Split (1)

train · ~4.74M rows (showing the first 1.13M)

text stringlengths 256 16.4k
Inner Limit in Normed Spaces by Open Balls Theorem Then the inner limit of $\left \langle{C_n}\right \rangle_{n \in \N}$ is: $\displaystyle \liminf_n \ C_n = \left\{{x: \forall \epsilon > 0: \exists N \in \mathcal N_\infty: \forall n \in N: x \in C_n + \epsilon B}\right\}$ where $B$ denotes the open unit ball of the space. Proof The proof is an immediate result of Inner Limit in Hausdorff Space by Open Neighborhoods since the arbitrary open sets can be here replaced by open balls. $\blacksquare$
HP 17bII+ Silver solver 09-14-2018, 09:20 PM Post: #1 HP 17bII+ Silver solver I'm very satisfied with my HP 17bII+ Silver, which I find very powerful, but also nice and not looking too complicated (no [f] and [g] functions on keys like the HP 12c or HP 35s that I used to use at work every day, but a really complete calculator except for trigs and complex numbers calculations). So the 17BII+ is my new everyday calculator. I don't come back to arguments where a good (HP) calculator is a perfect complement or subsitute to Excel. I chose the 17BII+ after having carefully studied the programs I use in my day to day work: - price and costs calculations: can be modeled in the solver, with 4 or 5 equations and share vars - margin calculations: built-in functions - time value of money: built-in functions - time functions: built-in functions For the few moments I need trigs or complex calculations, I always have Free42 or a real 35s / 15c not really far from me. After having rebuilt my work environment in the (so powerful) solver, I started to study the behavior of the solver, looking at loops (Σ function), conditional branchings (IF function), menu selection (S function), and the Get and Let functions (G(), L()). I googled a lot and found an interesting pdf file about the Solvers of the 19B, 17B, 17BII and - according to the author, the 17BII+ Silver. The file is here : http://www.mh-aerotools.de/hp/documents/...ET-LET.pdf I tried a few equations in the "Using New and Old Values" chapter, pages 4 and following. There I found lots of differences with my actual 17BII+ Silver, which I would like to share here with you. For instance, the following equation found page 4: Code: In the next example found page 5: Code: Then the equation : Code: I don't understand the first 2 cases. In the first one, A should be set to -B, not B, if not using the "old" value of A. In the second one, the solver does not use the "old" value, but it does not also solve the equation, as there is no defined solution. In the last case, I understand that the equation is evaluated twice before finding an answer. So the old value is used there, but not the way I could expect. I finally found one - and only one - way to use iterations in the solver, with the equation : Code: Note that neither A=G(A)+1 or A=1+G(A) or G(A)+2=A works. I'm not disappointed, as the solver is a really interesting and useful feature of the calculator, but I'm just surprised not having found more working cases of iterations, or a clear understanding of how the solver works. Comments are welcomed. Regards, Thibault 09-14-2018, 10:14 PM (This post was last modified: 09-14-2018 10:25 PM by rprosperi.) Post: #2 RE: HP 17bII+ Silver solver The 17BII+ (Silver Edition) is an excellent machine, in fact it has the best keyboard of any machine made today by HP, but the solver does have a bug, and there are also a few other smaller issues making the solver slightly inferior to the 17B/17BII/19B/19BII/27S version. See these 3 articles for details: http://www.hpmuseum.org/cgi-sys/cgiwrap/...ead=242551 http://www.hpmuseum.org/forum/thread-657...l#pid58685 http://www.hpmuseum.org/cgi-sys/cgiwrap/...ead=134189 Overall these are not dramatic issues, and once you understand the solver bug, you likely can create equations that can avoid the issue. I have most HP machines but the 17BII and 17BII+ are the ones I use most often for real work (vs. playing, exploring or following along interesting threads here). Edit: added 3rd link --Bob Prosperi 09-15-2018, 12:05 AM Post: #3 RE: HP 17bII+ Silver solver (09-14-2018 09:20 PM)pinkman Wrote: I'm not disappointed, as the solver is a really interesting and useful feature of the calculator, but I'm just surprised not having found more working cases of iterations, or a clear understanding of how the solver works. Thibault, when Kinpo built the 17bii+ calculator years ago (both the gold one and the silver one), they basically goofed the solver implementation. This has been discussed at length in the HPMuseum forum. The solvers on the 17b and 17bii work fine, just as you would expect them to. If you plan on making significant use of the solver and its incredible capabilities, forget the + and get an original 17b or 17bii. You won't be sorry. Also, get the manual (it's on the Museum DVD) Technical Applications for the HP-27s and HP-19b. It applies to the 17b as well. The Sigma function also works fine on the 17b and 17bii. 09-15-2018, 04:36 AM Post: #4 RE: HP 17bII+ Silver solver Thanks to both of you for the details, links and advice. I've read the threads carefully, I did not find them by myself first. It's the end of the night now, I'll try to make few testing later. 09-15-2018, 11:55 PM (This post was last modified: 09-15-2018 11:58 PM by rprosperi.) Post: #5 RE: HP 17bII+ Silver solver If you want to really explore the capabilities of the awesome Pioneer Solver, and confidently try to push it without worrying about using L() this way or that, I agree with Don, buy a 17BII and use that for the Solver stuff, but continue to use the 17BII+ for every day stuff. Here's a very nice 17BII for only $25 (shipping included, in US) and you can even find them cheaper if you're willing to wait: https://www.ebay.com/itm/HP-17Bll-Financ...3252533550 The 17BII is bug-free for solver use, while the 17BII+ has a much better LCD, readable in a wider range of lighting & use conditions. In case you haven't seen this yet, here's an example of what can be done with the solver: http://www.hpmuseum.org/forum/thread-2630.html --Bob Prosperi 09-16-2018, 09:59 PM Post: #6 RE: HP 17bII+ Silver solver Well I'll try to find one, even if I'm not in the US. I also want to continue using my actual 17bII+ at work, as the solver is powerful enough for me (for the moment), and it looks really good. Don and you Bob have done a lot to help understand what the solver can do, that's pretty good stuff. Regards, Thibault 09-16-2018, 10:23 PM Post: #7 RE: HP 17bII+ Silver solver Nah, Don and Gerson are the real Pioneer Solver Masters, I just have a good collection of links. Of all the various tools built-in to the various calculator models I've explored, the Pioneer Solver is easily the one that most exceeds it's initial apparent capability. This awesome tool must have been incredibly well-tested by the QA team. The fact that the sheer size (and audacity!) of Gerson's Trig formulas still return amazingly accurate results says more about the underlying design and code than any comments I could add. Enjoy exploring it, and when you've mastered it (or at least tamed it a bit), come back here and share some interesting Solver formulas. There are numerous folks here that enjoy entering the formulas and running some test cases. (well, I suppose "enjoy" is not really the right word about entering the formulas - I guess feel good about accomplishing it successfully is more accurate). When I see some of these long Solver equations, it reminds of TECO commands back in the PDP-11 days. --Bob Prosperi 09-16-2018, 11:05 PM Post: #8 RE: HP 17bII+ Silver solver If you feel like entering long formulas you can solve the 8-queens problem. ):0) Cheers Thomas 09-17-2018, 06:30 AM (This post was last modified: 09-18-2018 09:47 AM by Don Shepherd.) Post: #9 RE: HP 17bII+ Silver solver How about a nifty, elegant, simple number base conversion Solver equation for the 17b/17bii, courtesy Thomas Klemm: BC:ANS=N+(FROM-TO)$\times \Sigma$(I:0:LOG(N)$\div$LOG(TO):1:L(N:IDIV(N:TO))$\times$FROM^I) Note: either FROM or TO must be 10 unless you are doing HEX conversions 09-17-2018, 07:49 AM Post: #10 RE: HP 17bII+ Silver solver 09-17-2018, 09:17 AM (This post was last modified: 09-17-2018 10:28 AM by Don Shepherd.) Post: #11 RE: HP 17bII+ Silver solver (09-17-2018 07:49 AM)Thomas Klemm Wrote: Wow, I didn't realize that Thomas. Stunning. Don 09-17-2018, 10:24 AM Post: #12 RE: HP 17bII+ Silver solver As has already been written, unfortunately the solver in this calculator is flawed. Part of the problems comes from the fact that it evaluates the equation twice which leads to incrementing by two instead of one etc. But there seem to be more quirks. This is too bad, as the solver is a very capable tool (with G() and L()), while still easy and versatile to use (with the menu buttons). Accomplishing something similar (solve for one variable today, for another tomorrow) with modern tools like Excel is more complicated. Martin 09-17-2018, 12:48 PM Post: #13 RE: HP 17bII+ Silver solver (09-16-2018 10:23 PM)rprosperi Wrote: Nah, Don and Gerson are the real Pioneer Solver Masters, I just have a good collection of links. An obvious correction is in order here: Don, Gerson and Thomas Klemm are the real Pioneer Solver Masters... No disrespect intended by the omission. A review of past posts of significant solver formulas quickly reveals just how often all three of these three guys were the authors. --Bob Prosperi 09-28-2018, 01:33 PM Post: #14 RE: HP 17bII+ Silver solver Be sure I have great respect of everyone mentioned above Thanks for the advices, my new old 17bii is now in my hands ($29 on French TAS), I'm just waiting for the next rainy Sunday for pushing the solver to it's limits (mmh, to MY limits I guess). 09-28-2018, 07:47 PM (This post was last modified: 09-28-2018 07:48 PM by Jlouis.) Post: #15 RE: HP 17bII+ Silver solver Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S? TIA 09-29-2018, 01:37 AM Post: #16 RE: HP 17bII+ Silver solver (09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S? Check out the document by Martin Hepperle in this thread which lists the functions available for each calculator. The document is an excellent introduction to the Solver application. Solver GET-LET ~Mark Who decides? 09-29-2018, 02:25 AM Post: #17 RE: HP 17bII+ Silver solver (09-29-2018 01:37 AM)mfleming Wrote:(09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S? Thanks mfleming, I should have read the beginning of the manual. It is the same solver for the calculators of my question. I prefer to use the 19bII due the dedicated alpha keyboard, but the 27s is a pleasure to use too. Cheers 09-29-2018, 11:13 AM Post: #18 RE: HP 17bII+ Silver solver 09-29-2018, 11:53 AM (This post was last modified: 09-29-2018 05:12 PM by Don Shepherd.) Post: #19 RE: HP 17bII+ Silver solver (09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?Here is a 3-page cross-reference I created a few years ago of Solver functions present in the 19bii, 27s, and 17bii. solver 1.PDF (Size: 1.08 MB / Downloads: 33) solver 2.PDF (Size: 968.77 KB / Downloads: 21) solver 3.PDF (Size: 494.75 KB / Downloads: 19) 09-29-2018, 01:36 PM (This post was last modified: 09-29-2018 01:49 PM by Jlouis.) Post: #20 RE: HP 17bII+ Silver solver (09-29-2018 11:53 AM)Don Shepherd Wrote: Really thanks Dom for taking time to make these documents. This is what makes MoHC a fantastic community. Cheers JL Edited: Looks like the 19BII is a little more complete than the 27s and that the 17BII is the weaker one. User(s) browsing this thread: 1 Guest(s)
By definition supersymmetry transformations square to spacetime translations. In a superspace formalism the supersymmetry operator is constructed from the vector field $\partial_\theta$ with respect to the odd coordinates $\theta$. As this operator has to square to the vector field $\partial_x$ with respect to the even coordinates $x$, which is of dimension $1$, the vector field with respect to the odd coordinate has to be of dimension $1/2$ and so the odd coordinate as to be of dimension $-1/2$. Equivalently, a typical superfield is of the form $\phi + \theta \psi +...$ where $\phi$ is a scalar and $\psi$ a spinor. In $d$ spacetime dimensions, a scalar is of dimension $(d-2)/2$, a spinor is of dimension $(d-1)/2$ and so $\theta$ has to be of dimension $-1/2$.
The vector projection is of two types: Scalar projection that tells about the magnitude of vector projection and the other is the Vector projection which says about itself and represents the unit vector. If the vector veca is projected on vecb then Vector Projection formula is given below: \[\large proj_{b}\,a=\frac{\vec{a}\cdot\vec{b}}{\left\|\vec{b}\right\|^{2}}\;\vec{b}\] The Scalar projection formula defines the length of given vector projection and is given below: \[\large proj_{b}\,a=\frac{\vec{a}\cdot\vec{b}}{\left\|\vec{a}\right\|}\] Vector Projection Problems Below are problems based on vector projection which may be helpful for you. Question 1: Find the vector projection of $5\,\vec{i}-4\,\vec{j}+\vec{k}$ along the vector $3\,\vec{i}-2\,\vec{j}+4\,\vec{k}$ ? Solution: Given: $\vec{a}=5\,\vec{i}-4\,\vec{j}+\vec{k}$ $\vec{b}=3\,\vec{i}-2\,\vec{j}+4\,\vec{k}$ So, we have: $\vec{a}\cdot\vec{b}=\left(a_{1}\,a_{2}+b_{1}\,b_{2}+c_{1}\,c_{2}\right)$ = (5(3) + (-4)(-2) + (1)(4)) = 15 + 8 + 4 = 27 $\left\|\vec{b}\right\|=\sqrt{3^{2}+\left(-2\right)^{2}+\left(4\right)^{2}}$ $=\sqrt{9+4+16}$ $=\sqrt{29}\;\left\|\vec{b}\right\|^{2}$ = 29. The Vector projection is given by $proj_{b}\,a= \frac{\vec{a}\cdot\vec{b}}{\left\|\vec{b}\right\|^{2}}\vec{b}$ $=\frac{9}{29}\left(3,-2,4\right)$ $=\frac{1}{29}\left(27,-18,36\right)$ The Projection of a on b are: $\left(\frac{27}{29},\frac{-18}{29},\frac{36}{29}\right)$
The expression for the acceleration of a near-earth satellite as presented in the IERS Technical note is given by \begin{equation} \label{eq:problemeq} \tag{1} \frac{d^2\mathbf{r}}{dt^2} = \frac{GM_E}{c^2r^3} \left\{\left[2(\beta+\gamma)\frac{GM_E}{r} - \gamma \dot{\mathbf{r}} \cdot \dot{\mathbf{r}} \right] \mathbf{r} + 2(1+\gamma)(\mathbf{r}\cdot\dot{\mathbf{r}})\dot{\mathbf{r}} \right\}. \end{equation} We are working in the Parametrised Post-Newtonian formalism, hence the dimensionless constants $\beta,\gamma$. Now, I know what the Schwarzschild metric looks like in isotropic coordinates but I can't see where Eq. (\ref{eq:problemeq}) comes from. Another point to make is that in GR the dimensionless parameters $\beta,\gamma$ are equal to unity and when substituted above gives a well known formula for Schwarzschild precession which is given by \begin{equation} \label{eq:problemeq2} \tag{1} \frac{d^2\mathbf{r}}{dt^2} = \frac{GM_E}{c^2r^3} \left\{\left[4\frac{GM_E}{r} - \dot{\mathbf{r}} \cdot \dot{\mathbf{r}} \right] \mathbf{r} + 4(\mathbf{r}\cdot\dot{\mathbf{r}})\dot{\mathbf{r}} \right\}. \end{equation} Again, I don't recall this either. Any suggestions?
The Grassmann analytic continuation principle says that a real function $f(x,\xi)$ that depends on $n$ real variables $x=(x_1,\ldots,x_n)$ and $n$ nilpotent Grassmann numbers $\xi=(\xi_1,\ldots,\xi_n)$ has the following Taylor expansion containing only a finit ($2n?$) number of terms $$ f(x+\xi) = \sum_{\alpha}\frac{\xi^{\alpha}}{\alpha!}\frac{\partial^{\alpha}}{\partial x^{\alpha}}f(x) $$ where the $\xi_j \in \mathcal{G}_n$ which is the fermionic part of the supermanifold at hand are given by $$ \xi_j =\sum_{\|\epsilon\|>0}a_{\epsilon}^j\theta^{\epsilon}$$ The $\theta_i$ are the generators of $\mathcal{G}_n$, $\alpha=(\alpha_1,\ldots,\alpha_n)\in\mathbb{N}^n$, and $\xi^{\alpha}=\xi_1^{\alpha_1}\ldots\xi_n^{\alpha_n}$. How is this related to expanding a superfield in Grassmann coordinates? For example, a a scalar superfiled $\mathcal{F}(x,\theta_1,\theta_2)$ that depends on the ordinary spacetime coordinates as well as on two Grassmann coordinates can be expanded as $$\mathcal{F}(x,\theta_1,\theta_2) = A(x)+B(x)\theta_1+C(x)\theta_2+D(x)\theta_1\theta_2$$ Comparing this expansion to the Grassmann analytic continuation principle above I would think that in this specific case we have $n=2$, $\alpha=(\alpha_1,\alpha_2)$ and $\xi=\xi_1^{\alpha_1}\xi_2^{\alpha_2}$ as there are two Grassmann coordinates in addition to conventional spacetiem. The first thing I dont understand is why in the Grassmann analytic continuation principle the function is not expanded directly in the $\theta_i$ but the $\xi$ are used instead which makes it hard for me to see what is going on. Otherwise, I would have guessed that from setting $\alpha_1=0$ and $\alpha_2=0$ the coefficient $A(x)$ would correspond to $f(x)$, from setting $\alpha = (2,1)$ one obtains $B(x)=\frac{\partial f}{\partial\theta_1}$, from setting $\alpha = (1,2)$ $C(x)=\frac{\partial f}{\partial\theta_2}$ and from $\alpha = (1,1)$ one has $D(x)=\frac{\partial^2 f}{\partial\theta_1\partial\theta_2}$
ergm.ego Last updated: April 05, 2016 This tutorial is a joint product of the Statnet Development Team: Mark S. Handcock (University of California, Los Angeles) Carter T. Butts (University of California, Irvine) David R. Hunter (Penn State University) Steven M. Goodreau (University of Washington) Skye Bender de-Moll (Oakland) Pavel N. Krivitsky (University of Wollongong) Martina Morris (University of Washington) Open an R session, and set your working directory to the location where you would like to save this work. To install all of the CRAN packages in the statnet suite: install.packages('statnet')library(statnet) To install the ergm.ego, install.packages('ergm.ego') The ergm.ego package is designed to provide principled estimation of and statistical inference for Exponential-family Random Graph Models (“ERGMs”) from egocentrically sampled network data. In many empirical contexts, it is not feasible to collect a network census or even an adaptive (link-traced) sample. Even when one of these may be possible in practice, egocentrically sampled data are typically cheaper and easier to collect. Long regarded as the poor country cousin in the network data family, egocentric data contain a remarkable amount of information. With the right statistical methods, such data can be used to explore the properties of the complete networks in which they are embedded. The basic idea here is to combine what is observed, with assumptions, to define a class of models that represent a distribution of networks that are centered on the observed properties. The variation in these networks quantifies some of the uncertainty introduced by the assumptions. The package comprises: ergm package, but include the specific modifications needed in the egocentric data context. The package is designed to work with the other statnet packages. So, for example, once you have fit a model, you can use the summary and diagnostic functions from ergm, simulate to simulate complete network realizations from the model, the network descriptives from sna to explore the properities of the network, and you can use other R functions and packages as well after converting the network data structure into a data frame. Putting this all together, you can start with egocentric data, estimate a model, test the coefficients for statistical significance, assess the model goodness of fit, and simulate complete networks of any size from the model. The statistics in your simulated networks will be consistent with the appropriately scaled statistics from your sample for all of the terms that are represented in the model. The full technical details on ERGM estimation and inference from egocentrically sampled data are in a paper that is currently under review. The working paper can be found here. This tutorial provides a brief introduction to the key concepts. ERGMs represent a general class of models based in exponential-family theory for specifying the probability distribution for a set of random graphs or networks. Within this framework, one can—among other tasks—obtain maximum-likehood estimates for the parameters of a specified model for a given data set; test individual models for goodness-of-fit, perform various types of model comparison; and simulate additional networks with the underlying probability distribution implied by that model. The general form for an ERGM can be written as: \[ P(Y=y;\theta,x)=\frac{\exp(\theta^{\top}g(y,x))}{\kappa(\theta,x)}\qquad (1) \] where $Y$ is the random variable for the state of the network (with realization y), $g(y,x)$ is a vector of model statistics for network y, $\theta$ is the vector of coefficients for those statistics, and $\kappa(\theta)$ represents the quantity in the numerator summed over all possible networks (typically constrained to be all networks with the same node set as $y$). The model terms $g(y,x)$ are functions of network statistics that we hypothesize may be more or less common than what would be expected in a simple random graph (where all ties have the same probability). When working with egocentrically sampled network data, these statistics must be observed in the sample more details in section 4.2 A key distinction in model terms is dyad independence or dyad dependence. Dyad independent terms (like nodal homophily terms) imply no dependence between dyads—the presence or absence of a tie may depend on nodal attributes, but not on the state of other ties. Dyad dependent terms (like degree terms, or triad terms), by contrast, imply dependence between dyads. The design of an egocentric sample means that most observable statistics are dyad independent, but there are a few, like degree, that are dyad dependent. Network data are distinguished by having two units of analysis: the actors and the links between the actors. This gives rise to a range of sampling designs that can be classified into two groups: link tracing designs (e.g., snowball and respondent driven sampling) and egocentric designs. Egocentric network sampling comprises a range of designs developed specifically for the collection of network data in social science survey research. The design is (ideally) based on a probability sample of respondents (“egos”“) who, via interview, are asked to nominate a list of persons (”alters“) with whom they have a specific type of relationship (”tie“), and then asked to provide information on the characteristics of the alters and/or the ties. The alters are not recruited or directly observed. Depending on the study design, alters may or may not be uniquely identifiable, and respondents may or may not be asked to provide information on one or more ties among alters (the”alter" matrices). Alters could, in theory, also be present in the data as an ego or as an alter of a different ego; the likelihood of this depends on the sampling fraction. Egocentric designs sample egos using standard sampling methods, and the sampling of links is implemented through the survey instrument. As a result, these methods are easily integrated into population-based surveys, and, as we show below, inherit many of the inferential benefits. For the moment ergm.ego uses the minimal egocentric network study design, in which alters cannot be uniquely identified and alter matrices are not collected The minimal design is more common, and the data are more widely available, largely because it is less invasive and less time-consuming than designs which include identifiable alter matrices. However, deveopment of estimation where alter–alter matrices are available is being planned. Handcock and Gile (2010): Likelihood inference for partially observed networks, has egocentric data as a special case. Kosikinen and Robins (2010): Bayesian inference for partially observed networks, has egocentric data as a special case. Krivitsky and Morris (2015) Use design-based estimators for sufficient statistics of the ERGM of interest and then transfer their properties to the ERGM estimate. EgoStats”: currenly does not support alter–alter statistics or directed or bipartite networks. In particular, Also, let the $k$th attribute/covariate observed on ego $i$ and its alters as $e^e_{i,k}\equiv x_{i,k}$ and $e^a_{i,k}\equiv( x_{j,k})_{j\in y_i}$. Then, Egocentric ERGMs are specified the same way as plain ergm: via terms (e.g. nodematch) used to represent predictors on the right-hand size of equations used in: summary (to obtain measurements of network statistics on a dataset) ergm.ego (to estimate an ERGM) simulate (to simulate networks from an ERGM fit) The terms that can be used in an ERGM depend on the type of network being analyzed (directed or undirected, one-mode or two-mode (“bipartite”), binary or valued edges) and on the statistics that can be observed in the sample. Even if the whole population is egocentrically observed (i.e., $S=N$, a census), the alters are still not uniquely identifiable. This limits the kinds of network statistics that can be observed, and the ERGM terms that can be fit to such data. We turn to the notion of sufficiency to identify those that can be. We call a network statistic $g_{k}(\cdot,\cdot)$ egocentric if it can be expressed as \[g_{k}(y,x)\equiv \textstyle\sum_{i\in N} h_{k}(e_i) \] for some function $h_{k}(\cdot)$ of egocentric information associated with a single actor. The space of egocentric statistics includes dyadic-independent statistics that can be expressed in the general form of \[g_{k}(y,x)=\sum_{ij\in y} f_k(x_i,x_j) \] for some symmetric function $f_k(\cdot,\cdot)$ of two actors’ attributes; and some dyadic-dependent statistics that can be expressed as \[g_{k}(y,x)=\sum_{i\in N} f_k ({x_{i},(x_j)_{j\in y_i}})\] for some function $f_k(\cdot,\dotsb)$ of the attributes of an actor and their network neighbors. What is “egocentric” depends on available data.
Following up on his previous post, “Is Math a Gift From God?” — calculus students say, “No!” — Jason Rosenhouse has a new essay for your delectation, “Is God Like an Imaginary Number?” Again, the short answer is, “Nope.” The longer answer will take us into the history of mathematics, the role of mysticism in theology and the relationship between science and verbal description. Rosenhouse sets himself the task of fisking an essay in the religious periodical First Things, by a “Junior Fellow” of that publication named Amanda Shaw. Shaw’s thesis is that the notion of God is akin to that of an imaginary number, and moreover that the same closed-minded orthodoxy which rejected the latter from mathematics for oh so many years is unjustly keeping the former out of science. I find this stance to be, in a word, ironical: if you’re looking for dogmatism and condemnations of the heterodox, your search will be much more rewarding if you look among the people who reject scientific discoveries because they are inconsistent with a Bronze Age folk tale than if you search through science itself! Still, it’s a fun chance to talk about history and mathematics. PART A: COMPLEX NUMBERS As I described earlier, “imaginary” and “complex” numbers arise naturally when you think about the ordinary, humdrum “real numbers” — you know, fractions, decimals and all those guys — as lengths on a number line. In this picture, adding two numbers corresponds to sticking line segments end-to-end, multiplication means stretching or squishing (in general, scaling) line segments, and negation means flipping a segment over to lie on the opposite side of zero. Complex numbers appear when you ask the question, “What operation, when performed twice in succession upon a line segment, is equivalent to a negation?” Answer: rotating by a quarter-turn! Historically, mathematicians started getting into complex numbers when they tried to find better and better ways to solve real-number equations. Girolamo Cardano (1501–1576), also known as Jerome Cardan, posed the following problem: If some one says to you, divide 10 into two parts, one of which multiplied into the other shall produce […] 40, it is evident that this case or question is impossible. Nevertheless, we shall solve it in this fashion. Writing this in more modern algebraic notation, this is like saying [tex] x + y = 10 [/tex] and [tex] xy = 40 [/tex], which we can combine into one equation by solving for [tex] y [/tex], thusly: [tex] xy = x(10 – x) = 40.[/tex] In turn, shuffling the symbols around gives [tex] x^2 – 10x + 40 = 0,[/tex] which plugging into ye old quadratic formula yields [tex] x = \frac{10 \pm \sqrt{100 – 160}}{2}, [/tex] or, boiling it down, [tex] x = 5 \pm \sqrt{-15}. [/tex] Totally loony! Taking the square root of a negative number? Forsooth, thy brains are bubbled! Oh, wait, didn’t we just realize that we could maybe handle the square root of a negative number by moving into a two-dimensional plane of numbers? Yes, we did: that’s the prize our talk of flips and rotations won us! Continue reading Rosenhouse on Amanda Shaw
Let $X=[0,+ \infty[ $ and $d(x,y)=\|\frac{1}{1+x^2}- \frac{1}{1+y^2}\| $ 1) Show that $(X,d)$ and $(]0,1], d_{2})$ are homeomorphic (where $d_{2}=\|x-y\|$) 2) Is the space $(X,d)$ connected? compact? complete ? To do this it means we need to find any function such as: $f\colon (\mathbb{R}^+, d)\to (]0,1], d_{2})$ is bijective continuous and whose bijection is also continuous. Does it mean I can take $f(x)=\sin(x) $ which is bijective continuous and of continuous bijection? Is there a way to find the "neatest" homeomorphism?
I have a one-semester background in string theory (bosonic string theory, the NSR string, conformal field theory), but I have not taken any full length courses on supersymmetry and supergravity. I'm presently taking a second string theory course, but I'm forced to build the background I lack myself, in an attempt to understand the new material. So, here are some questions. My hope is that instead of just receiving answers, I will get some suggestions about specific books or reviews I could refer to, without getting carried away, to build some background for these topics. In D = 11 supergravity, there's a Majorana gravitino with 44 independent components, and the gamma-tracelessness condition gives $\frac{1}{2}(D-3)2^{[D/2]} = 128$ components. Here it is said that the factor of half accounts for the fact that "spinors have half as many physical states as components". What does this statement in quotes mean? The three form (totally antisymmetric) gauge field in D = 11 supergravity has a dimension of 9/2. How does dimensional analysis work here? Why does the gravitino field have a dimension of 5? What is a super-covariant spin connection? I read that it equals the usual spin connection with sermonic terms of the form $\psi\Gamma\psi$. Why is this called "super-covariant"? In order to get solutions like the M2 brane, the M5 brane, and the KK1 and KK6 solutions, one sets the fermions to zero (for bosonic backgrounds) and also imposes the BPS condition $\delta \psi = 0$. This condition is linear in derivatives. Is that sufficient to say that it is BPS? Why would quadratic derivatives result in non-BPS solutions? Finally, is there a fairly quick resource to learn everything that is required in the vielbein formalism of general relativity, to work with string theory/supergravity?This post imported from StackExchange Physics at 2015-02-13 11:43 (UTC), posted by SE-user leastaction
Search Now showing items 1-10 of 24 Production of Σ(1385)± and Ξ(1530)0 in proton–proton collisions at √s = 7 TeV (Springer, 2015-01-10) The production of the strange and double-strange baryon resonances ((1385)±, Ξ(1530)0) has been measured at mid-rapidity (\|y\|< 0.5) in proton–proton collisions at √s = 7 TeV with the ALICE detector at the LHC. Transverse ... Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV (Springer, 2015-05-20) The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ... Inclusive photon production at forward rapidities in proton-proton collisions at $\sqrt{s}$ = 0.9, 2.76 and 7 TeV (Springer Berlin Heidelberg, 2015-04-09) The multiplicity and pseudorapidity distributions of inclusive photons have been measured at forward rapidities ($2.3 < \eta < 3.9$) in proton-proton collisions at three center-of-mass energies, $\sqrt{s}=0.9$, 2.76 and 7 ... Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV (Springer, 2015-06) We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ... Measurement of pion, kaon and proton production in proton–proton collisions at √s = 7 TeV (Springer, 2015-05-27) The measurement of primary π±, K±, p and p¯¯¯ production at mid-rapidity (\|y\|< 0.5) in proton–proton collisions at s√ = 7 TeV performed with a large ion collider experiment at the large hadron collider (LHC) is reported. ... Two-pion femtoscopy in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV (American Physical Society, 2015-03) We report the results of the femtoscopic analysis of pairs of identical pions measured in p-Pb collisions at $\sqrt{s_{\mathrm{NN}}}=5.02$ TeV. Femtoscopic radii are determined as a function of event multiplicity and pair ... Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV (Springer, 2015-09) Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ... Charged jet cross sections and properties in proton-proton collisions at $\sqrt{s}=7$ TeV (American Physical Society, 2015-06) The differential charged jet cross sections, jet fragmentation distributions, and jet shapes are measured in minimum bias proton-proton collisions at centre-of-mass energy $\sqrt{s}=7$ TeV using the ALICE detector at the ... Centrality dependence of high-$p_{\rm T}$ D meson suppression in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (Springer, 2015-11) The nuclear modification factor, $R_{\rm AA}$, of the prompt charmed mesons ${\rm D^0}$, ${\rm D^+}$ and ${\rm D^{+}}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at a centre-of-mass ... K(892)$^0$ and $\Phi$(1020) production in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (American Physical Society, 2015-02) The yields of the K*(892)$^0$ and $\Phi$(1020) resonances are measured in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV through their hadronic decays using the ALICE detector. The measurements are performed in multiple ...
Show that the first excited state of 8-Be nucleus fits the experimental value of the rotational band $E(2^+) = 92 keV$ (this is the first excited state, which is 92 keV above the ground state). To do so model 8-Be to be made of 2 alpha particles $4.5 fm$ apart rotating about their center of mass. The method I used is: 1) Get the moment of inertia of the system. For two particles rotating each other (in the same plane) about its center of mass, one gets: $$I = m_1 r_1^2 + m_2 r_2^2 = 2m_{\alpha} (d/2)^2 = 6.728 \times 10^{-56}kgm^2$$ where: $$m_{\alpha} = 6.645 \times 10^{-27} kg$$ $$d = 4.5 \times 10^{-15}m$$ I've checked this value and it's correct. 2) Verify the experimental value for the rotational band by applying the rotational energy formula: $$E = J (J + 1) \frac{\hbar^2}{2I}$$ OK let's first get what's called "characteristic rotational energy": $\frac{\hbar^2}{2I}$ $$\frac{\hbar^2}{2I} = 8.256 \times 10^{-14} J= 0.516 MeV$$ where: $$\hbar = 1.054 \times 10^{-34}Js$$ $$1 eV = 1.6 \times 10^{-19}J$$ $2^+$ band has $J=2$ associated with it. So it's just a matter of plugging numbers in: $$E = J (J + 1) \frac{\hbar^2}{2I} = 3096 keV$$ Which is way off $92keV$. Where is my method gone wrong? Maybe the original question provided an incorrect experimental result. PS/ For anyone interested in rotational energy related to nucleus this is a good video to check: https://www.youtube.com/watch?v=rwdBnwznt3s
Computing R.M.S. Error RMS error measures the differences between values predicted by a model or an estimator and the values actually observed. Learning Objectives Define and compute root-mean-square error. Key Takeaways Key Points These individual differences are called residuals when the calculations are performed over the data sample that was used for estimation, and are called prediction errors when computed out-of-sample. The differences between values occur because of randomness or because the estimator doesn’t account for information that could produce a more accurate estimate. RMS error serves to aggregate the magnitudes of the errors in predictions for various times into a single measure of predictive power. In terms of a regression line, the error for the differing values is simply the distance of a point above or below the line. In general, about 68% of points on a scatter diagram are within one RMS error of the regression line, and about 95% are within two. Key Terms root-mean-square error: (RMS error) A frequently used measure of the differences between values predicted by a model or an estimator and the values actually observed. Root- mean -square (RMS) error, also known as RMS deviation, is a frequently used measure of the differences between values predicted by a model or an estimator and the values actually observed. These individual differences are called residuals when the calculations are performed over the data sample that was used for estimation, and are called prediction errors when computed out-of-sample. The differences between values occur because of randomness or because the estimator doesn’t account for information that could produce a more accurate estimate. Root-mean-square error serves to aggregate the magnitudes of the errors in predictions for various times into a single measure of predictive power. It is also a good measure of accuracy, but only to compare forecasting errors of different models for a particular variable and not between variables, as it is scale-dependent. RMS error is the square root of mean squared error (MSE), which is a risk function corresponding to the expected value of the squared error loss or quadratic loss. MSE measures the average of the squares of the “errors. ” The MSE is the second moment (about the origin) of the error, and thus incorporates both the variance of the estimator and its bias. For an unbiased estimator, the MSE is the variance of the estimator. Like the variance, MSE has the same units of measurement as the square of the quantity being estimated. Computing MSE and RMSE If [latex]\hat{\text{Y} }[/latex] is a vector of [latex]\text{n}[/latex] predictions, and [latex]\text{Y}[/latex] is the vector of the true values, then the (estimated) MSE of the predictor is as given as the formula: [latex]\displaystyle \text{MSE} = \frac{1}{\text{n}} \sum_{\text{i}=1}^\text{n} \left(\hat{\text{Y}_\text{i}} - \text{Y}_\text{i}\right)^2[/latex] This is a known, computed quantity given a particular sample (and hence is sample-dependent). RMS error is simply the square root of the resulting MSE quantity. RMS Error for the Regression Line In terms of a regression line, the error for the differing values is simply the distance of a point above or below the line. We can find the general size of these errors by taking the RMS size for them: [latex]\displaystyle \sqrt { \frac { { \left( \text{error}\ 1 \right) }^{ 2 }+{ \left(\text{error}\ 2 \right) }^{ 2 }+\cdots +{ \left( \text{error \text{n}} \right) }^{ 2 } }{ \text{n} } }[/latex]. This calculation results in the RMS error of the regression line, which tells us how far above or below the line points typically are. In general, about 68% of points on a scatter diagram are within one RMS error of the regression line, and about 95% are within two. This is known as the 68%-95% rule. Plotting the Residuals The residual plot illustrates how far away each of the values on the graph is from the expected value (the value on the line). Learning Objectives Differentiate between scatter and residual plots, and between errors and residuals Key Takeaways Key Points The sum of the residuals within a random sample is necessarily zero, and thus the residuals are necessarily not independent. The average of the residuals is always equal to zero; therefore, the standard deviation of the residuals is equal to the RMS error of the regression line. We see heteroscedasticity in a resitual plot as the difference in the scatter of the residuals for different ranges of values of the independent variable. Key Terms scatter plot: A type of display using Cartesian coordinates to display values for two variables for a set of data. residual: The difference between the observed value and the estimated function value. heteroscedasticity: The property of a series of random variables of not every variable having the same finite variance. Errors Versus Residuals Statistical errors and residuals are two closely related and easily confused measures of the deviation of an observed value of an element of a statistical sample from its “theoretical value. ” The error of an observed value is the deviation of the observed value from the (unobservable) true function value, while the residual of an observed value is the difference between the observed value and the estimated function value. A statistical error is the amount by which an observation differs from its expected value, the latter being based on the whole population from which the statistical unit was chosen randomly. For example, if the mean height in a population of 21-year-old men is 5′ 8″, and one randomly chosen man is 5′ 10″ tall, then the “error” is 2 inches. If the randomly chosen man is 5′ 6″ tall, then the “error” is [latex]-2[/latex] inches. The expected value, being the mean of the entire population, is typically unobservable, and hence the statistical error cannot be observed either. A residual (or fitting error), on the other hand, is an observable estimate of the unobservable statistical error. Consider the previous example with men’s heights and suppose we have a random sample of [latex]\text{n}[/latex] people. The sample mean could serve as a good estimator of the population mean, and we would have the following: The difference between the height of each man in the sample and the unobservable population mean is a statistical error, whereas the difference between the height of each man in the sample and the observable sample mean is a residual. Note that the sum of the residuals within a random sample is necessarily zero, and thus the residuals are necessarily not independent. The statistical errors on the other hand are independent, and their sum within the random sample is almost surely not zero. Residual Plots In scatter plots we typically plot an [latex]\text{x}[/latex]-value and a [latex]\text{y}[/latex]-value. To create a residual plot, we simply plot an [latex]\text{x}[/latex]-value and a residual value. The residual plot illustrates how far away each of the values on the graph is from the expected value (the value on the line). The average of the residuals is always equal to zero; therefore, the standard deviation of the residuals is equal to the RMS error of the regression line. As an example, consider the figure depicting the number of drunk driving fatalities in 2006 and 2009 for various states: The relationship between the number of drunk driving fatalities in 2006 and 2009 is very strong, positive, and linear with an [latex]\text{r}^2[/latex] (coefficient of determination) value of 0.99. The high [latex]\text{r}^2[/latex] value provides evidence that we can use the linear regression model to accurately predict the number of drunk driving fatalities that will be seen in 2009 after a span of 4 years. Considering the above figure, we see that the high residual dot on the residual plot suggests that the number of drunk driving fatalities that actually occurred in this particular state in 2009 was higher than we expected it would be after the 4 year span, based on the linear regression model. So, based on the linear regression model, for a 2006 value of 415 drunk driving fatalities we would expect the number of drunk driving fatalities in 2009 to be lower than 377. Therefore, the number of fatalities was not lowered as much as we expected they would be, based on the model. Considering the above figure, we see that the low residual plot suggests that the actual number of drunk driving fatalities in this particular state in 2009 was lower than we would have expected it to be after the 4 year span, based on the linear regression model. So, based on the linear regression model, for a 2006 value of 439 drunk driving fatalities we would expect the number of drunk driving fatalities for 2009 to be higher than 313. Therefore, this particular state is doing an exceptional job at bringing down the number of drunk driving fatalities each year, compared to other states. Advantages of Residual Plots Residual plots can allow some aspects of data to be seen more easily. We can see nonlinearity in a residual plot when the residuals tend to be predominantly positive for some ranges of values of the independent variable and predominantly negative for other ranges. We see outliers in a residual plot depicted as unusually large positive or negative values. We see heteroscedasticity in a resitual plot as the difference in the scatter of the residuals for different ranges of values of the independent variable. The existence of heteroscedasticity is a major concern in regression analysis because it can invalidate statistical tests of significance that assume that the modelling errors are uncorrelated and normally distributed and that their variances do not vary with the effects being modelled. Homogeneity and Heterogeneity By drawing vertical strips on a scatter plot and analyzing the spread of the resulting new data sets, we are able to judge degree of homoscedasticity. Learning Objectives Define, and differentiate between, homoscedasticity and heteroscedasticity. Key Takeaways Key Points When drawing a vertical strip on a scatter plot, the [latex]\text{y}[/latex]-values that fall within this strip will form a new data set, complete with a new estimated average and RMS error. This new data set can also be used to construct a histogram, which can subsequently be used to assess the assumption that the residuals are normally distributed. When various vertical strips drawn on a scatter plot, and their corresponding data sets, show a similar pattern of spread, the plot can be said to be homoscedastic (the prediction errors will be similar along the regression line ). A residual plot displaying homoscedasticity should appear to resemble a horizontal football. When a scatter plot is heteroscedastic, the prediction errors differ as we go along the regression line. Key Terms heteroscedasticity: The property of a series of random variables of not every variable having the same finite variance. homoscedastic: if all random variables in a sequence or vector have the same finite variance Vertical Strips in a Scatter Plot Imagine that you have a scatter plot, on top of which you draw a narrow vertical strip. The [latex]\text{y}[/latex]-values that fall within this strip will form a new data set, complete with a new estimated average and RMS error. This new data set can also be used to construct a histogram, which can subsequently be used to assess the assumption that the residuals are normally distributed. To the extent that the histogram matches the normal distribution, the residuals are normally distributed. This gives us an indication of how well our sample can predict a normal distribution in the population. Homoscedasticity Versus Heteroscedasticity When various vertical strips drawn on a scatter plot, and their corresponding data sets, show a similar pattern of spread, the plot can be said to be homoscedastic. Another way of putting this is that the prediction errors will be similar along the regression line. In technical terms, a data set is homoscedastic if all random variables in the sequence have the same finite variance. A residual plot displaying homoscedasticity should appear to resemble a horizontal football. The presence of this shape lets us know if we can use the regression method. The assumption of homoscedasticity simplifies mathematical and computational treatment; however, serious violations in homoscedasticity may result in overestimating the goodness of fit. In regression analysis, one assumption of the fitted model (to ensure that the least-squares estimators are each a best linear unbiased estimator of the respective population parameters) is that the standard deviations of the error terms are constant and do not depend on the [latex]\text{x}[/latex]-value. Consequently, each probability distribution for [latex]\text{y}[/latex] (response variable) has the same standard deviation regardless of the [latex]\text{x}[/latex]-value (predictor). When a scatter plot is heteroscedastic, the prediction errors differ as we go along the regression line. In technical terms, a data set is heteroscedastic if there are sub-populations that have different variabilities from others. Here “variability” could be quantified by the variance or any other measure of statistical dispersion. The possible existence of heteroscedasticity is a major concern in the application of regression analysis, including the analysis of variance, because the presence of heteroscedasticity can invalidate statistical tests of significance that assume that the modelling errors are uncorrelated and normally distributed and that their variances do not vary with the effects being modelled. Similarly, in testing for differences between sub-populations using a location test, some standard tests assume that variances within groups are equal.
Question: I want to determine a boolean vector $b \in \{0,1\}^n$ consisting of zeros and ones, but cannot access it directly. I can only call a black-box computer code which will take the dot product of $b$ with a real-valued vector $v \in \mathbb{R}^n$ of my choosing. I.e., access to $b$ is available through evaluation of the map $$v \mapsto b^T v.$$ How can I recover all of the entries of $b$ using as few of these dot products as possible? (maybe even just 1 dot product?) Below I detail a couple ideas I had which might work in theory, but which don't work in practice (I think). For concreteness, one may assume that $n \approx 1 \text{ million}$, and arithmetic is done in double precision floating point format. This question arose as a subproblem in a machine learning application. Idea 1: One idea I had is to use a vector with fast growing entries. Say, for example, $n=9$. Then we could use the vector $$v=\begin{bmatrix}1 & 10 & 100 & 1000 & \dots\end{bmatrix}^T.$$ One could then read off $b$ as the digits of $b^T v$. The problem with this solution is that the numbers grow so fast, that in finite precision computer arithmetic it will not work for large $n$. Idea 2: Another idea I had was to use a vector with entries that are algebraically independent. Then determining $b$ from $b^Tv$ is a subset sum problem. For example, if $n=3$ and $$v = \begin{bmatrix}\pi & e & 1\end{bmatrix}^T,$$ then $b^T v$ will take on one of a finite number of possibilities, $$b^T v \in \{\pi,~e,~1,~\pi+e,~\pi+1,~e+1,~\pi+e+1\}.$$ We can determine which of these is the case, thereby determining $b$. But this seems quite combinatorial, and therefore unfeasible for large $n$.
