14bobo$14bo12b2o$27bobo$27bo3$b2o$obo$2bo!
Extrementhusiast Posts: 1796 Joined: June 16th, 2009, 11:24 pm Location: USA
Glider + two-glider loaf/tub/block/blinker constellation lasts for over 10K gens:
Code: Select all
x = 16, y = 13, rule = B3/S23
3bobo$3b2o$4bo9bo$13bobo$4b2o8bo$4b2o3$6bo$5bobo$4bo2bo$5b2o$3o!
I Like My Heisenburps! (and others)
Entity Valkyrie Posts: 247 Joined: November 30th, 2017, 3:30 am
A glider synthesis of Sawtooth 311
x = 193, y = 140, rule = B3/S23 40bo$41bo$39b3o$72bo$70b2o$71b2o19$32bobo$33b2o$33bo30bo$63bo$63b3o2$ 75bo$74b2o$24bo49bobo$25b2o$24b2o$49bo$47b2o$48b2o2$2bo$obo$b2o7$34b2o $35b2o$34bo3$67b2o$67bobo$53b2o12bo$53bobo$53bo6$27b2o93b2o$26bobo93bo bo$28bo93bo2$53b3o$53bo74bobo$54bo73b2o$4bo124bo$4b2o172b2o$3bobo171b 2o$174b2o3bo$31b2o140bobo$30b2o98b2o43bo$32bo97bobo36bobo$130bo3bobo 10bo22b2o$135b2o11b2o20bo$135bo4bobo4b2o34bobo$115b2o21bobobobo6b2o20b o9b2o$116b2o21b2ob2o6b2o22bo9bo$115bo36bo19b3o2$120bo23b2ob2o23b3o5bo$ 120b2o21bobobobo24bo4bo$119bobo13b2ob2o5bobo25bo5b3o$134bobobobo$115bo 20bobo13b2o25b3o$116b2o12b2o19b2o22bo3bo$115b2o14b2o20bo22bo3bo$130bo 35bo4bo2b3o$164bobo2b2o$113b3o49b2o3b2o2b3o4bobo$115bo60bo4b2o$114bo 60bo6bo$179bo$126bo25bo18bo7b2o$125b2o23b2o19b2o5bobo$125bobo23b2o17bo bo$146b2o$147b2o$121b2o23bo$120b2o$122bo2$156b2o$142bobo10bobo$134bobo 5b2o13bo$134b2o7bo$135bo$132bo6bo$132b2o4bo$131bobo4b3o2$138b3o43bo$ 134bo3bo22bo20b2o$135bo3bo22b2o19b2o$133b3o25b2o13bobo$174bobobobo7bo$ 133b3o5bo25bobo5b2ob2o6bobo$135bo4bo24bobobobo15b2o3bo$63b3o68bo5b3o 23b2ob2o19b2o$63bo127b2o$64bo75b3o19bo$130bo9bo22b2o6b2ob2o$130b2o9bo 20b2o6bobobobo$129bobo34b2o4bobo$144bo20b2o$143b2o22bo12b2o$143bobo33b 2o$181bo2$139bo$138bo$138b3o2$137bo$136b2o$136bobo! Entity Valkyrie Posts: 247 Joined: November 30th, 2017, 3:30 am
This Simkin-Glider=Gun=like object actually produces two MWSS:
Code: Select all
x = 53, y = 17, rule = B3/S23
44b2o5b2o$44b2o5b2o2$47b2o$47b2o$12bo$12b3o$12bobo$14bo4$4b2o$4b2o2$2o
5b2o$2o5b2o!
mniemiec
Posts: 1055 Joined: June 1st, 2013, 12:00 am
Entity Valkyrie wrote:A glider synthesis of Sawtooth 311 ...
It's nice to have syntheses like this. Unfortunately, in this case, there are several pairs of gliders that would have had to pass through each other earlier (i.e. they would have already collided before this phase). To make sure this doesn't happen, it is usually a good idea to backtrack all the gilders a certain amount (e.g. far enough away that they are in four distinct clouds, one coming from each direction) and then run them to see if any unwanted interactions occur first.
Rhombic Posts: 1056 Joined: June 1st, 2013, 5:41 pm
This component (the reverse component would have been more useful). Found accidentally though.
Code: Select all
x = 12, y = 14, rule = B3/S23
11bo$9b3o$8bo$9bo$6b4o$6bo$2b2o3b3o$2b2o5bo$9bobo$2bo7b2o$bobo$bob2o$o
$2bo!
Code: Select all
x = 13, y = 15, rule = B3/S23
7bo$7b3o$10bo$2b2ob3o2bo$o2bobo2bob2o$2o4b3o3bo$9bobo$3b2o3b2ob2o$3b2o
2$3bo$2bobo$2bob2o$bo$3bo!
Extrementhusiast Posts: 1796 Joined: June 16th, 2009, 11:24 pm Location: USA
Switch engine turns two rows of beehives into two rows of table on tables:
Code: Select all
x = 88, y = 96, rule = B3/S23
13b2o$12bo2bo$13b2o6$21b2o$8b3o9bo2bo$9bo2bo8b2o$13bo$10bobo4$29b2o$
28bo2bo$29b2o2$bo$obo$obo$bo$37b2o$36bo2bo$37b2o2$9bo$8bobo$8bobo$9bo$
45b2o$44bo2bo$45b2o2$17bo$16bobo$16bobo$17bo$53b2o$52bo2bo$53b2o2$25bo
$24bobo$24bobo$25bo$61b2o$60bo2bo$61b2o2$33bo$32bobo$32bobo$33bo$69b2o
$68bo2bo$69b2o2$41bo$40bobo$40bobo$41bo$77b2o$76bo2bo$77b2o2$49bo$48bo
bo$48bobo$49bo$85b2o$84bo2bo$85b2o2$57bo$56bobo$56bobo$57bo5$65bo$64bo
bo$64bobo$65bo5$73bo$72bobo$72bobo$73bo!
I Like My Heisenburps! (and others)
KittyTac Posts: 533 Joined: December 21st, 2017, 9:58 am Extrementhusiast wrote:
Switch engine turns two rows of beehives into two rows of table on tables:
Code: Select all
x = 88, y = 96, rule = B3/S23
13b2o$12bo2bo$13b2o6$21b2o$8b3o9bo2bo$9bo2bo8b2o$13bo$10bobo4$29b2o$
28bo2bo$29b2o2$bo$obo$obo$bo$37b2o$36bo2bo$37b2o2$9bo$8bobo$8bobo$9bo$
45b2o$44bo2bo$45b2o2$17bo$16bobo$16bobo$17bo$53b2o$52bo2bo$53b2o2$25bo
$24bobo$24bobo$25bo$61b2o$60bo2bo$61b2o2$33bo$32bobo$32bobo$33bo$69b2o
$68bo2bo$69b2o2$41bo$40bobo$40bobo$41bo$77b2o$76bo2bo$77b2o2$49bo$48bo
bo$48bobo$49bo$85b2o$84bo2bo$85b2o2$57bo$56bobo$56bobo$57bo5$65bo$64bo
bo$64bobo$65bo5$73bo$72bobo$72bobo$73bo!
And then explodes. I wonder if there's a way to eat it at the end.
dvgrn Moderator Posts: 5878 Joined: May 17th, 2009, 11:00 pm Location: Madison, WI Contact: KittyTac wrote:
Extrementhusiast wrote:Switch engine turns two rows of beehives into two rows of table on tables...
And then explodes. I wonder if there's a way to eat it at the end.
Yeah, switch engine/swimmer eaters definitely aren't a problem:
Code: Select all
x = 96, y = 98, rule = B3/S23
13b2o$12bo2bo$13b2o6$8b3o10b2o$20bo2bo$8bo3bo8b2o$9b4o$12bo4$29b2o$28b
o2bo$29b2o2$bo$obo$obo$bo$37b2o$36bo2bo$37b2o2$9bo$8bobo$8bobo$9bo$45b
2o$44bo2bo$45b2o2$17bo$16bobo$16bobo$17bo$53b2o$52bo2bo$53b2o2$25bo$
24bobo$24bobo$25bo$61b2o$60bo2bo$61b2o2$33bo$32bobo$32bobo$33bo$69b2o$
68bo2bo$69b2o2$41bo$40bobo$40bobo$41bo$77b2o$76bo2bo$77b2o2$49bo$48bob
o$48bobo$49bo$85b2o$84bo2bo$85b2o2$57bo$56bobo$56bobo$57bo5$65bo$64bob
o$64bobo$65bo5$73bo$72bobo$72bobo17b2o$73bo18bo$93b3o$95bo!
#C [[ AUTOSTART STEP 9 THEME 2 ]]
kiho park
Posts: 50 Joined: September 24th, 2010, 12:16 am
I found this c/3 diagonal fuse while searching c/3 long barge crawler.
Code: Select all
x = 10, y = 11, rule = B3/S23:T40,27
8b2o$7bo2$6bobo$5bo2bo$4bobo$3bobo$2bobo$bobo$obo$bo! |
Prove that if $d_1$ and $d_2$ are metrics over $X$, $\tau_1$ and $\tau_2$ are the family of open subsets of their respective metric spaces then:
(i) $\implies$ (ii) $\iff$ (iii)
Where:
(i) There exists a $c>0$ that for any $x,y \in X$ we have $c \cdot d_1(x,y) \geq d_2(x,y)$
(ii) For any $x \in X$ and $r > 0 $, there exists a $r'>0$ where $B_{d_1}(x,r') \subset B_{d_2}(x,r)$
(iii) $\tau_2 \subset \tau_1$
If (i) holds, then for given $x \in X$ and $r>0$, define $r'=\frac{r}{c}$.
Then if $y \in B_{d_1}(x,r')$ then $d_1(x,y) < r'$. Also by (i): $$d_2(x,y) \le c\cdot d_1(x,y) < c \cdot r' = c \cdot \frac{r}{c}=r$$
so that $y \in B_{d_2}(x,r)$ , showing the inclusion. and so $\text{(i)} \implies \text{(ii)}$ holds.
The equivalence of (ii) and (iii) is quite obvious from the definitions. What does $O \in \tau_2$ mean? Also recall that open balls are open sets in their induced topologies. |
Let's focus on the technical note, but let's change the symbols for easy comparison. The note has two parts:Firstly, if we assume that the stock price is log normal with the following parameters:$S_t \sim \mathrm{LN}\left(\ln S_0+\mu t,\sigma^2 t\right)$then by definition, its log is normally distributed:$\ln S_t \sim \mathrm{N}\left(\ln S_0+\mu t,\...
In Hull's textbook, the stock price dynamics is lognormal: $S_T = S_0 \exp(\mu T - \frac{1}{2}\sigma^2T + \sigma W_T)$, where $W_t$ is a standard brownian motion. And so the mean of this is the mean of a lognormal random variable with the log mean as $\ln S_0 + \mu T - \frac{1}{2}\sigma^2T$ and the log standard deviation as $\sigma \sqrt{T}$, and so the ...
My answer isn’t really “quant”.I don’t think that “there is a way” to do it. The board of directors has to decide on the dividend as well as the amount of the dividends. So this hypothetical number can be anywhere between 0 and (almost) the company’s earnings.For a company like Berkshire, incomes can come from underlying holdings’ dividends as you notice,...
There isn't a single answer to this question. It strongly depends on your goals and why it is missing. If you have a long enough time-series, you will find large numbers of missing data points. The NYSE used to maintain a post for companies that did not trade weekly not so long ago.However, unless you have cause to believe there was a reason for it to ... |
Edit —
I've revised this answer to make some small improvements in the commands, to tidy up the commands for drawing the wires for instance, because it seemed worthwhile.
Flattering as it is to have this answer be the accepted one for the time being, I think I should point out that the
quantikz package (see Daftwullie's answer below) and the
qpic package (as pointed out in cnada's answer below) are both libraries with reasonably complete interfaces, and so better for people looking for a quick and simple solution. The code below is probably more suitable for people who are comfortable with TiKZ, and might like to tweak their circuit diagrams with TiKZ commands, but who wouldn't mind having some macros to streamline drawing their diagrams. — Nice as it might be to, for the moment I have no ambition to write a LaTeX package to make these macros available with a nice interface for all purposes (but anyone else who would like to is welcome if they give me some of the credit).
Snippet
See below for all of the code used to generate this example: the following commands are just the ones used to draw the circuit itself. (This snippet involves macros which I have defined for the purpose of this post, which I also define below.)
% define initial positions of the quantum wires
\xdef\dy{1.25}
\defwire (A) at (0);
\defwire (B) at ({-\dy});
\defwire (C) at ({-2*\dy});
% draw wires
\xdef\dt{0.8}
\drawwires [\dt] (15);
\node at ($(B-0)!0.5!(B-1)$) {$/$};
\node at ($(C-0)!0.5!(C-1)$) {$/$};
% draw gates
\gate (B-2) [H^{\otimes n}];
\ctrlgate (B-3) (C-3) [U];
\virtgate (A-3);
\gate (B-4) [\mathit{FT}^\dagger];
\ctrlgate (B-7) (A-7) [R];
\virtgate (C-7);
\gate (B-10) [\mathit{FT}];
\ctrlgate (B-11) (C-11) [U^\dagger];
\gate (B-12) [H^{\otimes n}];
\virtgate (A-12);
\meas (A-14) [Z];
% draw input and output labels
\inputlabel (A-0) [\lvert 0 \rangle];
\inputlabel (B-0) [\lvert 0 \rangle^{\otimes n}];
\inputlabel (C-0) [\lvert b \rangle];
\outputlabel (A-15) [\lvert 1 \rangle];
\outputlabel (B-15) [\lvert 0 \rangle^{\otimes n}];
\outputlabel (C-15) [\lvert x \rangle];
Result
Preamble
You will need a pre-amble which contains at least the
amsmath package, as well as the
tikz package. You may not need all of the
tikz libraries below, but they don't hurt. Be sure to include the commands involving layers.
\documentclass[a4paper,10pt]{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{shapes,arrows,calc,positioning,fit}
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
For the purposes of this post, I've defined some ad-hoc macros to make reading the coded circuit easier for public consumption. (The macro format is not exactly good LaTeX practise, but I define them this way in order for the syntax to be more easily read and for it to stand out.) The parameters for dimensions in these gates were chosen to look good in your sample-circuit, and were found by trial-and-error: you can change them to change the appearance of your circuit.
The first is a simple macro to draw a gate.
\def\gate (#1) [#2]{%
\node [
draw=black,fill=white, inner sep=0pt,
minimum width=2.5em, minimum height=2em, outer sep=1ex
] (#1) at (#1) {$#2$}%
}
The second is a macro to draw an 'invisible' gate. This is not really a command which is important for the circuit itself, but helps for the placement of background frames.
\def\virtgate (#1){%
\node [
draw=none, fill=none,
minimum width=2.5em, minimum height=2em, outer sep=1ex
] (#1) at (#1) {};
}
The third is a macro to draw a controlled gate. This command works well enough for your example circuit, but doesn't allow you to draw a CNOT. (Exercise for the reader proficient in TiKZ: make a
\CNOT command.)
\def\ctrlgate (#1) (#2) [#3]{%
\filldraw [black] (#1) circle (2pt) -- (#2);
\gate (#2) [#3]
}
The fourth is a macro to draw a "measurement" box. I think it is perfectly reasonable to want to specify an explicit basis or observable for the measurement, so I allow an argument to specify that.
\def\meas (#1) [#2]{%
\node [
draw=black, fill=white, inner sep=2pt,
label distance=-5mm, minimum height=2em, minimum width=2em
] (meas) at (#1) {};
\draw ($(meas.south) + (-.75em,1.5mm)$) arc (150:30:.85em);
\draw ($(meas.south) + (0,1mm)$) -- ++(.8em,1em);
\node [
anchor=north west, inner sep=1.5pt, font=\small
] at (meas.north west) {#2};
}
I define two short macros to produce the labels for the inputs and outputs of wires.
\def\inputlabel (#1) [#2]{%
\node at (#1) [anchor=east] {$#2$}
}
\def\outputlabel (#1) [#2]{%
\node at (#1) [anchor=west] {$#2$}
}
The macros above are all looking for co-ordinates at which to place the gates. I also define macros to define "wires", which have regularly spaced co-ordinates where gates can be located.The first is a macro which allows you to define a named wire (such as
A,
B,
x3,
etc.) and its vertical position in the circuit diagram (these diagrams are left-to-right by default, which you can change most easily using the
rotate option of the
tikzpicture environment.)
\def\defwire (#1) at (#2){%
\ifx\qmwires\empty
\edef\qmwires{#1}%
\else
\edef\qmwires{\qmwires,#1}%
\fi
\coordinate (#1-0) at ($(0,#2)$)%
}
Having defined a collection of wires, the following command then draws all of them, starting from the same left-most starting point and ending at the same right-most ending point, with increments by a fixed amount (given in the square brackets) and for a given number of time slices. This defines a sequence of 'time-slice' co-ordinates for each wire: for a wire
A, it defines the co-ordinates
A-0,
A-1, and so forth up until
A-t (where
t is the value of the second argument).
\def\drawwires [#1] (#2);{%
\xdef\u{0}
\foreach \t in {0,...,#2} {%
\foreach \l in \qmwires {%
\coordinate (\l-\t) at ($(\l-\u) + (#1,0)$);
\draw (\l-\u) -- (\l-\t);
}
\xdef\u{\t}
}
}
The final macro is one to draw a background frame for different stages in your circuit. It takes an argument specifying which gates (including the invisible virtual 'gates') are meant to belong to the frame.
\def\bgframe [#1]{%
\node [%
draw=black, fill=yellow!40!gray!30!white, fit=#1
] {}%
}
The circuit diagram itself
Now to begin drawing your circuit.
\begin{document}
\begin{tikzpicture}
We start by defining the relative positions of the wires. (For convenience, I do this using a macro to define the spacing between them, that I can quickly change to adjust the spacing.) Below, I define three wires:
A,
B, and
C.
\let\qmwires\empty
% define initial positions of the quantum wires
\xdef\dy{1.25}
\defwire (A) at (0);
\defwire (B) at ({-\dy});
\defwire (C) at ({-2*\dy});
We now draw the circuit, using the command to draw the wires and define the co-ordinates on the wire, and placing gates independently of one another according to those co-ordinates.
% draw circuit
\xdef\dt{0.8}
\drawwires [\dt] (15);
\node at ($(B-0)!0.5!(B-1)$) {$/$};
\node at ($(C-0)!0.5!(C-1)$) {$/$};
\gate (B-2) [H^{\otimes n}];
\ctrlgate (B-3) (C-3) [U];
\virtgate (A-3);
\gate (B-4) [\mathit{FT}^\dagger];
\ctrlgate (B-7) (A-7) [R];
\virtgate (C-7);
\gate (B-10) [\mathit{FT}];
\ctrlgate (B-11) (C-11) [U^\dagger];
\gate (B-12) [H^{\otimes n}];
\virtgate (A-12);
\meas (A-14) [Z];
% draw input and output labels
\inputlabel (A-0) [\lvert 0 \rangle];
\inputlabel (B-0) [\lvert 0 \rangle^{\otimes n}];
\inputlabel (C-0) [\lvert b \rangle];
\outputlabel (A-15) [\lvert 1 \rangle];
\outputlabel (B-15) [\lvert 0 \rangle^{\otimes n}];
\outputlabel (C-15) [\lvert x \rangle];
Annotations for the circuit
The rest of the circuit diagram is literally commentary. We can do this using a combination of plain-old TiKZ nodes, and the
\bgframe macro which I defined above. (Annotations are a little less predictable, so I don't have a good way of making them as systematic as the earlier parts of the circuit, so general TiKZ commands are a reasonable approach unless you know how to make your annotations uniform.)
First the annotations for the stages of the circuit:
% draw annotations
\node [minimum height=4ex] (annotate-1) at ($(A-3) + (0,1)$)
{\textit{Phase estimation}};
\node [minimum height=4ex] (annotate-2) at ($(A-7) + (0,1)$)
{\textit{$\smash{R(\tilde\lambda^{-1}})$ rotation}};
\node [minimum height=4ex] (annotate-3) at ($(A-11) + (0,1)$)
{\textit{Uncompute}};
\node (annotate-a) at ($(C-3) + (0,-1.25)$) {\textit{(a)}};
\node (annotate-b) at ($(C-7) + (0,-1.25)$) {\textit{(b)}};
\node (annotate-c) at ($(C-11) + (0,-1.25)$) {\textit{(c)}};
Next, the annotations for the registers, at the input:
\node (A-in-annotate) at ($(A-0) + (-3em,0)$) [anchor=east]
{\parbox{4.5em}{\centering Ancilla register $S$ }};
\node (B-in-annotate) at ($(B-0) + (-3em,0)$) [anchor=east]
{\parbox{4.5em}{\centering Clock \\ register $C$ }};
\node (C-in-annotate) at ($(C-0) + (-3em,0)$) [anchor=east]
{\parbox{4.5em}{\centering Input \\ register $I$ }};
Finally, the frames for the stages of the circuit.
% draw frames for stages of the circuit
\begin{pgfonlayer}{background}
\bgframe [(annotate-1)(B-2)(B-4)(C-3)];
\bgframe [(annotate-2)(B-7)(C-7)];
\bgframe [(annotate-3)(B-10)(B-12)(C-11)];
\end{pgfonlayer}
And that's the end of the circuit.
\end{tikzpicture}
\end{document} |
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 |
If yes could you explain why? Sorry if the question is trivial but I'm new to complex numbers and I see lots of examples where properties of real numbers are used in complex without to prove it . This really surprise me since complex numbers are a superset of real numbers. For example I saw in my book $z*1=z$ in $ \mathbb{C} $ I had to make the complex multiplication to be sure that it was true, because this 1 is in $\mathbb{C} $ (1,0)
First note that, for $z=a+ib$ we have $z\overline{z}=\sqrt{a^2+b^2}=|z|^2$ and also $\overline{z_1z_2}=\overline{z_1}\,\overline{z_2}.$
Therefore, $|zw|^2=(zw)\overline{zw}=zw\overline{z}\,\overline{w}=z\overline{z}w\overline{w}=|z|^2|w|^2.$ This implies $|zw|=|z||w|.$
Let $z=a+bi$ and $w=c+di$ where $(a,b,c,d)\in\mathbb{R}^{4}$. The module $|z|$ is defined by $|z|=\sqrt{a^{2}+b^{2}}$ and $|w|=\sqrt{c^{2}+d^{2}}$, thus $$\begin{align} |z||w| &=\sqrt{a^{2}+b^{2}}\sqrt{c^{2}+d^{2}}\\ &=\sqrt{a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2}} \end{align}$$
Now, consider $zw=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$. We have
$$\begin{align} |zw|&=\sqrt{(ac-bd)^{2}+(ad+bc)^{2}}\\ &=\sqrt{a^{2}c^{2}+b^{2}d^{2}-2abcd+a^{2}d^{2}+b^{2}c^{2}+2abcd}\\ &=\sqrt{a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}}\\ &=|z||w| \end{align}$$ |
As I understand it, the
proper time, $\tau$, between to events in spacetime is defined in terms of the spacetime interval $ds^{2}=\eta_{\mu\nu}dx^{\mu}dx^{\nu}$, such that $$d\tau =\sqrt{-ds^{2}}$$ (where we are using the "mostly +" signature with $c=1$).
Now, for
time-like intervals, for which $ds^{2}<0$, it is clear that proper time is well-defined since the quantity $\sqrt{-ds^{2}}$ is positive, and furthermore, one can always find a frame in which the two events occur at the same point in space, such that one can construct a worldline connecting the two events, along which an observer can travel, at rest with respect to both events, such that $d\tau =\sqrt{-ds^{2}}=dt$.
However, why is it the case that for
space-like, $ds^{2}>0$, and light-like intervals, $ds^{2}=0$, the notion of proper time is undefined (or perhaps ill-defined)?
For the
space-like case, I get that heuristically, one cannot construct a path between the two events along which an observer can travel and so in this sense proper time is meaningless, since a worldline connecting the events does not exist and so no clock can pass through both events. However, can this be seen purely by examining the definition of proper time in terms of the spacetime interval? Is it simply that the quantity $\sqrt{-ds^{2}}$ will become imaginary and so clearly cannot be used to represent any physical time interval?
Likewise, for a
light-like interval, only a beam of light can pass between both events and since there is no rest frame for light one cannot construct a frame in which a clock is at rest with respect to the beam and passes through both events. However, purely in terms of the spacetime interval, is it simply because the quantity $\sqrt{-ds^{2}}$ equals $0$, and so the notion of proper time is ill-defined since there is no invertible map between reference frames (here I'm thinking in terms of time dilation, $t =\gamma\tau$ and so for a light-like interval, $\gamma\rightarrow\infty$ meaning that the inverse relation $\tau =\frac{t}{\gamma}$ is ill-defined)?! |
Local Bases of a Point in a Topological Space
Recall from the Bases of a Topology page that if $(X, \tau)$ is a topological space then a base $\mathcal B$ of $\tau$ is a collection of subsets from $\tau$ such that each $U \in \tau$ is the union of some subcollection $\mathcal B^* \subseteq \mathcal B$ of $\mathcal B$, i.e., for all $U \in \tau$ we have that there exists a $\mathcal B^* \subseteq \mathcal B$ such that:(1)
We will now look at a similar definition called a local bases of a point $x$ in a topological space $(X, \tau)$.
Definition: Let $(X, \tau)$ be a topological space and let $x \in X$. A Local Base of the element $x$ is a collection of open neighbourhoods of $x$, $\mathcal B_x$ such that for all $U \in \tau$ with $x \in U$ there exists a $B \in \mathcal B_x$ such that $x \in B \subseteq U$. In other words, a local base of the point $x \in X$ is a collection of sets $\mathcal B_x$ such that in every open neighbourhood of $x$ there exists a base element $B \in \mathcal B_x$ contained in this open neighbourhood. Example 1
Consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology of open intervals on $\mathbb{R}$. Consider the point $0 \in \mathbb{R}$. One such local base of $0$ is the following collection:(2)
For example, if we consider the open set $U = (-1, 1) \cup (2, 3) \in \tau$ which contains $0$, then for $B = \left ( - \frac{1}{2}, \frac{1}{2} \right ) \in \mathcal B_0$ we see that $0 \in B \subseteq U$.
More generally, for any $x \in \mathbb{R}$, a local base of $x$ is(3)
This is because for any open set $U \in \tau$ containing $x$ there will be an open interval containing $x$ that is contained in $U$.
Example 2
For a different example, consider the set $X = \{ a, b, c, d, e \}$ and the topology $\tau = \{ \emptyset, \{a \}, \{a, b \}, \{a, c \}, \{a, b, c \}, \{a, b, c, d \}, X \}$.
What is a local base for the element $b \in X$? Let's first look at the sets in $\tau$ containing $b$. They are $U_1 = \{ a, b \}$, $U_2 = \{ a, b, c \}$, $U_3 = \{a, b, c, d \}$, and $U_4 = X$.
We see that $\mathcal B_b = \{ \{ b \} \}$ works as a local base of $b$ since:(4)
What is a local base for the element $c \in X$? The sets in $\tau$ containing $c$ are $U_1 = \{a, c \}$, $U_2 = \{a, b, c \}$, $U_3 = \{ a, b, c, d \}$, and $U_4 = X$.
We see that $\mathcal B_c = \{ \{ a, c \} \}$ works as a local base of $c$ since:(5) |
Table of Contents
Legendre Symbols
Suppose that we want to determine whether or not a quadratic congruence has solutions or not. We can determine just that utilizing what are called
Legendre symbols for quadratic congruences in the form $x^2 \equiv a \pmod p$ where p is an odd prime and $p \nmid a$:
Definition: Suppose p is an odd prime and $p \nmid a$. The Legendre symbol $(a/p)$ is defined by $(a/p) = 1$ if a is a quadratic residue (mod p) and $(a/p) = -1$ if a is a quadratic nonresidue (mod p). If p is not an odd prime, or if p divides a then the Legendre symbol is undefined. Example 1 Which of the following Legendre symbols are defined? A) $(9/5)$, B) $(52/13)$, C) $(19/2)$, and D) $(11/50)$. A)The Legendre symbol $(9/5)$ is defined since 5 is an odd prime and 5 does not divide 9. B)The Legendre symbol $(52/13)$ is NOT defined. 13 is an odd prime, but 13 divides 52. C)The Legendre symbol $(19/2)$ is NOT defined. 2 is an even prime. D)The Legendre symbol $(11/50)$ is NOT defined. 50 is not a prime. Properties of the Legendre Symbol Lemma 1: If a ≡ b (mod p), then the legendre symbol (a/p) = (b/p). Proof:This should come rather obvious. If $x^2 \equiv a \pmod p$ has solutions (or doesn't have solutions), then it follows that $x^2 \equiv b \pmod p$ should have solutions (or not) whenever $a \equiv b \pmod p$, the exact same ones.
To apply property one, suppose that we want to determine if the quadratic congruence $x^2 \equiv 5 \pmod {13}$ has solutions. This congruence is the same as $x^2 \equiv 18 \pmod {13}$. The Legendre symbol $(5/13)$ must be equal to the Legendre symbol $(18/15)$.
Lemma 2: If p does not divide a then (a 2/p) = 1. Proof:Property 2 implies that $x^2 \equiv a^2 \pmod p$, which obviously has solutions, more specifically, the least residue of a (mod p). Lemma 3: If p does not divide ab, then (ab/p) = (a/p)(b/p) Proof:Using Euler's Criterion, we know that $(a/p) = 1$ if $a^{(\frac{p-1}{2})} \equiv 1 \pmod p$ and that $(a/p) = -1$ if $a^{(\frac{p-1}{2})} \equiv -1 \pmod p$. Hence it follows that: Note that the left hand side of the congruence is either 1 or -1, and the right hand side of the congruence is the same. The only way that $(ab/p) \equiv (a/p)(b/p) \pmod p$ is when $(ab/p) = (a/p)(b/p)$ since p is restricted to being an odd prime. Other Properties of Legendre Symbols
There are many other rules that have been established regarding Legendre symbols, some of which are often necessary to evaluate Legendre symbols. We will state a few of them on this page and subsequently prove them all in future pages:
For the Legendre symbol (-1/p): For the Legendre symbol (2/p): For the Legendre symbol (3/p): For the Legendre symbol (6/p): The Quadratic Reciprocity Theorem states that $(p/q) = -(q/p)$ if $p \equiv q \equiv 3 \pmod {4}$ and $(p/q) = (q/p)$ otherwise. |
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Is the usual \(\leq\) ordering on the set \(\mathbb{R}\) of real numbers a total order?
So, yes.
1. Reflexivity holds
2. For any \\( a, b, c \in \tt{R} \\) \\( a \le b \\) and \\( b \le c \\) implies \\( a \le c \\)
3. For any \\( a, b \in \tt{R} \\), \\( a \le b \\) and \\( b \le a \\) implies \\( a = b \\)
4. For any \\( a, b \in \\tt{R} \\), we have either \\( a \le b \\) or \\( b \le a \\)
So, yes.
Perhaps, due to our interest in things categorical, we can enjoy (instead of Cauchy sequence methods) to see the order of the (extended) real line as the Dedekind-MacNeille_completion of the rationals. Matthew has told us interesting things about it before.
Hausdorff, on his part, in the book I mentioned here, says that any total order, dense, and without \( (\omega,\omega^*) \) gaps, has embedded the real line. I don't have a handy reference for an isomorphism instead of an embedding ("everywhere dense" just means dense here).
Perhaps, due to our interest in things categorical, we can enjoy (instead of Cauchy sequence methods) to see the order of the (extended) real line as the [Dedekind-MacNeille_completion of the rationals](https://en.wikipedia.org/wiki/Dedekind%E2%80%93MacNeille_completion#Examples). Matthew has told us interesting things about it [before](https://forum.azimuthproject.org/discussion/comment/16714/#Comment_16714).
Hausdorff, on his part, in the book I mentioned [here](https://forum.azimuthproject.org/discussion/comment/16154/#Comment_16154), [says](https://books.google.es/books?id=M_skkA3r-QAC&pg=PA85&dq=each+everywhere+dense+type&hl=en&sa=X&ved=0ahUKEwjLkJao-9DaAhWD2SwKHVrkBcIQ6AEIKTAA#v=onepage&q=each%20everywhere%20dense%20type&f=false) that any total order, dense, and without \\( (\omega,\omega^*) \\) [gaps](https://en.wikipedia.org/wiki/Hausdorff_gap), has embedded the real line. I don't have a handy reference for an isomorphism instead of an embedding ("everywhere dense" just means dense here).
I believe the hyperreal numbers give an example of a dense total order that embeds the reals without being isomorphic to it. (I can’t speak to the gaps condition though, and it’s just plausible that they’re isomorphic at the level of mere posets rather than ordered fields.)
I believe the [hyperreal numbers](https://en.wikipedia.org/wiki/Hyperreal_number) give an example of a dense total order that embeds the reals without being isomorphic to it. (I can’t speak to the gaps condition though, and it’s just plausible that they’re isomorphic at the level of mere posets rather than ordered fields.)
That's an interesting question, Jonathan.
That's an interesting question, Jonathan.
Jonathan Castello
I believe the hyperreal numbers give an example of a dense total order that embeds the reals without being isomorphic to it. (I can’t speak to the gaps condition though, and it’s just plausible that they’re isomorphic at the level of mere posets rather than ordered fields.)
In fact, while they are not isomorphic as lattices, they are in fact isomorphic as mere posets as you intuited.
First, we can observe that \(|\mathbb{R}| = |^\ast \mathbb{R}|\). This is because \(^\ast \mathbb{R}\) embeds \(\mathbb{R}\) and is constructed from countably infinitely many copies of \(\mathbb{R}\) and taking a quotient algebra modulo a free ultra-filter. We have been talking about quotient algebras and filters in a couple other threads.
Next, observe that all unbounded dense linear orders of cardinality \(\aleph_0\) are isomorphic. This is due to a rather old theorem credited to George Cantor. Next, apply the Morley categoricity theorem. From this we have that all unbounded dense linear orders with cardinality \(\kappa \geq \aleph_0\) are isomorphic. This is referred to in model theory as \(\kappa\)-categoricity.
Since the hypperreals and the reals have the same cardinality, they are isomorphic as unbounded dense linear orders.
Puzzle MD 1: Prove Cantor's theorem that all countable unbounded dense linear orders are isomorphic.
[Jonathan Castello](https://forum.azimuthproject.org/profile/2316/Jonathan%20Castello)
> I believe the hyperreal numbers give an example of a dense total order that embeds the reals without being isomorphic to it. (I can’t speak to the gaps condition though, and it’s just plausible that they’re isomorphic at the level of mere posets rather than ordered fields.)
In fact, while they are not isomorphic as lattices, they are in fact isomorphic as mere posets as you intuited.
First, we can observe that \\(|\mathbb{R}| = |^\ast \mathbb{R}|\\). This is because \\(^\ast \mathbb{R}\\) embeds \\(\mathbb{R}\\) and is constructed from countably infinitely many copies of \\(\mathbb{R}\\) and taking a [quotient algebra](https://en.wikipedia.org/wiki/Quotient_algebra) modulo a free ultra-filter. We have been talking about quotient algebras and filters in a couple other threads.
Next, observe that all [unbounded dense linear orders](https://en.wikipedia.org/wiki/Dense_order) of cardinality \\(\aleph_0\\) are isomorphic. This is due to a rather old theorem credited to George Cantor. Next, apply the [Morley categoricity theorem](https://en.wikipedia.org/wiki/Morley%27s_categoricity_theorem). From this we have that all unbounded dense linear orders with cardinality \\(\kappa \geq \aleph_0\\) are isomorphic. This is referred to in model theory as *\\(\kappa\\)-categoricity*.
Since the hypperreals and the reals have the same cardinality, they are isomorphic as unbounded dense linear orders.
**Puzzle MD 1:** Prove Cantor's theorem that all countable unbounded dense linear orders are isomorphic.
Hi Matthew, nice application of the categoricity theorem! One question if I may. You said:
In fact, while they are not isomorphic as lattices, they are in fact isomorphic as mere posets as you intuited.
But in my understanding the lattice and poset structure is inter-translatable as in here. Can two lattices be isomorphic and their associated posets not?
Hi Matthew, nice application of the categoricity theorem! One question if I may. You said:
> In fact, while they are not isomorphic as lattices, they are in fact isomorphic as mere posets as you intuited.
But in my understanding the lattice and poset structure is inter-translatable as in [here](https://en.wikipedia.org/wiki/Lattice_(order)#Connection_between_the_two_definitions). Can two lattices be isomorphic and their associated posets not?
(EDIT: I clearly have no idea what I'm saying and I should probably take a nap. Disregard this post.)
Can two lattices be isomorphic and their associated posets not?
Can two lattices be isomorphic and their associated posets not?
If two lattices are isomorphic preserving infima and suprema, ie limits, then they are order isomorphic.
The reals and hyperreals provide a rather confusing counter example to the converse. I am admittedly struggling with this myself, as it is highly non-constructive.
From model theory we have two maps \(\phi : \mathbb{R} \to\, ^\ast \mathbb{R} \) and \(\psi :\, ^\ast\mathbb{R} \to \mathbb{R} \) such that:
Now consider \(\{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \}\).
The hyperreals famously violate the Archimedean property. Because of this \(\bigwedge_{^\ast \mathbb{R}} \{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \}\) does not exist.
On the other than if we consider \( \bigwedge_{\mathbb{R}} \{ \psi(x) \, : \, x \in \mathbb{R} \text{ and } 0 < x \}\), that does exist by the completeness of the real numbers (as it is bounded below by \(\psi(0)\)).
