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 Physics , 2009, DOI: 10.1103/PhysRevLett.104.241803 Abstract: The COMPASS experiment at the CERN SPS has studied the diffractive dissociation of negative pions into the pi- pi- pi+ final state using a 190 GeV/c pion beam hitting a lead target. A partial wave analysis has been performed on a sample of 420000 events taken at values of the squared 4-momentum transfer t' between 0.1 and 1 GeV^2/c^2. The well-known resonances a1(1260), a2(1320), and pi2(1670) are clearly observed. In addition, the data show a significant natural parity exchange production of a resonance with spin-exotic quantum numbers J^PC = 1-+ at 1.66 GeV/c^2 decaying to rho pi. The resonant nature of this wave is evident from the mass-dependent phase differences to the J^PC = 2-+ and 1++ waves. From a mass-dependent fit a resonance mass of 1660 +- 10+0-64 MeV/c^2 and a width of 269+-21+42-64 MeV/c^2 is deduced.
 Physics , 2010, DOI: 10.1142/S0217751X11052347 Abstract: We present results from a Partial-Wave Analysis (PWA) of diffractive dissociation of 190 GeV/c $\pi^-$ into $\pi^-\pi^+\pi^-$ final states on nuclear targets. A PWA of the data sample taken during a COMPASS pilot run in 2004 on a Pb target showed a significant spin-exotic $J^{PC} = 1^{-+}$ resonance consistent with the controversial $\pi_1(1600)$, which is considered to be a candidate for a non-$q\bar{q}$ mesonic state. In 2008 COMPASS collected a large diffractive $\pi^-\pi^+\pi^-$ data sample using a hydrogen target. A first comparison with the 2004 data shows a strong target dependence of the production strength of states with spin projections $M = 0$ and 1.
 Physics , 2011, Abstract: The COMPASS fixed-target experiment at the CERN-SPS studies the structure and spectrum of hadrons. One important goal using hadron beams is the search for new states, in particular spin-exotic mesons and glueballs. As a first input to the puzzle, COMPASS observed a significant $J^{PC}$ spin-exotic signal in the 2004 pilot run data (190\,GeV/$c$ $\pi^{-}$ beam, Pb target) in three charged pion final states consistent with the disputed $\pi_1(1600)$. We started our hadron spectroscopy programme in 2008 by collecting very high statistics using a 190 GeV/$c$ negative pion beam scattered off a liquid hydrogen (proton) target. The current status and new results from the 2008 data on the search for the $\pi_1(1600)$ resonance with exotic $J^{PC}=1^{-+}$ quantum numbers obtained from partial-wave analyses of the $\rho\pi$ and $\eta'\pi$ decay channels are presented.
 Physics , 2010, Abstract: The COMPASS experiment at CERN is dedicated to light hadron spectroscopy with emphasis on the detection of new states, in particular the search for spin exotic states and glueballs. After a short pilot run in 2004 (190 GeV/c $\pi^{-}$ beam, Pb target) showing significant production strength of an exotic $J^{PC}=1^{-+}$ state at 1.66\,GeV/${\rm c^2}$, we have collected data with 190 GeV/c hadron beams on a liquid hydrogen target in 2008/09. The spectrometer features good coverage by calorimetry and our data provide excellent opportunity for simultaneous observation of new states in different decay modes. The diffractively produced $(3\pi)^{-}$ system for example can be studied in both modes $\pi^{-}p \rightarrow \pi^{-}\pi^{+}\pi^{-}p$ and $\pi^{-}~p \rightarrow \pi^{-}\pi^{0}\pi^{0}~p$. Reconstruction of charged and neutral mode rely on completely different parts of the apparatus. Observation of new st ates in both modes provides important checks. The results on diffractive dissociation into 3$\pi$ final states from the 2004 data recently published are discussed as well as the first comparison of neutral vs. charged mode based on a first partial wave analysis of 2008 data.
 Physics , 2011, DOI: 10.1063/1.3700539 Abstract: COMPASS is a multi-purpose fixed-target experiment at the CERN Super Proton Synchrotron aimed at studying the structure and spectrum of hadrons. One primary goal is the search for new hadronic states, in particular spin-exotic mesons and glueballs. We present recent results of partial-wave analyses of (3\pi)^- and \pi^-\eta' final states based on a large data set of diffractive dissociation of a 190 GeV/c \pi^- beam on a proton target in the squared four-momentum-transfer range 0.1 < t' < 1 (GeV/c)^2. We also show first results from a partial-wave analysis of diffractive dissociation of K^- into K^-\pi^+\pi^- final states are presented.
 Frank Nerling for the COMPASS Collaboration Physics , 2010, DOI: 10.1063/1.3483335 Abstract: The COMPASS experiment at CERN is designed for light hadron spectroscopy with emphasis on the detection of new states, in particular the search for exotic states and glue-balls. After a short pilot run in 2004 (190 GeV/c negative pion beam, lead target) showing significant production strength for an exotic $J^{PC}=1^{-+}$ state at 1.66\,GeV/${\rm c^2}$, we have collected data with a 190 GeV/c negative charged hadron beam on a proton (liquid hydrogen) and nuclear targets in 2008 and 2009. The spectrometer features good coverage by electromagnetic calorimetry, and our data provide excellent opportunity for simultaneous observation of new states in two different decay modes in the same experiment. The diffractively produced $(3\pi)^{-}$ system for example can be studied in both modes $\pi^{-}p \rightarrow \pi^{-}\pi^{+}\pi^{-}p$ and $\pi^{-}~p \rightarrow \pi^{-}\pi^{0}\pi^{0}~p$. Charged and neutral mode rely on completely different parts of the spectrometer. Observing a new state in both modes provides important cross-check. First results of a preliminary PWA performed on the 2008 data are presented.
 Physics , 2010, DOI: 10.1088/1742-6596/312/3/032017 Abstract: The COMPASS fixed-target experiment at CERN SPS is dedicated to the study of hadron structure and dynamics. In the physics programme using hadron beams, the focus is on the detection of new states, in particular the search for $J^{PC}$ exotic states and glueballs. After a short pilot run in 2004 (190 GeV/c negative pion beam, lead target), we started our hadron spectroscopy programme in 2008 by collecting an unprecedented statistics with a negative hadron beam (190 GeV/c) on a liquid hydrogen target. A similar amount of data with positive hadron beam (190 GeV/c) has been taken in 2009, as well as some additional data with negative beam on nuclear targets. The spectrometer features a large angular acceptance and high momentum resolution and also good coverage by electromagnetic calorimetry, crucial for the detection of final states involving $\pi^0$ or $\eta$. A first important result is the observation of a significant $J^{PC}$ spin exotic signal consistent with the disputed $\pi_1(1600)$ in the pilot run data. This result was recently published. We present an overview of the status of various ongoing analyses on the 2008/09 data.
 Physics , 2010, DOI: 10.1142/S0217751X11052785 Abstract: The COMPASS experiment at CERN is well designed for light-hadron spectroscopy with emphasis on the detection of new states, in particular the search for $J^{PC}$-exotic states and glueballs. We have collected data with 190 GeV/c charged hadron beams on a liquid hydrogen and nuclear targets in 2008/09. The spectrometer features good coverage by electromagnetic calorimetry and a RICH detector further provides $\pi$ / $K$ separation, allowing for studying final states involving neutral particles like $\pi^0$ or $\eta$ as well as hidden strangeness, respectively. We discuss the status of ongoing analyses with specific focus on diffractively produced $(\pi^0\pi^0\pi)^{-}$ as well as $(K\bar{K}\pi)^{-}$ final states.
 Physics , 2015, Abstract: The COmmon Muon and Proton Apparatus for Structure and Spectroscopy (COMPASS) is a multi-purpose fixed-target experiment at the CERN Super Proton Synchrotron (SPS) aimed at studying the structure and spectrum of hadrons. The two-stage spectrometer has a good acceptance for charged as well as neutral particles over a wide kinematic range and thus allows to access a wide range of reactions. Light mesons are studied with negative (mostly $\pi^-$) and positive ($p$, $\pi^+$) hadron beams with a momentum of 190 GeV/$c$. The spectrum of light mesons is investigated in various final states produced in diffractive dissociation reactions at squared four-momentum transfers to the target between 0.1 and 1.0 $(\text{GeV}/c)^2$. The flagship channel is the $\pi^-\pi^+\pi^-$ final state, for which COMPASS has recorded the currently largest data sample. These data not only allow to measure the properties of known resonances with high precision, but also to search for new states. Among these is a new resonance-like signal, the $a_1(1420)$, with unusual properties. Of particular interest is also the resonance content of the partial wave with spin-exotic $J^{PC} = 1^{-+}$ quantum numbers, which are forbidden for quark-antiquark states.
 Physics , 2012, Abstract: The COMPASS experiment at the CERN SPS investigates the structure and spectrum of hadrons by scattering high energetic hadrons and polarised muons off various fixed targets. During the years 2002-2007, COMPASS focused on nucleon spin physics using 160 GeV/c polarised $\mu^+$ beams on polarised deuteron and proton targets, including measurements of the gluon contribution to the nucleon spin using longitudinal target polarisation as well as studies of transverse spin effects in the nucleon on a transversely polarised target. One major goal of the physics programme using hadron beams is the search for new states, in particular the search for $J^{PC}$ exotic states and glue-balls. COMPASS measures not only charged but also neutral final-state particles, allowing for investigation of new objects in different reactions and decay channels. In addition COMPASS can measure low-energy QCD constants like, e.g. the electromagnetic polarisability of the pion. Apart from a few days pilot run data taken in 2004 with a 190 GeV/c $\pi^{-}$ beam on a Pb target, showing a significant spin-exotic $J^{PC}$ = $1^{-+}$ resonance at around 1660 MeV/$c^{2}$, COMPASS collected high statistics with negative and positive 190 GeV/$c$ hadron beams on a proton (H$_2$) and nuclear (Ni, Pb) targets in 2008 and 2009. We give a selected overview of the newest results and discuss the status of various ongoing analyses.
