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It looks like you're new here. If you want to get involved, click one of these buttons! One of the main lessons of category theory is that whenever you think about some kind of mathematical gadget, you should also think about maps between gadgets of this kind. For example, when you think about sets you should also think about functions. When you think about vector spaces you should also think about linear maps. And so on. We've been talking about various kinds of monoidal preorders. So, let's think about maps between monoidal preorders. As I explained in Lecture 22, a monoidal preorder is a crossbreed or hybrid of a preorder and a monoid. So let's think about maps between preorders, and maps between monoids, and try to hybridize those. We've already seen maps between preorders: they're called monotone functions: Definition. A monotone function from a preorder $(X,\le_X)$ to $(Y,\le_Y)$ is a function $f : X \to Y$ such that $$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ for all elements $x,x' \in X$. So, these functions preserve what a preorder has, namely the relation $\le$. A monoid, on the other hand, has an associative operation $\otimes$ and a unit element $I$. So, a map between monoids should preserve th0se! That's how this game works. Just to scare people, mathematicians call these maps "homomorphisms": Definition. A homomorphism from a monoid $ (X,\otimes_X,I_X) $ to a monoid $ (Y,\otimes_Y,I_Y) $ is a function $f : X \to Y$ such that: $$ f(x \otimes_X x') = f(x) \otimes_Y f(x') $$ for all elements $x,x' \in X$, and $$ f(I_X) = I_Y .$$ You've probably seen a lot of homomorphisms between monoids. Some of them you barely noticed. For example, the set of integers $\mathbb{Z}$ is a monoid with addition as $\otimes$ and the number $0$ as $I$. So is the set $\mathbb{R}$ of real numbers! There's a function that turns each integer into a real number: $$ i: \mathbb{Z} \to \mathbb{R} . $$It's such a bland function you may never have thought about it: it sends each integer to itself, but regarded as a real number. And this function is a homomorphism! What does that mean? Look at the definition. It means you can either add two natural numbers and then regard the result as a real number... or first regard each of them as a real number and then add them... and you get the same answer either way. It also says that integer $0$, regarded as a real number, is the real number we call $0$. Boring facts! But utterly crucial facts. Computer scientists need to worry about these things, because for them integers and real numbers (or floating-point numbers) are different data types, and $i$ is doing "type conversion". You've also seen a lot of other, more interesting homomorphisms between monoids. For example, the whole point of the logarithm function is that it's a homomorphism. It carries multiplication to addition: $$ \log(x \cdot x') = \log(x) + \log(x') $$ and it carries the identity for multiplication to the identity for addition: $$ \log(1) = 0. $$ People invented tables of logarithms, and later slide rules, precisely for this reason! They wanted to convert multiplication problems into easier addition problems. You may also have seen linear maps between vector spaces. A vector space gives a monoid with addition as $\otimes$ and the zero vector as $I$; any linear map between vector spaces then gives a homomorphism. Puzzle 80. Tell me a few more homomorphisms between monoids that you routinely use, or at least know. I hope I've convinced you: monotone functions between preorders are important, and so are homomorphisms between monoids. Thus, if we hybridize these concepts, we'll get a concept that's likely to be important. It turns out there are a few different ways! The most obvious way is simply to combine all the conditions. There are other ways, so this way is called "strict": Definition. A strict monoidal monotone from a monoidal preorder $ (X,\le_X,\otimes_X,I_X) $ to a monoidal preorder $ (Y,\le_Y,\otimes_Y,I_Y) $ is a function $f : X \to Y$ such that: $$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ and $$ f(x) \otimes_Y f(x') = f(x \otimes_X x') $$ for all elements $x,x' \in X$, and also $$ I_Y = f(I_X) . $$ For example, the homomorphism $$ i : \mathbb{Z} \to \mathbb{R} ,$$ is a strict monoidal monotone: if one integer is $\le$ another, then that's still true when we regard them as real numbers. So is the logarithm function. What other definition could we possibly use, and why would we care? It turns out sometimes we want to replace some of the equations in the above definition