Hence
$$\bigwedge_{\mathbb{R}} \{ \psi(x) \, : \, x \in \mathbb{R} \text{ and } 0 < x \} \neq \psi\left(\bigwedge_{^\ast\mathbb{R}} \{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \}\right)$$So \(\psi\) cannot be a complete lattice homomorphism, even though it is part of an order isomorphism.
However, just to complicate matters, I believe that \(\phi\) and \(\psi\) are a mere lattice isomorphism, preserving finite meets and joints.
> Can two lattices be isomorphic and their associated posets not?
If two lattices are isomorphic preserving *infima* and *suprema*, ie *limits*, then they are order isomorphic.
The reals and hyperreals provide a rather confusing counter example to the converse. I am admittedly struggling with this myself, as it is highly non-constructive.
From model theory we have two maps \\(\phi : \mathbb{R} \to\, ^\ast \mathbb{R} \\) and \\(\psi :\, ^\ast\mathbb{R} \to \mathbb{R} \\) such that:
- if \\(x \leq_{\mathbb{R}} y\\) then \\(\phi(x) \leq_{^\ast \mathbb{R}} \phi(y)\\)
- if \\(p \leq_{^\ast \mathbb{R}} q\\) then \\(\psi(q) \leq_{\mathbb{R}} \psi(q)\\)
- \\(\psi(\phi(x)) = x\\) and \\(\phi(\psi(p)) = p\\)
Now consider \\(\\{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \\}\\).
The hyperreals famously violate the [Archimedean property](https://en.wikipedia.org/wiki/Archimedean_property). Because of this \\(\bigwedge_{^\ast \mathbb{R}} \\{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \\}\\) does not exist.
On the other than if we consider \\( \bigwedge_{\mathbb{R}} \\{ \psi(x) \, : \, x \in \mathbb{R} \text{ and } 0 < x \\}\\), that *does* exist by the completeness of the real numbers (as it is bounded below by \\(\psi(0)\\)).
Hence
$$
\bigwedge_{\mathbb{R}} \\{ \psi(x) \, : \, x \in \mathbb{R} \text{ and } 0 < x \\} \neq \psi\left(\bigwedge_{^\ast\mathbb{R}} \\{ x \, : \, x \in \mathbb{R} \text{ and } 0 < x \\}\right)
$$
So \\(\psi\\) *cannot* be a complete lattice homomorphism, even though it is part of an order isomorphism.
However, just to complicate matters, I believe that \\(\phi\\) and \\(\psi\\) are a mere *lattice* isomorphism, preserving finite meets and joints. |
COMPARING QUANTITIES
Category : 7th Class
Learning Objectives:
RATIO
Ratio is comparison of two or more quantities of the same kind using division. It shown as a: b. The first term a is called antecedent and term b is called consequent.
IMPORTANT FACTS ABOUT RATIO:
\[e.g.\,\,4:7=\frac{4}{7}=\frac{4\times 3}{7\times 3}=\frac{12}{21}\]
\[\Rightarrow \,\,4:7=12:21\]
(i) Ratio cannot exist between height and weight.
(ii) We cannot write a ratio between the age of a student and marks obtained by him.
(b) Since ratio is a number, it has no units.
(c) To find the ratio of quantities of same kind, quantities should be in same unit.
e.g. To compare 3 : 5 and 2 : 3
\[\frac{3}{5}\]and\[\frac{2}{3}\]
LCM of 5 and 3 = 15
\[\frac{3}{5}=\frac{3}{5}\times \frac{3}{3}=\frac{9}{15}\]and \[\frac{2}{3}\times \frac{5}{5}=\frac{10}{15}\]
Since \[\frac{10}{15}>\frac{9}{15}\Rightarrow \frac{2}{3}>\frac{3}{5}\]
i.e. \[2:3>3:5\]
e.g. To convert the 15 : 35 in its lowest term
\[15=3\times 5\]
\[35=7\times 5\]
H.C.F.\[=5\]
Dividing both the terms by HCF
\[\frac{15\div 5}{35\div 5}=\frac{3}{7}=3:7\]
If a given quantity increases or decreases in the ratio\[a:b,\]then new quantity\[=\frac{b}{a}\]of the original quantity.
The fraction by which the original quantity is multiplied to get the new (increased) quantity is called the multiplying ratio (or factor).
\[\frac{\text{New}\left( \text{increased} \right)\text{quantity}}{\text{Original quantity}}=\text{Multiplying factor}\]
e.g. (i) To increase 24 kg in the ratio\[2:3.\]
New weight after increase\[=\frac{3}{2}\]of \[24=\frac{3}{2}\times 24=36\,kg.\]
(ii) To decrease Rs. 104 in the ratio \[8:5\]
New amount after decrease \[=\frac{5}{8}\times 104=Rs.\,\,65.\]
Note: Whenever we say that two numbers are in the ratio \[2:3,\] we can write them as \[2x\] and \[3x,\] similarly, three numbers in the ratio \[4:5:8\] can be written as \[4x,\,\,5x\] and \[8x.\] PROPORTION:
A proportion is an equation that states that two ratios are equal.
Four numbers are said to be in proportion if the ratio of the first two is equal to the ratio of the last two i.e. a, b, c and d are said to be in proportion if a : b = c : d.
This is expressed as a : b :: c : d. The first and fourth terms are called the extremes and second and third terms are called the means.
Product of extremes = Product of means
Three quantities, a b and c (of the same kind) are said to be in continued proportion
if \[a:b::b:c\]
\[\frac{a}{b}=\frac{b}{c}\Rightarrow {{b}^{2}}=ac\]
Here, b is called the mean proportional a and c are known as the first proportional and third proportional respectively.
Mean proportion between\[a\]and\[c=\sqrt{ac}\]
Note: In ratio both the terms should be of the same kind, but in proportion, the first two should be of the same kind and the last two should be of the same kind. Example:
(i) Determine the following numbers are in proportion or not 1.2, 2.7, 0.4, 0.9
Solution: Product of extremes\[=1.2\times 0.9=1.08\]
Product of means\[=2.7\times 0.4=1.08.\]
Product of extremes = Product of means
So these numbers are in proportion.
(ii) Find the fourth proportional to 4.8, 1.6, and 5.4.
Solution: Let the fourth proportion be \[x.\] Then 4.8, 1.6, 5.4, \[x\] are in proportion
\[\therefore \] Product of extremes = Product of means
i.e., \[4.8\times x=1.6\times 5.4\]
\[x=\frac{1.6\times 5.4}{4.8}\Rightarrow 1.8\]
(iii) Find the mean proportional between \[\frac{1}{4}\] and \[\frac{1}{25}\]
Solution: Let\[x\]be the mean proportional between \[\frac{1}{4}\] and \[\frac{1}{25}.\]
then \[{{x}^{2}}=\frac{1}{4}\times \frac{1}{25}=\frac{1}{100}\]
\[x=\sqrt{\frac{1}{100}}=\frac{1}{10}\]
\[\therefore \] \[\frac{1}{10}\] is the mean proportional between \[\frac{1}{4}\] and\[\frac{1}{25}.\]
(iv) Find the third proportional to i.e. 3.6, 1.8
Solution: Let the third proportional is\[x.\]Then 3.6 and 1.8 are in continued proportion.
\[\therefore \] \[\frac{3.6}{1.8}=\frac{1.8}{x}\]
\[3.6x=1.8\times 1.8\]
\[x=\frac{1.8\times 1.8}{3.6\times 10}\]
\[x=0.9\]
UNITARY METHOD
The method of finding the value of the required number of quantity by first finding the value of the unit quantity is called unitary method.
Example: A scooter consumes 28 liters of petrol in covering 2100 km. How much petrol will be needed to cover a distance of 3600 km? Solution: 2100 km can be covered in 28 liters of petrol
\[\therefore \] 1 km can be covered in\[\left( 28\times \frac{1}{2100} \right)\]liters
\[\therefore \] 3600 km can be covered in \[\frac{28}{2100}\times 3600\] liters = 48 liters.
PERCENTAGE
The word 'per cent' is an abbreviation of the Latin phrase 'per centum' which means per hundred or hundredths.
Thus, the term per cent means per hundred or for every hundred.
By a certain per cent we mean that many hundredths.
IMPORTANT FORMULAE
\[\text{Percentage}\,\text{increase}=\left( \frac{\text{Increase}\,\text{in}\,\text{quantity}}{\text{Original}\,\text{quantity}}\times 100 \right)%\]
\[\text{Percentage}\,\text{decrease}=\left( \frac{\text{Decrease}\,\text{in}\,\text{quantity}}{\text{Original}\,\text{quantity}}\times 100 \right)%\]
PROFIT AND LOSS
The cost price is abbreviated as C.P.
The selling price is abbreviated as S.P.
IMPORTANT FORMULAE Example: By selling a washing machine for Rs. 7200, Rajesh loses 10%. Find the Cost Price of the washing machine. Solution: Let cost price of washing machine be\[Rs.\,\,'x'\]
Given,
SP = Rs. 7200, % Loss = 10%
As % loss given, so CP > SP
\[\Rightarrow \,\,\text{SP}=\text{CP}-\text{Loss}\]
\[\Rightarrow \,\,7200=x-10%\,\,of\,x\]
\[\Rightarrow \,\,x-\frac{10}{100}x=7200\]
\[\Rightarrow \,\,\frac{9x}{10}=7200\]
\[\Rightarrow \,\,x=\frac{7200\times 10}{9}=8000\]
\[\therefore \] Cost price of washing machine = Rs. 8000
IMPORTANT FORMULAE Discount: Sometimes to increase the sales or dispose of the old stock, dealer or seller offers his goods at reduced prices. This reduction in price offered by dealer/seller is called discount. Marked Price (M.P.): The printed price or the tagged price of an article is called its marked price (M.P).
It is also called list price.
Discount is always calculated on M.P. of the article.
S.P. = Marked price\[-\]Discount.
For example, two discounts of 15% and 4% are equivalent to a single discount of
\[\left( 15+4-\frac{15\times 4}{100} \right)=19-\frac{60}{1000}=19-\frac{3}{5}=19-0.6=18.4%\]
SIMPLE INTEREST For example, a rate of '10% per annum', means Rs. 10 on Rs. 100 for 1 year.
IMPORTANT FORMULAE
S.I. = Simple interest, P = Principal amount, R = Rate of interest, T = Time
Example: At what rate percent by simple interest, will a sum of money double itself in 5 years 4 months? Solution: Let P = Rs.\[x\]
Amount A = Rs.\[2x\]
\[\therefore \] S.I. = A\[-\]P = Rs.\[2x\]\[-\]Rs.\[x\] = Rs.\[x\]
\[T=5\]years 4 months = \[5\frac{4}{12}\]years = \[5\frac{1}{3}\] years = \[\frac{16}{3}\] years
Let R be the rate percent per annum.
Using \[R=\frac{S.I.\times 100}{P\times T},\]
We get \[R=\frac{x\times 100}{x\times \frac{16}{3}}=\frac{300}{16}=18.75.\]
Hence required rate = 18.75% p.a.
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Suppose that $A$ is a list with $n$ elements where $n$ is even. I want to write a function that returns all pairs $(A_1,A_2)$ where the sets $A_1$ and $A_2$ each have length $\frac{n}{2}$, $A_1 \cap A_2=\emptyset$, and $A_1,A_2 \subset A$.
I'm assuming in A_1 A_2 pair, order matters:
set = Range[6];Transpose[{#, Reverse@#}] & @ Subsets[#, {Length[#]/2}] & @ set
{{{1, 2, 3}, {4, 5, 6}}, {{1, 2, 4}, {3, 5, 6}}, {{1, 2, 5}, {3, 4, 6}}, {{1, 2, 6}, {3, 4, 5}}, {{1, 3, 4}, {2, 5, 6}}, {{1, 3, 5}, {2, 4, 6}}, {{1, 3, 6}, {2, 4, 5}}, {{1, 4, 5}, {2, 3, 6}}, {{1, 4, 6}, {2, 3, 5}}, {{1, 5, 6}, {2, 3, 4}}, {{2, 3, 4}, {1, 5, 6}}, {{2, 3, 5}, {1, 4, 6}}, {{2, 3, 6}, {1, 4, 5}}, {{2, 4, 5}, {1, 3, 6}}, {{2, 4, 6}, {1, 3, 5}}, {{2, 5, 6}, {1, 3, 4}}, {{3, 4, 5}, {1, 2, 6}}, {{3, 4, 6}, {1, 2, 5}}, {{3, 5, 6}, {1, 2, 4}}, {{4, 5, 6}, {1, 2, 3}}}
If pair isn't ordered you can take halof of above or spare some memory doing:
With[{l = Length[#]}, Transpose[{#, Reverse@#2}] & @@ Partition[Subsets[#, {l/2}], Binomial[l, l/2]/2]] &@set
@Wizard, @Kuba, Sorry I do not have sufficient reputation for adding comments, therefore I take the liberty to post it as an Answer. Your solution result not only contains (A1,A2) but also contains (A2,A1), therefore // Take[#, Length @ #/2] & needs to be added in order to take only (A1,A2). :)
If order does not matter,
Clear[splitList]splitList[ list_?(VectorQ[#] && EvenQ[Length[list]] &)] := Module[{a1, a2, a3, len = Length[list]}, a1 = Subsets[list, {len/2}]; a2 = Complement[list, #] & /@ a1; a3 = Transpose[{a1, a2}]; Union[a3, SameTest -> (Sort[Sort /@ #1] === Sort[Sort /@ #2] &)]]splitList[{e1, e2, e3, e4, e5, e6}]
{{{e1, e2, e3}, {e4, e5, e6}}, {{e1, e2, e4}, {e3, e5, e6}}, {{e1, e2, e5}, {e3, e4, e6}}, {{e1, e2, e6}, {e3, e4, e5}}, {{e1, e3, e4}, {e2, e5, e6}}, {{e1, e3, e5}, {e2, e4, e6}}, {{e1, e3, e6}, {e2, e4, e5}}, {{e1, e4, e5}, {e2, e3, e6}}, {{e1, e4, e6}, {e2, e3, e5}}, {{e1, e5, e6}, {e2, e3, e4}}}
splitList[{j, a, l, b}]
{{{a, b}, {j, l}}, {{a, l}, {b, j}}, {{j, a}, {b, l}}}
I would suggest using a combination of
Subsets and
Complement, two basic set operations in Mathematica. Here is an example code, you should be able to transfer that to any sort of list on your own:
(*create list*)list = {1, 2, 3, 4, 5, 6, 7, 8};(*create all subsets of length n/2*)subsets = Subsets[list, {Length[list]/2}];(*create the corresponding complements*)complements = Complement[list, #] & /@ subsets; (*combine each subset with its corresponding complement*)output = Transpose@{subsets, complements};
output[[i]] then contains one specific pair of complements $(A_1,A_2)$. |
According to
Dynamic Hedging: Managing Vanilla and Exotic Options (Taleb, 1997), the Parkison volatility estimator has several meaningful properties. It is defined
$$P=\sqrt{\frac{1}{n}\sum_{i=1}^{n}\frac{1}{4\log\left(2\right)}\left(\log\left(\frac{S_{H,i}}{S_{L,i}}\right)\right)^{2}}$$
where $S_{H}$ and $S_{L}$ are the «close-to-close registered high and the registered low respectively in any particular time frame». From Taleb:
An important use of the Parkinson number is the assessment of the distribution of prices during the day as well as a better understanding of market dynamics. Comparing the Parkinson number and the periodically sampled volatility helps traders understand the mean reversion in the market as well as the distribution of stop-losses. Some clear rules can be derived from that information. Comparing the Parkinson number $P$ with the definition of periodically sampled historical volatility gives this result:
$$P=1.67\sigma'$$
Then Taleb adds:
Such measurement cannot be used to compare close-to-close volatility with intraday high/low. It can compare 24-hour high/low to data sampled every day at the same time. For markets, like most equities, which trade during the day only, it is better to use open-to-close volatility.
Whats $\sigma'$? It's defined as the noncentered volatility estimator:
$$\sigma'=\sqrt{\frac{1}{n}\sum_{t=1}^{n}x_{t}^{2}}$$
So Taleb suggests to set $x_{t}=\log\left(C_{t}\right)-\log\left(O_{t}\right)$ from a typical OHLC time series and then plot the ratio $z_{t}=P_{t}/\sigma'_{t}$: when $z_{t}>1.67$ we're in a mean reverting market, trending elsewhere. A figure shows that the Parkinson number ratio to the volatility is «strikingly convincing» because there seems to be a clear bias in favor of a wider high/low range than assumed by random walk when applying the ratio to U.S. Treasury bond futures from Aug-1992 to May-1995:
The problem arises when trying to reproduce such results. I downloaded many time series from Bloomberg, but everytime it seems that $P_{t}<1.67\sigma'_{t}$. Moreover, I picked even the same time series over the same period and my calculatiosn are really different:
$1.67$ seems a cap rather than a floor. So I'm going to share my
R snippet to see what's wrong with my code.
ohlc is the OHLC time series and I've loaded
quantmod and
magrittr packages. Then:
parkinson.vol <- TTR::volatility(OHLC = ohlc, n = 20, calc = 'parkinson')taleb.vol <- OpCl(ohlc) ^ 2 %>% runMean(n = 20) %>% sqrt() * sqrt(260)parkinson.ratio <- parkinson.vol / taleb.vol
Furthermore, Taleb says that:
Additional testing by the author shows the bias to be permanent in close to the 20 markets surveyed.
Are you able to reproduce Taleb's results? |
Gamma-function $\Gamma$-function
2010 Mathematics Subject Classification:
Primary: 33B15 Secondary: 33B2033D05 [MSN][ZBL]$\newcommand{\abs}[1]{\left|#1\right|}\newcommand{\Re}{\mathop{\mathrm{Re}}}\newcommand{\Im}{\mathop{\mathrm{Im}}}$
A transcendental function $\Gamma(z)$ that extends the values of the factorial $z!$ to any complex number $z$. It was introduced in 1729 by L. Euler in a letter to Ch. Goldbach, using the infinite product $$ \Gamma(z) = \lim_{n\rightarrow\infty}\frac{n!n^z}{z(z+1)\ldots(z+n)} = \lim_{n\rightarrow\infty}\frac{n^z}{z(1+z/2)\ldots(1+z/n)}, $$ which was used by L. Euler to obtain the integral representation (Euler integral of the second kind, cf. Euler integrals) $$ \Gamma(z) = \int_0^\infty x^{z-1}e^{-x} \rd x, $$ which is valid for $\Re z > 0$. The multi-valuedness of the function $x^{z-1}$ is eliminated by the formula $x^{z-1}=e^{(z-1)\ln x}$ with a real $\ln x$. The symbol $\Gamma(z)$ and the name gamma-function were proposed in 1814 by A.M. Legendre.
If $\Re z < 0$ and $-k-1 < \Re z < -k$, $k=0,1,\ldots$, the gamma-function may be represented by the Cauchy–Saalschütz integral: $$ \Gamma(z) = \int_0^\infty x^{z-1} \left( e^{-x} - \sum_{m=0}^k (-1)^m \frac{x^m}{m!} \right) \rd x. $$ In the entire plane punctured at the points $z=0,-1,\ldots $, the gamma-function satisfies a Hankel integral representation: \begin{equation} \label{eq1} \Gamma(z) = \frac{1}{e^{2\pi iz} - 1} \int_C s^{z-1}e^{-s} \rd s, \end{equation} where $s^{z-1} = e^{(z-1)\ln s}$ and $\ln s$ is the branch of the logarithm for which $0 < \arg\ln s < 2\pi$; the contour $C$ is represented in Figure 1. It is seen from the Hankel representation that $\Gamma(z)$ is a meromorphic function. At the points $z_n = -n$, $n=0,1,\ldots$ it has simple poles with residues $(-1)^n/n!$.
Contents Fundamental relations and properties of the gamma-function.
1) Euler's functional equation: $$ z\Gamma(z) = \Gamma(z+1), $$ or $$ \Gamma(z) = \frac{1}{z\ldots(z+n)}\Gamma(z+n+1); $$ $\Gamma(1)=1$, $\Gamma(n+1) = n!$ if $n$ is an integer; it is assumed that $0! = \Gamma(1) = 1$.
2) Euler's completion formula: $$ \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z}. $$ In particular, $\Gamma(1/2)=\sqrt{\pi}$; $$ \Gamma\left(n+\frac{1}{2}\right) = \frac{1.3\ldots(2n-1)}{2^n}\sqrt{\pi} $$ if $n>0$ is an integer; $$ \abs{\Gamma\left(\frac{1}{2} + iy\right)}^2 = \frac{\pi}{\cosh y\pi}, $$ where $y$ is real.
3) Gauss' multiplication formula: $$ \prod_{k=0}^{m-1} \Gamma\left( z + \frac{k}{m} \right) = (2\pi)^{(m-1)/2}m^{(1/2)-mz}\Gamma(mz), \quad m = 2,3,\ldots $$ If $m=2$, this is the Legendre duplication formula.
4) If $\Re z \geq \delta > 0$ or $\abs{\Im z} \geq \delta > 0$, then $\ln\Gamma(z)$ can be asymptotically expanded into the Stirling series: $$ \ln\Gamma(z) = \left(z-\frac{1}{2}\right)\ln z - z + \frac{1}{2}\ln 2\pi + \sum_{n=1}^m \frac{B_{2n}}{2n(2n-1)z^{2n-1}} + O\bigl(z^{-2m-1}\bigr), \quad m = 1,2,\ldots, $$ where $B_{2n}$ are the Bernoulli numbers. It implies the equality $$ \Gamma(z) = \sqrt{2\pi} z^{z-1/2} z^{-z} \left( 1 + \frac{1}{12}z^{-1} + \frac{1}{288}z^{-2} - \frac{139}{51840}z^{-3} - \frac{571}{2488320}z^{-4} + O\bigl(z^{-5}\bigr) \right). $$ In particular, $$ \Gamma(1+x) = \sqrt{2\pi} x^{x+1/2} e^{-x + \theta/12x}, \quad 0 < \theta < 1. $$ More accurate is Sonin's formula [So]: $$ \Gamma(1+x) = \sqrt{2\pi} x^{x+1/2} e^{-x + 1/12(x+\theta)}, \quad 0 < \theta < 1/2. $$
5) In the real domain, $\Gamma(x) > 0$ for $x > 0$ and it assumes the sign $(-1)^{k+1}$ on the segments $-k-1 < x < -k$, $k = 0,1,\ldots$ (Fig. b).
Figure: g043310b
The graph of the function $ $.
For all real $x$ the inequality $$ \Gamma\Gamma^{\prime\prime} > \bigl(\Gamma^\prime\bigr)^2 \geq 0 $$ is valid, i.e. all branches of both $\abs{\Gamma(x)}$ and $\ln\abs{\Gamma(x)}$ are convex functions. The property of logarithmic convexity defines the gamma-function among all solutions of the functional equation $$ \Gamma(1+x) = x\Gamma(x) $$ up to a constant factor (see also the Bohr–Mollerup theorem).
For positive values of $x$ the gamma-function has a unique minimum at $x=1.4616321\ldots$ equal to $0.885603\ldots$. The local minima of the function $\abs{\Gamma(x)}$ form a sequence tending to zero as $x\rightarrow -\infty$.
Figure: g043310c
The graph of the function $ $.
6) In the complex domain, if $\Re z > 0$, the gamma-function rapidly decreases as $\abs{\Im z} \rightarrow \infty$, $$ \lim_{\abs{\Im z} \rightarrow \infty} \abs{\Gamma(z)}\abs{\Im z}^{(1/2)-\Re z}e^{\pi\abs{\Im z}/2} = \sqrt{2\pi}. $$
7) The function $1/\Gamma(z)$ (Fig. c) is an entire function of order one and of maximal type; asymptotically, as $r \rightarrow \infty$, $$ \ln M(r) \sim r \ln r, $$ where $$ M(r) = \max_{\abs{z} = r} \frac{1}{\abs{\Gamma(z)}}. $$ It can be represented by the infinite Weierstrass product: $$ \frac{1}{\Gamma(z)} = z e^{\gamma z} \prod_{n=1}^\infty \left(\left( 1 + \frac{z}{n} \right) e^{-z/n} \right), $$ which converges absolutely and uniformly on any compact set in the complex plane ($\gamma$ is the Euler constant). A Hankel integral representation is valid: $$ \frac{1}{\Gamma(z)} = \frac{1}{2\pi i} \int_{C'} e^s s^{-z} \rd s, $$ where the contour $C'$ is shown in Fig. d.
Figure: g043310d
$ $
G.F. Voronoi [Vo] obtained integral representations for powers of the gamma-function.
In applications, the so-called poly gamma-functions — $k$th derivatives of $\ln\Gamma(z)$ — are of importance. The function (Gauss' $\psi$-function) $$ \psi(z) = \frac{\mathrm{d}}{\mathrm{d}z}\ln\Gamma(z) = \frac{\Gamma'(z)}{\Gamma(z)} = -\gamma + \sum_{n=0}^\infty \frac{z-1}{(n+1)(z+n)} = -\gamma + \int_0^1 \frac{1 - (1-t)^{z-1}}{t} \rd t $$ is meromorphic, has simple poles at the points $z=0,-1,\ldots$ and satisfies the functional equation $$ \psi(z+1) - \psi(z) = \frac{1}{z}. $$ The representation of $\psi(z)$ for $\abs{z}<1$ yields the formula $$ \ln\Gamma(1+z) = -\gamma z + \sum_{k=2}^\infty \frac{(-1)^k S_k}{k} z^k, $$ where $$ S_k = \sum_{n=1}^\infty n^{-k}. $$ This formula may be used to compute $\Gamma(z)$ in a neighbourhood of the point $z=1$.
For other poly gamma-functions see [BaEr]. The incomplete gamma-function is defined by the equation $$ I(x,y) = \int_0^y e^{-t}t^{x-1} \rd t. $$ The functions $\Gamma(z)$ and $\psi(z)$ are transcendental functions which do not satisfy any linear differential equation with rational coefficients (Hölder's theorem).
The exceptional importance of the gamma-function in mathematical analysis is due to the fact that it can be used to express a large number of definite integrals, infinite products and sums of series (see, for example, Beta-function). In addition, it is widely used in the theory of special functions (the hypergeometric function, of which the gamma-function is a limit case, cylinder functions, etc.), in analytic number theory, etc.
References
[An] A. Angot, "Compléments de mathématiques. A l'usage des ingénieurs de l'electrotechnique et des télécommunications", C.N.E.T. (1957) [BaEr] H. Bateman (ed.) A. Erdélyi (ed.), Higher transcendental functions, 1. The gamma function. The hypergeometric functions. Legendre functions, McGraw-Hill (1953) [Bo] N. Bourbaki, "Elements of mathematics. Functions of a real variable", Addison-Wesley (1976) (Translated from French) [JaEm] E. Jahnke, F. Emde, "Tables of functions with formulae and curves", Dover, reprint (1945) (Translated from German) [Ni] N. Nielsen, "Handbuch der Theorie der Gammafunktion", Chelsea, reprint (1965) [So] N.Ya. Sonin, "Studies on cylinder functions and special polynomials", Moscow (1954) (In Russian) [Vo] G.F. Voronoi, "Studies of primitive parallelotopes", Collected works, 2, Kiev (1952) pp. 239–368 (In Russian) [WhWa] E.T. Whittaker, G.N. Watson, "A course of modern analysis", Cambridge Univ. Press (1952) Comments
The $q$-analogue of the gamma-function is given by $$ \Gamma_q(z) = (1-q)^{1-z} \prod_{k=1}^\infty \frac{1-q^{k+1}}{1-q^{k+z}}, \quad z \neq 0,-1,-2,\ldots;\quad 0<q<1, $$ cf. [As]. Its origin goes back to E. Heine (1847) and D. Jackson (1904).
References
[Ar] E. Artin, "The gamma function", Holt, Rinehart & Winston (1964) [As] R. Askey, "The $q$-Gamma and $q$-Beta functions" Appl. Anal., 8 (1978) pp. 125–141 How to Cite This Entry:
Gamma-function.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Gamma-function&oldid=25650 |
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In this chapter we learned about left and right adjoints, and about joins and meets. At first they seemed like two rather different pairs of concepts. But then we learned some deep relationships between them. Briefly:
Left adjoints preserve joins, and monotone functions that preserve enough joins are left adjoints.
Right adjoints preserve meets, and monotone functions that preserve enough meets are right adjoints.
Today we'll conclude our discussion of Chapter 1 with two more bombshells:
Joins
are left adjoints, and meets are right adjoints.
Left adjoints are right adjoints seen upside-down, and joins are meets seen upside-down.
This is a good example of how category theory works. You learn a bunch of concepts, but then you learn more and more facts relating them, which unify your understanding... until finally all these concepts collapse down like the core of a giant star, releasing a supernova of insight that transforms how you see the world!
Let me start by reviewing what we've already seen. To keep things simple let me state these facts just for posets, not the more general preorders. Everything can be generalized to preorders.
In Lecture 6 we saw that given a left adjoint \( f : A \to B\), we can compute its right adjoint using joins:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$ Similarly, given a right adjoint \( g : B \to A \) between posets, we can compute its left adjoint using meets:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ In Lecture 16 we saw that left adjoints preserve all joins, while right adjoints preserve all meets.
Then came the big surprise: if \( A \) has all joins and a monotone function \( f : A \to B \) preserves all joins, then \( f \) is a left adjoint! But if you examine the proof, you'l see we don't really need \( A \) to have
all joins: it's enough that all the joins in this formula exist:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$Similarly, if \(B\) has all meets and a monotone function \(g : B \to A \) preserves all meets, then \( f \) is a right adjoint! But again, we don't need \( B \) to have
all meets: it's enough that all the meets in this formula exist:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ Now for the first of today's bombshells: joins are left adjoints and meets are right adjoints. I'll state this for binary joins and meets, but it generalizes.
Suppose \(A\) is a poset with all binary joins. Then we get a function
$$ \vee : A \times A \to A $$ sending any pair \( (a,a') \in A\) to the join \(a \vee a'\). But we can make \(A \times A\) into a poset as follows:
$$ (a,b) \le (a',b') \textrm{ if and only if } a \le a' \textrm{ and } b \le b' .$$ Then \( \vee : A \times A \to A\) becomes a monotone map, since you can check that
$$ a \le a' \textrm{ and } b \le b' \textrm{ implies } a \vee b \le a' \vee b'. $$And you can show that \( \vee : A \times A \to A \) is the left adjoint of another monotone function, the
diagonal
$$ \Delta : A \to A \times A $$sending any \(a \in A\) to the pair \( (a,a) \). This diagonal function is also called
duplication, since it duplicates any element of \(A\).
Why is \( \vee \) the left adjoint of \( \Delta \)? If you unravel what this means using all the definitions, it amounts to this fact:
$$ a \vee a' \le b \textrm{ if and only if } a \le b \textrm{ and } a' \le b . $$ Note that we're applying \( \vee \) to \( (a,a') \) in the expression at left here, and applying \( \Delta \) to \( b \) in the expression at the right. So, this fact says that \( \vee \) the left adjoint of \( \Delta \).
Puzzle 45. Prove that \( a \le a' \) and \( b \le b' \) imply \( a \vee b \le a' \vee b' \). Also prove that \( a \vee a' \le b \) if and only if \( a \le b \) and \( a' \le b \).
A similar argument shows that meets are really right adjoints! If \( A \) is a poset with all binary meets, we get a monotone function
$$ \wedge : A \times A \to A $$that's the
right adjoint of \( \Delta \). This is just a clever way of saying
$$ a \le b \textrm{ and } a \le b' \textrm{ if and only if } a \le b \wedge b' $$ which is also easy to check.
Puzzle 46. State and prove similar facts for joins and meets of any number of elements in a poset - possibly an infinite number.
All this is very beautiful, but you'll notice that all facts come in pairs: one for left adjoints and one for right adjoints. We can squeeze out this redundancy by noticing that every preorder has an "opposite", where "greater than" and "less than" trade places! It's like a mirror world where up is down, big is small, true is false, and so on.
Definition. Given a preorder \( (A , \le) \) there is a preorder called its opposite, \( (A, \ge) \). Here we define \( \ge \) by
$$ a \ge a' \textrm{ if and only if } a' \le a $$ for all \( a, a' \in A \). We call the opposite preorder\( A^{\textrm{op}} \) for short.
I can't believe I've gone this far without ever mentioning \( \ge \). Now we finally have really good reason.
Puzzle 47. Show that the opposite of a preorder really is a preorder, and the opposite of a poset is a poset. Puzzle 48. Show that the opposite of the opposite of \( A \) is \( A \) again. Puzzle 49. Show that the join of any subset of \( A \), if it exists, is the meet of that subset in \( A^{\textrm{op}} \). Puzzle 50. Show that any monotone function \(f : A \to B \) gives a monotone function \( f : A^{\textrm{op}} \to B^{\textrm{op}} \): the same function, but preserving \( \ge \) rather than \( \le \). Puzzle 51. Show that \(f : A \to B \) is the left adjoint of \(g : B \to A \) if and only if \(f : A^{\textrm{op}} \to B^{\textrm{op}} \) is the right adjoint of \( g: B^{\textrm{op}} \to A^{\textrm{ op }}\).
So, we've taken our whole course so far and "folded it in half", reducing every fact about meets to a fact about joins, and every fact about right adjoints to a fact about left adjoints... or vice versa! This idea, so important in category theory, is called
duality. In its simplest form, it says that things come on opposite pairs, and there's a symmetry that switches these opposite pairs. Taken to its extreme, it says that everything is built out of the interplay between opposite pairs.
Once you start looking you can find duality everywhere, from ancient Chinese philosophy:
to modern computers:
But duality has been studied very deeply in category theory: I'm just skimming the surface here. In particular, we haven't gotten into the connection between adjoints and duality!
This is the end of my lectures on Chapter 1. There's more in this chapter that we didn't cover, so now it's time for us to go through all the exercises. |
This 18 page article seems pretty good as a historical account of who was responsible for what.
In general, the push for rigor is usually in response to a failure to be able to demonstrate the kinds of results one wishes to. It's usually relatively easy to demonstrate that there exist objects with certain properties, but you need precise definitions to prove that no such object exists. The classic example of this is non-computable problems and Turing Machines. Until you sit down and say "this precisely and nothing else is what it means to be solved by computation" it's impossible to prove that something isn't a computation, so when people start asking "is there an algorithm that does $\ldots$?" for questions where the answer "should be" no, you suddenly need a precise definition. Similar things happened with real analysis.
In real analysis, as mentioned in an excellent comment, there was a shift in what people's conception of the notion of a function was. This broadened conception of a function suddenly allows for a number of famous "counter example" functions to be constructed. These often that require a reasonably rigorous understanding of the topic to construct or to analyze. The most famous is the everywhere continuous nowhere differentiable Weierstrass function. If you don't have a very precise definition of continuity and differentiability, demonstrating that that function is one and not the other is extremely hard. The quest for weird functions with unexpected properties and combinations of properties was one of the driving forces in developing precise conceptions of those properties.
Another topic that people were very interested in was infinite series. There are lots of weird results that can crop up if you're not careful with infinite series, as shown by the now famously cautionary theorem:
Theorem (Summation Rearrangement Theorem): Let $a_n$ be a sequence such that $\sum a_n$ converges conditionally. Then for every $x$ there is some $b_n$ that is a reordering of $a_n$ such that $\sum b_n=x$.
This theorem means you have to be very careful dealing with infinite sums, and for a long time people weren't and so started deriving results that made no sense. Suddenly the usual free-wheeling algebraic manipulation approach to solving infinite sums was no longer okay, because sometimes doing so changed the value of the sum. Instead, a more rigorous theory of summation manipulation, as well as concepts such as uniform and absolute convergence had to be developed.
Here's an example of an problemsurrounding an infinite product created by Euler:
Consider the following formula:
$$x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)$$
Does this expression even make sense? Assuming it does, does this equal $\sin(x)$ or $\sin(x)e^x$? How can you tell (notice that both functions have the same zeros as this sum, and the same relationship to their derivative)? If it doesn't equal $\sin(x)e^x$ (which it doesn't, it really does equal $\sin(x)$) how can we modify it so that it does?
Questions like this were very popular in the 1800s, as mathematicians were notably obsessed with infinite products and summations. However, most questions of this form require a very sophisticated understanding of analysis to handle (and weren't handled particularly well by the tools of the previous century). |
Proceedings of the Japan Academy, Series A, Mathematical Sciences Proc. Japan Acad. Ser. A Math. Sci. Volume 75, Number 7 (1999), 99-102. Infinitesimal locally trivial deformation spaces of compact complex surfaces with ordinary singularities Abstract
Let $S$ be a compact complex surface with
ordinary singularities. We denote by $\Theta_S$ the sheaf of germs of holomorphic tangent vector fields on $S$. In this paper we shall give a description of the cohomology $H^1(S, \Theta_S)$, which is called the infinitesimal locally trivial deformation space of $S$, using a 2-cubic hyper-resolution of $S$ in the sense of F. Guillén, V. Navarro Aznar et al. ([1]). As a by-product, we shall show that the natural homomorphisim $H^1(S, \Theta_S)\rightarrow H^1(X, \Theta_X(-\log D_X))$ is injective under some condition, where $X$ is the (non-singular) normal model of $S$, $D_X$ the inverse image of the double curve $D_S$ of $S$ by the normalization map $f\colon X\rightarrow S$, and $\Theta_X(-\log D_X)$ the sheaf of germs of logarithmic tangent vector fields along $D_X$ on $X$. Note that the cohomology $H^1(X, \Theta_X(-\log D_X))$ is nothing but the infinitesimal locally trivial deformation space of a pair $(X, D_X)$. Article information Source Proc. Japan Acad. Ser. A Math. Sci., Volume 75, Number 7 (1999), 99-102. Dates First available in Project Euclid: 23 May 2006 Permanent link to this document https://projecteuclid.org/euclid.pja/1148393857 Digital Object Identifier doi:10.3792/pjaa.75.99 Mathematical Reviews number (MathSciNet) MR1729852 Zentralblatt MATH identifier 0964.32026 Citation
Tsuboi, Shoji. Infinitesimal locally trivial deformation spaces of compact complex surfaces with ordinary singularities. Proc. Japan Acad. Ser. A Math. Sci. 75 (1999), no. 7, 99--102. doi:10.3792/pjaa.75.99. https://projecteuclid.org/euclid.pja/1148393857 |
$$3^5 + {1\over3^5}=?$$
My first instinct was to rewrite the second term as $3^{-5}$. Since the base is $3$, rewrite as $3^{5+-5}$. It simplifies to $3^0= 1$. Apparently this is incorrect. Can anyone please explain why?