Journal of Symbolic Logic J. Symbolic Logic Volume 45, Issue 2 (1980), 237-250. Constructible Models of Subsystems of ZF Abstract One of the main results of Godel [4] and [5] is that, if $M$ is a transitive set such that $\langle M, \epsilon \rangle$ is a model of ZF (Zermelo-Fraenkel set theory) and $\alpha$ is the least ordinal not in $M$, then $\langle L_\alpha, \epsilon \rangle$ is also a model of ZF. In this note we shall use the Jensen uniformisation theorem to show that results analogous to the above hold for certain subsystems of ZF. The subsystems we have in mind are those that are formed by restricting the formulas in the separation and replacement axioms to various levels of the Levy hierarchy. This is all done in $\S 1$. In $\S 2$ we proceed to establish the exact order relationships which hold among the ordinals of the minimal models of some of the systems discussed in $\S 1$. Although the proofs of these latter results will not require any use of the uniformisation theorem, we will find it convenient to use some of the more elementary results and techniques from Jensen's fine-structural theory of $L$. We thus provide a brief review of the pertinent parts of Jensen's works in $\S 0$, where a list of general preliminaries is also furnished. We remark that some of the techniques which we use in the present paper have been used by us previously in [6] to prove various results about $\beta$-models of analysis. Since $\beta$-models for analysis are analogous to transitive models for set theory, this is not surprising. Article information Source J. Symbolic Logic, Volume 45, Issue 2 (1980), 237-250. Dates First available in Project Euclid: 6 July 2007 Permanent link to this document https://projecteuclid.org/euclid.jsl/1183740556 Mathematical Reviews number (MathSciNet) MR569395 Zentralblatt MATH identifier 0436.03048 JSTOR links.jstor.org Citation Gostanian, Richard. Constructible Models of Subsystems of ZF. J. Symbolic Logic 45 (1980), no. 2, 237--250. https://projecteuclid.org/euclid.jsl/1183740556
Search Now showing items 1-10 of 11 Measurement of $W$ boson angular distributions in events with high transverse momentum jets at $\sqrt{s}=$ 8 TeV using the ATLAS detector (Elsevier, 2017-02) The $W$ boson angular distribution in events with high transverse momentum jets is measured using data collected by the ATLAS experiment from proton--proton collisions at a centre-of-mass energy $\sqrt{s}=$ 8 TeV at the ... Search for new resonances decaying to a $W$ or $Z$ boson and a Higgs boson in the $\ell^+ \ell^- b\bar b$, $\ell \nu b\bar b$, and $\nu\bar{\nu} b\bar b$ channels with $pp$ collisions at $\sqrt s = 13$ TeV with the ATLAS detector (Elsevier, 2017-01) A search is presented for new resonances decaying to a $W$ or $Z$ boson and a Higgs boson in the $\ell^+ \ell^- b\bar b$, $\ell\nu b\bar b$, and $\nu\bar{\nu} b\bar b$ channels in $pp$ collisions at $\sqrt s = 13$ TeV with ... Search for dark matter in association with a Higgs boson decaying to $b$-quarks in $pp$ collisions at $\sqrt{s} = 13$ TeV with the ATLAS detector (Elsevier, 2017-02) A search for dark matter pair production in association with a Higgs boson decaying to a pair of bottom quarks is presented, using 3.2 $fb^{-1}$ of $pp$ collisions at a centre-of-mass energy of 13 TeV collected by the ATLAS ... Measurement of the prompt $J/\psi$ pair production cross-section in pp collisions at $\sqrt{s} = 8$ TeV with the ATLAS detector (Springer, 2017-02) The production of two prompt $J/\psi$ mesons, each with transverse momenta $p_{\mathrm{T}}>8.5$ GeV and rapidity $\|y\| < 2.1$, is studied using a sample of proton-proton collisions at $\sqrt{s} = 8$ TeV, corresponding to ... Search for heavy resonances decaying to a $Z$ boson and a photon in $pp$ collisions at $\sqrt{s}=13$ TeV with the ATLAS detector (Elsevier, 2017-01) This Letter presents a search for new resonances with mass larger than 250 GeV, decaying to a $Z$ boson and a photon. The dataset consists of an integrated luminosity of 3.2 fb$^{-1}$ of $pp$ collisions collected at $\sqrt{s} ... Measurement of forward-backward multiplicity correlations in lead-lead, proton-lead, and proton-proton collisions with the ATLAS detector (American Physical Society, 2017-06) Two-particle pseudorapidity correlations are measured in $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV Pb+Pb, $\sqrt{s_{\rm{NN}}}$ = 5.02 TeV $p$+Pb, and $\sqrt{s}$ = 13 TeV $pp$ collisions at the LHC, with total integrated luminosities ... Performance of the ATLAS trigger system in 2015 (Springer, 2017-05) During 2015 the ATLAS experiment recorded $3.8 \mathrm{fb}^{-1}$ of proton--proton collision data at a centre-of-mass energy of $13 \mathrm{TeV}$. The ATLAS trigger system is a crucial component of the experiment, responsible ... Search for lepton-flavour-violating decays of the Higgs and $Z$ bosons with the ATLAS detector (Springer, 2017-02) Direct searches for lepton flavour violation in decays of the Higgs and $Z$ bosons with the ATLAS detector at the LHC are presented. The following three decays are considered: $H\to e\tau$, $H\to\mu\tau$, and $Z\to\mu\tau$. ... Measurement of the $t\bar{t}Z$ and $t\bar{t}W$ production cross sections in multilepton final states using 3.2 fb$^{-1}$ of $pp$ collisions at $\sqrt{s}$ =13 TeV with the ATLAS detector (Springer, 2017-01) A measurement of the $t\bar{t}Z$ and $t\bar{t}W$ production cross sections in final states with either two same-charge muons, or three or four leptons (electrons or muons) is presented. The analysis uses a data sample of ... A measurement of the calorimeter response to single hadrons and determination of the jet energy scale uncertainty using LHC Run-1 $pp$-collision data with the ATLAS detector (Springer, 2017-01) A measurement of the calorimeter response to isolated charged hadrons in the ATLAS detector at the LHC is presented. This measurement is performed with 3.2 nb$^{-1}$ of proton--proton collision data at $\sqrt{s}=7$ TeV ...
amp-mathml Displays a MathML formula. Required Script <script async custom-element="amp-mathml" src="https://cdn.ampproject.org/v0/amp-mathml-0.1.js"></script> Supported Layouts container Examples amp-mathml.amp.html This extension creates an iframe and renders a MathML formula. <amp-mathml layout="container" data-formula="\[x = {-b \pm \sqrt{b^2-4ac} \over 2a}.\]"> </amp-mathml> <amp-mathml layout="container" data-formula="\[f(a) = \frac{1}{2\pi i} \oint\frac{f(z)}{z-a}dz\]"> </amp-mathml> <amp-mathml layout="container" data-formula="$$ \cos(θ+φ)=\cos(θ)\cos(φ)−\sin(θ)\sin(φ) $$"> </amp-mathml> This is an example of a formula of <amp-mathml layout="container" inline data-formula="`x`"></amp-mathml>, <amp-mathml layout="container" inline data-formula="$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$"></amp-mathml> placed inline in the middle of a block of text. <amp-mathml layout="container" inline data-formula="$ \cos(θ+φ) $"></amp-mathml> This shows how the formula will fit inside a block of text and can be styled with CSS. Specifies the formula to render. If specified, the component renders inline ( inline-block in CSS). See amp-mathml rules in the AMP validator specification. You've read this document a dozen times but it doesn't really cover all of your questions? Maybe other people felt the same: reach out to them on Stack Overflow.Go to Stack Overflow Found a bug or missing a feature? The AMP project strongly encourages your participation and contributions! We hope you'll become an ongoing participant in our open source community but we also welcome one-off contributions for the issues you're particularly passionate about.Go to GitHub
I was recently thinking about some of my past math classes, and depending on the context I recall my professors would sometimes use the "$\equiv$" symbol in places where I'd feel "$=$" to be more appropriate. For example, since this would often be the case in my classes on differential equations and Fourier series, we would have (for $n \in \Bbb N, k \in \Bbb Z$) $$(-1)^{2n+1} \equiv -1$$ $$\sin(k\pi) \equiv 0$$ Is there a particular reason in this context why we would say "$\equiv$" instead of "$=$"? The latter feels more natural in this context, which makes me think that there's some reason my professors would use the former. I'm familiar with the notion of the "$\equiv$" symbol in the context of, say, elementary number theory (specifically modular arithmetic) where we might say $$10 \equiv 1 \pmod 3$$ which isn't saying "$10$ equals $1$", just that "$10$ is like $1$ in this context." But that doesn't seem to fit the case as with the first two statements - because I don't believe it is that $(-1)^{2n+1}$ is like $-1$, or that $\sin(k \pi)$ is like $0$, they $-1$ and $0$ respectively. are Am I just mistaken on this latter fact? Is there something I'm missing? What, precisely, is the difference between the two notations?
Find the radius of convergence and the convergence at the end points of the series: $$\sum_{n=1}^\infty(2+(-1)^n)^nx^n$$ This is what I did: $a_n=(2+(-1)^n)^n\Rightarrow R=\frac{1}{limsup\|a_n\|^\frac1n}\\ for\ n=2k \to lim(2+1)=3 \\ for\ n= 2k+1 \to lim(2-1)=1 \\ limsup \ a_n^{\frac1n}=3 \Rightarrow R=\frac13$ So the series converge at $(-\frac13 , \frac13)$ But now I don't really understand what I'm being asked for the end points. Am I supposed to check if the series really converge at the two end points ?
I have been recently learning a lot about Macdonald polynomials, which have been shown to have probabilistic interpretations, more precisely the eigenfunctions of certain Markov chains on the symmetric group. To make this post more educational, I will define these polynomials a bit. Consider the 2-parameter family of Macdonald operators (indexed by powers of the indeterminate $X$) for root system $A_n$, on a symmetric polynomial $f$ with $x = (x_1, \ldots, x_n)$: $$D(X;t,q) = a_\delta(x)^{-1} \sum_{\sigma \in S_n} \epsilon(\sigma) x^{\sigma \delta}\prod_{i=1}^n (1 + X t^{(\sigma \delta)_i} T_i),$$ (mathoverflow doesn't seem to parse $T_{q,x_i}$ in the formula above, so I had to use the shorter symbol $T_i$, which depends on q). where $\delta$ is the partition $(n-1,n-2,\ldots, 1,0)$, $a_\delta(x) = \prod_{1 \le i < j \le n} (x_i - x_j)$ is the Vandermonde determinant (in general $a_\lambda(x)$ is the determinant of the matrix $(a_i^{\lambda_j})_{i,j \in [n]}$). $x^{\sigma \delta}$ means $x_1^{(\sigma \delta)_1} x_2^{(\sigma \delta)_2} \ldots x_n^{(\sigma \delta)_n}$. Also $(\sigma \delta)_i$ denotes the $\sigma(i)$-th component of $\delta$, namely $n-i$. Finally the translation operator $T_i = T_{q,x_i}$ is defined as $$ T_{q,x_i}f(x_1, \ldots, x_n) = f(x_1, \ldots, x_{i-1}, q x_i , x_{i+1} ,\ldots, x_n).$$ I like to think of the translation operator as the quantized version of the differential operator $I + \partial_i$, where $q-1$ is analogous to the Planck constant(?). If we write $D(X;q,t) = \sum_{r=0}^n D_{n-r}(q,t) X^r$, then Macdonald polynomials $p_\lambda(q,t)$ are simply simultaneous eigenfunctions of these operators. When $q=t$ they become Schur polynomials, defined by $s_\lambda = a_{\delta +\lambda} / a_\delta$. When $q= t^\alpha$ and $t \to 1$, we get Jack symmetric polymomials, which are eigenfunctions of a Metropolis random walk on the set of all partitions that converge to the so-called Ewens sampling measure, which assigns probability proportional $\alpha^{\ell(\lambda)} z_\lambda^{-1}$. When $q = 0$, they become the Hall-Littlewood polynomials and when $t=1$ they become the monomial symmetric polynomials etc. I was told repeatedly by experts that Macdonald polynomials exhaust all previous symmetric polynomial bases in some sense. Does anyone know a theorem that says that every family of symmetric polynomial under some conditions can be obtained from Macdonald polynomials by specializing the $q$ and $t$?
Author: Zuguang Gu ( z.gu@dkfz.de ) Date: 2019-05-02 library(EnrichedHeatmap)load(system.file("extdata", "neg_cr.RData", package = "EnrichedHeatmap"))all_genes = all_genes[unique(neg_cr$gene)]all_tss = promoters(all_genes, upstream = 0, downstream = 1)mat_neg_cr = normalizeToMatrix(neg_cr, all_tss, mapping_column = "gene", w = 50, mean_mode = "w0") The object mat_neg_cr is a normalized matrix for regions showing significantnegative correlation between methylation and gene expression. The negativecorrelated regions (negCRs) are normalized to upstream 5kb and downstream 5kb of geneTSS with 50bp window by normalizeToMatrix() function. The value in thematrix is how much a window is covered by negCRs (values between 0 and 1). In the normalized matrix, each row corresponds to one gene and each columncorresponds to a window either on upstream of TSS or downstream of TSS. Forthe example of mat_neg_cr matrix, the first half columns correspond to theupstream of TSS and the last half columns correspond to downstream of TSS. Herewe compare following three different methods to order rows (which correspond to genes)in the normalized matrix. Rows are ordered by the enriched scores. For each row in the matrix, denote values in a certain row as $x$, indices 1, …, $n_1$ are for upstream windows, indices $n_1+1$, …, $n$ are for downstream windows and $n_2 = n - n_1$, the enriched score is calculated as the sum of $x$ weighted by distance to TSS (higher weight if the window is close to TSS). \[ \sum_{i=1}^{n_1}{x_i \cdot i/n_1} + \sum_{i=n_1+1}^n{x_i \cdot (n - i + 1)/n_2}\] Rows are ordered by hierarchical clustering with Euclidean distance. Rows are ordered by hierarchical clustering with closeness distance. For two rows in the normalized matrix, assume $a_1, a_2, …, a_{n_1}$ are the indices of windows for one gene which overlap with negCRs and $b_1, b_2, … b_{n_2}$ are the indices for the other gene which overlap with negCRs, the distance which is based on closeness of the overlapped windows in the two genes is defined as: \[ d_{closeness} = \frac{\sum_{i=1}^{n_1} \sum_{j=1}^{n_2} {\|a_i - b_j\|} }{n_1 \cdot n_2}\] So the closeness distance is basically the average distance of all pairs of negCR windows in the two genes. Euclidean distance between rows keeps unchanged when the matrix columns are permutated, while for closeness distance, the column order is also taken into account, which might be more proper for clustering normalized matrices because the columns correspond to relative distance to the target regions. Following three plots show heatmaps under different row ordering methods. EnrichedHeatmap(mat_neg_cr, name = "neg_cr", col = c("white", "darkgreen"), top_annotation = HeatmapAnnotation(enrich = anno_enriched(gp = gpar(col = "darkgreen"))), row_title = "by default enriched scores")EnrichedHeatmap(mat_neg_cr, name = "neg_cr", col = c("white", "darkgreen"), top_annotation = HeatmapAnnotation(enrich = anno_enriched(gp = gpar(col = "darkgreen"))), cluster_rows = TRUE, row_title = "by hierarchcal clustering + Euclidean distance\ndendrogram reordered by enriched scores")EnrichedHeatmap(mat_neg_cr, name = "neg_cr", col = c("white", "darkgreen"), top_annotation = HeatmapAnnotation(enrich = anno_enriched(gp = gpar(col = "darkgreen"))), cluster_rows = TRUE, clustering_distance_rows = dist_by_closeness, row_title = "by hierarchcal clustering + closeness distance\ndendrogram reordered by enriched scores") Generally, when the top annotation which summarises mean enrichment across genes is also added to the heatmap, ordering genes by enriched scores is not recommended because it provides redundant information as the top enriched annotation, and on the other hand, it fails to reveal spatial clusters as other two methods. Hierarchal clustering with Euclidean distance is good at clustering enrichment patterns, but since it does not take column order into account, thus, it still can be possible that two spatially close clusters are far separated in the heatmap. By using closeness distance, it clearly sorts and clusters the enrichment patterns. The row order, clustering method, distance method can all be self-adjusted by row_order, cluster_rows, clustering_method_rows, clustering_distance_rows arguments in EnrichedHeamtap()function. For how to properly set values for these arguments, users can go tothe help page of EnrichedHeatmap() or Heatmap() function.
TL; DR: obtaining a reasonably big single crystal of $\ce{ZnSO4 * H2O}$ seems to be a tricky task, whereas one can easily obtain $\ce{ZnSO4 * H2O}$ as a result of thermal dehydration of heptahydtare above $\pu{70 ^\circ C}$ in a form of melt, and above $\pu{120 ^\circ C}$ – as a polycrystalline solid material. I finally got across to suitable references to back up my comments. In nature zinc sulfate monohydrate occurs as mineral gunningite, always in a form of efflorescence deposition on the surface of sphalerite and other minerals and doesn't form single crystals [1]: Surface and ground waters carrying dissolved oxygen attack the sphalerite, yielding soluble sulphates of zinc [...]: $$\ce{ZnS + 2 O2 -> ZnSO4}$$ [...] In the presence of abundant water all of the soluble salts are removed. If, however, the conditions are such that the supply of water is restricted or evaporation takes place, the metal-bearing solutions become supersaturated with the consequent precipitation of a number of hydrates of zinc, manganese, and iron of which gunningite, $\ce{ZnSO4 * H2O}$, is one. Thermal dehydration among zinc sulfate hydrates has been studied with DTA in several publications. The most relevant and thorough investigation [2, p. 471] suggests the following scheme: $$\ce{ZnSO4 * 7 H2O (s) ->[\pu{37 ^\circ C}] ZnSO4 * 6 H2O (s) ->[\pu{70 ^\circ C}] ZnSO4 * H2O (l) ->[\pu{120 ^\circ C}] ZnSO4 * H2O (s) ->[\pu{285 ^\circ C}] ZnSO4 (s)}$$ Original text in German: Festgestellt werden vier endotherme Effekte: bei etwa $\pu{37 ^\circ C}$ verwandelt sich das Heptahydrat in Hexahydrat; bei etwa $\pu{70 ^\circ C}$ geht das Hexahydrat in Monohydrat über, wobei teilweise Schmelzen im Kristallwasser auftritt. Bei etwa $\pu{120 ^\circ C}$ siedet die Schmelze, Wasser scheidet sich ab, bis das feste Monohydrat zurückbleibt; bei etwa $\pu{285 ^\circ C}$ erfolgt der Übergang des Monohydrates in wasserfreies Salz. The tricky part is that monohydrate can be isolated in crystal form, but it initially isolates as a melt, and only above $\pu{120 ^\circ C}$ makes a polycrystalline solid phase (occurs as white powder or granules). Aqueous solubility $\omega_2$ for zinc sulfate $\ce{ZnSO4}$ in mass% of a solute ($\omega_2 = m_2/(m_1 + m_2)$, where $m_2$ is the mass of solute and $m_1$ the mass of water) as a function of temperature ($T$) from CRC Handbook [3, p. 5-172]: \begin{array}{r\|rrrrrrrr}T,\,\pu{^\circ C} & 0 & 10 & 20 & 25 & 30 & 40 & 50 & 60 & 70 & 80 & 90 & 100 \\\hline\omega_2,~\% & 29.1 & 32.0 & 35.0 & 36.6 & 38.2 & 41.3 & 43.0 & 42.1 & 41.0 & 39.9 & 38.8 & 37.6\end{array} With this in mind, what can be suggested is to prepare a saturated at $\pu{70 ^\circ C}$ solution, and then slowly increase the temperature (e.g. using a thermostat), hoping that zinc sulfate monohydrate starts to crystallize (using the decrease in solubility starting from $\pu{60 ^\circ C}$). According to [4, entry 3575, p. 470], solubility of $\ce{ZnSO4 * H2O}$ also decreases in this interval: $\pu{101 g}/\pu{100 g}~\ce{H2O}$ at $\pu{70 ^\circ C}$, and $\pu{87 g}/\pu{100 g}~\ce{H2O}$ at $\pu{105 ^\circ C}$. The difference is rather small, so I wouldn't expect rapidly-grown large crystals. References Jambor, J. L.; Boyle, R. W. The Canadian Mineralogist 1962, 7 (2), 209–218. Balarew, C.; Trendafelov, D.; Gerganova, M. Monatshefte für Chemie 1971, 102 (2), 465–473. DOI 10.1007/BF00909340 (in German). Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC handbook of chemistry and physics: a ready-reference book of chemical and physical data.; 2017; Vol. 97. Perry, D. L. Handbook of inorganic compounds, 2nd ed.; Taylor & Francis: Boca Raton, 2011. ISBN 978-1-4398-1461-1.
When choosing the public exponent e, it is stressed that $e$ must be coprime to $\phi(n)$, i.e. $\gcd(\phi(n), e) = 1$. I know that a common choice is to have $e = 3$ (which requires a good padding scheme) or $e=65537$, which is slower but safer. I also know that for two primes $p,q$, we have $\phi(pq) = (p - 1) (q - 1)$ Now, let me give a (simple) example: Say I choose $e = 3$, and two random primes $p = 5$ and $q = 13$. I can now compute $\gcd(3, \phi(5 \cdot 13)) = 3$. This reveals that $3$ and $\phi(n)$ are not coprime. I assume this could also happen for large values of $p$ and $q$, and likewise for another $e$. I therefore assume that the RSA algorithm must check that $\gcd(e, \phi(pq)) = 1$. But let's assume it doesn't. How does RSA become vulnerable if $\gcd(e, \phi(pq)) \neq 1$?
Every new study presents its own challenges. (I would have to say that one of the great things about being a biostatistician is the immense variety of research questions that I get to wrestle with.) Recently, I was approached by a group of researchers who wanted to evaluate an intervention. Actually, they had two, but the second one was a minor tweak added to the first. They were trying to figure out how to design the study to answer two questions: (1) is intervention $A$ better than doing nothing and (2) is $A^+$, the slightly augmented version of $A$, better than just $A$? It was clear in this context (and it is certainly not usually the case) that exposure to $A$ on one day would have no effect on the outcome under $A^+$ the next day (or vice versa). That is, spillover risks were minimal. Given this, the study was an ideal candidate for a cross-over design, where each study participant would receive both versions of the intervention and the control. This design can be much more efficient than a traditional RCT, because we can control for variability across patients. While a cross-over study is interesting and challenging in its own right, the researchers had a pretty serious constraint: they did not feel they could assign intervention $A^+$ until $A$ had been applied, which would be necessary in a proper cross-over design. So, we had to come up with something a little different. This post takes a look at how to generate data for and analyze data from a more standard cross-over trial, and then presents the solution we came up with for the problem at hand. Cross-over design with three exposures If we are free to assign any intervention on any day, one possible randomization scheme involving three interventions could look like this: Key features of this scheme are: (1) all individuals are exposed to each intervention over three days, (2) on any given day, each intervention is applied to one group of participants (just in case the specific day has an impact on the outcome), and (3) not every permutation is included (for example, $A$ does not immediately proceed $Control$ in any sequence), because the relative ordering of interventions in this case is assumed not to matter. (We might need to expand to six groups to rectify this.) Data simulation In this simulation, we will assume (1) that the outcome is slightly elevated on days two and three, (2) $A$ is an improvement over $Control$, (3) $A^+$ is an improvement over $A$, (4) there is strong correlation of outcomes within each individual, and (5) group membership has no bearing on the outcome. First, I define the data, starting with the different sources of variation. I have specified a fairly high intra-class coefficient (ICC), because it is reasonable to assume that there will be quite a bit of variation across individuals: vTotal = 1vAcross <- iccRE(ICC = 0.5, varTotal = vTotal, "normal")vWithin <- vTotal - vAcross### Definitionsb <- defData(varname = "b", formula = 0, variance = vAcross, dist = "normal")d <- defCondition(condition = "rxlab == 'C'", formula = "0 + b + (day == 2) * 0.5 + (day == 3) * 0.25", variance = vWithin, dist = "normal")d <- defCondition(d, "rxlab == 'A'", formula = "0.4 + b + (day == 2) * 0.5 + (day == 3) * 0.25", variance = vWithin, dist = "normal")d <- defCondition(d, "rxlab == 'A+'", formula = "1.0 + b + (day == 2) * 0.5 + (day == 3) * 0.25", variance = vWithin, dist = "normal") Next, I generate the data, assigning three groups, each of which is tied to one of the three treatment sequences. set.seed(39217)db <- genData(240, b)dd <- trtAssign(db, 3, grpName = "grp")dd <- addPeriods(dd, 3)dd[grp == 1, rxlab := c("C", "A", "A+")]dd[grp == 2, rxlab := c("A+", "C", "A")]dd[grp == 3, rxlab := c("A", "A+", "C")]dd[, rxlab := factor(rxlab, levels = c("C", "A", "A+"))]dd[, day := factor(period + 1)]dd <- addCondition(d, dd, newvar = "Y")dd ## timeID Y id period grp b rxlab day## 1: 1 0.9015848 1 0 2 0.2664571 A+ 1## 2: 2 1.2125919 1 1 2 0.2664571 C 2## 3: 3 0.7578572 1 2 2 0.2664571 A 3## 4: 4 2.0157066 2 0 3 1.1638244 A 1## 5: 5 2.4948799 2 1 3 1.1638244 A+ 2## --- ## 716: 716 1.9617832 239 1 1 0.3340201 A 2## 717: 717 1.9231570 239 2 1 0.3340201 A+ 3## 718: 718 1.0280355 240 0 3 1.4084395 A 1## 719: 719 2.5021319 240 1 3 1.4084395 A+ 2## 720: 720 0.4610550 240 2 3 1.4084395 C 3 Here is a plot of the treatment averages each day for each of the three groups: dm <- dd[, .(Y = mean(Y)), keyby = .(grp, period, rxlab)]ngrps <- nrow(dm[, .N, keyby = grp])nperiods <- nrow(dm[, .N, keyby = period])ggplot(data = dm, aes(y=Y, x = period + 1)) + geom_jitter(data = dd, aes(y=Y, x = period + 1), width = .05, height = 0, color="grey70", size = 1 ) + geom_line(color = "grey50") + geom_point(aes(color = rxlab), size = 2.5) + scale_color_manual(values = c("#4477AA", "#DDCC77", "#CC6677")) + scale_x_continuous(name = "day", limits = c(0.9, nperiods + .1), breaks=c(1:nperiods)) + facet_grid(~ factor(grp, labels = paste("Group", 1:ngrps))) + theme(panel.grid = element_blank(), legend.title = element_blank()) Estimating the effects To estimate the treatment effects, I will use this mixed effects linear regression model: \[Y_{it} = \alpha_0 + \gamma_{t} D_{it} + \beta_1 A_{it} + \beta_2 P_{it} + b_i + e_i\] where $Y_{it}$ is the outcome for individual $i$ on day $t$, $t \in (1,2,3)$. $A_{it}$ is an indicator for treatment $A$ in time $t$; likewise $P_{it}$ is an indicator for $A^+$. $D_{it}$ is an indicator that the outcome was recorded on day $t$. $b_i$ is an individual (latent) random effect, $b_i \sim N(0, \sigma_b^2)$. $e_i$ is the (also latent) noise term, $e_i \sim N(0, \sigma_e^2)$. The parameter $\alpha_0$ is the mean outcome on day 1 under $Control$. The $\gamma$’s are the day-specific effects for days 2 and 3, with $\gamma_1$ fixed at 0. $\beta_1$ is the effect of $A$ (relative to $Control$) and $\beta_2$ is the effect of $A^+$. In this case, the researchers were primarily interested in $\beta_1$ and $\beta_2 – \beta_1$, which is the incremental change from $A$ to $A^+$. library(lme4)lmerfit <- lmer(Y ~ day + rxlab + (1\|id), data = dd)rndTidy(lmerfit) ## term estimate std.error statistic group## 1: (Intercept) -0.14 0.08 -1.81 fixed## 2: day2 0.63 0.06 9.82 fixed## 3: day3 0.38 0.06 5.97 fixed## 4: rxlabA 0.57 0.06 8.92 fixed## 5: rxlabA+ 0.98 0.06 15.35 fixed## 6: sd_(Intercept).id 0.74 NA NA id## 7: sd_Observation.Residual 0.70 NA NA Residual As to why we would want to bother with this complex design if we could just randomize individuals to one of three treatment groups, this little example using a more standard parallel design might provide a hint: def2 <- defDataAdd(varname = "Y", formula = "0 + (frx == 'A') * 0.4 + (frx == 'A+') * 1", variance = 1, dist = "normal")dd <- genData(240)dd <- trtAssign(dd, nTrt = 3, grpName = "rx")dd <- genFactor(dd, "rx", labels = c("C","A","A+"), replace = TRUE)dd <- addColumns(def2, dd)lmfit <- lm(Y~frx, data = dd)rndTidy(lmfit) ## term estimate std.error statistic p.value## 1: (Intercept) -0.12 0.10 -1.15 0.25## 2: frxA 0.64 0.15 4.38 0.00## 3: frxA+ 1.01 0.15 6.86 0.00 If we compare the standard error for the effect of $A^+$ in the two studies, the cross-over design is much more efficient (i.e. standard error is considerably smaller: 0.06 vs. 0.15). This really isn’t so surprising since we have collected a lot more data and modeled variation across individuals in the cross-over study. Constrained cross-over design Unfortunately, the project was not at liberty to implement the three-way/three-day design just simulated. We came up with this approach that would provide some cross-over, but with an added day of treatment and measurement: The data generation is slightly modified, though the original definitions can still be used: db <- genData(240, b)dd <- trtAssign(db, 2, grpName = "grp")dd <- addPeriods(dd, 4)dd[grp == 0, rxlab := c("C", "C", "A", "A+")]dd[grp == 1, rxlab := c("C", "A", "A+", "A")]dd[, rxlab := factor(rxlab, levels = c("C", "A", "A+"))]dd[, day := factor(period + 1)]dd <- addCondition(d, dd, "Y") The model estimates indicate slightly higher standard errors than in the pure cross-over design: lmerfit <- lmer(Y ~ day + rxlab + (1\|id), data = dd)rndTidy(lmerfit) ## term estimate std.error statistic group## 1: (Intercept) 0.15 0.06 2.36 fixed## 2: day2 0.48 0.08 6.02 fixed## 3: day3 0.16 0.12 1.32 fixed## 4: day4 -0.12 0.12 -1.02 fixed## 5: rxlabA 0.46 0.10 4.70 fixed## 6: rxlabA+ 1.14 0.12 9.76 fixed## 7: sd_(Intercept).id 0.69 NA NA id## 8: sd_Observation.Residual 0.68 NA NA Residual Here are the key parameters of interest (refit using package lmerTest to get the contrasts). The confidence intervals include the true values ($\beta_1 = 0.4$ and $\beta_2 – \beta_1 = 0.6$): library(lmerTest)lmerfit <- lmer(Y ~ day + rxlab + (1\|id), data = dd)L <- matrix(c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, -1, 1), nrow = 2, ncol = 6, byrow = TRUE)con <- data.table(contest(lmerfit, L, confint = TRUE, joint = FALSE))round(con[, .(Estimate, `Std. Error`, lower, upper)], 3) ## Estimate Std. Error lower upper## 1: 0.462 0.098 0.269 0.655## 2: 0.673 0.062 0.551 0.795 Exploring bias A single data set does not tell us if the proposed approach is indeed unbiased. Here, I generate 1000 data sets and fit the mixed effects model. In addition, I fit a model that ignores the day factor to see if it will induce bias (of course it will). iter <- 1000ests <- vector("list", iter)xests <- vector("list", iter)for (i in 1:iter) { db <- genData(240, b) dd <- trtAssign(db, 2, grpName = "grp") dd <- addPeriods(dd, 4) dd[grp == 0, rxlab := c("C", "C", "A", "A+")] dd[grp == 1, rxlab := c("C", "A", "A+", "A")] dd[, rxlab := factor(rxlab, levels = c("C", "A", "A+"))] dd[, day := factor(period + 1)] dd <- addCondition(d, dd, "Y") lmerfit <- lmer(Y ~ day + rxlab + (1\|id), data = dd) xlmerfit <- lmer(Y ~ rxlab + (1\|id), data = dd) ests[[i]] <- data.table(estA = fixef(lmerfit)[5], estAP = fixef(lmerfit)[6] - fixef(lmerfit)[5]) xests[[i]] <- data.table(estA = fixef(xlmerfit)[2], estAP = fixef(xlmerfit)[3] - fixef(xlmerfit)[2])}ests <- rbindlist(ests)xests <- rbindlist(xests) The results for the correct model estimation indicate that there is no bias (and that the standard error estimates from the model fit above were correct): ests[, .(A.est = round(mean(estA), 3), A.se = round(sd(estA), 3), AP.est = round(mean(estAP), 3), AP.se = round(sd(estAP), 3))] ## A.est A.se AP.est AP.se## 1: 0.407 0.106 0.602 0.06 In contrast, the estimates that ignore the day or period effect are in fact biased (as predicted): xests[, .(A.est = round(mean(estA), 3), A.se = round(sd(estA), 3), AP.est = round(mean(estAP), 3), AP.se = round(sd(estAP), 3))] ## A.est A.se AP.est AP.se## 1: 0.489 0.053 0.474 0.057
Through the questions below, this post asks whether the concept of abelian subfactor is relevant. Remark : here abelian qualifies an inclusion of II$_1$ factors $(N \subset M)$, $N$ is not an abelian algebra. First some useful reminders about groups and lattices : Definitions: A lattice $(L, \wedge, \vee)$ is : $(\forall a,b,c \in L)$ Remark: Distributivity $\Rightarrow$ Modularity Let $G$ be a finite group and let $\mathcal{L}(G)$ be its lattice of subgroups, and $\mathcal{N}(G)$, of normal subgroups. Theorems : A finite group $G$ is Cycliciff $\mathcal{L}(G)$ is distributive (Ore 1938) Abelianiff $\mathcal{L}(G \times G)$ is modular (Lukacs-Palfy 1986) (see here thm2.3 p431 and thm6.5 p449) Remark : Of course, a cyclic group is abelian, and a direct product of abelian groups is abelian. Theorem : Every finite abelian groups is a direct product of finite cyclic groups. Theorem : $\mathcal{N}(G)$ is modular. Definition : $G$ is Dedekind if all its subgroups are normal. The abelian groups are Dedekind. A non-abelian Dedekind group is called Hamiltonian (for example the quaternion group $Q_8$). Remark : $G$ abelian implies $\mathcal{L}(G)$ modular, but the converse is false (see $Q_8$). All the subfactors $(N\subset M)$ are irreducible and finite index inclusions of II$_1$ factors. Let $(N\subset M)$ be a subfactor and $\mathcal{L}(N\subset M)$ its lattice of intermediate subfactors. Galois correspondence for group subfactors: $\mathcal{L}(R^G\subset R)$ $\leftrightarrow$ $\mathcal{L}(G)$ $\leftrightarrow$ $\mathcal{L}(R \subset R \rtimes G)$. Recall also that $(R^G \otimes R^H\subset R \otimes R) \simeq (R^{G \times H}\subset R)$ Definitions : A subfactor $(N\subset M)$ is Cyclicif $\mathcal{L}(N\subset M)$ is distributive. Abelianif $\mathcal{L}(N \otimes N \subset M \otimes M)$ is modular. Remark : here abelian qualifies the inclusion of factors $(N \subset M)$, $N$ is not an abelian algebra. Remark: $(R^G\subset R)$ is cyclic (resp. abelian) iff $G$ is cyclic (resp. abelian). Question 1a: Are the cyclic subfactors abelian ? Examples: If $(N\subset M)$ is $2$-supertransitive, then it is maximal, so cyclic. If also $[M:N]>2$ then $\mathcal{L}(N \otimes N \subset M \otimes M)$ is distributive (W prop5.1 p329), so modular, and then $(N\subset M)$ is abelian. All the maximal group-subgroup subfactors $(R^G\subset R^H)$ are abelian (see the corollary here). Let $(\otimes_{i \in I} A_i \subset \otimes_{i \in I} B_i)$ be the tensor product of the subfactors $(A_i \subset B_i)_{i \in I} $, with $I$ finite. Question 1b: Is a tensor product of abelian subfactors also abelian ? Question 1c: Is every abelian subfactor a tensor product of cyclic subfactors ? In this paper, T. Teruya introduced the notion of normal intermediate subfactors, generalizing exactly the notion of normal subgroups (see the post Jordan-Hölder theorem for subfactors for more details). Definitions : A subfactor $(N\subset M)$ is Dedekindif all its intermediate subfactors are normal. Hamiltonianif it is Dedekind and non-abelian. Question 2: Are the abelian subfactors Dedekind ? Problem : Find Hamiltonian subfactors not coming from group theory. Definition : A subfactor is basically abelian if $(N' \cap M_1)$ and $(M' \cap M_2)$ are abelian algebras. Remark : A group subfactor is abelian iff it is basically abelian. Question 3: Is a subfactor abelian iff it is basically abelian ? Remark : the implication $(\Leftarrow)$ is clear if the relative commutants deal with the tensor product. If the implication $(\Rightarrow)$ and the question 1a are true, then there is no non-trivial maximal Kac algebra ! (the original motivation for this post).
Can I be a pedant and say that if the question states that $\langle \alpha \vert A \vert \alpha \rangle = 0$ for every vector $\lvert \alpha \rangle$, that means that $A$ is everywhere defined, so there are no domain issues? Gravitational optics is very different from quantum optics, if by the latter you mean the quantum effects of interaction between light and matter. There are three crucial differences I can think of:We can always detect uniform motion with respect to a medium by a positive result to a Michelson... Hmm, it seems we cannot just superimpose gravitational waves to create standing waves The above search is inspired by last night dream, which took place in an alternate version of my 3rd year undergrad GR course. The lecturer talks about a weird equation in general relativity that has a huge summation symbol, and then talked about gravitational waves emitting from a body. After that lecture, I then asked the lecturer whether gravitational standing waves are possible, as a imagine the hypothetical scenario of placing a node at the end of the vertical white line [The Cube] Regarding The Cube, I am thinking about an energy level diagram like this where the infinitely degenerate level is the lowest energy level when the environment is also taken account of The idea is that if the possible relaxations between energy levels is restricted so that to relax from an excited state, the bottleneck must be passed, then we have a very high entropy high energy system confined in a compact volume Therefore, as energy is pumped into the system, the lack of direct relaxation pathways to the ground state plus the huge degeneracy at higher energy levels should result in a lot of possible configurations to give the same high energy, thus effectively create an entropy trap to minimise heat loss to surroundings @Kaumudi.H there is also an addon that allows Office 2003 to read (but not save) files from later versions of Office, and you probably want this too. The installer for this should also be in \Stuff (but probably isn't if I forgot to include the SP3 installer). Hi @EmilioPisanty, it's great that you want to help me clear out confusions. I think we have a misunderstanding here. When you say "if you really want to "understand"", I've thought you were mentioning at my questions directly to the close voter, not the question in meta. When you mention about my original post, you think that it's a hopeless mess of confusion? Why? Except being off-topic, it seems clear to understand, doesn't it? Physics.stackexchange currently uses 2.7.1 with the config TeX-AMS_HTML-full which is affected by a visual glitch on both desktop and mobile version of Safari under latest OS, \vec{x} results in the arrow displayed too far to the right (issue #1737). This has been fixed in 2.7.2. Thanks. I have never used the app for this site, but if you ask a question on a mobile phone, there is no homework guidance box, as there is on the full site, due to screen size limitations.I think it's a safe asssumption that many students are using their phone to place their homework questions, in wh... @0ßelö7 I don't really care for the functional analytic technicalities in this case - of course this statement needs some additional assumption to hold rigorously in the infinite-dimensional case, but I'm 99% that that's not what the OP wants to know (and, judging from the comments and other failed attempts, the "simple" version of the statement seems to confuse enough people already :P) Why were the SI unit prefixes, i.e.\begin{align}\mathrm{giga} && 10^9 \\\mathrm{mega} && 10^6 \\\mathrm{kilo} && 10^3 \\\mathrm{milli} && 10^{-3} \\\mathrm{micro} && 10^{-6} \\\mathrm{nano} && 10^{-9}\end{align}chosen to be a multiple power of 3?Edit: Although this questio... the major challenge is how to restrict the possible relaxation pathways so that in order to relax back to the ground state, at least one lower rotational level has to be passed, thus creating the bottleneck shown above If two vectors $\vec{A} =A_x\hat{i} + A_y \hat{j} + A_z \hat{k}$ and$\vec{B} =B_x\hat{i} + B_y \hat{j} + B_z \hat{k}$, have angle $\theta$ between them then the dot product (scalar product) of $\vec{A}$ and $\vec{B}$ is$$\vec{A}\cdot\vec{B} = \|\vec{A}\|\|\vec{B}\|\cos \theta$$$$\vec{A}\cdot\... @ACuriousMind I want to give a talk on my GR work first. That can be hand-wavey. But I also want to present my program for Sobolev spaces and elliptic regularity, which is reasonably original. But the devil is in the details there. @CooperCape I'm afraid not, you're still just asking us to check whether or not what you wrote there is correct - such questions are not a good fit for the site, since the potentially correct answer "Yes, that's right" is too short to even submit as an answer
Upcoming Events › Discrete Math Seminar Events Search and Views Navigation October 2019 For a graph $H$ and an integer $k\ge1$, the $k$-color Ramsey number $R_k(H)$ is the least integer $N$ such that every $k$-coloring of the edges of the complete graph $K_N$ contains a monochromatic copy of $H$. Let $C_m$ denote the cycle on $m\ge4 $ vertices. For odd cycles, Bondy and Erd\H{o}s in 1973 conjectured that for all $k\ge1$ and $n\ge2$, $R_k(C_{2n+1})=n\cdot 2^k+1$. Recently, this conjecture has been verified to be true for all fixed $k$ and all $n$ sufficiently large…Find out more » For a graph $H$, its homomorphism density in graphs naturally extends to the space of two-variable symmetric functions $W$ in $L^p$, $p\geq e(H)$, denoted by $t_H(W)$. One may then define corresponding functionals $\\|W\\|_{H}:=\|t_H(W)\|^{1/e(H)}$ and $\\|W\\|_{r(H)}:=t_H(\|W\|)^{1/e(H)}$ and say that $H$ is (semi-)norming if $\\|.\\|_{H}$ is a (semi-)norm and that $H$ is weakly norming if $\\|.\\|_{r(H)}$ is a norm. We obtain some results that contribute to the theory of (weakly) norming graphs. Firstly, we show that 'twisted' blow-ups of cycles, which include…Find out more » November 2019 It is a classic result that the maximum weight stable set problem is efficiently solvable for bipartite graphs. The recent bimodular algorithm of Artmann, Weismantel and Zenklusen shows that it is also efficiently solvable for graphs without two disjoint odd cycles. The complexity of the stable set problem for graphs without $k$ disjoint odd cycles is a long-standing open problem for all other values of $k$. We prove that under the additional assumption that the input graph is embedded in a…Find out more » Let $F$ be a graph. We say that a hypergraph $\mathcal H$ is an induced Berge $F$ if there exists a bijective mapping $f$ from the edges of $F$ to the hyperedges of $\mathcal H$ such that for all $xy \in E(F)$, $f(xy) \cap V(F) = \{x,y\}$. In this talk, we show asymptotics for the maximum number of edges in $r$-uniform hypergraphs with no induced Berge $F$. In particular, this function is strongly related to the generalized Turán function $ex(n,K_r, F)$, i.e., the…Find out more » December 2019 The notion of bounded expansion captures uniform sparsity of graph classes and renders various algorithmic problems that are hard in general tractable. In particular, the model-checking problem for first-order logic is fixed-parameter tractable over such graph classes. With the aim of generalizing such results to dense graphs, we introduce classes of graphs with structurally bounded expansion, defined as first-order interpretations of classes of bounded expansion. As a first step towards their algorithmic treatment, we provide their characterization analogous to the characterization of classes of bounded expansion via low…Find out more »
rencsee wrote: What is the remainder when the positive integer n is divided by 2? (1) When n is divided by 13, the remainder is 3 (2) n + 2 is a multiple of 7 $n \geqslant 1\,\,\,\left( * \right)$ $n\,\,\mathop = \limits^? \,\,{\text{even}}\,\,\,\,\,\left[ \begin{gathered} n\,\,{\text{even}}\,\, \Rightarrow \,\,\,{\text{remainder}} = 0\, \hfill \\ \,n\,\,{\text{odd}}\,\, \Rightarrow \,\,\,{\text{remainder}} = 1 \hfill \\ \end{gathered} \right]$ $\left( {1 + 2} \right)\,\,\,\left\{ \matrix{ \,n = 13Q + 3\,\,\,,\,\,\,\left( * \right)\,\,\,Q \ge 0\,\,{\mathop{\rm int}} \hfill \cr \,n + 2 = 7M\,\,\,,\,\,\,\left( * \right)\,\,\,M \ge 1\,\,{\mathop{\rm int}} \, \hfill \cr} \right.$ $7M = n + 2 = 13Q + 5\,\,\,\,\,\, \Rightarrow \,\,\,13Q + 5\,\,{\text{multiple}}\,\,{\text{of}}\,\,7\,\,\,$ $\left\{ \matrix{ \,Q = 5 \,\, \Rightarrow \,\,\,13Q + 5 = 70 = 7 \cdot 10 = n + 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr \,Q = 12 \,\, \Rightarrow \,\,\,13Q + 5 = 161 = 7 \cdot 23 = n + 2\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our high-level "quant" preparation starts here: https://gmath.net
I and my friend are thinking about a smooth analog of Rubik's cube. One idea is the following: Consider the 2-dimensional sphere $S^{2}$. We choose three parameters: $(L, H, \theta)$. Here $L$ is a ray that passes through the origin, $H$ is a plane that intersects with $L$, the sphere, and orthogonal to $L$. $\theta\in [0, 2\pi]$ is an angle that we will rotate the part of the sphere. Here is the picture for the action: Now I have some question about a group $G$ generated by this action on $S^{2}$. Is $G$ has finite dimensional Lie group? If it is, what is a dimension of the group? Is there any interesting and nontrivial relation among elements? For example, in $\mathrm{SO}(3)$, the composition of two rotations is also rotation. But our group clearly doesn't satisfy such things. Edit. As Elkies said, this is an infinite dimensional group, since it contains an infinite dimensional abelian group (which is generated by elements with fixed $L$ and varying $H\&\theta$. But we may think about some subgroups that seem to be finite dimensional, or even finite. Before we do that, there's some minor problem with the definition of $G$ that Jim Conant mentioned. I want to ignore the action that only moves measure 0 parts, such as an equator. To do this, for any two functions $g_{1}, g_{2}:S^{2} \to S^{2}$ we can define an equivalence relation as $$ g_{1}\sim g_{2} \text{ iff } \{x\in S^{2}\,:\, g_{1}(x) \neq g_{2}(x)\} \text{ has a measure 0} $$ Now consider the quotient of the group by this equivalence relation, then we will get more reasonable group. There's some reason why I'm thinking this reasonable - there are some subgroups that I want to think about. Assume that we only allow "semi-sphere turns". So we consider the subgroup of $G$ where the plane $H$ should pass the origin. Is this a finite-dimensional group? I think $\mathrm{SO}(3)$ is a codimension 1 subgroup of this semi-sphere-turn group, but I'm not sure about it. What will be the finite subgroups of this $G$? There's a natural but nontrivial way to construct a finite subgroup of $G$: to think realRubik's cube type puzzles. For example, put Rubik's cube in the center of the sphere and cur the sphere along the planes that correspond to the Rubik's cube's slice faces (I hope you understand what I mean), and turn the pieces along the axis of the cube. In other words, this is a subgroup that is isomorphic to the group of Rubik's cube. We can do the same thing with Pyraminx, Metaminx, Dogic, Skewb, or any regular polytope. I want to know if these covers all the possible finite subgroups of $G$. How about higher dimensional spheres, or other highly symmetric manifolds (such as a torus or hyperbolic spaces)?