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text: For this purpose, we introduce a natural topology on Milnor's K-groups [K.sup.M.sub.l](k) for a topological field k as the quotient topology induced by the joint determinant map and show that, in case of k = R or C, the natural topology on [K.sup.M.sub.l](k) is disjoint union of two indiscrete components or indiscrete topology, respectively. Some "extremal" examples Take any set X and let = {, X}. 7. Intuitively, this has the consequence that all points of the space are "lumped together" and cannot be distinguished by topological means. R := R R (cartesian product). [Justify your claims.] Is the subspace topology of a subset S Xnecessarily the indiscrete topology on S? Let Xbe a topological space with the indiscrete topology. R under addition, and R or C under multiplication are topological groups. The indiscrete topology on Xis de ned by taking Ëto be the collection consisting of only the empty set and X. The same argument shows that the lower limit topology is not ner than K-topology. SOME BASIC NOTIONS IN TOPOLOGY It is easy to see that the discrete and indiscrete topologies satisfy the re-quirements of a topology. If X is finite and has n elements then power set of X has _____ elements. In the discrete topology, one point sets are open. Then, clearly A\B= ;, but A\B= R 0 \R 0 = f0g. To learn more, see our tips on writing great answers. Let Ï be the collection all open sets on X. The largest topology contains all subsets as open sets, and is called the discrete topology. If Adoes not contain 7, then the subspace topology on Ais discrete. [Justify your claims.] TOPOLOGY TAKE-HOME CLAY SHONKWILER 1. Is there a difference between a tie-breaker and a regular vote? When k = R and l [greater than or equal to] 2, G either is an indiscrete space or has an indiscrete subgroup of index 2. This agrees with the usual notation for Rn. Removing just one element of the cover breaks the cover. $(0,1)$ is compact in discrete topology on $\mathbb R$. C The lower-limit topology (recall R with this the topology is denoted Râ). Why set of integer under indiscrete topology is compact? In parliamentary democracy, how do Ministers compensate for their potential lack of relevant experience to run their own ministry? Conclude that if T ind is the indiscrete topology on X with corresponding space Xind, the identity function 1 X: X 1!Xind is continuous for any topology T 1. 7. Let Bbe the collection of cartesian product of open intervals, (a;b) (c;d). (viii)Every Hausdorspace is metrizable. Let X be any set and let be the set of all subsets of X. contains) the other. The sets in the topology T for a set S are defined as open. valid topology, called the indiscrete topology. valid topology, called the indiscrete topology. Ø®ÓkqÂ\O¦K0¤¹@B Let V ï¬ zPU B 1 7 pzq. Then is a topology called the trivial topology or indiscrete topology. (b) Any function f : X â Y is continuous. 3. If we use the discrete topology, then every set is open, so every set is closed. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. In the indiscrete topology all points are limit points of any subset X of S which inclues points other than because the only open set containing a point p is the whole S which necessarily contains points of â¦ How/where can I find replacements for these 'wheel bearing caps'? In the discrete topology - the maximal topology that is in some sense the opposite of the indiscrete/trivial topology - one-point sets are closed, as well as open ("clopen"). because it closed and bounded. (Lower limit topology of R) Consider the collection Bof subsets in R: B:= Page 1. C The lower-limit topology (recall R with this the topology is denoted Râ). Proof. When $$\mathcal{T} = \{\emptyset, X\}$$, it is called the indiscrete topology on X. 5. , the indiscrete topology or the trivial topology on any set X. Use MathJax to format equations. Then Z = {Î±} is compact (by (3.2a)) but it is not closed. Then, clearly A\B= ;, but A\B= R 0 \R 0 = f0g. c.Let X= R, with the standard topology, A= R <0 and B= R >0. TOPOLOGY TAKE-HOME CLAY SHONKWILER 1. As for the indiscrete topology, every set is compact because there is only one possible open cover, namely the space itself. Proposition. Intersection of Topologies. The indiscrete topology is manifestly not Hausdorâµunless X is a singleton. Sierpinski Space 44 12. Is there any source that describes Wall Street quotation conventions for fixed income securities (e.g. There are also infinite number of indiscrete spaces. Confusion about definition of category using directed graph. Since that cover is finite already, every set is compact. (b) Suppose that Xis a topological space with the indiscrete topology. My professor skipped me on christmas bonus payment, How to gzip 100 GB files faster with high compression. Is it true that an estimator will always asymptotically be consistent if it is biased in finite samples? A The usual (i.e. Indiscrete topology is finer than any other topology defined on the same non empty set. The Discrete Topology Let Y = {0,1} have the discrete topology. corporate bonds)? 2. Why? Are they homeomorphic? Some "extremal" examples Take any set X and let = {, X}. A Topology on Milnor's Group of a Topological Field â¦ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Example: The indiscrete topology on X is Ï I = {â, X}. Let $X=\mathbb R$ with cofinite topology and $A=[0,1]$ with subspace topology - show $A$ is compact. Then Z is closed. and x 1. with the indiscrete topology. 38. Let X be the set of points in the plane shown in Fig. Ú Thanks for contributing an answer to Mathematics Stack Exchange! Here are four topologies on the set R. For each pair of topologies, determine whether one is a reï¬nement of (i.e. but the same set is not compact in indiscrete topology on $\mathbb R$ because it is not closed (because in indiscrete topolgy on $\mathbb R$ the closed sets is only $\phi$ and $\mathbb R$). Indiscrete Topology The collection of the non empty set and the set X itself is always a topology on X,â¦ Click here to read more. Proof. Partition Topology 43 6. standard) topology. However: (3.2d) Suppose X is a Hausdorï¬ topological space and that Z â X is a compact sub-space. (Lower limit topology of R) Consider the collection Bof subsets in R: B:= The standard topology on R induces the discrete topology on Z. Consider the set X=R with T x = the standard topology, let f be a function from X to the set Y=R, where f(x)=5, then the topology on Y induced by f and T x is. For this purpose, we introduce a natural topology on Milnor's K-groups [K.sup.M.sub.l](k) for a topological field k as the quotient topology induced by the joint determinant map and show that, in case of k = R or C, the natural topology on [K.sup.M.sub.l](k) is disjoint union of two indiscrete components or indiscrete topology, respectively. The properties verified earlier show that is a topology. Let Bbe the collection of cartesian product of open intervals, (a;b) (c;d). Ë is a topology on . So you can take the cover by those sets. A The usual (i.e. with the indiscrete topology. c.Let X= R, with the standard topology, A= R <0 and B= R >0. Such spaces are commonly called indiscrete, anti-discrete, or codiscrete. Then Xis compact. If Adoes not contain 7, then the subspace topology on Ais discrete. These sets all have in nite complement. space. The dictionary order topology on the set R R is the same as the product topology R d R, where R d denotes R in the discrete topology. Closed Extension Topology 44 13. By clicking âPost Your Answerâ, you agree to our terms of service, privacy policy and cookie policy. Proof We will show that C (Z). Page 1. If Xhas at least two points x 1 6= x 2, there can be no metric on Xthat gives rise to this topology. The standard topology on R n is Hausdorâµ: for x 6= y 2 R n ,letd be half the Euclidean distance â¦ X with the indiscrete topology is called an. The collection of the non empty set and the set X itself is always a topology on X, and is called the indiscrete topology on X. 2) ËË , ( Ë Ë power set of is a topology on and is called discrete topology on and the T-space Ë is called discrete topological space. Let X be any set and let be the set of all subsets of X. indiscrete). Where can I travel to receive a COVID vaccine as a tourist? Here, every sequence (yes, every sequence) converges to every point in the space. Example 2.4. The indiscrete topology on Xis de ned by taking Ëto be the collection consisting of only the empty set and X. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Finite Excluded Point Topology 47 14. is $(0,1)$ compact in indiscrete topology and discrete topolgy on $\mathbb R$, Intuition for the Discrete$\dashv$Forgetful$\dashv$Indiscrete Adjunction in $\mathsf{Top}$, $\mathbb{Q}$ with topology from $\mathbb{R}$ is not locally compact, but all discrete spaces are, Intuition behind a Discrete and In-discrete Topology and Topologies in between, Fixed points Property in discrete and indiscrete space. 10/3/20 5: 03. Making statements based on opinion; back them up with references or personal experience. Expert Answer 100% (1 rating) Previous question Next question Get more help from Chegg. 