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$$3^5 + {1\over3^5}=?$$
My first instinct was to rewrite the second term as $3^{-5}$. Since the base is $3$, rewrite as $3^{5+-5}$. It simplifies to $3^0= 1$. Apparently this is incorrect. Can anyone please explain why?
This is incorrect because the exponent rule that you were thinking of is: $$a^b\cdot a^c = a^{b + c}$$ So if you had $3^{5}\cdot 3^{-5}$ then you could use that rule: $$3^{5}\cdot 3^{-5} = 3^{0}$$.
As has already been pointed out $a^b+a^c\neq a^{b+c}$. That means in your case the best you can do is try and get a common denominator for both numbers and add them. Try and see if you know how.
There is not much you can do in terms of simplifying these exponents. You could rewrite $3^{5}$ as $\frac{3^{10}}{3^{5}}$ then simplify with common denominators to
$$\frac{3^{10}+1}{3^{5}}$$ |
In a certain fluid mechanics problem, the following equation for a stream function $f(R)$ must be solved
$$A^2 + f f'' = (f' )^2, \quad 0 < R < 1/2,$$
with boundary conditions $f(1/2) = -1$ and $f'(1/2) = 0$. $R = r^2/2$ is a re-scaled version of the radius. In this problem, the axial pressure gradient $A^2$ can be treated as an eigenvalue, which enforces the axis $R=0$ to be a stream line, thus providing the additional condition $f(0) = 0$.
I am able to identify a fairly simple solution $f = - \sin{(\pi R)}$ for which $A = \pi$ necessarily. I am amazed by the simplicity of the solution and I am unable to find it using standard techniques...
If, for instance, I ask Mathematica for a solution for $A \neq \pi$ (without boundary conditions), it gives:
\begin{align*} & f = -\frac{1}{2} e^{-e^{c_1} R-2 c_1-e^{c_1} c_2} \left(A^2 e^{2 c_1}-e^{2 e^{c_1} \left(c_2+R\right)}\right) \\ & f = \frac{1}{2} \left(e^{-e^{c_1} R-2 c_1-e^{c_1} c_2}-A^2 e^{e^{c_1} R+e^{c_1} c_2}\right) \end{align*}
which is far from amenable for imposing boundary conditions. Is there any simpler way to find the simple solution $f = -\sin{\pi R}$?
Thanks in advance!
Edit
I have just discovered that, if instead of $R = r^2/2$ you introduce $R = A r^2 /2$, then the problem reduces to
$$1 + f f'' = (f' )^2, \quad 0 < R < A/2,$$
whose solution Mathematica gives in a simpler way:
$$f = \pm a_1 \sinh[(r+a_2)/a_1] $$
or, equivalently,
$$f = \pm b_1 \sin[(r+b_2)/b_1] $$
where, I guess, $a_1$ and $a_2$ become complex valued once I impose the boundary conditions.
Now we are left with imposing the BCs, but I continue to have doubts about getting to this form of solution... |
The Method of Successive Approximations
Consider the initial value problem of solving the first order differential equation $\frac{dy}{dt} = f(t, y)$ with the initial condition $y(t_0) = y_0$. Recall that if $f$ and $\frac{\partial f}{\partial y}$ are both continuous on an open rectangle $I = (\alpha, \beta) \times (\gamma, \delta)$ and if $(t_0, y_0) \in I$ then for some interval $(t_0 - h, t_0 + h) \subseteq (\alpha, \beta)$ there exists a unique solution $y = \phi(t)$ to the initial value problem described above.
We are now going to look at a new technique of finding such unique solutions to first order differential equations known as the Method of Successive Approximations (or Picard's Iterative Method). We assume that we want to solve the differential equation $\frac{dy}{dt} = f(t, y)$ with the initial condition $y(0) = 0$. This assumption will make the calculations that follow much simpler, and furthermore, we can always transform the differential equation $\frac{dy}{dt} = f(y, t)$ with the initial condition $y(t_0) = y_0$ into a different differential equation with the initial condition $y(0) = 0$ using substituions (by letting $s = t - t_0$ and $v = y - y_0$ as you should verify).
The theorem for the existence of a unique solution $y = \phi (t)$ to this differential equation can be rephrased as such that if $f$ and $\frac{\partial f}{\partial y}$ are continuous on a rectangle $R$ such that $-a \leq t \leq a$ and $-b \leq y \leq b$ then there exists an interval $(-h, h) \subseteq (-a, a)$ such that a unique solution $y = \phi (t)$ exists. Suppose that these conditions hold. Then let $y = \phi (t)$ be the unique solution. Then:(1)
This is a function in terms of the variable $t$ only. Suppose that we integrate this function starting at $0$ to an arbitrary value of $t$. We then get that:(2)
We should note that if $y = \phi (t)$ satisfies the above integral equation, then it also satisfies the initial value problem from earlier since by The Fundamental Theorem of Calculus we have that $\frac{d \phi}{dt} = f(t, \phi(t)) = f(t, y)$ is a solution to the differential equation and $\phi (0) = \int_0^0 (s, \phi(s)) \: ds = 0$ shows that $y = \phi(t)$ satisfies the initial condition.
Now consider the simplest function:(3)
Clearly this function satisfies the initial condition (since $\phi_0(0) = 0$), though this function need not be a solution to our differential equation. So $\phi_0(t) = 0$ approximates the unique solution to our differential equation (though likely not that well). For a closer approximation to $y = \phi (t)$, consider the following functions obtained recursively:(4)
Notice that if for some $k = 0, 1, 2, ...$ we have that $\phi_k(t) = \phi_{k+1}(t)$, then $y = \phi_k(t)$ will be equal to the solution to our differential equation. This can easily be seen with some algebraic manipulation:(5)
Unfortunately, most of the time $\phi_k(t) \neq \phi_{k+1}(t)$ for any $k$. In such cases, we will want to consider the infinite sequence of functions $\phi_n$ that approximate the unique solution $y = \phi(t)$ to our initial value problem.(6)
The following theorem says that the limit of this sequence of approximations will be equal to our unique solution.
Theorem 1: If $\frac{dy}{dt} = f(t, y)$ is a first order differential equation with the initial condition $y(0) = 0$ and if $f$ and $\frac{\partial f}{\partial y}$ are both continuous on some rectangle $R$ for which $-a \leq t \leq a$ and $-b \leq y \leq b$ then $\lim_{n \to \infty} \phi_n = \lim_{n \to \infty} \int_0^t f(s, \phi_{n-1}(s)) \: ds = \phi(t)$ where $y = \phi(t)$ is the guaranteed unique solution contained in the interval $(-h, h) \subseteq (-a, a)$.
We will not prove Theorem 1 as the proof is a bit out of our scope. |
Title Stability of planar switched systems: the linear single input case Publication Type Journal Article Year of Publication 2002 Authors Boscain, U Journal SIAM J. Control Optim. 41 (2002), no. 1, 89-112 Abstract
We study the stability of the origin for the dynamical system $\\\\dot x(t)=u(t)Ax(t)+(1-u(t))Bx(t),$ where A and B are two 2 × 2 real matrices with eigenvalues having strictly negative real part, $x\\\\in {\\\\mbox{{\\\\bf R}}}^2$, and $u(.):[0,\\\\infty[\\\\to[0,1]$ is a completely random measurable function. More precisely, we find a (coordinates invariant) necessary and sufficient condition on A and B for the origin to be asymptotically stable for each function u(.). The result is obtained without looking for a common Lyapunov function but studying the locus in which the two vector fields Ax and Bx are collinear. There are only three relevant parameters: the first depends only on the eigenvalues of A, the second depends only on the eigenvalues of B, and the third contains the interrelation among the two systems, and it is the cross ratio of the four eigenvectors of A and B in the projective line CP1. In the space of these parameters, the shape and the convexity of the region in which there is stability are studied.
URL http://hdl.handle.net/1963/1529 DOI 10.1137/S0363012900382837 Stability of planar switched systems: the linear single input case
Research Group: |
I often see the right part of the vol smile referred to as 'call skew'. However, due to put/call parity, this also represents skew for ITM puts. Is there a reason behind this convention?
I am not familiar with the terms call and put skew but rather upside and downside skew. However, I can image why they are used by some people.
Short Answer
At-the-money and out-of-they-money options are usually more liquid than in-the-money options. I.e., on the upside (downside) calls (puts) have smaller spreads and give you a stronger signal of the mid market price at this strike. Consequently, you mainly use these instruments to construct your implied volatility smile.
Some Intuition
Here is a bit of (heavily simplified) intuition why this is the case. Assume you are a market maker quoting otherwise identical European call and put options. You have a calibrated fair/mid-market implied volatility smile and now need to determine the corresponding bid and offer prices you are willing to quote. Your spread should account for your risk exposures, and you arrive at the following formula
\begin{equation} s(T, K) = \alpha \frac{\partial V}{\partial \sigma}(T, K) + \beta \left| \frac{\partial V}{\partial S}(T, K) \right|. \end{equation}
Here, $\alpha$ is the spread you charge for one unit of vega and $\beta$ is the spread you charge for one unit of delta. Since the vega of European call and put options of the same strike are identical the spread difference stems from the delta. Since out-of-the-money options have a lower absolute delta, you are showing tighter quotes for them. |
Your assumption that the work done is zero is wrong.
Work is being done everywhere on the fluid by an external force: the pressure in the pipe. To see this, let's derive Bernoulli's Equation: $P_1 + 1/2 \rho v_1^2 + \rho g y_1 = P_2 + 1/2 \rho v_2^2 + \rho g y_2 $
Lets assume your pipe narrows like a nozzle and begin with conservation of energy and see what happens to a small section of
incompressible fluid of density $\rho$ and volume $V$:$$ W_{external} = \Delta K + \Delta U $$The pressure in the pipe is the external agent here causing work to be done. $W = F \cdot \Delta x = (P A) \Delta x$. If the fluid is flowing from high pressure to low pressure, in one direction the force will be with $\Delta x$:$$ W_1 = P_1 A_1 \Delta x_1$$
and opposite for the low pressure side:$$ W_2 = - P_2 A_2 \Delta x_2$$
Since the fluid is incompressible, $A\Delta x = V$ for any part of the pipe. Putting this into the work-energy relationship:$$ P_1 V - P_2 V = 1/2 (\rho V)v_2^2 - 1/2 (\rho V)v_1^2 + (\rho V)g y_2 - (\rho V) g y_1 $$Where the mass of the fluid is $m= \rho V$. Clearly the Volume cancels out and leaves you with Bernoulli's equation.
answer
Work is being done by whatever is supplying the pressure. This work is accounted for by the change in pressure in the fluid. When someone writes $P + 1/2 \rho v^2 + \rho g y = Constant$ what they are really writing is $W/V + K/V + U/V = E/V$, where $E/V$ is the energy per unit volume of fluid (J/m^3). So you see, the work done is hidden in the change in pressure.
additional edits: It is true to say that nozzles do no work in the same way its true normal forces do no work: their forces are always perpendicular to the direction of motion. It is wrong however to conclude that no work is done. It is true to say that the energy / volume doesn't change on a streamline. As long as that definition of energy includes work done on/by the fluid. So either write $\Delta K + \Delta U_{gravity} + \Delta U_{pressure} = 0$ or $\Delta K + \Delta U_{gravity} = \Sigma W_{pressure}$. But to say $\Delta K + \Delta U_{gravity} = 0$ is in general wrong. |
I am having trouble with a contradiction arising from some computation, and I cannot figure out at which point I make a mistake.
Consider a conformally flat metric $g_{\mu\nu}=e^{2\phi}\eta_{\mu\nu}$. Then, the ricci scalar of $g$ is not always 0, depending on $\phi$, as can be seen for example from the formulas here.
However, let us not consider a "conformal" change of variables $x'^\rho(x)$ such that, in the new coordinates, the metric is rescaled as such : $g'_{\mu\nu} = e^{-2\phi}g_{\mu\nu} = \eta_{\mu\nu}$. In other words, identify the conformal transformation which rescales the metric by $e^{-2\phi}$, and apply it as a change of variables.
However, since scalars remain unchanged under a change of variables (more precisely under a diffeomorphism), we should have that $R[g] = R[g'] = R[\eta] = 0$ (where $R[g]$ is the ricci scalar of the metric $g_{\mu\nu}$.
Now this is a problem since we saw that $R[g]$ need not be $0$ even if g is conformally flat. Thus the contradiction.
I don't understand at which step in my reasoning I have made a mistake. The only possibility that I see is that there is such conformal transformation that rescales the metric by $e^{-2\phi}$, but this seems very restrictive looking again at this.
I am sure that the mistake I made it's much more elemental, but I can't figure it out. |
The Annals of Applied Probability Ann. Appl. Probab. Volume 28, Number 5 (2018), 3184-3214. The size of the boundary in first-passage percolation Abstract
First-passage percolation is a random growth model defined using i.i.d. edge-weights $(t_{e})$ on the nearest-neighbor edges of $\mathbb{Z}^{d}$. An initial infection occupies the origin and spreads along the edges, taking time $t_{e}$ to cross the edge $e$. In this paper, we study the size of the boundary of the infected (“wet”) region at time $t$, $B(t)$. It is known that $B(t)$ grows linearly, so its boundary $\partial B(t)$ has size between $ct^{d-1}$ and $Ct^{d}$. Under a weak moment condition on the weights, we show that for most times, $\partial B(t)$ has size of order $t^{d-1}$ (smooth). On the other hand, for heavy-tailed distributions, $B(t)$ contains many small holes, and consequently we show that $\partial B(t)$ has size of order $t^{d-1+\alpha }$ for some $\alpha >0$ depending on the distribution. In all cases, we show that the exterior boundary of $B(t)$ [edges touching the unbounded component of the complement of $B(t)$] is smooth for most times. Under the unproven assumption of uniformly positive curvature on the limit shape for $B(t)$, we show the inequality $\#\partial B(t)\leq (\log t)^{C}t^{d-1}$ for all large $t$.
Article information Source Ann. Appl. Probab., Volume 28, Number 5 (2018), 3184-3214. Dates Received: September 2017 Revised: January 2018 First available in Project Euclid: 28 August 2018 Permanent link to this document https://projecteuclid.org/euclid.aoap/1535443246 Digital Object Identifier doi:10.1214/18-AAP1388 Mathematical Reviews number (MathSciNet) MR3847985 Zentralblatt MATH identifier 06974777 Citation
Damron, Michael; Hanson, Jack; Lam, Wai-Kit. The size of the boundary in first-passage percolation. Ann. Appl. Probab. 28 (2018), no. 5, 3184--3214. doi:10.1214/18-AAP1388. https://projecteuclid.org/euclid.aoap/1535443246 |
On the dual codes of skew constacyclic codes
1.
Universidad de Concepción, Escuela de Educación, Departamento de Ciencias Básicas, Los Ángeles, Chile
2.
Universidad de Concepción, Facultad de Ciencias Físicas y Matemáticas, Departamento de Matemática, Concepción, Chile
Let $\mathbb{F}_q$ be a finite field with $q$ elements and denote by $\theta : \mathbb{F}_q\to\mathbb{F}_q$ an automorphism of $\mathbb{F}_q$. In this paper, we deal with skew constacyclic codes, that is, linear codes of $\mathbb{F}_q^n$ which are invariant under the action of a semi-linear map $ \phi _{\alpha,\theta }:\mathbb{F}_q^n\to\mathbb{F}_q^n$, defined by $ \phi _{\alpha,\theta }(a_0,...,a_{n-2}, a_{n-1}): = (\alpha \theta (a_{n-1}),\theta (a_0),...,\theta (a_{n-2}))$ for some $\alpha \in \mathbb{F}_q\setminus\{0\}$ and $n≥2$. In particular, we study some algebraic and geometric properties of their dual codes and we give some consequences and research results on $1$-generator skew quasi-twisted codes and on MDS skew constacyclic codes.
Mathematics Subject Classification:Primary: 12Y05, 16Z05; Secondary: 94B05, 94B35. Citation:Alexis Eduardo Almendras Valdebenito, Andrea Luigi Tironi. On the dual codes of skew constacyclic codes. Advances in Mathematics of Communications, 2018, 12 (4) : 659-679. doi: 10.3934/amc.2018039
References:
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A. Blokhuis, A. A. Bruen and J. A. Thas,
Arcs in $ PG(n,q)$, MDS-codes and three fundamental problems of B. Segre - some extensions,
[4]
A. Blokhuis, A. A. Bruen and J. A. Thas,
On MDS-codes, arcs in $ PG(n,q)$ with $ q$ even, and a solution of three fundamental problems of B. Segre,
[5] [6] [7] [8]
D. Boucher and F. Ulmer, Codes as modules over skew polynomial rings,
[9]
D. Boucher and F. Ulmer, A note on the dual codes of module skew codes,
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J. W. P. Hirschfeld,
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T. Maruta, A geometric approach to semi-cyclic codes,
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L. Storme and J. A. Thas,
M.D.S. codes and arcs in $ PG(n,q)$ with $ q$ even: An improvement of the bounds of Bruen, Thas, and Blokhuis,
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show all references
References:
[1] [2] [3]
A. Blokhuis, A. A. Bruen and J. A. Thas,
Arcs in $ PG(n,q)$, MDS-codes and three fundamental problems of B. Segre - some extensions,
[4]
A. Blokhuis, A. A. Bruen and J. A. Thas,
On MDS-codes, arcs in $ PG(n,q)$ with $ q$ even, and a solution of three fundamental problems of B. Segre,
[5] [6] [7] [8]
D. Boucher and F. Ulmer, Codes as modules over skew polynomial rings,
[9]
D. Boucher and F. Ulmer, A note on the dual codes of module skew codes,
[10] [11] [12] [13] [14]
J. W. P. Hirschfeld,
[15] [16] [17] [18]
T. Maruta, A geometric approach to semi-cyclic codes,
[19] [20] [21]
L. Storme and J. A. Thas,
M.D.S. codes and arcs in $ PG(n,q)$ with $ q$ even: An improvement of the bounds of Bruen, Thas, and Blokhuis,
[22] [23] [24]
Generator Matrix 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 6 1 2 6 1 2 10 1 2 10 1 2 8 1 2 12 1 2 12 1 2 30 1 2 62 1 2 2 7 2 4 2 5 2 5 2 5 3 6 3
Generator Matrix 5 2 5 2 5 2 5 2 5 2 5 2 5 2 5 2 6 1 2 6 1 2 10 1 2 10 1 2 8 1 2 12 1 2 12 1 2 30 1 2 62 1 2 2 7 2 4 2 5 2 5 2 5 3 6 3
Generator Matrix 6 2 10 2 10 2 12 2 12 2 30 2 30 2 62 2
Generator Matrix 6 2 10 2 10 2 12 2 12 2 30 2 30 2 62 2
Polynomial
Polynomial
[1] [2] [3] [4] [5]
Ekkasit Sangwisut, Somphong Jitman, Patanee Udomkavanich.
Constacyclic and quasi-twisted Hermitian self-dual codes over finite fields.
[6]
Nuh Aydin, Yasemin Cengellenmis, Abdullah Dertli, Steven T. Dougherty, Esengül Saltürk.
Skew constacyclic codes over the local Frobenius non-chain rings of order 16.
[7]
David Grant, Mahesh K. Varanasi.
The equivalence of space-time codes and codes defined over finite fields and Galois rings.
[8]
Somphong Jitman, Ekkasit Sangwisut.
The average dimension of the Hermitian hull of constacyclic codes over finite fields of square order.
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Amita Sahni, Poonam Trama Sehgal.
Enumeration of self-dual and self-orthogonal negacyclic codes over finite fields.
[12] [13]
Minjia Shi, Daitao Huang, Lin Sok, Patrick Solé.
Double circulant self-dual and LCD codes over Galois rings.
[14] [15]
Nuh Aydin, Nicholas Connolly, Markus Grassl.
Some results on the structure of constacyclic codes and new linear codes over
[16]
Jérôme Ducoat, Frédérique Oggier.
On skew polynomial codes and lattices from quotients of cyclic division algebras.
[17] [18] [19] [20]
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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 ... |
Why is the coefficient of drag of a straight wing lower than the coefficient of drag of a swept back wing at higher supersonic speeds (above, say, Mach 2)?
Mark is right when he says that there is no induced drag at supersonic speeds, but it is an invitation for misunderstandings. Induced drag is replaced by lift wave drag, and all what happens is that aerodynamicists choose to use two different names for basically the same effect: Air gets pushed down.
As usual when I post a long answer, I was not quite happy with the existing answer(s). Now I have some time and try to give a better answer.
First, why does the straight wing work better only at very high Mach numbers (> 2.0)? Because at lower supersonic speed a swept wing gives overall better performance. The sweep angle $\varphi_0$ must be high enough to allow for a subsonic leading edge (Mach < $\frac{1}{cos\varphi_0}$). Then the flow around the leading edge is subsonic and creates a suction area when accelerating around the nose contour. This suction helps to reduce drag - after all, this same suction is why a subsonic airfoil in inviscid flow has no drag. Edward C. Polhamus did a lot of research on this at NACA Langley and published several papers with equations for calculating the suction force.
Once you fly faster than Mach 2, the sweep angle for a subsonic leading edge rapidly gets too high for acceptable subsonic flight, and an unswept wing becomes the better alternative since you need to accept a supersonic leading edge. Examples are the wing of the F-104 or the canard of the XB-70.
Rhombic airfoil in supersonic flow at zero angle of attack (own work). The plus sign denotes higher pressure, the minus sign lower pressure than ambient. By selecting a rhombic airfoil, the flow is very easy to determine because the pressure only changes when the local contour gradient changes. The two compression shocks create the typical sonic boom when arriving at the ground. Note that this airfoil already creates pressure drag even at zero lift. Any airfoil thickness greater than zero and any airfoil camber will cause this type of drag where the forward-facing areas see higher pressure and the rear-facing areas experience suction. This type of drag is called wave drag. It can only be minimized by minimizing the relative thickness of whatever is supposed fly at supersonic speed.
When the angle of attack is increased, this airfoil begins to create lift. Now the compression by the lower forward shock becomes stronger and that by the upper forward shock becomes weaker. The expansion fan is the same again on both sides, so the upper rear half experiences less pressure than the lower rear half. I tried to symbolize this by the amount of plus and minus signs:
Note that the pressure difference is constant over chord, so the center of pressure is at 50% of chord length. Note also that the lift vector is perpendicular to the chord line. Since lift is defined as the force perpendicular to the direction of the undisturbed air, supersonic lift always carries a drag component which is proportional to the angle of attack - there is no suction at the nose to alleviate this! The wave drag of the airfoil at zero angle of attack still comes on top, so we have a shape-dependent wave drag and a lift-dependent wave drag component. This lift-dependent wave drag replaces the induced drag of subsonic speeds. If we compare the magnitude of both, we find:
Subsonic: $c_{Di} = \frac{c^2_L}{\pi\cdot AR\cdot\epsilon}$
Supersonic formula for 2D flow: $c_{{DW}_L} = c_L\cdot\alpha$
This doesn't look so similar, so let's now express the angle of attack $\alpha$ by the lift coefficient divided by the lift curve slope:
$$\alpha = \frac{c_L}{c_{L\alpha}} = \frac{c_L}{\frac{4}{\sqrt{Ma^2-1}}\cdot\left(1 - \frac{\lambda}{2\cdot AR\cdot\sqrt{Ma^2-1}}\right)}$$
and the lift wave drag component becomes $c_{{DW}_L} = \frac{c^2_L}{\frac{4}{\sqrt{Ma^2-1}}\cdot\left(1 - \frac{\lambda}{2\cdot AR\cdot\sqrt{Ma^2-1}}\right)}$
Now let's compare the F-104 wing, which has an aspect ratio $AR$ of 2.45 and a taper ratio $\lambda$ of 0.385: If we plug in the parameters and adjust $\epsilon$ such that both sub- and supersonic lift-dependent drag coefficients agree, $\epsilon$ would need to be 0.89 at Mach 1.2, 0.58 at Mach 1.4 and 0.31 at Mach 2.0. The dramatic rise of lift wave drag over Mach is caused by the reduction in the lift curve slope over Mach.
For slender bodies the lift curve slope is $c_{L\alpha} = \frac{\pi\cdot AR}{2}$ and the lift wave drag component becomes $c_{{DW}_L} = 2\cdot\frac{c^2_L}{\pi\cdot AR}$. For slender bodies the supersonic $\epsilon$ is 0.5, regardless of Mach.
The important conclusions from this for wing selection are:
Sweep does no longer help once the leading edge is supersonic. Lift wave drag continues where induced drag drops off. Lift always causes drag. For a supersonic wing the aspect ratio is of minor importance.
Now back to the original question: Once the leading edge is supersonic, sweep is no longer helpful. Now the best wing is straight, because it will need the lowest wing area to create the required lift at subsonic speed. At supersonic speed its lower area will translate into lower friction drag, making it better than comparable delta or swept wings.
For a given lift coefficient, boundary layer drag is smaller for the straight wing configuration, and lift-induced drag is larger for straight wing. These conflicting effects may give an advantage to straight wing at low subsonic speeds.
At speeds of about Mach 0.8 you begin to get wave drag because the flow over the wing is not uniform, and in some regions you have supersonic flow. The wave drag is much less for swept wings, roughly in proportion to the aspect ratio. So when you are trying to push the plane to Mach 1, swept wings make your job easier.
At Mach 1 and a bit higher more and more of the wing region experiences supersonic flow, and the shock drag is dominant. Again, swept wings have a huge drag advantage (not to mention a control advantage).
But pretty soon, certainly by Mach 2, both wings have the same amount of wave drag, and the shock drag becomes less important as the region near Mach 1 disappears. And now we come to our first observation, about boundary layer and lift-induced drag roughly balancing. Except --
At supersonic flow there is no lift-induced drag
This is because the "wake cannot be felt upstream." More accurately, the penalty associated with turning the flow to generate lift is captured over the wing surface and what happens downstream of the wing cannot affect the flow over the wing because "knowledge" of what happens downstream propagates at the speed of sound.
So we have to remove, from our calculations for each type of wing, the lift-induced drag, which had been greater in the straight wing. And this makes the straight wing drag coefficient lower at moderately high supersonic speeds, compared to the swept wing.
I may be wrong here because fluid dynamics is famous for introducing subtle effects that nobody would have anticipated, but I believe this issue is understood and the trade-off mentioned is the heuristic reason. |
Discretizing the Weak Form Equations
This post continues our blog series on the weak formulation. In the previous post, we implemented and solved an exemplary weak form equation in the COMSOL Multiphysics software. The result was validated with simple physical arguments. Today, we will start to take a behind-the-scenes look at how the equations are discretized and solved numerically.
Our Simple Example
Recall our simple example of 1D heat transfer at steady state with no heat source, where the temperature T is a function of the position x in the domain defined by the interval 1\le x\le 5. With the boundary conditions that the outgoing flux should be 2 at the left boundary (x=1) and the temperature should be 9 at the right boundary (x=5), the weak form equation reads:
(1)
We now attempt to find a way to solve this equation numerically.
Basis Functions
To solve Eq. (1) numerically, we first divide the domain 1\le x\le 5 into four evenly spaced sub-intervals, or
mesh elements, bound by five nodal points x = 1, 2, \cdots, 5. Then, we can define a set of basis functions, or shape functions, \psi_{1L}(x), \psi_{1R}(x), \psi_{2L}(x), \psi_{2R}(x), \cdots, \psi_{4R}(x), as shown in the graph below, where there are two shape functions in each mesh element, represented by a solid line and a dashed line.
For example, in the first element (1 \le x \le 2), we have
(2)
\psi_{1L}(x) %26=%26 \left\{ \begin{array}{ll}
2-x \mbox{ for } 1 \le x \le 2,\\
0 \mbox{ elsewhere}
\end{array} \right \mbox{ (solid red line)}
\end{equation}
\psi_{1R}(x) %26=%26 \left\{ \begin{array}{ll}
x-1 \mbox{ for } 1 \le x \le 2, \\
0 \mbox{ elsewhere}
\end{array} \right \mbox{ (dashed red line)}
\end{equation}
We observe that each shape function is a simple linear function ranging from 0 to 1 within a mesh element, and vanishes outside of that mesh element.
Note: Of course, COMSOL Multiphysics allows shape functions formed with higher-order polynomials, not just linear functions. The choice of linear shape functions here is for visual clarity.
With this set of shape functions, we can approximate any arbitrary function defined in the domain 1\le x\le 5 by a simple linear combination of them:
(3)
where a_{1L}, a_{1R}, a_{2L}, a_{2R}, \cdots are some constant coefficients, one for each shape function. In the graph below, the arbitrary function u(x) is represented by the black curve. The cyan curve represents the approximation by the superposition of the shape functions (3). Each term on the right-hand side of Eq. (3) is plotted using the same color and line style as the graph above.
We see that in general the approximation (represented by the cyan curve) can be discontinuous across the boundary between adjacent mesh elements. In practice, many physical systems, including our simple example of heat conduction, are expected to have continuous solutions. For this reason, the default shape functions for most physics interfaces are
Lagrange elements, in which the shape function coefficients are constrained so that the solution is continuous across boundaries between adjacent elements. In this case, the approximation is simplified, as shown in the figure below,
where the cyan curve has been made continuous by setting the coefficients on each side of a mesh boundary to be equal: a_{1R} = a_{2L}, a_{2R} = a_{3L}, a_{3R} = a_{4L}. We also renamed the coefficients for brevity:
\begin{align}
a_1 %26\equiv a_{1L}\\
a_2 %26\equiv a_{1R} = a_{2L}\\
a_3 %26\equiv a_{2R} = a_{3L}\\
a_4 %26\equiv a_{3R} = a_{4L}\\
a_5 %26\equiv a_{4R}
\end{align}
\end{equation*}
We see that the continuity condition requires pairs of shape functions to share the same coefficient in making the approximation (3), which can now in turn be simplified by combining those pairs of shape functions into a new set of basis functions
ϕ1(x),ϕ2(x),⋯,ϕ5(x), with each function localized around a nodal point:
(4)
\begin{align}
\phi_1(x) \equiv \psi_{1L}(x) %26= \left\{ \begin{array}{ll}
2-x \mbox{ for } 1\:$\leq$\:\textit{x}\:$\leq$\:2,\\
0 \mbox{ elsewhere}
\end{array} \right
\\
\phi_2(x) \equiv \psi_{1R}(x) + \psi_{2L}(x) %26= \left\{ \begin{array}{lll}
x-1 \mbox{ for } 1\:$\leq$\:\textit{x}\:$\leq$\:2, \\
3-x \mbox{ for } 2\:\textless\:x\:$\leq$\:3, \\
0 \mbox{ elsewhere}
\end{array} \right
\\
\phi_3(x) \equiv \psi_{2R}(x) + \psi_{3L}(x) %26= \left\{ \begin{array}{lll}
x-2 \mbox{ for } 2\:$\leq$\:\textit{x}\:$\leq$\:3, \\
4-x \mbox{ for } 3\:\textless\:x\:$\leq$\:4, \\
0 \mbox{ elsewhere}
\end{array} \right
\\
\\ \cdot
\\ \cdot
\\ \cdot
\end{equation*}
As shown in the graph below, each new basis function is essentially a triangular-shaped, piecewise-linear function centered around a nodal point. Its value varies between 1 and 0 within the mesh element(s) adjacent to the nodal point, and vanishes everywhere else.
As discussed above, by choosing this new set of basis functions, we constrain the solution to be continuous across the boundary between adjacent mesh elements. Most physical systems satisfy this continuity constraint, including our simple heat transfer example here.
Now, with this new set of basis functions, the approximation (3) is simplified to
(5)
In the graph below, the arbitrary function u(x) is represented by the black curve. The cyan curve represents the approximation by the superposition of the new basis functions. Each term on the right-hand side of Eq. (5) is plotted using the same color scheme as the graph above.
As an aside, if the black curve represents the exact solution to some real modeling problem, then we see that the approximation is not very good, due to the coarseness of the mesh. Also, in general, the nodal point values a_1, a_2, \cdots are not required to lie on the exact solution, unless one is constrained to a known solution value (shown in a_5 as an example in the figure above). The discrepancy between the black and the cyan curves we see here represents the discretization error of the solution. In 2D and 3D models, there will also be a discretization error of the geometry. In my colleague Walter Frei’s blog post on meshing considerations, both types of errors are discussed in some detail. Due to these potential errors, a mesh refinement study is necessary to ensure the accuracy of modeling results.
We note that the approximation given by Eq. (5) (the cyan curve) is piecewise-linear. Thus, it’s impossible to evaluate its second derivatives numerically. As we have mentioned before, the weak formulation provides numerical benefits by reducing the order of differentiation in the equation system. In this case, only the first derivative is needed and it can be readily evaluated numerically. In a future blog entry, we will discuss an example of discontinuity in the material property that also benefits from the reduced order of differentiation.
Discretizing the Weak Form Equation in Two Steps
With the new set of basis functions defined above, we proceed to discretize the weak form equation (1) in two steps. First, the temperature function, T(x), can be approximated by the set of basis functions in the same way as in Eq. (5):
(6)
where a_1, a_2, \cdots , a_5 are unknown coefficients to be determined.
(7)
a_1 \int_1^5 \partial_x \phi_1(x) \partial_x \tilde{T}(x) \,dx + a_2 \int_1^5 \partial_x \phi_2(x) \partial_x \tilde{T}(x) \,dx + \cdots + a_5 \int_1^5 \partial_x \phi_5(x) \partial_x \tilde{T}(x) \,dx
\\
= -2 \tilde{T}_1 -\lambda_2 \tilde{T}_2 -\tilde{\lambda}_2 (a_5 -9)
\end{array}
where the temperature at the right boundary x=5, T_2, has been evaluated using the expression (6) and the fact that the basis functions are localized, leading to only one term, a_5 \phi_5(x=5) = a_5, contributing to T(x=5).
We see that there are six unknowns in the discretized version of the weak form equation (7): The five coefficients a_1, a_2, \cdots , a_5 and the one flux \lambda_2 at the right boundary. It is customary to call the unknowns
degrees of freedom. For example, here we say the (discretized) problem has “six degrees of freedom”.
To solve for the six unknowns, we need six equations. This leads to the second step of discretization. Recall from our first blog post that the role of the test functions is to sample the equation locally to clamp down the solution everywhere within the domain. Now we already have a set of localized functions, our basis functions \phi_1, \cdots, \phi_5, so we can just substitute them into the test function \tilde{T} in Eq. (7) to obtain the six equations we need.
Here is a table showing the six substitutions that will generate the six equations for us:
\tilde{T}(x) \tilde{\lambda}_2 \phi_1(x) 0 \phi_2(x) 0 \phi_3(x) 0 \phi_4(x) 0 \phi_5(x) 0 0 1
Since each of the basis functions is localized, each substitution yields an equation with a small number of terms. For example, the first substitution gives
a_1 \int_1^5 \partial_x \phi_1(x) \partial_x \phi_1(x) \,dx + a_2 \int_1^5 \partial_x \phi_2(x) \partial_x \phi_1(x) \,dx + \cdots + a_5 \int_1^5 \partial_x \phi_5(x) \partial_x \phi_1(x) \,dx
\\
= -2 \phi_1(x=1) -\lambda_2 \phi_1(x=5) -0 (a_5 -9)
\end{array}
We note that \phi_1 has non-trivial overlap only with itself and \phi_2. Therefore, only the first two terms on the left-hand side are non-zero. Also, \phi_1 is localized near the left boundary (x=1), so only the first term on the right-hand side remains. The equation now becomes
(8)
where we have evaluated the definite integrals on the left-hand side:
\begin{align}
\int_1^5 \partial_x \phi_1(x) \partial_x \phi_1(x) \,dx %26= 1\\
\int_1^5 \partial_x \phi_2(x) \partial_x \phi_1(x) \,dx %26= -1\\
\end{align}
\end{equation*}
and used the definition of \phi_1 on the right-hand side: \phi_1(x=1) = 1.
Similarly, the remaining five substitutions listed in the table above yield these equations:
(9)
\begin{align}
-a_1 + 2 a_2 -a_3 %26= 0\\
-a_2 + 2 a_3 -a_4 %26= 0\\
-a_3 + 2 a_4 -a_5 %26= 0\\
-a_4 + a_5 %26= -\lambda_2\\
0 %26= -(a_5 -9)\\
\end{align}
\end{equation*}
We now have six equations for our six unknowns and it is straightforward to verify that the solution matches what we have obtained using COMSOL Multiphysics software in the previous post. For example, the last equation immediately gives us a_5 = 9, and using the expression (6) for the temperature, we obtain its value at the right boundary:
\begin{align}
T(x=5) %26= a_1 \phi_1(x=5) + a_2 \phi_2(x=5) + \cdots + a_5 \phi_5(x=5)\\
%26= a_1 \cdot 0 + a_2 \cdot 0 + \cdots + a_5 \cdot 1\\
%26= 9\\
\begin{align}
\end{equation*}
This agrees with the fixed boundary condition that the temperature should be 9 at the right boundary. It is also easy to see that it is the term associated with the test function \tilde{\lambda}_2 that gives rise to the equation (0 = a_5 -9), as we would expect.