by inequalities! Definition. A lax monoidal monotone from a monoidal preorder $(X,\le_X,\otimes_X,I_X)$ to a monoidal preorder $(Y,\le_Y,\otimes_Y,I_Y)$ is a function $f : X \to Y$ such that: $$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ and $$ f(x) \otimes_Y f(x') \le_Y f(x \otimes_X x') $$ for all elements $x,x' \in X$, and also $$ I_Y \le_Y f(I_X). $$Fong and Spivak call this simply a monoidal monotone, since it's their favorite kind. But I will warn you that others call it "lax". We could also turn around those last two inequalities: Definition. An oplax monoidal monotone from a monoidal preorder $(X,\le_X,\otimes_X,I_X)$ to a monoidal preorder $(Y,\le_Y,\otimes_Y,I_Y)$ is a function $f : X \to Y$ such that: $$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ and $$ f(x) \otimes_Y f(x') \ge_Y f(x \otimes_X x') $$ for all elements $x,x' \in X$, and also $$ I_Y \ge_Y f(I_X). $$ You are probably drowning in definitions now, so let me give some examples to show that they're justified. The monotone function $$ i : \mathbb{Z} \to \mathbb{R} $$ has a right adjoint $$ \lfloor \cdot \rfloor : \mathbb{R} \to \mathbb{Z} $$which provides the approximation from below to the nonexistent inverse of $i$: that is, $ \lfloor x \rfloor $ is the greatest integer that's $\le x$. It also has a left adjoint $$ \lceil \cdot \rceil : \mathbb{R} \to \mathbb{Z} $$which is the best approximation from above to the nonexistent inverse of $i$: that is, $ \lceil x \rceil $ is the least integer that's $\ge x$. Puzzle 81. Show that one of the functions $ \lfloor \cdot \rfloor : \mathbb{R} \to \mathbb{Z} $, $ \lceil \cdot \rceil : \mathbb{R} \to \mathbb{Z} $ is a lax monoidal monotone and the other is an oplax monoidal monotone, where we make the integers and reals into monoids using addition. So, you should be sensing some relation between left and right adjoints, and lax and oplax monoidal monotones. We'll talk about this more! And we'll see why all this stuff is important for resource theories. Finally, for the bravest among you: Puzzle 82. Find a function between monoidal preorders that is both lax and oplax monoidal monotone but not strict monoidal monotone. In case you haven't had enough jargon for today: a function between monoidal preorders that's both lax and oplax monoidal monotone is called strong monoidal monotone.
I've heard a lot of definitions given for structural estimation. But it's never seemed entirely clear to me. Some times I've heard that what one person might call "reduced form" estimation should actually be called structural estimation. Sorry I don't have an example to illustrate, but I was wondering if somebody could clarify, hopefully with a link to a paper or some other source. What is structural estimation compared to reduced form estimation? Does the potential outcomes framework count as a structural equation? Structural estimation is a term coined by the Cowles commission which at the time seems to have been dominated by Haavelmo, Koopmans and a few others. The motto of the Cowles commission (after 1965) was: "Theory and Measurement". The phrase represents the underlying rationale of structural modelling, that measurement cannot be done without some kind of theory. To my knowledge, the phrase was first used by Koopmans in "Identification Problems in Economic Model Construction": Systems of structural equations may be composed entirely on the basis of economic "theory." By this term we shall understand the combination of (a) principles of economic of behavior derived from general observation--partly introspective, partly through interview or experience--of the motives of economic decisions,(b) knowledge of legal and institutional rules restricting individual behavior (tax schedules, price controls, reserve requirements, etc.), (c) technological knowledge, and (d) carefully constructed definitions of variables. Structural equations are then equations that come from an underlying economic (or physical, or legal) model. Structural estimation is precisely estimation which uses these equations to identify parameters of interest, and inform counter-factuals. Importantly, these parameters are usually taken to be invariant, and therefore counter-factuals taken from their estimates will be completely "correct". Counter-factuals were the main unit of interest to the Cowles commission. Koopmans also discusses reduced form estimation: By the reduced formof a complete set of linear structural equations... we mean the form obtained by solving for each of the dependent(i.e., nonlagged endogenous) variables, and in terms of transformed disturbances (which are linear functions of the disturbances in the original structural equations). The linearity is an artifact of the times (this was published in 1949!) but the point is that reduced-form equations are equations written in terms of economic variables which do not have a structural interpretation as defined above. So, a linear regression will be a reduced-form of some true structural model, because linear regression usually does not have a true economic interpretation. This does not mean that reduced form equations cannot be used to identify parameters in structural equations - in fact this is precisely how indirect inference works - just that they do not represent a deeper model of the data generating process. Reduced forms can (in principle) be used to identify structural parameters, in which cased you are still performing structural estimation, just through using the reduced form. Another way to look at this is that structural models are generally deductive, whereas reduced forms tend to be used as part of some greater inductive reasoning. For a comparison of this kind of Cowles commission structural modelling with Rubin causal modelling, check out this awesome set of slides by Heckman. For other resources I'd check out more of what Koopmans wrote, the book Structural Macroeconomics by DeJong and Dave, these lecture notes by Whited, this paper by Wolpin (written for the Cowles Foundation, in honour of Koopmans) and a response by Rust. Addendum: A simple example of reduced form and structural models. Suppose we were looking at data on the prices, $p_t$ and quantities, $q_t$ produced by a monopolist. The monopolist faces a series of unknown costs in the future, and a linear demand curve (this would really have to be justified). Let's say the $\hat q_t$ and $\hat p_t$ we observe are measured with some kinds of mean-zero error, $e_t$, and $v_t$ Noting that both price and quantity seem to be associated with changes in cost, a reduced form equation for this model might be: \begin{align} \hat q_t &= \gamma - \lambda c_t + \epsilon_t\\ \hat p_t &= \alpha + \beta c_t + \nu_t \end{align} Because this is a reduced form model, it needs no justification other than that it might work empirically. On the other hand, a structural model would start by specifying the demand curve (again to be strict this should start at the level of individual utility), and the monopolist's problem: \begin{align} \text{Demand curve: }&p_t=a-bq_t\\ \text{Producer's problem: }&\max E\left[\sum_{t=0}^\infty\delta^t (p_t-c_t)q_t(p_t)\right]\\ \text{Measurement equations: }&\hat q_t = q_t + e_t\\ &\hat p_t = p_t + v_t \end{align} From this further structural equations could be derived (structural because they are still representative of principles of economic behavior): \begin{align} \hat q_t&=\frac{a-c_t}{2b} +e_t\\ \hat p_t&=\frac{a+c_t}{2} + v_t\\ \end{align} This is a case where a reduced form equation will have a meaningful structural interpretation, as consistent estimates $\hat a$ and $\hat b$ can be formed: \begin{align} \hat a&= 2\hat\alpha \\ \hat b&= \frac{1}{2\hat \lambda} \end{align} Another case of identification of structural parameters from reduced forms is the logit model in the case of valuations with extreme value errors (see McFadden (1974)). In general it is unlikely a given reduced form model will have a structural interpretation.
AliPhysics 1168478 (1168478) Scripts used in the analysis Algorithms Corrections Monte-carlo code Mid-rapidity tracklet code for dN/deta Tasks Topical file AliAODForwardMult.h Per-event $ N_{ch}$ per $(\eta,\varphi)$ bin. file AliFMDEventPlaneFinder.h file AliForwardUtil.h Various utilities used in PWGLF/FORWARD. file AliLandauGaus.h Declaration and implementation of Landau-Gauss distributions. file AliLandauGausFitter.h Declaration and implementation of fitter of Landau-Gauss distributions to energy loss spectra. class AliForwardCreateResponseMatrices class AliForwardTriggerBiasCorrection class AliForwardUtil struct AliForwardUtil::Histos struct AliForwardUtil::RingHistos class AliLandauGaus class AliLandauGausFitter Code to do the multiplicity analysis in the forward psuedo-rapidity regions The code in this section defines methods to measure the charged-particle pseudorapidity density over $\|\eta\|<2$ using SPD tracklets. The code in this module constitutes tools for analysing SPD tracklet data for the charged-particle pseudorapidity density. It is based on Ruben's original code (see pwglf_fwd_spd_tracklet_1), but differs in some important aspects. This code also requires a pass of real-data (AliTrackletAODTask) and