2) ËË , ( Ë Ë power set of is a topology on and is called discrete topology on and the T-space Ë is called discrete topological space. In the discrete topology - the maximal topology that is in some sense the opposite of the indiscrete/trivial topology - one-point sets are closed, as well as open ("clopen"). If Xhas at least two points x 1 6= x 2, there can be no metric on Xthat gives rise to this topology. Expert Answer . Compact being the same as closed and bounded only works when $\mathbb{R}$ has the standard topology. So the equality fails. Let S Xand let T S be the subspace topology on S. Prove that if Sis an open subset of X, and if U2T S, then U2T. Let X be the set of points in the plane shown in Fig. That's because the topology is defined by every one-point set being open, and every one-point set is the complement of the union of all the other points. Previous question Next question Transcribed Image Text from this Question. If one considers on R the indiscrete topology in which the only open sets are the empty set and R itself, then int([0;1]) is the empty set. There are also infinite number of indiscrete spaces. 1.1.4 Proposition (a) Let Xbe a set with the co nite topology. 10/3/20 5: 03. It also converges to 7, e, 1;000;000, and every other real number. 1is just the indiscrete topology.) 3.Let (R;T 7) be the reals with the particular point topology at 7. The Discrete Topology Let Y = {0,1} have the discrete topology. ±.&£ïBvÙÚg¦m ûèÕùÜËò¤®Õþ±d«*üë6þ7Í£$DL»ÏÊêqbNÀ÷y°¡Èë$^'ÒBË¢KÊãRN$¤à½ôZð#{øEWùz]b2Áý@jíÍdº£à1v¾Ä$ÇáæáwÆ Then is a topology called the trivial topology or indiscrete topology. (In addition to X and we â¦ Proof. The same argument shows that the lower limit topology is not ner than K-topology. In (R;T indiscrete), the sequence 7;7;7;7;7;::: converges to Ë. Let R 2be the set of all ordered pairs of real numbers, i.e. 6. , the finite complement topology on any set X. Proposition 18. Similarly, if Xdisc is the set X equipped with the discrete topology, then the identity map 1 X: Xdisc!X 1 is continuous. 'øÈ÷¡àItþ:N#_ÉÂ#1NÄ]¤¸¬ F8µ$üù â¥n*q/öúyæMR«î«öjR(@Ï:,½PýT©mªUlºÆ¢Ã}Ø1Öé13&ô9ÐÁeQnÉ@ñß] 6J l¤ôÔ~¸KÚ¢ "çQ"ÔÈq#/C°Y0. This topology is called indiscrete topology on and the T-space Ë is called indiscrete topological space. In this, we use a set of axioms to prove propositions and theorems. X with the indiscrete topology is called an. 3.Let (R;T 7) be the reals with the particular point topology at 7. So the equality fails. space. 2.13.6. Let f : X !Y be the identity map on R. Then f is continuous and X has the discrete topology, but f(X) = R does not. but the same set is not compact in indiscrete topology on R because it is not closed (because in indiscrete topolgy on R the closed sets is only Ï and R). Don't one-time recovery codes for 2FA introduce a backdoor? 1.A product of discrete spaces is discrete, and a product of indiscrete spaces is indiscrete. If we use the indiscrete topology, then only â ,Rare open, so only â ,Rare closed and this implies that A â¦ Similarly, if Xdisc is the set X equipped with the discrete topology, then the identity map 1 X: Xdisc!X 1 is continuous. Then $$\tau$$ is called the indiscrete topology and $$(X, \tau)$$ is said to be an indiscrete space. Note that there is no neighbourhood of 0 in the usual topology which is contained in ( 1;1) nK2B 1:This shows that the usual topology is not ner than K-topology. The dictionary order topology on the set R R is the same as the product topology R d R, where R d denotes R in the discrete topology. Proposition. On the other hand, the union S x6=x 0 fxgequals Xf x 0g, which has complement fx 0g, so it is not open. In fact no infinite set in the discrete topology is compact. $$2^n$$ $$2^{n-1}$$ $$2^{n+1}$$ None of the given; The set of _____ of R (Real line) forms a topology called usual topology. 21 November 2019 Math 490: Worksheet #16 Jenny Wilson In-class Exercises 1. It su ces to show for all U PPpZq, there exists an open set V â¢R such that U Z XV, since the induced topology must be coarser than PpZq. If we thought for a moment we had such a metric d, we can take r= d(x 1;x 2)=2 and get an open ball B(x 1;r) in Xthat contains x 1 but not x 2. We are only allowing the bare minimum of sets, X and , to be open. ) (a) X has the discrete topology. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Zvi Rosen Applied Algebraic Topology Notes Vladimir Itskov 3.1. Review. (R Sorgenfrey)2 is an interesting space. Here are four topologies on the set R. For each pair of topologies, determine whether one is a reï¬nement of (i.e. This topology is called indiscrete topology on and the T-space Ë is called indiscrete topological space. 2. indiscrete topology 3. the subspace topology induced by (R, Euclidean) 4. the subspace topology induced by (R, Sorgenfrey) 5. the finite-closed topology 6. the order topology. 2 CHAPTER 1. This implies that A = A. Proof. Then Xis not compact. If A R contains 7, then the subspace topology on Ais also the particular point topology on A. This is the space generated by the basis of rectangles Chapter 2 Topology 2.1 Introduction Several areas of research in modern mathematics have developed as a result of interaction between two or more specialized areas. Thus openness is not a property determinable from the set itself; openness is a property of a set with respect to a topology. If we thought for a moment we had such a metric d, we can take r= d(x 1;x 2)=2 and get an open ball B(x 1;r) in Xthat contains x 1 but not x 2. X with the indiscrete topology is called an indiscrete topological space or simply an indiscrete space. Proposition 17. Note that there is no neighbourhood of 0 in the usual topology which is contained in ( 1;1) nK2B 1:This shows that the usual topology is not ner than K-topology. Let X = R with the discrete topology and Y = R with the indiscrete topol- ogy. 4. In the discrete topology any subset of S is open. 38. This question hasn't been answered yet Ask an expert. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Uncountable Particular Point Topology 44 11. (a) Let (X;T) be a topological space. rev 2020.12.10.38158, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Terminology: gis the genus of the surface = maximal number of â¦ Subscribe to this blog. is$(0,1)$compact in indiscrete topology and discrete topolgy on$\mathbb R$? $$2^n$$ $$2^{n-1}$$ $$2^{n+1}$$ None of the given; The set of _____ of R (Real line) forms a topology called usual topology. (R usual)2 = R2 usual. Usual Topology on $${\mathbb{R}^3}$$ Consider the Cartesian plane $${\mathbb{R}^3}$$, then the collection of subsets of $${\mathbb{R}^3}$$ which can be expressed as a union of open spheres or open cubed with edges parallel to coordinate axis from a topology, and is called a usual topology on $${\mathbb{R}^3}$$. !Nñ§UD AêÅ^SOÖÉ O»£ÔêeÎ/1TÏUèÍ5?.§Úx;©&Éaus^M(qê³S:S}ñ:]K¢é;í¶P¤1H8iTPÞ´×:bäàÖTÀçD3u^"(ÕêXIV´DØ ?§ÂQ4X¦Taðå«%x¸!iT 4K. 6. Example 2. (c) Any function g : X â Z, where Z is some topological space, is continuous. The indiscrete topology on X is the weakest topology, so it has the most compact sets. How to holster the weapon in Cyberpunk 2077? This is the next part in our ongoing story of the indiscrete topology being awful. Is it just me or when driving down the pits, the pit wall will always be on the left? Then $$\tau$$ is called the indiscrete topology and $$(X, \tau)$$ is said to be an indiscrete space. (This is the opposite extreme from the discrete topology. Then Z = {Î±} is compact (by (3.2a)) but it is not closed. For example, t Example 1.5. Indiscrete Topology 42 5. When should 'a' and 'an' be written in a list containing both? In particular, every point in X is an open set in the discrete topology. The indiscrete topology for S is the collection consisting of only the whole set S and the null set â . How do I convert Arduino to an ATmega328P-based project? contains) the other. Every sequence converges in (X, Ï I) to every point of X. Let Xbe a topological space with the indiscrete topology. In other words, for any non empty set X, the collection $$\tau = \left\{ {\phi ,X} \right\}$$ is an indiscrete topology on X, and the space $$\left( {X,\tau } \right)$$ is called the indiscrete topological space or simply an indiscrete space. It is called the indiscrete topology or trivial topology. The is a topology called the discrete topology. 1. The intersection of any two topologies on a non empty set is always topology on that set, while the unionâ¦ Click here to read more. Proof. TSLint extension throwing errors in my Angular application running in Visual Studio Code. Asking for help, clarification, or responding to other answers. [note: So you have 4 2 = 6 comparisons to make.] of X X X, and so on. Let X be the set of points in the plane shown in Fig. We sometimes write cl(A) for A. The is a topology called the discrete topology. (c) Any function g : X â Z, where Z is some topological space, is continuous. R := R R (cartesian product). (iii) The cofinite topology is strictly stronger than the indiscrete topology (unless card(X) < 2), but the cofinite topology also makes every subset of â¦ 1.1.4 Proposition (Limits of sequences are not unique.) Then Xis compact. 2 CHAPTER 1. [note: So you have 4 2 = 6 comparisons to make.] Let Xbe an in nite topological space with the discrete topology. 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And discrete topolgy on $\mathbb R$ containing both an ATmega328P-based project Framed... Than K-topology indiscrete topology on r fact no infinite set in the topology generated by the basis of a on. And a product of indiscrete topology on r balls, and so on Stack Exchange a... The smallest topology has two open sets, the pit wall will always be. Four topologies on the left yet Ask an expert ATmega328P-based project point in the discrete on... R } $has the standard topology, every sequence ( yes, every sequence ( yes, subset. Emptyset and X the Next part in our ongoing story of the indiscrete topology for S is since... Set of all ordered pairs of real numbers, i.e here are some simple examples discrete and indiscrete satisfy... } have the discrete topology, like other branches of pure mathematics, is an interesting..: Worksheet # 16 Jenny Wilson In-class Exercises 1 in fact no set. Back them up with references or personal experience of R with the indiscrete topology is than. 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text:
Lemma 10.131.7. Let $A \to B \to C$ be ring maps. Then there is a canonical exact sequence