Matrix Representation
(10)
\begin{array}{cccccc}
1 & -1 & 0 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 & 0 \\
0 & -1 & 2 & -1 & 0 & 0 \\
0 & 0 & -1 & 2 & -1 & 0 \\
0 & 0 & 0 & -1 & 1 & 1 \\
0 & 0 & 0 & 0 & 1 & 0
\end{array}
\right)
\left(
\begin{array}{c} a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \\ \lambda_2 \end{array}
\right)
= \left(
\begin{array}{c} -2 \\ 0 \\ 0 \\ 0 \\ 0 \\ 9 \end{array}
\right)
The matrix on the left-hand side is customarily called the
stiffness matrix and the vector on the right is called the load vector, due to the application of this technique in structural mechanics.
We notice two interesting facts about this matrix equation. First, there are a lot of zeros in the matrix (a so-called
sparse matrix). In a practical model where there are many more mesh elements than our four elements here, we can envision that most of the elements in the matrix will be zero. This is a direct benefit of choosing localized shape functions, and it lends itself to very efficient numerical methods to solve the equation system.
Second, the Lagrange multiplier \lambda_2 appears only in one equation (the last column of the matrix has only one non-zero element). The remaining five equations involve only the five unknown coefficients a_1, a_2, \cdots , a_5. Therefore, we can choose to solve for a_1, a_2, \cdots , a_5 using the five equations, without ever needing to solve for \lambda_2. As we briefly mentioned in the previous entry, in general, it is possible to choose not to solve for the Lagrange multiplier(s) in order to gain computation speed.
Summary and Next Topic in the Weak Form Series
Today, we reviewed the basic procedure for discretizing the weak form equation using our simple example. We took advantage of a set of localized shape functions in two steps:
Using them as a basis to approximate the real solution Substituting them one by one into the weak form equation to obtain the discretized system of equations
The resulting matrix equation is sparse, which can be solved efficiently using a computer.
In the previous blog post, when we implemented the weak form equation using COMSOL Multiphysics, the discretization was done under the hood without needing the user’s help. Next, we will show you how to inspect the stiffness matrix and load vector, as well as how to choose to solve for — or not to solve for — the Lagrange multiplier by using the
Weak Form PDE interface in the software. Comments (6) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
Map of Cape Cod
showing shores undergoing erosion (cliffed sections) in red, and shores characterized by marine deposition (barriers) in blue
[1]
Deposition is the geological process in which sediments, soil, and rocks are added to a landform or land mass. Wind, ice, and water, as well as sediment flowing via gravity, transport previously eroded sediment, which, at the loss of enough kinetic energy in the fluid, is deposited, building up layers of sediment.
Deposition occurs when the forces responsible for sediment transportation are no longer sufficient to overcome the forces of particle anaerobic conditions.
Contents Null-point hypothesis 1 Deposition of non-cohesive sediments 1.1 Deposition of cohesive sediments 1.2 The occurrence of null point theory 1.3 Applications for coastal planning and management 1.4 References 2 See also 3 Null-point hypothesis
The null-point hypothesis explains how sediment is deposited throughout a shore profile according to its grain size. This is due to the influence of hydraulic energy, resulting in a seaward-fining of sediment particle size, or where fluid forcing equals gravity for each grain size.
[2] The concept can also be explained as "sediment of a particular size may move across the profile to a position where it is in equilibrium with the wave and flows acting on that sediment grain". [3] This sorting mechanism combines the influence of the down-slope gravitational force of the profile and forces due to flow asymmetry, the position where there is zero net transport is known as the null point and was first proposed by Cornaglia in 1889. [3] Figure 1 illustrates this relationship between sediment grain size and the depth of the marine environment.
Figure 1. Illustrates the sediment size distribution over a shoreline profile, where finer sediments are transported away from high energy environments and settle out of suspension, or deposit in calmer environments. Coarse sediments are maintained in the upper shoreline profile and are sorted by the wave-generated hydraulic regime
The first principle underlying the null point theory is due to the gravitational force; finer sediments remain in the water column for longer durations allowing transportation outside the surf zone to deposit under calmer conditions. The gravitational effect, or settling velocity determines the location of deposition for finer sediments, whereas a grain's internal angle of friction determines the deposition of larger grains on a shore profile.
[3] The secondary principle to the creation of seaward sediment fining is known as the hypothesis of asymmetrical thresholds under waves; this describes the interaction between the oscillatory flow of waves and tides flowing over the wave ripple bedforms in an asymmetric pattern. [4] "The relatively strong onshore stroke of the wave forms an eddy or vortex on the lee side of the ripple, provided the onshore flow persists, this eddy remains trapped in the lee of the ripple. When the flow reverses, the eddy is thrown upwards off the bottom and a small cloud of suspended sediment generated by the eddy is ejected into the water column above the ripple, the sediment cloud is then moved seaward by the offshore stroke of the wave." [4] Where there is symmetry in ripple shape the vortex is neutralised, the eddy and its associated sediment cloud develops on both sides of the ripple. [4] This creates a cloudy water column which travels under tidal influence as the wave orbital motion is in equilibrium.
The Null-point hypothesis has been quantitatively proven in Akaroa Harbour, New Zealand, The Wash, U.K., Bohai Bay and West Huang Sera, Mainland China, and in numerous other studies; Ippen and Eagleson (1955), Eagleson and Dean (1959, 1961) and Miller and Zeigler (1958, 1964).
Deposition of non-cohesive sediments
Large grain sediments transported by either bed load or suspended load will come to rest when there is insufficient bed shear stress and fluid turbulence to keep the sediment moving,
[4] with the suspended load this can be some distance as the particles need to fall through the water column. This is determined by the grains downward acting weight force being matched by a combined buoyancy and fluid drag force [4] and can be expressed by: \frac{4}{3} \pi R^3 \rho_s g = \frac{4}{3} \pi R^3 \rho g + \frac{1}{2} \C_d \rho \pi R^2 w_s^2\,
Downward acting weight force = Upward-acting buoyancy force + Upward-acting fluid drag force
[4]
where:
π is the ratio of a circle's circumference to its diameter. R is the radius of the spherical object (in m), ρ is the mass density of the fluid (kg/m 3), g is the gravitational acceleration (m/s 2), C is the drag coefficient, and d w is the particle's settling velocity (in m/s). s
In order to calculate the drag coefficient, the grain's Reynolds number needs to be discovered, which is based on the type of fluid through which the sediment particle is flowing; laminar flow, turbulent flow or a hybrid of both. When the fluid becomes more viscous due to smaller grain sizes or larger settling velocities, prediction is less straight forward and it is applicable to incorporate Stokes Law(also known as the frictional force, or drag force) of settling.
[4] Deposition of cohesive sediments
Cohesion of sediment occurs with the small grain sizes associated with silts and clays, or particles smaller than 4ϕ on the phi scale.
[4] If these fine particles remain dispersed in the water column, Stokes law applies to the settling velocity of the individual grains, [4] although due to sea water being a strong electrolyte bonding agent, flocculation occurs where individual particles create an electrical bond adhering each other together to form flocs. [4] "The face of a clay platelet has a slight negative charge where the edge has a slight positive charge, when two platelets come into close proximity with each other the face of one particle and the edge of the other are electrostatically attracted." [4] Flocs then have a higher combined mass which leads to quicker deposition through a higher fall velocity, and deposition in a more shoreward direction than they would have as the individual fine grains of clay or silt. The occurrence of null point theory
Akaroa Harbour is located on Banks Peninsula, Canterbury, New Zealand, . The formation of this harbour has occurred due to active erosional processes on an extinct shield volcano, whereby the sea has flooded the caldera creating an inlet 16 km in length, with an average width of 2 km and a depth of -13 m relative to mean sea level at the 9 km point down the transect of the central axis.
[5] The predominant storm wave energy has unlimited fetch for the outer harbour from a southerly direction, with a calmer environment within the inner harbour, though localised harbour breezes create surface currents and chop influencing the marine sedimentation processess. [6] Deposits of loess from subsequent glacial periods have in filled volcanic fissures over millennia, [7] resulting in volcanic basalt and loess as the main sediment types available for deposition in Akaroa Harbour
Figure 2. Map of Akaroa Harbour showing a fining of sediments with increased bathymetry toward the central axis of the harbour. Taken from Hart et al. (2009) and the University of Canterbury under contract of Environment Canterbury.
[5]
Hart et al. (2009)
[5] discovered through bathymetric survey, sieve and pipette analysis of subtidal sediments, that sediment textures were related to three main factors: depth; distance from shoreline; and distance along the central axis of the harbour. Resulting in the fining of sediment textures with increasing depth and towards the central axis of the harbour, or if classified into grain class sizes, “the plotted transect for the central axis goes from silty sands in the intertidal zone, to sandy silts in the inner nearshore, to silts in the outer reaches of the bays to mud at depths of 6 m or more”. [5] See figure 2 for detail.
Other studies have shown this process of the winnowing of sediment grain size from the effect of hydrodynamic forcing; Wang, Collins and Zhu (1988)
[8] qualitatively correlated increasing intensity of fluid forcing with increasing grain size. "This correlation was demonstrated at the low energy clayey tidal flats of Bohai Bay (China), the moderate environment of the Jiangsu coast (China) where the bottom material is silty, and the sandy flats of the high energy coast of The Wash (U.K.)." This research shows conclusive evidence for the null point theory existing on tidal flats with differing hydrodynamic energy levels and also on flats that are both erosional and accretional.
Kirby R. (2002)
[9] takes this concept further explaining that the fines are suspended and reworked aerially offshore leaving behind lag deposits of mainly bivalve and gastropod shells separated out from the finer substrate beneath, waves and currents then heap these deposits to form chenier ridges throughout the tidal zone which tend to be forced up the foreshore profile but also along the foreshore. Cheniers can be found at any level on the foreshore and predominantly characterise an erosion-dominated regime. [9] Applications for coastal planning and management
The null point theory has been controversial in its acceptance into mainstream coastal science as the theory operates in dynamic equilibrium or unstable equilibrium, and many field and laboratory observations have failed to replicate the state of a null point at each grain size throughout the profile.
[3] The interaction of variables and processes over time within the environmental context causes issues; "the large number of variables, the complexity of the processes, and the difficulty in observation, all place serious obstacles in the way of systematisation, therefore in certain narrow fields the basic physical theory may be sound and reliable but the gaps are large" [10]
Geomorphologists, engineers, governments and planners should be aware of the processes and outcomes involved with the null point hypothesis when performing tasks such as beach nourishment, issuing building consents or building coastal defence structures. This is because sediment grain size analysis throughout a profile allows inference into the erosion or accretion rates possible if shore dynamics are modified. Planners and managers should also be aware that the coastal environment is dynamic and contextual science should be evaluated before implementation of any shore profile modification. Thus theoretical studies, laboratory experiments, numerical and hydraulic modelling seek to answer questions pertaining to littoral drift and sediment deposition, the results should not be viewed in isolation and a substantial body of purely qualitative observational data should supplement any planning or management decision.
[2] References ^ Oldale, Robert N. "Coastal Erosion on Cape Cod: Some Questions and Answers". U.S. Geological Survey. Retrieved 2009-09-11. ^ a b Jolliffe, I. P. (1978). "Littoral and offshore sediment transport". Progress in Physical Geography 2 (2): 264–308. ^ a b c d Horn, D. P. (1992). "A review and experimental assessment of equilibrium grain size and the ideal wave-graded profile" 108 (2). pp. 161–174. ^ a b c d e f g h i j k Masselink, G. Hughes, M. Knight, J. (2011). Sediments, boundary layers and transport: Coastal processes and geomorphology. pp. 105–147. ^ a b c d Hart, D. Todd, D. Nation, T. McWilliams, Z. (2009). Upper Akaroa Harbour seabed bathymetry and soft sediments: A baseline mapping study Coastal Research Report (Report). ^ Heuff, D. N., Spigel, R. H., and Ross, A. H. (2005). "Evidence of a significant wind‐driven circulation in Akaroa Harbour. Part 1: Data obtained during the September‐November, 1998 field survey". pp. 1097–1109. ^ Raeside, J.D. (1964). "Loess deposits of the South Island, New Zealand, and soils formed on them" 7. pp. 811–838. ^ Wang, Y.,Collins, M.B., and Zhu D. (1988). "A comparative study of open coast tidal flats: The Wash (U.K.), Bohai Bay and West Huang Sera (Mainland China)". pp. 120–134. ^ a b Kirby R. (2002). Distinguishing accretion from erosion-dominated muddy coasts. In T. W. Healy, Y. and Healy J. (Ed.), Muddy coasts of the world:Processes, deposits and function=. Elsevier. pp. 61–81). ^ Russell R. (1960). "Coastal erosion and protection: nine questions and answers" 3. See also
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Definition:Pi/Definition 1 Definition
Then $\pi$ can be defined as $\pi = \dfrac C {2r}$.
(It can be argued that $\pi = \dfrac C d$, where $d$ is the circle's diameter, is a simpler and more straightforward definition. However, the radius is, in general, far more immediately "useful" than the diameter, hence the above more usual definition in terms of circumference and radius.) Uniqueness of Pi
Note that $\pi$ is defined on a per-circle basis. For each circle with its own circumference $C$ and diameter $d$, $\pi$ is defined as the ratio between the two. It is conceivable, then, that $\pi$ has a different value for each circle. It is also true, however, that All Circles are Similar and thus proportional in size. Thus, the value of $\pi$ is consistent between any two circles, and the constancy of $\pi$ is proven.
The decimal expansion of $\pi$ starts:
$\pi \approx 3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
In binary:
$\pi \approx 11 \cdotp 00100 \, 10000 \, 11111 \, 1011 \ldots$ Also see Sources 1986: David Wells: Curious and Interesting Numbers... (previous) ... (next): $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$ 1997: David Wells: Curious and Interesting Numbers(2nd ed.) ... (previous) ... (next): $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$ 2008: Ian Stewart: Taming the Infinite... (previous) ... (next): Chapter $2$: The Logic of Shape: Archimedes |
TL;DR: Unless you assume people are unreasonably bad at judging car color, or that blue cars are unreasonably rare, the large number of people in your example means the probability that the car is blue is basically 100%.
Matthew Drury already gave the right answer but I'd just like to add to that with some numerical examples, because you chose your numbers such that you actually get pretty similar answers for a wide range of different parameter settings. For example, let's assume, as you said in one of your comments, that the probability that people judge the color of a car correctly is 0.9. That is: $$p(\text{say it's blue}|\text{car is blue})=0.9=1-p(\text{say it isn't blue}|\text{car is blue})$$and also$$p(\text{say it isn't blue}|\text{car isn't blue})=0.9=1-p(\text{say it is blue}|\text{car isn't blue})$$
Having defined that, the remaining thing we have to decide is: what is the prior probability that the car is blue? Let's pick a very low probability just to see what happens, and say that $p(\text{car is blue})=0.001$, i.e. only 0.1% of all cars are blue. Then the posterior probability that the car is blue can be calculated as:
\begin{align*}&p(\text{car is blue}|\text{answers})\\&=\frac{p(\text{answers}|\text{car is blue})\,p(\text{car is blue})}{p(\text{answers}|\text{car is blue})\,p(\text{car is blue})+p(\text{answers}|\text{car isn't blue})\,p(\text{car isn't blue})}\\&=\frac{0.9^{900}\times 0.1^{100}\times0.001}{0.9^{900}\times 0.1^{100}\times0.001+0.1^{900}\times0.9^{100}\times0.999}\end{align*}
If you look at the denominator, it's pretty clear that the second term in that sum will be negligible, since the relative size of the terms in the sum is dominated by the ratio of $0.9^{900}$ to $0.1^{900}$, which is on the order of $10^{58}$. And indeed, if you do this calculation on a computer (taking care to avoid numerical underflow issues) you get an answer that is equal to 1 (within machine precision).
The reason the prior probabilities don't really matter much here is because you have so much evidence for one possibility (the car is blue) versus another. This can be quantified by the
likelihood ratio, which we can calculate as:$$\frac{p(\text{answers}|\text{car is blue})}{p(\text{answers}|\text{car isn't blue})}=\frac{0.9^{900}\times 0.1^{100}}{0.1^{900}\times 0.9^{100}}\approx 10^{763}$$
So before even considering the prior probabilities, the evidence suggests that one option is already astronomically more likely than the other, and for the prior to make any difference, blue cars would have to be unreasonably, stupidly rare (so rare that we would expect to find 0 blue cars on earth).
So what if we change how accurate people are in their descriptions of car color? Of course, we could push this to the extreme and say they get it right only 50% of the time, which is no better than flipping a coin. In this case, the posterior probability that the car is blue is simply equal to the prior probability, because the people's answers told us nothing. But surely people do at least a little better than that, and even if we say that people are accurate only 51% of the time, the likelihood ratio still works out such that it is roughly $10^{13}$ times more likely for the car to be blue.
This is all a result of the rather large numbers you chose in your example. If it had been 9/10 people saying the car was blue, it would have been a very different story, even though the same ratio of people were in one camp vs. the other. Because statistical evidence doesn't depend on this ratio, but rather on the numerical difference between the opposing factions. In fact, in the likelihood ratio (which quantifies the evidence), the 100 people who say the car isn't blue exactly cancel 100 of the 900 people who say it is blue, so it's the same as if you had 800 people all agreeing it was blue. And that's obviously pretty clear evidence.
(Edit: As Silverfish pointed out, the assumptions I made here actually implied that whenever a person describes a non-blue car incorrectly, they will default to saying it's blue. This isn't realistic of course, because they could really say any color, and will say blue only some of the time. This makes no difference to the conclusions though, since the less likely people are to mistake a non-blue car for a blue one, the stronger the evidence that it is blue when they say it is. So if anything, the numbers given above are actually only a lower bound on the pro-blue evidence.) |
Gamma matrices are defined by the Clifford algebra
$$ \{\gamma^\mu, \gamma^\nu\}= 2g^{\mu\nu}\mathbb I_n \,. $$
So, you see the index $\mu$ in $\gamma^\mu$ runs from $0$ upto $D-1$ where $D$ is the number of spacetime dimensions. It does not mean $\gamma^\mu$ is a vector. The $\mu$ index here only tells you how many gamma matrices are there. The dimensionality of the matrices themselves is $n= 2^{[D/2]}$ where $[\cdot]$ gives you the integer part of a number. For example, in $(1+2)-$dimensions, $D=3$ and hence the Dirac matrices are $2^{[1.5]}= 2$ dimensional, which you recognize are the Pauli matrices. The labels of the entries of the gamma matrices are known as spinor indices. So, in 3 dimensions, for example, the $a,b$ in $\gamma^\mu_{ab}$ would run from $1$ to $2$.
What is a $4$-vector? It is something that transforms like a vector under Lorentz transformations $\Lambda$. Namely, $X^\mu$ is a vector if it transforms like
$$ X^\mu\to {\Lambda^\mu}_\nu X^\nu \,. $$
That's the definition! Just having a $4$-dimensional column vector with Greek indices labelling its entries does not make it a Lorentz vector. It needs to transform the right way.
Okay, so what is a spinor? A spinor is something that transforms like a spinor. Namely, $\psi$ is a spinor if it transforms, under a Lorentz transformation parametrized by $\omega_{\mu\nu}$, like
$$ \psi \to \Lambda_{1/2} \psi\, \qquad (\Rightarrow \overline\psi \to \overline\psi\ \Lambda_{1/2}^{-1}\ ) \,, $$
where $\Lambda_{1/2} = \exp{(-\frac i2 \omega_{\mu\nu} S^{\mu\nu})}$ and $S^{\mu\nu} = \frac i4 [\gamma^\mu, \gamma^\nu]$ generates an $n-$dimensional representation of the Lorentz algebra.
Let's make a remark on why we use something like $\overline \psi =
\psi^\dagger \gamma^0$. Well, because we want to construct bilinear
Lorentz scalars like $\psi^\dagger \psi$, but $\psi^\dagger \psi$ is
not a Lorentz scalar precisely because the matrix $\Lambda_{1/2}$ is
not unitary. Under a Loretz transformation,
$$ \psi^\dagger \to \psi^\dagger \Lambda_{1/2}^\dagger \ne
\psi^\dagger \Lambda_{1/2}^{-1}\,.$$
However, we notice an interesting property of the gamma matrix $\gamma^0$.
$$ \boxed{ \Lambda_{1/2}^\dagger \gamma^0 = \gamma^0
\Lambda_{1/2}^{-1} }$$
This immediately tells us that defining something like $\overline \psi
\equiv \psi^\dagger \gamma^0$ will do the job.
$$ \overline \psi \to (\psi^\dagger \Lambda_{1/2}^\dagger)\gamma^0 =
\psi^\dagger \gamma^0 \Lambda_{1/2}^{-1} = \overline\psi
\Lambda_{1/2}^{-1} $$
Because of this special property of $\gamma^0$, now we have that
$\overline\psi\psi\to \overline\psi\psi$.
You can check that the gamma matrices also satisfy the relation
$$ \Lambda_{1/2}^{-1} \gamma^\mu_{ab} \Lambda_{1/2} = {\Lambda^\mu}_\nu \gamma^\nu_{ab}\,. $$
Understand that this is not a transformation of the gamma matrices under a Lorentz transformation. Gamma matrices are fixed constant matrices that form the basis of an algebra. They do not transform. The above is just a property of the gamma matrices due to them being generators of a particular representation of the Lorentz algebra.
However, this relation allows you to take the $\mu$ index in $\gamma^\mu$ "seriously". Because, due to this you can immediately see that under a Lorentz transformation, the current $J^\mu := \overline\psi \gamma^\mu \psi= \overline\psi^a \gamma^\mu_{ab} \psi^b$ indeed transforms like a vector.
$$ J^\mu \to {\Lambda^\mu}_\nu J^\nu \,.$$ |
I am trying to reconcile data that I have found in one publication (Allen 1969) with data that I found in another publication (George 2003) that synthesized this data. The data is root respiration rate, it was originally measured at $27\ ^\circ C$.
Approach
I am trying to convert a rate of oxygen consumed as volume per mass of root per time to carbon dioxide produced as mass per unit mass per time.
In the appendix table, George 2003 reports the range of root respiration rates, converted to $15\ ^\circ C$ and standard units:
$$[11.26, 22.52] \frac{\mathrm{nmol CO}_2}{\mathrm{g}\ \mathrm{s}}$$
In the original publication Allen (1969), root respiration was measured at $27\ ^\circ $C. The values can be found in table 3 and figure 2. The data include a minimum (Group 2 Brunswick, NJ plants) and a maximum (Group 3 Newbery, South Carolina), which I assume are the ones used by George 2003:
$$[27.2, 56.2] \frac{\mu\mathrm{L}\ \mathrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$$
Step 1
Transformed George 2003 measurements back to the measurement temperature using a rearrangement of equation 1 from George, the standardized temperature of $15\ ^\circ $C stated in the Georgeh table legend, and Q$_{10} = 2.075$ from George 2003, and the measurement temperature of $27\ ^\circ $C reported by Allen 1969:
$$R_T = R_{15}[\exp(\ln(Q_{10})(T- 15))/10]$$
$$[11.26, 22.52] * exp(log(2.075)*(27 - 15)/10)$$
Now we have the values that we would have expected to find in the Allen paper, except that the units need to be converted back to the original:
$$[27.03,54.07] \mathrm{nmol CO}_2\ \mathrm{g}^{-1}\mathrm{s}^{-1}$$
Step 2: convert the units Required constants: inverse density of $\mathrm{O}_2$ at $27^\circ C$: $\frac{7.69 \times 10^5\ \mu\mathrm{L}\ \mathrm{O}_2}{\mathrm{g}\ \mathrm{O}_2}$ first assume that Allen converted to sea level pressure (101 kPa), although maybe they were measured at elevation (Allen may have worked at \~{} 900 kPa near Brevard, NC) molar mass of $\mathrm{O}_2$: $\frac{32\mathrm{g}\ \mathrm{O}_2}{\mathrm{mol}}$ treat 10mg, which is in the unit of root mass used by Allen, as a unit of measurement for simplicity
Now convert $$[27.03,54.07] \mathrm{nmol CO}_2\ \mathrm{g}^{-1}\mathrm{s}^{-1}$$ to units of $\frac{\mu\mathrm{L}\ \textrm{O}_2}{10\mathrm{mg}\ \mathrm{root}\ \mathrm{h}}$. The expected result is the original values reported by Allen: $[27.2, 56.2] \frac{\mu\mathrm{L}\ \mathrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$
$$[27.03, 54.07]\ \frac{\mathrm{nmol}\ \mathrm{CO}_2}{\mathrm{g}\ \mathrm{root}\ \mathrm{s}} \times \frac{1\ \mathrm{g}}{100\times10\mathrm{mg}} \times \frac{3600\ \mathrm{s}}{\mathrm{h}} \times \frac{3.2 \times 10^{-8}\ \mathrm{g}\ \mathrm{O}_2}{\mathrm{nmol}\ \mathrm{O}_2}\times \frac{7.69\times10^5\ \mu\mathrm{L}\ \mathrm{O}_2}{\mathrm{g}\ \mathrm{O}_2}$$
Result:
$$[23.8, 47.8] \frac{\mu\mathrm{L}\ \textrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$$
These are the units reported in the Allen paper, but they appear to be underestimates . Since the ratio of observed:expected values are different, it is not likely that Q$_{10}$ or the atmospheric pressure at time of measurement would explain this error.
Question
Am I doing something wrong?
Reference 1: Allen, 1969, Racial variation in physiological characteristics of shortleaf pine roots., Silvics Genetics 18:40-43 Reference 2: George et al 2003, Fine-Root Respiration in a Loblolly Pine and Sweetgum Forest Growing in Elevated CO2. New Phytologist, 160:511-522
Footnote 1: The values from reference 2 are adjusted from the $15^\circ C$ reference temperature to the $27^\circ C$ in reference 1 using the Ahhrenius equation, but I am off by an order of magnitude so I do not think that this is relevant:
$$R_T = R_{15}[\exp(\ln(Q_{10})(T- 15))/10]$$
$$[26.9, 54.0] = [11.2, 22.5] * exp(log(2.075)*(27 - 15)/10)$$
note: I have been updating the equation based errors pointed out by Mark and rcollyer, but the problem remains |
The Existence/Uniqueness of Solutions to Higher Order Linear Differential Equations
We will now begin to look at methods to solving higher order differential equations.
Definition: An $n^{\mathrm{th}}$ Order Differential Equation is a differential equation containing an $n^{\mathrm{th}}$ derivative and can be written in the form $\frac{d^n y}{dt^n} = f \left (t, y, \frac{dy}{dt}, ..., \frac{d^{(n-1)}y}{dt^{(n-1)}} \right )$.
Expanded out, we have that an $n^{\mathrm{th}}$ order
linear differential equation has the form:
Of course we can also represent an $n^{\mathrm{th}}$ order linear differential equation using Lagrange notation:(2)
Here it is important to note that the notation $y^{(k)}$ does NOT mean $y$ raised to the $k^{\mathrm{th}}$ power, but instead, the $k^{\mathrm{th}}$ derivative of $y$.
Now recall from The Existence/Uniqueness of Solutions to First Order Linear Differential Equations and The Existence/Uniqueness of Solutions to Second Order Linear Differential Equations that provided that the coefficient functions were continuous on an interval $I$ containing the point $t_0$, then there existed a unique solution $y = \phi (t)$ throughout the interval $I$ that satisfied the initial conditions $y(t_0) = y_0$ (and in the case of second order linear differential equations, also satisfied $y'(t_0) = y_0'$). The following theorem is a higher order analogue to those last two theorems mentioned.
Theorem 1: Let $p_1, p_2, ..., p_n, g$ all be continuous functions on an open interval $I$ such that $t_0 \in I$. Then the $n^{\mathrm{th}}$ order linear differential equation $\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = g(t)$ with the initial conditions $y(t_0) = y_0$, $y'(t_0) = y_0'$ has a unique solution $y = \phi (t)$ throughout $I$. Once again, it is important to stress that Theorem 1 above is simply an extension to the Theorems on the existence and uniqueness of solutions to first order and second order linear differential equations. Furthermore, for this theorem to apply, we must have that coefficient in front of the $\frac{d^n}{dt^n}$ term is $1$.
Let's look at an example of verifying that a unique solution to a higher order linear differential equation exists.
Example 1 Show that there exists a unique solution to the third order linear differential equation $t\frac{dy^3}{dt^3}+ 3 \frac{d^2y}{dt^2} + \sin t \frac{dy}{dt} + e^t y = 0$ with the initial conditions $y(1) = 1$, $y'(1) = 1$, $y''(1) = 2$.
We first divide each term by $t$ ($t \neq 0$) to get our third order linear differential equation in the appropriate form:(3)
Note that the functions $p_1(t) = \frac{3}{t}$, $p_2(t) = \frac{\sin t}{t}$, and $p_3(t) = \frac{e^t}{t}$ are all continuous for $t \neq 0$. However, we have that $t_0 = 1 \neq 0$, and so in fact we have that a unique solution $y = \phi(t)$ exists to this third order linear differential equation through the interval $(0, \infty)$.
Example 2 For what intervals do solutions to the differential equation $(t - 1) y^{(4)} + \frac{1}{t - 3} y^{(3)} - \ln t y = 4$ exist?
We divide both sides of the equation by $t - 1$ to get:(4)
Note that $\frac{1}{(t - 1)(t - 3)}$ is continuous for $t \neq 1$, $t \neq 3$, $\frac{\ln t}{(t - 1)}$ is continuous for $t > 0$ and $t \neq 1$ and $\frac{4}{(t - 1)}$ is continuous for $t \neq 1$. Therefore, solutions exist on the interval $(0, 1)$, $(1, 3)$, and $(3, \infty)$. |
And a given solution for his languages ${L}_{\mathrm{End}}(M_2)$ and ${L}_{\mathrm{PDA}}(M_2)$ with $ \mathrm{L}_{\mathrm{End}}\left(\mathrm{M}_{2}\right)=\left\{\mathrm{a}^{\mathrm{n}} \mathrm{b}^{\mathrm{m}} x\mid \mathrm{n}, \mathrm{m} \in \mathbb{N}^{+} \wedge x \in\{\mathrm{b}, \mathrm{c}\}^{*} \wedge| x |=\mathrm{m}\right\} $
and
$ \mathrm{L}_{\mathrm{PDA}} = \{a^*\} $
My problem is that I do not understand how to come up with this solution. If I had a DFA, it would not be a problem to find the language, but here i have to find two, and I have no idea how find them. |
I have some very basic questions about stabilizers.
What I understood:
To describe a state $|\psi \rangle$ that lives in an $n$-qubit Hilbert space, we can either give the wavefunction (so the expression of $|\psi\rangle$), either give a set of commuting observable that $|\psi\rangle$ is an eigenvector with $+1$ eigenvalue.
We define a stabilizer $M$ of $|\psi \rangle$ as a tensor product of $n$ Pauli matrices (including the identity) that verifies $M |\psi \rangle = |\psi\rangle$.
And (apparently) we need $n$ stabilizers to fully define a state.
The things I don't understand: How can a stabilizer necessarily be a product of Pauli matrices?
With $n=1$, I take $|\psi \rangle = \alpha | 0 \rangle + \beta |1 \rangle$, excepted for specific values of $\alpha$ and $\beta$, this state is only an eigenvector of $I$ (not of the other pauli matrices). But saying $I$ is the stabilizer doesn't give me which state I am working for.
How can we need only $n$ stabilizers to fully define a state?
With $n$ qubits we have $2^n$ dimensional Hilbert space. I thus expect to have $2^n$ stabilizers, not $n$ to fully describe a state.
I am looking for a simple answer. Preferably an answer based on the same materials as my question, if possible. I am really a beginner in quantum error correction.
I learned these things within a 1-hour tutorial, so I don't have references for which book I learned this from. It is what I understood (maybe badly) from the professor talking. |
Basic Theorems Regarding the Closure of Sets in a Topological Space
Recall from the The Closure of a Set in a Topological Space that if $(X, \tau)$ is a topological space and $A \subseteq X$ then the closure of $A$ denoted $\bar{A}$ is defined to be the smallest closed set containing $A$.
We also proved the very important fact that the closure of $A$ is equal to the union of $A$ with its set of accumulation points $A'$, that is, $\bar{A} = A \cup A'$.
We will now look at some basic theorems regarding the closure of sets in a topological space.
Theorem 1: Let $(X, \tau)$ be a topological space and let $A, B \subseteq X$. If $A \subseteq B$ then $\bar{A} \subseteq \bar{B}$. Proof:Suppose that $A \subseteq B$. Let $x \in \bar{A}$. Since $\bar{A} = A \cup A'$ we have that $x \in A \cup A'$. If $x \in A$ then since $A \subseteq B$ we have that $x \in B$ and $B \subseteq \bar{B}$ so $x \in \bar{B}$. If $x \in A'$ then we've already proven that $A' \subseteq B'$ and $B' \subseteq \bar{B}$ (since $\bar{B} = B \cup B'$) so $x \in \bar{B}$. In both cases we see that $x \in \bar{A}$ implies that $x \in \bar{B}$, so $\bar{A} \subseteq \bar{B}$. $\blacksquare$
In general if $\bar{A} \subseteq \bar{B}$ then we cannot conclude that $A \subseteq B$. For example, consider the set $X = \{ a, b, c, d \}$ and the topology $\tau = \{ \emptyset, \{a \}, \{ b \}, \{a , b \}, \{a, b, c \}, X \}$. Then the closed sets are:(1)
Let $A = \{c, d \}$ and $B = \{ a, c \}$. Then $\bar{A} = \{ c, d \}$ and $\bar{B} = \{ a, c, d \}$, so $\bar{A} \subseteq \bar{B}$. However $A \not \subseteq B$!
Theorem 2: Let $(X, \tau)$ be a topological space and let $A, B \subseteq X$. Then $\bar{A} \cup \bar{B} = \overline{A \cup B}$. Proof:Let $x \in \bar{A} \cup \bar{B}$. Then: So $\bar{A} \cup \bar{B} \subseteq \overline{A \cup B}$. Similarly, if $x \in \overline{A \cup B}$ then: So $\overline{A \cup B} \subseteq \bar{A} \cup \bar{B}$. Hence we conclude that $\bar{A} \cup \bar{B} = \overline{A \cup B}$. $\blacksquare$
Theorem 3: Let $(X, \tau)$ be a topological space and let $A, B \subseteq X$. Then $\bar{A} \cap \bar{B} \supseteq \overline{A \cap B}$. Proof:Let $x \in \overline{A \cap B}$. Then $x \in (A \cap B) \cup (A \cap B)' = (A \cap B) \cap (A' \cap B')$. Let $C = A' \cap B'$. Then: So $x \in (\bar{A} \cap \bar{B}) \cap ([A \cup B'] \cap [B \cup A'])$, i.e., $x \in \bar{A} \cap \bar{B}$, so $\bar{A} \cap \bar{B} \supseteq \overline{A \cap B}$. $\blacksquare$ |
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
That seems like what I need to do, but I don't know how to actually implement it... how wide of a time window is needed for the Y_{t+\tau}? And how on earth do I load all that data at once without it taking forever?
And is there a better or other way to see if shear strain does cause temperature increase, potentially delayed in time
Link to the question: Learning roadmap for picking up enough mathematical know-how in order to model "shape", "form" and "material properties"?Alternatively, where could I go in order to have such a question answered?
@tpg2114 For reducing data point for calculating time correlation, you can run two exactly the simulation in parallel separated by the time lag dt. Then there is no need to store all snapshot and spatial points.
@DavidZ I wasn't trying to justify it's existence here, just merely pointing out that because there were some numerics questions posted here, some people might think it okay to post more. I still think marking it as a duplicate is a good idea, then probably an historical lock on the others (maybe with a warning that questions like these belong on Comp Sci?)
The x axis is the index in the array -- so I have 200 time series
Each one is equally spaced, 1e-9 seconds apart
The black line is \frac{d T}{d t} and doesn't have an axis -- I don't care what the values are
The solid blue line is the abs(shear strain) and is valued on the right axis
The dashed blue line is the result from scipy.signal.correlate
And is valued on the left axis
So what I don't understand: 1) Why is the correlation value negative when they look pretty positively correlated to me? 2) Why is the result from the correlation function 400 time steps long? 3) How do I find the lead/lag between the signals? Wikipedia says the argmin or argmax of the result will tell me that, but I don't know how
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
Because I don't know how the result is indexed in time
Related:Why don't we just ban homework altogether?Banning homework: vote and documentationWe're having some more recent discussions on the homework tag. A month ago, there was a flurry of activity involving a tightening up of the policy. Unfortunately, I was really busy after th...
So, things we need to decide (but not necessarily today): (1) do we implement John Rennie's suggestion of having the mods not close homework questions for a month (2) do we reword the homework policy, and how (3) do we get rid of the tag
I think (1) would be a decent option if we had >5 3k+ voters online at any one time to do the small-time moderating. Between the HW being posted and (finally) being closed, there's usually some <1k poster who answers the question
It'd be better if we could do it quick enough that no answers get posted until the question is clarified to satisfy the current HW policy
For the SHO, our teacher told us to scale$$p\rightarrow \sqrt{m\omega\hbar} ~p$$$$x\rightarrow \sqrt{\frac{\hbar}{m\omega}}~x$$And then define the following$$K_1=\frac 14 (p^2-q^2)$$$$K_2=\frac 14 (pq+qp)$$$$J_3=\frac{H}{2\hbar\omega}=\frac 14(p^2+q^2)$$The first part is to show that$$Q \...