If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor. Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$: given a homomorphism $\phi$, we can define an action of $\Gamma$ on $G$: $\gamma\circ g=\phi(\gamma)g$; the number $d_i$ is the largest number such that every $g\in G$ is moved by one of the generators by at least $d_i$ in the word metric). Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space. The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to a hyperbolic group is finite (the latter fact follows because the asymptotic cone of a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree). Update: To make the question more concrete, consider one of the easiest (comparing to the other statements) finiteness results about Diophantine equations. Let $P(x,y)$ be a homogeneous polynomial. If the degree of $P$ is at least 3 and $P$ is not a product of two polynomials with integer coefficients, then for every integer $m\ne 0$ the equation $P(x,y)=m$ has only finitely many integer solutions. It is Thue's theorem. Note that for degree 2 the statement is false because of the Pell equation $x^2-2y^2=1$. The standard proof of Thue's theorem is this. Let the degree of $P$ be 3 (the general case is similar). Represent $P$ as $d(x-ay)(x-by)(x-cy)$ where $a,b,c$ are the roots none of which is rational by assumption. Then we should have $\|(x/y-a)\|\cdot \|(x/y-b)\|\cdot \|(x/y-c)\|=O(1)/\|y^3\|$ for infinitely many integers $x,y$. Then the right hand side can be made arbitrarily small. Note that if one of the factors in the left hand side is small, the other factors are $O(1)$ (all roots are different). Hence we have that $\|x/y-a\|=O(1)/y^3$ (or the same with $b$ or $c$). But for all but finitely many $x,y$ we have $\|x/y-a\|\ge C/y^{5/2+\epsilon}$ for any $\epsilon>0$ by another theorem of Thue (a "bad" approximation property of algebraic numbers), a contradiction. The question is then: is there an asymptotic geometry proof of the Thue theorem.
Recruiting is the essence of College Football. It is one of the most important factors, if not the most important factor that determines how good a season will be. With this in mind, I dug up some data from 247Sports to see if we could use public recruiting event data to see if we could predict where a player was committing to, or at the very least see if a school of (my) interest was doing a good job at recruiting. Summary of 2018 Recruiting On average, recruits receive 13 offers. As we would expect, the higher the recruit is ranked, the more offers he might receive. In the chart below I grouped recruits by 100s according to their rank. During the recruiting process, students get invited to schools (both officially and unofficialy), coaches visit students, and studentets attend summer camps. There are other events in the process but these 4 are the most consistent events documented in our source. So This analysis will be based on just those. MEAN 2.32 School Visits, 6.19 Unoficial Visits, 1.97 Official Visits, 3.04 Coach Visits As we can see, the top 100 tend to get the most love, but that doesn’t mean a ranked 999 recruit doesn’t get any attention. Top FIVE Schools Looking back at the 2018 class, Clemson, Georgia, Ohio State, Texas and USC finished in the top 5. Georgia putting 200+ offers is the first thing that I noticed. UGA dropped over 200 offers when the average of the remianing Top 5 schools was 100. So much for OKG (Our Kind of Guy) for Coach Kirby huh. Second thing I noticed was Ohio State dominating in getting recruits to their campus on visits and School camps. Modeling Using the recruiting event data, let’s see if we can create a model to help us gauge where a recruit will end up playing. Again, the recruiting events tracked are: Offer School Camp Unofficial Visit Official Visit Coach Visit After extensive data manipulations, I create at table where each row represents a unique recruit and school that interacted with him. The metric columns (other than rank, in_state, and won) are the sum of events. name actionSchool rank offer schoolCamp uvisit cvisit ovisit commits decommits in_state won <chr> <fct> <dbl> <int> <int> <int> <int> <int> <int> <int> <dbl> <int>1 Aaron Brule Cincinnati Bearcats 496 1 0 0 0 0 0 0 0 02 Aaron Brule Georgia Bulldogs 496 1 0 1 0 0 1 1 0 03 Aaron Brule Georgia Tech Yellow Jackets 496 1 0 0 0 0 0 0 0 04 Aaron Brule Mississippi State Bulldogs 496 1 1 0 0 1 1 0 0 15 Aaron Brule Oklahoma State Cowboys 496 1 0 0 0 1 0 0 0 06 Aaron Brule TCU Horned Frogs 496 1 0 0 0 1 0 0 0 0 The Method I’m going to use a Logistic Regression Model, which would output the probability of a recruit commiting to a School. $ P = h_{\theta(X)} = \sigma(\theta^{T}) X $ Where the logit is: $logit[\pi(X)] = \beta{1}X_{SC} + \beta_{2}X_{oVisit}+ \beta_{3}X_{uVisit}+ \beta{4}X_{cVisit}+ \beta_{5}X_{inState} $ The Prediction Call:glm(formula = won ~ schoolCamp + uvisit + cvisit + ovisit + in_state, family = binomial(link = "logit"), data = training, control = list(maxit = 50))Deviance Residuals: Min 1Q Median 3Q Max -2.8262 -0.1343 -0.1343 -0.1343 3.0703 Coefficients: Estimate Std. Error z value Pr(>\|z\|) (Intercept) -4.70429 0.12495 -37.648 < 2e-16 *schoolCamp 0.52197 0.18353 2.844 0.00445 uvisit 0.23327 0.05058 4.612 3.99e-06 *cvisit -0.13102 0.08994 -1.457 0.14518 ovisit 4.22311 0.14542 29.040 < 2e-16 in_state 1.29659 0.17052 7.604 2.88e-14 ---Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1(Dispersion parameter for binomial family taken to be 1) Null deviance: 3143 on 7079 degrees of freedomResidual deviance: 1620 on 7074 degrees of freedomAIC: 1632Number of Fisher Scoring iterations: 7 From the coefficients above, we can see why coaches (and fans mostly) get excited when recruits do official visits. This is not too surprising as the NCAA only allows 5 official visits per student. Unofficial visits are good indicators as well, althought they are not quite as strong. Summer Camps on the other hand seem to be very effective in the recruiting journey. Also interesting to me is the strength of the in_state coefficient. As a Texas fan I’ve heard much talk about keeping in state athletes home. The fact that this is the second strongest coefficient shows Texas isn’t the only school emphasizing on this point. Now the question.. how good is this model? Overall, accuracy is at 95.56%, but I must admit this data is heavily skewed towards not winning a recruit. predict<-predict(model,testing,type='response')table(testing$won, predict > 0.5) FALSE TRUE 0 5486 44 1 218 152 As the last part of this analysis, I tried using the top 5 school (Texas, Georgia, USC, OSU, and Clemson) to see if they carried could help improve the regression analysis. Unfortunatley after several hours of feature engineering I couldn’t improve the model. The simplest regression including schools as factors, shows Texas is the only school that can brag about being a significant factor. > summary(model)Call:glm(formula = won ~ schoolCamp + uvisit + ovisit + in_state + TEX + USC + CLE + OSU + GA, family = binomial(link = "logit"), data = training, control = list(maxit = 50))Deviance Residuals: Min 1Q Median 3Q Max -3.3269 -0.1288 -0.1288 -0.1288 3.0969 Coefficients: Estimate Std. Error z value Pr(>\|z\|) (Intercept) -4.78724 0.12926 -37.035 < 2e-16 *schoolCamp 0.36336 0.18779 1.935 0.0530 . uvisit 0.27680 0.05157 5.368 7.98e-08 ovisit 4.23948 0.14461 29.317 < 2e-16 in_state 1.26161 0.17415 7.245 4.34e-13 TEX 1.10482 0.45379 2.435 0.0149 USC -0.11298 0.55735 -0.203 0.8394 CLE 0.62129 0.54551 1.139 0.2547 OSU 0.76816 0.46793 1.642 0.1007 GA 0.01940 0.41965 0.046 0.9631 ---Signif. codes: 0 ‘*’ 0.001 ‘’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1(Dispersion parameter for binomial family taken to be 1) Null deviance: 3165.2 on 7079 degrees of freedomResidual deviance: 1588.7 on 7070 degrees of freedomAIC: 1608.7Number of Fisher Scoring iterations: 7 Conclusion Recruiting is tough for both recruits and schools (and fans!). Predicting where a player will commit and enroll is perhaps just as tough. I do think Texas is doing an outstanding job at recruiting, and unlike Georgia, Texas is being selective in who gets an offer. The idea of getting the best players from within the state worked well for the Texas 2018 recruiting class. However, not all states are as big as Texas, and not all states have the same football talent as Texas so it might be hard for other schools copy this model and be succesful.
From Sipser, Language A is mapping reducible to language B, written $A \leq_m B$, if there is a computable function $f : \Sigma^* \rightarrow \Sigma^*$, where for every $w$, $w ∈ A \Leftrightarrow f(w) \in B$. The function $f$ is called the reduction from A to B. A function is injective if every element in the co-domain is mapped to by at most one element of the domain. I'm confused on how $w \notin A$ are handled. It seems that for $w \in A$ every element must map to a unique $b \in B$. Because $w \in A \Leftarrow f(w) \in B $. New question then, is there any use to having an injective reduction? I've noticed that many proofs of NP-hardness seem to have reductions that are injective (at least on $w \in A$). If reductions are not injective ~~ Why don't we simply define our reduction function $R$ as follows, check if $w \in A$, If $w \in A$ then map to a constant $b \in B$. Else, map to a constant $c \notin B$?
Starting, generally, with a commutative ring $R$ and a rank $2$ projectivemodule $P$ given with a quadratic form $\varphi$ we can form its Clifford algebra$C(P)$. Its even part $S:=C^+(P)$ is then a commutative $R$ algebra of rank $2$.If (which we may assume locally) $P=Re_1+Re_2$ we have that $S$ has basis$1,e_1e_2=:h$ and$h^2=e_1e_2e_1e_2=e_1(\langle e_1,e_2\rangle-e_1e_2)e_2=\langle e_1,e_2\rangle h-e_1^2e_2^2=\langle e_1,e_2\rangle h-\varphi(e_1)\varphi(e_2)$,where $\langle-,-\rangle$ is the bilinear form associated to $\varphi$, giving an explicitquadratic algebra over $R$. Furthermore, $C^-(P)=P$ and it thus becomes an$C^+(P)$-module, explicitly$h\cdot e_1=e_1e_2e_1=\langle e_1,e_2\rangle e_1-e_1^2e_2=\langle e_1,e_2\rangle e_1-\varphi(e_1)e_2$. Furthermore,putting $L:=S/R$ we get an isomorphism $\gamma\colon \Lambda^2_RP \to L$ given by $u\land v \mapsto\overline{uv}$ (note that $u^2=\varphi(u)$ which maps to zero in $L$). Note for futureuse that, putting $[u,v]:=\gamma(u\land v)$, we also have $[[u,v]u,u]=\varphi(u)[u,v]$, wherethe left hand side is well-defined as$[([u,v]+r)u,u]=[[u,v]u,u]+r[u,u]=[[u,v]u,u]$. We have therefore constructedfrom $(P,\varphi)$ a triple $(S,P,\gamma)$, where $S$ is a quadratic (i.e., projective ofrank $2$) $R$-algebra, $P$ an $S$-module projective of rank $2$ as $R$-moduleand an isomorphism $\gamma\colon \Lambda^2_RP \to S/R$ of $R$-modules. Conversely, given such atriple $(S,P,\gamma)$ there is a unique quadratic form $\varphi$ on $P$ such that$[[u,v]u,u]=\varphi(u)[u,v]$ (where again the left hand side is well-defined). It iseasily verified that these constructions are inverse to each other. In case $R=\mathbb Z$ I hope this gives the usual association between forms and modules over orders in quadratic number fields (but I admit shamelessly that I haven't checked it). When $R=k[t]$ I again hope this gives what we want. In terms of algebraic group schemes and torsors we have the following situation.If the quadratic form is perfect, we have that $S$ is an \'etale covering andhence corresponds to an element of $H^1(\mathrm{Spec}R,\mathbb Z/2)$. It is theimage under $O_2 \to \mathbb Z/2$ of the torsor in $H^1(\mathrm{Spec}R,O_2)$corresponding to $\varphi$. As $\mathbb Z/2$ also acts as an automorphism group of$SO_2$ (by conjugation in $O_2$) we can use the torsor in$H^1(\mathrm{Spec}K,\mathbb Z/2)$ to twist $SO_2$ to get $SO(\varphi)$, the connectedcomponent of $O(\varphi)$. Torsors over this group corresponds to isomorphism classesof pairs consisting of a rank $2$ quadratic $R$-forms $\phi$ and an $R$-isomorphism$C^+(\phi)\simeq S$. We have an alternative description of $SO(\varphi)$: We can consider theunits $T:=S^\ast$ of $S$ as an algebraic group and have a (surjective)norm map $T \to \mathbb G_m$ and then $SO(\varphi)$ is the kernel of this norm map.Using the exact sequence $1\to SO(\varphi)\to T\to\mathbb G_m\to1$ we see that an $SO(\varphi)$-torsoris given by a projective $S$-module $P$ of rank $1$ together with thechoice of an isomorphism $\Lambda^2_RP\simeq R$. [Added] Overcoming some of my laziness I did the calculation in the integralcase ($R=\mathbb Z$ although the only thing we use is that $2$ is invertible):Start with the quadratic form $Cx^2+Bxy+Ay^2$ (the switch between $A$ and $C$ isto make my definition come out the same as the formula of the question). We thenhave $h^2=Bh-AC$ so that $h=\frac{B-\sqrt{\Delta}}{2}$. We have $h\cdot e_1=Be_1-Ce_2$ and$h\cdot e_2=Ae_1$. This is just an abstract module over $S$ but we can make it afractional ideal by mapping $e_2$ to $A$, then $e_1$ maps to $h$ so that thefractional ideal is indeed $(A,\frac{B-\sqrt{\Delta}}{2})$. [Added later] One could say that the association of the ideal $(A,h)$ to thequadratic form $Cx^2+Bxy+Ay^2$ over the ring $R[h]/(h^2-Bh+AC)$ is an answer tothe question which workds in all classical cases ($R=\mathbb Z$ and $R=k[T]$). The reason that this looks simpler than the traditional ($R=\mathbb Z$) answer is that we let the presentation of the quadratic order depend on thequadratic form itself. Usually we have fixed the quadratic order and want to consider all forms with this fixed quadratic order as associated order. This means that we should fix some normal form for the order and then express the ideal in terms of this normal. When $R=\mathbb Z$ orders are in bijection with their discriminants $\Delta$ which are actual integers (as the discriminant is well-determined modulo squares of units). A normal form for the order is $\mathbb Z+\mathbb Z\sqrt{\Delta}$ or $\mathbb Z+\mathbb Z(\sqrt{\Delta}+1)/2$ dependingon whether $\Delta$ is even or odd. A curious feature of the form $(A,\frac{B-\sqrt{\Delta}}{2})$ of the ideal is that the distinction between the odd and even case is not apparent. However, the crucial thing is that it expresses a $\mathbb Z$-basis for the ideal in terms of the canonical form of the quadratic order. The case of $R=k[T]$ for $k$ a field of odd characteristic is somewhat deceptiveas the discriminant $\Delta=B^2-4AC$ is not quite an invariant of the quadratic orderas there are units different from $1$ that are squares (except when $k=\mathbb Z/3$!). Hence the formula $(A,\frac{B-\sqrt{\Delta}}{2})$ is not quite of the same nature as for the $\mathbb Z$ case as there is a very slight dependence of $\Delta$ on the form (and not just on the order). We could fix that by choosing coset representatives for the squares as a subgroup of $k^\ast$ and then the formula for the ideal would would take the form $(A,\frac{B-\lambda\sqrt{\Delta}}{2})$ where $\Delta$ nowhas been normalised so as to have its top degree coefficient to be a coset representative. The case when $k$ has characteristic $2$ is more complicated. We get and orderof the form $k[T][h]/(h^2+gh+f)$ but the question of a normal form is trickier.The discriminant of the order (in the sense of the discriminant of the traceform) is equal to $g^2$ and as all elements of $k$ are squares we can normalisethings so that $g$ is monic. However, the order is not determined by itsdiscriminant. This can be seen already in the unramified case when $g=1$ when anormal form for $f$ is that it contain no monomials of even degree and theconstant term is one of a chosen set of coset representives for the subgroup$\{\lambda^2+\lambda\}$ of $k$. We can decide to rather arbitrarily fix a generator $H$ forevery order with $H^2=GH+F$ where one sensible first normalisation is that $G$be monic (for which we have to assume that $k$ is perfect). Then we have that $h=H+a$, with $a\in k[T]$, and the ideal would be$(A,H+a)$. There is a particular (arguably canonical) choice of $H$: We assume $G$ is monicand then can write uniquely $G=G_1G_2\cdots G_n$ with $G_i$ monic and$G_{i+1}\|G_i$. The normal form is then that $F$ have the form$F=G_1F_1+G_1^2G_2F_2+\cdots+G_1^2G_2^2\cdots G_nF_n+G^2F'$ where $\deg F_i<\deg G_i$ and$F'$ contain no square monomials and its constant term belongs to a chosen set ofcoset representatives for $\{\lambda^2+\lambda\}$ in $k$. One further comment relating to the classical formulas. When passing from afractional ideal to a quadratic form one classically divides by the norm of theideal (as is done in KConrad's reply). This means that the constructed form isprimitive, i.e., the ideal generated by its values is the unit ideal. Hence ifone starts with a quadratic form, passes to the ideal and then back to aquadratic form one does not end up with the same form if the form is notprimitive. Rather the end form is the "primitivisation" where one has dividedthe form by a generator for its ideal of values. This of course only makes senseif the base ring is a PID. Even for a general Dedekind ring if one wants to workwith primitive forms one has to accept quadratic forms that take values ingeneral rank $1$ modules (i.e., fractional ideals). The approach above makes another choice. It deals only with $R$-valued forms butaccepts non-primitive ones. This would seem to lead to a contradiction asthe classical construction leads to a bijection between modules and primitiveforms and the above leads to a bijection between modules and arbitraryforms. There is no contradiction however (phew!) as the above construction leadsto smaller orders than the classical one in the non-primitive case. Classically what one really works with (when $R$ is a PID) are lattices in $L$(the fraction field of $S$), where a lattice $M$ is a finitely generated$R$-submodule of $K$ containing a $K$-basis ($K$ the fraction field of $R$) for$L$. The order one associates to $M$ is the subring of $L$ stabilising $M$. Whenthe quadratic form $\varphi$ is non-primitive $C^+(P)$ is not equal to thiscanonically associated order but is a proper suborder by it. Dividing by agenerator for the ideal of values gives us a primitive form for which $C^+(P)$is equal to the canonical order. Finally there is a particular miracle that occurs for lattices in quadraticextensions (of the fraction field of a Dedekind ring $R$), $M$ as a module over its stabilisation order isprojective. This is why classes of primitive forms with fixed order are inbijection with the class group of the order.
I want to understand what quantum entanglement is and what role does it play in quantum error correction. NOTE:As per the suggestions of @JamesWootton and @NielDeBeaudrap, I have asked a separate question for the classical analogy here. Quantum Computing Stack Exchange is a question and answer site for engineers, scientists, programmers, and computing professionals interested in quantum computing. It only takes a minute to sign up.Sign up to join this community Classical correlations between variables occur when the variables appear random, but whose values are found to systematically agree (or disagree) in some way. However, there will always be someone (or something) that 'knows' exactly what the variables are doing in any given case. Entanglement between variables is the same, except for the last part. The randomness is truly random. Random outcomes are completely undecided until the time of measurement. But somehow the variables, though they may be separated by galaxies, still know to agree. So what does this mean for error correction? Let's start by thinking about error correction for a simple bit. When storing a classical bit, the kinds of errors you need to worry about are things like bit flips and erasures. So something might make your 0 become a 1, or vice-versa. Or your bit might wander off somewhere. To protect the information, we can ensure that our logical bits (the actual information we want to store) are not just concentrated on single physical bits. Instead, we spread it out. So we could use a simple repetition encoding, for example, where we copy our information across many physical bits. This lets us still get our information out, even if some of the physical bits have failed. This is the basic job of error correction: we spread our information out, to make it hard for errors to mess it up. For qubits, there are more kinds of error to worry about. For example, you may know that qubits can be in superposition states, and that measurements change these. Unwanted measurements are therefore another source of noise, caused by the environment interacting with (and so in some sense 'looking at' our qubits). This type of noise is known as decoherence. So how does this affect things? Suppose we use the repetition encoding with qubits. So we replace the $\|0\rangle$ in our desired logical qubit state with $\|000...000\rangle$, repeated across many physical qubits, and replace the $\|1\rangle$ with $\|111...111\rangle$. This again protects against bit flips and erasures, but it makes it even easier for stray measurements. Now the environment measure whether we have $\|0\rangle$ or $\|1\rangle$ by looking at any of many qubits. This will make the effect of decoherence much stronger, which is not what we wanted at all! To fix this, we need to make it hard for decoherence to disturb our logical qubit information, just as we made it hard for bit flips and erasures. To to this, we have to make it harder to measure our logical qubit. Not too hard that we can't do it whenever we want to, of course, but too hard for environment to do easily. This means ensuring that measuring a single physical qubit should tell us nothing about the logical qubit. In fact, we must make it so that a whole bunch of qubits need to be measured and their results compared to extract any information about the qubit. In some sense, it is a form of encryption. You need enough pieces of the puzzle to have any idea what the picture is. We could try to do this classically. Information could be spread out in complex correlations among many bits. By looking at enough of the bits and analyzing the correlations, we can we extract some information about the logical bit. But this would not be the only way to get this information. As I mentioned before, classically there is always that someone or something that already knows everything. It doesn't matter whether it is a person, or just the patterns in the air caused when the encryption was carried out. Either way, the information exists outside of our encoding, and this is essentially an environment that knows everything. Its very existence means that decoherence has occurred to an irreparable degree. So that's why we need entanglement. With it, we can hide the information away using correlations in the true and unknowable random outcomes of quantum variables. Entanglement is a natural part of quantum information and quantum computation. If it isn't present --- if you try to do things in such a way that entanglement does not arise --- then you get no benefit from quantum computation. And if a quantum computer is doing something interesting, it will produce a lot of entanglement, at least as a side-effect. However, this does not mean that entanglement is "what makes quantum computers go". Entanglement is like the spinning gears of a machine: nothing happens if they aren't turning, but that doesn't mean that having those gears spin quickly is enough to make the machine do what you want. (Entanglement is a primitive resource in this way for communication, but not for computation as far as anyone has seen.) This is as true for quantum error correction as it is for computation. Like all forms of error correction, quantum error correction works by distributing information around a larger system, in particular in the correlations of certain measurable pieces of information. Entanglement is just the usual way in which quantum systems become correlated, so it should come as no surprise that a good quantum error correction code then involves a lot of entanglement. But that doesn't mean that trying to "pump your system full of entanglement", like some sort of helium balloon, is something which is useful or meaningful to do to protect quantum information. While quantum error correction is sometimes described vaguely in terms of entanglement, more important is how it involves parity checks using different 'observables'. The most important tool for describing this is the stabiliser formalism. The stabiliser formalism can be used to describe some states with large amounts of entanglement, but more importantly it allows you to reason about multi-qubit properties ("observables") fairly easily. From that perspective, one can come to understand that quantum error correction is much more closely related to low-energy many-body physics of spin-Hamiltonians, than just entanglement in general. There is no classical equivalent to entanglement. Entanglement is perhaps best understood using Dirac (bra-ket) notation. Each qubit can be in the (ket) state $\left\|0\right>$ or in the state $\left\|1\right>$ or in a superposition $\alpha\left\|0\right> + \beta \left\|1\right>$ where $\alpha$ and $\beta$ are complex numbers that fulfill $\|\alpha\|^2 + \|\beta\|^2 = 1$. If you have two qubits, the basis states of the 2-qubit system are $\left\|0\right> \otimes \left\|0\right>$, $\left\|0\right> \otimes \left\|1\right>$, $\left\|1\right> \otimes \left\|0\right>$, and $\left\|1\right> \otimes \left\|1\right>$. To simplify the notation, physicists often write these as $\left\|00\right>$, $\left\|01\right>$, $\left\|10\right>$, and $\left\|11\right>$. So being in state $\left\|01\right>$ means that the first qubit is in state $\left\|0\right>$ and the second qubit is in state $\left\|1\right>$. Now consider a superposition of the kind $\alpha \left\|01\right> + \beta \left\|10\right>$. This means that the first qubit is in state $\left\|0\right>$ with probability $\|\alpha\|^2$ and in state $\left\|1\right>$ otherwise, whilst the second qubit is always in the opposite state that the first one is in: The two particles are entangled. It is unimportant that in this example, the entangled qubits happen to be in opposite states: They might as well be in the same state and still be entangled. What matters is that their states are not independent from each other. This has caused major headaches for physicists because it means that qubits (or the particles carrying them) cannot simultaneously have strictly local properties and be governed by a concept called realism (reflect their states as intrinsic property). Einstein famously called the resulting paradox (if you still assume locaility and realism) "spooky action at a distance." Entanglement does not play a special role in quantum error correction: Error correction must work for every state in the computational basis (which does not have entanglement). Then it automatically works also for superpositions of these states (which may be entangled states). For a certain class of codes called pure, the presence of entanglement is a necessary and sufficient requirement for quantum error correction, i.e. to correct all errors affecting up to a certain number of subsystems. Recall the Knill-Laflamme conditions for a quantum error correcting code to be able to detect a certain set of errors $\{E_\alpha\}$: pick any orthonormal basis $\|i_\mathcal{Q}\rangle$ that spans the code-space. Then the error $E_\alpha$ can be detected if and only if \begin{equation} \langle i_\mathcal{Q} \| E_\alpha \| j_\mathcal{Q} \rangle = \delta_{ij} C(E_\alpha). \quad\quad\quad (1) \end{equation} Note that $C(E_\alpha)$ is a constant that only depends on the specific error $E_\alpha$, but not on $i$ and $j$. (This means that the error $E_\alpha$ affects all states in the code subspace in the same way). In the case of $C(E_\alpha) \propto \operatorname{tr}(E_\alpha)$, the code if called pure. Many stabilizer codes considered are of this form, not however Kitaev's toric code. Let us assume an error-model where we are only interested in on how many subsystems our errors act. If our code can detect all errors $E_\alpha$ that act on at most $(d-1)$ subsystems nontrivially, the code is said to have distance $d$. As a consequence, any combination of errors that affect up to $\lfloor(d-1)/2\rfloor$ subsystems can be corrected. What follows is that for pure codes of distance $d$, every vector lying in the code subspace must be maximally entangled across any bipartition whose smaller subsystem has at most size $(d-1)$: to see this, note that for every error $E_\alpha \neq \mathbb{1}$ and vector $\|v_\mathcal{Q} \rangle$ in the subspace (our ONB was chosen arbitrary), one has that \begin{equation} \langle E \rangle = \operatorname{tr}(E \|v_\mathcal{Q} \rangle\langle v_\mathcal{Q}\|) = \langle v_\mathcal{Q} \|E_\alpha\|v_\mathcal{Q} \rangle = \operatorname{tr}(E) = 0. \end{equation} Thus all observables on at most $(d-1)$ parties are vanishing, and all reduced density matrices on $(d-1)$ parties must be maximally mixed. This implies that $\|v_\mathcal{Q}\rangle$ is maximally entangled for any choice of $(d-1)$ parties vs. the rest. Addendum (for the sufficiency):As an equivalent definition to Eq. (1): All errors $E_\alpha$ acting on less than $d$ places can be detected, if and only if for all $\|v\rangle, \|w\rangle$ in the code subspace following condition holds, \begin{equation} \langle v\| E_\alpha \| v\rangle = \langle w \| E_\alpha\| w\rangle. \end{equation} In the case of pure codes, above expression vanishes. It follows that if one has a subspace where every state is maximally entangled for all bipartitions of (d-1) parties vs. the rest, then it is a pure code of distance $d$. tl;dr: For a large distance $d$, a pure code consists of highly entangled states. It is a necessary and sufficient requirement for the existence of the code. Addendum: we looked into this question further, details can be found in the paper Quantum Codes of Maximal Distance and Highly Entangled Subspaces. There is a trade-off: the more errors a quantum code can correct, the more entangled must every vector in the code space be. This makes sense, because if the information where not distributed amongst many particles, the environment - by reading a few qubits - could recover the message in the code space. This would then necessarily destroy the coded message, due to the no-cloning theorem. Thus a high distance needs high entanglement. Here is a way to think about the role of entanglement in quantum codes which I think is complementary to Felix Hubers response. Suppose that we take a maximally entangled state $\|\Psi\rangle_{RQ}$ and record the $Q$ system into some quantum error-correcting code. Suppose the code records $Q$ into subsystems $S_1,S_2,S_3$ such that erasure of any one subsystem can be corrected for (I've taken a simple example, but generalizations are possible). Then, there is an entropic way of thinking about the error correction conditions (as compared to the more algebraic Knill-Laflamme conditions). Specifically, if $$I(R:S_3) = 0$$ Then it follows that $Q$ can be recovered from $S_1S_2$. See for example arXiv:quant-ph/0112106 for a nice presentation of this fact. Using this entropic approach to error correction there are fairly direct routes to understanding entanglement in codes. For instance, we can prove that, $$I(S_1S_2:S_3) \geq 2\log d_R$$ as follows. First we write out this mutual information in terms of its definition, $$I(S_1S_2:S_3) = S(S_1S_2) + S(S_3) - S(S_1S_2S_3)$$ We'll introduce a purifying system $X$, so that the state on $RS_1S_2S_3X$ is pure. Then using purity we can write $$I(S_1S_2:S_3) = S(S_3XR) + S(S_3) - S(XR)$$ Note that since we can recover $Q$ from $S_1S_2$, $I(R:S_3X)=I(R:X)=0$. Using this in the above $$I(S_1S_2:S_3) = S(S_3\|X) + S(S_3)$$ Finally, we can bound the right hand side here below by $2 \log d_R$. The intuition behind how we can do this is that $S_3$ is ``significant'' in the sense that there is a set of shares (say $S_1$) which itself reveals no information about $Q$, but together with $S_3$ allows $Q$ to be recovered. Given this, we expect $S_3$ must carry $2\log d_R$ of entropy, since transferring it can be used to establish $2\log d_R$ worth of entanglement. A similar intuition appears in arXiv:quant-ph/0608223. More precisely we consider the quantity $I(R:S_1S_3) - I(R:S_1)$, which some basic manipulations reveal $$I(R:S_1S_3) - I(R:S_1) = S(S_3\|S_1) + S(S_3\|XS_2) \leq S(S_3) + S(S_3\|X)$$ But then we notice $I(R:S_1S_3) \geq 2\log d_R$ since $S_1S_3$ allows recovery of $Q$, while $I(R:S_1)=0$ by the entropic error correction condition. This lower bounds $S(S_3) + S(S_3\|X)$ and so lower bounds $I(S_1S_2:S_3)$.
Assume, an extension of the lambda calculus with terms $t$ and values $v$ is defined in big-step operational semantics with evaluation relation $t \Downarrow v$. It is intuitive to assume that $\beta$-equivalence holds, e.g. (λx. t) unit $\equiv_\beta$ t when x is not free in t It is however unclear to me, how $\equiv_\beta$ can be precisely defined in such a setting: Obviously, there is no small-step relation $\rightarrow_\beta$ that can be used to express a single reduction step, since the evaluation relates terms and values. On the other hand, beta-equivalent terms do not evaluate to syntactically equal terms (e.g. when comparing abstractions which already are values). So how does one define this equivalence in such a case?
I am teaching a course on knots for the first time, and this led me to play with an approach which I have not seen in the literature. I would be surprised if no one had used it before, so I am asking whether this approach has a name, and if there are any relevant references. The basic idea is like this. Let $\Lambda$ be the group freely generated by $\chi$ and $\rho$ subject to $\chi^2=\rho^4=1$. We will formulate everything in terms of finite sets with $\Lambda$-action. Let $U\subset\mathbb{R}^2\subset S^2$ be a link universe, i.e. a planar projection of a link, without any information about undercrossings or overcrossings (for the moment). Suppose for simplicity that there are no isolated circles. Let $A$ be the collection of pairs consisting of an arc in the diagram, together with a choice of direction for that arc. Define $\chi(a)$ to be the same arc with the opposite direction. Define $\rho(a)$ to be the arc obtained by "rotating $a$ through a quarter turn anticlockwise around the starting point", in an evident sense. This gives a $\Lambda$-action on $A$, and many of the standard things can be formulated in terms of this. The following three facts are enough to show that the combinatorial structure arises from a diagram in $S^2$: The subgroup $\langle\rho\rangle$ acts freely The stabiliser of every point is contained in the kernel of the map $c\colon\Lambda\to\{\pm 1\}$ given by $c(\chi)=c(\rho)=-1$. If $B\subseteq A$ is a $\Lambda$-orbit with $\|B\|=n$ and $\|B/\langle{\rho^{-1}\chi}\rangle\|=m$ then $4m=n+8$. (This is because we can use $B$ to construct a CW structure on the ambient $S^2$, and then calculate its Euler characteristic.) The components are the orbits for $\langle\rho^2\chi\rangle$, crossing information is given by a map $\omega\colon A\to\{\pm 1\}$ with $\omega\rho=-\omega$, an orientation is a map $\delta\colon A\to\{\pm 1\}$ with $\delta\rho^2=\delta\chi=-\delta$, a Reidemeister move of type $k\in\{1,2,3\}$ can be performed whenever you have an element $a$ with $(\rho^{-1}\chi)^k(a)=a$, a connected sum splitting is given by an element $a$ with $(\rho^{-1}\chi)^i(a)=(\rho\chi)^j(a)\neq a$ for some $i,j>0$, and so on.
How can I make the symbol of product operator (pi caps) making the extremes remain the same above and below the symbol? (I do not want the extremes on the right side of the symbol) I wanna write this symbol with that conditions in the simplest possible way. Please be aware that the default behaviour in displaymath mode is for the limits to appear above and below the symbol: \[ \prod_{i = 1}^{n} a_{i} \quad \sum_{i = 1}^{n} a_{i} \quad \lim_{n \to \infty} n^{2}\] The only exception to this is integrals: \[ \int_{0}^{\infty} x^{2} \, dx\] The default behaviour in inline math mode is for the limits to appear to the side, e.g. $\prod_{i = 1}^{n} a_{i} \quad \sum_{i = 1}^{n} a_{i} \quad \lim_{n \to \infty} n^{2}$ The reason for this is to avoid irregular spacing between lines and excessive white space. Compare: With: However, to force the limits to appear above and below in inline mathmode you can use \limits as suggested by Christian Hupfer. $\prod\limits_{i = 1}^{n} a_{i} \quad \sum\limits_{i = 1}^{n} a_{i}\quad \lim\limits_{n \to \infty} n^{2}$ And to force the limits to appear to the side in displaymath mode, you can use \nolimits \[ \prod\nolimits_{i = 1}^{n} a_{i} \quad \sum\nolimits_{i = 1}^{n} a_{i} \quad \lim\nolimits_{n \to \infty} n^{2}\] Do you mean this (which is mathematical a little bit incorrect ;-)) The usage of \prod\limits^{upper}_{lower} will produce the product operator with the limits set below and above. Omitting \limits will generate the limits set aside, if \displaystyle isn't used. In displayed equations the limits are set below and above in amsmath, since the sumlimits option is set by default. If this is not desired, use nosumlimits (valid for \sum, \prod, \coprod, \bigoplus and \bigotimes etc operators, but not for integrals) The 'correct' usage should follow the guideline that limits above and below should be used only in purely mathematical context, not inline with other text, since the line height is stretched here. \documentclass{article}\usepackage{mathtools}\begin{document}$n! = \prod\limits^{n}_{k=1} k$ versus $n! = \prod^{n}_{k=1} k$However $n!=\displaystyle\prod^{n-1}_{k=0} (n-k)$ is the better definition\end{document}
Find the differential of each of the following expressions, ie. zap each of the following with $d$: (mbZapWithD) Find the differential of special functions. $$f=3x-5z^2+2xy$$ $$g=\sin^2(\omega t)$$ $$h=\frac{\mu B}{k_B T}$$ $$j=\exp\left(\frac{\mu B}{k_B T}\right)$$ $$k=\ln\left(e^{\frac{\mu B}{k_B T}} +e^{-\frac{\mu B}{k_B T}}\right)$$ (mbDifferentials) Finding the differential abstractly. Find the differential of $R$ where\ldots $R(B,C) = B^2 + C^2$ $R(B,C) = BC$ $R(B,C) = e^{B^2+C^2}$ $R(B,C) = e^{S(B,C)}$ $R(B,C) = ST$, where $S = S(B,C)$ and $T = T(B,C)$ (mbDerivVSDiff) Find the differential of special functions. Consider $V=-A\cos(kx)$, where $A$ is a constant with dimensions of electrostatic potential. Find $dV$ and $\frac{dV}{dx}$. Discuss the differences and similarities between the two. What are the dimensions of $k$, $V$, $dx$, $dV$, and $\frac{dV}{dx}$? (mbMathematicaContours) Plot contours, Change of Variable. On this problem, you will use Mathematica to graph and interpret some functions following the steps below. Mathematica can be found in the computers on campus, or you can download a student license for free, or even use online tools like sandbox.open.wolframcloud.com. Please include both your graphs and your source code to make grading easier! Create a contour plot of the function $G = A^4 + B^2$. Play around with the options until you have a graph that has all of the following features: The axes are labeled. The contours are numbered. The plot is titled. There are 8 contours. The colored shading on the contours is different from the default. Describe how your contour plot changes if you modify the plot range ( i.e.,zooming in or out)? Suppose that $5A = 4U + 3V$ and $5B = 4V - 3U$. Make a contour plot of $G$ in terms of the new variables $U$ and $V$. Label and present your new graph as described above. Make a contour plot of $G$ in terms of the variables $A$ and $V$. Label and present your new graph as described above. Explain why all three of the contour plots you have made are valid representations of $G$. What is different about the three graphs? (CoffeeAndBagels) A thermo-like system. Requires Clairaut's theorem. \newcommand\juice{\ensuremath{\mathcal{O}}} \newcommand\cocoa{\ensuremath{\mathcal{C}}} \newcommand\priceorange{\ensuremath{\mathcal{P_O}}} \newcommand\pricecocoa{\ensuremath{\mathcal{P_C}}} In economics, the term utilityis roughly related to overall happiness. Many things affect your happiness, including the amount of money you have and the amount of coffee you drink. We cannot directly measure your happiness, but we canmeasure how much money you are willing to give up in order to obtain coffee or bagels. If we assume you choose wisely, we can thus determine that your happiness increases when you decrease your amount of money by that amount in exchange for increasing your coffee consumption. Thus money is a (poor) measure of happiness or utility. Money is also a nice quantity because it is conserved—just like energy! You may gain or lose money, but you always do so by a transaction. (There are some exceptions to the conservation of money, but they involve either the Fed, counterfeiters, or destruction of cash money, and we will ignore those issues.) In this problem, we will assume that you have bought all the coffee and bagels you want (and no more), so that your happiness has been maximized. Thus you are in equilibrium with the coffee shop. We will assume further that you remain in equilibrium with the coffee shop at all times, and that you can sell coffee and bagels back to the coffee shop at cost.\footnote{Yes, this is ridiculous. It would be slightly less ridiculous if we were talking about nations and commodities, but also far less humorous.} Thus your savings $S$ can be considered to be a function of your bagels $B$ and coffee $C$. In this problem we will also discuss the prices $P_B$ and $P_C$, which you may notassume are independent of $B$ and $C$. It may help to imagine that you have The prices of bagels and coffee $P_B$ and $P_C$ have derivative relationships between your savings and the quantity of coffee and bagels that you have. What are the units of these prices? What is the mathematical definition of $P_C$ and $P_B$? Write down the total differential of your savings, in terms of $B$, $C$, $P_B$ and $P_C$. Use the equality of mixed partial derivatives (Clairut's theorem) to find a relationship between $P_B$, $P_C$, $B$ and $C$. Write this relationship mathematically, and also describe in words what it means. Solve for the total differential of your net worth. Once again use Clairut's theorem considering second derivatives of $W$ to find a different partial derivative relationship between $P_B$, $P_C$, $B$ and $C$.