$C \otimes _ B \Omega _{B/A} \to \Omega _{C/A} \to \Omega _{C/B} \to 0$

of $C$-modules.

Proof. We get a diagram (10.131.4.1) by putting $R = A$, $S = C$, $R' = B$, and $S' = C$. By Lemma 10.131.6 the map $\Omega _{C/A} \to \Omega _{C/B}$ is surjective, and the kernel is generated by the elements $\text{d}(c)$, where $c \in C$ is in the image of $B \to C$. The lemma follows. $\square$

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## Newcomb’s Paradox and Bell Theorem

Today I watched a lovely presentation of Newcomb’s paradox by Norm Wildberger http://www.youtube.com/watch?v=aR5GYeZkgvY

and I could not help but notice the deep relationship with Bell theorem. In particular Bell result shows clearly the solution for Newcomb’s paradox and I am interrupting the series to present the solution. Since I did not follow Newcomb’s paradox developments, I don’t know if my take on it is new or not.

So what is Newcomb’s paradox http://en.wikipedia.org/wiki/Newcomb%27s_paradox ? This paradox was popularized by Martin Gardner and is described in his book: “The Colossal Book of Mathematics” on page 580. Let me quote its description from this book to achieve maximum clarity:

Two closed boxes, B1 and B2, are on a table. B1 contains $1,000. B2 contains either nothing or$1 million. You do not know which. You have an irrevocable choice between to actions:

1. Take what is in both boxes.
2. Take only what is in B2.

At some time before the test a superior Being has made a prediction about what you will decide. It is not necessary to assume determinism. You only need be persuaded that the Being’s predictions are “almost certainly” correct. If you like, you can think of the Being as God, but the paradox is just as strong if you regard the Being as a superior intelligence from another planet or a supercomputer capable of probing your brain and making highly accurate predictions about your decisions. If the Being expects you to choose both boxes, he has left B2 empty. If he expects you to take only B2, he has put $1 million in it. (If he expects you to randomize your choice by say, flipping a coin, he has left B2 empty.) In all cases B1 contains$1,000. You understand the situation fully, the Being knows you understand, you know that he knows, and so on.

What should you do? Clearly it is not to your advantage to flip a coin, so that you must decide on your own. The paradox lies in the disturbing fact that a strong argument can be made for either decision. Both arguments cannot be right. The problem is to explain why one is wrong.

Let us look first at the argument for taking only B2. You believe the Being is an excellent predictor. If you take both boxes, the Being almost certainly will have anticipated your action and have left B2 empty. You will get only the $1,000 in B1. On the other hand, if you take only B2, the Being, expecting that, almost certainly will have placed$1 million in it. Clearly it is to your advantage to take only B2.

Convincing? Yes, but the Being made his prediction, say a week ago, and then left. Either he put the $1 million in B2, or he did not. If the money is already there, it will stay there whatever you choose. It is not going to disappear. If it is not already there, it is not going to suddenly appear if you choose only what is in the second box. It is assumed that no “backwards causality” is operating; that is, your present actions cannot influence what the Being did last week. So why not take both boxes and get everything that is there? If B2 is filled, you get$1,001,000. If it is empty, you get at least $1,000. If you are so foolish as to take only B2, you know you cannot get more than$1 million, and there is even a slight possibility of getting nothing. Clearly it is to your advantage to take both boxes!

So here is the solution.  Let us call the first solution (pick only B2): trust in correlations, and the second solution (pick both B1 and B2): trust in local realism.

If you trust correlations, you obviously should pick only B2 and the video above has a nice mathematical formulation of why and when you should choose this. If you trust local realism clearly you should pick both boxes. Now here is where Bell’s Theorem comes into play. As originally written, the paradox is based on a mathematical contradiction: local realism imposes a limit on the maximum possible correlation and Bell inequalities captures precisely this limit. As stated, Newcomb’s Paradox violates Bell’s theorem.

If the correlations are higher than Bell inequality limit, this cannot be achieved with classical objects like money. Therefore the physically correct solution is to pick both boxes.

On the other hand we can introduce a variation to the problem by populate the two boxes with quantum objects instead of money. Suppose we have may identically prepared copies of the quantum objects and the “Being” perform a weak ago half measurement in an EPR-B type experiment. Then the choice for the player is to either perform the other half (and get \$1 million if the correlation is above Bell limit) or to perform a random intermediate measurement (to be discarded) and then complete the other half measurement and this reduces the correlation into Bell’s limit. What happens in this case is the violation of this: “If the money is already there, it will stay there whatever you choose. It is not going to disappear. If it is not already there, it is not going to suddenly appear if you choose only what is in the second box.” because unperformed measurements have no pre-defined values in quantum mechanics.

Therefore both solutions are valid in their range of validity and the hidden contradiction in the original paradox was the incompatibility between local realism and correlations which Bell theorem explains. The original setup of the paradox is a physical and a mathematical impossibility as shown by Bell theorem.

### UPDATE

I did some reading to see other people's take on this paradox, and I found a paper by Eric Cavalcanti http://philsci-archive.pitt.edu/4872/1/Newcomb_preprint.pdf who had the same idea as mine, but presented it in a physics context with emphasis on the Bayesian approach.

After the paradox was introduced, for the two solutions of the paradox, two schools of thought developed: Causal Decision Theory and Evidential Decision Theory (see http://fitelson.org/few/few_05/egan.pdf)

The way to think about the paradox is to rename "Being" as "a potential Con Artist" and then assert the degree of trust into his ability to predict your mind. If the laws of physics prevents him to achieve the correlation, then pick both boxes, if not, go with the solution dictated by the Evidential Decision Theory and pick one box.

Bell theorem sets a limit on the maximum achievable correlations and Cavalcanti showed explicitly that the standard Bell analysis for local hidden variable theories (LHV) corresponding to local realism is same as the analysis advocated by the Causal Decision Theory. This being the case, Bell inequalities set up the maximum achievable correlations for local realism.

The domain of validity of the Evidential Decision Theory is limited by the Bell limit for classical resources, or by the Tsirelson limit for quantum resources.

In the case of this paradox, Bell limit demands p_a = p_b = 50% where p_a and p_b are the probability of "a potential Con Artist" to correctly predict your choices (see the above video for definition)

By Bell theorem, p_a,b > 50% OR p_a,b < 50% demands that "a potential Con Artist" is definitely a "Con Artist" and if you follow the course of action suggested by Evidential Decision Theory I have a bridge to sell you.


text:
cos(pi/4)= 1/√2 (With Explanation)

-

The value of cos(π/4) is equal to cos(45) = 1/√2 = √2/2. In decimal form, this value is cos 45 degrees is approximately equal to 0.707.

Why is cos(pi/4)= 1/√2 ?