Okay. I guess we'll have to see what people say but my guess is the unclear part is what constitutes homework itself. We've had discussions where some people equate it to the level of the question and not the content, or where "where is my mistake in the math" is okay if it's advanced topics but not for mechanics
Part of my motivation for wanting to write a revised homework policy is to make explicit that any question asking "Where did I go wrong?" or "Is this the right equation to use?" (without further clarification) or "Any feedback would be appreciated" is not okay
@jinawee oh, that I don't think will happen.
In any case that would be an indication that homework is a meta tag, i.e. a tag that we shouldn't have.
So anyway, I think suggestions for things that need to be clarified -- what is homework and what is "conceptual." Ie. is it conceptual to be stuck when deriving the distribution of microstates cause somebody doesn't know what Stirling's Approximation is
Some have argued that is on topic even though there's nothing really physical about it just because it's 'graduate level'
Others would argue it's not on topic because it's not conceptual
How can one prove that$$ \operatorname{Tr} \log \cal{A} =\int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr}e^{-s \mathcal{A}},$$for a sufficiently well-behaved operator $\cal{A}?$How (mathematically) rigorous is the expression?I'm looking at the $d=2$ Euclidean case, as discuss...
I've noticed that there is a remarkable difference between me in a selfie and me in the mirror. Left-right reversal might be part of it, but I wonder what is the r-e-a-l reason. Too bad the question got closed.
And what about selfies in the mirror? (I didn't try yet.)
@KyleKanos @jinawee @DavidZ @tpg2114 So my take is that we should probably do the "mods only 5th vote"-- I've already been doing that for a while, except for that occasional time when I just wipe the queue clean.
Additionally, what we can do instead is go through the closed questions and delete the homework ones as quickly as possible, as mods.
Or maybe that can be a second step.
If we can reduce visibility of HW, then the tag becomes less of a bone of contention
@jinawee I think if someone asks, "How do I do Jackson 11.26," it certainly should be marked as homework. But if someone asks, say, "How is source theory different from qft?" it certainly shouldn't be marked as Homework
@Dilaton because that's talking about the tag. And like I said, everyone has a different meaning for the tag, so we'll have to phase it out. There's no need for it if we are able to swiftly handle the main page closeable homework clutter.
@Dilaton also, have a look at the topvoted answers on both.
Afternoon folks. I tend to ask questions about perturbation methods and asymptotic expansions that arise in my work over on Math.SE, but most of those folks aren't too interested in these kinds of approximate questions. Would posts like this be on topic at Physics.SE? (my initial feeling is no because its really a math question, but I figured I'd ask anyway)
@DavidZ Ya I figured as much. Thanks for the typo catch. Do you know of any other place for questions like this? I spend a lot of time at math.SE and they're really mostly interested in either high-level pure math or recreational math (limits, series, integrals, etc). There doesn't seem to be a good place for the approximate and applied techniques I tend to rely on.
hm... I guess you could check at Computational Science. I wouldn't necessarily expect it to be on topic there either, since that's mostly numerical methods and stuff about scientific software, but it's worth looking into at least.
Or... to be honest, if you were to rephrase your question in a way that makes clear how it's about physics, it might actually be okay on this site. There's a fine line between math and theoretical physics sometimes.
MO is for research-level mathematics, not "how do I compute X"
user54412
@KevinDriscoll You could maybe reword to push that question in the direction of another site, but imo as worded it falls squarely in the domain of math.SE - it's just a shame they don't give that kind of question as much attention as, say, explaining why 7 is the only prime followed by a cube
@ChrisWhite As I understand it, KITP wants big names in the field who will promote crazy ideas with the intent of getting someone else to develop their idea into a reasonable solution (c.f., Hawking's recent paper) |
Advances in Differential Equations Adv. Differential Equations Volume 23, Number 9/10 (2018), 725-750. Well-posedness for the Cauchy problem of the Klein-Gordon-Zakharov system in five and higher dimensions Abstract
We study the Cauchy problem of the Klein-Gordon-Zakharov system in spatial dimension $d \ge 5$ with initial datum $(u, \partial_t u, n, \partial_t n)|_{t=0} \in H^{s+1}(\mathbb{R}^d) \times H^s(\mathbb{R}^d) \times \dot{H}^s(\mathbb{R}^d) \times \dot{H}^{s-1} (\mathbb{R}^d)$. The critical value of $s$ is $s_c=d/2-2$. By $U^2, V^2$ type spaces, we prove that the small data global well-posedness and scattering hold at $s=s_c$ in $d \ge 5$.
Article information Source Adv. Differential Equations, Volume 23, Number 9/10 (2018), 725-750. Dates First available in Project Euclid: 13 June 2018 Permanent link to this document https://projecteuclid.org/euclid.ade/1528855477 Mathematical Reviews number (MathSciNet) MR3813998 Zentralblatt MATH identifier 06973943 Subjects Primary: 35Q55: NLS-like equations (nonlinear Schrödinger) [See also 37K10] 35B40: Asymptotic behavior of solutions 35A01: Existence problems: global existence, local existence, non-existence 35A02: Uniqueness problems: global uniqueness, local uniqueness, non- uniqueness Citation
Kato, Isao; Kinoshita, Shinya. Well-posedness for the Cauchy problem of the Klein-Gordon-Zakharov system in five and higher dimensions. Adv. Differential Equations 23 (2018), no. 9/10, 725--750. https://projecteuclid.org/euclid.ade/1528855477 |
The Eigenvalues of a Matrix
Often times when we're dealing with linear operators we can use an $n \times n$ matrix $A$ to describe such an operator. Sometimes such a matrix will have eigenvalues which are critically important in applications to linear operators. We formally define an eigenvalue of a matrix below.
Definition: If $A$ is an $n \times n$ matrix, then $\lambda$ is an eigenvalue of $A$ if $Av = \lambda v$ for some nonzero column vector $v$. The vectors $v$ which satisfy this equation are called the corresponding Eigenvectors to the eigenvalue $\lambda$.
For example, consider one of the simplest of matrices, the $2 \times 2$ identity matrix $I = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$, and consider the equation $Iv = \lambda v$. We note that $Iv = v$, so this equation reduces to $v = \lambda v$, so $\lambda = 1$ is an eigenvalue of the identity matrix, and every vector is an eigenvector corresponding to the eigenvalue $\lambda$.
For another example, consider the $3 \times 3$ matrix $A = \begin{bmatrix} 3 & 1 & 1\\ 0 & 2 & 0\\ 0 & 0 & 1 \end{bmatrix}$. The eigenvalues of this matrix are $\lambda_1 = 1$, $\lambda_2 = 2$, and $\lambda_3 = 3$. Let's prove this.
First consider the eigenvalue $\lambda_1 = 1$. We need to show that there exists a nonzero vector $v$ such that $Av = 1v = v$, that is, consider the equation:(1)
From this matrix equation above we have that:(2)
We have that the second equation implies that $v_2 = 0$. Meanwhile, the third equation implies that $v_3$ can equal any number, say $v_3 = a \in \mathbb{R}$. Thus equation 1 implies that $3v_1 + a = v_1$, so $2v_1 = -a$ and $v_1 = \frac{-a}{2}$. Therefore, the corresponding eigenvectors to the eigenvalue $\lambda_1 = 1$ are $\begin{bmatrix} \frac{-a}{2}\\ 0\\ a \end{bmatrix}$.
Now let's consider the eigenvalue $\lambda_2 = 2$ and consider the following equation:(3)
From this matrix equation above we have the following system of equations:(4)
The third equation above implies that $v_3 = 0$. Furthermore, the second equation implies that $v_2$ can be any number, say $v_2 = b \in \mathbb{R}$. The first equation implies that $3v_1 + b = 2v_1$ so $v_1 = -b$. Therefore, the corresponding eigenvectors to the eigenvalue $\lambda_2 = 2$ are $\begin{bmatrix} -b \\ b\\ 0 \end{bmatrix}$.
Lastly, consider the the eigenvalue $\lambda_3 = 3$ and consider the following equation:(5)
From this matrix equation above we have the following system of equation:(6)
The second and third equations imply that $v_2 = 0$ and $v_3 = 0$. Plugging this into the first equation and we get that $3v_1 = 3v_1$, so $v_1$ can be any number, say $v_1 = c \in \mathbb{R}$, and so the corresponding eigenvectors for the eigenvalue $\lambda_3 = 3$ are $\begin{bmatrix} c \\ 0\\ 0 \end{bmatrix}$.
Note that in the previous example, the eigenvalues of $A$ were the entries on the main diagonal. For triangular matrices, this is always true as summarized with the following theorem.
Theorem 1: If $A$ is an $n \times n$ triangular matrix, then the eigenvalues of $A$ are precisely the entries on the main diagonal of $A$.
We will soon look at some more methods for quickly computing the eigenvalues associated with a matrix. |
How can I track a fixed point $P=(x_P, y_P)$ from a moving robot?
Coordinates of $P$ are relative to the state/pose of the robot (x axis looks forward the robot and y axis is positive on the right of the robot). Suppose that the initial robot state/pose is at $S_{R}=(x_R, y_R, \theta_R)$. The next frame (namely after $\Delta t$) with the applied control $(v, \omega)$ the robot is at state $S_{R'}=(x_{R'}, y_{R'}, \theta_{R'})$.
Where (I set the axes as OpenCV):
$x_{R'} = x_R + v cos(\theta_R) \Delta t $
$y_{R'} = y_R + v sin(\theta_R) \Delta t $
$\theta_{R'} = \theta_{R} + \omega\Delta t$
The question is: which are the coordinates $(x_P', y_P')$ of the same point $P$ relative to $S_{R'}$?
As visible in the picture, I know the transformation from the initial state to the next state of the robot and the coordinate of P in reference to the initial state
$$ t = \begin{pmatrix} cos(\theta_{R'}) & -sin(\theta_{R'}) & x_{R'}\\ sin(\theta_{R'}) & cos(\theta_{R'}) & y_{R'}\\ 0 & 0 & 1\\ \end{pmatrix} $$
Please correct me if I made some mistakes!
Thank you, any help is appreciated. |
Introduction
This page is a description of a Hill-type muscle model implemented in OpenSim that is based on the work of Thelen (2003). The model was modified by Matt Millard, Ajay Seth, and Peter Loan, and it is implemented in OpenSim as OpenSim::Thelen2003Muscle. In previous versions of OpenSim (2.4 and earlier), there were numerical errors in the implementation.
Background
The force-producing properties of muscle are complex and nonlinear (see McMahon (1984) for review) (Fig. 1). For simplicity, lumped-parameter, dimensionless muscle models, capable of representing a range of muscles with different architectures, are most commonly used in the dynamic simulation of movement (Zajac, 1989). In a complex musculoskeletal model, the model can be actuated by 50 or more muscle–tendon units, each of which is represented as a Hill-type contractile element in series with a tendon.
The Thelen2003Muscle model uses a standard equilibrium muscle model based on the Hill model. The muscle–tendon complex consists of three components: a contractile element (CE), a parallel element (PE), and a series element (SE). The muscle force generated is a function of three factors: the activation value (a), the normalized length of the muscle unit, and the normalized velocity of the muscle unit. The functions describing the force generated by a muscle as its length varies are called the active length curve (AL) for the contractile element and the passive length curve (PL) for the parallel element.
The parameters used to characterize each muscle are maximum isometric force, optimal muscle fiber length, tendon slack length, maximum contraction velocity, and pennation angle. For a table of muscle and tendon parameters, consult Anderson and Pandy (1999). During a forward dynamic simulation, the muscle force is calculated using two states: the activation value and the muscle fiber length.
Figure 1 from Thelen (2003).
Implementation
Using Newton's third law (under the assumption that the muscle and tendon units are massless), the differential equation of the muscle–tendon element is
\[ f_{iso} \left(a\left(t\right)f_{AL}\left(l^M\,\right)f_{v}\big(\dot{l}^M\,\big) + f_{PL}\left(l^M\,\right)\right) cos\alpha - f_{iso}\,f_{SE}\left(l^T\,\right) = 0\]
Rearranging the above equation and solving for \( f_{v}\big(\dot{l}^{M}\,\big)\) yields
\[ f_{v}\big(\dot{l}^{M}\,\big) = \frac{\frac{f_{SE}\left(l^{T}\right)}{cos\alpha} - f_{PL}\left(l^{M}\right)}{a\left(t\right)f_{AL}\left(l^{M}\,\right)}\]
\[ \dot{l}^{M} = f_{v}^{-1} \left\{ \frac{\frac{f_{SE}\left(l^T\right)}{cos\alpha} - f_{PL}\left(l^{M}\right)}{a\left(t\right)f_{AL}\left(l^{M}\,\right)} \right\}\]
\( a\left(t\right) \rightarrow 0\)
\( f_{AL}\left(l^{M}\,\right) \rightarrow 0\)
\( \alpha \rightarrow 90°\)
\( f_{v}\big(\dot{l}^{M}\,\big) \leq 0\) or \( f_{v}\big(\dot{l}^{M}\,\big) \geq F^{m}_{len}\)
This implementation has been modified from the model presented in Thelen (2003) to avoid some of the numerical singularities:
\( a\left(t\right) \rightarrow a_{min} \gt 0\) A modified activation dynamic equation is used (MuscleFirstOrderActivationDynamicModel) that smoothly approaches a minimum value that is greater than zero.
\( f_{AL}\left(l^{M}\,\right) \rightarrow 0\) The active-force–length curve of the Thelen muscle is a Gaussian, which is always greater than 0.
\( \alpha \rightarrow 90°\) This singularity cannot be removed without changing the first equation, and still exists in the present Thelen2003Muscle implementation.
\( f_{v}\big(\dot{l}^{M}\,\big) \leq 0\) or \( f_{v}\big(\dot{l}^{M}\,\big) \geq F^{m}_{len}\) Equation 6 in Thelen (2003) has been modified so that \( V^{M}\) is linearly extrapolated when \( F^{M} \lt 0\) (during a concentric contraction) and when \( F^{M} \gt 0.95 F^{M}_{len}\) (during an eccentric contraction). These two modifications make the force–velocity curve invertible. The original force–velocity curve published by Thelen was not invertible.
A unilateral constraint has been implemented to prevent the fiber from approaching a fiber length that is smaller than 0.01*optimal fiber length, or a fiber length that creates a pennation angle greater than the maximum pennation angle specified by the pennation model. Note that this unilateral constraint does not prevent the muscle fiber from becoming shorter than is physiologically possible (i.e., shorter than approximately half the normalized fiber length). References
McMahon, T.A. (1984) Muscles, Reflexes, and Locomotion. Princeton University Press, Princeton, New Jersey.
Thelen, D.G. (2003) Adjustment of muscle mechanics model parameters to simulate dynamic contractions in older adults. ASME Journal of Biomechanical Engineering, 125(1):70–77. Download PDF here. Zajac, F.E. (1989) Muscle and tendon: properties, models, scaling, and application to biomechanics and motor control. Critical Reviews in Biomedical Engineering, 17(4):359–411. |
Consider a random sample from exponential distribution with mean $\frac{1}{\theta}$. I have to prove that $nX_{(1)}$ is not consistent for $\frac{1}{\theta}$ . A sufficient condition for consistency is
$$\lim_{n \to \infty}E(\hat\theta)=\frac{1}{\theta}\quad ;\quad \lim_{n \to \infty}\operatorname{Var}(\hat\theta)=0$$
Since $X_{(1)}$ follow $\exp(n\theta)$, the second condition, was seen to be violated as its variance $\frac{1}{\theta^{2}}$ doesn't tend to zero. Is this enough to prove that $nX_{(1)}$ is inconsistent?
Can i say that since $X_{(1)}$ converge in probability to $\frac{1}{n\theta}$, $nX_{(1)}$ converge in probability to $\frac{1}{\theta}$. But this does not prove what i want.
Im confused about the right method to be used. |
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Constraining the magnitude of the Chiral Magnetic Effect with Event Shape Engineering in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2018-02)
In ultrarelativistic heavy-ion collisions, the event-by-event variation of the elliptic flow $v_2$ reflects fluctuations in the shape of the initial state of the system. This allows to select events with the same centrality ...
First measurement of jet mass in Pb–Pb and p–Pb collisions at the LHC
(Elsevier, 2018-01)
This letter presents the first measurement of jet mass in Pb-Pb and p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV and 5.02 TeV, respectively. Both the jet energy and the jet mass are expected to be sensitive to jet ...
First measurement of $\Xi_{\rm c}^0$ production in pp collisions at $\mathbf{\sqrt{s}}$ = 7 TeV
(Elsevier, 2018-06)
The production of the charm-strange baryon $\Xi_{\rm c}^0$ is measured for the first time at the LHC via its semileptonic decay into e$^+\Xi^-\nu_{\rm e}$ in pp collisions at $\sqrt{s}=7$ TeV with the ALICE detector. The ...
D-meson azimuthal anisotropy in mid-central Pb-Pb collisions at $\mathbf{\sqrt{s_{\rm NN}}=5.02}$ TeV
(American Physical Society, 2018-03)
The azimuthal anisotropy coefficient $v_2$ of prompt D$^0$, D$^+$, D$^{*+}$ and D$_s^+$ mesons was measured in mid-central (30-50% centrality class) Pb-Pb collisions at a centre-of-mass energy per nucleon pair $\sqrt{s_{\rm ...
Dielectron production in proton-proton collisions at √s=7 TeV
(Springer, 2018-09-12)
The first measurement of e^+e^− pair production at mid-rapidity (|ηe| < 0.8) in pp collisions at √s=7 TeV with ALICE at the LHC is presented. The dielectron production is studied as a function of the invariant mass (m_ee ...
Anisotropic flow of identified particles in Pb-Pb collisions at √sNN=5.02 TeV
(Springer, 2018-09-03)
The elliptic (v_2), triangular (v_3), and quadrangular (v-4) flow coefficients of π^±, K^±, p+p¯,Λ+Λ¯,K^0_S , and the ϕ-meson are measured in Pb-Pb collisions at √s_NN=5.02 TeV. Results obtained with the scalar product ...
Azimuthally-differential pion femtoscopy relative to the third harmonic event plane in Pb–Pb collisions at √sNN = 2.76 TeV
(Elsevier, 2018-06-22)
Azimuthally-differential femtoscopic measurements, being sensitive to spatio-temporal characteristics of the source as well as to the collective velocity fields at freeze out, provide very important information on the ...
Inclusive J/ψ production at forward and backward rapidity in p-Pb collisions at √sNN=8.16 TeV
(Springer Berlin Heidelberg, 2018-07-25)
Inclusive J/ψ production is studied in p-Pb interactions at a centre-of-mass energy per nucleon-nucleon collision 𝑠NN‾‾‾‾√=8.16 TeV, using the ALICE detector at the CERN LHC. The J/ψ meson is reconstructed, via its ...
Inclusive J/ψ proInclusive J/ψ production in Xe–Xe collisions at √sNN = 5.44 TeVduction in Xe–Xe collisions at √sNN = 5.44 TeV
(Elsevier, 2018-08-31)
Inclusive J/ψ production is studied in Xe–Xe interactions at a centre-of-mass energy per nucleon pair of TeV, using the ALICE detector at the CERN LHC. The J/ψ meson is reconstructed via its decay into a muon pair, in the ...
Neutral pion and η meson production at midrapidity in Pb-Pb collisions at √sNN=2.76 TeV
(American Physical Society, 2018-10-04)
Neutral pion and η meson production in the transverse momentum range 1 < p_T < 20 GeV/c have been measured at midrapidity by the ALICE experiment at the Large Hadron Collider (LHC) in central and semicentral Pb-Pb collisions ... |
Let $G$ be an Abelian group. And let H={$g\in G |\mathrm{order}(g) < \infty$}
I need to prove that H is a normal subgroup in G.
I know that if i prove it's a subgroup, it will be normal as well, since G is Abelian. But i wonder about it being a subgroup.
I know this has to work: $$\forall h_1, h_2\in H,\hspace{1cm}h_1*h_2^{-1}\in H$$
But I wonder, is it enough to just show that the order of the product is still finite? Is there any special way to show the order is finite? Or is it just trivial?
If anyone can help clearing this out, I would really appreciate it. |
Still at some of my earliest encounters with the elementary concepts of Set Theory, I understand what the
Power set of a set is but I have a hard time wrapping my head around what the definition implies in the case of a set like $\Bbb R^2$.
I understand that any function $f:\Bbb R \rightarrow \Bbb R$ is a subset of $\Bbb R^2$
$$\lbrace (x,f(x)): x\in\Bbb R\rbrace \subseteq \Bbb R^2$$
and I can also see why this extends to any set of points in the Cartesian plane
$$\lbrace (x,y): x\in\Bbb R \wedge y \in\Bbb R\rbrace \subseteq \Bbb R^2$$
but then, when I think about what this could mean, my brain starts to hurt a bit. If I can think about
something and then write that something down or make a sketch of it, it becomes a collection of points in a plane ergo, by $\lbrace (x,y): x\in\Bbb R \wedge y \in\Bbb R\rbrace \subseteq \Bbb R^2$, it is an element of $\mathscr P(\Bbb R^2)$.
Does that imply
any drawing, text, plan, map or even this post could be considered a subset of $\Bbb R^2$ such that $\mathscr P(\Bbb R^2)$ contains any $2$-dimensional "object"?
Since I can make a sketch of most of the things I can think about, does that mean $\mathscr P(\Bbb R^2)$ contains anything that I could think about and sketch and even anything I could sketch but will never think about as well?
Thanks for taking the time to provide your input, I appreciate it a lot! |
Is $S^4$ diffeomorfhic to $S^2\times S^2$?
Moreover. Is $S^n$ diffeomorphic to some cross product of manifolds $X\times Y$ for $n\geq2$?
Is there a elemental topological invariant to let me see this?
Any suggestions are welcome! thanks
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Is $S^4$ diffeomorfhic to $S^2\times S^2$?
Moreover. Is $S^n$ diffeomorphic to some cross product of manifolds $X\times Y$ for $n\geq2$?
Is there a elemental topological invariant to let me see this?
Any suggestions are welcome! thanks
The second homotopy group, or the second homology group of $S^2\times S^2$ is isomorphic to $\mathbb Z\times\mathbb Z$, which that of $S^4$ is zero, so the two spaces are not homeomorphic.
Suppose that $S^n\cong X\times Y$, and fix an homeo. Pick points $x\in X$ and $y\in Y$, and let $r$ be the point in $S^k$ corresponding to $(x,y)$. The Künneth formula for relative homology tells us then that $$H_p(S^n,S^n\setminus r)\cong H_p(X\times Y,X\times Y\setminus\{(x,y)\}\cong\bigoplus_{p'+p''=p}H_{p'}(X,X\setminus x)\otimes H_{p''}(Y,Y\setminus y),$$ taking coefficients in any field, for all $p\geq0$.
If $X$ and $Y$ are manifolds of dimensions $n'$ and $n''$, then this implies that $H_{n'-1}(S^n,S^n\setminus r)\neq0$ and $H_{n''-1}(S^n,S^n\setminus r)\neq0$, because in the direct sum decomposition above there are non-zero terms. Moreover, we know thhat $H_{n-1}(S^n,S^n\setminus r)\neq0$.
Since the sphere is a manifold, this is only possible if the set $\{n,n',n''\}$ has only two elements. Since $n=n'+n''$, one of $n'$ or $n''$ must be zero. Suppose $n'=0$. Since $X$ and $Y$ must be connected, because $S^n$ is, we see that $X$ is point.
You can probably push this to $X$ and $Y$ any space, as in https://mathoverflow.net/questions/60375/is-r3-the-square-of-some-topological-space |
A Vicious Pikeman (Easy) Note that this is an easier version of the problem pikemanhard.
One of the reasons for this somewhat peculiar theory was the finding of ancient pike, a combat spear. Scientists have found many of these throughout the years. They come with a special symbol carved into them, usually the symbol of the tribe. This one didn’t have a symbol carved into it, it had pseudo code for Fenwick trees, as well as a config file for some kind of editor. Scientists are unsure which editor it might have been, but they believe it was some version of the closed Emacs beta.
Instead of looking for more evidence, the archeologists started speculating what strategy the pikemen used in these programming contests. They must have been well prepared, since this guy had algorithms carved into his spear. The contest rules were probably as follows: When submiting a solution to the judge, the time in minutes from contest start was added to a penalty counter. So in order to plan his problem solving, a pikeman must have been good at approximating the number of minutes required to solve each problem.
You are given a number of problems which were designed for a contest in which the pikeman participated. For each problem, you are given the estimated time in minutes for solving the problem. Calculate the maximum number of problems a pikeman can solve in the contest, and the minimum penalty he can get, under the assumptions that these estimations are correct. You may assume that the pikemen are very efficient: submissions are always correct, and after submitting a problem they start solving the next problem immediately.
Input
Input starts with two integers on a single line $1 \le N \le 10^4$ and $1 \le T \le 10^{9}$, the number of problems in the ancient contest and the total length of the contest in minutes. Then follows a line with four integers $1 \le A, B, C, t_0 \le 10^6$, where $t_0$ $(t_0\leq C)$ specifies the time in minutes required for solving the first problem, and the rest of the times $t_1, \dots , t_{N-1}$ are given by:\begin{equation*} t_ i = ((At_{i-1}+B) \text {mod } C) + 1, i \in [1,N-1] \end{equation*}
Output
Output should consist of two integers: the maximum number of problems a pikeman can solve within the time limit, and the total penalty he will get for solving them. As the penalty might be huge, print it modulo $1\, 000\, 000\, 007$. Print them on the same line, separated by a single space.
Sample Input 1 Sample Output 1 1 3 2 2 2 1 1 1
Sample Input 2 Sample Output 2 2 10 2 2 2 2 2 4 |
The Interior of Open Sets in a Topological Space
Recall from The Interior Points of Sets in a Topological Space page that if $(X, \tau)$ is a topological space and $A \subseteq X$ then a point $a \in A$ is said to be an interior point of $A$ if there exists an open neighbourhood $U$ of $a$ contained in $A$, that is, there exists a $U \in \tau$ with $a \in U$ such that:(1)
We called the set of all interior points of $A$ the interior of $A$ and denoted it by $\mathrm{int} (A)$. Equivalently, we saw that $\mathrm{int} (A)$ is the largest open set contained in $A$.
We will now look at a very useful theorem which will tell us that a set is open if and only if $A = \mathrm{int} (A)$.
Theorem 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $A$ is open if and only if every $a \in A$ is an interior point of $A$, i.e., $A = \mathrm{int} (A)$. Proof:$\Rightarrow$ Suppose that $A$ is an open set. Then $A \in \tau$ and so for every element $a \in A$ let $U = A \in \tau$ so that: Hence every point $a \in A$ is an interior point of $A$. $\Leftarrow$ Now suppose that every $a \in A$ is an interior point of $A$. Then for every $a \in A$ there exists a an open subset $U_a \in \tau$ such that: So $\displaystyle{A = \bigcup_{a \in A} U_a}$ with is a union of arbitrary open sets which is open. Therefore $A$ is open. $\blacksquare$ |
The Weak Topology Induced by W ⊆ X♯
Recall from The Weak Topology Induced by F page that if $X$ is a nonempty set and if $\mathcal F$ is a collection of functions defined on $X$ then for each $\epsilon > 0$, $F \subseteq \mathcal F$ finite, and $x \in X$ we defined:(1)
We then proved that the collection $\{ N_{\epsilon, F}(x) : \epsilon > 0, \: F \subseteq \mathcal F \: \mathrm{finite}, \: x \in X \}$ is a base for some topology $\tau$ on $X$ and we defined that topology to be the weak topology induced by $\mathcal F$.
Now let $X$ be a linear space. Recall that $X^{\sharp}$ is the collection of all linear operators on $X$. If $W \subseteq X^{\sharp}$ then we can consider the $W$-weak topology on $X$.
The following proposition tells us that the only $W$-weakly continuous functions are those functions in $W$.
Theorem 1: Let $X$ be a linear space and let $W \subseteq X^{\sharp}$. Then a linear functional $\varphi \in X^{\sharp}$ is continuous with respect to the $W$-weak topology on $X$ if and only if $\varphi \in W$. Proof:$\Rightarrow$ Let $\varphi : X \to \mathbb{C}$ be a linear functional that is continuous with respect to the $W$-weak topology on $X$. Consider the following set: Then the above set is the inverse image of the unit open ball in $\mathbb{C}$. Since $\varphi$ is continuous (with respect to the $W$-weak topology on $X$), the above set is open and also contains $0$. So there exists an element of the base that contains $0$ and is contained in this set. That is, there exists an $\epsilon > 0$ and a finite set $F \subseteq W$ for which: Therefore, if $| \psi (x) | < \epsilon$ for all $\psi \in F$ we have that $| \varphi (x) | < 1$. Now let $x \in \bigcap_{\psi \in F} \ker \psi$. Then $\psi (x) = 0$ for every $\psi \in F$. Hence: Hence $|\varphi (tx)| < 1$ for every $t > 0$ and: This shows that $| \varphi(x) | = 0$, so $\varphi (x) = 0$. Hence $x \in \ker \varphi$. So: From the theorem on the Expressing a Linear Functional as a Linear Combination of Other Linear Functionals page, we must have that $\varphi$ is a linear combination of the linear functionals in $F$. So $\varphi \in W$. $\Leftarrow$ Suppose that $\varphi \in W$. Since the $W$-weak topology is the weakest topology which makes all of the functions in $W$ continuous, we have that $\varphi$ is continuous with respect to the $W$-weak topology. $\blacksquare$ |
One disadvantage of the fact that you have posted 5 identical answers (1, 2, 3, 4, 5) is that if other users have some comments about the website you created, they will post them in all these place. If you have some place online where you would like to receive feedback, you should probably also add link to that. — Martin Sleziak1 min ago
BTW your program looks very interesting, in particular the way to enter mathematics.
One thing that seem to be missing is documentation (at least I did not find it).
This means that it is not explained anywhere: 1) How a search query is entered. 2) What the search engine actually looks for.
For example upon entering $\frac xy$ will it find also $\frac{\alpha}{\beta}$? Or even $\alpha/\beta$? What about $\frac{x_1}{x_2}$?
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Is it possible to save a link to particular search query? For example in Google I am able to use link such as: google.com/search?q=approach0+xyz Feature like that would be useful for posting bug reports.
When I try to click on "raw query", I get curl -v https://approach0.xyz/search/search-relay.php?q='%24%5Cfrac%7Bx%7D%7By%7D%24' But pasting the link into the browser does not do what I expected it to.
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If I copy-paste search query into your search engine, it does not work. For example, if I copy $\frac xy$ and paste it, I do not get what would I expect. Which means I have to type every query. Possibility to paste would be useful for long formulas. Here is what I get after pasting this particular string:
I was not able to enter integrals with bounds, such as $\int_0^1$. This is what I get instead:
One thing which we should keep in mind is that duplicates might be useful. They improve the chance that another user will find the question, since with each duplicate another copy with somewhat different phrasing of the title is added. So if you spent reasonable time by searching and did not find...
In comments and other answers it was mentioned that there are some other search engines which could be better when searching for mathematical expressions. But I think that as nowadays several pages uses LaTex syntax (Wikipedia, this site, to mention just two important examples). Additionally, som...
@MartinSleziak Thank you so much for your comments and suggestions here. I have took a brief look at your feedback, I really love your feedback and will seriously look into those points and improve approach0. Give me just some minutes, I will answer/reply to your in feedback in our chat. — Wei Zhong1 min ago
I still think that it would be useful if you added to your post where do you want to receive feedback from math.SE users. (I suppose I was not the only person to try it.) Especially since you wrote: "I am hoping someone interested can join and form a community to push this project forward, "
BTW those animations with examples of searching look really cool.
@MartinSleziak Thanks to your advice, I have appended more information on my posted answers. Will reply to you shortly in chat. — Wei Zhong29 secs ago
We are open-source project hosted on GitHub: http://github.com/approach0Welcome to send any feedback on our GitHub issue page!
@MartinSleziak Currently it has only a documentation for developers (approach0.xyz/docs) hopefully this project will accelerate its releasing process when people get involved. But I will list this as a important TODO before publishing approach0.xyz . At that time I hope there will be a helpful guide page for new users.
@MartinSleziak Yes, $x+y$ will find $a+b$ too, IMHO this is the very basic requirement for a math-aware search engine. Actually, approach0 will look into expression structure and symbolic alpha-equivalence too. But for now, $x_1$ will not get $x$ because approach0 consider them not structurally identical, but you can use wildcard to match $x_1$ just by entering a question mark "?" or \qvar{x} in a math formula. As for your example, enter $\frac \qvar{x} \qvar{y} $ is enough to match it.
@MartinSleziak As for the query link, it needs more explanation, technologically the way you mentioned that Google is using, is a HTTP GET method, but for mathematics, GET request may be not appropriate since it has structure in a query, usually developer would alternatively use a HTTP POST request, with JSON encoded. This makes developing much more easier because JSON is a rich-structured and easy to seperate math keywords.
@MartinSleziak Right now there are two solutions for "query link" problem you addressed. First is to use browser back/forward button to navigate among query history.
@MartinSleziak Second is to use a computer command line 'curl' to get search results from particular query link (you can actually see that in browser, but it is in developer tools, such as the network inspection tab of Chrome). I agree it is helpful to add a GET query link for user to refer to a query, I will write this point in project TODO and improve this later. (just need some extra efforts though)
@MartinSleziak Yes, if you search \alpha, you will get all \alpha document ranked top, different symbols such as "a", "b" ranked after exact match.
@MartinSleziak Approach0 plans to add a "Symbol Pad" just like what www.symbolab.com and searchonmath.com are using. This will help user to input greek symbols even if they do not remember how to spell.
@MartinSleziak Yes, you can get, greek letters are tokenized to the same thing as normal alphabets.
@MartinSleziak As for integrals upper bounds, I think it is a problem on a JavaScript plugin approch0 is using, I also observe this issue, only thing you can do is to use arrow key to move cursor to the right most and hit a '^' so it goes to upper bound edit.
@MartinSleziak Yes, it has a threshold now, but this is easy to adjust from source code. Most importantly, I have ONLY 1000 pages indexed, which means only 30,000 posts on math stackexchange. This is a very small number, but will index more posts/pages when search engine efficiency and relevance is tuned.
@MartinSleziak As I mentioned, the indices is too small currently. You probably will get what you want when this project develops to the next stage, which is enlarge index and publish.
@MartinSleziak Thank you for all your suggestions, currently I just hope more developers get to know this project, indeed, this is my side project, development progress can be very slow due to my time constrain. But I believe its usefulness and will spend my spare time to develop until its publish.
So, we would not have polls like: "What is your favorite calculus textbook?" — GEdgar2 hours ago
@GEdgar I'd say this goes under "tools." But perhaps it could be made explicit. — quid1 hour ago
@quid I think that the type of question mentioned in GEdgar's comment is closer to book-recommendations which are valid questions on the main. (Although not formulated like that.) I also think that his comment was tongue-in-cheek. (Although it is a bit more difficult for me to detect sarcasm, as I am not a native speaker.) — Martin Sleziak57 mins ago
"What is your favorite calculus textbook?" is opinion based and/or too broad for main. If at all it is a "poll." On tex.se they have polls "favorite editor/distro/fonts etc" while actual questions on these are still on-topic on main. Beyond that it is not clear why a question which software one uses should be a valid poll while the question which book one uses is not. — quid7 mins ago
@quid I will reply here, since I do not want to digress in the comments too much from the topic of that question.
Certainly I agree that "What is your favorite calculus textbook?" would not be suitable for the main. Which is why I wrote in my comment: "Although not formulated like that".
Book recommendations are certainly accepted on the main site, if they are formulated in the proper way.
If there will be community poll and somebody suggests question from GEdgar's comment, I will be perfectly ok with it. But I thought that his comment is simply playful remark pointing out that there is plenty of "polls" of this type on the main (although ther should not be). I guess some examples can be found here or here.
Perhaps it is better to link search results directly on MSE here and here, since in the Google search results it is not immediately visible that many of those questions are closed.
Of course, I might be wrong - it is possible that GEdgar's comment was meant seriously.
I have seen for the first time on TeX.SE. The poll there was concentrated on TeXnical side of things. If you look at the questions there, they are asking about TeX distributions, packages, tools used for graphs and diagrams, etc.
Academia.SE has some questions which could be classified as "demographic" (including gender).
@quid From what I heard, it stands for Kašpar, Melichar and Baltazár, as the answer there says. In Slovakia you would see G+M+B, where G stand for Gašpar.
But that is only anecdotal.
And if I am to believe Slovak Wikipedia it should be Christus mansionem benedicat.
From the Wikipedia article: "Nad dvere kňaz píše C+M+B (Christus mansionem benedicat - Kristus nech žehná tento dom). Toto sa však často chybne vysvetľuje ako 20-G+M+B-16 podľa začiatočných písmen údajných mien troch kráľov."
My attempt to write English translation: The priest writes on the door C+M+B (Christus mansionem benedicat - Let the Christ bless this house). A mistaken explanation is often given that it is G+M+B, following the names of three wise men.
As you can see there, Christus mansionem benedicat is translated to Slovak as "Kristus nech žehná tento dom". In Czech it would be "Kristus ať žehná tomuto domu" (I believe). So K+M+B cannot come from initial letters of the translation.
It seems that they have also other interpretations in Poland.