simulated (AliTrackletAODMCTask) ESDs, but the output is not near-final histograms but an array of data structures (AliAODTracklet) stored on the output AOD. The data structure contains basic information on each tracklet Tracklet structures from simulations contain in addition During the AOD production, no cuts, except those defined for the re-reconstruction are imposed. In this way, the AOD contains the minimum bias information on tracklets for all events, And for AODs corresponding to simulated data, we can also In this way, we do a single pass of ESDs for real and simulated data, and we can then process the generated AODs with various cuts imposed. The AODs are generally small enough that they can be processed locally and quickly (for example using ProofLite). This scheme allows for fast turn-around with the largest possible flexibility. The final charged-particle pseudorapidity density is produced by an external class (AliTrackletdNdeta2). Other differences to Ruben's code is that the output files are far more structured, allowing for fast browsing of the data and quality assurance. The analysis requires real data and simulated data, anchored to the real data runs being processed. For both real and simulated data, the analysis progresses through two steps: A pass over ESD plus clusters to generate an AOD branch containing a TClonesArray of AliAODTracklet objects. In this pass, there are no selections imposed on the events. In this pass, the SPD clusters are reprocesed and the tracklets are re-reconstructed. In this pass, we also form so-called injection events. In these events, a real cluster is removed and a new cluster put in at some other location in the detector. The tracklets of the event is then reconstructed and stored. This procedure is repeated as many times as possible. The injection events are therefore superpositions of many events - each with a real cluster removed and replaced by a fake cluster. The injection events are used later for background estimates. When processing simulated data, the tracklets are also inspected for their origin. A tracklet can have three distinct classes of origins: The last class is the background from combinations of clusters that does not correspond to true particles. This background must be removed from the measurements. The second class, tracklets from secondaries, also form a background, but these tracklets are suppressed by cuts on the sum-of-square residuals \[ \Delta = \left[\frac{\Delta_{\theta}^2\sin^2\theta}{\sigma_{\theta}^2}+ \frac{(\Delta_{\phi}-\delta_{\phi})^2}{\sigma_{\phi}^2}\right] \] \[ \frac{d^2N_X}{d\eta d\mathrm{IP}_z}\quad, \] where $ X$ is $ M$ or $ I$ for real data, or $ M',I',P',S',C'$ or $ G'$ for simulated data. For each of these tracklet samples, except $ G'$, we also form the 3-dimensional differential $\Delta$ distributions \[ \frac{d^3N_X}{d\eta d\mathrm{IP}_z d\Delta}\quad, \] which are later used to estimate the background due to wrong combinations of clusters into tracks. Once both the real and simulated data has passed these two steps, we combine the to data sets into the final measurement. The final measurement is given by \[ R = \frac{G'}{(1-\beta')M'}(1-\beta)M, \] where \[ \beta' = \frac{C'}{M'}\quad\mathrm{and}\quad \beta = k\beta'\quad. \] Here, $ k$ is some scaling derived from the 3-dimensional differential $\Delta_M$ and $\Delta_{M'}$ distributions . There are classes for containing data, classes that represent analysis tasks, and classes that perform calculations, as well as specialized classes for analysis of simulation (MC) output. The classes AliAODTracklet and AliAODMCTracklet stores individual tracklet parameters. The difference between the two are that AliAODMCTracklet also stores the PDG code(s) and transverse momentum (momenta) of the mother primary particle (which may be the particle it self). The pass over the ESD is done by the classes AliTrackletAODTask and AliTrackletAODMCTask. These tasks generated the array of tracklets in the AOD events. The difference between the two is that AliTrackletAODMCTask inspects and groups each tracklet according to it's origin, and create pseudo-tracklets corresponding to the generated primary, charged particles. \[ \frac{d^2N_X}{d\eta d\mathrm{IP}_z}\quad, \] and \[ \frac{d^3N_X}{d\eta d\mathrm{IP}_z d\Delta}\quad. \] The first task does this for $ X=M$ and $ I$, while the second and thhird tasks does this for $ X=M',I',P',S',C'$ and $ G'$. The third task reweighs all tracklets according to the particle specie(s) and transverse momentum (momenta) of the mother primary particle(s). AliTrackletWeights defines