Recall that π (“pi”) radians are equal to 180 degrees. Therefore, pi/4 radians are equal to 180/4 = 45 degrees. We want to find the values of cosine 45 degrees. For this, we can use the definition of cos. The ratio cos is defined as the ratio of the adjacent side to the hypotenuse in a right-angled triangle. Consider the right-angled triangle ABC as shown below:

In the above image, the triangle is right-angled at vertex B. Both of the angles A and C are equal to 45 degrees each. This means that the above triangle is an isosceles triangle meaning that the two sides AB and BC are equal. Let us denote the two sides as x. We now apply the Pythagoras theorem which tells us that, AC^2 = AB^2 + BC^2 AC^2 = x^2 + x^2 AC^2 = 2x^2 \implies AC=x\sqrt{2}

We now use the definition of the cosine of an angle. Notice that the side BC is adjacent to angle C and that AC is the hypotenuse of our triangle. We have that, \begin{align*}cos(C) &= cos(45) \\ &= \frac{Adjacent}{Hypotenuse}\\ &= \frac{BC}{AC} \\ &= \frac{x}{x\sqrt{2}} \\ &= \frac{1}{\sqrt{2}} \\ &= \frac{\sqrt{2}}{2}\end{align*}

Summary
Article Name
cos(pi/4)= 1/√2 (With Explanation)
Description
The value of cos(π/4) is equal to cos(45) = 1/√2 = √2/2. In decimal form, this value is cos 45 degrees is approximately equal to 0.707.
Publisher Name
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10.6 Lattice structures in crystalline solids  (Page 8/29)

 Page 8 / 29

If we know the edge length of a unit cell of an ionic compound and the position of the ions in the cell, we can calculate ionic radii for the ions in the compound if we make assumptions about individual ionic shapes and contacts.

The edge length of the unit cell of LiCl (NaCl-like structure, FCC) is 0.514 nm or 5.14 Å. Assuming that the lithium ion is small enough so that the chloride ions are in contact, as in [link] , calculate the ionic radius for the chloride ion.

Note: The length unit angstrom, Å, is often used to represent atomic-scale dimensions and is equivalent to 10 −10 m.

Solution

On the face of a LiCl unit cell, chloride ions contact each other across the diagonal of the face:

Drawing a right triangle on the face of the unit cell, we see that the length of the diagonal is equal to four chloride radii (one radius from each corner chloride and one diameter—which equals two radii—from the chloride ion in the center of the face), so d = 4 r . From the Pythagorean theorem, we have:

${a}^{2}+{a}^{2}={d}^{2}$

which yields:

${\left(0.514\phantom{\rule{0.2em}{0ex}}\text{nm}\right)}^{2}+{\left(0.514\phantom{\rule{0.2em}{0ex}}\text{nm}\right)}^{2}={\left(4r\right)}^{2}=16{r}^{2}$

Solving this gives:

$r=\phantom{\rule{0.2em}{0ex}}\sqrt{\frac{{\left(0.514\phantom{\rule{0.2em}{0ex}}\text{nm}\right)}^{2}+{\left(0.514\phantom{\rule{0.2em}{0ex}}\text{nm}\right)}^{2}}{16}}\phantom{\rule{0.2em}{0ex}}=\text{0.182 nm}\phantom{\rule{0.2em}{0ex}}\left(\text{1.82 Å}\right)\phantom{\rule{0.2em}{0ex}}{\text{for a Cl}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}\text{radius}.$

The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. The radius of the chloride ion is 1.82 Å.

The radius of the potassium ion is 1.33 Å.

It is important to realize that values for ionic radii calculated from the edge lengths of unit cells depend on numerous assumptions, such as a perfect spherical shape for ions, which are approximations at best. Hence, such calculated values are themselves approximate and comparisons cannot be pushed too far. Nevertheless, this method has proved useful for calculating ionic radii from experimental measurements such as X-ray crystallographic determinations.

X-ray crystallography

The size of the unit cell and the arrangement of atoms in a crystal may be determined from measurements of the diffraction of X-rays by the crystal, termed X-ray crystallography    . Diffraction is the change in the direction of travel experienced by an electromagnetic wave when it encounters a physical barrier whose dimensions are comparable to those of the wavelength of the light. X-rays are electromagnetic radiation with wavelengths about as long as the distance between neighboring atoms in crystals (on the order of a few Å).

When a beam of monochromatic X-rays strikes a crystal, its rays are scattered in all directions by the atoms within the crystal. When scattered waves traveling in the same direction encounter one another, they undergo interference , a process by which the waves combine to yield either an increase or a decrease in amplitude (intensity) depending upon the extent to which the combining waves’ maxima are separated (see [link] ).

to what volume must 8.32 NaOH be diluted to its analytical concentration 0.20 M
weight in mg 1.76 mole of I
Sheriza
the types of hydrocarbons
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aliphatic and aromatic hydrocarbons
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reeza
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reeza
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5, -2 & -2
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is a smallest particle of a chemical element that can exist
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An atom is the smallest part of an element dat can take part in chemical reaction.
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text:
## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) The force exerted on the flat end of each cylinder is $2.9\times 10^4~N$ (b) 30 football players would need to hang from the lower cylinder to pull the two cylinders apart.
(a) The pressure difference between the inside of the cylinders and the outside of the cylinders is $1~atm$. We can find the force exerted on the flat end of each cylinder. $F = \Delta P~A$ $F = \Delta P~\pi~r^2$ $F = (1.013\times 10^5~N/m^2)(\pi)(0.30~m)^2$ $F = 2.9\times 10^4~N$ The force exerted on the flat end of each cylinder is $2.9\times 10^4~N$ (b) We find the weight of one football player. $weight = (100~kg)(9.80~m/s^2) = 980~N$ We then find the number $n$ of football players. $n\times ~weight = 2.9\times 10^4~N$ $n = \frac{2.9\times 10^4~N}{weight}$ $n = \frac{2.9\times 10^4~N}{980~N}$ $n = 30~players$ 30 football players would need to hang from the lower cylinder to pull the two cylinders apart.


text:
# 关于/usr/include/下各种unistd的区别

unistd是unix standard的简称，主要封装了一些glibc函数和系统调用相关的定义（个人理解）。具体参考这里：https://zh.wikipedia.org/wiki/Unistd.h

/usr/include/unistd.h  1

/usr/include/linux/unistd.h 2

/usr/include/asm/unistd.h  3

/usr/include/asm/unistd_32.h  4

/usr/include/asm/unistd_64.h 5

/usr/include/bits/unistd.h  6

1中include了6，如果编译时开了FORTIFY，那么会用6中的函数，对危险函数做一个wrapper。1 6 这些是和glibc有关的。

2中仅仅是include了3。3中根据OS的bits选择是4还是5。2 3 4 5 这些都是从kernel来的。

stackoverflow的介绍言简意赅：

linux/unistd.h actually points to asm/unistd.h, which in turn points to either asm/unistd_32.h or asm/unistd_64.h, which is where system call numbers are defined and presented to user space depending on the system's architecture. These come from the kernel.

bits/unistd.h is a collection of macros that augment unistd.h (mostly stuff to help prevent buffer overflows), which is conditionally included via:

/* Define some macros helping to catch buffer overflows.  */
#if __USE_FORTIFY_LEVEL > 0 && defined __extern_always_inline
# include <bits/unistd.h>
#endif
In essence, the only POSIX required header is in fact, just unistd.h, the rest are either extensions, or definitions from the kernel. So, just including unistd.h is all you have to worry about doing, everything you need will be pulled in depending on your architecture and whatever build options you've selected.

https://stackoverflow.com/questions/2948696/what-is-the-difference-between-the-various-unistd-h-under-usr-include-in-linux/2949262#2949262


text:
# Poisson Distribution

• Sep 14th 2008, 10:37 PM
lllll
Poisson Distribution
Let $X$ be a Poisson distribution. Show that $P\{X=k\}$ increases as it approaches to $\lambda$, reaching it's max when k is the largest integer less then $\lambda$ then decrease as $k > \lambda$

Hint: Consider $\frac{P\{X=k\}}{P\{X=k-1\}}$.

since the Poisson distribution is defined as:

$\frac{\lambda^k}{k!}e^{-k}$ I was thinking of using derivatives to max out the equation with respect to k, but realized that due to the factorial I can't since it's not continuous.

I'm not sure I'm doing this right, but if I apply the hint I would get:

$\left( \frac{\lambda^k}{k!}e^{-k} \right) \left( \frac{k!}{\lambda^k}e^{-k} \right) = \left( \frac{(k-1)!}{k!} \right) \left( \frac{\lambda^k}{\lambda^{k-1}} \right) = \frac{\lambda}{k}$

now I'm not sure what I have to do
• Sep 14th 2008, 11:17 PM
CaptainBlack
Quote:

Originally Posted by lllll
Let $X$ be a Poisson distribution. Show that $P\{X=k\}$ increases as it approaches to $\lambda$, reaching it's max when k is the largest integer less then $\lambda$ then decrease as $k > \lambda$

Hint: Consider $\frac{P\{X=k\}}{P\{X=k-1\}}$.

since the Poisson distribution is defined as:

$\frac{\lambda^k}{k!}e^{-k}$ I was thinking of using derivatives to max out the equation with respect to k, but realized that due to the factorial I can't since it's not continuous.

I'm not sure I'm doing this right, but if I apply the hint I would get:

$\left( \frac{\lambda^k}{k!}e^{-k} \right) \left( \frac{k!}{\lambda^k}e^{-k} \right) = \left( \frac{(k-1)!}{k!} \right) \left( \frac{\lambda^k}{\lambda^{k-1}} \right) = \frac{\lambda}{k}$

now I'm not sure what I have to do

$\frac{P\{X=k\}}{P\{X=k-1\}}=\left( \frac{(k-1)!}{k!} \right) \left( \frac{\lambda^k}{\lambda^{k-1}} \right) = \frac{\lambda}{k}$.