STANDARD DEVIATION This post is a part of [GMAT MATH BOOK ]created by: walkeredited by: bb , Bunuel -------------------------------------------------------- Get The Official GMAT Club's App - GMAT TOOLKIT 2 . The only app you need to get 700+ score! [iOS App ] [Android App ] --------------------------------------------------------Definition Standard Deviation (SD, or STD or $\sigma$) - a measure of the dispersion or variation in a distribution, equal to the square root of variance or the arithmetic mean (average) of squares of deviations from the arithmetic mean . $variance = \frac{\sum(x_i - x_{av})^2}{N}$ $\sigma = \sqrt{variance} = \sqrt{\frac{\sum(x_i - x_{av})^2}{N}}$ In simple terms, it shows how much variation there is from the "average" (mean). It may be thought of as the average difference from the mean of distribution, how far data points are away from the mean. A low standard deviation indicates that data points tend to be very close to the mean, whereas high standard deviation indicates that the data are spread out over a large range of values.Properties $\sigma \ge 0$; $\sigma = 0$ only if all elements in a set is equal; Let standard deviation of $\{x_i\}$ be $\sigma$ and mean of the set be $\m$: Standard deviation of $\{\frac{x_i}{a}\}$ is $\sigma^'=\frac{\sigma}{a}$. Decrease/increase in all elements of a set by a constant percentage will decrease/increase standard deviation of the set by the same percentage. Standard deviation of $\{x_i+a\}$ is $\sigma^'=\sigma$. Decrease/increase in all elements of a set by a constant value DOES NOT decrease/increase standard deviation of the set. if a new element $y$ is added to $\{x_i\}$ set and standard deviation of a new set $\{\{x_i\},y\}$ is $\sigma^'$, then: 1) $\sigma^'>\sigma$ if $\|y-\m\|>\sigma$ 2) $\sigma^'=\sigma$ if $\|y-\m\|=\sigma$ 3) $\sigma^'<\sigma$ if $\|y-\m\|<\sigma$ 4) $\sigma^'$ is the lowest if $y=\m$Tips and Tricks GMAC in majority of problems doesn't ask you to calculate standard deviation. Instead it tests your intuitive understanding of the concept. In 90% cases it is a faster way to use just average of $\|x_i-x_{av}\|$ instead of true formula for standard deviation, and treat standard deviation as "average difference between elements and mean ". Therefore, before trying to calculate standard deviation, maybe you can solve a problem much faster by using just your intuition. Advance tip. Not all points contribute equally to standard deviation. Taking into account that standard deviation uses sum of squares of deviations from mean, the most remote points will essentially contribute to standard deviation. For example, we have a set A that has a mean of 5. The point 10 gives $(10-5)^2=25$ in sum of squares but point 6 gives only $(6-5)^2=1$. 25 times the difference! So, when you need to find what set has the largest standard deviation, always look for set with the largest range because remote points have a very significant contribution to standard deviation. ExamplesExample #1Q : There is a set $\{67,32,76,35,101,45,24,37\}$. If we create a new set that consists of all elements of the initial set but decreased by 17%, what is the change in standard deviation?Solution: We don't need to calculate as we know rule that decrease in all elements of a set by a constant percentage will decrease standard deviation of the set by the same percentage. So, the decrease in standard deviation is 17%.Example #2Q : There is a set of consecutive even integers. What is the standard deviation of the set? (1) There are 39 elements in the set. (2) the mean of the set is 382.Solution: Before reading Data Sufficiency statements, what can we say about the question? What should we know to find standard deviation? "consecutive even integers" means that all elements strictly related to each other. If we shift the set by adding or subtracting any integer, does it change standard deviation (average deviation of elements from the mean)? No. One thing we should know is the number of elements in the set, because the more elements we have the broader they are distributed relative to the mean. Now, look at DS statements, all we need it is just first statement. So, A is sufficient.Example #3Q : Standard deviation of set $\{23,31,76,45,16,55,54,36\}$ is 18.3. How many elements are 1 standard deviation above the mean?Solution: Let's find mean: $\m=\frac{23+31+76+45+16+55+54+36}{8}=42$ Now, we need to count all numbers greater than 42+18.3=60.3. It is one number - 76. The answer is 1.Example #4Q : There is a set A of 19 integers with mean 4 and standard deviation of 3. Now we form a new set B by adding 2 more elements to the set A. What two elements will decrease the standard deviation the most? A) 9 and 3 B) -3 and 3 C) 6 and 1 D) 4 and 5 E) 5 and 5Solution: The closer to the mean, the greater decrease in standard deviation. D has 4 (equal our mean) and 5 (differs from mean only by 1). All other options have larger deviation from mean.Normal distribution It is a more advance concept that you will never see in GMAT but understanding statistic properties of standard deviation can help you to be more confident about simple properties stated above. In probability theory and statistics, the normal distribution or Gaussian distribution is a continuous probability distribution that describes data that cluster around a mean or average. Majority of statistical data can be characterized by normal distribution. $\m-\sigma<x<\m+\sigma$ covers 68% of data $\m-2\sigma<x<\m+2\sigma$ covers 95% of data $\m-3\sigma<x<\m+3\sigma$ covers 99% of dataOfficial GMAC Books:The Official Guide , 12th Edition: DT #9; DT #31; PS #199; DS #134; The Official Guide , 11th Edition: DT #31; PS #212; Generated from [GMAT ToolKit 2 ]Resources Bunuel's post with PS SD-problems: [PS Standard Deviation Problems ] Bunuel's post with DS SD-problems: [DS Standard Deviation Problems ] -------------------------------------------------------- Get The Official GMAT Club's App - GMAT TOOLKIT 2 . The only app you need to get 700+ score! [iOS App ] [Android App ] -------------------------------------------------------- Attachment: Math_SD_graph.png [ 6.04 KiB \| Viewed 141006 times ] Attachment: Math_SD_graph_low.png [ 5.92 KiB \| Viewed 140893 times ] Attachment: Math_SD_graph_high.png [ 7.06 KiB \| Viewed 141007 times ] Attachment: Math_SD_normal.png [ 7.67 KiB \| Viewed 141131 times ] Attachment: Math_icon_std.png [ 2.67 KiB \| Viewed 139667 times ] _________________ HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ \| Limited GMAT/GRE Math tutoring in Chicago
"The Mathematical Ninja is currently on sabbatical. Leave a message after the tone... or else!" Oh dear! How are we going to figure out $e^e$ now? Let alone $e^{-\frac{1}{e}}$? We'll just have to roll up our sleeves and get our thinking hats on, that's all. OK, $e^e$ First of all,Read More → The Mathematical Ninja peered at the problem sheet: Given that $(1+ax)^n = 1 - 12x + 63x^2 + \dots$, find the values of a and n Barked: “$n=-8$ and $a=\frac{3}{2}$.” The student sighed. “I get no marks if I just write down the answer.” Snarled: “You get noRead More → Via @markritchings, an excellent logs problem: If $a = \log_{14}(7)$ and $b = \log_{14}(5)$, find $\log_{35}(28)$ in terms of $a$ and $b$. One of the reasons I like this puzzle is that I did it a somewhat brutal way, and once I had the answer, a much neater way jumpedRead More → “I beg your pardon?!” yelled the Mathematical Ninja. The terribly well-dressed gentleman stood his ground. “I said, sensei, I would work $0.8^{10}$ out differently.” A sarcastic laugh. “This, I have to see!” “Well, $8^{10} = 2^{30}$, which is about $10^{9}$.” “About.” “Obviously, we can do better with the binomial: $2^{10}$Read More → It took the Mathematical Ninja a little longer than normal; the student had managed to rummage around in her bag and lay a finger on the calculator before simultaneously feeling her arm pulled away by a lasso and hearing "0.3805. Or, as a one-off, since the question is asking forRead More → “Here’s a quick one,” suggested a fellow tutor. “Prove that $2^{50} < 3^{33}$.” Easy, I thought: but I knew better than to say it aloud. First approach “I know that $9 > 8$,” I said, checking on my fingers. “So if $2^3 < 3^2$, then $2^{150} < 3^{100}$ and $2^{50}Read More → The student, at the third time of asking, navigated the perilous straits of negative powers and fractions of $\pi$ and came to rest, exhausted, on the answer: "$r^3 = \frac{500}{\pi}$," he said. The Mathematical Ninja stopped poking him with the foam sword (going soft? perhaps. Or perhaps this student neededRead More → The student swam away, thinking almost as hard as he was swimming. The cube root of four? The square root was easy enough, he could do that in his sleep. But the cube root? OK. Breathe. It's between 1 and 2, obviously. What's 1.5 cubed? The Mathematical Ninja isn't goingRead More → A professor - according to Reddit - asked their class how many people you'd need to have in a room to be absolutely certain two of them would have Social Security numbers1 ending in the same four digits (in the same order). 10001, obviously. How about a probability of 99.9%?Read More → Glancing over sample papers for the new GCSE, I stumbled on this: Zahra mixes 150g of metal A and 150g of metal B to make 300g of an alloy. Metal A has a density of $19.3 \unit{g/cm^3}$. Metal B has a density of $8.9 \unit{g/cm^3}$. Work out the density ofRead More →
Content Access the CASTELLI-KURUCZ ATLAS The Castelli AND Kurucz 2004 Stellar Atmosphere Models The atlas contains about 4300 stellar atmosphere models for a wide range of metal abundances, effective temperatures and gravities. These LTE models with no convective overshooting computed by Fiorella Castelli, have improved opacities and abundances upon previously used by Kurucz (1990). The main improvements from previous opacity distribution functions listed in Castelli & Kurucz 2003 (IAU Symposium 210, Modelling of Stellar Atmospheres, Uppsala, Sweden, eds. N.E. Piskunov, W.W. Weiss. and D.F. Gray, 2003, ASP-S210) are: 1- the replacement of the solar abundances from Andres & Grevesse (1989, GCA,53,197; AG89) with those from Grevesse & Sauval (1998 Space. Sci. Rev.,85,161; GS98). See Table 2 of Castelli & Kurucz 2004. 2- The replacement of the TiO lines provided by Kurucz (1993) with the TiO lines from Schwenke (1998, Faraday Discuss., 109,321). Addition of the H_2 O lines (Partridge & Schwenke 1997, J. Chem. Phys., 106, 4618) and of the HI-HI and HI-H+ quasi-molecular absorptions near 1600 A and 1400 A (Allard et al. 1998, A&A, 335,1124), as distributed by Kurucz 1999a, 1999b. Extended molecular list and corrected previous bugs in Kurucz line lists. These models are computed with the same wavelength resolution and a smaller temperature resolution than the Kurucz 1993 models. All of the models have the same number of plane parallel layers from $log(\tau_{Ross})=-6.875 \pm \ 2.00$ in steps of $\Delta[log(\tau_{Ross})] = 0.125$, computed assuming a pure mixing-length convection (no overshooting) with 1/Hp=1.25. As before the microturbulent velocity used is 2 $km \ s^{-1}$ The ATLAS9 installed in TRDS is from "The Grids of ATLAS9-ODFNEW models and fluxes" from Fiorella Castelli's web page (http://wwwuser.oat.ts.astro.it/castelli/grids.html) created on January 2007. These grids are also available from Dr. R. Kurucz website. The ATLAS9 includes models for abundances [M/H]=0.0, -0.5, -1.0, -1.5, -2.0, -2.5, +0.5, +0.2 and gravity range from log_g= 0.0 to +5.0 in steps of +0.5. The range in effective temperature from 3500 K to 50000 K is covered with an uneven grid (see Table 1a). The model spectra cover the ultraviolet (1000 Å) to infrared (10 μm) spectral range with non-uniform wavelength spacing (see Table 1b). TABLE 1a: Grid of temperatures for the models Temperature Range (K) Grid Step (K) 3000 - 13000 250 13000 - 50000 1000 TABLE 1b: Wavelength coverage for the models Wavelength Range (microns) Grid Step (A) 0.10 - 0.29 10 0.29 - 1.00 20 1.00 - 1.60 50 1.60 - 3.20 100 3.20 - 8.35 200 8.35 - 10.0 400 The TRDS version of the ATLAS9 The atlas is divided into 8 independent subdirectories, according to metal abundance. Within each subdirectory, the stellar atmosphere models are given in fits table format. Each table consists of 12 different columns, the first contains the wavelength grid and the rest containing the spectrum of a star with the same effective temperature but different gravity, ranging from log_g= 0.0 to +5.0. Columns filled with zeros indicate that the model spectrum for that particular metal abundance, effective temperature and gravity combination is not covered by the atlas. , where " cksmh_ttttt.fits ck", for Castelli & Kurucz, are the first two letters of the atlas; " smh" is the metal abundance of the model (mh) with its sign (s); and " ttttt" is the model's effective temperature, using four or five digits depending on the value. For example, a model with an effective temperature of 5000 K with [M/H]= -0.5 is indicated by ttttt= 5000, s= m, mh= 05or ckm05_5000.fits.Whereas a one with an effective temperature of 5000K,with [ M/H]= +3.5is indicated by ttttt= 5000, s= p, mh= 35and have the name ckm05_5000.fits. Within each individual table file, each column is named "gyy" where "yy" corresponds to 10log_g. For example, log_g= +0.5 and log_g= +4.0 models are located in columns named g05 and g40, respectively. See the appendix for an example of a standard header of a table file. Physical fluxes of the spectra are given in FLAM surface flux units, i.e. $ergs cm^{-2} s^{-1} A^{-1}$. These flux units differ from those in the Castelli & Kurucz tables by a factor of $3.336 x 10^{-19} \lambda^{2} /4\pi$, i.e. are converted from $ergs \ cm^{-2} s^{-1} Hz^{-1}str^{-1}$ to $ergs\ cm^{-2} s^{-1} A^{-1}$ by mutiplying the Castelli & Kurucz values by $3.336 x 10^{-19} * \lambda^{2} /4\pi$, where lambda is in Angstroms. To convert to observed flux at Earth, multiply by a factor of ($(R/D)^2 $where R is the stellar radius, and D is the distance to Earth. The names of the files located in each metal abundance subdirectory are listed in the README file located in each subdirectory. The range in gravity covered by the models for the different temperatures is also indicated. Use with Pysynphot or Astropy Synphot Pysynphot or Astropy Synphot permit the selection of spectra within one of many columns in a single FITS table file using the "pysynphot. Icat ()" function. The syntax is "sp = pysynphot. Icat ('ck04models', t,m,g) where "t" is the effective temperature, "m" is the metal abundance [M/H], and "g" is the log gravity. Please note that the model spectra in the atlas are in surface flux units. Thus, if the number of counts or the calculated absolute flux is needed, the model spectrum must be renormalized appropriately. One can do this in with the "sp.renorm()" function or the "sp.normalize()" function in pysynphot Astropy Synphot . A list of solar abundance stars of different spectral types and luminosity classes together with their closest Castelli & Kurucz 2004 model spectrum is presented in Table 2. The physical parameters, $T_{eff}$ and log_g, characterizing each of the O stars are taken from Martins, Schaerer, & Hiller's compilation of stellar parameters of Galactic O stars (Martins, Scharer & Hiller 2005,A&A,436,1049). The physical parameters for later stars are taken from Schmidt-Kaler's compilation of physical parameters of stars (Schmidt-Kaler 1982, Landolt-Bornstein VI/2b). For the later, the U-B and B-V colors of the closest model agree with the characteristic color of each star (see Schmidt-Kaler 1982) to better than 0.06 magnitude. For the cool end, K5I, M0I and M2I stars, the physical parameters are taken from Leusque et al 2006, ApJ 645, 1102. TABLE 2: Suggested models for specific stellar types Type T_{eff} log_g Kurucz model O3V 44852 +3.92 ckp00_45000[g45] O4V 42857 +3.92 ckp00_43000[g45] O5V 40862 +3.92 ckp00_41000[g45] O5.5V 39865 +3.92 ckp00_40000[g40] O6V 38867 +3.92 ckp00_39000[g40] O6.5V 37870 +3.92 ckp00_38000[g40] O7V 36872 +3.92 ckp00_37000[g40] O7.5V 35874 +3.92 ckp00_36000[g40] O8V 34877 +3.92 ckp00_35000[g40] O8.5 33879 +3.92 ckp00_34000[g40] O9V 32882 +3.92 ckp00_33000[g40] O9.5 31884 +3.92 ckp00_32000[g40] B0V 30000 +3.90 ckp00_30000[g40] B1V 25400 +3.90 ckp00_25000[g40] B3V 18700 +3.94 ckp00_19000[g40] B5V 15400 +4.04 ckp00_15000[g40] B8V 11900 +4.04 ckp00_12000[g40] A0V 9520 +4.14 ckp00_9500[g40] A1V 9230 +4.10 ckp00_9250[g40] A3V 8270 +4.20 ckp00_8250[g40] A5V 8200 +4.29 ckp00_8250[g40] F0V 7200 +4.34 ckp00_7250[g40] F2V 6890 +4.34 ckp00_7000[g40] F5V 6440 +4.34 ckp00_6500[g40] F8V 6200 +4.40 ckp00_6250[g45] G0V 6030 +4.39 ckp00_6000[g45] G2V 5860 +4.40 ckp00_5750[g45] G5V 5770 +4.49 ckp00_5750[g45] G8V 5570 +4.50 ckp00_5500[g45] K0V 5250 +4.49 ckp00_5250[g45] K2V 4780 +4.5 ckp00_4750[g45] K4V 4560 +4.5 ckp00_4500[g45] K5V 4350 +4.54 ckp00_4250[g45] K7V 4060 +4.5 ckp00_4000[g45] M0V 3850 +4.59 ckp00_3750[g45] M2V 3580 +4.64 ckp00_3500[g45] M4V 3370 +4.80 ckp00_3500[g50] M5V 3240 +4.94 ckp00_3500[g50] M6V 3050 +5.00 ckp00_3500[g50] B0III 29000 +3.34 ckp00_29000[g35] B5III 15000 +3.49 ckp00_15000[g35] G0III 5850 +2.94 ckp00_5750[g30] G5III 5150 +2.54 ckp00_5250[g25] K0III 4750 +2.14 ckp00_4750[g20] K5III 3950 +1.74 ckp00_4000[g15] M0III 3800 +1.34 ckp00_3750[g15] O5I 40300 +3.34 ckp00_40000[g45] O6I 39000 +3.24 ckp00_39000[g40] O8I 34200 +3.24 ckp00_34000[g40] BOI 26000 +2.84 ckp00_26000[g30] B5I 13600 +2.44 ckp00_14000[g25] AOI 9730 +2.14 ckp00_9750[g20] A5I 8510 +2.04 ckp00_8500[g20] F0I 7700 +1.74 ckp00_7750[g20] F5I 6900 +1.44 ckp00_7000[g15] G0I 5550 +1.34 ckp00_5500[g15] G5I 4850 +1.14 ckp00_4750[g10] K0I 4420 +0.94 ckp00_4500[g10] K5I 3850 +0.34 ckp00_3750[g05] M0I 3650 +0.14 ckp00_3750[g00] M2I 3450 -0.06 ckp00_3500[g00] Appendix Below is an example of a standard header for the table files in the CDBS version of Castelli & Kurucz 2004 atlas. In this example the name of the file is ckp00_8000.fits and contains all the models for a star of metallicity log_Z= 0.0 and effective temperature $T_{eff}$= 8000 K. Models cover a range of gravities from log_g= +1.0 (g10 in the header) to log_g= +5.0 (g50 in the header). Models for gravities log_g= +0.0 and +0.5 are not available for this particular metallicity and effective temperature combination, and therefore do not appear listed in the header. Their corresponding columns (g00 and g05) are filled with zeros. The models are in FLAM surface flux units, i.e.$ergs \ cm^2\ s^{-1}\ {\it \unicode{xC5}}^{-1}$ Header for table file ckp00_8000.fits 1 TEFF i 8000 2 LOG_Z d 0. This is actually metal abundance 3 HISTORY t File created by F.R.Boffi 4 HISTORY t ATLAS9 model atmospheres by Castelli and Kurucz (2004). 5 HISTORY t Wavelength is in Angstrom. 6 HISTORY t Fluxes tabulated in units of erg/s/cm^2/A 7 HISTORY t (after converting original units into "flam", 8 HISTORY t as described in README file and the SYNPHOT manual) 9 HISTORY t and are surface fluxes. To transform to observed 10 HISTORY t fluxes multiply by (R/D)^2 where R is the 11 HISTORY t radius of the star and D the distance. 12 HISTORY t Each column in the table represents the 13 HISTORY t spectrum of a star for the same metallicity 14 HISTORY t and effective temperature but different gravity. 15 HISTORY t Gravities range from log_g = +0.0 (g00 in the column 16 HISTORY t header) to log_g = +5.0 (g50 in the column header). and not metallicity
Prove that if $ab \equiv 0 \pmod p$, where p is a prime number, then $a \equiv 0 \pmod p$ or $b \equiv 0 \pmod p$. All I have right now is that the prime divisibility property may help with the then part of this problem. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community Prove that if $ab \equiv 0 \pmod p$, where p is a prime number, then $a \equiv 0 \pmod p$ or $b \equiv 0 \pmod p$. All I have right now is that the prime divisibility property may help with the then part of this problem. The ideal $(p) \subset \mathbb Z$ is prime, thus if $ab \in (p)$, then $a \in (p)$ or $b \in (p)$. In other words: $ab \equiv 0 \pmod p \implies ab=pk \implies p\|a$ or $p\|b \implies a \equiv 0 \pmod p$ or $b \equiv 0 \pmod p$. Remember that in the integers, for a prime $\;p\;\;,\;\;\;p\|a\iff a=kp\;,\;\;a,k\in\Bbb Z\;$ , and then $\;a=0\pmod p\iff p\|a\iff a=kp\;$ , so by what you wrote in your second comment below your question: $$ab=0\pmod p\iff p\|ab\iff p\|a\;\;\text{or}\;\;p\|b\iff $$ $$a=kp\;\;\text{or}\;\;b=mp\;,\;\;k,m\in\Bbb Z\iff a=0\pmod 0\;\;\text{or}\;\;b=0\pmod p$$
Let $M$ be a complete Riemannian 3-manifold and $\gamma \subset M$ a simple closed curve that bounds a least-area disc $D$ - a disc that minimizes the area among all discs bounded by $\gamma$. Let $Conv(\gamma)$ be the convex hull of $\gamma$, i.e, the smallest convex subset of $M$ containing $\gamma$. When $M = \mathbb R^3$ (with the standard Euclidean metric), it is a well-known result that $D \subseteq Conv(\gamma)$. Question 1: What (topologcial, metric) conditions need to be satisfied on $M$, along with possible restrictions to $\gamma$, so that $D \subseteq Conv(\gamma)$ holds true ? I only know of partial results to this question, such as when $M$ is equipped with a complete metric of bounded sectional curvature and positive injectivity radius, then the above will hold true for all sufficiently small simple closed curves $\gamma$. This is a consequence of a result proven by Joel Hass and Peter Scott in this paper (Theorem 6.3). I am very interested in whether this has been generalized even further as of today. Note also that when $M = \mathbb R^3$ and $X \subset M$ is any subset, $Conv(X)$ is simply the set of all convex combinations of points in $X$. Orangeskid, a user over at stackexchange has brought to my attention in his answer to a similiar question I posed a week ago that this, along with the convexity of the norm function, is enough to show that \begin{equation} \text{Diam}(Conv(X)) = \text{Diam}(X) \end{equation} always holds true. This means that on $\mathbb R^3$, taking the convex hull of a set does not increase the diameter. Question 2: What (topological, metric) conditions need to be satisfied on $M$, so that $\text{Diam}(Conv(X)) = \text{Diam}(X)$ holds true for sufficiently nice subsets $X \subset M$ ? The problem here lies in the fact that a general $3$-manifolds lacks a vector space structure, thus both the feasible description of a convex hull, as well as the "convexity" of the metric are not available any more. Any help is appreciated.
Hamiltonian Gromov-Witten theory(see Mundet-Tian paper) corresponds to a new type of Symplectic vortex equations: Such type of models gives a connection to Hitchin-Kobayashi correspondence and Floer theory(for example Donaldson's book: Floer homology groups in Yang–Mills theory). Question: Take the vortex equations with values in a symplectic manifold $(M, \omega)$. We assume that $(M, \omega)$ is equipped with a Hamiltonian action by a compact Lie group $G$ that is generated by an equivariant moment map $\mu : M \to \mathfrak g$. The symplectic vortex equations have the form $$\overline ∂_{J,A}(u) = 0, \; \; ∗F_A + \mu(u) = \tau\in \mathcal Z(\mathfrak g)$$ Here $ P\to Σ$ is a principal $G$-bundle over a compact Riemann surface, $u : P → M $ is an equivariant smooth function, and $A$ is a connection on $P$. Now how can we extend the study of symplectic vortex equations to logarithmic setting. In fact if we assume $M$ is a log-symplectic manifold, then Gualtieri- Li-Pelayo and Ratiu developed log symplectic reduction and log moment map. So the study of Log symplectic vortex equations in Hamiltonian Log GW theory or its Hitchin-Kobayashi type correspondencis natural to be considered. Is there any progress about it. Any reference or comment is welcomed
I'm trying to solve this particular equation $\frac{\partial u}{\partial t} = \frac{\partial}{\partial x} \big[D_{i}(x)\frac{\partial u}{\partial x} \big] + S(x,t)$ where the $i$ index denotes different layers. Let's say that $ i \in \{1,2,3\}$. The interface conditions are: $-D_1(x)\frac{\partial u}{\partial x}\bigg\|_{x=x_1^-} = -D_2(x)\frac{\partial u}{\partial x}\bigg\|_{x=x_1^+}$ $-D_2(x)\frac{\partial u}{\partial x}\bigg\|_{x=x_2^-} = -D_3(x)\frac{\partial u}{\partial x}\bigg\|_{x=x_2^+}$ and at the boundaries: $\frac{\partial u}{\partial x}\bigg\|_{x=0}=0$ $\frac{\partial u}{\partial x}\bigg\|_{x=L}=0$ with initial conditions: $u(x,t_0) = u_0$ where $[0;L]$ denotes the spatial domain over which I want to solve my problem. Every single layer has its own length, let's define with $L_1,L_2$ and $L_3$ and $L = \sum_i L_i$. I'm using the MOL approach in order to obtain a continous time-discrete space formulation. Moreover let's assume that the source term $S(x,t)$ is not zero within the layers 1 and 3, but it's null within the second layer. My current workflow has been: 1 - Divide every signel layer into $N_i$ points with the FDM method 2 - Treat the interfaces points as "common" points; therefore the total number of discretized points will be $N_1+N_2+N_3-2$ 3 - At the interfaces points ($N_1$ and $N_1+N_2-1$) enforce the interface BCs. In $N_1$ I am treating that point as part of the layer 1, while in $N_1+N_2-1$ I'm treating that point as part of the layer 2 (is that correct?) 4 - At the end points I'm enforcing the null-flux conditions. I am able to solve the equation, but I'm experiencing an increase of the overall mass in the system. As far as I know this is a common problem when dealing with conservation of mass problems and FDM; theoretically I could adopt the FVM method in order to avoid this behaviour, but I was courious to understand if I was doing right in this way or wrong. Understand if it was up to the numerical approach or up to some kind of mine bug. Moreover: the term $S(x,t)$ I'm assuming to be non-zero in $[1;N_1]$ and $[N_1+N_2;N_1+N_2+N_3-2]$ Thanks
Can I be a pedant and say that if the question states that $\langle \alpha \vert A \vert \alpha \rangle = 0$ for every vector $\lvert \alpha \rangle$, that means that $A$ is everywhere defined, so there are no domain issues? Gravitational optics is very different from quantum optics, if by the latter you mean the quantum effects of interaction between light and matter. There are three crucial differences I can think of:We can always detect uniform motion with respect to a medium by a positive result to a Michelson... Hmm, it seems we cannot just superimpose gravitational waves to create standing waves The above search is inspired by last night dream, which took place in an alternate version of my 3rd year undergrad GR course. The lecturer talks about a weird equation in general relativity that has a huge summation symbol, and then talked about gravitational waves emitting from a body. After that lecture, I then asked the lecturer whether gravitational standing waves are possible, as a imagine the hypothetical scenario of placing a node at the end of the vertical white line [The Cube] Regarding The Cube, I am thinking about an energy level diagram like this where the infinitely degenerate level is the lowest energy level when the environment is also taken account of The idea is that if the possible relaxations between energy levels is restricted so that to relax from an excited state, the bottleneck must be passed, then we have a very high entropy high energy system confined in a compact volume Therefore, as energy is pumped into the system, the lack of direct relaxation pathways to the ground state plus the huge degeneracy at higher energy levels should result in a lot of possible configurations to give the same high energy, thus effectively create an entropy trap to minimise heat loss to surroundings @Kaumudi.H there is also an addon that allows Office 2003 to read (but not save) files from later versions of Office, and you probably want this too. The installer for this should also be in \Stuff (but probably isn't if I forgot to include the SP3 installer). Hi @EmilioPisanty, it's great that you want to help me clear out confusions. I think we have a misunderstanding here. When you say "if you really want to "understand"", I've thought you were mentioning at my questions directly to the close voter, not the question in meta. When you mention about my original post, you think that it's a hopeless mess of confusion? Why? Except being off-topic, it seems clear to understand, doesn't it? Physics.stackexchange currently uses 2.7.1 with the config TeX-AMS_HTML-full which is affected by a visual glitch on both desktop and mobile version of Safari under latest OS, \vec{x} results in the arrow displayed too far to the right (issue #1737). This has been fixed in 2.7.2. Thanks. I have never used the app for this site, but if you ask a question on a mobile phone, there is no homework guidance box, as there is on the full site, due to screen size limitations.I think it's a safe asssumption that many students are using their phone to place their homework questions, in wh... @0ßelö7 I don't really care for the functional analytic technicalities in this case - of course this statement needs some additional assumption to hold rigorously in the infinite-dimensional case, but I'm 99% that that's not what the OP wants to know (and, judging from the comments and other failed attempts, the "simple" version of the statement seems to confuse enough people already :P) Why were the SI unit prefixes, i.e.\begin{align}\mathrm{giga} && 10^9 \\\mathrm{mega} && 10^6 \\\mathrm{kilo} && 10^3 \\\mathrm{milli} && 10^{-3} \\\mathrm{micro} && 10^{-6} \\\mathrm{nano} && 10^{-9}\end{align}chosen to be a multiple power of 3?Edit: Although this questio... the major challenge is how to restrict the possible relaxation pathways so that in order to relax back to the ground state, at least one lower rotational level has to be passed, thus creating the bottleneck shown above If two vectors $\vec{A} =A_x\hat{i} + A_y \hat{j} + A_z \hat{k}$ and$\vec{B} =B_x\hat{i} + B_y \hat{j} + B_z \hat{k}$, have angle $\theta$ between them then the dot product (scalar product) of $\vec{A}$ and $\vec{B}$ is$$\vec{A}\cdot\vec{B} = \|\vec{A}\|\|\vec{B}\|\cos \theta$$$$\vec{A}\cdot\... @ACuriousMind I want to give a talk on my GR work first. That can be hand-wavey. But I also want to present my program for Sobolev spaces and elliptic regularity, which is reasonably original. But the devil is in the details there. @CooperCape I'm afraid not, you're still just asking us to check whether or not what you wrote there is correct - such questions are not a good fit for the site, since the potentially correct answer "Yes, that's right" is too short to even submit as an answer
Let $G = ( V_G, E_G )$ be a graph. Let $E_H \subseteq E_G$. The subgraph of $G$ edge-induced by $E_H$ is $H = ( V_H, E_H)$, where $V_H = \{ v \in V_G : \exists ( u, w ) \in E_H\ v = u \lor v = w \}$ Let $k_1 \leq \|E_G\|$ and $k_2 \leq \|V_G\|$ be given in input. I would like to determine the number $k_3$ of edge-induced subgraphs of $G$ having $k_1$ edges and $k_2$ nodes. Clearly, I don't want to enumerate all the (exponentially many) subgraphs of $G$ having $k_1$ edges. Questions: Is it possible to determine $k_3$ in time polynomial in $\|E_G\|$? What if $G$ is 3-regular? What if $G$ is 3-regular planar? What if $G$ is 3-regular planar bipartite? Which is the best known algorithm to compute $k_3$? $\longleftarrow$ Added on 30/08/2012
Allowed by whom? There is no Central Graph Administration that decides what you can and cannot do. You can define objects in any way that's convenient for you, as long as you're clear about what the definition is. If zero-weighted edges are useful to you, then use them; just make sure your readers know that's what you're doing.The reason you don't usually ... Dijkstra relies on one "simple" fact: if all weights are non-negative, adding an edge can never make a path shorter. That's why picking the shortest candidate edge (local optimality) always ends up being correct (global optimality).If that is not the case, the "frontier" of candidate edges does not send the right signals; a cheap edge might lure you down a ... Claim: Yes, that statement is true.Proof Sketch: Let $T_1,T_2$ be two minimal spanning trees with edge-weight multisets $W_1,W_2$. Assume $W_1 \neq W_2$ and denote their symmetric difference with $W = W_1 \mathop{\Delta} W_2$.Choose edge $e \in T_1 \mathop{\Delta} T_2$ with $w(e) = \min W$, that is $e$ is an edge that occurs in only one of the trees and ... If edge weights are integers in $\{0,1,\ldots,K\}$, you can implement Dijkstra's to run in $O(K\|V\|+\|E\|)$ time, following @rrenaud's suggestion. Here is a more explicit explanation.At any time, the (finite) keys in the priority queue are in some range $\{D,D+1,\ldots,D+K\}$, where $D$ is the value of the last key removed from the priority queue. (Every key ... It depends on the context. In general yes, edges of zero and even negative weight may be allowed. In some specific cases the edge weights might be required to be non-negative or strictly positive (for instance, Dijkstra's algorithm requires weights to be non-negative). It is NP-complete to even decide whether any path exists.It is clearly possible to verify any given path is a valid path in the given graph. Thus the bounded-length problem is in NP, and so is its subset, the any-path problem.Now, to prove NP-hardness of the any-path problem (and thus of the bounded-length problem), let's reduce SAT-CNF to this problem:... A Continuous-time Markov Chain can be represented as a directed graph with constant non-negative edge weights. An equivalent representation of the constant edge-weights of a directed graph with $N$ nodes is as an $N \times N$ matrix. The Markov property (that the future states depend only on the current state) is implicit in the constant edge weights (or ... There are exponentially many such routes.Think of a sequence of $n$ diamonds. At each diamond, you can go either left or right, independently of what you do at all other diamonds. This leads to $2^n$ paths, each of which is non-intersecting. Now the complete graph on those vertices contains all of these paths, plus some more, so this is a lower-bound on ... A path of length $n$ consists of $n$ line segments in the plane. You want to find all intersections between these line segments. This is a standard problem that has been studied in depth in the computer graphics literature.A simple algorithm is the following: for each pair of line segments, check whether they intersect (using a standard geometric ... You've refined your problem some more in the comments. To be more specific, you have a DAG with all edges flowing away from the source $s$ and towards the sink $t$ (that is, all edges are on a path from $s$ to $t$). You want to find the minimum cut between two pieces of the DAG, where the first piece is connected to $s$, and the second connected to $t$. For ... in the first picture: the right graph has a unique MST, by taking edges $(F,H)$ and $(F,G)$ with total weight of 2.Given a graph $G=(V,E)$ and let $M=(V,F)$ be a minimum spanning tree (MST) in $G$.If there exists an edge $e=\{v,w\}\in E \setminus F$ with weight $w(e)=m$ such that adding $e$ to our MST yields a cycle $C$, and let $m$ also be the lowest ... When edges connect more than two nodes, you don't have a graph, you have a hypergraph. More precisely, since transitions are oriented (you're starting from a digraph) and there are probability on each transition, you have a weighted hyperdigraph. I'm not sure having this term will help you that much: as data structures go, this isn't that much of a classic.... Markov Chains come in two flavors: continuous time and discrete time.Both continuous time markov chains (CTMC) and discrete time markov chains (DTMC) are represented as directed weighted graphs.For DTMC's the transitions always take one unit of "time." As a result, there is no choice for what your weight on an arc should be-- you put the probability of ... The setting you suggest is not clear enough to determine what kind of an automaton you are looking for.A short explanation regarding the types of automata:A weighted automaton is typically an automaton with a weight function on the edges (or states). Then, a run is assigned a weight according to the weight it traverses. There are many different semantics ... Your question was already asked before it seems, but got no explicit examples. I try to give these here.First note the question only makes sense if we consider a node $u$, and there exist spanning trees starting with $u$. The algorithms of Prim and Kruskal make choices in a greedy way. Once the choice is made, it will not be reconsidered. We show how this ... There is no direct relationship between the diameter of a (minimum) spanning tree and the total cost of the tree1. Consider the following example:The spanning tree on the left (whose edges are highlighted in red) is minimum. Its total cost is 7 and the diameter is equal to 5. In contrast, the spanning tree on the right is not minimum (since its total cost ... This is NP-hard, so it's very unlikely that a polynomial-time algorithm exists.Given any instance $G=(V, E)$ of Hamiltonian Path, create a new graph $G'=(V', E')$ in which every vertex $v \in V$ becomes a pair of vertices $v_+, v_-$ connected by an edge in $G'$. All of these edges should also be added to $F$. Then for each $(u, v) \in E$, add the ... Here is the original statement in CLRS.Assume that we have a connected, undirected graph $G$ with a weight function $w: E\to\Bbb R$, and we wish to find a minimum spanning tree for $G$.It is pretty good to understand "a weight function $w:E\rightarrow \mathbb{R}$" as "an edge has a weight". In fact, that is how I would interpret that notation in a rush ... I contacted one of the authors (Kevin Wayne; thanks) of the textbook "Algorithms, 4th Edition" and got a hint:Try adding "t-joins" or "perfect matching" to your web searches.Following this, I found the following two lecture notes:Shortest Path Algorithms Luis Goddyn, Math 408: Using Edmonds' Minimum Weight Perfect Matching Algorithm to solve shortest ... I would suggest the following approach.Maintain a data structure $H$ of $(i,j, g(i,j))$ triples so that you can efficiently find and remove a triple $(i,j,w)$ that minimises $w$.Maintain a partition $P$ of nodes $V = \{1,2,\dotsc,N\}$.Maintain a tree $T$.We will maintain the invariant that $T$ contains some edges of the tree that we are trying to ... Your conceptual difficulty stems from not distinguishing between TSP and Weighted Hamiltonian Cycle. These are usually discussed as if they are the same problem, but they're not.In Weighted Hamiltonian Cycle, we are given a graph with nonnegative edge weights and we wish to determine the minimum-weight Hamiltonian cycle, i.e., the minimum-weight cycle that ... First of all, bear in mind that the Longest Path Problem (LPP) is a NP-complete problem whereas finding the Shortest Path Problem (SPP) is a problem known to be in P. The proof for the NP-completeness of LPP is trivial and consists of a reduction from the Hamiltonian Circuit (HC) problem which is already known to be NP-complete.In other words, if you could ... Based on your above comments with @Gilles, what you describe is just a higher order markov model. For example an $n$th order markov model, is a model which assumes$$ P(x_t \| x_{t-1}, x_{t-2}, \ldots x_1) = P(x_t \| x_{t-1}, x_{t-2}, \ldots x_{t-n}).$$If $n$ is not fixed you have a variable order markv model. Here is a slightly simpler argument that also works for other matroids. (I saw this question from another one.)Suppose that $G$ has $m$ edges. Without loss of generality, assume that the weight function $w$ takes on values in $[m]$, so we have a partition of $E$ into sets $E_i := w^{-1}(i)$ for $i\in [m]$. We can do induction on the number $j$ of non-... I assume here that $K$ is an integer and the edge weights are integral. Otherwise it doesn't really buy you anything, you can always rescale weights so that the min edge has cost $1$ and the max has cost $K$, so the problem is identical to the standard shortest path problem.Algorithm/proof sketch: Implement the priority queue in this kind of crazy way as ... IntroductionThis answer is in two parts.The first is an analysis of the problem mixed with a sketch of thealgorithm to solve it. As it is my first version, it is detailed, butresults in an algorithm that is a bit more complex than needed.It is followed by a pseudo-code version of the algorithm, written sometime later, as the algorithm was clearer. ... Adding a constant amount to each edge length can change the shortest path for the simple reason that it increases the length of a path with many edges by more than it increases the length of a path with only a few edges.For a simple case, consider the graph with vertices $\{a,b,c\}$ and edges $\{ab,bc,ac\}$, where $ab$ and $bc$ have length $1$ ... Thus I'm still curious if there are other more efficient approaches to solving the problem.If you take the complement graph $\overline{G}$, then your problem corresponds to a coloring problem. Cover by cliques in $G$ is the same as covering by independent set in $\overline{G}$.The problem is para-NP-hard in the unweighted case and the problem is just ... As the other answers note, you're perfectly free to consider (or exclude from consideration) weighted graphs with zero-weight edges.That said, in my experience, the usual convention in most applications of weighted graphs is to make no distinction between a zero-weight edge and the absence of an edge. One reason for this is that, typically, weighted ...
\defl{Binomial Distribution has the following properties:} There are a fixed number of trials or observations, $n$, determined in advance. Each trial can take on one of two possible outcomes, labeled ''success'' and ''failure''. Each trial's outcome is determined independently of all the other trials. The probability of a success and that of a failure remains the same from one trial to the next, and is denoted by $\pi$ and $1-\pi$, respectively. \defm{Binomial Distribution:} \[ P(X=x)=f(x)={n \choose x}\pi^{x}(1-\pi)^{n-x} \] $X$ is Binomially distributed $x$ the number of successes, where $x = 0,1,\ldots,n$ $n$ the number of trials $\pi$ the probability of success $1-\pi$ the probability of failure $\mu = E(X) = n\pi$ $\sigma^2 = V(X)= n\pi(1-\pi)$ \defs{Examples} The number of heads out of 2 coin tosses. Assume that the probability of SET, Securities Exchange of Thailand, will end the day on the positive side is constant and each day's outcome, positive or not, is independent of what occurred on prior days. The number of times the SET increases in 2 weeks, 10 working days. The number of women in a group of 20 randomly selected people.
Exams can be easily created in LaTeX by means of the class exam.cls. This class makes it straightforward to typeset questions, and it sets a 1in margin in all paper sizes and provides special commands to write and compute grades. This article explains how to edit with exam.cls. Contents Let's see a simple working example of the exam class: \documentclass{exam} \usepackage[utf8]{inputenc} \begin{document} \begin{center} \fbox{\fbox{\parbox{5.5in}{\centering Answer the questions in the spaces provided. If you run out of room for an answer, continue on the back of the page.}}} \end{center} \vspace{5mm} \makebox[\textwidth]{Name and section:\enspace\hrulefill} \vspace{5mm} \makebox[\textwidth]{Instructor’s name:\enspace\hrulefill} \begin{questions} \question Is it true that $x^n + y^n = z^n$ if $x,y,z$ and $n$ are positive integers?. Explain. \question Prove that the real part of all non-trivial zeros of the function $\zeta(z)$ is $\frac{1}{2}$ \question Compute \[\int_{0}^{\infty} \frac{\sin(x)}{x}\] \end{questions} To use the exam class you must put the line \documentclass{exam} on top of your .tex file. This will enable the package's exam-related commands, and set the page format to allow margins for corrections. The syntax of the questions environment is very similar to that of the itemize and enumerate environments. Each question is typed by putting the command \question before it. The other commands in this example are not specific to the exam class, but may be useful to create a quick header for your exam. In the previous section, a basic example showing how to create question was presented. Questions can be further customized, and this section explains how. If the students are required to answer the exam in the space provided, that space can be manually set or evenly distributed. See the example below: \begin{questions} \question Is it true that $x^n + y^n = z^n$ if $x,y,z$ and $n$ are positive integers?. Explain. \vspace{\stretch{1}} \question Prove that the real part of all non-trivial zeros of the function $\zeta(z)$ is $\frac{1}{2}$ \vspace{\stretch{1}} \question Compute \[\int_{0}^{\infty} \frac{\sin(x)}{x}\] \vspace{\stretch{1}} \end{questions} \clearpage In this example the command \vspace{\stretch{1}} after each question equally distributes the available space. The command \clearpage inserts a page break point to continue typing questions in a new page. If you want to manually assign the space to each question, use the command \vspace{} and in between the braces write the units of space you need. For instance, \vspace{1in} inserts a 1-inch vertical space. Check the documentation about lenghts in LaTeX for a list of available units. If your questions have several parts focused on some subtopics you can use the environments parts, subparts, subsubparts and the corresponding commands \part, \subpart and \subsubpart. See the next example: \begin{questions} \question Given the equation $x^n + y^n = z^n$ for $x,y,z$ and $n$ positive integers. \begin{parts} \part For what values of $n$ is the statement in the previous question true? \vspace{\stretch{1}} \part For $n=2$ there's a theorem with a special name. What's that name? \vspace{\stretch{1}} \part What famous mathematician had an elegant proof for this theorem but there was not enough space in the margin to write it down? \vspace{\stretch{1}} \begin{subparts} \subpart Who actually proved the theorem? \vspace{\stretch{1}} \subpart How long did actually take to solve this problem? \vspace{\stretch{1}} \end{subparts} \end{parts} \question Prove that the real part of all non-trivial zeros of the function $\zeta(z)$ is $\frac{1}{2}$ ... \end{questions} The environments parts and subparts provide question-like nested lists. Jut like in questions you can set manually the vertical spacing. There are four environments to create multiple choice questions. \question Which of these famous physicists invented time? \begin{oneparchoices} \choice Stephen Hawking \choice Albert Einstein \choice Emmy Noether \choice This makes no sense \end{oneparchoices} \question Which of these famous physicists published a paper on Brownian Motion? \begin{checkboxes} \choice Stephen Hawking \choice Albert Einstein \choice Emmy Noether \choice I don't know \end{checkboxes} In this example, two different environments are used to list the possible choices for multiple-choice questions. oneparchoices labels the choices with upper case letters and prints them horizontally. If you want the choices to be printed in a list-like format, the environment choices is the right choice. checkboxes prints check boxes before each choice. If you need the choices to be printed horizontally use the environment oneparcheckboxes instead. Another important feature of the exam class is that it provides commands to make grading the exams easier. You can add a parameter to each \question or \part to print the number of points you attain by correctly answering it \begin{questions} \question Given the equation $x^n + y^n = z^n$ for $x,y,z$ and $n$ positive integers. \begin{parts} \part[10] For what values of $n$ is the statement in the previous question true? \vspace{\stretch{1}} \part[10] For $n=2$ there's a theorem with a special name. What's that name? \vspace{\stretch{1}} \part[10] What famous mathematician had an elegant proof for this theorem but there was not enough space in the margin to write it down? \vspace{\stretch{1}} \end{parts} \question[20] Compute \[\int_{0}^{\infty} \frac{\sin(x)}{x}\] \vspace{\stretch{1}} \end{questions} The additional parameter inside brackets after a question or a part represents the number of points assigned to it. You can change the appearance and the place where the points are printed, see the reference guide for additional commands. Sometimes it's convenient to include half points as value for parts of a questions. You can do this and then print then the value of the whole question. See the example below: \documentclass[addpoints]{exam} \usepackage[utf8]{inputenc} \begin{document} \begin{questions} \question Given the equation $x^n + y^n = z^n$ for $x,y,z$ and $n$ positive integers. \begin{parts} \part[5] For what values of $n$ is the statement in the previous question true? \vspace{\stretch{1}} \part[2 \half] For $n=2$ there's a theorem with a special name. What's that name? \vspace{\stretch{1}} \part[2 \half] What famous mathematician had an elegant proof for this theorem but there was not enough space in the margin to write it down? \vspace{\stretch{1}} \end{parts} \droptotalpoints \question[20]... \end{questions} The command \half adds half points to a question. The command \droptotalpoints prints the total number of points for the last question. For this last command to work you must add the option [addpoints] to the document class statement. It is possible to add bonus questions, this extra points will later show up in the grading table. Adding bonus questions and parts is actually as simple as creating regular questions and parts. \begin{questions} \question Given the equation $x^n + y^n = z^n$ for $x,y,z$ and $n$ positive integers. \begin{parts} \part[5] For what values of $n$ is the statement in the previous question true? \vspace{\stretch{1}} \part[2 \half] For $n=2$ there's a theorem with a special name. What's that name? \vspace{\stretch{1}} \bonuspart[2 \half] What famous mathematician had an elegant proof for this theorem but there was not enough space in the margin to write it down? \vspace{\stretch{1}} \end{parts} \droptotalpoints \question[20] Compute \[\int_{0}^{\infty} \frac{\sin(x)}{x}\] \vspace{\stretch{1}} \bonusquestion[30] Prove that the real part of all non-trivial zeros of the function $\zeta(z)$ is $\frac{1}{2}$ \vspace{\stretch{1}} \end{questions} The commands \bonusquestion and \bonuspart print "(bonus)" next to the point value of the question. A table that show the points of each question can be printed with a special command. There are three commands to print a table of grades: \gradetable \bonusgradetable \combinedgradetable These commands take two extra parameters, each parameter inside brackets. [h] for a horizontal table or [v] for a vertical table. [questions] to index the points by question and [pages] to list the points by page number. There is no support for other languages than English in the exam class. Nevertheless, it's easy to translate the default words for those in your local language. The next snippet shows how to translate the example presented in the previous sections to Spanish. ss[addpoints]{exam} \usepackage[utf8]{inputenc} \usepackage[spanish]{babel} \pointpoints{punto}{puntos} \bonuspointpoints{punto extra}{puntos extra} \totalformat{Pregunta \thequestion: \totalpoints puntos} \chqword{Pregunta} \chpgword{Página} \chpword{Puntos} \chbpword{Puntos extra} \chsword{Puntos obtenidos} \chtword{Total} ... The rest of the document would be exactly the same shown in previous examples. The commands typed here change the default words in the exam class. \pointpoints{punto}{puntos} \bonuspointpoints{punto extra}{puntos extra} \totalformat{Pregunta \thequestion: \totalpoints puntos} \droptotalpoints. In the example it prints \chqword{Pregunta} chpgword for \chpword for \chbpword for \chsword for \chtword for Placing and formatting the points mark for questions These commands can be typed in the preamble to change the format of the whole document or right before a question to change the format from that question down to the next formatting command or the end of the document. \poinstmargin. The point values are printed in the left margin. Use \nopointsmargin to revert this command to the default format. \pointsmarginright. The point values are printed in the right margin. The command \nopoinstmarginright will revert to the normal behaviour. \bracketedpoints. Uses brackets instead of parentheses around the point values. \boxedpoints. Draws a box around the point values. Changing default names in Grade Tables The commands depend on the format and the information displayed on the table. The h and v within each command mean horizontal or vertical orientation. If the command is preceded by a b means it changes the format in a bonus table, if the command is preceded by a c means it works on combined tables. For instance, to change the word "Score" in a vertical oriented bonus table for the words "Points Awarded" you should use \bvsword{Points Awarded}. Below a table with the default values is shown. horizontal vertical grades table \hpgword{Page:} \hpword{Points:} \hsword{Score:} \htword{Total} \vpgword{Page} \vpword{Points} \vsword{Score} \vtword{Total:} bonus points table \bhpgword{Page:} \bhpword{Bonus Points:} \bhsword{Score:} \bhtword{Total} \bvpgword{Page} \bvpword{Bonus Points} \bvsword{Score} \bvtword{Total:} combined table \chpgword{Page:} \chpword{Points:} \chbpword{Bonus Points:} \chsword{Score:} \chtword{Total} \cvpgword{Page} \cvpword{Points} \cvbpword{Bonus Points} \cvsword{Score} \cvtword{Total:} For more information see:
For $n\in\mathbb{N}$ and $q\in(0,1)$, define $$f_{n}(q):=\sum_{i_{1},i_{2},\dots,i_{n}=1}^{\infty}\frac{q^{i_1+i_2+\dots+i_n}}{(1-q^{i_1+i_2})(1-q^{i_2+i_3})\dots(1-q^{i_{n-1}+i_n})(1-q^{i_n+i_1})}.$$ I would like to know something about the asymptotic behavior of $f_{n}(q)$, as $n\to\infty$. Some remarks on possible ways to find an answer: I believe that having, for instance, a recurrence rule for $f_{n}(q)$ or a generating function would be helpful. So far, however, I wasn't able to find anything. Perhaps there is a relation between $f_{n}(q)$ and Jacobi theta functions (or some other special functions from the $q$-world) which could be helpful as well. Maybe the expression for $f_{n}(q)$ is familiar to someone. I would appreciate any information concerning the sequence (even if it does not yield the asymptotic formula). Many thanks.