the interface used for reqeighing the data. To produce the AODs with the tracklet information in, one needs to run a train with a task of the class AliTrackletAODTask (or AliTrackletAODMCTask for simulated data) and a task of the class AliSimpleHeaderTask in it. This is most easily done using the TrainSetup (Using the TrainSetup facility) derived class TrackletAODTrain. For example for real data from run 245064 of LHC15o using the first physics pass runTrain --name=LHC15o_245064_fp_AOD \ --class=TrackletAODTrain.C \ --url="alien:///alice/data/2015/LHC15o?run=245064&pattern=pass_lowint_firstphys//AliESDs.root&aliphysics=last,regular#esdTree" or for simulated data from the LHC15k1a1 production anchored to run 245064 runTrain --name=LHC15k1a1_245064_fp_AOD \ --class=TrackletAODTrain.C \ --url=alien:///alice/sim/2015/LHC15k1a1?run=245064&pattern=/AliESDs.root&aliphysics=last,regular&mc#esdTree (note the addition of the option "&mc" to the URL argument) In both cases a sub-directory - named of the name argument - of the current directory is created. In that sub-directory there are scripts for merging the output, downloading results, and downloading the generated AODs. It is highly recommended to download the generated AODs to your local work station to allow fast second step analysis. To download the AODs, go to the generated sub-directory an run the DownloadAOD.C script. For example, for the real data analysis of run 245064 of LHC15o, one would do (cd LHC15o_245064_fp_AOD && root -l -b -q DownloadAOD.C) and similar for the analysis of the simulated data. To produce the histograms for the final charged-particle pseudorapidity density , one needs to run a train with a task of the class AliTrackletAODdNdeta (or AliTrackletAODMCdNdeta for simulated data) in it. This is most easily done using the TrainSetup derived class TrackletAODdNdeta. For example for real data from run 245064 of LHC15o where we store the AODs generated above on the grid runTrain --name=LHC15o_245064_fp_dNdeta \ --class=TrackletAODdNdeta.C \ --url="alien:///alice/cern.ch/user/a/auser/LHC15o_245064_fp_dNdeta/output?run=245064&pattern=* /AliAOD.root&aliphysics=last,regular#aodTree" or for simulated data from the LHC15k1a1 production anchored to run 245064 runTrain --name=LHC15k1a1_245064_fp_dNdeta \ --class=TrackletAODdNdeta.C \ --url=alien:///alice/cern.ch/user/a/auser/LHC15k1a1_245064_fp_AOD/output?run=245064&pattern=* /AliAOD.root&aliphysics=last,regular&mc#esdTree (note the addition of the option "&mc" to the URL argument) If we had downloaded the AODs, we can use ProofLite to do this step runTrain --name=LHC15o_245064_fp_dNdeta \ --class=TrackletAODdNdeta.C \ --url="lite:///${PWD}/LHC15o_245064_fp_dNdeta?pattern=AliAOD_.root#aodTree" and similar for simulated data runTrain --name=LHC15k1a1_245064_fp_dNdeta \ --class=TrackletAODdNdeta.C \ --url="lite:///${PWD}/LHC15k1a1_245064_fp_dNdeta?pattern=AliAOD_.root&mc#aodTree" (note the addition of the option "&mc" to the URL argument) The final result is obtained by runnin the class AliTrackletdNdeta2 over the histograms from both real data and simulations. As an example, suppose we ran or histogram production on the Grid and have downloaded the merged results into LHC15o_245064_fp_dNdeta/root_archive_000245064/AnalysisResult.root (using LHC15o_245064_fp_dNdeta/Download.C) and LHC15k1a1_245064_fp_dNdeta/root_archive_245064/AnalysisResult.root (using LHC15k1a1_245064_fp_dNdeta/Download.C). Then we should do (see AliTrackletdNdeta2::Run for more information on arguments) Alternatively one can use the script Post.C to do this. The script is used like. where simFile is the simulation data input file (or directory) realFile is the real data input file (or directory) outDir is the output directory (created) process are the processing options visualize are the visualisation options nCentralities is the maximum number of centrality bins Processing options: 0x0001 Do scaling by unity 0x0002 Do scaling by full average 0x0004 Do scaling by eta differential 0x0008 Do scaling by fully differential 0x0010 Correct for decay of strange to secondary 0x1000 MC closure test Visualization options: 0x0001 Draw general information 0x0002 Draw parameters 0x0004 Draw weights 0x0008 Draw dNch/deta 0x0010 Draw alphas 0x0020 Draw delta information 0x0040 Draw backgrounds 0x0100 Whether to make a PDF 0x0200 Whether to pause after each plot 0x0400 Draw in landscape 0x0800 Alternative markers By default, each plot will be made and the process paused. To advance, simple press the space-bar. If we had made the histograms using ProofLite, we should do