Therefore:

$\frac{P\{X=k\}}{P\{X=k-1\}}\ge 1$ when $k\le \lambda$

and:

$\frac{P\{X=k\}}{P\{X=k-1\}} < 1$ when $k > \lambda$

RonL


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# Consumer and Producer Surplus

 Site: Saylor Academy Course: ECON101: Principles of Microeconomics Book: Consumer and Producer Surplus
 Printed by: Guest user Date: Monday, December 6, 2021, 2:55 AM

## 1. LEARNING OBJECTIVES

• Explain, calculate, and illustrate consumer surplus
• Explain, calculate, and illustrate producer surplus
• Explain, calculate, and illustrate social surplus

Source: Lumen Learning, https://courses.lumenlearning.com/wmopen-microeconomics/chapter/consumer-producer-surplus/

## 2. Demand, Supply and Efficiency

The familiar demand and supply diagram holds within it the concept of allocative efficiency. One typical way that economists define efficiency is when it is impossible to improve the situation of one party without imposing a cost on another. Conversely, if a situation is inefficient, it becomes possible to benefit at least one party without imposing costs on others.

Efficiency in the demand and supply model has the same basic meaning: the economy is getting as much benefit as possible from its scarce resources and all the possible gains from trade have been achieved. In other words, the optimal amount of each good and service is being produced and consumed.

## 3. Consumer Surplus, Producer Surplus, Social Surplus

Consider a market for tablet computers, as shown in Figure 1. We usually think of demand curves as showing what quantity of some product consumers will buy at any price, but a demand curve can also be read the other way. If we choose a quantity of output, the demand curve shows the maximum price consumers would be willing to pay for that quantity. According to the demand curve in Figure 1, if producers wanted to sell a quantity of 20 million tablets, some customers are willing to pay $90 each (see point J.) In other words, a tablet is worth$90 to those customers.

Figure 1. Consumer and Producer Surplus. The somewhat triangular area labeled by F in the graph shows the area of consumer surplus, which shows that the equilibrium price in the market was less than what many of the consumers were willing to pay. The somewhat triangular area labeled by G shows the area of producer surplus, which shows that the equilibrium price received in the market was more than what many of the producers were willing to accept for their products.

However, that doesn't mean that those customers will end up paying $90. Figure 1 shows that the equilibrium price is$80 and the equilibrium quantity is 28 million tablets. At that price, each customer who would have been willing to pay $90 for a tablet is getting a good deal. We all know what a good deal is - it's when you get something for less than you think it's worth. We don't have to stop there. If suppliers chose to produce only 14 tables (as shown in point K), we can look at Figure 1 and2 up to the demand curve to see that some customers would have been willing to pay about$115 for a tablet at this quantity produced. What that means is that this subset of customers got an even better deal at the equilibrium price.

The demand curve shows what consumers are willing to pay for any given quantity of tablets. In other words, the height of the demand curve at any quantity shows what some consumers think those tablets are worth. We can formalize this idea of how good a deal consumers get on a transaction using the concept of consumer surplus.

Since a demand curve traces consumers' willingness to pay for different quantities, we can define the gain to consumers as the difference between what they would have been willing to pay and the price that they actually paid. At point J, consumers were willing to pay $90, but they were able to purchase tablets at the equilibrium price of$80, so they gained $10 of extra value on each tablet. This is exactly analogous to the "profit" Bill earned from buying apples that we described in the previous page of reading. If we add up the gains at every quantity, we can measure the consumer surplus as the area under the demand curve up to the equilibrium quantity and above the equilibrium price. In Figure 1, the consumer surplus is the area labeled F. The supply curve shows the quantity that firms are willing to supply at each price. For example, point K in Figure 1 illustrates that firms would have been willing to supply a quantity of 14 million tablets at a price of$45 each. Those producers were instead able to charge the equilibrium price of $80, clearly receiving an extra benefit beyond what they required to supply the product. The amount that a seller is paid for a good minus the seller's actual cost is called producer surplus. In Figure 1, producer surplus is the area labeled G - that is, the area between the market price and the segment of the supply curve below the equilibrium. To summarize, producers created and sold 28 tablets to consumers. Both producers and consumers benefited. The value of the tablets is the area under the demand curve up to the equilibrium quantity. The cost to produce that value is the area under the supply curve. The new value created by the transactions, i.e. the net gain to society, is the area between the supply curve and the demand curve, that is, the sum of producer surplus and consumer surplus. This sum is called social surplus, also referred to as economic surplus or total surplus. In Figure 1 we show social surplus as the area F + G. Social surplus is larger at the equilibrium quantity and price than it would be at any other quantity. This is what economists mean when they say that market equilibrium is (perfectly) allocatively efficient. At the efficient level of output, it is impossible to produce greater consumer surplus without reducing producer surplus, and it is impossible to produce greater producer surplus without reducing consumer surplus. In other words, the consumer and producers gains from exchange are maximized at the equilibrium point. ## 4. TRY IT If Kay is willing to pay$25,000 for a truck and finds the model he wants for $22,500, Kay has • a consumer surplus of$2,500.00.
• a producer surplus of $2,500.00. • a social surplus of$25,000.00.

### 4.1. Answer to Try It

• a consumer surplus of $2,500.00. Correct. The difference between what Kay is willing to pay and what he actually pays is his consumer surplus: ($25,000 - $22,500 =$2,500)
• a producer surplus of $2,500.00. Incorrect Kay is not a producer but a consumer and therefore the difference between what Kay is willing to pay and what he actually pays is his consumer surplus. • a social surplus of$25,000.00.
Incorrect Kay is not a producer but a consumer and therefore the difference between what Kay is willing to pay and what he actually pays is his consumer surplus.

In this video, you'll consider the holiday market for Santa hats. The market is efficient and both consumer and producer surplus are maximized at the equilibrium point of $5. If the government establishes a price ceiling, a shortage results, which also causes the producer surplus to shrink, and results in inefficiency called deadweight loss. If government implements a price floor, there is a surplus in the market, the consumer surplus shrinks, and inefficiency produces deadweight loss. ## 6. EXAMPLE: CALCULATE CONSUMER SURPLUS Figure 2. Consumer and producer surpluses are shown as the area where consumers would have been willing to pay a higher price for a good or the price where producers would have been willing to sell a good. In the sample market shown in the graph, equilibrium price is$10 and equilibrium quantity is 3 units. The consumer surplus area is highlighted above the equilibrium price line. This area can be calculated as the area of a triangle.

Recall that to find the area of a triangle, you will need to know its base and height. Refer to the following example if you need a refresher.

Figure 3. The area of a triangle.

Let's apply the calculation for the area of a triangle to our example market to see the added value that consumers will get for this item at the equilibrium price in our sample market.

Step 1: Define the base and height of the consumer surplus triangle.

The base of the consumer surplus triangle is 3 units long. Be careful when you define the height of this triangle, it is tempting to say it is 25, can you see why it isn't? The height is determined by the distance from the equilibrium price line and where the demand curve intersects the vertical axis. The height of the triangle begins at $10 and ends at$25, so it will be $25 –$10 = $15 $b=3$ $h=15$ Step 2: Apply the values for base and height to the formula for the area of a triangle. $\begin{array}{c} \mathrm{A}=\dfrac{1}{2} b \times h \\ \mathrm{~A}=\dfrac{1}{2} 3 \times 15 \\ \mathrm{~A}=\dfrac{1}{2} 45 \\ \mathrm{~A}=\dfrac{45}{2}=22.5 \end{array}$ ## 7. TRY IT Consider the market for corn flour below. What is the producer surplus? $6,000 $2,000$4,000

Consider the market for corn flour below. What is the social surplus?

 $4,000$12,000 $6,000 ### 7.1. Answers to Try It 1. •$6,000
Incorrect The producer surplus is the area above the supply curve below the market price ($3). The area is [$(3-1) x 2000 ]/2= $2,000. Make sure you use the right formula to calculate the producer surplus. Don't confuse it with the social surplus that is the area below the demand curve and above the supply curve or the producer surplus + the consumer surplus.. •$2,000
Correct The producer surplus is the area above the supply curve below the market price ($3). The area is [$(3-1) x 2000 ]/2= $2,000. •$4,000
Incorrect The producer surplus is the area above the supply curve below the market price ($3). The area is [$(3-1) x 2000 ]/2= $2,000. Make sure you use the right formula to calculate the producer surplus. Don't confuse it with the consumer surplus that is the area below the demand curve and above the market price line. 2. •$4,000
Incorrect The social surplus is the area area below the demand curve and above the supply curve or the producer surplus + the consumer surplus. Make sure you use the right formula to calculate the producer and consumer surplus.
• $12,000 Incorrect The social surplus is the area area below the demand curve and above the supply curve or the producer surplus + the consumer surplus. Make sure you use the right formula to calculate the producer and consumer surplus. •$6,000
Correct. The social surplus is the area area below the demand curve and above the supply curve or the producer surplus + the consumer surplus.($3). The producer surplus is [$(3-1) x 2000]/2= $2,000. The consumer surplus is [($7-3) * 2000] 12 = $4000. The social surplus =$6,000

## 8. TRY IT

This next question allows you to get as much practice as you need, as you can click the link at the top of the question ("Try another version of this question') to get a new version of the question. Practice until you feel comfortable with this concept.