Depth 2 circuits require exponential size to compute addition since a depth 2 circuit must be either DNF or CNF and it is easy to verify that there are exponentially many minterms and maxterms. Warning: the part below is buggy. See the comments under the answer. The way I count it, addition can be done in depth 3. Assume $a_i$ and $b_i$ are the $i$th bits of the two numbers, where $0$ is the index of the LSB and $n$ of the MSB. Let us compute the $i$th bit of the sum, $s_i$ in the standard way with carry look ahead: $$s_i = a_i \oplus b_i \oplus c_i$$ where $\oplus$ is XOR and $c_i$ is the carry computed as: $$c_i = \bigvee_{j\mid j < i} (g_j \wedge p_j)$$ and $g_j$ means that the $j$th location "generated" the carry: $$g_j = (a_j \wedge b_j)$$ and $p_j$ means that the carry gets propagated from $j$ to $i$: $$p_j = \bigwedge_{k\mid j < k < i} (a_j \vee b_j)$$ Counting the depth, $p_j$ is depth 2, and $c_i$ is depth 3. While it would seem that $s_i$ is depth 4 or 5, it really is also just depth 3 since it is a bounded fanin computation of depth 3 circuits so one may push the top two levels down using de-Morgan formulas, while blowing the circuit size by a polynomial amount.
Rocky Mountain Journal of Mathematics Rocky Mountain J. Math. Volume 44, Number 4 (2014), 1145-1160. Elliptic curves coming from Heron triangles Abstract Triangles having rational sides $a, b, c$ and rational area $Q$ are called \textit{Heron triangles}. Associated to each Heron triangle is the quartic \[ v^2=u(u-a)(u-b)(u-c). \] The Heron formula states that $Q=\sqrt{P(P-a)(P-b)(P-c)}$ where $P$ is the semi-perimeter of the triangle, so the point $(u, v)=(P, Q)$ is a rational point on the quartic. Also, the point of infinity is on the quartic. By a standard construction, it can be proved that the quartic is equivalent to the elliptic curve \[ y^2=(x+ a\, b)(x+ b \,c)(x+c\, a). \] The point $(P, Q)$ on the quartic transforms to \[ (x, y)= \bigg(\frac{-2 a b c }{a + b +c} ,\frac{ 4 Q a b c}{(a+ b+ c)^2}\bigg) \] on the cubic, and the point of infinity goes to $(0, a b c)$. Both points are independent, so the family of curves induced by Heron triangles has rank $\geq 2$. In this note we construct subfamilies of rank at least $3$, $4$ and $5$. For the subfamily with rank $\geq 5$, we show that its generic rank is exactly equal to $5$, and we find free generators of the corresponding group. By specialization, we obtain examples of elliptic curves over $\Q$ with rank equal to $9$ and $10$. This is an improvement of results by Izadi et al., who found a subfamily with rank $\ge 3$ and several examples of curves of rank $7$ over $\Q$. Article information Source Rocky Mountain J. Math., Volume 44, Number 4 (2014), 1145-1160. Dates First available in Project Euclid: 31 October 2014 Permanent link to this document https://projecteuclid.org/euclid.rmjm/1414760946 Digital Object Identifier doi:10.1216/RMJ-2014-44-4-1145 Mathematical Reviews number (MathSciNet) MR3274341 Zentralblatt MATH identifier 1305.14017 Citation Dujella, Andrej; Peral, Juan Carlos. Elliptic curves coming from Heron triangles. Rocky Mountain J. Math. 44 (2014), no. 4, 1145--1160. doi:10.1216/RMJ-2014-44-4-1145. https://projecteuclid.org/euclid.rmjm/1414760946
The Annals of Statistics Ann. Statist. Volume 28, Number 3 (2000), 922-947. Maximum likelihood estimation of smooth monotone and unimodal densities Abstract We study the nonparametric estimation of univariate monotone and unimodal densities usingthe maximum smoothed likelihood approach. The monotone estimator is the derivative of the least concave majorant of the distribution correspondingto a kernel estimator.We prove that the mapping on distributions $\Phi$ with density $\varphi$, $$\varphi \mapsto \text{the derivative of the least concave majorant of $\Phi},$$ is a contraction in all $L^P$ norms $(1 \leq p \leq \infty)$, and some other “distances” such as the Hellinger and Kullback–Leibler distances. The contractivity implies error bounds for monotone density estimation. Almost the same error bounds hold for unimodal estimation. Article information Source Ann. Statist., Volume 28, Number 3 (2000), 922-947. Dates First available in Project Euclid: 12 March 2002 Permanent link to this document https://projecteuclid.org/euclid.aos/1015952005 Digital Object Identifier doi:10.1214/aos/1015952005 Mathematical Reviews number (MathSciNet) MR1792794 Zentralblatt MATH identifier 1105.62332 Subjects Primary: 62G07: Density estimation Citation Eggermont, P. P. B.; LaRiccia, V. N. Maximum likelihood estimation of smooth monotone and unimodal densities. Ann. Statist. 28 (2000), no. 3, 922--947. doi:10.1214/aos/1015952005. https://projecteuclid.org/euclid.aos/1015952005
In this paper we study the L p-Minkowski problem for p=鈭n鈭1, which corresponds to the critical exponent in the Blaschke鈥揝antalo inequality. We first obtain volume estimates for general solutions, then establish a priori estimates for rotationally symmetric solutions by using a Kazdan鈥揥arner type obstruction. Finally we give sufficient conditions for the existence of rotationally symmetric solutions by a blow-up analysis. We also include an existence result for the L p-Minkowski problem which corresponds to the super-critical case of the Blaschke鈥揝antalo inequality. We consider the asymptotics of the Turaev-Viro and the Reshetikhin-Turaev invariants of a hyperbolic $3$-manifold, evaluated at the root of unity $exp(2\pi\sqrt{-1}/r)$ instead of the standard $exp(\pi\sqrt{-1}/r)$. We present evidence that, as $r$ tends to $\infty$, these invariants grow exponentially with growth rates respectively given by the hyperbolic and the complex volume of the manifold. This reveals an asymptotic behavior that is different from that of Witten's Asymptotic Expansion Conjecture, which predicts polynomial growth of these invariants when evaluated at the standard root of unity. This new phenomenon suggests that the Reshetikhin-Turaev invariants may have a geometric interpretation other than the original one via $SU(2)$ Chern-Simons gauge theory. For almost all Riemannian metrics (in the C1 Baire sense) on a closed manifold M^{n+1}, 3 \leq n + 1 \leq 7, we prove that there is a sequence of closed, smooth, embedded, connected minimal hypersurfaces that is equidistributed in M. This gives a quantitative version of a result by Irie and the first two authors, that established density of minimal hypersurfaces for generic metrics. The main tool is the Weyl Law for the Volume Spectrum proven by Liokumovich and the first two authors . Folded concave penalization methods have been shown to enjoy the strong oracle property for high-dimensional sparse estimation. However, a folded concave penalization problem usually has multiple local solutions and the oracle property is established only for one of the unknown local solutions. A challenging fundamental issue still remains that it is not clear whether the local optimum computed by a given optimization algorithm possesses those nice theoretical properties. To close this important theoretical gap in over a decade, we provide a unified theory to show explicitly how to obtain the oracle solution via the local linear approximation algorithm. For a folded concave penalized estimation problem, we show that as long as the problem is localizable and the oracle estimator is well behaved, we can obtain the oracle estimator by using the one-step local linear approximation. In addition, once the oracle estimator is obtained, the local linear approximation algorithm converges, namely, it produces the same estimator in the next iteration. The general theory is demonstrated by using four classical sparse estimation problems, that is, sparse linear regression, sparse logistic regression, sparse precision matrix estimation, and sparse quantile regression.
First it is easy to see that if $ a_i \leq0 $ at optimal solution we have $x_i = 0$, and we can omit the variable $x_i$ so let's first assume $a_i$ are distinct hence (WLOG by rearranging x_i) $$ 0 < a_1 < a_2 < a_3 ... < a_n $$ Now let assume at the Unique optimal solution, say $(x^_1 , x^_2,..., x^_n)$, we have $\sum\limits_{i = 1}^nx^_i = \epsilon,$ obviously based our assumption $ \epsilon > 0. $ So one can rewrite problem equivalently as$$ \begin{array}{ll} \text{maximize} & \sum\limits_{i = 1}^na_ix_i\\ \text{subject to} & \\ & \sum\limits_{i = 1}^nx_i = \epsilon \\ & 0\leq x_1,x_2,...,x_n \leq 1\end{array} $$ This is a linear programing. A simple analysis reveals that the optimal solution is $(0,0,..0, \epsilon -j,1,1,..,1)$. Where numbers of $1$s is $j$ and $ 0 \leq x^_{n-j} = \epsilon - j < 1.$ Clarification about the index $j:$ depending $\epsilon$ the index $j$ can be any number within the set $\{0,1,2,..,n\}$, for example if $\epsilon < 1$ then optimal solution is $(0,0,..0,\epsilon)$ i.e., $j=0$, or if $\epsilon = \frac{3}{2}$ then the optimal solution is $(0,0,..0,\frac{1}{2},1)$ i.e., $j=1$, or if $\epsilon =n$ then optimal solution is $(1,1,...,1)$ i.e., $j=n$. Now to finish problem we only need find $\epsilon$ (or $x^_{n-j}$)! To End this, back to original problem we know at optimal solution we have $$ x_1 = x_2 = x_{n-j-1} = 0, ~ x_{n-j+1} = x_{n-j+2}=...=x_{n} =1 $$ so we only need to maximize $a_{n-j} x_{n-j} - (x_{n-j} + j )^2 $ over the interval constraint $ x_{n-j} \in [0,1]$, Hence, $x^*_{n-j} = \max \{ \frac{a_{n-j}}{2} - j , 0 \}.$ Therefore in general the closed formula is obtained if $a_I$ are distinct. For the case $a_i$ are not distinct, one can merge all variables with same $a_i$! and consider similar approach to get result.
Since $On⊂L⊆V$, properties of ordinals that depend on the absence of a function or other structure (i.e. $\Pi_1^{ZF}$ formulas) are preserved when going down from $V$ to $L$. Hence initial ordinals of cardinals remain initial in L. Regular ordinals remain regular in $L$. Weak limit cardinals become strong limit cardinals in $L$ because the generalized continuum hypothesis holds in $L$. Weakly inaccessible cardinals become strongly inaccessible. Weakly Mahlo cardinals become strongly Mahlo. And more generally, any large cardinal property weaker than 0# (see the list of large cardinal properties) will be retained in $L$. Let $\kappa$ be a Mahlo cardinal in $V$, i.e. let $\kappa$ be inaccessible such that $S:= \{ \alpha \in \kappa \mid \alpha \text{ is regular} \}$ is stationary in $\kappa$. First, note that $L \models \kappa \text{ is inaccessible}$. Indeed, if $L \models \kappa \text{ is not a cardinal}$, then there is some $\mu < \kappa$ and some $f \in L$ such that $L \models f \colon \mu \to \kappa \text{ is surjective}$. However, this is a $\Sigma_{0}$ property and hence, in $V$, $f \colon \mu \to \kappa$ is surjective. Contradiction. Since $\kappa > \omega$ (as an ordinal), this also yields that $L \models \kappa \text{ is uncountable}$. Repeating this argument with cofinal $f \colon \mu \to \kappa$ yields that $L \models \kappa \text{ is regular}$. Since $L \models \operatorname{GCH}$, it now suffices to prove that $L \models \kappa \text{ is a limit cardinal}$. This is trivial, because for any ordinal $\gamma < \kappa$, we have that $(\gamma^{+})^V < \kappa$ and since cardinals in $V$ are cardinals in $L$, this proves $$L \models \forall \gamma < \kappa \exists \gamma < \mu < \kappa \colon \mu \text{ is a cardinal}.$$ Now let $T := \{ \alpha \in \kappa \mid L \models \alpha \text{ is regular} \}$. By the argument given above, any $\alpha$ that is regular in $V$ is regular in $L$ and hence $S \subseteq T$. Suppose that $L \models \kappa \text{ is not Mahlo}$. Then there is some $C \subseteq \kappa$ such that $L \models C \text{ is club in } \kappa \text{ and } C \cap T = \emptyset$. Being club is a $\Sigma_0$, property and hence $C \subseteq \kappa$ is club in $\kappa$. Since $S \subseteq T$, we have that $C \cap S = \emptyset$ and hence $S$ is not stationary in $V$. This is a contradiction and we therefore must have that $L \models T \text{ is stationary}$. Thus, $\kappa$ remains Mahlo in $L$. Let $C(k)$ be the set of club subsets of $k.$ Let $R(k)=\{l\in k: l=cf(l)\}.$ Observe that for any set $S\subset On,$ if $S\in L$ then (1) $\forall a\in On \;[\;\{b\cap S :b\in a\} \in L\;], \; \text {and}$ (2) $\forall a\in On\;[\;a=\cup (a\cap S)\iff L\Vdash (a= \cup (a\cap S)\;].$ (3) Also observe that $\forall a\in On\;[a=\|a\|\implies L\Vdash a=\|a\|.)]$ Let $C(k)$ be the set of club subsets of $k.$ From (1) and (2) we have $ C(k)\supset (C(k))^L.$ Let $R(k)=\{l<k: l=\|l\|\}.$ From (3), we have $R(k)\subset (R(k))^L.$ So for any $c\in (C(k))^L$ we have $c\in C(k), $ so $\emptyset\ne c\cap R(k)\subset c\cap (R(k))^L.$ Remark. If $k=\|k\|>\omega$ and $R(k)$ is stationary in $k,$ then $k$ must be weakly inaccessible. Obviously $k$ can't be a successor cardinal. If $k$ is singular then (i): If $k>cf(k)=\omega $ let $f(n):\omega \to k$ be a co-final strictly increasing map with $f(0)=\omega.$ Let $g(n)=f(n)+1$ for $n\in \omega.$ Then $\{g(n):n\in \omega\}$ is club in $k$ and contains no cardinals. (ii): If $k>cf (k)=l>\omega,$ let $S\subset k$ with $\|S\|=l$ and $\cup S=k.$ Let $C=\{a\in k: l<a=\cup (a\cap S)\}.$ It is easy to show that $C$ is club in $k.$ For $a\in C$ we have $cf (a)\leq \|a\cap S\|\leq l<a.$ So $C$ has no regular members.
In my research I came upon a recursively defined sequence, and I'm pretty sure it converges to $\sqrt{2}$ though I can't prove it easily. I don't think it is a difficult question but I'm not sure. Consider the following sequence of functions over $\mathbb{R}$, where it makes sense: $f_0(x)=0$, $\displaystyle f_{n+1}(x)=\frac{1}{2(x-f_n(x))}\hspace{1cm}$ ($n\geq 0)$. Now let us define the sequence $(x_n)$ by $\displaystyle x_n:=\max\left\{y\in[0,\sqrt{2}[\,:\quad y=\frac{1}{2y}+f_n(y)\right\}$. Question: is is true that $x_n\rightarrow\sqrt{2}$ when $n\rightarrow +\infty$? Numerical evidence strongly suggest that, and it completely makes sense with the problem it originated from. The issue is that the functions $f_n$ have more and more poles as $n$ grows, and there is no function it converges to. It looks like the set of the poles of $f_n$ tends to be dense in $[-\sqrt{2},\sqrt{2}]$ when $n\rightarrow +\infty$, and $f_n$ is always decreasing outside of the poles. The poles seem to accumulate more around $\pm\sqrt{2}$ than around $0$. For visual reference, one can see a graph of $f_{10}$ here In advance, thank you for your interest/time. Edit: I added the fact that I'm only interested in the $y\in[0,\sqrt{2}[$. I don't care what happens outside the interval since then it's trivial.
Such a dramatic change would indeed make it harder to reach space, in a precisely defined sense. The easiest measure of how 'hard' it is to get so space is the escape velocity, which is determined by the planet's mass $M$ and radius $R$ as$$v_\text{esc}=\sqrt{\frac{2GM}R}.$$For a planet like the one you mention, the escape velocity goes up by a factor of $\sqrt{17/2}\approx3$ with respect to the Earth's escape velocity $v_\text{E}$. The reason this matters is that the total mass $M$ of a rocket that will send a payload of mass $m$ into space depends exponentially on the escape velocity. (More accurately, on the change in velocity required, $\Delta v$, which is of the order of $v_\text{esc}$.) This relation is known as the Tsiolkovsky rocket equation and it is one of the fundamental principles of rocket science; it states that$$M=m\exp\left(\frac{v_\text{esc}}{v_\text{exh}}\right),$$where $v_\text{exh}$ is the exhaust velocity. Because of this exponential dependence, if $v_\text{esc}$ goes up from $v_\text{E}$ by a factor of three, the rocket mass will increase by a factor of$$\left(e^{v_\text{E}/v_\text{exh}}\right)^{3},$$which can be a lot more than three. To put some numbers in, $v_\text{E}\approx11.2 \,\text{km}/\text s$ whereas liquid propellants can get up to about $v_\text{exh}\approx 5 \,\text{km}/\text s$. When exponentials are involved the details do matter, but if you put this in you get an overall factor of the order of$$(e^2)^3\approx 400.$$ This would mean, for example, that a behemoth like the 3,000-ton Saturn V would be able to transport about 100kg of payload - about the size of a 'minisatellite', instead of the 45 tons of a fully-fledged Apollo mission. Now, there are a number of ways to get around this restriction, of which many are technological but some are physical. The most obvious one, to me, is that the size of the atmosphere will also change. This obviously depends on the amount and composition of the exoplanet's atmosphere, but with all other things equal, a more massive planet will compress its atmosphere into a thinner layer. This change on length scale is linear: as a first approximation, it is inversely proportional to the surface gravitational acceleration, $$g_0=\frac{GM}{R^2}$$which is about 4 times that of Earth for the exoplanet in question. Thus, if all other things were equal - if the atmospheric composition and surface pressure were the same as Earth's - then you'd only have to go up, say, 40 km for low exoplanet orbit, instead of the ~160 km of low Earth orbit. This is important, because it radically reduces the $\Delta v$ required to get into orbit, and this goes again into the exponential dependence of the Tsiolkovsky equation. The $\Delta v$ to get to a height $h$ is roughly$$\Delta v=\sqrt{\frac{2GM}R-\frac{2GM}{R+h}}=\sqrt{\frac{2GMh}{R+h}},$$and this has now gone down, slightly, for Kepler-10c. (You still need to accelerate to stay in orbit, but that depends on the planet's rate of rotation which is yet another completely unknown variable.) To summarize, then, it will indeed be harder to go to space from such a planet, but under certain circumstances it may be easier to get to orbit. The problem with all this, though, is that details - about the specifics of the planet and its atmosphere, and also about what you want to do - do matter, because of the exponential dependence, which is hard to understand until you run into a good number of walls like this one. As Phil Frost mentions, xkcd what-if is a good place to read about this, but in general, it pays to sit up and pay attention when a variable of interest is on the exponent.
If $F$ is continuously differentiable and smooth, you can use the gradient to walk along the contour. In this way you could walk on the contour starting at $x_1$ (tracing both directions of the contour) and see if you reach $x_2$. In particular, if you're currently at a point $x \in \mathcal{C}$ on the contour, then the gradient $\nabla F(x)$ tells you what direction to walk if you want to stay on the contour: you should walk in the direction $(\nabla F(x))^\perp$, where we define $(a,b)^\perp = (b,-a)$. (In other words, you should walk in the direction perpendicular to the gradient.) In particular, when $\epsilon$ is sufficiently small, $x + \epsilon \cdot (\nabla F(x))^\perp$ will be approximately on the contour. This is because $\nabla F(x)$ points in the direction that most increases $F$, and if we're on a contour and the contour is a line, $\nabla F(x)$ points in a direction perpendicular to that contour. (See, e.g., https://math.stackexchange.com/q/599488/14578.) So in principle if the contour was a simple line (whether closed or open), the following would be sufficient: start at $x_1$ and walk in one direction and see if you eventually reach $x_1$. In parallel, start at $x_2$ and walk in the other direction and see if you eventually reach $x_2$. In practice there are two complications. First, the contour is curved, so the gradient only helps you walk in a direction that is approximately along the contour -- but maybe not exactly. So, I suggest a refined way to walk at the contour. At you're currently at the point $x$, first to the point $x' = x + \epsilon \cdot (\nabla F(x))^\perp$. The point $x'$ might not be exactly on the contour, but it'll probably be close. Next, use gradient descent starting at $x'$ to find a point on the contour that is close to $x'$. If $F$ is everywhere positive, you might do this by minimizing the objective function $\Phi(x'') = F(x'') + c \cdot \\|x''-x'\\|^2$ where $c>0$ is a sufficiently small constant; otherwise, you can use $\Phi(x'') = F(x'')^2 + c \cdot \\|x''-x'\\|^2$. If you have a better way to find the nearest point on the contour, go ahead nad use that. Finally, walk to this point $x''$. This gives you an adjacent point on the contour, that's approximately distance $\epsilon$ away from where you started. Repeat this. The second complication is that the contour might not be a simple line; there might be places where two or more lines meet at an intersection. At this point, $F$ won't be differentiable and the gradient might not be computable or meaningful. You'll have to figure out how to detect that situation; perhaps you can use your domain knowledge about $F$ to help you detect that. If you have no other ideas, one possibility is to pick a dozen random points in the neighborhood of $x$, project each one to the nearest point on the contour, check the gradient, and see if all of them seen consistent with a single nearly-straight line or whether their appear to be multiple lines converging. Once you detect an intersection point, you can use depth-first search (basically, explore all directions out of the intersection point) to traverse the connected component -- then it reduces to the problem of checking whether two vertices in an undirected graph are in the same connected component, and that can be done using DFS. This uses techniques that are quite similar to that proposed in my answer to your previous question on a 3D version of this situation.
Yes, both Theorem 1 and Theorem 2 have constructive proofs. In the following, I will work with rational numbers, but the same argumentswork for any ordered field. (Note that in constructive logic, fields areautomatically discrete -- i.e., any two elements of a field are either equalor not. Therefore, I don't think that $\mathbb{R}$ is a field for anyreasonable constructive definition of $\mathbb{R}$. Perhaps $\mathbb{R}$ issome sort of "pro-field" in the same way as Laurent series over a field are;to be frank, I don't care enough about $\mathbb{R}$ to find out.) I set $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $. Nonnegative matrices My main tool will be the following result from linear optimization: Theorem 3. Let $n\in\mathbb{N}$, $n^{\prime}\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A$ be an $m\times n$-matrix. Let $A^{\prime}$ be an $m\times n^{\prime}$-matrix. Then, exactly one of the following two assertions holds: Assertion L1: There exist two vectors $x\in\mathbb{Q}^{n}$ and $x^{\prime }\in\mathbb{Q}^{n^{\prime}}$ such that $x>0$, $x^{\prime}\geq0$ and $Ax+A^{\prime}x^{\prime}=0$. Assertion L2: There exists a vector $y\in\mathbb{Q}^{m}$ such that $y^{T}A\geq0$, $y^{T}A^{\prime}\geq0$ and $y^{T}A\neq0$. Theorem 3 is Theorem 2.5l in my Elementary derivations of some results oflinear optimization (where I prove it for $\mathbb{R}$ instead of$\mathbb{Q}$, but the proof works over any ordered field). It seems to be due to Motzkin; it is actually a particular case of what is called "solvability of f)" on the EoM page for the Motzkin transposition theorem (note that the general case is easily seen to follow from this particular case). It is one of several results that are roughly equivalent to Farkas's lemma or the duality of linear programming. Its constructive proof (there are probably much better sources than my notes -- perhaps Motzkin's thesis?) is more or less based on Fourier-Motzkin elimination. Recall that if $a=\left( a_{1},a_{2},\ldots,a_{n}\right) ^{T}\in\mathbb{Q}^{n}$ and $b=\left( b_{1},b_{2},\ldots,b_{n}\right) ^{T}\in\mathbb{Q}^{n}$ are two vectors of the same size, then $a<b$ means"$a_{i}<b_{i}$ for each $i\in\left\{ 1,2,\ldots,n\right\}$", whereas $a\leq b$ means "$a_{i}\leqb_{i}$ for each $i\in\left\{ 1,2,\ldots,n\right\} $". Thus,saying "$a<b$" is not the same as saying"$a\leq b$ and $a\neq b$". The pedant in mewants to add that it isn't even stronger, because the vector $0_{0}=\left({}\right) ^{T}\in\mathbb{Q}^{0}$ (with no entries at all) satisfies$0_{0}<0_{0}$ but not $0_{0}\neq0_{0}$. Note also that if two vectors $a$ and $b$ satisfy $a\geq b$, then $a^{T}\geqb^{T}$. The same holds for $\leq$, $>$ and $<$. With this out of our way, let's state a few simple lemmas: Lemma 4. Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A\in \mathbb{Q}^{n\times m}$ be a matrix whose entries are nonnegative. Let $v\in\mathbb{Q}^{m}$ be such that $v\geq0$. Then, $Av\geq0$. Proof of Lemma 4. Clear from the definition of $Av$. Lemma 5. Let $n$ be a positive integer. Let $A\in\mathbb{Q}^{n\times n}$ be a matrix whose entries are nonnegative. Let $y\in\mathbb{Q}^{n}$ and $z\in\mathbb{Q}^{n}$ be such that $y\geq0$, $z\geq0$, $Ay\geq y$ and $Az<z$. Then, $y=0$. Proof of Lemma 5. Assume the contrary. Thus, $y\neq0$. From $Az<z$, we obtain $z>Az\geq0$ (by Lemma 4). Write $y$ as $y=\left(y_{1},y_{2},\ldots,y_{n}\right) ^{T}$, and write $z$ as $z=\left(z_{1},z_{2},\ldots,z_{n}\right) ^{T}$. The coordinates $z_{1},z_{2},\ldots,z_{n}$ of $z$ are positive (since $z>0$), while the coordinates$y_{1},y_{2},\ldots,y_{n}$ of $y$ are nonnegative (since $y\geq0$), and atleast one of them is positive (since $y\neq0$). Hence, $\max\left\{y_{i}/z_{i}\ \mid\ i\in\left\{ 1,2,\ldots,n\right\} \right\} $ is awell-defined positive rational number. Denote this number by $\mu$. Thus, foreach $j\in\left\{ 1,2,\ldots,n\right\} $, we have $y_{j}/z_{j}\leq\mu$, sothat\begin{equation}y_{j}\leq z_{j}\mu.\tag{1} \label{pf.l5.0}\end{equation} Now, write the $n\times n$-matrix $A$ in the form $A=\left( a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$. Then, $a_{i,j}\geq0$ for all $i$ and $j$(since all entries of $A$ are nonnegative). We have $Ay\geq y$; in otherwords,\begin{equation}\sum_{j=1}^{n}a_{i,j}y_{j}\geq y_{i}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{ 1,2,\ldots,n\right\} .\tag{2} \label{pf.l5.1}\end{equation}Also, $Az<z$; in other words,\begin{equation}\sum_{j=1}^{n}a_{i,j}z_{j}<z_{i}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{ 1,2,\ldots,n\right\} .\tag{3} \label{pf.l5.2}\end{equation}Hence, for each $i\in\left\{ 1,2,\ldots,n\right\} $, we have\begin{align}y_{i} & \leq\sum_{j=1}^{n}a_{i,j}\underbrace{y_{j}}_{\substack{\leq z_{j}\mu\\\text{(by \eqref{pf.l5.0})}}}\ \ \ \ \ \ \ \ \ \ \left( \text{by\eqref{pf.l5.1}}\right) \\& \leq\underbrace{\sum_{j=1}^{n}a_{i,j}z_{j}}_{\substack{<z_{i}\\\text{(by\eqref{pf.l5.2})}}}\mu<z_{i}\mu\ \ \ \ \ \ \ \ \ \ \left( \text{since }\mu\text{ is positive}\right) .\end{align}Thus, $y_{i}/z_{i}<\mu$ for each $i\in\left\{ 1,2,\ldots,n\right\} $. Hence,$\max\left\{ y_{i}/z_{i}\ \mid\ i\in\left\{ 1,2,\ldots,n\right\} \right\}<\mu$. This contradicts $\mu=\max\left\{ y_{i}/z_{i}\ \mid\ i\in\left\{1,2,\ldots,n\right\} \right\} $. This completes the proof of Lemma 5. Proof of Theorem 1. The entries of the matrix $A$ are nonnegative; thus, thesame holds for the matrix $A^{T}$. As usual, let $I_{n}$ denote the $n\times n$ identity matrix. Theorem 3(applied to $n$, $n$, $I_{n}$ and $A-I_{n}$ instead of $m$, $n^{\prime}$, $A$and $A^{\prime}$) yields that we are in one of the following two cases: Case 1: There exist two vectors $x\in\mathbb{Q}^{n}$ and $x^{\prime}\in\mathbb{Q}^{n}$ such that $x>0$, $x^{\prime}\geq0$ and $I_{n}x+\left(A-I_{n}\right) x^{\prime}=0$. Case 2: There exists a vector $y\in\mathbb{Q}^{n}$ such that $y^{T}I_{n}\geq0$, $y^{T}\left( A-I_{n}\right) \geq0$ and $y^{T}I_{n}\neq0$. Let us first consider Case 1. In this case, there exist two vectors$x\in\mathbb{Q}^{n}$ and $x^{\prime}\in\mathbb{Q}^{n}$ such that $x>0$,$x^{\prime}\geq0$ and $I_{n}x+\left( A-I_{n}\right) x^{\prime}=0$. Considerthese $x$ and $x^{\prime}$. The vector $x^{\prime}$ has nonnegativecoordinates (since $x^{\prime}\geq0$). From $I_{n}x+\left( A-I_{n}\right)x^{\prime}=0$, we obtain $0=I_{n}x+\left( A-I_{n}\right) x^{\prime}=x+Ax^{\prime}-x^{\prime}$, so that $Ax^{\prime}-x^{\prime}=-x<0$ (since$x>0$). In other words, $Ax^{\prime}<x^{\prime}$. Hence, $x^{\prime}\neq0$.Thus, there exists a nonzero vector $v\in\mathbb{Q}^{n}$ with nonnegativecoordinates such that $Av<v$ (namely, $v=x^{\prime}$). Hence, Theorem 1 isproven in Case 1. Let us now consider Case 2. In this case, there exists a vector $y\in\mathbb{Q}^{n}$ such that $y^{T}I_{n}\geq0$, $y^{T}\left( A-I_{n}\right)\geq0$ and $y^{T}I_{n}\neq0$. Denote this vector $y$ by $u$. Thus,$u\in\mathbb{Q}^{n}$ satisfies $u^{T}I_{n}\geq0$, $u^{T}\left( A-I_{n}\right) \geq0$ and $u^{T}I_{n}\neq0$. We have $u^{T}=u^{T}I_{n}\geq0$ andthus $u\geq0$. Also, $u^{T}=u^{T}I_{n}\neq0$ and thus $u\neq0$. Moreover,$u^{T}A-u^{T}=u^{T}\left( A-I_{n}\right) \geq0$ and thus $u^{T}A\geq u^{T}$.Taking transposes, we find $\left( u^{T}A\right) ^{T}\geq\left(u^{T}\right) ^{T}=u$, so that $u\leq\left( u^{T}A\right) ^{T}=A^{T}u$. Inother words, $A^{T}u\geq u$. Next, Theorem 3 (applied to $n$, $n$, $I_{n}$ and $A^{T}-I_{n}$ instead of$m$, $n^{\prime}$, $A$ and $A^{\prime}$) yields that we are in one of thefollowing two subcases: Subcase 2.1: There exist two vectors $x\in\mathbb{Q}^{n}$ and $x^{\prime}\in\mathbb{Q}^{n}$ such that $x>0$, $x^{\prime}\geq0$ and $I_{n}x+\left(A^{T}-I_{n}\right) x^{\prime}=0$. Subcase 2.2: There exists a vector $y\in\mathbb{Q}^{n}$ such that$y^{T}I_{n}\geq0$, $y^{T}\left( A^{T}-I_{n}\right) \geq0$ and $y^{T}I_{n}\neq0$. Let us first consider Subcase 2.1. In this subcase, there exist two vectors$x\in\mathbb{Q}^{n}$ and $x^{\prime}\in\mathbb{Q}^{n}$ such that $x>0$,$x^{\prime}\geq0$ and $I_{n}x+\left( A^{T}-I_{n}\right) x^{\prime}=0$.Consider these $x$ and $x^{\prime}$. From $I_{n}x+\left( A^{T}-I_{n}\right)x^{\prime}=0$, we obtain $0=I_{n}x+\left( A^{T}-I_{n}\right) x^{\prime}=x+A^{T}x^{\prime}-x^{\prime}$, so that $A^{T}x^{\prime}-x^{\prime}=-x<0$(since $x>0$). In other words, $A^{T}x^{\prime}<x^{\prime}$. Hence, Lemma 5(applied to $A^{T}$, $u$ and $x^{\prime}$ instead of $A$, $y$ and $z$) yields$u=0$. This contradicts $u\neq0$. This contradiction shows that Subcase 2.1cannot happen. Hence, we are in Subcase 2.2. Thus, there exists a vector $y\in\mathbb{Q}^{n}$such that $y^{T}I_{n}\geq0$, $y^{T}\left( A^{T}-I_{n}\right) \geq0$ and$y^{T}I_{n}\neq0$. Consider this $y$. We have $y^{T}=y^{T}I_{n}\geq0$, so that$y\geq0$. We have $y^{T}=y^{T}I_{n}\neq0$, so that $y\neq0$. Thus, $y$ is anonzero vector in $\mathbb{Q}^{n}$ with nonnegative coordinates (since$y\geq0$). We have $y^{T}A^{T}-y^{T}=y^{T}\left( A^{T}-I_{n}\right) \geq0$,so that $y^{T}A^{T}\geq y^{T}$. Taking transposes, we obtain $\left(y^{T}A^{T}\right) ^{T}\geq\left( A^{T}\right) ^{T}$. This rewrites as$Ay\geq y$. Hence, there exists a nonzero vector $v\in\mathbb{Q}^{n}$ withnonnegative coordinates such that $Av\geq v$ (namely, $v=y$). Hence, Theorem 1is proven in Case 2. We have now proven Theorem 1 in both Cases 1 and 2; thus we are done. Irreducible nonnegative matrices Next, we take aim at Theorem 2. For any $n\times m$-matrix $A\in\mathbb{Q}^{n\times m}$ and any $i\in\left\{1,2,\ldots,n\right\} $ and $j\in\left\{ 1,2,\ldots,m\right\} $, we let$A_{i,j}$ denote the $\left( i,j\right) $-th entry of $A$. We will need the following equivalent conditions for irreducibility: Theorem 6. Let $n$ be a positive integer. Let $A\in\mathbb{Q}^{n\times n}$ be an $n\times n$-matrix whose entries are nonnegative. Then, the following statements are equivalent: Statement 1: The matrix $A$ is irreducible. Statement 2: For any $i\in\left\{ 1,2,\ldots,n\right\} $ and $j\in\left\{ 1,2,\ldots,n\right\} $, there exists some $k\in\mathbb{N}$ such that $\left( A^{k}\right) _{i,j}>0$. Statement 3: There exists some $m\in\mathbb{N}$ such that all entries of the matrix $A^{0}+A^{1}+\cdots+A^{m}$ are positive. Statement 4: If $\left\{ U,V\right\} $ is a partition of $\left\{ 1,2,\ldots,n\right\} $ into two nonempty subsets $U$ and $V$, then there exist $u\in U$ and $v\in V$ satisfying $A_{u,v}>0$. Theorem 6 is well-known (it is the algebraic analogue of the equivalence ofdifferent definitions of "strong connectivity" for a directed graph), and theproof you will find in the literature (e.g., on Markov chains, I believe) isalready constructive. All I will need is the implication from Statement 1 toStatement 3. We will also need the following simple lemma: Lemma 7. Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A\in \mathbb{Q}^{n\times m}$ be a matrix whose entries are positive. Let $v\in\mathbb{Q}^{m}$ be such that $v\geq0$ and $v\neq0$. Then, $Av>0$. Proof of Lemma 7. This is similar to Lemma 4, except that each coordinate of$Av$ is a sum of nonnegative addends containing at least one positiveaddend, and thus is positive. Proof of Theorem 2. The matrix $A$ is irreducible. Thus, by Theorem 6 (moreprecisely, by the implication $\left( \text{Statement 1}\right)\Longrightarrow\left( \text{Statement 3}\right) $), there exists some$m\in\mathbb{N}$ such that all entries of the matrix $A^{0}+A^{1}+\cdots+A^{m}$ are positive. Consider this $m$, and set $B=A^{0}+A^{1}+\cdots+A^{m}\in\mathbb{Q}^{n\times n}$. Then, all entries of the matrix $B$ are positive. Theorem 1 shows that there exists a nonzero vector $v\in\mathbb{Q}^{n}$ withnonnegative coordinates such that either $Av\geq v$ or $Av<v$. Consider such a$v$, and denote it by $w$. Thus, $w\in\mathbb{Q}^{n}$ is a nonzero vector withnonnegative coordinates and satisfies either $Aw\geq w$ or $Aw<w$. Define a vector $v\in\mathbb{Q}^{n}$ by $v=Bw$. We shall show that $v$ is anonzero vector with positive coordinates such that either $Av>v$ or $Av=v$or $Av<v$. This will clearly prove Theorem 2. We have $w\geq0$ (since the vector $w$ has nonnegative coordinates) and$w\neq0$ (since $w$ is nonzero). Hence, Lemma 7 (applied to $n$, $B$ and $w$instead of $m$, $A$ and $v$) yields that $Bw>0$. Thus, $v=Bw>0$. In other words,$v$ is a vector with positive coordinates. Hence, $v$ is nonzero (since$n>0$). We have $B=A^{0}+A^{1}+\cdots+A^{m}$. Thus, $AB=A^{1}+A^{2}+\cdots+A^{m+1}=BA$. Now, recall that either $Aw\geq w$ or $Aw<w$. Hence, we are in one ofthe following three cases: Let us first consider Case 1. In this case, we have $Aw\geq w$ and $Aw\neq w$.In other words, $Aw-w\geq0$ and $Aw-w\neq0$. Hence, Lemma 7 (applied to $n$, $B$and $Aw-w$ instead of $m$, $A$ and $v$) yields $B\left( Aw-w\right) >0$. From$v=Bw$, we obtain\begin{equation}Av-v=\underbrace{AB}_{=BA}w-Bw=BAw-Bw=B\left( Aw-w\right) >0.\end{equation}In other words, $Av>v$. Thus, in Case 1, we have proven on our goal (namely,that either $Av>v$ or $Av=v$ or $Av<v$). Let us next consider Case 2. In this case, we have $Aw=w$. Now, from $v=Bw$,we obtain $Av=\underbrace{AB}_{=BA}w=B\underbrace{Aw}_{=w}=Bw=v$. Thus, inCase 2, we have proven on our goal (namely, that either $Av>v$ or $Av=v$ or$Av<v$). Let us finally consider Case 3. In this case, we have $Aw<w$. Hence, $w-Aw>0$.Therefore, $w-Aw\geq0$ and $w-Aw\neq0$. Hence, Lemma 7 (applied to $n$, $B$ and$w-Aw$ instead of $m$, $A$ and $v$) yields $B\left( w-Aw\right) >0$. From $v=Bw$,we obtain\begin{equation}v-Av=Bw-\underbrace{AB}_{=BA}w=Bw-BAw=B\left( w-Aw\right) >0.\end{equation}In other words, $Av<v$. Thus, in Case 3, we have proven on our goal (namely,that either $Av>v$ or $Av=v$ or $Av<v$). We have now proven our goal in each of the three Cases 1, 2 and 3. Thus, wealways have either $Av>v$ or $Av=v$ or $Av<v$. As we have said, this provesTheorem 2.