"A tradition in Poland and German-speaking Catholic areas is the writing of the three kings' initials (C+M+B or C M B, or K+M+B in those areas where Caspar is spelled Kaspar) above the main door of Catholic homes in chalk. This is a new year's blessing for the occupants and the initials also are believed to also stand for "Christus mansionem benedicat" ("May/Let Christ Bless This House").
Depending on the city or town, this will be happen sometime between Christmas and the Epiphany, with most municipalities celebrating closer to the Epiphany."
BTW in the village where I come from the priest writes those letters on houses every year during Christmas. I do not remember seeing them on a church, as in Najib's question.
In Germany, the Czech Republic and Austria the Epiphany singing is performed at or close to Epiphany (January 6) and has developed into a nationwide custom, where the children of both sexes call on every door and are given sweets and money for charity projects of Caritas, Kindermissionswerk or Dreikönigsaktion[2] - mostly in aid of poorer children in other countries.[3]
A tradition in most of Central Europe involves writing a blessing above the main door of the home. For instance if the year is 2014, it would be "20 * C + M + B + 14". The initials refer to the Latin phrase "Christus mansionem benedicat" (= May Christ bless this house); folkloristically they are often interpreted as the names of the Three Wise Men (Caspar, Melchior, Balthasar).
In Catholic parts of Germany and in Austria, this is done by the Sternsinger (literally "Star singers"). After having sung their songs, recited a poem, and collected donations for children in poorer parts of the world, they will chalk the blessing on the top of the door frame or place a sticker with the blessing.
On Slovakia specifically it says there:
The biggest carol singing campaign in Slovakia is Dobrá Novina (English: "Good News"). It is also one of the biggest charity campaigns by young people in the country. Dobrá Novina is organized by the youth organization eRko. |
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 |
Comparable Topologies on a Set
For an arbitrary set $X$ there are many different topologies that can be obtained. Sometimes these topologies are comparable in the sense that one of the topologies is contained in the other.
Definition: Let $X$ be a set and $\tau_1$ and $\tau_2$ be two topologies defined on $X$. If either $\tau_1 \subseteq \tau_2$ or $\tau_1 \supseteq \tau_2$ then $\tau_1$ and $\tau_2$ are said to be Comparable. If $\tau_1 \subseteq \tau_2$ then $\tau_2$ is said to be a Finer or Stronger topology than $\tau_1$ (or Strictly Finer/Stronger if also $\tau_1 \neq \tau_2$) and $\tau_1$ is said to be a Coarser or Weaker topology than $\tau_2$ (or Strictly Coarser/Weaker if also $\tau_1 \neq \tau_2$). Example 1
Consider the set $X = \{ a, b, c, d, e \}$ and the nested topology $\tau_1 = \{ \emptyset, \{a \}, \{a, b \}, \{a, b, c \}, \{a, b, c, d \}, X \}$.
Now consider the following topology:(1)
So $\tau_1 \subset \tau_2$, so the finer topology is $\tau_1$ and the coarser topology is $\tau_2$. Now consider the this topology:(2)
So $\tau_1 \subset \tau_2 \subset \tau_3$. Between these three topologies, $\tau_1$ is the coarsest and $\tau_3$ is the finest.
Example 2
For another example, consider the set of real numbers $\mathbb{R}$ with the usual topology $\tau_1$ of open intervals. Recall from The Lower and Upper Limit Topologies on the Real Numbers page that the lower limit topology is the topology, denote it $\tau_2$, generated by all unions of elements in $\{ [a, b) : a, b \in \mathbb{R}, a \leq b \}$ and the upper limit topology is the topology, denote it $\tau_3$, generated by all unions of elements in $\{ (a, b] : a, b \in \mathbb{R}, a, b \}$.
Notice that all open intervals can be written as the union of open sets in the lower limit topology and in the upper limit topology since for the interval $(c, d)$ we have that:(3)
Therefore $\tau_1 \subset \tau_2$ and $\tau_1 \subset \tau_3$. Hence, the lower and upper limit topologies on $\mathbb{R}$ are both finer than the usual topology on $\mathbb{R}$. |
This problem builds on the one in May on calculating Pi. This brilliant man Archimedes managed to establish that $3\frac{10}{71} < \pi < 3\frac{1}{7}$.
He needed to be able to calculate square roots first so that he could calculate the lengths of the sides of the polygons which he used to get his approximation for $\pi$. How did he calculate square roots? He didn't have a calculator but needed to work to an appropriate degree of accuracy. To do this he used what we now call numerical roots.
How might he have calculated $\sqrt{3}$?
This must be somewhere between 1 and 2. How do I know this?
Now calculate the average of $\frac{3}{2}$ and 2 (which is 1.75)- this is a second approximation to $\sqrt 3$.
i.e. we are saying that a better approximation to $\sqrt 3$ is $$\frac{(\frac{3}{n} + n)}{2}$$ where n is an approximation to $\sqrt 3$ .
We then repeat the process to find the new (third) approximation to $\sqrt{3}$ $$\sqrt{3} \approx {(3 / 1.75 + 1.75) \over {2}} = 1.73214... $$
to find a forth approximation repeat this process using 1.73214 and so on...
How many approximations do I have to make before I can find $\sqrt{3}$ correct to five decimal places.
Why do you think it works?
Will it always work no matter what I take as my first approximation and does the same apply to finding other roots? |
Section 2.1 of "Exegeses on linear models" (p. 3) describes a local linear model based on a second-order Taylor expansion.
How would I go about fitting the model?
I see that the first two terms reduce to linear main effects, quadratic main effects, and pairwise interactions. Those alone I know I can fit with OLS or Gaussian maximum likelihood after centering the data around some $x_0$. But I don't know how to handle the model when the other two terms are included.
Notably, Venables' examples do not make explicit use of this model. Is it even possible to simultaneously identify $\sigma$, $\gamma$, and $\delta$? Or is there another reason why I shouldn't fit this fairly general model? Am I missing Venables' point?
Edit: in light of Glen_b's answer this is probably moving the goalposts a little, but ideally an answer would describe how to fit that particular functional form, or why it's impossible or otherwise a bad idea to try. And for reference here it is: $$ Y = \beta_0 + \sum_i \beta_i (X_i - x_{0i}) + \sum_{i,j} \beta_{i,j} (X_i - x_{0i})(X_j - x_{0j}) + \left( \sigma + \sum_i \gamma_i (X_i - x_{0i}) \right) Z + \delta Z^2 $$ where $i$ and $j$ index predictors $X$, $x_0$ is some point for centering the data, and $Z$ is a standard Gaussian error term |
According to the KMV
1 model, the standard deviation for the default of the \(i\)th borrower \(\sigma_{Di}\) is given by \(\sigma_{Di}=\sqrt{(EDF)(1-EDF)}\) where \(EDF\) is the expected default frequency.
Specifically, let \(p\) be the probability of complete repayment of the loan, and \((1-p)\) be the probability of default, or the expected default frequency (EDF)
Probability of repayment
A Bernoulli trial
2, repayment \(p\) returns 1 and default \((1-p)\) returns 0
Expected outcome:
$$ E(\mbox{payoff}) = 1 \times p + 0 \times (1-p)=p$$
Standard deviation: \begin{align*}\sigma & =\sqrt{(1-p)^2 \times p + (0-p)^2 \times (1-p)} \\
& =\sqrt{(1-2p + p^2) \times p + p^2 \times (1-p)} \\ & =\sqrt{(p-2p^2 + p^3) + (p^2 -p^3)} \\ & =\sqrt{p-p^2} \\ & =\sqrt{p(1-p)} \end{align*} Probability of default
A Bernoulli trial
3, repayment (EDF) returns 0 and default (1-EDF) returns 1
Expected outcome:
$$ E(\mbox{payoff}) = 0 \times EDF + 1 \times (1 – EDF)= 1 – EDF $$
Standard deviation: \begin{align*}\sigma & =\sqrt{(0 – (1 – EDF)^2 \times EDF + (1 – (1 – EDF))^2 \times (1 – EDF)} \\
& =\sqrt{(-1 + EDF)^2 \times EDF + EDF^2 \times (1-EDF)} \\ & =\sqrt{(1-2EDF + EDF^2) \times EDF + EDF^2 – EDF^3} \\ & =\sqrt{EDF – 2EDF^2 + EDF^3 + EDF^2 – EDF^3} \\ & =\sqrt{EDF – EDF^2} \\ & =\sqrt{EDF(1-EDF)} \end{align*} Notes The KMV model is named after Stephen Kealhofer, John McQuown and Oldrich Vasicek. The model was acquired by Moody’s Analytics in 2002 Saunders and Cornett (2010, p330) KMV References Saunders A and M Cornett, (2010), Financial institutions management: a risk management approach, 7 ed., McGraw-Hill |
People like prisoner puzzles. Here is the hardest one I have ever seen. I heard it from Prof. Boris Bukh on the social hour of Maths Department couple of weeks ago, whom heard it on a maths conference. Since recently I am quite into Cryptography, especially setting up communication protocols, I would like to log the puzzle in my blog.
The storyline is like The Shawshank Redemption. You are locked into one of the cells in the jail. Unfortunately, you do not know how many prisoners there are in the jail. You do not even know whether you are the only one! The evil-hearted yet mathematically inclined warden wants to play a game with you. The rules are as follows.
On the first day, you can write up a communication protocol. The warden will make identical photocopies of your protocol and distribute them among other prisoners. During the following days, everyday, each prisoner will be given one black card and one white card, and they should turn in one of them to the warden according to your protocol. The rest of the cards will be discarded. The warden would line up all prisoners in a circle in his mind and imagine each of them passes his or her chosen card to the next person clockwise. After going through all this in his mind, he will redistribute the cards to the prisoners accordingly. Notice, the warden could line up the people by any order he likes. Moreover, the orders could probably differ from day to day. One day, you should call for a stop and immediately tell the number of prisoners in this jail. All prisoners including yourself will be released if you got the number correct.
Remember, the evil-hearted warden knows every detail of your protocol. So all odds are always against you.
The following is the solution I came up.
Denote the number of prisoners as p. First of all, we want to draft the protocol so that all prisoners know an upper bound of p, denoted as P. Before we say anything about the protocol, let us take a look at the significance of knowing such an upper bound.
If all prisoners know an upper bound P, then they can broadcast messages. To be precise, if one turns in a card of some color he chooses, say c. Everyone else should keep turning in white cards until he receives a black card, and then keep turning in black cards regardless of what cards are received. Then after P days, all prisoners will receive the the cards of color c. (Think about why this is true.) Now everyone knows the color c. Hence they succeeded in broadcasting a single bit. If the protocol says something about coding messages by colors, one can broadcast any message to others. Note that prisoners might not know who has sent the messages.
Though direct message is hard to realize in this situation, broadcasting would be good enough for us to figure out p.
Now, let us see how prisoners cooperatively figure out an upper bound P. We call it the first stage.
In the first stage, all prisoners will be labeled as either visible or hidden. At the beginning, only I am labeled as visible, all the rest are labeled as hidden. We break down the following days into several phases. At the beginning of the kth phase, all prisoners know the upper bound of the current visible prisoners P_k. For instance, P_1=1. Now, all hidden people broadcast the a single bit of message telling the others there are hidden people. The broadcasting is done similarly, i.e., whoever receives the black card once should keep sending black cards. The broadcasting can be done within P_k days because there are at most these many visible prisoners.
There are two possible turnouts. If no visible prisoner receives this message, then all prisoners are visible, and immediately all prisoners know P=P_k. Otherwise, on the next day, all visible prisoners turn in black cards. Whoever receives a black card changes his label to visible if it was hidden. The number of prisoners who change their labels is at most P_k. Thus we can use P_{k+1}=2P_k in the k+1th phase as now there are at most P_k+P_k visible prisoners. Since at least one prisoner changes his label in each phase, eventually, all prisoners will become visible. Hence this process will terminate at a certain phase and provides us an upper bound P.
After the first stage, comes the second stage.
In the second stage, we will partition prisoners by clans and try refining the clans as much as possible. During the second stage, each prisoner knows the total number of clans, the names of each clan and which clan he belongs to. At the beginning, I myself alone consists the first clan and the rest of prisoners consist the second clan. This initial set up is written into the protocol, so everyone knows it.
The following is how I refine the clans. I can broadcast a message that asks the members in a certain clans if any one has turned in a black card on a given day. Prisoners in this certain clan will broadcast a message of a single bit to indicate their answer. And then I broadcast a message that asks if any one in this clan has turned in a white card on that day. If both answers are ‘yes’, this means there are two people in this clan turned in different card on that given day. Then I will demand this clan to divide into two smaller clans according to the color of their cards turned in on that given day. I can also do the same thing to the cards they received on that given day as well.
So in the second phase, I can constantly refine the clans by asking questions clan by clan and day by day. Since the number of clans can not exceed P, this refining process will stop and the number of clans will stabilize. Moreover, once it stabilizes, prisoners in each clan will turn in the same cards and receive the same cards as well. Otherwise, some day I will interrogate the members in this clan and separate them into smaller clans.
Now we can assume the prisoners in each clans always turn in the same cards and receive the same cards. This is amazing!
Suppose c_i is the number of prisoners in the ith clan C_i for i=1,\ldots, n. Assume I am in C_1. Then c_1=1. Initially the values of all other c_is are unknown. Our goal is to find out each c_i, though it is enough to find \sum_{i=1}^nc_i. Notation. [n] is the set of all naturals from 1 to n. If I\subset [n], c_I := \{c_i: i\in I\} and \mathrm{span}(c_I) be the set of all linear combination of \{c_i: i\in I\}.
For r = 1,\ldots, n-1, we claim that for all r-element subset I\subset [n] and i\in I, c_i can be written as a linear combination of \{c_j: j\in[n]-I\}. In abbreviation we shall write c_i\in \mathrm{span}(c_{[n]-I}). Note when we say ‘some variable can be written as a linear combination of other variables’, we mean all prisoners know this information. In other words, all prisoners keep track of these linear dependencies along the way. We will omit the details of how prisoners would share this information in the following proof and leave them to the readers.
Before we prove the claim, let us take a look at the consequence if what we claimed is true. When r=n-1, it means c_2, \ldots, c_n can be written as a linear combination of c_1=1. Thus we know c_i for all i\in[n]. Hence we know the total number of prisoners.
Now we prove the claim by induction on r. If r=1, we shall prove c_i\in \mathrm{span}(c_{[n]-\{i\}}) for all i\in[n]. This is relatively easy. Let the prisoners in the clan C_i send out black cards. Some of the other clans will receive those black cards. Thus c_i is the sum of the size of those clans which receive the black cards.
Suppose the claim is true for some r. Now we shall prove for every r+1-element subset I\subset [n] and i\in I, c_i\in \mathrm{span}(c_J), where J=[n]-I.
Fix some i_0\in I. By the induction hypothesis, we know for all i\in I with i\neq i_0, we have
c_i = \alpha_ic_{i_0}+\beta_i(c_J), where \beta_i(c_J)\in \mathrm{span}(c_J). This is also true for i=i_0. Because we may set \alpha_{i_0}=1 and \beta_{i_0}=0.
Let the index set I_+ be the set \{i\in I: \alpha_i > 0\} and I_-=I-I^+. As i_0\in I_+, I_+ is non-empty.
Now we demand all clans indexed by I^+ send out black cards. Suppose the clans indexed by I_+'\subset I_+ who send out black cards receive white cards and the clans indexed by I_-'\subset I_- and J'\subset J who send out white cards receive black cards. Hence we have the following equation.
\sum_{i\in I'_+}c_i = \sum_{i\in I'_-}c_i+\sum_{j\in J'}c_j
Hence
\sum_{i\in I'_+}\left(\alpha_ic_{i_0}+\beta_i(c_J)\right)=\sum_{i\in I'_-}\left(\alpha_ic_{i_0}+\beta_i(c_J)\right)+\sum_{j\in J'}c_j
The \alpha-coefficient on the left hand side is positive. Meanwhile it is nonpositive on the right hand side. Thus by the equation above we can solve for c_{i_0} in terms of a linear combination of c_J. Hence we can write c_i in terms of a linear combination of c_J for all i\in I. [qed] |
We know that: $$\frac 1{2!}+\frac 1{3!}+\frac 1{4!}+\frac 1{5!}+\frac 1{6!}+\cdots =e-2\approx0.71828$$ But I am getting the above sum as $1,$ as shown below: \begin{align} S & = \frac 1{2!}+\frac 1{3!}+\frac 1{4!}+\frac 1{5!}+\frac 1{6!}+\cdots \\[10pt] & = \frac 1{2!} + \frac {3-2}{3!} +\fra...
Perhaps the chaos-theory and chaotic-systems tags should be merged? The second has no official description but I can hardly imagine what the difference should be.
The list of proposals on the 2016 thread that are still open: Proposal to rename the "adjoint" tag Proposal to join the "chaos theory" and "chaotic systems" tags Proposal to change the name of the "divisors" tag Proposal to make the "compactification" tag a synonym of the "compactness" tag Pr...
Please merge chaotic-systems and chaos-theory. I am active in these tags and the respective field and I fail to see a meaningful difference between them, let alone a need for a distinction. This already got 10 upvotes last year.
I propose creating relation-composition tag and making it a synonym of function-composition. I think that if composition of functions is important enough to have its own tag, then so is composition of relations. But it would probably be better to have both topics under the same tag. We definitel...
Maple shows that $$ \sum_{0 \le k \le m} \frac{2^k}{(k+1)} = -i/2\pi -2\,{2}^{m} \left( 1/4\,{\it \Phi} \left( 2,1,m \right) -1 /4\,{m}^{-1}-1/2\, \left( m+1 \right) ^{-1} \right) $$ where $\Phi$ denotes Lerch's transcendent. How can we prove this? I have checked a few books but haven't got a cl...
maybeit's clearer than having relations just as a synonym an mentioned in the tag-wiki...?)
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1. Measurement of the top quark mass with lepton+jets final states using $$\mathrm {p}$$ p $$\mathrm {p}$$ p collisions at $$\sqrt{s}=13\,\text {TeV} $$ s=13TeV
The European Physical Journal C, ISSN 1434-6044, 11/2018, Volume 78, Issue 11, pp. 1 - 27
The mass of the top quark is measured using a sample of $${{\text {t}}\overline{{\text {t}}}$$ tt¯ events collected by the CMS detector using proton-proton...
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Journal Article
2. Measurement of prompt and nonprompt $$\mathrm{J}/{\psi }$$ J / ψ production in $$\mathrm {p}\mathrm {p}$$ p p and $$\mathrm {p}\mathrm {Pb}$$ p Pb collisions at $$\sqrt{s_{\mathrm {NN}}} =5.02\,\text {TeV} $$ s NN = 5.02 TeV
The European Physical Journal C, ISSN 1434-6044, 04/2017, Volume 77, Issue 4, pp. 1 - 27
Abstract This paper reports the measurement of $$\mathrm{J}/{\psi }$$ J / ψ meson production in proton–proton ( $$\mathrm {p}\mathrm {p}$$ p p ) and...
Journal Article
3. Study of the underlying event in top quark pair production in $$\mathrm {p}\mathrm {p}$$ p p collisions at 13 $$~\text {Te}\text {V}$$ Te
The European Physical Journal C, ISSN 1434-6044, 02/2019, Volume 79, Issue 2
Journal Article
4. Search for heavy Majorana neutrinos in same-sign dilepton channels in proton-proton collisions at $\sqrt{s} =$ 13 TeV
Journal of High Energy Physics (Online), ISSN 1029-8479, 06/2018, Volume 2019, Issue 1
JHEP 01 (2019) 122 A search is performed for a heavy Majorana neutrino (N), produced by leptonic decay of a W boson propagator and decaying into a W boson and...
Physics - High Energy Physics - Experiment | Beyond Standard Model | Hadron-Hadron scattering (experiments) | PHYSICS OF ELEMENTARY PARTICLES AND FIELDS
Physics - High Energy Physics - Experiment | Beyond Standard Model | Hadron-Hadron scattering (experiments) | PHYSICS OF ELEMENTARY PARTICLES AND FIELDS
Journal Article
European Physical Journal C, ISSN 1434-6044, 11/2018, Volume 78, Issue 11
Journal Article
6. Measurement of the weak mixing angle using the forward–backward asymmetry of Drell–Yan events in $$\mathrm {p}\mathrm {p}$$ p p collisions at 8 $$\,\text {TeV}$$ TeV
The European Physical Journal C, ISSN 1434-6044, 09/2018, Volume 78, Issue 9
Journal Article
European Physical Journal C, ISSN 1434-6044, 04/2017, Volume 77, Issue 4
Journal Article
8. Measurement of the top quark mass with lepton+jets final states using $\mathrm {p}$$ $$\mathrm {p}$ collisions at $\sqrt{s}=13\,\text {TeV}
European Physical Journal. C, Particles and Fields, ISSN 1434-6044, 11/2018, Volume 78, Issue 11
The mass of the top quark is measured using a sample of $\mathrm{t\overline{t}}$ events containing one isolated muon or electron and at least four jets in the...
PHYSICS OF ELEMENTARY PARTICLES AND FIELDS
PHYSICS OF ELEMENTARY PARTICLES AND FIELDS
Journal Article
Physical Review Letters, ISSN 0031-9007, 06/2015, Volume 115, Issue 1, p. 012301
The second-order azimuthal anisotropy Fourier harmonics, nu(2), are obtained in p-Pb and PbPb collisions over a wide pseudorapidity (.) range based on...
PLUS AU COLLISIONS | ANISOTROPIC FLOW | PROTON-PROTON | PHYSICS, MULTIDISCIPLINARY | ECCENTRICITIES | Correlation | Large Hadron Collider | Anisotropy | Dynamics | Collisions | Luminosity | Charged particles | Dynamical systems
PLUS AU COLLISIONS | ANISOTROPIC FLOW | PROTON-PROTON | PHYSICS, MULTIDISCIPLINARY | ECCENTRICITIES | Correlation | Large Hadron Collider | Anisotropy | Dynamics | Collisions | Luminosity | Charged particles | Dynamical systems
Journal Article
10. Combination of inclusive and differential tt¯ charge asymmetry measurements using ATLAS and CMS data at s√=7 and 8 TeV
ISSN 1029-8479, 2018
mass spectrum: (2top) | experimental results | signature | statistical analysis: Bayesian | mass dependence | CMS | CERN LHC Coll | ATLAS | top: pair production | data analysis method | 7000 GeV-cms8000 GeV-cms | p p: colliding beams | p p: scattering | charge: asymmetry: measured | final state: ((n)jet lepton)
Journal Article
11. Measurement of the weak mixing angle using the forward–backward asymmetry of Drell–Yan events in p p collisions at 8 TeV
European Physical Journal C, ISSN 1434-6044, 09/2018, Volume 78, Issue 9, p. 701
A measurement is presented of the effective leptonic weak mixing angle ( ) using the forward-backward asymmetry of Drell-Yan lepton pairs ( and ) produced in...
Journal Article
12. Study of high-p(T) charged particle suppression in PbPb compared to pp collisions at root s(NN)=2.76 TeV
EUROPEAN PHYSICAL JOURNAL C, ISSN 1434-6044, 03/2012, Volume 72, Issue 3
Journal Article
13. Evidence for transverse-momentum- and pseudorapidity-dependent event-plane fluctuations in PbPb and p Pb collisions
Physical Review C - Nuclear Physics, ISSN 0556-2813, 09/2015, Volume 92, Issue 3
Journal Article
14. Measurements of the $$\mathrm {p}\mathrm {p}\rightarrow \mathrm{Z}\mathrm{Z}$$ pp→ZZ production cross section and the $$\mathrm{Z}\rightarrow 4\ell $$ Z→4ℓ branching fraction, and constraints on anomalous triple gauge couplings at $$\sqrt{s} = 13\,\text {TeV} $$ s=13TeV
The European Physical Journal C, ISSN 1434-6044, 2/2018, Volume 78, Issue 2, pp. 1 - 29
Four-lepton production in proton-proton collisions, $$\mathrm {p}\mathrm {p}\rightarrow (\mathrm{Z}/ \gamma ^*)(\mathrm{Z}/\gamma ^*) \rightarrow 4\ell $$...
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Journal Article |
My question is as follows.
Do the Chern classes as defined by Grothendieck for smooth projective varieties coincide with the Chern classes as defined with the aid of invariant polynomials and connections on complex vector bundles (when the ground field is $\mathbf{C}$)?
I suppose GAGA is involved here. Could anybody give me a reference where this is shown as detailed as possible? Or is the above not true?
Some background on my question:
Let $X$ be a smooth projective variety over an algebraically closed field $k$. For any integer $r$, let $A^r X$ be the group of cycles of codimension $r$ rationally equivalent to zero. Let $AX=\bigoplus A^r X$ be the Chow ring.
Grothendieck proved the following theorem on Chern classes.
There is a unique "theory of Chern classes", which assigns to each locally free coherent sheaf $\mathcal{E}$ on $X$ an $i$-th Chern class $c_i(\mathcal{E})\in A^i(X)$ and satisfies the following properties:
C0. It holds that $c_0(\mathcal{E}) = 1$. C1. For an invertible sheaf $\mathcal{O}_X(D)$ on $X$, we have that $c_1(\mathcal{O}_X(D)) = [D]$ in $A^1(X)$. C2. For a morphism of smooth quasi-projective varieties $f:X\longrightarrow Y$ and any positive integer $i$, we have that $f^\ast(c_i(\mathcal{E})) =c_i(f^\ast(\mathcal{E}))$. C3. If $$0\longrightarrow \mathcal{E}^\prime \longrightarrow \mathcal{E} \longrightarrow \mathcal{E}^{\prime\prime} \longrightarrow 0$$ is an exact sequence of vector bundles on $X$, then $c_t(\mathcal{E}) = c_t(\mathcal{E}^\prime)c_t(\mathcal{E}^{\prime\prime})$ in $A(X)[t]$.
So that's how it works in algebraic geometry. Now let me sketch the complex analytic case.
Let $E\longrightarrow X$ be a complex vector bundle. We are going to associate certain cohomology classes in $H^{even}(X)$ to $E$. The outline of this construction is as follows.
Step 1. We choose a connection $\nabla^E$ on $E$; Step 2. We construct closed even graded differential forms with the aid of $\nabla^E$; Step 3. We show that the cohomology classes of these differential forms are independent of $\nabla^E$.
Let us sketch this construction. Let $k= \textrm{rank}(E)$. Let us fix an invariant polynomial $P$ on $\mathfrak{gl}_k(\mathbf{C})$, i.e. $P$ is invariant under conjugation by $\textrm{GL}_k(\mathbf{C})$.
Let us fix a connection $\nabla^E$ on $E$. We denote denote its curvature by $R^E = (\nabla^E)^2$. One shows that $$R^E \in \mathcal{C}^\infty(X,\Lambda^2(T^\ast X)\otimes \textrm{End}(E)).$$ That is, $R^E$ is a $2$-form on $X$ with values in $\textrm{End}(E)$. Define $$P(E,\nabla^E) = P(-R^E/{2i\pi}).$$ (This is well-defined.)
The Chern-Weil theorem now says that:
The even graded form $P(E,\nabla^E)$ is a smooth complex differential form which is closed. The cohomology class of $P(E,\nabla^E)$ is independent of the chosen connection $\nabla^E$ on $E$.
Choosing $P$ suitably, we get the Chern classes of $E$ (by definition). These are cohomology classes. In order for one to show the equivalence of these "theories" one is forced to take the leap from the Chow ring to the cohomology ring.
How does one choose $P$? You just take $P(B) = \det(1+B)$ for a matrix.
Motivation: If one shows the equivalence of these two theories one gets "two ways" of "computing" the Chern character. |
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The co-production and co-facilitation of recovery-focused education programmes is one way in which service users may be meaningfully involved as partners.
Objectives:
To evaluate the impact of a clinician and peer co-facilitated information programme on service users’ knowledge, confidence, recovery attitudes, advocacy and hope, and to explore their experience of the programme.
Methods:
A sequential design was used involving a pre–post survey to assess changes in knowledge, confidence, advocacy, recovery attitudes and hope following programme participation. In addition, semi-structured interviews with programme participants were completed. Fifty-three participants completed both pre- and post-surveys and twelve individuals consented to interviews.
Results:
The results demonstrated statistically significant changes in service users’ knowledge about mental health issues, confidence and advocacy. These improvements were reflected in the themes which emerged from the interviews with participants (n = 12), who reported enhanced knowledge and awareness of distress and wellness, and a greater sense of hope. In addition, the peer influence helped to normalise experiences for participants, while the dual facilitation engendered equality of participation and increased the opportunity for meaningful collaboration between service users and practitioners.
Conclusions:
The evaluation highlights the potential strengths of a service user and clinician co-facilitated education programme that acknowledges and respects the difference between the knowledge gained through self-experience and the knowledge gained through formal learning.
In this paper, we summarise and critique a network meta-analysis (NMA) of antidepressant efficacy and tolerability for paediatric depression and an accompanying editorial. Although we agree that many of the extant studies are flawed, this meta-analysis showed clear efficacy of fluoxetine in the NMA, and for sertraline and escitalopram in pairwise analyses. Consequently, these papers underestimate the benefits of antidepressants for paediatric depression, and provide support for current practice guideline, which recommends the use of an antidepressant if the patient does not respond to psychotherapy. In these circumstances, fluoxetine should be the first choice, with escitalopram and sertraline as alternatives.
Declaration of interest
D.A.B. receives royalties from Guilford Press, has or will receive royalties from the electronic self-rated version of the C-SSRS from eResearch Technology, Inc., is on the editorial board of UpToDate, and is a reviewer for Healthwise. R.D.G. serves as an expert witness for the US Department of Justice, Pfizer, Wyeth and GSK; and is the founder of Adaptive Testing Technologies. P.W. receives personal fees from Lundbeck and Takeda. B.D. reports a licensing agreement with Lundbeck for a psychosocial treatment manual for depression. No other disclosures were reported.
Low-Reynolds-number polymer solutions exhibit a chaotic behaviour known as ‘elastic turbulence’ when the Weissenberg number exceeds a critical value. The two-dimensional Oldroyd-B model is the simplest constitutive model that reproduces this phenomenon. To make a practical estimate of the resolution scale of the dynamics, one requires the assumption that an attractor of the Oldroyd-B model exists; numerical simulations show that the quantities on which this assumption is based are bounded. We estimate the Lyapunov dimension of this assumed attractor as a function of the Weissenberg number by combining a mathematical analysis of the model with direct numerical simulations.
We consider the time dependence of a hierarchy of scaled$L^{2m}$-norms$D_{m,\unicode[STIX]{x1D714}}$and$D_{m,\unicode[STIX]{x1D703}}$of the vorticity$\unicode[STIX]{x1D74E}=\unicode[STIX]{x1D735}\times \boldsymbol{u}$and the density gradient$\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}$, where$\unicode[STIX]{x1D703}=\log (\unicode[STIX]{x1D70C}^{\ast }/\unicode[STIX]{x1D70C}_{0}^{\ast })$, in a buoyancy-driven turbulent flow as simulated by Livescu & Ristorcelli (J. Fluid Mech., vol. 591, 2007, pp. 43–71). Here,$\unicode[STIX]{x1D70C}^{\ast }(\boldsymbol{x},t)$is the composition density of a mixture of two incompressible miscible fluids with fluid densities$\unicode[STIX]{x1D70C}_{2}^{\ast }>\unicode[STIX]{x1D70C}_{1}^{\ast }$, and$\unicode[STIX]{x1D70C}_{0}^{\ast }$is a reference normalization density. Using data from the publicly available Johns Hopkins turbulence database, we present evidence that the$L^{2}$-spatial average of the density gradient$\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}$can reach extremely large values at intermediate times, even in flows with low Atwood number$At=(\unicode[STIX]{x1D70C}_{2}^{\ast }-\unicode[STIX]{x1D70C}_{1}^{\ast })/(\unicode[STIX]{x1D70C}_{2}^{\ast }+\unicode[STIX]{x1D70C}_{1}^{\ast })=0.05$, implying that very strong mixing of the density field at small scales can arise in buoyancy-driven turbulence. This large growth raises the possibility that the density gradient$\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}$might blow up in a finite time.
The present study evaluated the impact on psychosocial outcome of parallel clinician and peer-led information programmes for people with a diagnosis of schizophrenia and bipolar disorder and for family members within an Irish context.
Methods
A sequential mixed method design was used. Quantitative data were collected using pre- and post-programme questionnaires followed by an integrated qualitative component involving semi-structured interviews after the programme. The questionnaires assessed knowledge, attitudes towards recovery, hope, support, advocacy and well-being. Interviews with participants, facilitators and project workers explored their experiences and views of the programme.
Findings
While a number of the questionnaires did not show a statistically significant change, findings from the interviews suggest that the1 programmes had a number of positive outcomes, including increases in perceived knowledge, empowerment and support. Participants in both programmes valued the opportunity to meet people in similar circumstances, share their experiences, learn from each other and provide mutual support.
Conclusion
The EOLAS programmes offer a novel template for communication and information sharing in a way that embodies the principles of collaboration and offers users and families a meaningful opportunity to become involved in service design, delivery and evaluation.
The EOLAS programme is a peer and clinician-led mental health information programme on recovery from mental health difficulties, specifically for people with a diagnosis of schizophrenia spectrum or bipolar disorders, their family members and significant others.
Method
This article, the first of a two part series, outlines the background to and the rationale behind the EOLAS programme, and traces the participatory process used to inform the development and implementation of the pilot phase of the project. The aims of the programme, and the overarching principles that guided its development, delivery and evaluation, including the set-up of the project steering group are outlined and discussed.
Findings
Two separate programmes, one for family members and one for service users were designed. In addition, participant and facilitator handbooks were developed for each programme, including a training programme for facilitators.
Conclusion
Central to a recovery oriented service is the involvement of service users and families in the design and delivery of services. EOLAS is one potential model for achieving this aim.
Setting priorities in the field of infectious diseases requires evidence-based and robust baseline estimates of disease burden. Therefore, the European Centre for Disease Prevention and Control initiated the Burden of Communicable Diseases in Europe (BCoDE) project. The project uses an incidence- and pathogen-based approach to measure the impact of both acute illness and sequelae of infectious diseases expressed in disability-adjusted life years (DALYs). This study presents first estimates of disease burden for four pathogens in Germany. The number of reported incident cases adjusted for underestimation served as model input. For the study period 2005–2007, the average disease burden was estimated at 33 116 DALYs/year for influenza virus, 19 115 DALYs/year for Salmonella spp., 8708 DALYs/year for hepatitis B virus and 740 DALYs/year for measles virus. This methodology highlights the importance of sequelae, particularly for hepatitis B and salmonellosis, because if omitted, the burden would have been underestimated by 98% and 56%, respectively.
The issue of intermittency in numerical solutions of the 3D Navier–Stokes equations on a periodic box${[0, L] }^{3} $is addressed through four sets of numerical simulations that calculate a new set of variables defined by${D}_{m} (t)= {({ \varpi }_{0}^{- 1} {\Omega }_{m} )}^{{\alpha }_{m} } $for$1\leq m\leq \infty $where${\alpha }_{m} = 2m/ (4m- 3)$and${[{\Omega }_{m} (t)] }^{2m} = {L}^{- 3} \int \nolimits _{\mathscr{V}} {\vert \boldsymbol{\omega} \vert }^{2m} \hspace{0.167em} \mathrm{d} V$with${\varpi }_{0} = \nu {L}^{- 2} $. All four simulations unexpectedly show that the${D}_{m} $are ordered for$m= 1, \ldots , 9$such that${D}_{m+ 1} \lt {D}_{m} $. Moreover, the${D}_{m} $squeeze together such that${D}_{m+ 1} / {D}_{m} \nearrow 1$as$m$increases. The values of${D}_{1} $lie far above the values of the rest of the${D}_{m} $, giving rise to a suggestion that a depletion of nonlinearity is occurring which could be the cause of Navier–Stokes regularity. The first simulation is of very anisotropic decaying turbulence; the second and third are of decaying isotropic turbulence from random initial conditions and forced isotropic turbulence at fixed Grashof number respectively; the fourth is of very-high-Reynolds-number forced, stationary, isotropic turbulence at up to resolutions of$409{6}^{3} $.
Limited Reaction Processing (LRP) is a new technique which combines Rapid Thermal Processing (RTP) and Chemical Vapor Deposition (CVD). The added temperature control provided in rapid thermal processing enables the use of substrate temperature as a reaction switch. In addition, rapid thermal technology has been shown to provide other advantages for chemical vapor deposition of Si and III–V materials. Results are presented for group IV materials including epitaxial Si, SiGe alloys, SiO2 , and polysilicon. MOSFETs have been demonstrated and sensitive tests of interface quality are presented, paving the way for future bipolar transistor fabrication. III–V materials such as GaAs, AlGaAs, InGaAs have been grown. GaAs electron mobilities are the best reported for material grown using trimethylarsenic. As-ambient rapid thermal anneals of GaAs have also been performed.
Limited reaction processing (LRP), a new technique which provides precise control of thermally driven surface reactions, was used to grow multilayer structures composed of semiconductors and insulators. Results are presented for group IV-based materials including epitaxial Si, SiGe alloys, SiO2, and polysilicon. III–V materials such as GaAs, AlGaAs, and InGaAs have also been successfully grown. A number of diagnostic techniques were used to define the advantages and capabilities of LRP, including TEM, SIMS and AES. In addition, some preliminary device results are presented.
In an effort to extend the performance limits of semiconductors, devices based on heterojunctions rather than homojunctions are being investigated with great interest. Heterojunctions allow certain device design constraints to be relaxed because the charge distribution, electric field, and potential can be tailored extensively, permitting better device structures to be utilized. Silicon technology today enjoys a firm grip on a large portion of the electronics industry due in part to its superior material properties.