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# Math Help - Physical Science Algebraic Problem Related to Motion

1. ## Physical Science Algebraic Problem Related to Motion

Would anybody be able to explain this problem and its equations for me? I don't understand it at ALL.

1)A ball is thrown upwards from a 100 m high cliff at 7.5 m/sec. If you ignore air resistance
A) What is its velocity after 7.0 sec?
B) How long until it is at its high point?
C) Where is it (Vertical height related to the ground) after 7.0 sec
D) When does it reach the ground?
E) What is the total distance moved?
F) WHat is the ball's speed just before hitting the ground?

2) A car accelerates uniformly in a straight line from rest at the rate of 2.3 m/s^2. What is the speed of the car after it has traveled 55m. Answer in meters per second.

Please, I really need somebody to explain this to me ASAP. I don't necessarily need the answers, but I NEED the equations! I just don't understand! And none of these are trick questions.

2. ## Re: Physical Science Algebraic Problem Related to Motion

Originally Posted by Psyche
Would anybody be able to explain this problem and its equations for me? I don't understand it at ALL.

1)A ball is thrown upwards from a 100 m high cliff at 7.5 m/sec. If you ignore air resistance
A) What is its velocity after 7.0 sec?
B) How long until it is at its high point?
C) Where is it (Vertical height related to the ground) after 7.0 sec
D) When does it reach the ground?
E) What is the total distance moved?
F) WHat is the ball's speed just before hitting the ground?

2) A car accelerates uniformly in a straight line from rest at the rate of 2.3 m/s^2. What is the speed of the car after it has traveled 55m. Answer in meters per second.

Please, I really need somebody to explain this to me ASAP. I don't necessarily need the answers, but I NEED the equations! I just don't understand! And none of these are trick questions.
Kinematics equations for uniformly accelerated motion in the vertical direction under the influence of gravity ...

$v = v_0 - gt$

$\Delta y = v_0 t - \frac{1}{2}gt^2$

$v_f^2 = v_0^2 -2g \Delta y$

3. ## Re: Physical Science Algebraic Problem Related to Motion

Okay. I know 1a goes with eq #1, but what about the others, and how would you plug the numbers in? I'm sorry. I'm so confused. Haha

4. ## Re: Physical Science Algebraic Problem Related to Motion

B) How long until it is at its high point?

$v = 0$ at the top of the ball's trajectory

C) Where is it (Vertical height related to the ground) after 7.0 sec

use the $\Delta y$ equation with $y_0 = 100 m$

D) When does it reach the ground?

set $\Delta y = -100$ , solve for t

E) What is the total distance moved?

2(displacement up) + 100

F) WHat is the ball's speed just before hitting the ground?

|v| when it hits the ground (use the solution for part D)

6. ## Re: Physical Science Algebraic Problem Related to Motion

Originally Posted by Psyche
go to the link and learn about motion in one dimension ...

http://www.physicsclassroom.com/class/1DKin/U1L6a.cfm

7. ## Re: Physical Science Algebraic Problem Related to Motion

Well i found this site is very useful defined the theory
Motion in One Dimension

8. ## Re: Physical Science Algebraic Problem Related to Motion

Originally Posted by Psyche
We can take the third kinematics equations given by skeeter and modify it a bit to meet our needs. We are told the car begins at rest ( $v_0=0$) and accelerates uniformly at $a=2.3\text{ }\frac{\text{m}}{\text{s}^2}$ and undergoes a displacement of $\Delta x=55\text{ m}$, so the equation becomes:

$v_f^2=2a\Delta x$

$\left|v_f\right|=\sqrt{2a\Delta x}$

Now just plug in the given data to find the magnitude of the final velocity vector, which is the speed.

Suppose we did not have the formula handy. We can find it by reasoning as follows:

We know the distance traveled is the average velocity times the time traveled:

$\Delta x=\frac{v_0+v_f}{2}t\:\therefore\:t=\frac{2\Delta x}{v_f+v_0}$

We know uniform acceleration is the change in velocity divided by the change in time:

$a=\frac{v_f-v_0}{t}\:\therefore\:t=\frac{v_f-v_0}{a}$

We have 2 expression for time t, so equate them to get:

$\frac{2\Delta x}{v_f+v_0}=\frac{v_f-v_0}{a}$

Cross-multiply:

$v_f^2-v_0^2=2a\Delta x$

$v_f^2=2a\Delta x+v_0^2$


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# Math Help - Integrating Factor

1. ## Integrating Factor

I have this equation:

$y'+ytanx=sinx$

The integrating factor give me $f=e^xcosx$

Factor replacement and give me:

$e^xcosxdy+e^xcosx(ytanx-senx)dx=0$ but the partial derivates give me:

$Q_x=cosxe^x-senxe^x$ and $Q_y=e^xsenx$

Thanks

2. 1- What is $senx$ ?
2- I think if you read the title of your thread, you will know how to solve it ..


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December 8, 2018

In CS 2110 at Georgia Tech, we use the Gameboy Advance to demonstrate how C code can interact directly with hardware. On a high level, students create games by reading button states directly from memory-mapped registers and write directly to a special region of memory called the “video buffer,” which the video controller continuously reads and draws on the screen.

But my favorite example of students poking with the GBA hardware introduces them to an important real-life example: Direct Memory Access (DMA)! The GBA has a DMA controller which amounts to a fast memcpy() in hardware which some additional switches to flip. (See tonc for more details on GBA DMA.)

Students use DMA as follows:

// provided boilerplate
typedef struct {
const volatile void *src;
volatile void *dst;
volatile uint32_t cnt;
} DMA_CONTROLLER;

#define DMA ((volatile DMA_CONTROLLER *) 0x040000B0)

#define DMA_DST_INC (0 << 21)
#define DMA_DST_DEC (1 << 21)
#define DMA_DST_FIX (2 << 21)

#define DMA_SRC_INC (0 << 23)
#define DMA_SRC_DEC (1 << 23)
#define DMA_SRC_FIX (2 << 23)

#define DMA_ON (1 << 31)

// STUDENT CODE:
// Turns the whole screen red
void redscreen(void) {
volatile u16 color = RED;
DMA[3].src = &color;
DMA[3].dst = videoBuffer;
// dimensions of gba screen
DMA[3].cnt = DMA_ON | DMA_SRC_FIX | DMA_DST_INC | (240 * 160);
}

Notice that the student code above which invokes DMA does not call a function we can mock out in an autograder, like how a student’s malloc() could call a fake sbrk() — it’s just hitting memory at some weird address.

You might think: well this is easy you dingus, just make the DMA macro point to some chunk of memory which you can check after their code returns! However, GBA DMA normally stops the CPU while it makes the copy, and that would not. Let’s look at a less trivial example of DMA:

void redsquare(int width, int height) {
volatile u16 color = RED;
for (int row = 0; row < height; row++) {
DMA[3].src = &color;
DMA[3].dst = videoBuffer + row * 240;
// dimensions of gba screen
DMA[3].cnt = DMA_ON | DMA_SRC_FIX | DMA_DST_INC | width;
}
}

If we used the “magic DMA buffer” idea mentioned above, later DMA transfers would clobber src/dst/cnt values from earlier transfers. That is, we would only see the DMA results for the last row of the red square. So that ain’t gonna work.

Consequently, before this semester, we never autograded DMA. Instead, we would grep students’ code for for loops to see if they were manipulating the videoBuffer pixel-by-pixel1 without DMA and then would make sure the emulator showed the right image. But my goal in 2110 is to autograde as much as possible, especially because TAs would’ve needed to grade this timed lab during or right before their final exams, so I decided to try to adapt my favorite homework from my Operating Systems class into an autograder.

# Abusing Virtual Memory to Solve Your Life Problems

What’s a good way to make a memory access halt execution of a program, much like how setting the DMA_ON bit in the control register makes the DMA controller halt the CPU while the copy runs? A page fault!

We can make DMA accesses pagefault by making the DMA macro point to some page marked as unwritable in the page table. Then our page fault handler runs and can read what they wrote! Except it’s not that simple: when an instruction pagefaults, it doesn’t actually run, so we can’t see what students wrote to our fake DMA page. In other words, we can see what address the student tried to hit, but not what they wrote (yet). So instead, our page fault handler needs to mark the fake DMA page as writable in the page table, so that when the hardware re-tries the faulting instruction, it actually has an effect.

But this re-introduces the original problem: how do we know what they wrote? Indeed, if our page fault handler makes the fake DMA page subsequently writable, then later DMA accesses won’t pagefault, meaning we’ll miss them and they’ll clobber the written data in the fake DMA page. We’re back to square one!