X Search Filters Format Subjects Library Location Language Publication Date Click on a bar to filter by decade Slide to change publication date range 1. Measurement of the ratio of the production cross sections times branching fractions of B c ± → J/ψπ ± and B± → J/ψK ± and ℬ B c ± → J / ψ π ± π ± π ∓ / ℬ B c ± → J / ψ π ± $$ \mathrm{\mathcal{B}}\left({\mathrm{B}}_{\mathrm{c}}^{\pm}\to \mathrm{J}/\psi {\pi}^{\pm }{\pi}^{\pm }{\pi}^{\mp}\right)/\mathrm{\mathcal{B}}\left({\mathrm{B}}_{\mathrm{c}}^{\pm}\to \mathrm{J}/\psi {\pi}^{\pm}\right) $$ in pp collisions at s = 7 $$ \sqrt{s}=7 $$ TeV Journal of High Energy Physics, ISSN 1029-8479, 1/2015, Volume 2015, Issue 1, pp. 1 - 30 The ratio of the production cross sections times branching fractions σ B c ± ℬ B c ± → J / ψ π ± / σ B ± ℬ B ± → J / ψ K ± $$ \left(\sigma... B physics \| Branching fraction \| Hadron-Hadron Scattering \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory B physics \| Branching fraction \| Hadron-Hadron Scattering \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory Journal Article Monthly Notices of the Royal Astronomical Society, ISSN 0035-8711, 2018, Volume 480, Issue 3, pp. 3224 - 3242 The C-Band All-Sky Survey (C-BASS) is an all-sky full-polarization survey at a frequency of 5 GHz, designed to provide complementary data to the all-sky... Surveys \| Instrumentation: miscellaneous \| Diffuse radiation \| Methods: data analysis \| Cosmic microwave background \| Radio continuum: general \| ROTATION \| ANOMALOUS MICROWAVE EMISSION \| methods: data analysis \| GALACTIC SYNCHROTRON EMISSION \| diffuse radiation \| radio continuum: general \| surveys \| BAYESIAN COMPONENT SEPARATION \| POLARIZED EMISSION \| instrumentation: miscellaneous \| SPECTRAL INDEX \| ASTRONOMY & ASTROPHYSICS \| cosmic microwave background \| FULL-SKY \| RADIO-CONTINUUM EMISSION \| PROBE \| MHZ Surveys \| Instrumentation: miscellaneous \| Diffuse radiation \| Methods: data analysis \| Cosmic microwave background \| Radio continuum: general \| ROTATION \| ANOMALOUS MICROWAVE EMISSION \| methods: data analysis \| GALACTIC SYNCHROTRON EMISSION \| diffuse radiation \| radio continuum: general \| surveys \| BAYESIAN COMPONENT SEPARATION \| POLARIZED EMISSION \| instrumentation: miscellaneous \| SPECTRAL INDEX \| ASTRONOMY & ASTROPHYSICS \| cosmic microwave background \| FULL-SKY \| RADIO-CONTINUUM EMISSION \| PROBE \| MHZ Journal Article 3. The "C" students guide to scholarships : a creative guide to finding scholarships w hen your grades suck and your parents are broke! 2012, ISBN 0768936152, vii, 208 Book 4. Association of Serum Interleukin 6 and C-Reactive Protein in Childhood With Depression and Psychosis in Young Adult Life: A Population-Based Longitudinal Study JAMA Psychiatry, ISSN 2168-622X, 10/2014, Volume 71, Issue 10, pp. 1121 - 1128 IMPORTANCE: Longitudinal studies have linked the systemic inflammatory markers interleukin 6 (IL-6) and C-reactive protein (CRP) with the risk of developing... RISK-FACTORS \| NONAFFECTIVE PSYCHOSIS \| PSYCHIATRY \| AUTOIMMUNE-DISEASES \| DOUBLE-BLIND \| SCHIZOPHRENIA \| MAJOR DEPRESSION \| CORONARY-HEART-DISEASE \| SURVEY REPLICATION \| SEVERE INFECTIONS \| NATIONAL COMORBIDITY SURVEY \| Depression - epidemiology \| Prospective Studies \| Humans \| Risk Factors \| Male \| Biomarkers - blood \| Psychotic Disorders - blood \| Psychotic Disorders - epidemiology \| Psychiatric Status Rating Scales \| Interleukin-6 - blood \| England - epidemiology \| Adolescent \| Female \| Interview, Psychological \| C-Reactive Protein - analysis \| Child \| Depression - blood \| C-reactive protein \| Research \| Psychoses \| Depression in children \| Risk factors \| Interleukin-6 RISK-FACTORS \| NONAFFECTIVE PSYCHOSIS \| PSYCHIATRY \| AUTOIMMUNE-DISEASES \| DOUBLE-BLIND \| SCHIZOPHRENIA \| MAJOR DEPRESSION \| CORONARY-HEART-DISEASE \| SURVEY REPLICATION \| SEVERE INFECTIONS \| NATIONAL COMORBIDITY SURVEY \| Depression - epidemiology \| Prospective Studies \| Humans \| Risk Factors \| Male \| Biomarkers - blood \| Psychotic Disorders - blood \| Psychotic Disorders - epidemiology \| Psychiatric Status Rating Scales \| Interleukin-6 - blood \| England - epidemiology \| Adolescent \| Female \| Interview, Psychological \| C-Reactive Protein - analysis \| Child \| Depression - blood \| C-reactive protein \| Research \| Psychoses \| Depression in children \| Risk factors \| Interleukin-6 Journal Article 5. Total synthesis of hibispeptin A via Pd-catalyzed C(sp 3 )-H arylation with sterically hindered aryl iodides Organic Letters, ISSN 1523-7060, 12/2014, Volume 16, Issue 24, pp. 6488 - 6491 To access the key Ile-Hpa pseudodipeptide motif in hibispeptins, a series of bidentate carboxamide-based auxiliary groups have been explored to facilitate the... C(SP)-H BONDS \| CYCLIC PEPTIDE \| COUPLING REACTIONS \| INTRAMOLECULAR AMINATION \| NATURAL-PRODUCTS \| BOND FUNCTIONALIZATION \| CHEMISTRY, ORGANIC \| ALPHA-AMINO-ACIDS \| UNACTIVATED C(SP)-H \| C-H FUNCTIONALIZATION \| HIBISCUS-SYRIACUS \| Peptides, Cyclic - isolation & purification \| Hydrogen Bonding \| Peptides, Cyclic - chemical synthesis \| Plant Roots - chemistry \| Hibiscus - chemistry \| Iodides - chemistry \| Molecular Structure \| Catalysis \| Palladium - chemistry \| Peptides, Cyclic - chemistry C(SP)-H BONDS \| CYCLIC PEPTIDE \| COUPLING REACTIONS \| INTRAMOLECULAR AMINATION \| NATURAL-PRODUCTS \| BOND FUNCTIONALIZATION \| CHEMISTRY, ORGANIC \| ALPHA-AMINO-ACIDS \| UNACTIVATED C(SP)-H \| C-H FUNCTIONALIZATION \| HIBISCUS-SYRIACUS \| Peptides, Cyclic - isolation & purification \| Hydrogen Bonding \| Peptides, Cyclic - chemical synthesis \| Plant Roots - chemistry \| Hibiscus - chemistry \| Iodides - chemistry \| Molecular Structure \| Catalysis \| Palladium - chemistry \| Peptides, Cyclic - chemistry Journal Article 6. Functional Suppression by$FoxP3^{+}CD4^{+}CD25^{high}$Regulatory T Cells during Acute Hepatitis C Virus Infection The Journal of Infectious Diseases, ISSN 0022-1899, 1/2008, Volume 197, Issue 1, pp. 46 - 57 Background. Infection with hepatitis C virus (HCV) is characterized by impairment of viral effector T cell responses and a high propensity for viral... T lymphocytes \| Transcription factors \| Infectious diseases \| RNA \| Antigen presenting cells \| Viruses \| Infections \| Transponders \| Hepatitis C \| Hepacivirus \| IMMUNE-RESPONSE \| INFECTIOUS DISEASES \| ACTIVATION \| IN-VITRO PROLIFERATION \| MULTIPLE-SCLEROSIS \| DENDRITIC CELLS \| DYSFUNCTION \| CORE PROTEIN \| FOXP3 \| VIRAL CLEARANCE \| LYMPHOCYTES \| Acute Disease \| T-Lymphocytes, Regulatory - classification \| Hepacivirus - immunology \| Viremia \| Humans \| Middle Aged \| Male \| Case-Control Studies \| T-Lymphocytes, Regulatory - immunology \| Hepatitis C - immunology \| Fibrinogens, Abnormal \| Forkhead Transcription Factors - metabolism \| Adolescent \| Aged, 80 and over \| Adult \| Female \| Hepatitis C - physiopathology \| CD24 Antigen - immunology \| Immunity, Cellular \| Longitudinal Studies \| Interleukin-2 Receptor alpha Subunit - immunology \| T-Lymphocytes, Regulatory - virology \| Immunosuppression \| Patient outcomes \| Physiological aspects \| Development and progression \| Research \| T cells T lymphocytes \| Transcription factors \| Infectious diseases \| RNA \| Antigen presenting cells \| Viruses \| Infections \| Transponders \| Hepatitis C \| Hepacivirus \| IMMUNE-RESPONSE \| INFECTIOUS DISEASES \| ACTIVATION \| IN-VITRO PROLIFERATION \| MULTIPLE-SCLEROSIS \| DENDRITIC CELLS \| DYSFUNCTION \| CORE PROTEIN \| FOXP3 \| VIRAL CLEARANCE \| LYMPHOCYTES \| Acute Disease \| T-Lymphocytes, Regulatory - classification \| Hepacivirus - immunology \| Viremia \| Humans \| Middle Aged \| Male \| Case-Control Studies \| T-Lymphocytes, Regulatory - immunology \| Hepatitis C - immunology \| Fibrinogens, Abnormal \| Forkhead Transcription Factors - metabolism \| Adolescent \| Aged, 80 and over \| Adult \| Female \| Hepatitis C - physiopathology \| CD24 Antigen - immunology \| Immunity, Cellular \| Longitudinal Studies \| Interleukin-2 Receptor alpha Subunit - immunology \| T-Lymphocytes, Regulatory - virology \| Immunosuppression \| Patient outcomes \| Physiological aspects \| Development and progression \| Research \| T cells Journal Article 7. Vitamins C and E for prevention of pre-eclampsia in women with type 1 diabetes (DAPIT): a randomised placebo-controlled trial Lancet, The, ISSN 0140-6736, 2010, Volume 376, Issue 9737, pp. 259 - 266 Summary Background Results of several trials of antioxidant use during pregnancy have not shown a reduction in pre-eclampsia, but the effect in women with... Internal Medicine \| SUPPLEMENTATION \| MEDICINE, GENERAL & INTERNAL \| PREGNANT-WOMEN \| RISK \| OXIDATIVE STRESS \| Pregnancy \| Young Adult \| Pre-Eclampsia - prevention & control \| Antioxidants - therapeutic use \| Ascorbic Acid - therapeutic use \| Oxidative Stress \| Humans \| Vitamin E - therapeutic use \| Adult \| Female \| Diabetes Mellitus, Type 1 - complications \| Pre-Eclampsia - physiopathology \| Prevention \| Vitamin E \| Preeclampsia \| Vitamin C \| Dosage and administration \| Diabetes \| Risk factors \| Hypertension \| Studies \| Nutrition \| Womens health \| Pathogenesis \| Infant mortality \| Autoimmune diseases \| Pharmaceuticals Internal Medicine \| SUPPLEMENTATION \| MEDICINE, GENERAL & INTERNAL \| PREGNANT-WOMEN \| RISK \| OXIDATIVE STRESS \| Pregnancy \| Young Adult \| Pre-Eclampsia - prevention & control \| Antioxidants - therapeutic use \| Ascorbic Acid - therapeutic use \| Oxidative Stress \| Humans \| Vitamin E - therapeutic use \| Adult \| Female \| Diabetes Mellitus, Type 1 - complications \| Pre-Eclampsia - physiopathology \| Prevention \| Vitamin E \| Preeclampsia \| Vitamin C \| Dosage and administration \| Diabetes \| Risk factors \| Hypertension \| Studies \| Nutrition \| Womens health \| Pathogenesis \| Infant mortality \| Autoimmune diseases \| Pharmaceuticals Journal Article MONTHLY NOTICES OF THE ROYAL ASTRONOMICAL SOCIETY, ISSN 0035-8711, 04/2019, Volume 484, Issue 4, pp. 5377 - 5388 The C-Band All-Sky Survey (C-BASS) is an all-sky full-polarization survey at a frequency of 5 GHz, designed to provide data complementary to the all-sky... DESIGN \| methods: data analysis \| POLARIMETRY \| ASTRONOMY & ASTROPHYSICS \| radio continuum: general \| IMPLEMENTATION \| instrumentation: polarimeters DESIGN \| methods: data analysis \| POLARIMETRY \| ASTRONOMY & ASTROPHYSICS \| radio continuum: general \| IMPLEMENTATION \| instrumentation: polarimeters Journal Article 9. HERSCHEL/HIFI SPECTRAL MAPPING of C+, CH+, and CH in ORION BN/KL: The PREVAILING ROLE of ULTRAVIOLET IRRADIATION in CH+ FORMATION Astrophysical Journal, ISSN 0004-637X, 09/2016, Volume 829, Issue 1, p. 15 The CH+ ion is a key species in the initial steps of interstellar carbon chemistry. Its formation in diverse environments where it is observed is not well... molecular processes \| ISM: clouds \| ISM: lines and bands \| ISM: molecules \| astrochemistry \| ISM: abundances \| PHYSICAL CONDITIONS \| DIFFUSE INTERSTELLAR-MEDIUM \| C-12/C-13 ISOTOPE RATIO \| MEUDON PDR CODE \| KL HOT CORE \| VIBRATIONALLY EXCITED H-2 \| MOLECULAR LINE EMISSION \| STAR-FORMING REGIONS \| ROTATIONAL-EXCITATION \| ASTRONOMY & ASTROPHYSICS \| PHOTON-DOMINATED REGIONS molecular processes \| ISM: clouds \| ISM: lines and bands \| ISM: molecules \| astrochemistry \| ISM: abundances \| PHYSICAL CONDITIONS \| DIFFUSE INTERSTELLAR-MEDIUM \| C-12/C-13 ISOTOPE RATIO \| MEUDON PDR CODE \| KL HOT CORE \| VIBRATIONALLY EXCITED H-2 \| MOLECULAR LINE EMISSION \| STAR-FORMING REGIONS \| ROTATIONAL-EXCITATION \| ASTRONOMY & ASTROPHYSICS \| PHOTON-DOMINATED REGIONS Journal Article 10. The C-Band All-Sky Survey (C-BASS): Constraining diffuse Galactic radio emission in the North Celestial Pole region Monthly Notices of the Royal Astronomical Society, ISSN 0035-8711, 02/2019, Volume 485, Issue 2, pp. 2844 - 2860 The C-Band All-Sky Survey (C-BASS) is a high sensitivity all-sky radio survey at an angular resolution of 45 arcmin and a frequency of 4.7 GHz. We present a... Surveys \| Radiation mechanisms: non-thermal \| Diffuse radiation \| Radio continuum: ISM \| Radiation mechanisms: thermal \| radiation mechanisms: non-thermal \| ANOMALOUS MICROWAVE EMISSION \| radiation mechanisms: thermal \| DUST EMISSION \| 1420 MHZ \| diffuse radiation \| radio continuum: ISM \| surveys \| PROBE WMAP OBSERVATIONS \| CENTIMETER-WAVE CONTINUUM \| SPECTRAL INDEX \| ASTRONOMY & ASTROPHYSICS \| BRIGHT SAMPLE \| SYNCHROTRON EMISSION \| COMPONENT SEPARATION \| POLARIZATION Surveys \| Radiation mechanisms: non-thermal \| Diffuse radiation \| Radio continuum: ISM \| Radiation mechanisms: thermal \| radiation mechanisms: non-thermal \| ANOMALOUS MICROWAVE EMISSION \| radiation mechanisms: thermal \| DUST EMISSION \| 1420 MHZ \| diffuse radiation \| radio continuum: ISM \| surveys \| PROBE WMAP OBSERVATIONS \| CENTIMETER-WAVE CONTINUUM \| SPECTRAL INDEX \| ASTRONOMY & ASTROPHYSICS \| BRIGHT SAMPLE \| SYNCHROTRON EMISSION \| COMPONENT SEPARATION \| POLARIZATION Journal Article The New England Journal of Medicine, ISSN 0028-4793, 04/2010, Volume 362, Issue 14, pp. 1282 - 1291 It has been hypothesized that oxidative stress may underlie the development of preeclampsia. In this large, multicenter, randomized, double-blind trial... WOMEN \| MEDICINE, GENERAL & INTERNAL \| OXIDATIVE STRESS \| CLINICAL-TRIALS \| RANDOMIZED-TRIAL \| RISK \| PROTEINURIA \| RATIO \| PREECLAMPSIA \| Double-Blind Method \| Humans \| Pregnancy Trimester, First \| Pregnancy \| Young Adult \| Pre-Eclampsia - prevention & control \| Antioxidants - therapeutic use \| Hypertension, Pregnancy-Induced - prevention & control \| Ascorbic Acid - therapeutic use \| Treatment Failure \| Vitamin E - therapeutic use \| Adult \| Female \| Oxidative Stress - drug effects \| Pregnancy Complications - prevention & control \| Parity \| Drug Combinations \| Pregnancy Outcome \| Drugs \| Prevention \| Diet therapy \| Vitamin E \| Vitamin C \| Adverse and side effects \| Hypertension in pregnancy \| Health aspects \| Mineral oils \| Blood pressure \| Womens health \| Preeclampsia WOMEN \| MEDICINE, GENERAL & INTERNAL \| OXIDATIVE STRESS \| CLINICAL-TRIALS \| RANDOMIZED-TRIAL \| RISK \| PROTEINURIA \| RATIO \| PREECLAMPSIA \| Double-Blind Method \| Humans \| Pregnancy Trimester, First \| Pregnancy \| Young Adult \| Pre-Eclampsia - prevention & control \| Antioxidants - therapeutic use \| Hypertension, Pregnancy-Induced - prevention & control \| Ascorbic Acid - therapeutic use \| Treatment Failure \| Vitamin E - therapeutic use \| Adult \| Female \| Oxidative Stress - drug effects \| Pregnancy Complications - prevention & control \| Parity \| Drug Combinations \| Pregnancy Outcome \| Drugs \| Prevention \| Diet therapy \| Vitamin E \| Vitamin C \| Adverse and side effects \| Hypertension in pregnancy \| Health aspects \| Mineral oils \| Blood pressure \| Womens health \| Preeclampsia Journal Article
One of the many lovely things about Big MathsJam is that I’ve found My People - I’ve made several very dear friends there, introduced others to the circle, and get to stay in touch with other maths fans through the year. It’s golden. Adam Atkinson is one of those dearRead More → Dear Uncle Colin, I’m given that $0 \le x \lt 180^o$, and that $\cos(x) + \sin(x) = \frac{1}{2}$. I have to find $p$ and $q$ such that $\tan(x) = -\frac{p + \sqrt{q}}{3}$. Where do I even start? - Some Identity Needing Evaluation Hi, SINE, and thanks for your message! ThereRead More → So far in the Dictionary of Mathematical Eponymy, I’ve not picked anyone properly famous. I mean, if you’re a keen recreational mathematician, you’ll have heard of Collatz or Banach; a serious mathematician might know about Daubechies, and a chess enthusiast would conceivably have come across Elo. But everyone has heardRead More → Dear Uncle Colin, I’m told that $5\times 2^x + 1$ (with $x$ a non-negative integer) is a square number - how do I find $x$? - A Baffling Equation. Logs? Hi, ABEL, and thanks for your message! We’re looking for a square number - let’s call it $y^2$ - that’sRead More → Aaaages ago, @vingaints tweeted: This is pretty wild. It feels like what the Basis Representation Theorem is for Integers but for Rational Numbers. Hmm - trying to prove it now. Feels like a tough one. Need to work some examples! https://t.co/tgcy8iaXHa pic.twitter.com/tgcy8iaXHa — Ving Aints (@vingAints) September 18, 2018 InRead More → Dear Uncle Colin, Suppose Team 1 beats Team 2 by a score of 10-7, and Team 2 beats Team 3 by a score of 10-4. How would we predict the score of a match between Team 1 and Team 3? - Make A Team Calculation Happen Hi, MATCH, and thanksRead More → “That looks straightforward,” I thought. “I’ll keep on looking at this geometry puzzle.” Nut-uh. A standard pack of 52 cards is shuffled. The cards are turned over one at a time, and you guess whether each will be red or black. How many correct guesses do you expect to make?Read More → In this month's Wrong, But Useful, we're joined by @sheena2907, who is Sheena in real life. We discuss: Sheena's Number of the Podcast: 3,212 Board Games - Number Fluxx, Prime Climb Magic: The Gathering is undecidable! Oxbridge Time surprises The oddness of the Fibonacci sequence The heights of women BigRead More → Dear Uncle Colin, How would you work out $\br{\frac{4}{5}}^{10}$ in your head? - Probability Estimation Needed, Relic Of Slide-rule Era Hi, PENROSE, and thanks for your message! That’s really two questions: how I would do it, and how the Mathematical Ninja would do it. Me Me, I’d probably use theRead More → Like everyone else on Twitter, I’m a sucker for a nice-looking question, and @cshearer41 is a reliable source of such things. I particularly liked this one: There are two equilateral triangles inside this semicircle. What’s the area of the larger one? pic.twitter.com/Nvy01z2j5f — Catriona Shearer (@Cshearer41) November 7, 2018 StraightRead More →
The time energy uncertainty principle arises from the fact that a function can be described as a sum of terms in a conjugate variable, as is the case of any fourier transform. In this case the variables are frequency and time. The time-energy uncertainty does not seem to have any more fundamental meaning than this, unlike the Heisenberg position-momentum uncertainty principle. As an excited state decays it is necessary to write the wavefunction in a time dependent form. If a pure state has energy $E$ and wavefuntion $\psi$ we can write the time-dependent (or total) wavefunction as $\displaystyle \varphi=\psi e^{-iE_0/\hbar}$ (where $i=\sqrt{-1}$ and $\hbar =h/2\pi$ ) and is a stationary state because the probability of finding it in that state is $\|\varphi\|^2=\|\psi\|^2$ which is a constant. However, if the state decays, because of interactions between electrons/nuclei within a molecule or with nearby molecules (collisions for example), and if the lifetime is $\tau$ then the probability of being in the excited state at time $t$ is $\displaystyle \|\varphi\|^2=\|\psi\|^2 e^{-t/\tau}$ which means that $$ \varphi(t)=\psi e^{-iE_0t/\hbar-t/2\tau}$$ expressing this as a product $\displaystyle \varphi=\psi e^{-iE_0t/\hbar}e^{-t/2\tau}$ and using the mathematical relationship $e^{ix}=\cos(ix)+i\sin(ix)$ shows that the real part of $\varphi$ oscillates due to the cosine term and decays away due to the exponential term in $\tau$. i.e a damped oscillation. The calculation is easier by making some substitutions, for example the frequency is $\omega_0=E_0/\hbar$ and decay rate constant $\gamma=1/\tau$ then $\displaystyle \varphi=\psi \cos(\omega_0 t)e^{-\gamma/2}$. As the excited state is decaying the frequency of the emitted radiation is not monochromatic, which it would be if the state did not decay, but has a frequency distribution that is related to the exponential decay by a fourier transform. Put another way, the exponential decay can be constructed out of a set of waves each with a different frequency and amplitude, just as any normal function can be constructed as a fourier series. This means that we can find a function $A$ such that $$\varphi(t) =\int A(\omega)e^{i\omega t}d\omega$$ where $$A(\omega) \approx \frac{1}{i(\omega-\omega_o)+\gamma/2}+\frac{1}{i(\omega+\omega_o)+\gamma/2}$$ The second fraction approaches zero as $\omega \to \omega_0$, i.e close to the transition frequency (or transition energy) and so is insignificant compared to the first term where the frequency difference tends to zero. The intensity of the transition is $A(\omega)^A(\omega)$ where indicates the complex conjugate giving, $$I(\omega)= \frac{\gamma/2}{(\omega-\omega_0)^2+(\gamma/2)^2}$$ where the intensity is normalised to unity over all frequencies. The full width at half maximum of the transition is $$ \Delta\omega =\gamma , \text{ or} \qquad \Delta E\tau =\hbar $$ which is the 'time-energy' uncertainty after substituting back for the energy.
Model Description NOTE: formulae are not displayed correctly since our wikimedia update. model description Contents 1 Model description 1.1 Overview 1.2 Basic model description 1.3 Introduction of variable climate sensitivities 1.4 Updated carbon cycle 1.5 Other additional capabilities compared to MAGICC4.2 Model description Overview The 'Model for the Assessment of Greenhouse Gas Induced Climate Change' (MAGICC) is a simple/reduced complexity climate model. MAGICC was originally developed by Tom Wigley (National Centre for Atmospheric Research, Boulder, US, and University of Adelaide, Australia) and Sarah Raper (Manchester Metropolitan University, UK) in the late 1980s and continuously developed since then. It has been one of the widely used climate models in various IPCC Assessment Reports. The latest version, MAGICC6, is co-developed by Malte Meinshausen (Potsdam Institute for Climate Impact Research, Germany, and the University of Melbourne, Australia). These pages provide an extensive model description, sourced from a 2011 publication in Atmospheric Chemistry & Physics (M. Meinshausen, S. Raper and T. Wigley, 2011). This Page provides a detailed description of MAGICC6 and itsdifferent modules (see Fig-A1 below). A basic model description is given, while subsections describe MAGICC's carbon cycle, the atmospheric-chemistryparameterizations and derivation of non-CO 2 concentrations, radiative forcing routines, and the climate module to get fromradiative forcing to hemispheric (land and ocean, separately) toglobal-mean temperatures (climate model), aswell as oceanic heat uptake. Finally, details are provided on the implementation scheme for the upwelling-diffusion-entrainment oceanclimate module. A technical upgrade is that MAGICC6 has been re-coded in Fortran95,updated from previous Fortran77 versions. Nearly all of the MAGICC6 code is directly based on the earlierMAGICC versions programmed by Wigley and Raper ( 1987, 1992, 2001). Basic model description MAGICC has a hemispherically averaged upwelling-diffusion ocean coupled to an atmosphere layer and a globally averaged carbon cycle model. As with most other simple models, MAGICC evolved from a simple global average energy-balance equation. The energy balance equation for the perturbed climate system can be written as:<math>\Delta Q_G = \lambda_G \Delta T_G + \frac{d H}{d t}\label{eq_globalenergybalance}</math> 1) where <math>\Delta Q_G</math> is the global-mean radiative forcing at the top of the troposphere. This extra energy influx is partitioned into increased outgoing energy flux and heat content changes in the ocean <math>\frac{d H}{d t}</math>. The outgoing energy flux is dependent on the global-mean feedback factor, <math>\lambda_G</math>, and the surface temperature perturbation <math>\Delta T_G</math>. While MAGICC is designed to provide maximum flexibility in order to match different types of responses seen in more sophisticated models, the approach in MAGICC's model development has always been to derive the simple equations as much as possible from key physical and biological processes. In other words, MAGICC is as simple as possible, but as mechanistic as necessary. This process-based approach has a strong conceptual advantage in comparison to simple statistical fits that are more likely to quickly degrade in their skill when emulating scenarios outside the original calibration space of sophisticated models. The main improvements in MAGICC6 compared to the version used in the IPCC AR4 are briefly highlighted in this section (Note that there is an intermediate version, MAGICC 5.3, described in Wigley et al., 2009). The options introduced to account for variable climate sensitivities are described in Sect. introduction of variable climate sensitivities. With the exception of the updated carbon cycle routines updated carbon cycle, the MAGICC 4.2 and 5.3 parameterizations are covered as special cases of the 6.0 version, i.e., the IPCC AR4 version, for example, can be recovered by appropriate parameter settings. Introduction of variable climate sensitivities Climate sensitivity (<math>\Delta T_{2x}</math>) is a useful metric to compare models and is usually defined as the equilibrium global-mean warming after a doubling of CO<math>_2</math> concentrations. In the case of MAGICC, the equilibrium climate sensitivity is a primary model parameter that may be identified with the eventual global-mean warming that would occur if the CO<math>_2</math> concentrations were doubled from pre-industrial levels. Climate sensitivity is inversely related to the feedback factor <math>\lambda</math>:<math>\label{eq_climatesensitivity}\Delta T_{2x} = \frac{\Delta Q_{2x}}{\lambda}</math> 2) where <math>\Delta T_{2x}</math> is the climate sensitivity, and <math>\Delta Q_{2x}</math> the radiative forcing after a doubling of CO<math>_2</math> concentrations (see energy balance Eq. A45). The (time- or state-dependent) effective climate sensitivity (<math>S^t</math>)(Murphy and Mitchell, 1995) is defined using the transient energy balance Eq. (1) and can be diagnosed from model output for any part of a model run where radiative forcing and ocean heat uptake are both known and their sum is different from zero, so that:<math>\label{eq_effective_climatesensitivity} S^t = \frac{\Delta Q_{2x}}{\lambda^t} = \Delta Q_{2x} \frac{\Delta T_{G}^t}{\Delta Q^t - \frac{d H}{dt}\|^t}</math> 3) where <math>\Delta Q_{2x}</m> is the model-specific forcing for doubled CO<math>_2</math> concentration, <math>\lambda_t</math> is the time-variable feedback factor, <math>\Delta Q^t</math> the radiative forcing, <math>\Delta T_{GL}^t</math> the global-mean temperature perturbation and <math>\frac{dH}{dt}\|^t</math> the climate system's heat uptake at time <math>t</math>. By definition, the traditional (equilibrium) climate sensitivity (<math>\Delta T_{2x}</math>) is equal to the effective climate sensitivity <math>S^t</math> at equilibrium (<math>\frac{dH}{dt}\|^t</math>=0) after doubled (pre-industrial) CO<math>_2</math> concentration. If there were only one globally homogenous, fast and constant feedback process, the diagnosed effective climate sensitivity would always equal the equilibrium climate sensitivity <math>\Delta T_{2x}</math>. However, many CMIP3 AOGCMs exhibit variable effective climate sensitivities, often increasing over time (e.g. models CCSM3, CNRM-CM3, GFDL-CM2.0, GFDL-CM2.1, GISS-EH - see Figs. (B1, B2, B3 in http://www.atmos-chem-phys.net/11/1417/2011/acp-11-1417-2011.html). This is consistent with earlier results of increasing effective sensitivities found by (Senior and Mitchell (2000);Raper et al. (2000)) for the HadCM2 model. In addition, some models present significantly higher sensitivities for higher forcing scenarios (1pctto4x) than for lower forcing scenarios (1pctto2x) (e.g. ECHAM5/MPI-OM and GISS-ER, see Fig.1 In order to better emulate these time-variable effective climate sensitivities, this version of MAGICC incorporates two modifications: Firstly, an amended land-ocean heat exchange formulation allows effective climate sensitivities to increase on the path to equilibrium warming. In this formulation, changes in effective climate sensitivity arise from a geometrical effect: spatially non-homogenous feedbacks can lead to a time-variable effective global-mean climate sensitivity, if the spatial warming distributions change over time. Hence, by modifying land-ocean heat exchange in MAGICC, the spatial evolution of warming is altered, leading to changes in effective climate sensitivities (Raper, 2004) given that MAGICC has different equilibrium sensitivities over land and ocean. Secondly, the climate sensitivities, and hence the feedback parameters, can be made explicitly dependent on the current forcing at time <math>t</math>. Both amendments are detailed in the Revised land-ocean heat formulation, and Accounting for climate-state dependent feedbacks sections. Although these two amendments both modify the same diagnostic, i.e., the time-variable effective sensitivities in MAGICC, they are distinct: the land-ocean heat exchange modification changes the shape of the effective climate sensitivity's time evolution to equilibrium, but keeps the equilibrium sensitivity unaffected. In contrast, making the sensitivity explicitly dependent on the forcing primarily affects the equilibrium sensitivity value. Note that time-varying effective sensitivities are not only empirically observed in AOGCMs, but they are necessary here in order for MAGICC to accurately emulate AOGCM results. Alternative parameterizations to emulate time-variable climate sensitivities are possible, e.g.~assuming a dependence on temperatures instead of forcing, or by implementing indirect radiative forcing effects that are most often regarded as feedbacks see Section 6.2 in http://www.atmos-chem-phys.net/11/1417/2011/acp-11-1417-2011.html. However, this study chose to limit the degrees of freedom with respect to time-variable climate sensitivities given that a clear separation into three (or more) different parameterizations seemed unjustified based on the AOGCM data analyzed here. Updated carbon cycle MAGICC's terrestrial carbon cycle model is a globally integrated box model, similar to that in Harvey (1989) and Wigley (1993). The MAGICC6 carbon cycle can emulate temperature-feedback effects on the heterotrophic respiration carbon fluxes. One improvement in MAGICC6 allows increased flexibility when accounting for CO 2 fertilization. This increase in flexibility allows a better fit to some of the more complex carbon cycle models reviewed in C<math>^4</math>MIP(Friedlingstein, 2006)(see terrestrial carbon cycle). Another update in MAGICC6 relates to the relaxation in carbon pools after a deforestation event. The gross CO 2 emissions related to deforestation and other land use activities are subtracted from the plant, detritus and soil carbon pools (see Fig. terrestrial carbon cycle]. While in previous versions only the regrowth in the plant carbon pool was taken into account to calculate the net deforestation, MAGICC6 now includes an effective relaxation/regrowth term for all three terrestrial carbon pools (see terrestrial carbon cycle). The original ocean carbon cycle model used a convolution representation (Wigley, 1991) to quantify the ocean-atmosphere CO<math>_2</math> flux. A similar representation is used here, but modified to account for nonlinearities. Specifically, the impulse response representation of the Princeton 3D GFDL model (Sarmiento, 1992) is used to approximate the inorganic carbon perturbation in the mixed layer (for the impulse response representation see, Joos, 1991). The temperature sensitivity of the sea surface partial pressure is implemented based on Takahashi et al. (1993) as given in Joos et al. (2001). For details on the updated carbon cycle routines, see the The carbon cycle. Other additional capabilities compared to MAGICC4.2 Five additional amendments to the climate model have been implemented in MAGICC6 compared to the MAGICC4.2 version that has been used in IPCC AR4. Aerosol indirect effects It is now possible to account directly for contributions from black carbon, organic carbon and nitrate aerosols to indirect (i.e., cloud albedo) effects (Twomey, 1977). The first indirect effect, affecting cloud droplet size and the second indirect effect, affecting cloud cover and lifetime, can also be modeled separately. Following the convention in IPCC AR4 (Forster et al., 2007), the second indirect effect is modeled as a prescribed change in efficacy of the first indirect effect. See Tropospheric aerosols for details. Depth-variable ocean with entrainment Building on the work by Raper et al. (2007), MAGICC6 includes the option of a depth-dependent ocean area profile with entrainment at each of the ocean levels (default, 50 levels) from the polar sinking water column. The default ocean area profile decreases from unity at the surface to, for example, 30<math>%</math>, 13<math>%</math> and 0<math>%</math> at depths of 4000, 4500 and 5000 m. Although comprehensive data on depth-dependent heat uptake profiles of the CMIP3 AOGCMs were not available for this study, this entrainment update provides more flexibility and allows for a better simulation of the characteristic depth-dependent heat uptake as observed in one analyzed AOGCM, namely HadCM2 (Raper et al., 2000). Vertical mixing depending on warming gradient Simple models, including earlier versions of MAGICC, sometimes overestimated the ocean heat uptake for higher warming scenarios when applying parameter sets chosen to match heat uptake for lower warming scenarios, see e.g. Fig. 17b in Harvey et al. (1997). A strengthened thermal stratification and hence reduced vertical mixing might contribute to the lower heat uptake for higher warming cases. To model this effect, a warming-dependent vertical gradient of the thermal diffusivity is implemented here(seeDepth-dependent ocean with entrainment). Forcing efficacies Since the IPCC TAR, a number of studies have focussed on forcing efficacies, i.e., on the differences in surface temperature response due to a unit forcing by different radiative forcing agents with different geographical and vertical distributions (Joshi et al., 1997). This version of MAGICC includes the option to apply different efficacy terms for the different forcings agents (see the efficacies section for details and supplement pdf in http://www.atmos-chem-phys.net/11/1417/2011/acp-11-1417-2011.html for default values). Radiative forcing patterns Earlier versions of MAGICC used time-independent (but user-specifiable) ratios to distribute the global-mean forcing of tropospheric ozone and aerosols to the four atmospheric boxes, i.e., land and ocean in both hemispheres. This model structure and the simple 4-box forcing patterns are retained as it is able to capture a large fraction of the forcing agent characteristics of interest here. However, we now use patterns for each forcing individually, and allow for these patterns to vary over time. For example, the historical forcing pattern evolutions for tropospheric aerosols are based on results from Hansen et al. (2005), which are interpolated to annual values and extrapolated into the future using hemispheric emissions. Additionally, MAGICC6 now incorporates forcing patterns for the long-lived greenhouse gases as well, although these patterns are assumed to be constant in time and scaled with global-mean radiative forcing (supplement pdf in http://www.atmos-chem-phys.net/11/1417/2011/acp-11-1417-2011.html for details on the default forcing patterns and time series).
Complex Impedance Complex impedances are commonly used quantities in the analysis of AC power systems. A complex impedance is represented by the following relation: [math] Z = R + jX \, [/math] Where [math] Z \, [/math] is the complex impedance ([math] \Omega [/math]) [math] R \, [/math] is the resistance ([math] \Omega [/math]) [math] X \, [/math] is the reactance ([math] \Omega [/math]) [math] j \, [/math] is the complex component, i.e. [math] \sqrt{-1} \, [/math]) For more details about why complex quantities are used in electrical engineering, see the article on complex electrical quantities. Complex Arithmetic The manipulation of complex impedances follow the rules of complex arithmetic. Series Impedances Two impedances in series can be combined by simply adding the individual real and complex terms (i.e. resistance and reactance components). For example, given: [math] Z_{1} = R_{1} + jX_{1} \, [/math] [math] Z_{2} = R_{2} + jX_{2} \, [/math] Then, [math] Z_{1} + Z_{2} = R_{1} + R_{2} + j \left( X_{1} + X_{2} \right) \, [/math] Parallel Impedances Two impedances in parallel can be combined according to the following standard relation: [math] Z_{1} \|\| Z_{2} = \frac{Z_{1} Z_{2}}{Z_{1} + Z_{2}} \, [/math] However, note that the multiplication and division of complex numbers is more involved than simply multiplying or dividing the real and complex terms: Multiplication:involves multiplying cross-terms, i.e. [math] Z_{1} \times Z_{2} = \left( R_{1} + jX_{1} \right) \left( R_{2} + jX_{2} \right) \, [/math] [math] = R_{1} R_{2} + j^{2} X_{1} X_{2} +j \left( R_{1} X_{2} \right) +j \left( X_{1} R_{2} \right) \, [/math] [math] = R_{1} R_{2} - X_{1} X_{2} +j \left( R_{1} X_{2} + X_{1} R_{2} \right) \, [/math] Division:involves multiplying by the complex conjugate of the denominator, i.e [math] \frac{Z_{1}}{Z_{2}} = \frac{\left( R_{1} + jX_{1} \right)}{\left( R_{2} + jX_{2} \right)} \, [/math] [math] = \frac{\left( R_{1} + jX_{1} \right)}{\left( R_{2} + jX_{2} \right)} \times \frac{\left( R_{2} - jX_{2} \right)}{\left( R_{2} - jX_{2} \right)} \, [/math] [math] = \frac{R_{1} R_{2} + X_{1} X_{2} +j \left( R_{1} X_{2} - X_{1} R_{2} \right)}{\left( R_{2}^{2} + X_{2}^{2} \right)} \, [/math]
[Sponsors] November 24, 2004, 04:52 test cases # 1 Guest Posts: n/a Hello, For purposes of my MSc Thesis I am now doing a set of test benchmarks (test cases) for code validation. I found really nice test case with an analytical solution as presented above: It has been taken from: Amara, "Vorticity–velocity–pressure formulation for Navier–Stokes equations", Comput Visual Sci 6: 47–52 (2004) Unfortunately author of this paper did not provide enough information about this test case. They call it as "Academic Test" - and I am not sure about how to specify boundary conditions for it. It seems like a driven cavity type flow with different velocities on all the edges, but I am not sure. Anybody know that test case and is able to help me a little bit to find other reference article for it, or give an idea about how to specify conditions of the flow here? (I am working with primitive variable SIMPLE solver, no stream function and vorticity). Best Regards Maciej Matyka <A HREF="http://panoramix.ift.uni.wroc.pl/~maq/eng">http://panoramix.ift.uni.wroc.pl/~maq/eng</a> ps. If you have any other idea about interesting test cases other than Driven Cavity and Couette Flow (those two I have done so far), please give me a hint. I am working now with code: in primitive variables, for incompressible flow, without free surface, without temperature convection terms. November 24, 2004, 08:27 Re: test cases # 4 Guest Posts: n/a It's Taylor's array of vortices. General time-dependent solution is as follows: $$u(x,y,t)=-\cos(nx) \sin (ny) \exp (-2n^2\nu t)$$ $$v(x,y,t)=-\sin(nx) \cos (ny) \exp (-2n^2\nu t)$$ $$p(x,y,t)=-{1/4} (\cos(2nx) + \cos (2ny)) \exp (-4n^2\nu t)$$ Your equations are for $t \to \infty$. Of course, these are nice analytic solutions, but fundamentally, they are too simple. These might be just a starting point for validations of your code. The flow has no shear whatsoever, i.e.: $$\sigma_{12}=\sigma_{21}=0$$. This is one of the reasons you need more thorough test cases. Good luck. MMG Thread Tools Search this Thread Display Modes Similar Threads Thread Thread Starter Forum Replies Last Post critical error during installation of openfoam Fabio88 OpenFOAM Installation 21 June 2, 2010 03:01 OF 1.6 \| Ubuntu 9.10 (64bit) \| GLIBCXX_3.4.11 not found piprus OpenFOAM Installation 22 February 25, 2010 13:43 FSI test cases Kirikou Main CFD Forum 8 November 6, 2006 09:49 Test Cases for Axial Flow Turbine shraman Main CFD Forum 0 July 25, 2006 04:43 Test cases of different levels of complexity quarkz Main CFD Forum 1 September 15, 2005 18:47
Edit for clarity: How does Mathematica's function Series know that Gamma[x] has a pole at x=0? When Series is called to expand near x=0, it gives the proper 1/x term in its Laurent expansion. I need to copy this behavior of Series to user-defined functions. Series[Gamma[x],{x,0,1}] 1/x-EulerGamma+1/12 (6 EulerGamma^2+[Pi]^2) x+O[x]^2 In the package I'm writing I have several complicated functions that have simple poles near isolated points. These functions are to be evaluated only when its arguments are numerical. Otherwise, for the sake of brevity of the output, these functions remain unevaluated and is supposed to act like a special function. Simple example: $$f(x) = \frac{\cos(x)}{x}$$ Task: I would like to be able to carry out a Series expansion in $x$; especially around $x=0$ where it should yield the $1/x$ pole term in the Laurent series: Here is what I did: SetAttributes[f,NumericFunction];f[x_?NumericQ] := 1/x Cos[x];Derivative[n_][f][x_] := Derivative[n][1/# Cos[#] &][x]; So now, I test this, and try to obtain the expansion near $x=\pi/2$. Series[f[x], {x, Pi/2, 2}] which works, but around $x=0$, Series[f[x], {x, 0, 2}] I get an error. So, how can I tell Series that at $x=0$, the function f starts at $1/x$? I need Series to behave just as it would if I gave it the full functional form explicitly: Series[1/x Cos[x], {x, 0, 2}]
We say that the uniformly elliptic operator $$Lu\ =\ -\sum_{i,j=1}^na^{ij}u_{x_ix_j}\ +\ \sum_{i=1}^nb^iu_{x_i}\ +\ cu$$ satisfies the weak maximum principleif for all $u\in C^2(U)\cap C(\bar{U})$ $$\left\{\begin{array}{rl} Lu \leq 0 & \mbox{in } U\\ u \leq 0 & \mbox{on } \partial U \end{array}\right.$$ implies that $u\leq 0$ in $U$. Suppose that there exists a function $v\in C^2(U)\cap C(\bar{U})$ such that $Lv \geq 0$ in $U$ and $v > 0$ on $\bar{U}$. Show the $L$ satisfies the weak maximum principle. (Hint: Find an elliptic operator $M$ with no zeroth-order term such that $w := u/v$ satisfies $Mw \leq 0$ in the region $\{u > 0\}$. To do this, first compute $(v^2w_{x_i})_{x_j}$.) This is from PDE Evans, 2nd edition: Chapter 6, Exercise 12. A question has been asked already about this problem, but my question is not considered a duplicate of it. That other question asks to solve the problem altogether; my question about the problem is merely finding the elliptic operator $M$, which is not explained thoroughly in the other question. How can I construct the elliptic operator $M$? I am following the hint given by computing $$(v^2w_{x_i})_{x_j}=2vv_{x_j}w_{x_i}+v^2w_{x_i x_j}.$$ Now, I do not know what to do here honestly, but I thought about saying $$Mw = -\sum_{i,j=1}^n v^2 w_{x_i x_j} - \sum_{i=1}^n 2vv_{x_j}w_{x_i}-cu,$$ so that $Mw \le 0$, when considering that "$w:=u/v$ satisfies $Mw\le 0$ in the region $\{u > 0\}$". As soon as I receive confirmation that my $M$ is fine, then from this point on I can complete the exercise on my own.