Every geometry textbook has formulas for the circumference ($C = 2 \pi r$) and area ($A = \pi r^2$) of a circle. But where do these come from? How can we prove them? Well, the first is more a definition than a theorem: the number $\pi$ is usually defined as the ratio of a circle’s circumference to its diameter: $\pi = C/(2r)$. Armed with this, we can compute the area of a circle. Archimedes’ idea (in 260 BCE) was to approximate this area by looking at regular $n$-sided polygons drawn inside and outside the circle, as in the diagram below. Increasing $n$ gives better and better approximations to the area. Look first at the inner polygon. Its perimeter is slightly less than the circle’s circumference, $C = 2 \pi r$, and the height of each triangle is slightly less than $r$. So when reassembled as shown, the triangles form a rectangle whose area is just under $C/2\cdot r = \pi r^2$. Likewise, the outer polygon has area just larger than $\pi r^2$. As $n$ gets larger, these two bounds get closer and closer to $\pi r^2$, which is therefore the circle’s area. Archimedes used this same idea to approximate the number $\pi$. Not only was he working by hand, but the notion of “square root” was not yet understood well enough to compute with. Nevertheless, he was amazingly able to use 96-sided polygons to approximate the circle! His computation included impressive dexterity with fractions: for example, instead of being able to use $\sqrt{3}$ directly, he had to use the (very close!) approximation $\sqrt{3} > 265/153$. In the end, he obtained the bounds $ 3\frac{10}{71} < \pi < 3\frac{1}{7} $, which are accurate to within 0.0013, or about .04%. (In fact, he proved the slightly stronger but uglier bounds $3\frac{1137}{8069} < \pi < 3\frac{1335}{9347}$. See this translation and exposition for more information on Archimedes’ methods.) These ideas can be pushed further. Focus on a circle with radius 1. The area of the regular $n$-sided polygon inscribed in this circle can be used as an approximation for the circle’s area, namely $\pi$. This polygon has area $A_n = n/2 \cdot \sin(360/n)$ (prove this!). What happens when we double the number of sides? The approximation changes by a factor of $$\frac{A_{2n}}{A_n} = \frac{2\sin(180/n)}{\sin(360/n)} = \frac{1}{\cos(180/n)}.$$ Starting from $A_4 = 2$, we can use the above formula to compute $A_8,A_{16},A_{32},\ldots$, and in the limit we find that $$\pi = \frac{2}{\cos(180/4)\cdot\cos(180/8)\cdot\cos(180/16)\cdots}.$$ Finally, recalling that $\cos(180/4) = \cos(45) = \sqrt{\frac{1}{2}}$ and $\cos(\theta/2) = \sqrt{\frac{1}{2}(1+\cos\theta)}$ (whenever $\cos(\theta/2) \ge 0$), we can rearrange this into the fun infinite product $$\frac{2}{\pi} = \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}} \cdots$$ (which I found at Mathworld). (It’s ironic that this formula for a circle uses so many square roots!)
Evaluating the Determinant of a 2×2 Matrix A determinant is a real number that can be very useful in mathematics because it has multiple applications, such as calculating area, volume, and other quantities. Here, we will use determinants to reveal whether a matrix is invertible by using the entries of a square matrix to determine whether there is a solution to the system of equations. Perhaps one of the more interesting applications, however, is their use in cryptography. Secure signals or messages are sometimes sent encoded in a matrix. The data can only be decrypted with an invertible matrix and the determinant. For our purposes, we focus on the determinant as an indication of the invertibility of the matrix. Calculating the determinant of a matrix involves following the specific patterns that are outlined in this section. A General Note: Find the Determinant of a 2 × 2 Matrix The determinant of a [latex]2\text{ }\times \text{ }2[/latex] matrix, given is defined as Notice the change in notation. There are several ways to indicate the determinant, including [latex]\mathrm{det}\left(A\right)[/latex] and replacing the brackets in a matrix with straight lines, [latex]\|A\|[/latex]. Example 1: Finding the Determinant of a 2 × 2 Matrix Find the determinant of the given matrix. Solution [latex]\begin{array}{l}\mathrm{det}\left(A\right)=\|\begin{array}{cc}5& 2\\ -6& 3\end{array}\|\hfill \\ =5\left(3\right)-\left(-6\right)\left(2\right)\hfill \\ =27\hfill \end{array}[/latex] Using Cramer’s Rule to Solve a System of Two Equations in Two Variables We will now introduce a final method for solving systems of equations that uses determinants. Known as Cramer’s Rule, this technique dates back to the middle of the 18th century and is named