The Boltzmann transport equation has been used to calculate range anddamage distributions in multilayer targets of general interest for semiconductor fabrication. A comprehensive review of the calculations will be presented, with particular emphasis on how large angle scattering events and channeling phenomena may be included. Examples of the quality of fit between the theory and experiment show that difficult phenomena (such as residual channeling) can be reasonable modelled. |
Parseval's Identity for Inner Product Spaces
Recall from the Hilbert Bases (Orthonormal Bases) for Hilbert Spaces page that if $H$ is a Hilbert space then a sequence $(x_n)_{n=1}^{\infty}$ of orthonormal points in $H$ is said to be a Hilbert basis (or orthogonal basis) of $H$ if every $y \in H$ can be written as:(1)
We will now state and prove Parseval's identity for Hilbert spaces. The result is very similar to Bessel's inequality but is stronger.
Theorem 1 (Parseval's Identity for Hilbert Spaces): Let $H$ be a Hilbert space and let $(x_n)_{n=1}^{\infty}$ be a Hilbert basis of $H$. Then for every $y \in H$, $\displaystyle{\sum_{n=1}^{\infty} |\langle y, x_n \rangle|^2 = \| y \|^2}$. Proof:Let $y \in H$. From the theorem on the Bessel's Inequality for Inner Product Spaces page we have that the series $\displaystyle{\sum_{n=1}^{\infty} |\langle y, x_n \rangle|^2 \leq \| y \|^2}$ and so the series converges. By the theorem on the Convergence Criterion for Series in Hilbert Spaces page we also have that the series $\displaystyle{\sum_{n=1}^{\infty} \langle y, x_n \rangle x_n}$ converges to some $z \in H$, and since $(x_n)_{n=1}^{\infty}$ is a Hilbert basis we have that: Now since the norm $\| cdot \|$ is a continuous function we have that by The Pythagorean Identity for Inner Product Spaces that: |
Let $f$ and $g$ be uniformly continuous on A. Then given $\epsilon >0$ there exists a $\delta_{1} > 0$ such that if $|x-y| < \delta_{1}, \forall x,y \in A$, then $|f(x)-f(y)| < \frac{\epsilon}{2M}$. There also exists a $\delta_{2} > 0$ such that if $|x-y| < \delta_{2}, \forall x,y \in A$, then $|g(x)-g(y)| < \frac{\epsilon}{2M}$. Since $f$ and $g$ are both bounded on a, there exists $M_{1}>0$ such that $|f(x)| \leq M_{1}, \forall x,y \in A$ and there exists $M_{2}>0$ such that $|g(x)| \leq M_{2}, \forall x,y \in A$. Let $M=\left\{M_{1}, M_{2}\right\}$. Then $|f(x)| \leq M$ and $|g(x) < M$ for all $x \in A$. Let $\delta=\left\{\delta_{1}, \delta_{2}\right\}$. So if $|x-y|<\delta$ then $|f(x)-f(y)| < \frac{\epsilon}{2M}$ and $|g(x)-g(y)| < \frac{\epsilon}{2M}$. Now consider $|f(x)g(x) - f(y)g(y)| = |f(x)g(x) - g(x)f(y) + g(x)f(y)-f(y)g(y)|$. Then, $|g(x)||f(x)-f(y)| + |f(y)||g(x)-g(y)| \leq M|f(x)-f(y)|+M|g(x)-g(y)|< M\frac{\epsilon}{2M} + M\frac{\epsilon}{2M} = \epsilon$. Does this look right to anyone?
And now for something slightly different:
Since $f,g$ are bounded on $A$, their ranges lie in some compact set $[-B,B]$. Since $[-B,B]^2$ is compact, we see that multiplication $\cdot : [-B,B]^2 \to \mathbb{R}$ is uniformly continuous.
Since $f,g$ are uniformly continuous separately, we see that the map $p(x) = (f(x),g(x))$ is also uniformly continuous.
Since the composition of uniformly continuous maps is again uniformly continuous (this follows almost immediately from the definition), we see that $\cdot \circ p$ is uniformly continuous.
The following version of your proof works out the essentials:
For any sort of multiplication one has $$uv-u_0v_0=v(u-u_0)+u_0(v-v_0)\ .$$ If all quantities appearing here are bounded in absolute value by some constant $M>0$ it follows that $$|uv-u_0v_0|\leq |v||u-u_0|+|u_0||v-v_0|\leq M\bigl(|u-u_0|+|v-v_0|\bigr)\ .\tag{1}$$ Given an $\epsilon>0$ choose a $\delta>0$ that is sufficient for both $f$ and $g$ and tolerance $\epsilon':={\epsilon\over 2M}$. From $(1)$ it then follows that this $\delta$ is sufficient for $fg$ and tolerance $\epsilon$. |
Dark Energy Survey Year 1 Results: Cross-correlation between DES Y1 galaxy weak lensing and SPT+Planck CMB weak lensing Abstract
We cross-correlate galaxy weak lensing measurements from the Dark Energy Survey (DES) year-one (Y1) data with a cosmic microwave background (CMB) weak lensing map derived from South Pole Telescope (SPT) and Planck data, with an effective overlapping area of 1289 deg$$^{2}$$. With the combined measurements from four source galaxy redshift bins, we reject the hypothesis of no lensing with a significance of $$10.8\sigma$$. When employing angular scale cuts, this significance is reduced to $$6.8\sigma$$, which remains the highest signal-to-noise measurement of its kind to date. We fit the amplitude of the correlation functions while fixing the cosmological parameters to a fiducial $$\Lambda$$CDM model, finding $$A = 0.99 \pm 0.17$$. We additionally use the correlation function measurements to constrain shear calibration bias, obtaining constraints that are consistent with previous DES analyses. Finally, when performing a cosmological analysis under the $$\Lambda$$CDM model, we obtain the marginalized constraints of $$\Omega_{\rm m}=0.261^{+0.070}_{-0.051}$$ and $$S_{8}\equiv \sigma_{8}\sqrt{\Omega_{\rm m}/0.3} = 0.660^{+0.085}_{-0.100}$$. These measurements are used in a companion work that presents cosmological constraints from the joint analysis of two-point functions among galaxies, galaxy shears, and CMB lensing using DES, SPT and Planck data.
Authors: Publication Date: Research Org.: Argonne National Lab. (ANL), Argonne, IL (United States); Oak Ridge National Lab. (ORNL), Oak Ridge, TN (United States); Lawrence Berkeley National Lab. (LBNL), Berkeley, CA (United States); Brookhaven National Lab. (BNL), Upton, NY (United States); SLAC National Accelerator Lab., Menlo Park, CA (United States); Fermi National Accelerator Lab. (FNAL), Batavia, IL (United States) Sponsoring Org.: USDOE Office of Science (SC), High Energy Physics (HEP) (SC-25) Contributing Org.: DES; SPT OSTI Identifier: 1487048 Report Number(s): arXiv:1810.02441; FERMILAB-PUB-18-513-AE 1697154 DOE Contract Number: AC02-07CH11359 Resource Type: Journal Article Journal Name: TBD Additional Journal Information: Journal Name: TBD Country of Publication: United States Language: English Subject: 79 ASTRONOMY AND ASTROPHYSICS Citation Formats
Omori, Y., and et al.
Dark Energy Survey Year 1 Results: Cross-correlation between DES Y1 galaxy weak lensing and SPT+Planck CMB weak lensing. United States: N. p., 2018. Web.
Omori, Y., & et al.
Dark Energy Survey Year 1 Results: Cross-correlation between DES Y1 galaxy weak lensing and SPT+Planck CMB weak lensing. United States.
Omori, Y., and et al. Thu . "Dark Energy Survey Year 1 Results: Cross-correlation between DES Y1 galaxy weak lensing and SPT+Planck CMB weak lensing". United States. https://www.osti.gov/servlets/purl/1487048.
@article{osti_1487048,
title = {Dark Energy Survey Year 1 Results: Cross-correlation between DES Y1 galaxy weak lensing and SPT+Planck CMB weak lensing}, author = {Omori, Y. and et al.}, abstractNote = {We cross-correlate galaxy weak lensing measurements from the Dark Energy Survey (DES) year-one (Y1) data with a cosmic microwave background (CMB) weak lensing map derived from South Pole Telescope (SPT) and Planck data, with an effective overlapping area of 1289 deg$^{2}$. With the combined measurements from four source galaxy redshift bins, we reject the hypothesis of no lensing with a significance of $10.8\sigma$. When employing angular scale cuts, this significance is reduced to $6.8\sigma$, which remains the highest signal-to-noise measurement of its kind to date. We fit the amplitude of the correlation functions while fixing the cosmological parameters to a fiducial $\Lambda$CDM model, finding $A = 0.99 \pm 0.17$. We additionally use the correlation function measurements to constrain shear calibration bias, obtaining constraints that are consistent with previous DES analyses. Finally, when performing a cosmological analysis under the $\Lambda$CDM model, we obtain the marginalized constraints of $\Omega_{\rm m}=0.261^{+0.070}_{-0.051}$ and $S_{8}\equiv \sigma_{8}\sqrt{\Omega_{\rm m}/0.3} = 0.660^{+0.085}_{-0.100}$. These measurements are used in a companion work that presents cosmological constraints from the joint analysis of two-point functions among galaxies, galaxy shears, and CMB lensing using DES, SPT and Planck data.}, doi = {}, journal = {TBD}, number = , volume = , place = {United States}, year = {2018}, month = {10} } Figures / Tables: i s(z) for the 4 tomographic bins for Metacalibration. The black line shows the CMB lensing kernel. |
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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 ... |
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Isomorphisms are very important in mathematics, and we can no longer put off talking about them. Intuitively, two objects are 'isomorphic' if they look the same. Category theory makes this precise and shifts the emphasis to the 'isomorphism' - the
way in which we match up these two objects, to see that they look the same.
For example, any two of these squares look the same after you rotate and/or reflect them:
An isomorphism between two of these squares is a
process of rotating and/or reflecting the first so it looks just like the second.
As the name suggests, an isomorphism is a kind of morphism. Briefly, it's a morphism that you can 'undo'. It's a morphism that has an inverse:
Definition. Given a morphism \(f : x \to y\) in a category \(\mathcal{C}\), an inverse of \(f\) is a morphism \(g: y \to x\) such that
and
I'm saying that \(g\) is 'an' inverse of \(f\) because in principle there could be more than one! But in fact, any morphism has at most one inverse, so we can talk about 'the' inverse of \(f\) if it exists, and we call it \(f^{-1}\).
Puzzle 140. Prove that any morphism has at most one inverse. Puzzle 141. Give an example of a morphism in some category that has more than one left inverse. Puzzle 142. Give an example of a morphism in some category that has more than one right inverse.
Now we're ready for isomorphisms!
Definition. A morphism \(f : x \to y\) is an isomorphism if it has an inverse. Definition. Two objects \(x,y\) in a category \(\mathcal{C}\) are isomorphic if there exists an isomorphism \(f : x \to y\).
Let's see some examples! The most important example for us now is a 'natural isomorphism', since we need those for our databases. But let's start off with something easier. Take your favorite categories and see what the isomorphisms in them are like!
What's an isomorphism in the category \(\mathbf{3}\)? Remember, this is a free category on a graph:
The morphisms in \(\mathbf{3}\) are paths in this graph. We've got one path of length 2:
$$ f_2 \circ f_1 : v_1 \to v_3 $$ two paths of length 1:
$$ f_1 : v_1 \to v_2, \quad f_2 : v_2 \to v_3 $$ and - don't forget - three paths of length 0. These are the identity morphisms:
$$ 1_{v_1} : v_1 \to v_1, \quad 1_{v_2} : v_2 \to v_2, \quad 1_{v_3} : v_3 \to v_3.$$ If you think about how composition works in this category you'll see that the only isomorphisms are the identity morphisms. Why? Because there's no way to compose two morphisms and get an identity morphism unless they're both that identity morphism!
In intuitive terms, we can only move from left to right in this category, not backwards, so we can only 'undo' a morphism if it doesn't do anything at all - i.e., it's an identity morphism.
We can generalize this observation. The key is that \(\mathbf{3}\) is a poset. Remember, in our new way of thinking a
preorder is a category where for any two objects \(x\) and \(y\) there is at most one morphism \(f : x \to y\), in which case we can write \(x \le y\). A poset is a preorder where if there's a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(x = y\). In other words, if \(x \le y\) and \(y \le x\) then \(x = y\). Puzzle 143. Show that if a category \(\mathcal{C}\) is a preorder, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(g\) is the inverse of \(f\), so \(x\) and \(y\) are isomorphic. Puzzle 144. Show that if a category \(\mathcal{C}\) is a poset, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then both \(f\) and \(g\) are identity morphisms, so \(x = y\).
Puzzle 144 says that in a poset, the only isomorphisms are identities.
Isomorphisms are a lot more interesting in the category \(\mathbf{Set}\). Remember, this is the category where objects are sets and morphisms are functions.
Puzzle 145. Show that every isomorphism in \(\mathbf{Set}\) is a bijection, that is, a function that is one-to-one and onto. Puzzle 146. Show that every bijection is an isomorphism in \(\mathbf{Set}\).
So, in \(\mathbf{Set}\) the isomorphisms are the bijections! So, there are lots of them.
One more example:
Definition. If \(\mathcal{C}\) and \(\mathcal{D}\) are categories, then an isomorphism in \(\mathcal{D}^\mathcal{C}\) is called a natural isomorphism.
This name makes sense! The objects in the so-called 'functor category' \(\mathcal{D}^\mathcal{C}\) are functors from \(\mathcal{C}\) to \(\mathcal{D}\), and the morphisms between these are natural transformations. So, the
isomorphisms deserve to be called 'natural isomorphisms'.
But what are they like?
Given functors \(F, G: \mathcal{C} \to \mathcal{D}\), a natural transformation \(\alpha : F \to G\) is a choice of morphism
$$ \alpha_x : F(x) \to G(x) $$ for each object \(x\) in \(\mathcal{C}\), such that for each morphism \(f : x \to y\) this naturality square commutes:
Suppose \(\alpha\) is an isomorphism. This says that it has an inverse \(\beta: G \to F\). This \(beta\) will be a choice of morphism
$$ \beta_x : G(x) \to F(x) $$ for each \(x\), making a bunch of naturality squares commute. But saying that \(\beta\) is the inverse of \(\alpha\) means that
$$ \beta \circ \alpha = 1_F \quad \textrm{ and } \alpha \circ \beta = 1_G .$$ If you remember how we compose natural transformations, you'll see this means
$$ \beta_x \circ \alpha_x = 1_{F(x)} \quad \textrm{ and } \alpha_x \circ \beta_x = 1_{G(x)} $$ for all \(x\). So, for each \(x\), \(\beta_x\) is the inverse of \(\alpha_x\).
In short: if \(\alpha\) is a natural isomorphism then \(\alpha\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\).
But the converse is true, too! It takes a
little more work to prove, but not much. So, I'll leave it as a puzzle. Puzzle 147. Show that if \(\alpha : F \Rightarrow G\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\), then \(\alpha\) is a natural isomorphism.
Doing this will help you understand natural isomorphisms. But you also need examples!
Puzzle 148. Create a category \(\mathcal{C}\) as the free category on a graph. Give an example of two functors \(F, G : \mathcal{C} \to \mathbf{Set}\) and a natural isomorphism \(\alpha: F \Rightarrow G\). Think of \(\mathcal{C}\) as a database schema, and \(F,G\) as two databases built using this schema. In what way does the natural isomorphism between \(F\) and \(G\) make these databases 'the same'. They're not necessarily equal!
We should talk about this. |
Consider the space of newforms $S^{\mathrm{new}}_k(\Gamma_1(q))$ of weight $k$ and level $q$ for the congruence subgroup $\Gamma_1(q)$ of $\mathrm{SL}_2(\mathbb{Z})$; for simplicity's sake, let's assume that $q$ is prime. Then for $k \geq 2$, it is known via Riemann-Roch that $$\dim S^{\mathrm{new}}_k(\Gamma_1(q)) = \frac{k - 1}{24} (q^2 - 1) + E(q,k)$$ for an error term $E(q,k)$. This error term can be calculated explicitly (though not particularly neatly): see Theorem 13 of http://www.math.ubc.ca/~gerg/papers/downloads/DSCFN.pdf. So for $k \geq 2$, it is certainly possible to determine $\dim S^{\mathrm{new}}_k(\Gamma_1(q))$ precisely.
For $k = 1$, on the other hand, no such precise equations seem to exist, as the method used to prove the $k \geq 2$ case breaks down. Instead, it is conjectured (see Conjecture 2.1 of http://arxiv.org/pdf/0906.4579v1) that $$\dim S^{\mathrm{new}}_1(\Gamma_1(q)) = \frac{q - 2}{2} h(K_q) + O_{\varepsilon}(q^{\varepsilon}),$$ for any $\varepsilon > 0$ with the error term is uniform in $q$, and where $h(K_q)$ is the class number of $\mathbb{Q}(\sqrt{-q})$; here the leading term comes from the dihedral modular forms, while the error term is due to the others (icosahedral etc.).
Now note that the leading term in the formula for $S^{\mathrm{new}}_k(\Gamma_1(q))$ for $k \geq 2$ vanishes when $k = 1$, so if that formula where to be valid for $k = 1$, we would be left with the error term $E(q,k)$, which we can explicitly compute.
Question: Is there a reason why we should not expect $\dim S^{\mathrm{new}}_1(\Gamma_1(q)) = E(q,1)$? Obviously a quick check on Magma or Sage should prove that this is not the case, but unfortunately I don't have either installed.
If not, is there any chance that we will one day find a closed form for $\dim S^{\mathrm{new}}_1(\Gamma_1(q))$? |
Research Open Access Published: Interior regularity criterion for incompressible Ericksen-Leslie system Boundary Value Problems volume 2017, Article number: 62 (2017) Article metrics
652 Accesses
Abstract
An interior regularity criterion of suitable weak solutions is formulated for the Ericksen-Leslie system of liquid crystals. Such a criterion is point-wise, with respect to some appropriate norm of velocity
u and the gradient of d, and it can be viewed as a sort of simply sufficient condition on the local regularity of suitable weak solutions. Introduction and main results
In this paper, we investigate the local regularity of weak solutions to the following 3D incompressible Ericksen-Leslie liquid crystal system:
with the initial boundary conditions
where \(u, d, P\) denote the velocity of the fluid, the uniaxial molecular direction, and the pressure, respectively, the \(i,j\)th element of \(\nabla d\odot\nabla d\) is \(\partial_{i}d^{k}\partial_{j}d^{k}\), \(d_{0}(x)\) is a unit vector, \(\Omega\subset \mathbb {R}^{3}\) is a smooth domain. Additionally, \(f(d)=\nabla F(d)\), and \(F(d)=\frac{1}{\zeta^{2}}( \vert d \vert ^{2}-1)^{2}, \zeta\) is a small number, formally speaking, as \(\zeta\to0, d\) tends to a unit vector.
The dynamic flows of liquid crystals have been successfully described by the Ericksen-Leslie theory [1–4]. System (1.1a)-(1.1c) is a coupled system of the Navier-Stokes equations with a parabolic system. It is Leray [5] and Hopf [6] that established the global existence of weak solutions to the 3D Navier-Stokes; however, the regularity of the weak solutions is still an open problem. Since the regularity of weak solutions to the 3D Navier-Stokes equations is hard to get, some related conditions or criteria for the regularity of the weak solutions are considered, such as the well-known Serrin type criterion [7] and the Beale-Kato-Majda type criterion [8]. Furthermore, based on the suitable weak solutions, some point-wise sufficient regularity criteria were imposed in [9–12].
The global existence of suitable weak solutions to system (1.1a)-(1.1c) was established in [13, 14] by Lin and Liu; however, noticing that system (1.1a)-(1.1c) contains the 3D Navier-Stokes equations as a subsystem, the uniqueness and regularity of these weak solutions are not known. In this paper, we would extend some point-wise sufficient conditions, which guarantee the local regularity of weak solutions for 3D Navier-Stokes equations, to the Ericksen-Leslie system (1.1a)-(1.1c). We would like to mention that when \(f(d)\) in system (1.1a)-(1.1c) is replaced by \(- \vert \nabla d \vert ^{2}d\), the global existence of weak solutions to the resulting system in three dimensions has only been known under the additional assumption that \(d_{3}\geq0\) or small initial data (see [15, 16]). Without these conditions, the general existence of weak solutions is still open. However, the Serrin type criterion and the Beale-Kato-Majda type criterion still hold true even for a weak solution (if it exists) (see [17, 18]).
The suitable weak solution established in [14] can be stated as below.
Definition 1.1
Suitable weak solutions in \(\Omega\times(0, T)\subset\mathbb {R}^{3}\times(0,\infty)\)
A pair \((u, d)\) is called a suitable weak solution to system (1.1a)-(1.1c) and (1.2) in an open set \(\mathcal {O}\subset\mathbb {R}^{3}\times(0,\infty)\) (we set \(\mathcal {O}_{t}=\mathcal {O}\cap(\mathbb {R}^{3}\times\{t\} )\)), if it satisfies the following properties:
\((u, d)\) is a weak solution in the sense of distribution;
\(u\in L^{\infty}(0,T;L^{2}(\Omega ))\cap L^{2}(0,T;H^{1}(\Omega)), d\in L^{\infty}(0,T;H^{1}(\Omega))\cap L^{2}(0,T;H^{2}(\Omega))\), or generally, there exist constants \(E_{1}, E_{2}\), such that$$\begin{aligned} &\int_{\mathcal {O}_{t}} \bigl[ \vert u \vert ^{2}+ \vert \nabla d \vert ^{2}+F(d) \bigr]\,\mathrm{ d}x< E_{1},\\ &\int\!\int_{\mathcal {O}} \bigl[ \vert \nabla u \vert ^{2}+ \bigl\vert \Delta d-f(d) \bigr\vert ^{2}+F(d) \bigr]\,\mathrm{ d}x\,\mathrm{ d}t< E_{2}; \end{aligned}$$
for any \(\varphi\in C_{c}^{\infty}(\mathcal {O})\), more specifically, for any \(\varphi\in C_{c}^{\infty}(B(x_{0}, R)\times(t_{0}-R^{2}, t_{0}))\), the following generalized energy inequality holds$$\begin{aligned} &\int_{B(x_{0}, R)} \bigl( \vert u \vert ^{2}+ \vert \nabla d \vert ^{2} \bigr) \varphi\,\mathrm{ d}x+2 \int _{t_{0}-R^{2}}^{t} \int_{B(x_{0}, R)} \bigl( \vert \nabla u \vert ^{2}+ \bigl\vert \nabla^{2} d \bigr\vert ^{2} \bigr)\varphi\,\mathrm{ d}x \,\mathrm{ d}\tau \\ &\quad\leq \int_{t_{0}-R^{2}}^{t} \int_{B(x_{0}, R)} \bigl\{ \bigl( \vert u \vert ^{2}+ \vert \nabla d \vert ^{2} \bigr) (\varphi _{t}+\Delta \varphi)+ \bigl( \vert u \vert ^{2}+ \vert \nabla d \vert ^{2}+2P \bigr)u\cdot\nabla\varphi \bigr\} \,\mathrm{ d}x\,\mathrm{ d}\tau \\ &\qquad{}+2 \int_{t_{0}-R^{2}}^{t} \int_{B(x_{0}, R)} \bigl((u\cdot\nabla) d\nabla d\nabla \varphi-\nabla f(d):\nabla d\varphi \bigr)\,\mathrm{ d}x\,\mathrm{ d}\tau. \end{aligned}$$(1.3)
In the following, we can take \(Q((x_{0}, t_{0}), R)\equiv B(x_{0}, R)\times(t_{0}-R^{2}, t_{0})\), \(B(x_{0}, R)\equiv\{y\in\mathbb {R}^{3}| \vert y-x_{0} \vert < R\}, z_{0}\equiv(x_{0}, t_{0})\) for simplicity.
We now state our main result of this paper.
Theorem 1.2 Then there is a positive number \(\varepsilon=\varepsilon(s,l)\), such that if then \(z_{0}\) is a regular point of \((u,\nabla d)\), i. e. \((u,\nabla d)\) is Hölder continuous in \(Q(z_{0},r)\), for some \(r\in(0, R]\).
Throughout this paper, we use
c to denote a generic positive constant which can be different from line to line. Preliminaries
As the preparation for proving Theorem 1.2, we first give two auxiliary lemmas.
Lemma 2.1 We have where Proof Step 1. For (1.1a), we choose the test function \(w=\chi\nabla q\), for any \(\chi\in C_{c}^{\infty}((t_{0}-\rho^{2}, t_{0})), q\in C_{c}^{\infty}(B(x_{0}, \rho))\), then it yields
It follows from \(\nabla\cdot u=0\) that
Therefore, for a.e. \(t\in(t_{0}-\rho^{2}, t_{0})\), we have
Step 2. Approximate p with \(p_{1}\) by confining q in \(W^{2, 3}(B(x_{0}, \rho))\).
Set \(p_{1}\in L^{\frac{3}{2}}(Q(z_{0}, \rho))\) such that, for a.e. \(t\in(t_{0}-\rho ^{2}, t_{0})\),
for any \(q(\cdot, t)\in W^{2, 3}(B(x_{0}, \rho))\), and \(q(\cdot, t)=0 \text{ on } \partial B(x_{0}, \rho)\). The existence of \(p_{1}\) is established due to the Lax-Milgram theorem with appropriate approximating process on
u and d (see [11]).
Next, choose \(q_{0}(\cdot, t)\in W^{2, 3}(B(x_{0}, \rho))\), such that, for a.e. \(t\in(t_{0}-\rho^{2}, t_{0})\),
Then, by the Calderon-Zygmund inequality, it yields
Therefore, it follows from (2.3) and the Hölder inequality that
which yields \(\int_{Q(z_{0}, \rho)} \vert p_{1}(\cdot, t) \vert ^{\frac{3}{2}}\,\mathrm{ d}z\leq c\rho^{2}C(z_{0}, \rho; u, \nabla d)\).
Step 3. Estimates for the remainder \(p-p_{1}\).
By the harmonic property, one can get
while
Step 4. Estimates for p.
We have
□
We denote
Lemma 2.2 Under the assumptions of Theorem 1.2, we have where \(q=2l(\frac{3}{s}+\frac{2}{l}-\frac{3}{2})\), and \(q'=\frac{q}{q-1}\). Proof
With the help of the Hölder and Sobolev embedding inequalities, one gets
where \(\lambda s+2\mu+6\gamma=3, \lambda+\mu+\gamma=1\). Substituting
v by u and ∇ d, respectively, then one can get the summation
Therefore, by choosing appropriate parameters \(\lambda=\frac{1}{2s(\frac{3}{s}+\frac{2}{l}-\frac{3}{2})}\), \(\mu=\frac {\frac{3}{s}+\frac{3}{l}-2}{2(\frac{3}{s}+\frac{2}{l}-\frac{3}{2})}\), \(\gamma= \frac{\frac{2}{s}+\frac{1}{l}-1}{2(\frac{3}{s}+\frac{2}{l}-\frac{3}{2})}\), and integrating from \(t_{0}-\rho^{2}\) to \(t_{0}\) with the variable
t, it follows from the Hölder and Young inequalities that
where \(\kappa=\frac{3l}{s}+2-l\) as in Theorem 1.2, and in the last step, we used the fact that \(\mu q\leq1, H(\rho)\leq A(\rho)\). □
Proof of Theorem 1.2
Due to the induction argument as Proposition 2.6 in [10] or Lemma 2.2 in [19] (the parabolic version of the Campanato criterion), to get the desired consequence, it suffices to prove \(C(\theta^{k})+D(\theta^{k})< \epsilon_{0}\) for some small \(\epsilon_{0}\). Here
θ is a small number, which will be chosen later.
From the generalized energy inequality, it is easy to check that, for \(\rho\in(0, R]\),
where in the last step we have used \(\frac{\epsilon^{\frac{2}{3q}}}{\theta^{\frac{4}{3}}}(G(\rho)+1)^{\frac{2}{3}}\leq c[\epsilon^{\frac{1}{q}}+\frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}(G(\rho)+1)]\). Now choosing
θ and ϵ such that \(c\theta<\frac{1}{4}\) and \(c\frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}<\frac{1}{4}\), then it yields \(G(\theta\rho)\leq\frac{1}{2}G(\rho)+c\frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}\). Iterating the above process, we obtain \(G(\theta^{k} \rho)\leq\frac{1}{2^{k}}G(\rho)+c\frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}\), therefore,
For \(C(\theta^{k}\rho)\), by Lemma 2.2, we have
where in the last step we use the fact that \(\epsilon^{\frac{1}{q}}\leq \frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}\) for
ϵ small enough. With these inequalities in hand, for fixed ρ and \(\epsilon_{0}\), we can choose \(k_{0}\) large enough such that \(c\frac{1}{2^{k_{0}}}G(\rho)<\frac{\epsilon_{0}}{4}\), and choose ϵ small enough, such that \(c\frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}<\frac{\epsilon_{0}}{4}\). With these prerequisites and (3.1)-(3.2), it follows that \(D(\theta^{k}\rho)+C(\theta^{k}\rho)<\epsilon_{0}\). References 1.
Ericksen, JL: Conservation laws for liquid crystals. Trans. Soc. Rheol.
5, 23-34 (1961) 2.
Ericksen, JL: Liquid crystals with variable degree of orientation. Arch. Ration. Mech. Anal.
113, 97-120 (1991) 3.
Leslie, FM: Some constitutive equations for liquid crystals. Arch. Ration. Mech. Anal.
28, 265-283 (1968) 4.
Leslie, FM: Theory of flow phenomenum in liquid crystals. Adv. Liq. Cryst.
4, 1-81 (1979) 5.
Leray, J: Sur le mouvement d’un liquide visqueux emplissant l’espace. Acta Math.
63, 183-248 (1934) 6.
Hopf, E: Uber die Aufangswertaufgabe für die hydrodynamischen Grundgleichungen. Math. Nachr.
4, 213-231 (1951) 7.
Serrin, J: On the interior regularity of weak solutions of the Navier-Stokes equations. Arch. Ration. Mech. Anal.
9, 187-195 (1962) 8.
Beale, JT, Kato, T, Majda, A: Remarks on the breakdown of smooth solutions for the 3-D Euler equation. Commun. Math. Phys.
94, 61-66 (1984) 9.
Cafferelli, L, Kohn, R, Nirenberg, L: Partial regularity of suitable weak solutions of Navier-Stokes equations. Commun. Pure Appl. Math.
35, 771-831 (1982) 10.
Ladyzhenskaya, OA, Seregin, G: On partial regularity of suitable weak solutions to the three-dimensional Navier-Stokes equations. J. Math. Fluid Mech.
1, 356-387 (1999) 11.
Seregin, G: On the number of singular points of weak solutions to the Navier-Stokes equations. Commun. Pure Appl. Math.
54, 1019-1028 (2001) 12.
Zajaczkowski, W, Seregin, G: Sufficient condition of local regularity for the Navier-Stokes equations. J. Math. Sci.
143, 2869-2874 (2007) 13.
Lin, FH, Liu, C: Nonparabolic dissipative systems modeling the flow of liquid crystals. Commun. Pure Appl. Math.
48, 501-537 (1995) 14.
Lin, FH, Liu, C: Partial regularities of the nonlinear dissipative systems modeling the flow of liquid crystals. Discrete Contin. Dyn. Syst.
2, 1-23 (1996) 15.
Lin, FH, Wang, CY: Global existence of weak solutions of the nematic liquid crystal flow in dimension three. Commun. Pure Appl. Math.
69, 1532-1571 (2016) 16.
Ma, WY, Gong, HJ, Li, JK: Global strong solutions to incompressible Ericksen-Leslie system in \(\mathbb {R}^{3}\). Nonlinear Anal.
109, 230-235 (2014) 17.
Huang, T, Wang, CY: Blow up criterion for nematic liquid crystal flows. Commun. Partial Differ. Equ.
37, 875-884 (2012) 18.
Hong, MC, Li, JK, Xin, ZP: Blow up critera of strong solutions to the Ericksen-Leslie system in \(\mathbb{R}^{3}\). Commun. Partial Differ. Equ.
39, 1284-1328 (2014) 19.
Escauriaza, L, Seregin, G, Sverak, V: \(L_{3,\infty}\)-solutions of the Navier-Stokes equations and backward uniqueness. Russ. Math. Surv.
58(2), 211-250 (2003) Acknowledgements
The authors thank Dr. Huajun Gong and Dr. Jinkai Li for helpful discussions and suggestions. Ma is supported by Fostering Talents of NSFC-Henan Province (U1404102) and NSFC (No. 11501174, 11626090). Feng is supported by NSFC (No. 61401283, 11601342, 61472257), GDPSTPP (No. 2013B040403005) and (No. GCZX-A1409).
Additional information Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Both authors contributed equally in writing this paper. They both read and approved the final manuscript.
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This is the question:
The solid generated by rotating the region inside the ellipse with equation $$ \left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1 $$ around the $x$-axis is called an
ellipsoid.
(a)Show that the ellipsoid has volume $\displaystyle \frac{4}{3} \pi a b^2.$
(b)What is the volume if the ellipse is rotated around the $y$-axis.
(I want to accomplish this using integrals and basic plane geometry. To give you an idea of how much I know about integrals I'm $4$ weeks into my calculus II course, which is my first exposure to integrals)
I started by drawing an ellipse on a Cartesian plane. The ellipse went from $a$ to $b$, I then rotated this ellipse around the $x$-axis to get an ellipsoid. Now to get the volume I have to find a cross-sectional area, so I noticed that the ellipsoid is really made up of a bunch of circles stacked along the $x$-axis. Where I'm stuck at right now is how I can find the radius of these circles, which would give me my integrand. |
User talk:WikiSysop Contents 1 Test external Link 29th January 2015 2 Test Copy&Paste HTML 3 Test Asymptote 3.1 Test January 12th 2015 3.2 Tests November 1th 3.3 Tests November 17th 3.4 Tests November 4th 3.5 Tests October 27th 3.6 Previous tests 4 Test Cite Extension 5 Test MathJax 6 Pages A-Z 7 Recent Changes Test external Link 29th January 2015 Test Copy&Paste HTML
Test-copy-paste Test Test
Test Asymptote Test January 12th 2015 Case 1 modified
Tests November 1th Case 1 modified
Case 1
Tests November 17th Case 1
Tests November 4th Case 1
Case 2
Case 3
Tests October 27th Case 1
Case 2
Case 3
Case 4
Case 5
[asy] pair A,B,C,X,Y,Z; A = (0,0); B = (1,0); C = (0.3,0.8); draw(A--B--C--A); X = (B+C)/2; Y = (A+C)/2; Z = (A+B)/2; draw(A--X, red); draw(B--Y,red); draw(C--Z,red); [/asy]
Previous tests
Test Cite Extension
Example: Cite-Extension
Test MathJax
\begin{align} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{align}
\[ \frac{1}{(\sqrt{\phi \sqrt{5}}-\phi) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\ldots} } } } \]
Some Text \( \frac{1}{(\sqrt{\phi \sqrt{5}}-\phi) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\ldots} } } } \)
Some Text \[ \frac{1}{(\sqrt{\phi \sqrt{5}}-\phi) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\ldots} } } } \]
Alphabetically ordered index of all pages
List of previous changes on EOM
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How to Cite This Entry:
WikiSysop.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=WikiSysop&oldid=36277
Test January 12th 2015 |
I wanted to better understand dfa. I wanted to build upon a previous question:Creating a DFA that only accepts number of a's that are multiples of 3But I wanted to go a bit further. Is there any way we can have a DFA that accepts number of a's that are multiples of 3 but does NOT have the sub...
Let $X$ be a measurable space and $Y$ a topological space. I am trying to show that if $f_n : X \to Y$ is measurable for each $n$, and the pointwise limit of $\{f_n\}$ exists, then $f(x) = \lim_{n \to \infty} f_n(x)$ is a measurable function. Let $V$ be some open set in $Y$. I was able to show th...
I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...
Consider a non-UFD that only has 2 units ( $-1,1$ ) and the min difference between 2 elements is $1$. Also there are only a finite amount of elements for any given fixed norm. ( Maybe that follows from the other 2 conditions ? )I wonder about counting the irreducible elements bounded by a lower...
How would you make a regex for this? L = {w $\in$ {0, 1}* : w is 0-alternating}, where 0-alternating is either all the symbols in odd positions within w are 0's, or all the symbols in even positions within w are 0's, or both.
I want to construct a nfa from this, but I'm struggling with the regex part |
Yes, but it may not be valid. The extrapolation will be valid for about 0.1 * PBL Height using the Log-Wind Profile
You will need:
PBL Height. A second Wind speed (within 0.1*PBL Height) Surface Sensible Heat Flux Surface Latent Heat Flux Potential Temperature
You can use the last three variables to calculate the Monin-Obukhov Length (MOL). Then use the MOL to calculate $\psi$:
If L>0 $\psi=\frac{z}{L}$
If L=0 $\psi=0$
If L<0 $\psi=2\ln(\frac{1+x}{2})+\ln(\frac{1+x^2}{2})-2\tan^{-1}(x)+\frac{\pi}{2}$
where $x=(1-15\frac{z}{L})^{\frac{1}{4}}$
Now that you have $\psi$, you can derive the friction velocity and surface roughness length using the log-wind profile and $\psi(\frac{z}{L})$. Once you have those two, you can extrapolate wind speed 600 meters up, provided the surface layer is that high. |
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Isomorphisms are very important in mathematics, and we can no longer put off talking about them. Intuitively, two objects are 'isomorphic' if they look the same. Category theory makes this precise and shifts the emphasis to the 'isomorphism' - the
way in which we match up these two objects, to see that they look the same.