Let E be a holomorphic bundle over algebra surface X, let $H$ be a Hermitian metric of $E$, recall the Hermitian-Yang-mills equation is $\wedge F_H=\lambda.1$. Let $H_t$ be Hermitian metrics over $E$ parametrized by $t$, Donaldson in [1] consider the following flow equation: \begin{equation} H_t^{-1}\frac{\partial H_t}{\partial t}=-2i(\wedge F_{H_t}-\lambda.1),\;\;H_t\|_{t=0}=H_0, \label{flow} \end{equation} for some initial metric $H_0$. In [1], page 15, there is a note: If $E$ is indecomposable and has a solution $K$ to the Hermitian-Yang-mills equation, then for any initial condition $H_0$, the corresponding solution $H_t$ of the flow equation converges in $\mathcal{C}^{\infty}$ to $K$ as $t\to\infty$. In addition, consider the distance function $\sigma$ between two metric $H_t,K$, $\sigma(H_t,K):=Tr(H_t^{-1}K)+Tr(K^{-1}H_t)-2\;\mathrm{rank}\;E$, then we have a bound \begin{equation} \\|(\frac{\partial}{\partial t}+\Delta)\sigma(K,H_t)\\|_{L^1}\leq -const.\\|\sigma(K,H_t)\\|_{L^1} \end{equation} and $\sigma$ decays exponentially. My question is how to verify these two claims: (A)If a solution $K$ exists, then the flow convergence to the solution in $\mathcal{C}^{\infty}$. (B)This convergence is exponentially decays. [1] S.Donaldson, Anti Self-Dual Yang-Mills Connections over Complex Algebraic Surfaces and Stable Vector Bundles
I have found the following claim made very clearly at least once in the published literature (see below): Let $P$ be a linear partial differential operator defined on an open set $\Omega \subset \mathbb{R}^{n+1}$, strictly hyperbolic with respect to the level surfaces of the first coordinate, which I will denote by $t$. Let $u$ be a distribution in $H^s_\mathrm{loc}(\Omega)$ for some $s \in \mathbb{R}$, such that $Pu=0$. Then the restriction (i.e. pullback) of $u$ to each of the level sets $\Sigma_{\tau} := \Omega \cap \{ t = \tau \}$ is itself in $H^s_\mathrm{loc}(\Sigma_\tau)$. In fact, a more general statement may be found in Section 2 of: Bao, Gang; Symes, William W., A trace theorem for solutions of linear partial differential equations, Math. Methods Appl. Sci. 14, No.8, 553-562 (1991). ZBL0754.35023. There, the authors say that the statement boxed above follows from standard energy estimates, and refer to e.g. Taylor's book on pseudo-differential operators for the arguments. However, I meet the following difficulties when attempting to fill in all details: Ostensibly, the "standard energy estimates" discussed in Taylor's book (and elsewhere, e.g. in Hörmander's) pertain distributional solutions which (in the case $\Omega = \mathbb{R}^{n+1}$) are in spaces $\bigcap_{j=0}^{m-1} C^j([\tau_1,\tau_2]; H^{s-j}(\mathbb{R}^n))$, where $m$ is the order of $P$. I, on the other hand, am interested in distributions (locally) in $H^s$ in both time and space. In any case, as far as I am aware, the usual energy estimate controls the evolution in time of the sum of (appropriate) "spatial" Sobolev norms of (appropriate) $t$-derivatives of $u$ restricted to the level sets of $t$. Specifically, in the case $Pu=0$ the energy estimate says that the supremum of this quantity over a compact interval $[\tau_0,\tau_1]$ is bounded by a constant times its value at $\tau_0$. To prove that the restriction of $u$ to the $t=\tau_0$ surface is (locally) in $H^s$, presumably we would like to control the $H^s$ norm of this restriction and place an upper bound on it, but this does not seem to be the scenario in the energy inequality. By the way, I am not doubting that $u$ can be pulled back to a spacelike hypersurface, as a distribution. I know that this is the case because $WF(u)$ is disjoint from the normal bundle of the hypersurface whenever $Pu$ is smooth. Update I might have the beginning of an argument, but this leads to a further question about Bochner–Sobolev spaces whose answer seems clear in the case of positive Sobolev order $s$, but not so much otherwise. I will explain schematically and in particular assuming for simplicity and definiteness that $P$ is of second order and that $\Omega = \mathbb{R}^{n+1}$ (so that also $\Sigma_\tau \cong \mathbb{R}^n$). Namely, let $u \in H^s_\mathrm{loc}(\mathbb{R}^{n+1})$ be in the kernel of $P$, and let $\phi \in C_\mathrm{c}^\infty(\mathbb{R}^{n+1})$. Let $\psi \in C_\mathrm{c}^\infty(\mathbb{R}^{n+1})$ be such that $\psi = 1$ on a neighbourhood of $\mathrm{supp}{\phi}$. Then $P(\phi u) = [P,\phi] u = [P,\phi] (\psi u)$, and the commutator is a first-order differential operator. Hence, $\phi u$ solves the PDE with non-zero right-hand side but vanishing Cauchy data on $\Sigma_{T_0}$ for any sufficiently small $T_0$. Now, if $u$ were smooth then the standard energy estimates would say that for any $T_1 > T_0$ there exists a constant $C>0$ such that$$ () \quad \mathcal{E}_s(\tau; \phi u) \leq C \int_{T_0}^{T_1}\\| [P,\phi] (\psi u) (t,\cdot) \\|^2_{H^{s-1}(\mathbb{R}^n)} \, \mathrm{d}t $$where the energy of order $s \in \mathbb{R}$ and at time $\tau$ is defined as$$\mathcal{E}_s(\tau; \phi u) = \\| (\phi u)\|_{\Sigma_\tau} \\|^2_{H^s(\mathbb{R}^n)} + \\| \partial_t (\phi u)\|_{\Sigma_\tau} \\|^2_{H^{s-1}(\mathbb{R}^n)}.$$If we could provide a lower bound for the right-hand side of $()$, in terms of the norm of $\psi u$ in $H^s(\mathbb{R}^{n+1})$, then presumably we would be close to being done (by a suitable density argument from $C^\infty$). Now I would like to say something along these lines:$$ () \quad \int_{T_0}^{T_1}\\| [P,\phi] (\psi u) (t,\cdot) \\|^2_{H^{s-1}(\mathbb{R}^n)} \, \mathrm{d}t \leq C_1 \\| [P,\phi] (\psi u) \\|^2_{H^{s-1}(\mathbb{R}^{n+1})} \leq C_2 \\| \psi u \\|^2_{H^{s}(\mathbb{R}^{n+1})}$$where the second inequality would be due to $[P,\phi]$ being differential of order $1$. But how can I justify the first inequality? It seems almost obvious in the case of integer $s \geq 1$ – we are simply ignoring those multi-indices with a non-zero entry corresponding to $t$, which would otherwise contribute to the $H^{s-1}(\mathbb{R}^{n+1})$ norm. For general $s \in \mathbb{R}$, I'm not sure. Notice that the left-hand side of $()$ is the square of the norm in the Bochner–Sobolev space $L^2([T_0,T_1]; H^{s-1}(\mathbb{R}^{n}))$. Update 2 Perhaps proving the first inequality in $(*)$ is not so difficult when the Sobolev order $s$ is real and $s \geq 1$. For then it seems we can use the Fourier transform, as follows. Set $\ell := s-1$ and $g := [P,\phi](\psi u) \in C^\infty_\mathrm{c}(\mathbb{R}^{n+1})$, and denote partial Fourier transforms with respect to $t$ and the $x$-variables in $\mathbb{R}^n$ by $\mathcal{F}_t$ and $\mathcal{F}_x$ respectively, while the full Fourier transform of $g$ will be $\hat{g}$. Then, using that $\mathcal{F}_t \mathcal{F}_x g = \hat{g}$, and the Plancherel theorem: \begin{align} \int_{T_0}^{T_1} \\| g(t,\cdot) \\|^2_{H^{\ell}(\mathbb{R}^n)} \, \mathrm{d}t &\leq \int_{\mathbb{R}} \\| g(t,\cdot) \\|^2_{H^{\ell}(\mathbb{R}^n)} \, \mathrm{d}t \\ &= \int_{\mathbb{R}} \left( \int_{\mathbb{R}^n} (1 + \|k\|^2)^\ell \|[\mathcal{F}_x g](t,k)\|^2 \, \mathrm{d}^n k\right) \mathrm{d}t \\ &= \int_{\mathbb{R}} \left( \int_{\mathbb{R}^n} (1 + \|k\|^2)^\ell \|[\mathcal{F}_t^{-1} \hat{g}](t,k)\|^2 \, \mathrm{d}^n k\right) \mathrm{d}t \\ &= \int_{\mathbb{R}^n} (1 + \|k\|^2)^\ell \left( \int_{\mathbb{R}} \|[\mathcal{F}_t^{-1} \hat{g}](t,k)\|^2 \, \mathrm{d}t \right) \mathrm{d}^n k \\ &= \int_{\mathbb{R}^n} (1 + \|k\|^2)^\ell \left( \int_{\mathbb{R}} \|\hat{g}(\tau,k)\|^2 \, \mathrm{d}\tau \right) \mathrm{d}^n k \\ &\leq \int_{\mathbb{R}^{n+1}} (1 + \|\tau\|^2 + \|k\|^2)^\ell\|\hat{g}(\tau,k)\|^2 \, \mathrm{d}\tau \, \mathrm{d}^n k \\ &= \\| g \\|^2_{H^{\ell}(\mathbb{R}^{n+1})}. \end{align*} Of course, I used the fact that $\ell \geq 0$ (i.e. $s \geq 1$) in the final inequality above.
I find the Frobenius Method quite beautiful, and I would like to be able to apply it. In particular there are three questions in my text book that I have attempted. In each question my limited understanding has stopped me. Only one of these questions (the last) is assigned homework. The rest are examples I found interesting. 1) $ L[y] = xy'' + 2xy' +6e^xy = 0 $ (1) The wikipedia article begins by saying that the Frobenius method is a way to find solutions for ODEs of the form $ x^2y'' + xP(x) + Q(x)y = 0 $ To put (1) into that form I might multiply across by x, giving me $ x^2y'' + x[2x]y' + [6xe^x]y = 0 $ (2) But is that OK? The first step in the method seems to be dividing by $ x^2 $, so can't I just leave the equation in its original form? I'll assume I can. Now we let $ y_1 = \sum _{n=0}^{\infty} a_n x^{r+n} $ then, $ y_1' = \sum _{n=0}^{\infty} (r+n)a_nx^{r+n-1} $ and, $ y_1'' = \sum _{n=0}^{\infty} (r+n)(r+n-1)a_nx^{r+n-2} $ substituting into (2) we get, $ x\sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-2} + 2x\sum _{n=0}^{\infty}(r+n)a_nx^{r+n-1} + 6e^x\sum _{n=0}^{\infty}a_nx^{r+n} = 0 $ But now what? I am aware that $ 6e^x = 6\sum _{n=0}^{\infty}x^n/n! $, but my powers stop there. Can I multiply the two series together? I would have to multiply each term in one series by every term in the other, and I don't know how to deal with that. The text provides no worked examples in which P(x) or Q(x) are not polynomials... so for now my work stops here. 2) $ L[y] = x(x-1)y'' + 6x^2y' + 3y = 0 $ Again, I will leave the question in its original form, rather than try to get that x^2 in front (I realise I am not checking that the singular point is a regular singular point, but checking the answer in the back of the book, x = 1 and x = 0 are indeed regular points). With two regular singular points, I expect I will get 2 sets of answers: one near x = 1 and the other near x = 0. Is it enough to just proceed with one case and then the next? I will assume so, and begin with the case close to x = 0. Again, letting $ y_1 = \sum _{n=0}^{\infty} a_n x^{r+n} $, and taking the appropriate derivatives, we find by substitution, $ x(x-1)\sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-2} + 6x^2\sum _{n=0}^{\infty}(r+n)a_nx^{r+n-1} + 3\sum _{n=0}^{\infty}a_nx^{r+n} = 0 $ $ x^2\sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-2} - x\sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-2} + 6x^2\sum _{n=0}^{\infty}(r+n)a_nx^{r+n-1} + 3\sum _{n=0}^{\infty}a_nx^{r+n} = 0 $ $ \sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n} - \sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=0}^{\infty}6(r+n)a_nx^{r+n+1} + \sum _{n=0}^{\infty}3a_nx^{r+n} = 0 $ we shift the indexes of the above sums, so that everything will be in terms of the same power of x. $ \sum _{n=1}^{\infty}(r+n-1)(r+n-2)a_{n-1}x^{r+n-1} - \sum _{n=0}^{\infty}(r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=2}^{\infty}6(r+n-2)a_{n-2}x^{r+n-1} + \sum _{n=1}^{\infty}3a_{n-1}x^{r+n-1} = 0 $ we synchronise the indexes in order to group like terms, by extracting early terms from each series, $ r(r-1)a_0x^r + \sum _{n=2}^{\infty}(r+n-1)(r+n-2)a_{n-1}x^{r+n-1} - r(r-1)a_0x^{r-1} - r(r+1)a_1x^r - \sum _{n=2}^{\infty}(r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=2}^{\infty}6(r+n-2)a_{n-2}x^{r+n-1} + 3a_0x^{r-1} + \sum _{n=2}^{\infty}3a_{n-1}x^{r+n-1} = 0 $ rearranging, we get $ r(r-1)a_0x^r - r(r-1)a_0x^{r-1} - r(r+1)a_1x^r + 3a_0x^{r-1} + \sum _{n=2}^{\infty}(r+n-1)(r+n-2)a_{n-1}x^{r+n-1} - \sum _{n=2}^{\infty}(r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=2}^{\infty}6(r+n-2)a_{n-2}x^{r+n-1} + \sum _{n=2}^{\infty}3a_{n-1}x^{r+n-1} = 0 $ At this point I expect the indicial equation to emerge, and I expect it to be similar to an Euler Equation. That is, I expect a polynomial that I can solve to get two 'exponents at the singularity'. Unfortunately, I cannot see an indicial equation and am at a loss to know precisely why. 3) $ L[y] = xy'' + y = 0 $ Finally we come to the assigned question, which I have been able to manipulate into an almost final form. Again, letting $ y_1 = \sum _{n=0}^{\infty} a_n x^{r+n} $, taking derivatives, and substituting into L, we get $ x\sum _{n=0}^{\infty} (r+n)(r+n-1)a_nx^{r+n-2} + \sum _{n=0}^{\infty} a_n x^{r+n} = 0 $ $ \sum _{n=0}^{\infty} (r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=0}^{\infty} a_n x^{r+n} = 0 $ Now shifting indexes, $ \sum _{n=0}^{\infty} (r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=1}^{\infty} a_{n-1} x^{r+n-1} = 0 $ and extracting the $ 0^{th} $ term of the first sum, $ r(r-1)a_0x^{r-1} + \sum _{n=1}^{\infty} (r+n)(r+n-1)a_nx^{r+n-1} + \sum _{n=1}^{\infty} a_{n-1} x^{r+n-1} = 0 $ $ r(r-1)a_0x^{r-1} + \sum _{n=1}^{\infty}[(r+n)(r+n-1)a_n + a_{n-1}]x^{r+n-1} = 0 $ And voila! We have an indicial term with solutions $r_1 = 1$ and $r_2 = 0$, and a recurrence relation. From my text I expect that $y_1 = \|x\|^{r_1}[1+\sum _{n=0}^{\infty}a_nx^n]$ and since $ r_1 - r_2 \in \mathbb{Z} $, $y_2 = ay_1ln\|x\| + \|x\|^{r_2}[1 + \sum _{n=0}^{\infty}b_nx^n]$ to find $a_n$ we observe the recurrence relation with $ r = r_1 = 1 $, $ (r+n)(r+n-1)a_n + a_{n-1} $ $ a_n = -a_{n-1}/n(n+1) $ so, $ a_1 = -a_0/21 $ $ a_2 = -a_1/23 = a_0/32121 = a_0/3!2! $ $ a_3 = -a_2/34 = -a_0/4!3! $ and in general, $ a_n = (-1)^na_0/n!(n+1)! $ so we have $ y_1 = \|x\| + \sum _{n=0}^{\infty} (-1)^na_0x^{n+1}/n!(n+1)! $ Not so easily done with r = r_2 = 0, I'm afraid... since the relation becomes $ a_n = -a_{n-1}/n(n-1) $, which means we can't have a_1 for fear of division by zero. Never the less, starting at n = 2 we get, $ a_2 = -a_1/21 $ $ a_3 = -a_2/23 = a_1/32121 = a_1/3!2! $ $ a_4 = -a_3/3*4 = -a_1/4!3! $ and in general, $ a_n = (-1)^{n-1}a_1/n!(n-1)! $ so we have $ y_2 = ay_1ln\|x\| + 1 + \sum _{n=0}^{\infty} (-1)^{n-1}a_1x^{n+1}/n!(n-1)! $ Which I feel may not be correct... and even if it is, how should one man solve for a in a single lifetime? Thanks everyone for looking at this. I want to stress that I am not just a student looking for help in his homework: I would really like to understand this method because it appeals to me. I particularly like the way we extract the indicial expression from the sums, in order to synchronise them. That is so cool. And how you get 1 recurrence relation that you can use for both solutions: neat. PS sorry if my Latex is not perfect? I'm just getting started with it. questions taken from "Elementary Differential Equations and Boundary Value Problems" by William E. Boyce and Richard C. DiPrima (9th ed), from sectin 5.6 pp 290
Let $X$ be a topological space and $V$ and $N$ are subspaces of $X$ such that $N$ deformation retracts onto $V$. I want to show that $X-V$ deformation retracts onto $X-N$. So i need to construct a retraction $r':X-V\longrightarrow X-N$ and show that there exists a homotopy between $i'\circ r'$ and $id'$ where $i':X-N\rightarrow X-V$ is the inclusion and $id'$ is the identity of $X-V$. So the main problem is to find $r'$, i know that we should use the retraction $r:N\longrightarrow V$ at some step but i don't see how to do it. thank you for your help! The proposition stated is false. But I had the same idea and came up with this: The deformation retract can be viewed as a homotopy $R(t,x)$ such that $R(0,x)=id$ and $R(1,x)=r$. Note that fixing $x$ defines a collection of paths $$\gamma_x =R(t,x)$$ from $x$ to $r(x)$. Suppose that the retract has the additional property, that there exists a set $\beta \in N$ (imagine the boundary) together with a continuous function $\phi : N -V \to \beta$, such that $\gamma_{\phi(x)}$ crosses $x$ exactly once. Then $X-V$ deformation retracts onto $(X-N)\cup \beta$ by $$R^(t,x)=\gamma_{\phi(x)} ((1-t_0)(1-t)) $$ where $t_0$ is the number in $I$, such that $\gamma_{\phi(x)}(t_0)=x$. (Define $R^(t,x)=id$ outside $N$).
Let $k$ be a field. What is an explicit power series $f \in k[[t]]$ that is transcendental over $k[t]$? I am looking for elementary example (so there should be a proof of transcendence that does not use any big machinery). MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.Sign up to join this community Let $k$ be a field. What is an explicit power series $f \in k[[t]]$ that is transcendental over $k[t]$? I am looking for elementary example (so there should be a proof of transcendence that does not use any big machinery). If $k$ has characteristic zero, then $\displaystyle e^t = \sum_{n \ge 0} \frac{t^n}{n!}$ is certainly transcendental over $k[t]$; the proof is essentially by repeated formal differentiation of any purported algebraic relation satisfied by $e^t$. Edit: Let me fill in a few details. Given a polynomial $P$ in $e^t$ of degree $d$ where each coefficient is a polynomial in $k[t]$ of degree at most $m$, the possible terms that appear in any formal derivative of $P$ lie in a vector space of dimension $(m+1)(d+1)$, so by taking at least $(m+1)(d+1)$ formal derivatives we obtain too many linear relationships between the terms $t^k e^{nt}$. The coefficient of $e^{dt}$ in particular eventually dominates all other coefficients. Eisenstein proved (actually, stated) in 1852 that if $f=\sum a_n z^n$ is an algebraic power series with rational coefficients, there exist positive integers $A$ and $B$ such that $A a_n B^n$ are integers for all $n$. In particular, as Eisenstein himself remarks, only finitely many prime numbers appear in the denominators of the coefficients of $f$. For example, $e^z$, $\log(1+z)$, etc., are transcendental. How about $\sum t^{n!}$? Doesn't a "sea-of-zeroes" argument show it can't be algebraic? Coming back to the lacunary series, I would prefer the series $f(z)=\sum_{k\ge0}z^{d^k}$, where $d>1$ is an integer, because it is the classical example in Mahler's method; this function satisfies the functional equation $f(z^d)=f(z)-z$. I simply copy Ku.Nishioka's argument from her book "Mahler functions and transcendence" (Theorem 1.1.2). Assume that $f(z)$ is algebraic over $\mathbb C(z)$, hence satisfies an \emph{irreducible} equation $f(z)^n+a_{n-1}(z)f(z)^{n-1}+\dots+a_0(z)=0$ where the coefficients $a_j(z)\in\mathbb C(z)$. Substituting $z^d$ for $z$ and using $f(z^d)=f(z)-z$ we obtain $f(z)^n+(-nz+a_{n-1}(z^d))f(z)^{n-1}+\dots=0$. The left-hand sides of both polynomial relations for $f(z)$ must coincide because of the irreducibility. This in particular implies that $a_{n-1}(z)=-nz+a_{n-1}(z^d)$. Letting $a_{n-1}(z)=a(z)/b(z)$ where $a$ and $b$ are two coprime polynomials we see that $a(z)b(z^d)=-nzb(z)b(z^d)+a(z^d)b(z)$. Since $a(z^d)$ and $b(z^d)$ are coprime, $b(z^d)$ must divide $b(z)$. This is possible if only $\deg b(z)=0$, that is, $b(z)=b$ is a nonzero constant. Then $a(z)=-bnz+a(z^d)$ and comparing the degrees of both sides we see that $a(z)$ is constant as well, hence $nz=0$, a contradiction. over the rationals every power serie with integer coefficents not periodic is tracendent, over Fp a power serie is algebraic iff the secuence of coeficient is p automatic there is a article od jp allouch tracendence of formal series with it information.
In general, this is a hard problem. For classes of expressions for which there is a canonical form, such as polynomials (in any number of variables), the answer is straightforward: two expressions are equal if and only if they have the same canonical form. (A canonical form means that for each expression there is one and only one canonical expression to which it is equal, and there is an effective procedure for calculating the canonical form of any given expression.) Then you can get an algorithm for comparing two expressions: calculate the canonical form for each expression and check to see if the canonical forms are identical. Algebra students learn to do exactly this in order to decide themselves if two polynomials are equal. (Students of arithmetic learn an analogous method for deciding if two arithmetic expressions are equal, for example converting the expression $2\cdot(3+4)$ into the canonical form $14$; this algorithm is a subroutine of the one that reduces polynomial expressions to canonical form.) A canonical form for polynomials is to combine all the like terms, list the terms in descending order of degree, with the terms of equal degree listed in lexicographic order by the variables they contain, or something of that sort. Calculating a canonical form for an arbitrary polynomial is not a difficult matter. It is the sort of thing a competent programmer can produce in a couple of hours; or as several other people here have suggested you could put the solutions into a computer algebra system, which will contain exactly this sort of algorithm for several different sorts of expressions. But for more general expressions it can be extremely difficult, or even impossible, to decide of the two expressions are equal. There is no canonical form, and recognizing when two particular expressions are equal can be a major theorem. For example, consider the expressions $\cos 2x$ and $\left(\cos x\right)^2 - \left(\sin x\right)^2$. These are equal, but not obviously so. Or for a more difficult example, consider the two expressions $$0$$ and $$\sum_{a,b,c > 1\atop n>2} I(a^n + b^n - c^n)$$ (where $I(x)$ denotes the function which has $I(0)=1$ and $I(x)=0$ for $x\ne 0$). It was conjectured for some time that these two expressions were equal, but the proof turned out to be somewhat tricky. The previous paragraph was a joke, but it is a serious joke: a substantial part of mathematics is precisely how to perform such calculations and to recognize when two different-seeming expressions are equal. Euler is famous (among other things) for recognizing that $e^{ix}$ and $\cos x + i\sin x$ are equal expressions.Leaving aside jokes, a theorem of Daniel Richardson says that for a fairly small, fairly natural class of expressions, there is no method that can reliably determine equality in all cases. So to get an answer, you need to be more specific about what your question is. If you only need to compare polynomials, the answer is fairly straightforward. If your expressions are more complicated than that, there may or may not be an answer; it depends on what is in them. [ Addendum: I see that you have added comments saying that you are only interested in polynomials, and that you want to know if your idea of substituting test values for $x$ is sound. It is sound, but plugging in one number is not enough, even for the simples polyomials. The polynomials $x+1, 3x-1$, and $3-x$ all have the same value at $x=1$. But you can easily avoid false positives by checking $n+1$ different values for an $n$th-degree polynomial. A polynomial of degree $n$ is completely determined by its values at $n+1$ points, so if two polynomials of degree $n$ agree at $n+1$ different points you can be sure they are identical; it does not even matter which $n+1$ values you sample. (In the example above, any value of $x$ other than $x=1$ is sufficient to distinguish the three polynomials.) Similarly if the polynomial has three variables $x$, $y$, and $z$, of degrees $n_x, n_y, $ and $n_z$, it suffices to select $n_x+1$ values for $x$, $n_y+1$ values for $y$, and$n_z+1$ values for $z$, and then check all $(n_x+1)(n_y+1)(n_z+1)$ triples of those values. ]
According to remark 6.14 in Shigeru Mukai's An introduction to invariants and moduli (unfortunately, the page is not available on Google Books, so I explain it below), the GIT-quotient of an affine variety $X$ by a reductive group $G$ with respect to a nontrivial character $\chi: G \to \mathbb G_m$ may be considered as first taking an affine quotient $\operatorname{Spec} (\mathbb k[X]^{G_\chi})$ by $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$, then taking a quotient by $\mathbb G_m=G / G_\chi$ via replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings. As a result, I wonder why it is so much better then just taking an affine quotient $\operatorname{Spec}(\mathbb k[X]^G)$ if $G$ has much more characters like $\mathrm{GL}_{k_1} \times \ldots \times \mathrm{GL}_{k_r}$. It seems that replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings is a correct way to take a quotient by $\mathbb G_m$, but the first step looks fairly stupid. Update: Yes, as Will Sawin said, the question is approximately why one does not iterate it (why there is no need/why nothing better happens)? Details of Mukai's description. For simplicity, work over an algebraically closed field $\mathbb k$ of characteristic $0$. Let $X$ be an affine space or any other affine irreducible algebraic variety with $\operatorname{Pic} X=0$, and let $G$ be a connected reductive group acting on $X$. Then a $G$-linearized line bundle $\mathcal L$ is equivalent to a character $\chi: G \to \mathbb G_m$, so Mumford's GIT quotient is$$X //_{\mathcal L} G:=\operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \Gamma(X, \mathcal L^{\otimes n}) \right) = \operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \mathbb k[X]^G_{\chi^n} \right).$$ Here $$\mathbb k[X]^G_{\chi^n}=\{ f \in \mathbb k[X]: f(gx)=\chi(g)^nf(x) \}$$ are $\chi^n$-semi-invariants. If $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$, then $\mathbb k[X]^{G_\chi}$ has an action of $\mathbb G_m=G / G_\chi$, so is graded -- exactly by $(\mathbb k[X]^{G_\chi})_n=\mathbb k[X]^G_{\chi^n}$. From this follows Mukai's description of $X //_{\mathcal L} G$. Generality of Mumford's GIT-quotient. By the way, I believe that I have seen somewhere that any geometric quotient arises as an open subset in Mumford's GIT-quotient for some line bundle or something like this. Could you give me a reference?
A recent question on the notion and notation of multiplicative integrals ( What is the standard notation for a multiplicative integral? ) induced me to play with the Riemann products of the Gamma function, in order to evaluate the multiplicative integral of $\Gamma(x)$, exploiting the multiplicative formula. I will, however, put the question mainly in terms of a standard integral; and I will also use the factorial function $x!=\Gamma(x+1)$ instead (that seems to be more appreciated here). Consider the multiplicative formula for $x!$: $$x!=(2\pi)^{-\frac{m-1}{2}}\, m^{x+\frac{1}{2}}\, \big( \frac{x}{m} \big)!\,\big( \frac{x-1}{m} \big)!\dots \big( \frac{x-m+1}{m} \big)!\, \,$$ For $x=m\in\mathbb{N}$ we get, using the Stirling asymptotics for $m!$: $$\prod_{k=1}^{m}\big(\frac{k}{m} \big)!\sim (2\pi)^{\frac{m}{2}}e^{-m} $$ Take a logarithm; divide by $m$ and let $m\to\infty$: one finds $$\int_0^1\log(x!)\, dx=\frac{1}{2}\log(2\pi )-1,$$ or, as a multiplicative integral $$\prod_0^1 (x!\, dx)=\frac{\sqrt{2\pi}}{e}.$$ Now the question: How to evaluate the above integralby means of standard integralcalculus? I guess it's feasible, but how? Otherwise, it would be a remarkable case of an integral that one can only (edit: or say "more easily") evaluate directly from the definition of Riemann sums, like one does e.g. with $x^2$ in introductory calculus courses.
Suppose that $K/\mathbb{Q}_p$ is a finite extension and $k_K$ the residue field of $K$. Let $A/K$ be an abelian variety with good reduction. Suppose that $E\to\mathrm{End}^0_K(A)$ is an inclusion of a number field that sends $1$ to the identity. Denote by $T_\ell A$ the Tate module for a prime $\ell\neq p$ and let $V_\ell A=T_\ell A\otimes_{\mathbb{Z}_\ell}\mathbb{Q}_\ell$. The Galois representation $$\rho_\ell:\mathrm{Gal(\bar{K}/K)}\to \mathrm{Aut}(V_\ell A)$$ is then $E_\ell=E\otimes \mathbb{Q}_\ell$-linear. We can decompose $E_\ell=\prod_\lambda E_\lambda$ where $\lambda$ run through the places of $E$ dividing $\ell$ and $E_\lambda$ is the corresponding completion of $E$. Then by $E_\ell$-linearity, $V_\ell A=\prod_\lambda V_\lambda$ as $E_\ell[\mathrm{Gal}(\bar{K}/K)]$-modules. Let $\mathrm{Frob}_K$ be a lift of the Frobenius element. I want to prove that each $E_\lambda$-characteristic polynomial of $\rho_\lambda(\mathrm{Frob}_K)$ is the image of some common polynomial $P_0\in E[X]$ via $E[X]\hookrightarrow E_\lambda[X]$. I believe this follows from a result by Shimura from "Algebraic Number Fields and Symplectic Discontinuous Groups", Prop. 11.09, but I want to understand the argument of his proof. It is well-known that the action of the lift of the Frobenius element is obtained via the Frobenius endomorphism $\pi$ of the reduction $\tilde{A}/k_K$, so the $\mathbb{Q}_\ell$-characteristic polynolmial of $\rho_\ell(\mathrm{Frob}_K)$ has rational coefficients. Then, we need to find an $E[\pi]$-module $U$ so that $V_\ell A\simeq U\otimes \mathbb{Q}_\ell$ as $E[\pi]\otimes \mathbb{Q}_\ell$-modules. But how to get this $U$?
The electrical power grid operates in equilibrium. The power supply from the generator matches the demand levied by the end-users. The flow of power to enable this demand-supply equilibrium over the transmission lines (T-lines) is governed by the following conditions. Thermal limitation of T-lines Voltage drop limitation Steady-state stability limitation Thermal Limitation of T-lines Current carrying conductors are designed to carry a limited amount of current. When it is operated beyond its design rating, it starts to sag. Excessive sagging of T-lines leads to the infringement of safe electrical clearances. Power flow over short lines (less than 80km) are limited by their thermal ratings. Voltage Drop Limitation On long T-lines (greater than 300km), voltage drop is an issue. Unless compensated for, power flow should be monitored and regulated to minimize the drop in voltage. Steady-State Stability Limitation In AC systems, the power delivered from point a to point b is given by: P = \frac{3V_aV_b}{X_s}\sin\delta Watts Although it makes sense to use \delta=90^{\circ} to maximize the power flow, it is never done, however. This is because the generators connected to the grid may fall out of sync and may trip for a small disturbance in the system. \delta at the generation is typically maintained between 30 to 45deg. Steady-state stability limitation too, like voltage drop, does not allow loading of long T-lines (greater than 300km) to approach the thermal rating of the conductor. In the modern power systems, Flexible AC Transmission devices and High Voltage DC systems are playing a crucial role in controlling and maximizing the power flow over the transmission lines. In the near future, it is possible to have the power system operate close to its limitations while operating reliably. Resourceful Link See the power flow in action here. Your browser will need java capability. Very informative page.
We have to do it without calculus or any complex inequality. Level of complexity is that we cannot even use the AM-GM inequality. So I tried, $$(\sin\theta-\cos\theta)^2\geq0$$ $$1-2\sin\theta\cos\theta\geq0$$ $$\frac12\geq\sin\theta\cos\theta$$ Reverting back to the previous step, $$(\sin\theta+\cos\theta)^2\geq4\sin\theta\cos\theta$$ I am stuck here, please help. Edit: Sorry, but we can only use knowledge upto class 10th. Which includes, $\sin^2\theta+\cos^2\theta=1$ etc. $\sin(90^{\circ}-\theta)=\cos\theta$ etc. And basic trig ratios. Hint: We have $$\sin\theta+\cos\theta=\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right).$$Clearly we have $$-1\leq\sin(\theta+\pi/4)\leq1.$$ Update: To answer the edited question (that is, not using the sum formula for $\sin(\theta+\pi/4)$, we have $$(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2=2\cdot\underbrace{(\sin^2\theta+\cos^2\theta)}_1+2(\sin\theta)(\cos\theta)-2(\sin\theta)(\cos\theta)=2.$$ Since $x^2\geq0$ for all real $x$, we can subtract $(\sin\theta-\cos\theta)^2$ from both sides to obtain $$2-(\sin\theta-\cos\theta)^2\geq0.$$Equivalently, this can be written $$(\sin\theta-\cos\theta)^2\leq2.$$This just means $$\|\sin\theta-\cos\theta\|\leq\sqrt{2}.$$ By definition of absolute value, this says $$-\sqrt{2}\leq\sin\theta-\cos\theta\leq\sqrt{2}.$$ There are easier ways, but we can use AM-GM. By the Triangle Inequality we have $\|\sin\theta+\cos\theta\|\le \|\sin\theta\|+\|\cos\theta\|$. By AM-GM we have $\|\sin\theta\|+\|\cos\theta\|\le \frac{1}{\sqrt{2}}\|\sin\theta\|\|\cos\theta\|$. This is equal to $\sqrt{2}\|\sin(2\theta)\|$, which is $\le \sqrt{2}$. Remark: You were almost finished, for $4\sin\theta\cos\theta=2\sin(2\theta)$. And $2\sin(2\theta)$ has absolute value $\le 2$. Using $$\displaystyle (\sin \phi+\cos \phi)^2+(\sin \phi-\cos \phi)^2 = 2$$ Now $$\displaystyle (\sin \phi-\cos \phi)^2 = 2-(\sin \phi-\cos \phi)^2$$ Using $$\bf{Square\; Quantity\geq 0}$$ So $$\displaystyle (\sin \phi-\cos \phi)^2\geq0$$ So $$\displaystyle 2-\displaystyle (\sin \phi+\cos \phi)^2\geq 0$$ OR we get $$\displaystyle (\sin \phi+\cos \phi)^2\leq \left(\sqrt{2}\right)^2$$ So we get $$\displaystyle-\sqrt{2} \leq (\sin \phi+\cos \phi)\leq \sqrt{2}$$ A (simple?) answer: Since $\sin$ and $\cos$ are bounded, there is $r>0$ such that $$\left\|\sin\theta+\cos\theta\right\|\leq r$$ Replacing $\theta$ by $-\theta$ gives $$\left\|\sin\theta-\cos\theta\right\|\leq r$$ Multiplying together $$\left\|\sin^2\theta-\cos^2\theta\right\|\leq r^2$$ But the LHS is bounded by $2$, so picking $r$ smallest possible, we get that $r^2\leq 2$. So $r\leq\sqrt{2}$. You could use the R-Formula (which can be derived using trigonometric identities) to combine $\sin \theta + \cos \theta$ into $$\sqrt{2}\sin \left(\theta + \frac{\pi}{4}\right)$$ and then use the fact that $-1 \le \sin x \le 1$. Hint: Write $$\cos(b)=\sin(\frac{\pi}{2}-b)$$Now use the formula of $\sin(a)+\sin(b)$ 1st PROOF: $\sin \pi/4 =\cos \pi/4 =1/\sqrt 2$. Therefore $\|\sin x + \cos x\|=\sqrt 2 \|\sin x \cos \pi/4 +\cos x \sin \pi/4\|=\sqrt 2 \|\sin (x+\pi/4)\|\le \sqrt 2$.Equality iff $\|\sin (x+\pi/4)\|=1.$ E.g. $x= \pi/4.$...............2nd PROOF: $(\sin x +\cos x)^2 \le (\sin x +\cos x)^2+(\sin x - \cos x)^2 =2$. Equality iff $\sin x - \cos x =0$ and $\|\sin x=1/\sqrt 2\|.$ E.g. $x=\pi/4.$
Dear Uncle Colin, I have a pair of parametric equations giving $x$ and $y$ each as a function of $t$. I'm happy with the first derivative being $\diff{y}{t} \div \diff{x}{t}$, but I struggle to find the second derivative. How would I do that? - Can't Handle An Infinitesimal Nuance Hi,Read More → Eagle-eyed friend of the blog @robjlow spotted an error in Uncle Colin's last answer. As I'm forever telling my students, making errors is how you learn; Rob has graciously delivered a lesson for us all. Thanks for keeping me honest! Recently, Uncle Colin gave a couple of ways to seeRead More → Dear Uncle Colin, What is $\lim_{x \to \infty} \left\{ \sqrt{x^2 + 3x} - x\right\}$? - Raging Over Obnoxious Terseness Hi, ROOT, and thanks for your very brief question. My approach would be to split up the square root and use either a binomial expansion or completing the square, as follows:Read More → In this month's installment of Wrong, But Useful, our special guest co-host is @mathsjem (Jo Morgan in real life) from the indispensable resourceaholic.com. We start by talking about resourceaholic.com and how Jo manages to fit such a punishing blog schedule around being a nearly-full-time maths teacher. Colin wonders how writingRead More → "Arr, that be a scurvy-lookin' expression!" said the Mathematical Pirate. "A quartic on the top and a quadratic on the bottom. That Ninja would probably try to factorise and do it all elegant-like." "Is that not the point?" "When you got something as 'orrible as that, it's like puttin' lipstickRead More → Dear Uncle Colin, Help! My calculator is broken and I need to solve - or at least approximate - $0.1 = \frac{x}{e^x - 1}$! How would you do it? -- Every $x$ Produces Outrageous Numbers, Exploring New Techniques Hi, ExPONENT, and thanks for your message! That's a bit of aRead More → "Sensei! I have a problem!" The Mathematical Ninja nodded. "Bring it on." "There's a challenge! Someone has picked a five-digit integer and cubed it to get 6,996,364,932,376. I know it ends with a six, and I could probably get the penultimate digit with a bit of work... I just wonderedRead More → Dear Uncle Colin, When I differentiate $y=2x^2 + 7x + 2$ and apply the $nx^{n-1}$ rule, why do I only apply it to the $2x^2$ and the $7x$ but not the 2? -- Nervous Over Rules, Mathematically A Liability Hi, NORMAL, and thanks for your message! There are several waysRead More → "Isn't it somewhere around $\phi$?" asked the student, brightly. "That number sure crops up in a lot of places!" The Mathematical Ninja's eyes narrowed. "Like shells! And body proportions! And arrawk!" Hands dusted. The Mathematical Ninja stood back. "The Vitruvian student!" The student arrawked again as the circular machine heRead More →
Search Now showing items 1-10 of 26 Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV (Elsevier, 2017-12-21) We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ... Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV (American Physical Society, 2017-09-08) The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ... Online data compression in the ALICE O$^2$ facility (IOP, 2017) The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ... Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV (American Physical Society, 2017-09-08) In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ... J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV (American Physical Society, 2017-12-15) We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($\|y\| < 0.9$) ... Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions (Nature Publishing Group, 2017) At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ... K$^{}(892)^{0}$ and $\phi(1020)$ meson production at high transverse momentum in pp and Pb-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 2.76 TeV (American Physical Society, 2017-06) The production of K$^{}(892)^{0}$ and $\phi(1020)$ mesons in proton-proton (pp) and lead-lead (Pb-Pb) collisions at $\sqrt{s_\mathrm{NN}} =$ 2.76 TeV has been analyzed using a high luminosity data sample accumulated in ... Production of $\Sigma(1385)^{\pm}$ and $\Xi(1530)^{0}$ in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV (Springer, 2017-06) The transverse momentum distributions of the strange and double-strange hyperon resonances ($\Sigma(1385)^{\pm}$, $\Xi(1530)^{0}$) produced in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV were measured in the rapidity ... Charged–particle multiplicities in proton–proton collisions at $\sqrt{s}=$ 0.9 to 8 TeV, with ALICE at the LHC (Springer, 2017-01) The ALICE Collaboration has carried out a detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and ... Energy dependence of forward-rapidity J/$\psi$ and $\psi(2S)$ production in pp collisions at the LHC (Springer, 2017-06) We present ALICE results on transverse momentum ($p_{\rm T}$) and rapidity ($y$) differential production cross sections, mean transverse momentum and mean transverse momentum square of inclusive J/$\psi$ and $\psi(2S)$ at ...
Answer parallelogram matches (f) square matches (e) trapezoid matches (d) circle matches (b) rectangle matches (c) triangle matches (a) Work Step by Step A=b$\times$h is the formula for the area of a parallelogram. A=s$^2$ is the formula for the area of a square. A=$\frac{1}{2}$ $\times$h$\times$(b+B) is the formula for the area of a trapezoid. A=$\pi$$\times$r$^2$ is the formula for the area of a circle. A=l$\times$w is the formula for area of a rectangle. A=$\frac{1}{2}$ $\times$b$\times$h is the formula for the area of a triangle.
\[\] Exponential Distribution has the following properties: Equals the distance between successive occurances or arrivals of a Poisson process with mean $\lambda > 0$ $\lambda$ is the average number of occurances or arrivals per unit of time (length, space, etc.) $\frac{1}{\lambda}$ is the average time between occurrences or arrivals. \defl{Exponential Distribution:} \[f(x) = \lambda e^{-{\lambda}x} \] \[F(x) = 1- e^{-{\lambda}x} \] $0 \leq x < \infty$ $\mu = E(X) =\frac{1}{\lambda}$ $\sigma^2 = V(X) = \frac{1}{\lambda^2}$ The amount of time until the next customer at The Pizza Customer will arrive. The amount of time until the DVD player will break. The exponential distribution is very useful for determining length of warranty. The amount of time until the next person will arrive at a specific ATM
so let's say we have: $f''+4f=\frac{1}{\sin(2x)}$ Homogeneous Problem: We use Euler-Ansatz: $char(\lambda)=\lambda^2+4 \Rightarrow \lambda=\pm 2i$ So we get $f_h(x)=\hat{A}e^{2ix}+\hat{B}e^{-2ix}=A\cos(2x)+B\sin(2)$ Particular Problem: We use variaton of constants. From the homogeneous solution we get the basis $\{\cos(2x),\sin(2x)\}$. We get [I changed A and B to $u_1$ and $u_2$] I: $-4u_1\cos(2x)+4u_2\cos(2x)=0$ II:$-4u_1\sin(2x)+4u_2\sin(2x)=\frac{1}{\sin(2x)}$ so: $\begin{pmatrix}\cos(2x)&\sin(2x)\\-2\sin(x)&2\cos(x)\end{pmatrix}=\begin{pmatrix}u_1\\u_2\end{pmatrix}=\begin{pmatrix}0\\\frac{1}{\sin(2x)}\end{pmatrix}$ whereas the particular solution will be $y_p(x)=U_1\cos(2x) + U_2\sin(2x)$ with $U_i$ being the $u_i$'s integrated. Question: Let me also quickly tell you about the idea I have about the whole thing we are doing here. Basicaly, the solution to a ODE is a vector space, so by solving the homogeneous problem, we have a nice vector space. We then only need to somehow adjust that vectorspace s.t. it "works" with the inhomogenity. We can do that by "changign" the coefficients. What I don't see yet is why we set I=0$ and $II=\frac{1}{\sin(2x)}$. Could someone please elaborate? Why can't I do $I=\frac{1}{\sin(2x)}, II=\frac{1}{\sin(2x)}$? Could I also do $I=\frac{1}{\sin(2x)}, II=0$?