for its innovator, the Swiss mathematician Gabriel Cramer (1704–1752), who introduced it in 1750 in Introduction à l’Analyse des lignes Courbes algébriques. Cramer’s Rule is a viable and efficient method for finding solutions to systems with an arbitrary number of unknowns, provided that we have the same number of equations as unknowns. Cramer’s Rule will give us the unique solution to a system of equations, if it exists. However, if the system has no solution or an infinite number of solutions, this will be indicated by a determinant of zero. To find out if the system is inconsistent or dependent, another method, such as elimination, will have to be used. To understand Cramer’s Rule, let’s look closely at how we solve systems of linear equations using basic row operations. Consider a system of two equations in two variables. We eliminate one variable using row operations and solve for the other. Say that we wish to solve for [latex]x[/latex]. If equation (2) is multiplied by the opposite of the coefficient of [latex]y[/latex] in equation (1), equation (1) is multiplied by the coefficient of [latex]y[/latex] in equation (2), and we add the two equations, the variable [latex]y[/latex] will be eliminated. [latex]\begin{array}\text{ }b_{2}a_{1}x+b_{2}b_{1}y=b_{2}c_{1} \hfill& \text{Multiply }R_{1}\text{ by }b_{2} \\−b_{1}a_{2}x−b_{1}b_{2}y=−b_{1}c_{2} \hfill& \text{Multiply }R_{2}\text{ by }−b_{2} \\ \text{______________________} \\ b_{2}a_{1}x−b_{1}a_{2}x=−b_{2}c_{1}−b_{1}c_{2}\end{array}[/latex] Now, solve for [latex]x[/latex]. Similarly, to solve for [latex]y[/latex], we will eliminate [latex]x[/latex]. Solving for [latex]y[/latex] gives Notice that the denominator for both [latex]x[/latex] and [latex]y[/latex] is the determinant of the coefficient matrix. We can use these formulas to solve for [latex]x[/latex] and [latex]y[/latex], but Cramer’s Rule also introduces new notation: [latex]D:[/latex] determinant of the coefficient matrix [latex]{D}_{x}:[/latex] determinant of the numerator in the solution of [latex]x[/latex][latex]x=\frac{{D}_{x}}{D}[/latex] [latex]{D}_{y}:[/latex] determinant of the numerator in the solution of [latex]y[/latex][latex]y=\frac{{D}_{y}}{D}[/latex] The key to Cramer’s Rule is replacing the variable column of interest with the constant column and calculating the determinants. We can then express [latex]x[/latex] and [latex]y[/latex] as a quotient of two determinants. A General Note: Cramer’s Rule for 2×2 Systems Cramer’s Rule is a method that uses determinants to solve systems of equations that have the same number of equations as variables. Consider a system of two linear equations in two variables. The solution using Cramer’s Rule is given as If we are solving for [latex]x[/latex], the [latex]x[/latex] column is replaced with the constant column. If we are solving for [latex]y[/latex], the [latex]y[/latex] column is replaced with the constant column. Example 2: Using Cramer’s Rule to Solve a 2 × 2 System Solve the following [latex]2\text{ }\times \text{ }2[/latex] system using Cramer’s Rule. Solution Solve for [latex]x[/latex]. Solve for [latex]y[/latex]. The solution is [latex]\left(2,-3\right)[/latex]. Try It 1 Use Cramer’s Rule to solve the 2 × 2 system of equations.
Consider a static, complete information game with 2 players. Strategy sets are $S_1=\{U,D\},S_2=\{l,m,r\}$. Payoffs are irrelevant to this question as I am trying to get the concept of rationalizability correct. Suppose I want to verify whether $m$ is a rationalizable strategy for player 2. Then, I want to ask the following question: $\exists \sigma_1=(q^,1-q^)\in\Delta(S_1)$ such that for $\sigma^_2=(0,1,0),$ $u_2(\sigma_1,\sigma^_2)\geq u_2(\sigma_1,\sigma_2)$ for all $\sigma_2\in\Delta(S_2)?$ Now, suppose I have payoff matrix such that I could find $(q^,1-q^)$ such that it satisfies both: (1) $u_2(\sigma_1,\sigma^_2)\geq u_2(\sigma_1,(1,0,0))$ (2) $u_2(\sigma_1,\sigma^_2)\geq u_2(\sigma_1,(0,0,1))$. This means, I could find a valid range of $q^$ such that for player 2, choosing $m$ provides a weakly better payoff for her compared to the degenerate (i.e. pure) strategies of $l$ or $r$. My question is: If I could find such $q^$ that satisfies both (1),(2), then I do not have to check for any other strategy profiles in $\Delta(S_2)$, that is any convex combo of $(1,0,0)$ and $(0,0,1)$? My intuition is that for any $\alpha\in[0,1]$, I could simple have: (1)' $\alpha u_2(\sigma_1,\sigma^_2)\geq \alpha u_2(\sigma_1,(1,0,0))$ (2)' $(1-\alpha)u_2(\sigma_1,\sigma^_2)\geq (1-\alpha)u_2(\sigma_1,(0,0,1))$. and show that (1)'+(2)'implies the degenerate strategy $m$ for player 2 is a best response to some belief, $\sigma_1\in\Delta(S_1)$. Hence, the bottom line is (1),(2) is sufficient, and I do not have to check the convex combo of the two other pure strategies.