For example, any two of these squares look the same after you rotate and/or reflect them:
An isomorphism between two of these squares is a
process of rotating and/or reflecting the first so it looks just like the second.
As the name suggests, an isomorphism is a kind of morphism. Briefly, it's a morphism that you can 'undo'. It's a morphism that has an inverse:
Definition. Given a morphism \(f : x \to y\) in a category \(\mathcal{C}\), an inverse of \(f\) is a morphism \(g: y \to x\) such that
and
I'm saying that \(g\) is 'an' inverse of \(f\) because in principle there could be more than one! But in fact, any morphism has at most one inverse, so we can talk about 'the' inverse of \(f\) if it exists, and we call it \(f^{-1}\).
Puzzle 140. Prove that any morphism has at most one inverse. Puzzle 141. Give an example of a morphism in some category that has more than one left inverse. Puzzle 142. Give an example of a morphism in some category that has more than one right inverse.
Now we're ready for isomorphisms!
Definition. A morphism \(f : x \to y\) is an isomorphism if it has an inverse. Definition. Two objects \(x,y\) in a category \(\mathcal{C}\) are isomorphic if there exists an isomorphism \(f : x \to y\).
Let's see some examples! The most important example for us now is a 'natural isomorphism', since we need those for our databases. But let's start off with something easier. Take your favorite categories and see what the isomorphisms in them are like!
What's an isomorphism in the category \(\mathbf{3}\)? Remember, this is a free category on a graph:
The morphisms in \(\mathbf{3}\) are paths in this graph. We've got one path of length 2:
$$ f_2 \circ f_1 : v_1 \to v_3 $$ two paths of length 1:
$$ f_1 : v_1 \to v_2, \quad f_2 : v_2 \to v_3 $$ and - don't forget - three paths of length 0. These are the identity morphisms:
$$ 1_{v_1} : v_1 \to v_1, \quad 1_{v_2} : v_2 \to v_2, \quad 1_{v_3} : v_3 \to v_3.$$ If you think about how composition works in this category you'll see that the only isomorphisms are the identity morphisms. Why? Because there's no way to compose two morphisms and get an identity morphism unless they're both that identity morphism!
In intuitive terms, we can only move from left to right in this category, not backwards, so we can only 'undo' a morphism if it doesn't do anything at all - i.e., it's an identity morphism.
We can generalize this observation. The key is that \(\mathbf{3}\) is a poset. Remember, in our new way of thinking a
preorder is a category where for any two objects \(x\) and \(y\) there is at most one morphism \(f : x \to y\), in which case we can write \(x \le y\). A poset is a preorder where if there's a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(x = y\). In other words, if \(x \le y\) and \(y \le x\) then \(x = y\). Puzzle 143. Show that if a category \(\mathcal{C}\) is a preorder, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then \(g\) is the inverse of \(f\), so \(x\) and \(y\) are isomorphic. Puzzle 144. Show that if a category \(\mathcal{C}\) is a poset, if there is a morphism \(f : x \to y\) and a morphism \(g: x \to y\) then both \(f\) and \(g\) are identity morphisms, so \(x = y\).
Puzzle 144 says that in a poset, the only isomorphisms are identities.
Isomorphisms are a lot more interesting in the category \(\mathbf{Set}\). Remember, this is the category where objects are sets and morphisms are functions.
Puzzle 145. Show that every isomorphism in \(\mathbf{Set}\) is a bijection, that is, a function that is one-to-one and onto. Puzzle 146. Show that every bijection is an isomorphism in \(\mathbf{Set}\).
So, in \(\mathbf{Set}\) the isomorphisms are the bijections! So, there are lots of them.
One more example:
Definition. If \(\mathcal{C}\) and \(\mathcal{D}\) are categories, then an isomorphism in \(\mathcal{D}^\mathcal{C}\) is called a natural isomorphism.
This name makes sense! The objects in the so-called 'functor category' \(\mathcal{D}^\mathcal{C}\) are functors from \(\mathcal{C}\) to \(\mathcal{D}\), and the morphisms between these are natural transformations. So, the
isomorphisms deserve to be called 'natural isomorphisms'.
But what are they like?
Given functors \(F, G: \mathcal{C} \to \mathcal{D}\), a natural transformation \(\alpha : F \to G\) is a choice of morphism
$$ \alpha_x : F(x) \to G(x) $$ for each object \(x\) in \(\mathcal{C}\), such that for each morphism \(f : x \to y\) this naturality square commutes:
Suppose \(\alpha\) is an isomorphism. This says that it has an inverse \(\beta: G \to F\). This \(beta\) will be a choice of morphism
$$ \beta_x : G(x) \to F(x) $$ for each \(x\), making a bunch of naturality squares commute. But saying that \(\beta\) is the inverse of \(\alpha\) means that
$$ \beta \circ \alpha = 1_F \quad \textrm{ and } \alpha \circ \beta = 1_G .$$ If you remember how we compose natural transformations, you'll see this means
$$ \beta_x \circ \alpha_x = 1_{F(x)} \quad \textrm{ and } \alpha_x \circ \beta_x = 1_{G(x)} $$ for all \(x\). So, for each \(x\), \(\beta_x\) is the inverse of \(\alpha_x\).
In short: if \(\alpha\) is a natural isomorphism then \(\alpha\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\).
But the converse is true, too! It takes a
little more work to prove, but not much. So, I'll leave it as a puzzle. Puzzle 147. Show that if \(\alpha : F \Rightarrow G\) is a natural transformation such that \(\alpha_x\) is an isomorphism for each \(x\), then \(\alpha\) is a natural isomorphism.
Doing this will help you understand natural isomorphisms. But you also need examples!
Puzzle 148. Create a category \(\mathcal{C}\) as the free category on a graph. Give an example of two functors \(F, G : \mathcal{C} \to \mathbf{Set}\) and a natural isomorphism \(\alpha: F \Rightarrow G\). Think of \(\mathcal{C}\) as a database schema, and \(F,G\) as two databases built using this schema. In what way does the natural isomorphism between \(F\) and \(G\) make these databases 'the same'. They're not necessarily equal!
We should talk about this. |
Stationary solutions of neutral stochastic partial differential equations with delays in the highest-order derivatives
a).
College of Mathematical Sciences, Tianjin Normal University, Tianjin 300387, China
b).
Department of Mathematical Sciences, School of Physical Sciences, The University of Liverpool, Liverpool, L69 7ZL, UK
In this work, we shall consider the existence and uniqueness of stationary solutions to stochastic partial functional differential equations with additive noise in which a neutral type of delay is explicitly presented. We are especially concerned about those delays appearing in both spatial and temporal derivative terms in which the coefficient operator under spatial variables may take the same form as the infinitesimal generator of the equation. We establish the stationary property of the neutral system under investigation by focusing on distributed delays. In the end, an illustrative example is analyzed to explain the theory in this work.
Keywords:Stochastic functional differential equation of neutral type, strongly continuous or $ c_0$ semigroup, resolvent operator, stationary solution. Mathematics Subject Classification:60H15, 60G15, 60H05. Citation:Kai Liu. Stationary solutions of neutral stochastic partial differential equations with delays in the highest-order derivatives. Discrete & Continuous Dynamical Systems - B, 2018, 23 (9) : 3915-3934. doi: 10.3934/dcdsb.2018117
References:
[1]
A. Bátkai and S. Piazzera,
[2]
E. B. Davies,
[3]
G. Di Blasio, K. Kunisch and E. Sinestrari,
$ L^2$-regularity for parabolic partial integrodifferential equations with delay in the highest-order derivatives,
[4] [5]
J. Hale and S. Lunel,
[6] [7] [8]
J. L. Lions and E. Magenes,
[9]
K. Liu,
Uniform $ L^2$-stability in mean square of linear autonomous stochastic functional differential equations in Hilbert spaces,
[10] [11]
K. Liu,
Norm continuity of solution semigroups of a class of neutral functional differential equations with distributed delay,
[12]
C. Prévôt and M. Röckner,
[13]
H. Tanabe,
[14]
H. Tanabe,
show all references
References:
[1]
A. Bátkai and S. Piazzera,
[2]
E. B. Davies,
[3]
G. Di Blasio, K. Kunisch and E. Sinestrari,
$ L^2$-regularity for parabolic partial integrodifferential equations with delay in the highest-order derivatives,
[4] [5]
J. Hale and S. Lunel,
[6] [7] [8]
J. L. Lions and E. Magenes,
[9]
K. Liu,
Uniform $ L^2$-stability in mean square of linear autonomous stochastic functional differential equations in Hilbert spaces,
[10] [11]
K. Liu,
Norm continuity of solution semigroups of a class of neutral functional differential equations with distributed delay,
[12]
C. Prévôt and M. Röckner,
[13]
H. Tanabe,
[14]
H. Tanabe,
[1]
Nicholas J. Kass, Mohammad A. Rammaha.
Local and global existence of solutions to a strongly damped wave equation of the $ p $-Laplacian type.
[2]
Jiří Neustupa.
On $L^2$-Boundedness of a $C_0$-Semigroup generated by the perturbed oseen-type operator arising from flow around a rotating body.
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Peng Mei, Zhan Zhou, Genghong Lin.
Periodic and subharmonic solutions for a 2$n$th-order $\phi_c$-Laplacian difference equation containing both advances and retardations.
[5]
Annalisa Cesaroni, Serena Dipierro, Matteo Novaga, Enrico Valdinoci.
Minimizers of the $ p $-oscillation functional.
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Chengxiang Wang, Li Zeng, Wei Yu, Liwei Xu.
Existence and convergence analysis of $\ell_{0}$ and $\ell_{2}$ regularizations for limited-angle CT reconstruction.
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Ilwoo Cho, Palle Jorgense.
Free probability on $ C^{*}$-algebras induced by hecke algebras over primes.
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Piotr Bizoń, Dominika Hunik-Kostyra, Dmitry Pelinovsky.
Stationary states of the cubic conformal flow on $ \mathbb{S}^3 $.
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Convergence of a finite volume scheme for a stochastic conservation law involving a $Q$-brownian motion.
[15]
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On the stability of $\vartheta$-methods for stochastic Volterra integral equations.
[16]
Shihu Li, Wei Liu, Yingchao Xie.
Large deviations for stochastic 3D Leray-$ \alpha $ model with fractional dissipation.
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The Relative Complement and Complement of a Set The Relative Complement of a Set
Definition: If $A$ is a set, then the Relative Complement of $A$ with respect to $B$ denoted $B \setminus A$ is the set of elements contained $B$ but that are not contained in $A$, that is, $B \setminus A = \{ x : x \in B \: \mathrm{and} \: x \not \in A \}$.
For example, consider the sets $A = \{1, 3, 5, 7, 9 \}$ and $B = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \}$. Then:(1)
We can represent the relative complement of two sets graphically using Venn diagrams as illustrated below:
We will now state some relatively simple results regarding the relative complement of two sets that the reader is invited to prove if they've never seen the results before:
Theorem 1: Let $A$ and $B$ be sets. Then: a) $(A \setminus B) \cup B = A \cup B$. b) $(A \setminus B) \cap B = \emptyset$. The Complement of a Set
Definition: If $A$ is a set, then the Complement of $A$ denoted $A^c$ is the set of elements that are not contained in $A$, that is, $A^c = \{ x : x \not \in A \}$.
In general, if $A$ is a set without any context then the complement of $A$ may not be well defined. For example, if we consider the set of real numbers $\mathbb{R}$ then what exactly is $\mathbb{R}^c$? By definition, it is the set of all elements that are not real numbers. Hence, we could say that all cats or all potatoes are contained in $\mathbb{R}^c$, but of course, this sort of inclusion probably isn't the most useful in mathematics.
That said, it must be important to have some sort of context in mind when talking about the complement of a set. If we consider the set of real numbers as a
Universal Set, then it is clear from context that the complement of $\mathbb{R}^+$ of nonnegative real numbers is equal to the set of negative real numbers. Hence if a universal set $U$ is understood with respect to $A$, then:
We can represent the complement of a set $A$ with respect to some universal set $U$ with the following Venn diagram: |
In order to enable an iCal export link, your account needs to have an API key created. This key enables other applications to access data from within Indico even when you are neither using nor logged into the Indico system yourself with the link provided. Once created, you can manage your key at any time by going to 'My Profile' and looking under the tab entitled 'HTTP API'. Further information about HTTP API keys can be found in the Indico documentation.
Additionally to having an API key associated with your account, exporting private event information requires the usage of a persistent signature. This enables API URLs which do not expire after a few minutes so while the setting is active, anyone in possession of the link provided can access the information. Due to this, it is extremely important that you keep these links private and for your use only. If you think someone else may have acquired access to a link using this key in the future, you must immediately create a new key pair on the 'My Profile' page under the 'HTTP API' and update the iCalendar links afterwards.
Permanent link for public information only:
Permanent link for all public and protected information:
byProf.Maarten Golterman(San Francisco State University)
→Europe/Rome
Aula Seminari (LNF)
Aula Seminari
LNF
Via Enrico Fermi, 4000044 Frascati (Roma)
Description
We review our recent determination of the strong coupling $\alpha_s$ from the OPAL data for non-strange hadronic tau decays. We find that $\alpha_s(m^2_\tau) =0.325\pm 0.018$ using fixed-order perturbation theory, and $\alpha_s(m^2_\tau)=0.347\pm 0.025$ using contour-improved perturbation theory. These errors are larger than those claimed in previous determinations of $\alpha_s$ from tau decays. The reasons for this are several: (1) The spectral-data moments used in the standard analysis are likely to have an unreliable perturbative expansion, and OPE corrections have not been treated systematically; (2) Violations of quark-hadron duality have been ignored; (3) The nominally more precise ALEPH data are flawed because the correlation matrix is incomplete. We thus consider our determination to supersede all previous values. |
berylium? really? okay then...toroidalet wrote:I Undertale hate it when people Emoji movie insert keywords so people will see their berylium page.
A forum where anything goes. Introduce yourselves to other members of the forums, discuss how your name evolves when written out in the Game of Life, or just tell us how you found it. This is the forum for "non-academic" content.
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
When xq is in the middle of a different object's apgcode. "That's no ship!"
Airy Clave White It Nay
When you post something and someone else posts something unrelated and it goes to the next page.
Also when people say that things that haven't happened to them trigger them.
Also when people say that things that haven't happened to them trigger them.
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
Huh. I've never seen a c/posts spaceship before.drc wrote:"The speed is actually" posts
Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace!
It could be solved with a simple PM rather than an entire post.Gamedziner wrote:What's wrong with them?drc wrote:"The speed is actually" posts
An exception is if it's contained within a significantly large post.
I hate it when people post rule tables for non-totalistic rules. (Yes, I know some people are on mobile, but they can just generate them themselves. [citation needed])
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
OK this is a very niche one that I hadn't remembered until a few hours ago.
You know in some arcades they give you this string of cardboard tickets you can redeem for stuff, usually meant for kids. The tickets fold beautifully perfectly packed if you order them one right, one left - zigzagging. When people fold them randomly in any direction giving a clearly low density packing with loads of strain, I just think
You know in some arcades they give you this string of cardboard tickets you can redeem for stuff, usually meant for kids.
The tickets fold beautifully perfectly packed if you order them one right, one left - zigzagging.
When people fold them randomly in any direction giving a clearly low density packing with loads of strain, I just think
omg why on Earth would you do that?!Surely they'd have realised by now? It's not that crazy to realise? Surely there is a clear preference for having them well packed; nobody would prefer an unwieldy mess?!
Also when I'm typing anything and I finish writing it and it just goes to the next line or just goes to the next page. Especially when the punctuation mark at the end brings the last word down one line. This also applies to writing in a notebook: I finish writing something but the very last thing goes to a new page.
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: ... you were referencing me before i changed it, weren't you? because I had fit both of those.A for awesome wrote: When people put non- spectacularly-interesting patterns, questions, etc. in their signature.
ON A DIFFERENT NOTE.
When i want to rotate a hexagonal file but golly refuses because for some reason it calculates hexagonal patterns on a square grid and that really bugs me because if you want to show that something has six sides you don't show it with four and it makes more sense to have the grid be changed to hexagonal but I understand Von Neumann because no shape exists (that I know of) that has 4 corners and no edges but COME ON WHY?! WHY DO YOU REPRESENT HEXAGONS WITH SQUARES?!
In all seriousness this bothers me and must be fixed or I will SINGLEHANDEDLY eat a universe.
EDIT: possibly this one.
EDIT 2:
IT HAS BEGUN.
HAS
BEGUN.
Last edited by 83bismuth38 on September 19th, 2017, 8:25 pm, edited 1 time in total.
Actually, I don't remember who I was referencing, but I don't think it was you, and if it was, it wasn't personal.83bismuth38 wrote:... you were referencing me before i changed it, weren't you? because I had fit both of those.A for awesome wrote: When people put non- spectacularly-interesting patterns, questions, etc. in their signature.
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
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: oh okay yeah of course sureA for awesome wrote:Actually, I don't remember who I was referencing, but I don't think it was you, and if it was, it wasn't personal.83bismuth38 wrote:... you were referencing me before i changed it, weren't you? because I had fit both of those.A for awesome wrote: When people put non- spectacularly-interesting patterns, questions, etc. in their signature.
but really though, i wouldn't have cared.
When someone gives a presentation to a bunch of people and you
knowthat they're getting the facts wrong. Especially if this is during the Q&A section.
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
When you watch a boring video in class but you understand it perfectly and then at the end your classmates dont get it so the teacher plays the borinh video again
Airy Clave White It Nay
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact:
when scientists decide to send a random guy into a black hole hovering directly above Earth for no reason at all.
hit; that random guy was me.
hit; that random guy was me.
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact:
When I see a "one-step" organic reaction that occurs in an exercise book for senior high school and simply takes place under "certain circumstance" like the one marked "?" here but fail to figure out how it works even if I have prepared for our provincial chemistry olympiadEDIT: In fact it's not that hard.Just do a Darzens reaction then hydrolysis and decarboxylate.
Current status: outside the continent of cellular automata. Specifically, not on the plain of life.
An awesome gun firing cool spaceships:
An awesome gun firing cool spaceships:
Code: Select all
x = 3, y = 5, rule = B2kn3-ekq4i/S23ijkqr4eikry2bo$2o$o$obo$b2o!
When there's a rule with a decently common puffer but it can't interact with itself
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
When that oscillator is just
When you're sooooooo close to a thing you consider amazing but miss... not sparky enough.
When you're sooooooo close to a thing you consider amazing but miss...
Airy Clave White It Nay
People posting tons of "new" discoveries that have been known for decades, showing that they've not observed standard netiquette by reading the forums a while before posting, nor done the most minimal research about whether things have been already known, despit repeated posts about where to find such resources (e.g. jslife, wiki, Life lexicon, etc.).
People posting tons of useless "new" discoveries that take longer to post than to find (e.g. "look what happens when I put this blinker next to this beehive"). Newbies with attitudes, who think they know more than people who have been part of the community for years or even decades. Posts where the quoted text is substantially longer than added text. Especially "me too" posts. People whose signatures are longer than the actual text of their posts. People whose signatures include graphics or pattern files, especially ones that are just human-readable text. Improper grammar, spelling, and punctuation (although I've gotten used to that; long-term use of the internet has made me rather fluent in typo, both reading and writing). Imperfect English is not unreasonable from people for whom English is not a primary language, but from English speakers, it is a symptom of sloppiness that can also manifest in other areas.
People posting tons of useless "new" discoveries that take longer to post than to find (e.g. "look what happens when I put this blinker next to this beehive").
Newbies with attitudes, who think they know more than people who have been part of the community for years or even decades.
Posts where the quoted text is substantially longer than added text. Especially "me too" posts.
People whose signatures are longer than the actual text of their posts.
People whose signatures include graphics or pattern files, especially ones that are just human-readable text.
Improper grammar, spelling, and punctuation (although I've gotten used to that; long-term use of the internet has made me rather fluent in typo, both reading and writing). Imperfect English is not unreasonable from people for whom English is not a primary language, but from English speakers, it is a symptom of sloppiness that can also manifest in other areas.
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X That's G U S T A V O right theremniemiec wrote:People posting tons of "new" discoveries that have been known for decades, showing that they've not observed standard netiquette by reading the forums a while before posting, nor done the most minimal research about whether things have been already known, despit repeated posts about where to find such resources (e.g. jslife, wiki, Life lexicon, etc.). People posting tons of useless "new" discoveries that take longer to post than to find (e.g. "look what happens when I put this blinker next to this beehive"). Newbies with attitudes, who think they know more than people who have been part of the community for years or even decades.
Also, when you walk into a wall slowly and carefully but you hit your teeth on the wall and it hurts so bad.
Airy Clave White It Nay |
Your task is to take an array of numbers and a real number and return the value at that point in the array. Arrays start at \$\pi\$ and are counted in \$\pi\$ intervals. Thing is, we're actually going to interpolate between elements given the "index". As an example:
Index: 1π 2π 3π 4π 5π 6πArray: [ 1.1, 1.3, 6.9, 4.2, 1.3, 3.7 ]
Because it's \$\pi\$, we have to do the obligatory trigonometry, so we'll be using cosine interpolation using the following formula:
\${\cos(i \mod \pi) + 1 \over 2} * (\alpha - \beta) + \beta\$
where:
\$i\$ is the input "index" \$\alpha\$ is the value of the element immediately before the "index" \$\beta\$ is the value of the element immediately after the "index" \$\cos\$ takes its angle in radians Example
Given [1.3, 3.7, 6.9], 5.3:
Index 5.3 is between \$1\pi\$ and \$2\pi\$, so 1.3 will be used for
before and 3.7 will be used for
after. Putting it into the formula, we get:
\${\cos(5.3 \mod \pi) + 1 \over 2} * (1.3 - 3.7) + 3.7\$
Which comes out to 3.165
Notes Input and output may be in any convenient format You may assume the input number is greater than \$\pi\$ and less than
array length* \$\pi\$
You may assume the input array will be at least 2 elements long. Your result must have at least two decimal points of precision, be accurate to within 0.05, and support numbers up to 100 for this precision/accuracy. (single-precision floats are more than sufficient to meet this requirement)
Happy Golfing! |
I am fitting a gaussian mixture to financial data. My mixture density is given by:
$f(l)=πϕ(l;μ_1,σ^2_1)+(1−π)ϕ(l;μ_2,σ_2^2)$
I calculated the skewness of the data already. Now, I want to look at the skewness of the fitted gaussian mixture. Since I used ML (EM algorithm) and not method of moments, the moments will not be the same. I know this. But I don't know how to calculate the skewness of the mixed gaussian? I want to have a theoretical derived formula, so I mean, I don't want to calculate this empirical by taking the fitted values and do e.g. skew(...) in R. I will do this to control myself, but first I want to have the theoretical formula for it. I could not find it (I googled for skewness mixture density and so.)
I know that the the skewness is given by
$ \gamma_1 = \operatorname{E}\Big[\big(\tfrac{X-\mu}{\sigma}\big)^{\!3}\, \Big] = \frac{\mu_3}{\sigma^3} = \frac{\operatorname{E}\big[(X-\mu)^3\big]}{\ \ \ ( \operatorname{E}\big[ (X-\mu)^2 \big] )^{3/2}} $
So what is the Skewness of a mixture gaussian? How can I derive it? A mathematical derivation would be great. I have estimated both densities and I have the estimates for μ and σ. I want a formula in what I can insert those values to get the skewness of the mixed density. Then I will control it empirically with skew(...) in R.
I know this for the kurtosis: and I want to have this for skewness and - this would be great- a derivation of it? |
The Initial and Final Segment Topologies
Recall from the Topological Spaces page that a set $X$ an a collection $\tau$ of subsets of $X$ together denoted $(X, \tau)$ is called a topological space if:
$\emptyset \in \tau$ and $X \in \tau$, i.e., the empty set and the whole set are contained in $\tau$. If $U_i \in \tau$ for all $i \in I$ where $I$ is some index set then $\displaystyle{\bigcup_{i \in I} U_i \in \tau}$, i.e., for any arbitrary collection of subsets from $\tau$, their union is contained in $\tau$. If $U_1, U_2, ..., U_n \in \tau$ then $\displaystyle{\bigcap_{i=1}^{n} U_i \in \tau}$, i.e., for any finite collection of subsets from $\tau$, their intersection is contained in $\tau$.
On The Topology of Open Intervals on the Set of Real Numbers page we saw that if $\tau = \emptyset \cup \mathbb{R} \cup \{ (-n, n) : n \in \mathbb{Z}, n \geq 1 \}$ then $(X, \tau)$ is a topological space.
We will now look at two topologies known as the initial segment and final segment topologies.
The Initial Segment Topology
Definition: Consider the set of natural numbers $\mathbb{N}$. The collection of subsets $\tau = \{ \emptyset, \mathbb{N} \} \cup \{ \{1, 2, ..., n \} : n \in \mathbb{N} \}$ from $\mathbb{N}$ is called the Initial Segment Topology.
Let's verify that $(\mathbb{N}, \tau)$ is indeed a topology.
For the first condition, clearly $\emptyset, \mathbb{N} \in \tau$ by the definition of $\tau$.
For the second condition, notice that:(1)
Therefore any arbitrary union $\displaystyle{\bigcup_{i \in I} U_i}$ where $U_i \in \tau$ for all $i \in I$ will be the "largest" subset in the collection (or $\mathbb{N}$ itself) with respect to the nesting above, and hence is contained in $\tau$.
For the third condition, any finite intersection $\displaystyle{\bigcap_{i=1}^{n} U_i}$ where $U_i \in \tau$ for all $i \in \{1, 2, ..., n \}$ will be the "smallest" subset in the collection with respect to the nesting above, and hence is contained in $\tau$.
Therefore $(\mathbb{N}, \tau)$ is a topological space.
The Final Segment Topology
Definition: Consider the set of natural numbers $\mathbb{N}$. The collection of subsets $\tau = \{ \emptyset, \mathbb{N} \} \cup \{ \{n, n+1, ...\} : n \in \mathbb{N} \}$ from $\mathbb{N}$ is called the Final Segment Topology.
Let's verify that $\tau$ is indeed a topology.
For the first condition, we have that once again $\emptyset, \mathbb{N} \in \tau$ by the definition of $\tau$.
For the second condition, notice that:(2)
Therefore, any arbitrary union $\displaystyle{\bigcup_{i \in I} U_i}$ for $U_i \in \tau$ for all $i \in I$ is the "largest" subset in the union (or $\mathbb{N}$) with respect to the nesting above, so it is contained in $\tau$.
For the third condition, any finite intersection $\displaystyle{\bigcap_{i=1}^{n} U_i}$ for $U_i \in \tau$ for all $i \in \{1, 2, ... n \}$ is the "smallest" subset in the intersection with respect to the nesting above, so it is contained in $\tau$. |
I know what a Taylor series expansion is and I know how to find the Lagrange remainder but what does it mean intuitively? I need an explanation of what the Lagrange remainder represents in terms of the Taylor series expansion without proofs, strictly intuition.
Taking your comment about “any remainder” in mind, BCLC’s comment is effectively the answer you seem to be looking for, i.e. the remainder $R_n(x)$ is simply the difference between the $n^{th}$ order Taylor polynomial $T_n(x)$ and the function it is approximating. That is, given $T_n(x) = \sum_{i=0}^{n} \frac{f^{(i)}( c)}{i!}(x-c)^i$, the remainder is:
$$ R_n(x) = f(x) - T_n(x) $$
There are various forms for this remainder, one of which is the Lagrange remainder you mentioned. Wikipedia groups the various forms under “Taylor’s theorem,” and provides a more detailed discussion about the technical details.
Broadly speaking, for applications, having an expression for the remainder allows one to, e.g.:
Determine the error on a given approximation (i.e. for a given number of terms used in the approximation, determine the error); Determine the number of terms required to get an approximation that will be within some desired error (i.e. to obtain a value of the function accurate to a given number of decimal places, determine the number of terms $n$); Determine the largest interval over which the error in a given approximation will be below a desired level (i.e. fix the number of terms and the error, determine the interval).
One can also use the remainder to prove that the limit (i.e. as $n \to \infty$) of the sequence of partial sums of $T_n(x)$ converges to $f(x)$, by showing that $R_n(x) \to 0$, in the same limit, i.e.:
$$ \begin{aligned} f(x) & = T_n(x) + R_n(x) \\[5pt] % \Leftrightarrow \qquad \lim_{n \to \infty} f(x) & = \lim_{n \to \infty} T_n(x) + \lim_{n \to \infty} R_n(x) \\[5pt] % = \qquad f(x) & = \lim_{n \to \infty} T_n(x) + 0 \end{aligned} $$
Although, in practice, it is usually impractical to work with $R_n(x)$ directly (even using Taylor’s inequality) and other methods may be simpler (e.g. showing that the series satisfies the D.E. and initial condition which define the function; for example $\frac{df}{dx} = f$ and $f(0) = 1$ for $f(x) = e^x$, or $(1+x)f = rf$ and $f(0) = 1$ for $f(x) = (1+x)^r$, etc.).
Hope this gives you a non-technical sense for the remainder. |
Differential and Integral Equations Differential Integral Equations Volume 25, Number 1/2 (2012), 117-142. Some new well-posedness results for the Klein-Gordon-Schrödinger system Abstract
We consider the Cauchy problem for the 2D and 3D Klein-Gordon Schrödinger system. In 2D we show local well posedness for Schrödinger data in $H^s$ and wave data in $H^{\sigma} \times H^{\sigma -1}$ for $s=-1/4 \, +$ and $\sigma = -1/2$, whereas ill posedness holds for $s < - 1/4$ or $\sigma < -1/2$, and global well-posedness for $s\ge 0$ and $s-\frac{1}{2} \le \sigma < s+ \frac{3}{2}$. In 3D we show global well posedness for $s \ge 0$, $ s - \frac{1}{2} < \sigma \le s+1$. Fundamental for our results are the studies by Bejenaru, Herr, Holmer and Tataru [2], and Bejenaru and Herr [3] for the Zakharov system, and also the global well-posedness results for the Zakharov and Klein-Gordon-Schrödinger system by Colliander, Holmer and Tzirakis [5].
Article information Source Differential Integral Equations, Volume 25, Number 1/2 (2012), 117-142. Dates First available in Project Euclid: 20 December 2012 Permanent link to this document https://projecteuclid.org/euclid.die/1356012829 Mathematical Reviews number (MathSciNet) MR2906550 Zentralblatt MATH identifier 1249.35309 Citation
Pecher, Hartmut. Some new well-posedness results for the Klein-Gordon-Schrödinger system. Differential Integral Equations 25 (2012), no. 1/2, 117--142. https://projecteuclid.org/euclid.die/1356012829 |
Continuity of Complex Functions
Recall from the Limits of Complex Functions page that if $A \subseteq \mathbb{C}$, $z_0$ is an accumulation point of $\mathbb{C}$, and $f : A \to \mathbb{C}$ then the limit of $f$ as $z$ approaches $z_0$ equals $A$ denoted $\displaystyle{\lim_{z \to z_0} f(z) = A}$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $z \in A$ and $\mid z - z_0 \mid < \delta$ then $\mid f(z) - A \mid < \epsilon$.
With the definition of a limit of a complex function defined we can now also define the concept of continuity of a function $f$ at a point $z_0$.
Definition: Let $A \subseteq \mathbb{C}$ and let $f : A \to \mathbb{C}$. Then $f$ is said to be Continuous at $z_0 \in A$ if $\displaystyle{\lim_{z \to z_0} f(z) = f(z_0)}$. If $f$ is continuous for every point $z_0 \in A$ then $f$ is said to be Continuous on $A$. If $f$ is not continuous at $z_0$ then $f$ is said to be Discontinuous at $z_0$.
Many of the basic functions that we come across will be continuous functions. For example, all polynomials are continuous on $\mathbb{C}$, i.e., $f(z) = a_0 +a_1z^1 + a_2z^2 + ... + a_nz^n$ is continuous at every $z_0 \in \mathbb{C}$.
For a more complicated example, consider the following function:(1)
This is a rational function. Notice that the numerator of this function is simply a polynomial and is continuous at every $z_0 \in \mathbb{C}$. Problems arise when the denominator equals $0$ though for which $f$ becomes undefined. Notice that $1 + z^2 = 0$ if and only if $z = \pm i$. So $f$ is actually discontinuous at $z = \pm i$ since $f(\pm i)$ is not even defined!
We will now state some of the familiar properties regarding continuous functions.
Proposition 1: Let $f : A \to \mathbb{C}$. Then $f$ is continuous at $z_0$ if and only if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $f(B(z_0, \delta)) \subseteq B(f(z_0), \epsilon)$. Proof:We have that $f$ is continuous at $z_0$ if and only if $\displaystyle{\lim_{z \to z_0} f(z) = f(z_0)}$ if and only if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $|z - z_0| < \delta$ then $|f(z) - f(z_0)| < \epsilon$. Equivalently, $f$ is continuous at $z_0$ if and only if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $z \in B(z_0, \delta)$ then $f(z) \in B(f(z_0), \epsilon)$, that is:
Proposition 1 is a useful classification for continuous functions. It states that a function $f$ is continuous at $z_0$ provided that for any $\epsilon > 0$ we can find a $\delta > 0$ (possibly depending on $\epsilon$) for which whenever we are $\delta$ close to $z_0$ we can guarantee that $f$ is $\epsilon$ close to $f(z_0)$.
Theorem 2: Let $f$ and $g$ be continuous functions at $z_0 \in \mathbb{C}$. Then: a) $f + g$ is continuous at $z_0$. b) $f - g$ is continuous at $z_0$. c) $fg$ is continuous at $z_0$. d) $\displaystyle{\frac{f}{g}}$ is continuous at $z_0$ provided that $g(z_0) \neq 0$ The proofs of (a) - (d) are omitted but follow from the limit laws for complex functions. |
Eigenvalues and Eigenvectors Examples 5
Recall from the Eigenvalues and Eigenvectors page that the number $\lambda \in \mathbb{F}$ is said to be an eigenvalue of the linear operator $T \in \mathcal L (V)$ if $T(u) = \lambda u$ for some nonzero vector $u \in V$. The nonzero vectors $u$ such that $T(u) = \lambda u$ are called eigenvectors corresponding to the eigenvalue $\lambda$.
We will now look at some examples regarding eigenvalues of linear operators and eigenvectors corresponding to eigenvalues.
Example 1 Let $U$ and $W$ be nonzero subspaces of $V$ such that $V = U \oplus W$. Let $P$ be a linear operator on $V$ defined by $P(u + w) = u$ for all vectors $u \in U$ and for all vectors $w \in W$. Find all eigenvalues of $P$ and the corresponding eigenvectors, and verify that these eigenvectors are indeed associated with these eigenvalues. We will linear operators $P$ of this form later on.
Since $V = U \oplus W$ then for every vector $v \in V$ we have that $v = u + w$ where $u \in U$ and $w \in W$. We want to find numbers $\lambda \in \mathbb{F}$ such that:(1)
From the equation above, we have that:(2)
From the second equation we have that $\lambda = 0$ or $w = 0$.
If $\lambda = 0$, then we have that $u = 0$. Therefore $\lambda_1 = 0$ is an eigenvalue of $T$ and the corresponding set of eigenvectors is $\{ v = w : w \neq 0 \}$. To verify that this set of vectors are eigenvectors for $\lambda_1 = 0$, we note that if $v = w$ where $w \in W$, then we have that:(3)
Now suppose that instead $w = 0$. Then our system of equations reduces down to:(4)
We note that $u \neq 0$, so therefore $\lambda = 1$, and so $\lambda_2 = 1$ is an eigenvalue of $T$ and the corresponding set of eigenvectors is $\{ v = u: \in U : u \neq 0 \}$. To verify that this set of vectors are eigenvectors for $\lambda_2 = 1$, we note that if $v = u$ where $u \in U$, then we have that:(5)
Example 2 Prove that the right shift operator $T \in \mathcal L (\mathbb{F}^{\infty})$ defined by $T(x_1, x_2, ...) = (0, x_1, x_2, ... )$ has no eigenvalues.
Let $u = (x_1, x_2, ...) \in \mathbb{F}^{\infty}$. We want to find $\lambda \in \mathbb{F}$ such that:(6)
From the equation above, we have that:(7)
The first equation implies that $\lambda = 0$ or $x_1 = 0$. If $\lambda = 0$, then $x_1 = x_2 = ... = 0$, which is not a nonzero vector in $\mathbb{F}^{\infty}$.
If instead $x_1 = 0$, then this implies that $0 = \lambda x_2$. Therefore $\lambda = 0$ (which we've already seen cannot happen) or $x_2 = 0$. So if $x_2 = 0$, then once again, we have that $0 = \lambda x_3$, etc…, and we get that $x_1 = x_2 = x_3 = ... = 0$ which is not a nonzero vector in $\mathbb{F}^{\infty}$.
Therefore $T$ has no eigenvalues.
Example 3 Find the eigenvalues and corresponding eigenvectors of the left shift operator $T \in \mathcal L (\mathbb{F}^{\infty})$ defined by $T(x_1, x_2, ...) = (x_2, x_3, ... )$.
Let $u = (x_1, x_2, ...) \in \mathbb{F}^{\infty}$. We want to find $\lambda \in \mathbb{F}$ such that:(8)
The equation above gives us the following system of equations:(9)
Note that we can choose $x_1$ however we'd like. We get that:(10)
Therefore we have that each $\lambda \in \mathbb{F}$ is an eigenvalue of the left shift operator. The set of corresponding eigenvectors is for each $\lambda$ is $\{ (x_1, \lambda x_1, \lambda^2 x_1, ...) : x_1 \in \mathbb{F} \}$. |
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