Context Suppose we have a grid-based game where a unit has a range parameter that serves as the upperbound of the sum of the costs of his movements in a single turn. Moving orthogonally to an adjacent tile costs 1, and moving diagonally to an adjacent tile costs $\sqrt{2}$. So a unit with range 4 can move up to 2 tiles diagonally (costs about 2.8284) and one more tile orthogonally if he wishes. But he cannot move 3 tiles diagonally since its cost (about 4.2426) is strictly greater than 4. Prompt Suppose we were to use an approximation of $\sqrt{2}$ for the diagonal cost. My question is: at what point will this approximation fails in terms of the context above? i.e. if $x = \sqrt{2}$, and $\bar{x}$ its approximation, $\bar{x} = 1.414$, say. What's the smallest $k \in \mathbb{N}$ for which $\lceil k x\rceil \neq \lceil k \bar{x}\rceil$? By using a spreadsheet (dropbox), I get the following results: x bar k------------------1.5 71.4 51.41 171.414 991.4142 169 So if our grid is sufficiently large to allow for more than 99 diagonal movement, then we should use 1.4142 instead of 1.414, say. But there's also an immediate extension: Suppose we know that our grid won't exceed 300 diagonals, say. Then what's the least precise (in terms of the number of decimal places) we can be? i.e. Given a $K \in \mathbb{N}$, find the least precise $\bar{x}$ for which $k > K$, where $k$ is as defined above.
Short Answer Set Method -> {"MethodOfLines", "DifferentiateBoundaryConditions" -> {True, "ScaleFactor" -> 1}} inside NDSolve will resolve the problem. It's not necessary to set "ScaleFactor" to 1, it just needs to be a not-that-small positive number. Long Answer The answer for this problem is hidden in this obscure tutorial. I'll try my best to retell it in a easier to understand way. Let's consider the following simpler initial-boundary value problem (IBVP) of heat equation that suffers from the same issue: $$\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$$$$u(0,x)=x(1-x)$$$$u(t,0)=0,\ \frac{\partial u}{\partial x}\bigg\|_{x=1}=0$$$$t>0,\ 0\leq x\leq 1$$ Clearly $u(0,x)=x(1-x)$ and $\frac{\partial u}{\partial x}\bigg\|_{x=1}=0$ is inconsistent. When you solve it with NDSolve / NDSolveValue, ibcinc warning will be spit out: tend = 1; xl = 0; xr = 1; With[{u = u[t, x]}, eq = D[u, t] == D[u, x, x]; ic = u == (x - xl) (xr - x) /. t -> 0; bc = {u == 0 /. x -> xl, D[u, x] == 0 /. x -> xr};] sol = NDSolveValue[{eq, ic, bc}, u, {t, 0, tend}, {x, xl, xr}] NDSolveValue::ibcinc and further check shows the boundary condition (b.c.) $\frac{\partial u}{\partial x}\bigg\|_{x=1}=0$ isn't satisfied at all: Plot[D[sol[t, x], x] /. x -> xr // Evaluate, {t, 0, tend}, PlotRange -> All] Why does this happen? The best way for explaination is re-implementing the method used by NDSolve in this case i.e. method of lines, all by ourselves. As mentioned in the document, method of lines is a numeric method that discretizes the partial diffential equation (PDE) in all but one dimension and then integrating the semi-discrete problem as a system of ordinary differential equations (ODEs) or differential algebraic equations (DAEs). Here I discretize the PDE with 2nd order centered difference formula: $$f'' (x_i)\simeq\frac{f (x_{i}-h)-2 f (x_i)+f (x_{i}+h)}{h^2}$$ Clear@dx formula = eq /. {D[u[t, x], t] -> u[x]'[t], D[u[t, x], x, x] -> (u[x - dx][t] - 2 u[x][t] + u[x + dx][t])/dx^2} points = 5; dx = (xr - xl)/(points - 1); ode = Table[formula, {x, xl + dx, xr - dx, dx}] {u[1/4][t] == 16 (u[0][t] - 2 u[1/4][t] + u[1/2][t]), u[1/2][t] == 16 (u[1/4][t] - 2 u[1/2][t] + u[3/4][t]), u[3/4][t] == 16 (u[1/2][t] - 2 u[3/4][t] + u[1][t])} I've chosen a very coarse grid for better illustration. The initial condition (i.c.) should also be discretized: odeic = Table[ic /. u[t_, x_] :> u[x][t] // Evaluate, {x, xl, xr, dx}] {u[0][0] == 0, u[1/4][0] == 3/16, u[1/2][0] == 1/4, u[3/4][0] == 3/16, u[1][0] == 0} We still need to deal with b.c.. The Dirichlet b.c. doesn't need disrcetization: bcnew1 = bc[[1]] /. u[t_, x_] :> u[x][t] u[0][t] == 0 The Neumann b.c. contains derivative of $x$ so we need to discretize it with one-sided difference formula: $$f' (x_n)\simeq \frac{f (x_{n}-2h)-4 f (x_{n}-h)+3 f (x_n)}{2 h}$$ bcnew2 = bc[[2]] /. D[u[t, x_], x_] :> (u[x - 2 dx][t] - 4 u[x - dx][t] + 3 u[x][t])/(2 dx) 2 (u[1/2][t] - 4 u[3/4][t] + 3 u[1][t]) == 0 "OK, 5 unknowns, 5 equations, we can now solve the system with any ODE solver! Just as !" Sadly you were wrong if you thought this statement is correct, because: NDSolve does Though {ode, odeic, bcnew1, bcnew2} is already a solvable system, it's not a set of ODEs, but DAEs. Notice here ODE refers to explicit ODE i.e. the coefficient of derivative term can't be $0$. Clearly, bcnew1 and bcnew2 doesn't explicitly contain derivative of $t$. Though NDSolve is able to handle this DAE system directly, it doesn't solve the PDE in this way by default. Instead, it'll try to transform the DAE system to an explicit ODE system, probably because its ODE solver is generally stronger than the DAE solver (at least now). So, how does NDSolve transform the DAE sytem to ODE system? "That's simple! Just eliminate some of the variables with bcnew1 and bcnew2! " Yeah this is a possible method, but not the one implemented in NDSolve. NDSolve has chosen a method that may be rather unusual at first glance. It mixes the original b.c. with its 1st order derivative respects to $t$. For our specific problem, the b.c. becomes: odebc1 = D[#, t] + scalefactor1 # & /@ bcnew1 scalefactor1 u[0][t] + u[0]'[t] == 0 odebc2 = D[#, t] + scalefactor2 # & /@ bcnew2 2 scalefactor2 (u[1/2][t] - 4 u[3/4][t] + 3 u[1][t]) + 2 (u[1/2]'[t] - 4 u[3/4]'[t] + 3 u[1]'[t]) == 0 Where scalefactor1 and scalefactor2 are properly chosen coefficients. It's not hard to notice this approach is systematic and easy to implement, and I guess that's the reason why NDSolve chooses it for transforming algebraic equation to ODE. Nevertheless, this method has its disadvantage. The generated b.c. is equivalent to the original b.c., only if the original b.c. is continuous and i.c. is consistent with the b.c.. Let's use odebc1 as example. In our case, bc[[1]] is continuous, and it's consistent with ic, so it can be easily rebuilt from odebc1: DSolve[{odebc1, odeic[[1]]}, u[0][t], t] {{u[0][t] -> 0}} However, if the i.c. is something that isn't consistent with bc[[1]], for example u[0][0] == 1, the b.c. rebuilding from odebc1 will become: DSolve[{odebc1, u[0][0] == 1}, u[0][t], t] {{u[0][t] -> E^(-scalefactor1 t)}} It's no longer equivalent to bc[[1]], but when scalefactor1 is a large positive number, this b.c. will converge to the original one. Now here comes the key point. As stated in the document: With the default "ScaleFactor" value of Automatic, a scaling factor of 1 is used for Dirichlet boundary conditions and a scaling factor of 0 is used otherwise. i.e. scalefactor2 will be set to 0. Guess what b.c. will be rebuilt in this way?: With[{scalefactor2 = 0}, DSolve[{D[#, t] + scalefactor2 # & /@ bc[[2]], D[ic, x] /. x -> xr}, D[u[t, x], x] /. x -> xr, t]] {{Derivative[0, 1][u][t, 1] -> -1}} It's a completely different b.c.. Back to the problem mentioned in the question, we can analyse its b.c. with the same method as above: bcInQuestion = (D[u[t, x], x] /. x -> 1) == c1 (u[t, 1] - c2); icInQuestion = u[0, x] == c3; With[{sf = 0}, DSolve[{D[#, t] + sf # & /@ bcInQuestion, D[icInQuestion, x] /. x -> 1}, D[u[t, x], x] /. x -> 1, t]] /. u[0, _] :> c3 {{Derivative[0, 1][u][t, 1] -> -c1 c3 + c1 u[t, 1]}} We see c1 is still in the rebuilt b.c., while c2 is completely killed by the derivation, that's why OP found "editing c1 does affect the graph, while editing c2 does not". OK, then why does NDSolve choose such a strange setting for scaling factor? The document explains as follows: There are two reasons that the scaling factor to multiply the original boundary condition is zero for boundary conditions with spatial derivatives. First, imposing the condition for the discretized equation is only a spatial approximation, so it does not always make sense to enforce it as exactly as possible for all time. Second, particularly with higher-order spatial derivatives, the large coefficients from one-sided finite differencing can be a potential source of instability when the condition is included. … but personally I think this design is just too lazy: why not make NDSolve choose a non-zero scaling factor at least when ibcinc warning pops up and the order of spatial derivatives isn't too high (this is usually the case, the differential order of most of PDEs in practise is no higher than 2 )? Anyway, now we know how to fix the issue. Just choose a positive scaling factor: Clear[c2] c1 = -10; (c2=10;) c3 = 20; q[t, x] = 100000; heat = ParametricNDSolveValue[{1591920 D[u[t, x], t] == .87 D[u[t, x], x, x] + q[t, x], (D[u[t, x], x] /. x -> 0) == 0, (D[u[t, x], x] /. x -> 1) == c1 (u[t, 1] - c2), u[0, x] == c3}, u, {t, 0, 600}, {x, 0, 1}, c2, Method -> {"MethodOfLines", "DifferentiateBoundaryConditions" -> {True, "ScaleFactor" -> 1}}]; Plot[heat[#][300, x] & /@ Range[10, 50, 10] // Evaluate, {x, 0.9, 1}, PlotRange -> All, AxesLabel -> {"x(m)", "T"}] Now c2 influences the solution. Young has also solved the problem with the new-in- v10 "FiniteElement" method, I guess it's probably because b.c.s are imposed in a completely different way when "FiniteElement" method is chosen, but I'd like not to talk too much about it given I'm still in v9 and haven't look into "FiniteElement".
This article provides answers to the following questions, among others: How can the microstructure fraction of a steel be determined from the iron-carbon phase diagram? How is the phase fraction of a steel determined from the iron-carbon phase diagram? Introduction For many applications it is important to know exactly what microstructure or phase fractions a steel is composed of at a certain carbon content. This ultimately necessitates a calculation. In order to carry out this, however, the entire iron-carbon phase diagram must be considered. Therefore, the complete phase diagram of the metastable system is briefly described below, before the determination of the microstructure and phase fractions is finally explained. Up to now, the iron-carbon phase diagram has only been considered up to a carbon content of 2 % (steel part). At higher carbon concentrations, further phase transformations occur, which lead to a different microstructure. Such ferrous materials are then no longer referred to as steels but as cast iron. In the corresponding article on cast iron, the microstructure formation of such materials is described in more detail. In principle, however, the iron-carbon phase diagram of the metastable system ends at a carbon content of 6.67 %, since the microstructure consists of 100 % cementite. Chemically, the cementite consists of three iron atoms (each with an atomic mass of 56 u) and one carbon atom (with an atomic mass of 12 u). Thus, the mass-related carbon content in the cementite is 6.67 %: \begin{align} &\underline{\text{carbon content}} = \frac{12u}{12u+3 \cdot 56u} \cdot 100 \text{ %} = \underline{6.67 \text{ %}} \\[5px] \end{align} Determination of the microstructure fractions and phase fractions In principle, the microstructure and phase fractions are determined by applying the lever rule. The lever arms must always be pulled to the corresponding microstructural or phase boundaries. In the following, the microstructure and phase fractions at room temperature for an hyper- and hypoeutectoid steel will be determined as an example. Hypoeutectoid steels In a hypoeutectoid (hypoperlitic) steel, the microstructure consists of ferrite and pearlite grains at room temperature. In order to determine the respective microstructure fractions, the lever arms are drawn accordingly from the considered state point up to the ferrite phase region (at 0 % carbon) and to the pearlite limit (at 0.8 % carbon). For a steel with, for example, 0.3 % carbon, this results in a ferrite content of 62.5 % at room temperature and a corresponding pearlite content of 37.5%: \begin{align} &\underline{\text{ferrite}} = \frac{0.8-0.3}{0.8} \cdot 100 \text{ %} = \underline{62.5 \text{ %}} \\[5px] &\underline{\text{pearlite}} = \frac{0.3}{0.8} \cdot 100 \text{ %} = \underline{37.5 \text{ %}} \\[5px] \end{align} Due to the lever rule, there is generally a linear relationship between the carbon content and the microstructure fractions. For a hypoeutectoid steel, the fraction of pearlite increases steadily with a higher carbon content until it finally reaches 100 % at 0.8 % carbon. Accordingly, the ferrite content decreases to 0 %. The explicit relationship is shown in a microstructure diagram below the phase diagram. The term microstructure fraction (“grain fraction”) should not be confused with the term phase fraction at this point! After all, the microstructural component pearlite consists of a phase mixture consisting of ferrite as well as cementite. The steel can thus also be characterized by the phase components ferrite and cementite instead of the microstructural components ferrite and pearlite. The procedure for determining the phase fractions is basically identical, but it must be noted that the lever arms must then be drawn up to the respective phase boundaries ferrite and cementite. For the hypoeutectoid steel with 0.3 % carbon, the total phase content of ferrite is 95.5%. The remaining 4.5 % is finally accounted for by the cementite phase: \begin{align} &\underline{\text{ferrite}} = \frac{6.67-0.3}{6.67} \cdot 100 \text{ %} = \underline{95.5 \text{ %}} \\[5px] &\underline{\text{cementite}} = \frac{0.3}{6.67} \cdot 100 \text{ %} = \underline{4.5 \text{ %}} \\[5px] \end{align} Hypereutectoid steels The microstructure constituents of hypereutectoid steels can be determined in the same way as for hypoeutectoid steels. The lever arms are pulled to the respective microstructure components of the pearlite (at 0.8 % carbon) and the grain boundary cementite (at 6.67 % carbon). For a steel with, for example, 1.4 % carbon, this results in a pearlite fraction of around 89.8 % at room temperature and a corresponding grain boundary cementite fraction of 10.2 %: \begin{align} &\underline{\text{pearlite}} = \frac{6.67-1.4}{6.67-0.8} \cdot 100 \text{ %} = \underline{89.8 \text{ %}} \\[5px] &\underline{\text{grain boundary cementite}} = \frac{1.4-0.8}{6.67-0.8} \cdot 100 \text{ %} = \underline{10.2 \text{ %}} \\[5px] \end{align} For a hypereutectoid steel, the proportion of pearlite decreases steadily with increasing carbon content to a minimum of 78.5 % (at 2.06 % carbon). Accordingly, the fraction of grain boundary cementite increases to a maximum of 21.5 %. The more detailed relationship is shown in the corresponding microstructure diagram below the iron-carbon phase diagram. For a hypereutectoid steel, too, the term microstructure fraction must again be distinguished from the term phase fraction. Finally, the phase cementite is not only at the grain boundaries but also in the pearlite microstructure, which also consists of ferrite. The corresponding phase proportions of ferrite and cementite can be determined after pulling the lever arms to the respective phase boundaries. For the hypereutectoid steel with 1.4 % carbon, the total phase fraction of ferrite is 79.0 %. The remaining 21.0 % are finally accounted for by the cementite phase: \begin{align} &\underline{\text{ferrite}} = \frac{6.67-1.4}{6.67} \cdot 100 \text{ %} = \underline{79.0 \text{ %}} \\[5px] &\underline{\text{cementite}} = \frac{1.4}{6.67} \cdot 100 \text{ %} = \underline{21.0 \text{ %}} \\[5px] \end{align}
Chakradhar, Sreekanth RP and Nagabhushana, BM and Chandrappa, GT and Ramesh, KP and Rao, JL (2004) EPR of $Ni^{2+}$ ions in macro porous nanocrystalline $CaSiO_3$ ceramic powders. In: DAE Solid State Physics Symposium, Dec 26 - 30, 2004, Amritsar, India. PDF NiAbstract.pdf Download (101kB) Abstract Porous wollastonite $(CaSiO_3)$ and $Ni^{2+}$ doped (1, 2, 2.5 3 and 5 mol%) wollastonite ceramic powders have been synthesized by a novel self-propagating, gas producing solution combustion process. The solution combustion method renders a low-temperature synthetic route to prepare fine-grained wollastonite powders with better sintering properties. Single phases of $\beta-CaSiO_3$ and $\alpha-CaSiO_3$ were observed at 950 °C and 1200 °C respectively. The samples calcined at 950 °C has low porosity, however the porosity increases with increase in calcinations temperature. The agglomeration of the solution combustion product was determined by SEM and the average agglomerated particle size of calcined $CaSiO_3$ is in the range 1 to 10 \mu m. The EPR spectrum of $Ni^{2+}$ ions in $CaSiO_3$ exhibits a symmetric absorption at g = 2.23 \pm 0.01. Owing to the moderately high spin orbit coupling, the isotropic g factor of octahedrally coordinated $Ni^{2+}$ departs from the free electron g value, $g_e$. The number of spins participating in resonance (N) and the paramagnetic susceptibilities (\chi) have been evaluated from EPR data as a function of $Ni^{2+}$ content. The effect of alkali ions (Li, Na and K) on the EPR spectra, have also been studied and the spectra exhibit a marked alkali effect. The results obtained from these studies have been discussed in detail. Item Type: Conference Poster Keywords: EPR;Ni2+ ions;macro porous;nanocrystalline CaSiO3;ceramic powders Department/Centre: Division of Physical & Mathematical Sciences > Physics Depositing User: Sreekanth Chakradhar Ph.D., Dr. R. P. Date Deposited: 22 Nov 2007 Last Modified: 19 Sep 2010 04:17 URI: http://eprints.iisc.ac.in/id/eprint/2594 Actions (login required) View Item
(Sorry was asleep at that time but forgot to log out, hence the apparent lack of response) Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference
Determining the short-circuit rating of the breaker requires knowledge of the prospective fault current at the panelboard. Following section provides a simple method to obtain this current magnitude. 3-phase circuit breakers Let’s start with calculating the prospective current from the nearest transformer. Fault current on the secondary side of the 3-phase transformer: I= \frac{FLA100}{\%Z} Amps where, FLA = Full load amps of the transformer = \frac{Apparant power (S_3\phi)} {\sqrt3 Line voltage} %Z = Per unit impedance of transformer The short circuit amps obtained from this method ignores the source impedance. If you do have this impedance then use the following equation. I = \frac{FLA*100}{\%Z + \%Zsource} Amps Ofcourse, you will be more accurate with the result now. However, source impedance is not easily available. This piece of information is typically obtained from the utility serving you. %Zsource encompasses the impedance of the entire power system, calculated upto the primary side of the indicated transformer. 1-Phase Circuit Breaker The procedure in determining the short-circuit rating of a 1-phase circuit-breaker in a 1-phase system is same as above except for the modification to the full load current equation (of the 1-phase transformer.)FLA=\frac{S_1\phi}{V_{line}} Figure 1: Calculation of short circuit rating of a single phase breaker The X/R factor When selecting a circuit breaker be wary of choosing breakers that will become marginally rated i.e. fault currents are 80 – 100% of the breaker rating. In circuits with X/R ratio greater than 15 (i.e. highly inductive circuits), the magnitude of prospective fault current might be greater than the calculated current. Why? Well, in AC systems, fault currents have asymmetrical waveform. The asymmetry (due to DC offset) gradually tapers off to become symmetric. See figure below. The rate at which the asymmetry decays varies as a function of the circuit X/R ratio. The higher the X/R ratio of a circuit, the higher the fault current magnitude a circuit breaker has to handle. So, assuming that you are oblivious of the system X/R at the point where you are installing the circuit breaker and that the high current magnitude will persist for a while, it is prudent to select circuit breakers such that the calculated fault currents are only 80% (or less) than its short-circuit rating. Refer to National Electric Code for sizing the breaker.
103 0 1. Homework Statement From Resnick and Halliday: A metal plate 8 cm on a side carries a total charge of 60 microC. Using the infinite plate approximation, calculate the electric field 0.5 mm above the surface of the plate near the plate's center. 2. Homework Equations (1) [tex]E = \frac{\sigma}{\epsilon_0}[/tex] or (2) [tex]E = \frac{\sigma}{2\epsilon_0}[/tex] [itex]\sigma[/itex] being charge density. 3. The Attempt at a Solution I got a bit confused here. Equation (1) should be used when the plate is a conductor, and equation (2) should be used when the plate is an insulator, according to the explanation in the book. However, when I use equation (1) to calculate, I get 106MN/C, and in the answers in the back of the book it says it should be 53 MN/C. Either I used the wrong equation, or I should have used half of the charge. What went wrong? And honestly, I really did not understand why there are 2 different equations for each situation (conductor, insulator). I tried deriving them on my own but no success in gaining real understanding, so I'd be really glad if someone could shed some light on the subject.
Search Now showing items 1-5 of 5 Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV (Springer, 2015-05-20) The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ... Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV (Springer, 2015-06) We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ... Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV (Springer, 2015-09) Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ... Coherent $\rho^0$ photoproduction in ultra-peripheral Pb-Pb collisions at $\mathbf{\sqrt{\textit{s}_{\rm NN}}} = 2.76$ TeV (Springer, 2015-09) We report the first measurement at the LHC of coherent photoproduction of $\rho^0$ mesons in ultra-peripheral Pb-Pb collisions. The invariant mass and transverse momentum distributions for $\rho^0$ production are studied ... Inclusive, prompt and non-prompt J/ψ production at mid-rapidity in Pb-Pb collisions at √sNN = 2.76 TeV (Springer, 2015-07-10) The transverse momentum (p T) dependence of the nuclear modification factor R AA and the centrality dependence of the average transverse momentum 〈p T〉 for inclusive J/ψ have been measured with ALICE for Pb-Pb collisions ...
Is the volume of solid bounded below by the sphere $$\rho=2 \cos(\phi)$$ and above by the cone $$z=\sqrt{x^2+y^2}$$ will be gotten by the following integration ? in cylindrical form : $$V =\int_{0}^{2\pi}d\theta\int_{0}^{1}\int_{1+\sqrt{1-r^2}}^{r} r \,dz \, dr$$ (since the projection of their intersection will be a unit circle) in spherical form : $$V =\int_{0}^{2\pi}d\theta\int_{\pi/4}^{\pi/2}\int_{0}^{2 \cos(\phi)}\rho^2 \sin(\phi )d\rho \, d\phi$$ But they are not giving me the same answer .. the first gives me $$-\pi$$ while the second one gives me $$\pi/3$$ If I try to get the volume by logic then it is the volume of the unit sphere when we subtract from it the volume of the cone(with unit circle base - height is 2 is the diameter of sphere) : $$4\pi/3 - 2 \pi/3= 2\pi/3$$ which is totally different from the above answers ! Note the solution manual wrote the final answer $$\pi/3$$ but I want to know what is wrong with my logic and integration in cylindrical..
When I discuss a question with my colleague, he said he has a feeling that the symmetrization of functions should be unique up to a coefficient. However, I keep being critical at the point. Specifically, symmetrization(standard) of a multivariable function is $$Sf(x_1,\dots,x_n)=\frac{1}{n!}\sum_{\sigma\in S_n}f(\sigma(x_1,...,x_n))$$ Here, $S_n$ is the symmetric group of n elements with order $\|n!\|$. $\sigma(x_1,...,x_n)$ is a short notation for $(x_{\sigma(1)},\dots,x_{\sigma(n)})$ since $\sigma$ is a permutation acting on these subscripts. Then the question can be rephrased as the following: Let $S$ be the symmetrization operation for $n-$variable functions as the above formula, if $T$ is another symmetrization operation which is not trivial (say the constant operator, $Tf=c$, where $c$ is some constant) for $n-$variable functions, which satisfies $T(Tf)=Tf$, then $Sf-Tf=0$. (For this case, there cannot be extra coefficient since the operation should be projective ($T\circ T=T$)). Or if $T$ is a projection operator of $n-$variable function such that for all permutation $\sigma\in S_n$, $\sigma(Tf):=(Tf)(\sigma(x_1,...,x_n))=Tf$. Then $Tf=Sf$. Can someone show this thing, find a proof in any books, or just find a counter example? Appreciated.
Topological proof of Benoist-Quint's orbit closure theorem for $ \boldsymbol{ \operatorname{SO}(d, 1)} $ Department of Mathematics, Yale University, New Haven, CT 06520, USA We present a new proof of the following theorem of Benoist-Quint: Let $ G: = \operatorname{SO}^\circ(d, 1) $, $ d\ge 2 $ and $ \Delta<G $ a cocompact lattice. Any orbit of a Zariski dense subgroup $ \Gamma $ of $ G $ is either finite or dense in $ \Delta \backslash G $. While Benoist and Quint's proof is based on the classification of stationary measures, our proof is topological, using ideas from the study of dynamics of unipotent flows on the infinite volume homogeneous space $ \Gamma \backslash G $. Mathematics Subject Classification:Primary: 37A17; Secondary: 22E40. Citation:Minju Lee, Hee Oh. Topological proof of Benoist-Quint's orbit closure theorem for $ \boldsymbol{ \operatorname{SO}(d, 1)} $. Journal of Modern Dynamics, 2019, 15: 263-276. doi: 10.3934/jmd.2019021 References: [1] [2] [3] [4] [5] S. G. Dani and G. A. Margulis, [6] [7] [8] [9] G. A. Margulis, Problems and conjectures in rigidity theory, in [10] G. A. Margulis and G. M. Tomanov, Invariant measures for actions of unipotent groups over local fields on homogeneous spaces, [11] [12] [13] [14] [15] [16] [17] N. A. Shah, Invariant measures and orbit closures on homogeneous spaces for actions of subgroups generated by unipotent elements, in [18] [19] show all references References: [1] [2] [3] [4] [5] S. G. Dani and G. A. Margulis, [6] [7] [8] [9] G. A. Margulis, Problems and conjectures in rigidity theory, in [10] G. A. Margulis and G. M. Tomanov, Invariant measures for actions of unipotent groups over local fields on homogeneous spaces, [11] [12] [13] [14] [15] [16] [17] N. A. Shah, Invariant measures and orbit closures on homogeneous spaces for actions of subgroups generated by unipotent elements, in [18] [19] [1] [2] [3] [4] [5] [6] [7] [8] [9] S. A. Krat. On pairs of metrics invariant under a cocompact action of a group. [10] François Gay-Balmaz, Cesare Tronci, Cornelia Vizman. Geometric dynamics on the automorphism group of principal bundles: Geodesic flows, dual pairs and chromomorphism groups. [11] [12] [13] [14] [15] [16] [17] [18] Andres del Junco, Daniel J. Rudolph, Benjamin Weiss. Measured topological orbit and Kakutani equivalence. [19] [20] 2018 Impact Factor: 0.295 Tools Article outline [Back to Top]
In p-adic properties of modular schemes and modular forms Katz formulates the following base change theorem as Theorem 1.7.1 Let $n\geq 3$ and $\overline{\mathcal{M}}_n$ be the compactified moduli scheme of elliptic curves with level-$n$-structure over $\mathbb{Z}[\frac1n]$. Let $K$ be any $\mathbb{Z}[\frac1n]$-module. Then the morphism $$K\otimes H^0(\overline{\mathcal{M}}_n; \omega^{\otimes k}) \to H^0(\overline{\mathcal{M}}_n; K\otimes \omega^{\otimes k})$$ is an isomorphism for $k\geq 2$. The key step is to show that $H^1(\overline{\mathcal{M}}_n; \omega^{\otimes k})$ vanishes for $k\geq 2$. He claims that this follows from $\omega^{\otimes k}$ having degree strictly greater than $2g-2$ on every connected component of $\overline{\mathcal{M}}_n \otimes \mathbb{Z}[\frac1n, \zeta_n]$ (where $g$ is the genus of such a connected component) by Riemann-Roch. I am familiar with this kind of Riemann-Roch argument for projective curves over a field, where it can be found in Hartshorne in nice cases and in more general cases in Liu's book. But Katz uses it in the case that the base is $\mathbb{Z}[\frac1n]$. What kind of Riemann-Roch argument works here?
Let us consider the simplest isoperimetric inequality. Consider a smooth simple closed curve given by $r=\rho(\theta)$ in polar coordinates, where $\rho(\theta)>0$ can be regarded as a smooth periodic function with period $2\pi$. As $dA=rdrd\theta$, the area of the region $\Omega$ enclosed by the closed curve is \begin{align} A=\iint_{\Omega}rdrd\theta =\frac{1}{2}\int_0^{2\pi}\rho(\theta)^2 d\theta. \end{align} On the other hand, it is an easy exercise in calculus that the length of the curve is given by \begin{align} L=&\int_{0}^{2\pi}\sqrt{\rho(\theta)^2+\rho'(\theta)^2}d\theta. \end{align} By the isoperimetric inequality, we have $L^2\ge 4\pi A$, so we must have \begin{equation} 2\pi \int_{0}^{2\pi} \rho(\theta)^2 d\theta \le \left(\int_{0}^{2\pi}\sqrt{\rho(\theta)^2+\rho'(\theta)^2}d\theta\right)^2. \end{equation} This looks similar to the Wirtinger's inequality, which states that \begin{equation} \begin{split} \int_{0}^{2\pi}\rho'(\theta)^2d\theta \ge \int_{0}^{2\pi} \left(\rho(\theta)-\overline \rho\right)^2 d\theta = \int_{0}^{2\pi} \rho(\theta)^2 d\theta-\frac{1}{2\pi} \left(\int_{0}^{2\pi}\rho(\theta)d\theta\right)^2. \end{split} \end{equation} So \begin{align} 2\pi \int_{0}^{2\pi} \rho(\theta)^2 d\theta \le& \left(\int_{0}^{2\pi}\rho(\theta)d\theta\right)^2+2\pi\int_{0}^{2\pi}\rho'(\theta)^2d\theta, \end{align} which isn't quite what I want. It doesn't seem right to me that we can apply Wirtinger's inequality directly to prove it because the equality case in Wirtinger's inequality holds when $\rho(\theta)=C+a\cos \theta+ b\sin \theta$, but the equality in our inequality can only hold when $\rho$ is constant the equality in our inequality doesn't hold in this case. (However, by geometric consideration, the equality does hold for, say, $\rho(\theta)=\sin \theta$, but only if we restrict $\theta$ to $[0, \pi]$.) This is where I am stuck. So my question is: Can we show this inequality without using the isoperimetric inequality (say by Fourier analysis or using Wirtinger's inequality more carefully)? Can it be used to prove (at least a simple case of) the isoperimetric inequality? If not, why? (If nothing else, at least I obtain an inequality on circle :-) ) To elaborate further, it is well-known that we can apply Wirtinger's inequality (or Fourier type argument) to prove the isoperimetric inequality on the plane. Indeed, the Wirtinger's inequality and the isoperimetric inequality are equivalent (e.g. Osserman's paper on isoperimetric inequality). Usually, these kinds of proofs involve shifting the center of mass to $0$, applying the Green's theorem and the Wirtinger's (or Cauchy-Schwarz) inequality on some combination of two functions (say $x(s), y(s)$). So as a subquestion, why is there no such argument involving only a single function (say $\rho(\theta)$)?
(→Stacking) (31 intermediate revisions by 2 users not shown) Line 1: Line 1: − + + + + + * [[Siril:Tutorial_import\|Convert your images in the FITS format Siril uses (image import)]] * [[Siril:Tutorial_import\|Convert your images in the FITS format Siril uses (image import)]] * [[Siril:Tutorial_sequence\|Work on a sequence of converted images]] * [[Siril:Tutorial_sequence\|Work on a sequence of converted images]] * [[Siril:Tutorial_preprocessing\|Pre-processing images]] * [[Siril:Tutorial_preprocessing\|Pre-processing images]] − * [[Siril:Tutorial_manual_registration\|Registration ( + * [[Siril:Tutorial_manual_registration\|Registration (alignment)]] * → '''Stacking''' * → '''Stacking''' − ==Stacking== + ==Stacking== − The final + + The final to do with Siril is to stack the images. Go to the "stacking" tab, indicate if you want to stack all images, only selected images or the best images regarding the value of FWHM previously computed. + Sum Stacking + is the stack . The + Stacking + is to + , + + and + is very . − + : + . − + + + + the in the + , is . + + to + the . + + is + the the . + + + + + + + + + + + + + + + + + + + + + + + [[File:Siril stacking result.png\|700px]] [[File:Siril stacking result.png\|700px]] + + + + [[File:Siril inal_result.png\|700px]] [[File:Siril inal_result.png\|700px]] − The images above + + The images above the result in Siril the . Note the improvement of the signal-to-noise ratio regarding the result given for one frame in the previous [[Siril:Tutorial_preprocessing\|step]] + + + . + + Now should start the process of the image with crop, background extraction (to remove gradient), and some other processes to enhance your image. To see processes available in Siril please visit this [[Siril:Manual\|page]]. Now should start the process of the image with crop, background extraction (to remove gradient), and some other processes to enhance your image. To see processes available in Siril please visit this [[Siril:Manual\|page]]. + + − + + − [[ + + [[:. \|]] − + Latest revision as of 10:34, 13 September 2016 Siril processing tutorial Convert your images in the FITS format Siril uses (image import) Work on a sequence of converted images Pre-processing images Registration (Global star alignment) → Stacking Stacking The final step to do with Siril is to stack the images. Go to the "stacking" tab, indicate if you want to stack all images, only selected images or the best images regarding the value of FWHM previously computed. Siril proposes several algorithms for stacking computation. Sum Stacking This is the simplest algorithm: each pixel in the stack is summed using 32-bit precision, and the result is normalized to 16-bit. The increase in signal-to-noise ratio (SNR) is proportional to [math]\sqrt{N}[/math], where [math]N[/math] is the number of images. Because of the lack of normalisation, this method should only be used for planetary processing. Average Stacking With Rejection Percentile Clipping: this is a one step rejection algorithm ideal for small sets of data (up to 6 images). Sigma Clipping: this is an iterative algorithm which will reject pixels whose distance from median will be farthest than two given values in sigma units ([math]\sigma_{low}[/math], [math]\sigma_{high}[/math]). Median Sigma Clipping: this is the same algorithm except than the rejected pixels are replaced by the median value of the stack. Winsorized Sigma Clipping: this is very similar to Sigma Clipping method but it uses an algorithm based on Huber's work [1] [2]. Linear Fit Clipping: this is an algorithm developed by Juan Conejero, main developer of PixInsight [2]. It fits the best straight line ([math]y=ax+b[/math]) of the pixel stack and rejects outliers. This algorithm performs very well with large stacks and images containing sky gradients with differing spatial distributions and orientations. These algorithms are very efficient to remove satellite/plane tracks. Median Stacking This method is mostly used for dark/flat/offset stacking. The median value of the pixels in the stack is computed for each pixel. As this method should only be used for dark/flat/offset stacking, it does not take into account shifts computed during registration. The increase in SNR is proportional to [math]0.8\sqrt{N}[/math]. Pixel Maximum Stacking This algorithm is mainly used to construct long exposure star-trails images. Pixels of the image are replaced by pixels at the same coordinates if intensity is greater. Pixel Minimum Stacking This algorithm is mainly used for cropping sequence by removing black borders. Pixels of the image are replaced by pixels at the same coordinates if intensity is lower. In the case of NGC7635 sequence, we first used the "Winsorized Sigma Clipping" algorithm in "Average stacking with rejection" section, in order to remove satellite tracks ([math]\sigma_{low}=4[/math] and [math]\sigma_{high}=3[/math]). The output console thus gives the following result: 14:33:06: Pixel rejection in channel #0: 0.181% - 1.184% 14:33:06: Pixel rejection in channel #1: 0.151% - 1.176% 14:33:06: Pixel rejection in channel #2: 0.111% - 1.118% 14:33:06: Integration of 12 images: 14:33:06: Pixel combination ......... average 14:33:06: Normalization ............. additive + scaling 14:33:06: Pixel rejection ........... Winsorized sigma clipping 14:33:06: Rejection parameters ...... low=4.000 high=3.000 14:33:07: Saving FITS: file NGC7635.fit, 3 layer(s), 4290x2856 pixels 14:33:07: Execution time: 9.98 s. 14:33:07: Background noise value (channel: #0): 9.538 (1.455e-04) 14:33:07: Background noise value (channel: #1): 5.839 (8.909e-05) 14:33:07: Background noise value (channel: #2): 5.552 (8.471e-05) After that, the result is saved in the file named below the buttons, and is displayed in the grey and colour windows. You can adjust levels if you want to see it better, or use the different display mode. In our example the file is the stack result of all files, i.e., 12 files. The images above picture the result in Siril using the Auto-Stretch rendering mode. Note the improvement of the signal-to-noise ratio regarding the result given for one frame in the previous step (take a look to the sigma value). The increase in SNR is of [math]21/5.1 = 4.11 \approx \sqrt{12} = 3.46[/math] and you should try to improve this result adjusting [math]\sigma_{low}[/math] and [math]\sigma_{high}[/math]. Now should start the process of the image with crop, background extraction (to remove gradient), and some other processes to enhance your image. To see processes available in Siril please visit this page. Here an example of what you can get with Siril: Peter J. Huber and E. Ronchetti (2009), Robust Statistics, 2nd Ed., Wiley Juan Conejero, ImageIntegration, Pixinsight Tutorial
How to Model Moisture Flow in COMSOL Multiphysics® Computing laminar and turbulent moisture flows in air is both flexible and user friendly with the Moisture Flow multiphysics interfaces and coupling in the COMSOL Multiphysics® software. Available as of version 5.3a, this comprehensive set of functionality can be used to model coupled heat and moisture transport in air and building materials. Let’s learn how the Moisture Flow interface complements existing functionality, while highlighting its benefits. Modeling Heat and Moisture Transport Modeling the transport of heat and moisture through porous materials, or from the surface of a fluid, often involves including the surrounding media in the model in order to get accurate estimates of the conditions at the material surfaces. In the investigations of hygrothermal behavior of building envelopes, food packaging, and other common engineering problems, the surrounding medium is probably moist air (air with water vapor). Moist air is the environing medium for applications such as building envelopes (illustration, left) and solar food drying (right). Right image by ArianeCCM — Own work. Licensed under CC BY-SA 3.0, via Wikimedia Commons. When considering porous media, the moisture transport process, which includes capillary flow, bulk flow, and binary diffusion of water vapor in air, depends on the nature of the material. In moist air, moisture is transported by diffusion and advection, where the advecting flow field in most cases is turbulent. Computing heat and moisture transport in moist air requires the resolution of three sets of equations: The Navier-Stokes equations, to compute the airflow velocity field \mathbf{u} and pressure p The energy equation, to compute the temperature T The moisture transport equation, to compute the relative humidity \phi These equations are coupled together through the pressure, temperature, and relative humidity, which are used to evaluate the properties of air (density \rho(p,T,\phi); viscosity \mu(T,\phi); thermal conductivity k(T,\phi); and heat capacity C_p(T,\phi)); molecular diffusivity D(T) and through the velocity field used for convective transport. With the addition of the Moisture Flow multiphysics interface in version 5.3a, COMSOL Multiphysics defines all three of these equations in a few steps, as shown in the figure below. Using the Moisture Flow Multiphysics Interface Whenever studying the flow of moist air, two questions should be asked: Does the flow depend on moisture distribution? Does the nature of the flow require the use of a turbulence model? If the answer is “yes” for at least one of these questions, then you should consider using the Moisture Flow multiphysics interfaces, found under the Chemical Species Transport branch. The Moisture Flow group under the Chemical Species Transport branch of the Physics Wizard , with the single-physics interfaces and coupling node added with each version of the Moisture Flow predefined multiphysics interface. The Laminar Flow version of the multiphysics interface combines the Moisture Transport in Air interface with the Laminar Flow interface and adds the Moisture Flow coupling. Similarly, each version under Turbulent Flow combines the Moisture Transport in Air interface and the corresponding Turbulent Flow interface and adds the Moisture Flow coupling. Besides providing a user-friendly way to define the coupled set of equations of the moisture flow problem, the multiphysics interfaces for turbulent flow handle the moisture-related turbulence variables required for the fluid flow computation. Automatic Coupling Between Single-Physics Interfaces One advantage of using the Moisture Flow multiphysics interface is its usability. When adding the Moisture Flow node through the predefined interface, an automatic coupling of the Navier-Stokes equations is defined for the fluid flow and the moisture transport equations by the software (center screenshot in the image below) by using the following variables: The density and dynamic viscosity in the Navier-Stokes equations, which depend on the relative humidity variable from the Moisture Transportinterface through a mixture formula based on dry air and pure steam properties (left screenshot below) The velocity field and absolute pressure variables from the Single-Phase Flowinterface, which are used in the moisture transport equation (right screenshot below) Support for Turbulent Fluid Flow The performance of the Moisture Flow multiphysics interface is especially attractive when dealing with a turbulent moisture flow. For turbulent flows, the turbulent mixing caused by the eddy diffusivity in the moisture convection is automatically accounted for by the COMSOL® software by enhancing the moisture diffusivity with a correction term based on the turbulent Schmidt number . The Kays-Crawford model is the default choice for the evaluation of the turbulent Schmidt number, but a user-defined value or expression can also be entered directly in the graphical user interface. Selection of the model for the computation of the turbulent Schmidt number in the user interface of the Moisture Flow coupling. In addition, for coarse meshes that may not be suitable for resolving the thin boundary layer close to walls, Wall functions can be selected or automatically applied by the software. The wall functions are such that the computational domain is assumed to be located at a distance from the wall, the so-called lift-off position, corresponding to the distance from the wall where the logarithmic layer meets the viscous sublayer (or would meet it if there was no buffer layer in between). The moisture flux at the lift-off position, g_{wf}, which accounts for the flux to and from the wall, is automatically defined by the Moisture Flow interface, based on the relative humidity. Approximation of the flow field and the moisture flux close to walls when using wall functions in the turbulence model for fluid flow. Note that the Low-Reynolds and Automatic options for Wall Treatment are also available for some of the RANS models. For more information, read this blog post on choosing a turbulence model. Mass Conservation Across Boundaries By using the Moisture Flow interface, an appropriate mass conservation is granted in the fluid flow problem by the Screen and Interior Fan boundary conditions. A continuity condition is also applied on vapor concentration at the boundaries where the Screen feature is applied. For the Interior Fan condition, the mass flow rate is conserved in an averaged way and the vapor concentration is homogenized at the fan outlet, as shown in the figure below. Average mass flow rate conservation across a boundary with the Interior Fan condition. Example: Modeling Evaporative Cooling with the Moisture Flow Interface Let’s consider evaporative cooling at the water surface of a glass of water placed in a turbulent airflow. The Turbulent Flow, Low Reynolds k-ε interface, the Moisture Transport in Air interface, and the Heat Transfer in Moist Air interface are coupled through the Nonisothermal Flow, Moisture Flow, and Heat and Moisture coupling nodes. These couplings compute the nonisothermal airflow passing over the glass, the evaporation from the water surface with the associated latent heat effect, and the transport of both heat and moisture away from this surface. By using the Automatic option for Wall treatment in the Turbulent Flow, Low Reynolds k-ε interface, wall functions are used if the mesh resolution is not fine enough to fully resolve the velocity boundary layer close to the walls. Convective heat and moisture fluxes at lift-off position are added by the Nonisothermal Flow and Moisture Flow couplings. The temperature and relative humidity solutions after 20 minutes are shown below, along with the streamlines of the airflow velocity field. Temperature (left) and relative humidity (right) solutions with the streamlines of the velocity field after 20 minutes. The temperature and relative humidity fields have a strong resemblance here, which is quite natural since the fields are strongly coupled and since both transport processes have similar boundary conditions, in this case. In addition, heat transfer is given by conduction and advection while mass transfer is described by diffusion and advection. The two transport processes originate from the same physical phenomena: conduction and diffusion from molecular interactions in the gas phase while advection is given by the total motion of the bulk of the fluid. Also, the contribution of the eddy diffusivity to the turbulent thermal conductivity and the turbulent diffusivity originate from the same physical phenomenon, which adds further to the similarity of the temperature and moisture field. Next Steps Learn more about the key features and functionality included with the Heat Transfer Module, and add-on to COMSOL Multiphysics: Read the following blog posts to learn more about heat and moisture transport modeling: How to Model Heat and Moisture Transport in Porous Media with COMSOL® How to Model Heat and Moisture Transport in Air with COMSOL® Get a demonstration of the Nonisothermal Flow and Heat and Moisture couplings in these tutorial models: Comments (2) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.