My solution was to use two fake DMA pages and alternate between them. The DMA macro always points to an unwritable page, but when an instruction faults, we change the page it’s trying to hit to writable. (Changing the page to which the DMA macro points will not change the page that the faulting instruction hits when the hardware retries the instruction because the old page will still be in some register, unchanged.) This solves our problem because every DMA access will pagefault, and we can see the data that the user wrote on DMA access $t$ when DMA access $t+1$ pagefaults.

So the pseudocode for the tester would look like:

page fault handler:
make sure this is a fake DMA-related page fault
read what the student wrote to the fake DMA page on
the last faulting access. log it
mark the page to which the DMA macro points as
writable so the faulting instruction can run
set the DMA macro to point to the alternate page and
set it as unwritable

tester code:
set up page fault handler
run student code
check fake DMA page for data written by student on
the final access
exit

A visualization for all this follows:

On Linux, you can map pages with mmap(), change their permissions with mprotect(), and catch page faults by registering a SIGSEGV handler. And to make sure the compiler loads the address of the fake DMA page from memory every time we dereference the DMA macro, we can define the DMA macro such that it expands to a volatile *volatile. That is, a pointer to volatile memory that itself is volatile, like

#define DMA ((volatile DMA_CONTROLLER *volatile) fake_dma_page)

# Simulating DMA

Simulating DMA seems simple enough. When the the segfault handler or cleanup code reads what the student wrote to the fake DMA page, we see if they set the DMA_ON bit in the control register, and if they did, simulate it!

But there’s a complication here too. Returning to the earlier example,

void redscreen(void) {
volatile u16 color = RED;
DMA[3].src = &color;
DMA[3].dst = videoBuffer;
DMA[3].cnt = DMA_ON | DMA_SRC_FIX | 38400;
}

we see the student (correctly) sets the source register to the address of a variable on the stack. But we can only check to see that they set DMA_ON in the control register after their function returns because there are no later fake DMA segfaults — this is true for the last DMA access in general. But accessing memory in the stack frame of a function which has returned is undefined behavior in C, because it’s gonna get clobbered by subsequent function calls! Yikes!

## Speculative Copying, aka Copy Like a Madman

The instant the user sets the DMA_ON bit in the control register, we need to take a snapshot of memory around where the source register points. However, we have a few problems with when this copy needs to happen:

1. We only know they’re writing to the control register, not that they’re setting the DMA_ON bit
2. Similarly, we don’t know how much they’re copying
3. We don’t how if they’re using source decrement or source increment. (Source decrement means subtracting 2, or 4 if 32-bit DMAing is enabled, from the source address instead of adding after each one of the N transfers.)

We can fix all three of these issues by simply copying super aggressively when the user sets the control register, even if we don’t know what flags they set or size they requested. In particular, if they copy more than 38,400 pixels (shorts), they’re doing the problem wrong anyway (since that’s bigger than the actual videoBuffer!), so that provides an upper bound. And to handle source decrement, we can copy the 76,800 bytes on the left-hand and right-hand side of the source pointer, just in case.

But now we have a new problem: what if, for example, they’re using source decrement and they set the source pointer near the end of the heap? Then we’re gonna segfault when trying to “snapshot” memory. The hack I found is to fork() and then copy until failure. Pseudocode:

zero out source buffer

fork()
copy left-hand side of src pointer byte-by-byte
until I segfault, stopping at the max copy size

fork()
copy right-hand side of src pointer byte-by-byte
until I segfault, stopping at the max copy size

wait()
wait()

So if they copy too far, instead of a segfault in the mystery tester, they’ll get black pixels (note the zeroing out of the source buffer at the top).

This approach seems like it would be slow because of the sheer amount of copying, but it actually seems to perform pretty well. 🤷

Now that we’ve got GBA DMA to work on x86, we still have to autograde it. How? Let’s make up an assignment in which the student simply draws an image in the top left of the GBA screen. That is, they have to complete:

void drawImage3(const u16 *image, int width, int height) {
// Student writes this code
for (int r = 0; r < height; r++) {
DMA[3].src = image + width * r;
DMA[3].dst = videoBuffer + 240 * r;
DMA[3].cnt = DMA_ON | width;
}
}

What exactly is “right” here? Is this

void drawImage3(const u16 *image, int width, int height) {
for (int r = 0; r < height; r++)
for (int c = 0; c < width; c++)
videoBuffer[240 * r + c] = image[width * r + c];
}

void drawImage3(const u16 *image, int width, int height) {
for (int r = 0; r < height; r++) {
for (int c = 0; c < width; c++) {
DMA[3].src = image + width * r + c;
DMA[3].dst = videoBuffer + 240 * r + c;
DMA[3].cnt = DMA_ON | 1;
}
}
}

or this

void drawImage3(const u16 *image, int width, int height) {
for (int r = height - 1; r >= 0; r++) {
DMA[3].src = image + width * r + width - 1;
DMA[3].dst = videoBuffer + 240 * r + width - 1;
DMA[3].cnt = DMA_ON | DMA_SRC_DEC | DMA_DST_DEC | width;
}
}

strange example? I settled on the following:

1. Require they use DMA to draw pixels on the screen
2. Require they use the minimum number of DMA calls possible without violating #1
3. For each test input, have two separate tests:
1. One simulates DMA and tests that the videoBuffer is correct. This awards credit for not using DMA or misusing it (e.g. the single-pixel DMA example above) but still getting it working
2. The other compares the log of DMA calls with the expected, but does its best to allow perfectly valid cases like the double-decrementing one right above even though it does everything backwards from the simpler approach

## Regions of the Screen

For the second test above, I ended up taking the log of their DMA calls and converting each call to a generalized form of some ordered quadruplet $(i_\text{start}, i_\text{end}, v_\text{start}, v_\text{end}) \in \mathbb{Z}^4$, where the first two coordinates are the start/end indices in the image of the copy, and the next two are the same except for the videoBuffer.

Then I normalized each copy in both the actual and expected lists of copies using the following two steps: First, if $i_\text{start} > i_\text{end}$ for any transfer $(i_\text{start}, i_\text{end}, v_\text{start}, v_\text{end})$, then I converted the quadruplet to $(i_\text{end}, i_\text{start}, v_\text{end}, v_\text{start})$ since that represents the same copy. Then I sorted the list of copies by start index.

After normalizing both the actual and expected copies, I can compare them copy-by-copy in order and display a (hopefully) helpful error message to the student on the first one that does not match.

# Example Tester

If you’re interested, the GitHub repository for all this DMA stuff has a complete example of a tester which uses the strategies I rambled about above. Beware, the student half, which runs a GBA emulator, requires the CS 2110 GBA toolchain, but that’s not really the fun part — the grader is!

1. One student infamously evaded our grep checking by unrolling their loop by hand. That’s hundreds of lines and we loved it ↩


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# 'Closure' of CFLs under complementation and intersection

Consider two context-free languages $L_1, L_2$. Of course, $L_1 - L_2, L_1\cap L_2, \bar{L}_1$, etc. are not necessarily context-free, but they are context-sensitive (the second is easy, the other two I think follow from Immerman-Szelepcsenyi (if I spelled that right)). However, there's no nice structure to context-sensitive languages (e.g., pumping/Ogden's lemmas), so I was wondering if the inclusion is any better--is the 'closure' of CFLs under the standard operations (well, only $\cap$ and complementation) a proper subset of CSLs, and if so, does it have any natural description?

Thanks much.

-
The title makes this sort of look like a question about introductory CS... By 'closure' I mean the smallest set containing the set of CFLs that is closed under those operations. –  alpoge Jan 10 '11 at 16:39

Grammars involving the usual context-free operations plus intersection are called conjunctive grammars. Adding negation (in addition to intersection) gives boolean grammars.

Alexander Okhotin has done quite a bit of recent work on the closure properties of the languages (sets of strings) specified by these sorts of grammars. He also has a paper showing how to parse these languages using a variant on the Lang-Tomita GLR algorithm.

As you note, CFLs are strictly included in conjunctive grammars. Surprisingly, it appears that the strictness of the inclusion relationship between conjunctive and boolean grammars is still an open problem (#6). The last inclusion is also strict: there are context-sensitive languages which cannot be specified by a boolean grammar.

-
This looks right. Thanks! –  alpoge Jan 10 '11 at 18:35
It seems conjunctive grammars also strictly contain the set of $2$-intersection languages with complementation (or whatever one might call them), since in the unary case the set of conjunctive languages contains nonregular languages (which are obviously not in the previous set). Though conjunctive/boolean grammars do indeed answer the question. Thanks again! –  alpoge Jan 10 '11 at 19:28
Would An infinite hierarchy of intersections of context-free languages (Liu Weiner 1973) answer your question? They prove that the intersection of $k$ CFL forms a class strictly contained in the intersection of $k +1$ CFL.
Hm. I suppose that probably shows that the 'closure' I'm talking about is pretty complicated at least. What I really want to know is about $2$-intersection languages--well, specifically, $2$-intersection languages 'adjoined' with the co-CFL languages, so to speak. –  alpoge Jan 10 '11 at 16:29