1. The rupee/coin changing machine at a bank has a flaw. It gives 10 ten rupee notes if you put a 100 rupee note and 10 one rupee coins if you insert a 10 rupee note but gives 10 hundred rupee notes when you put a one rupee coin!Sivaji, after being ruined by his rivals in business is left with a one rupee coin and discovers the flaw in the machine by accident. By using the machine repeatedly, which of the following amounts is a valid amount that Sivaji can have when he gets tired and stops at some stage (assume that the machine has an infinite supply of notes and coins): a. 26975 b. 53947 c. 18980 d. 33966 Explanation: The process works like this: Rs.1 Coin ⇒ 10 × 100 = Rs.1000 Rs.100 ⇒ 10 × 10 Rs.10 ⇒ 1 × 10 Sivaji gets more money when he inserts a rupee coin only. For each rupee coin he gets his money increased by 1000 times. Suppose he inserted 1 rupee coin and got 1000 rupees and again converted this into coins. So he ends up with 1000 coins. Now of this, he inserts one coin, he gets 1000. So he has 1999 with him. Now if he inserts another coin, he has 1998 + 1000 = 2998. Now each of these numbers are in the form of 999n + 1. So option B can be written as 54 × 999 + 1. 2. Seven movie addicts- Guna, Isha, Leela, Madhu, Rinku, Viji and Yamini attend a film festival. Three films are shown, one directed by Rajkumar Hirani ,one by S.Shankar,and one by Mani Ratnam. Each of the film buffs sees only one of the three films. The films are shown only once, one film at a time. The following restrictions must apply :- Exactly twice as many of the film buffs sees the S.shankar film as see the Rajkumar Hirani film.- Guna and Rinku do not see the same film as each other.- Isha and Madhu do not see same film as each other.- Viji and Yamini see the same film as each other.- Leela sees the S.Shankar film.- Guna sees either the Rajkumar Hirani film or the Mani Ratnam film.Which one of the following could be an accurate matching of the film buffs to films ?(A) Guna: the S.Shankar film; Isha: the Mani Ratnam film; Madhu: the S.Shankar film(B) Guna: the Mani Ratnam film; Isha: the Rajkumar Hirani film; Viji: the Rajkumar Hirani film(C) Isha : the S.Shankar film; Rinku: the Mani Ratnam film; Viji: the Rajkumar Hirani film(D) Madhu: the Mani Ratnam film; Rinku: the Mani Ratnam film; Viji: the Mani Ratnam film a. A b. C c. D d. B Explanation: Guna × Rinku Isha × Madhu (Viji + Yamini) Leela - Film: Shankar Guna = RKH/Mani Ratnam The following options are possible: RKH Shankar Mani Ratnam 1 2 4 2 4 1 Option A: Guna should not watch Shankar's Film. So ruled out Option B: RKH Shankar Mani Ratnam Isha _ Guna Viji _ _ Option C: RKH Shankar Mani Ratnam Viji Isha Rinku Yamini Leela _ Option D: RKH Shankar Mani Ratnam Guna Leela Madhu _ Isha Rinku _ _ Viji and Yamini 3. Which of the following numbers must be added to 5678 to give a remainder of 35 when divided by 460? a. 955 b. 980 c. 797 d. 618 Explanation: 5678 - 35 + (one of the answer option) should be divisible by 460. Only option C satisfies. 4. Find the probability that a leap year chosen at random will have 53 Sundays. a. 1/7 b. 2/7 c. 1/49 d. 3/7 Explanation: A leap year has 366 day which is 52 full weeks + 2 odd days. Now these two odd days may be (sun + mon), (mon + tue), .... (Sat + sun). Now there are total 7 ways. Of which Sunday appeared two times. So answer 2/7 5. An ant starts moving on the mesh shown below along the wires towards a food particle.If the ant is at the bottom-left corner of cell A and the food is at the top-right corner of cell F, then find the number of optimal routes for the ant. a. 13884156 b. 3465280 c. 4368 d. 6748 Explanation: (Please read "Counting" to understand this question: Click here ) Total ways to move from A to the junction: There are 13 upward ways, 3 right side ways this Ant can move. Now these 16 ways may be in any order. So number of ways of arrangements = $\dfrac{{16!}}{{13! \times 3!}}$ = 560 Similarly, from the junction to F, Total 12 upward ways and 5 right-side ways. These 17 ways can be in any order. So Total ways = $\dfrac{{17!}}{{12! \times 5!}}$ = 6188 Total ways to move from A to F = 560 × 6188 = 3465280 6. You have been given a physical balance and 7 weights of 52, 50, 48, 44, 45, 46 and 78 kgs. Keeping weights on one pan and object on the other, what is the maximum you can weigh less than 183 kgs. a. 180 b. 181 c. 182 d. 178 Explanation: 52+50+78 = 180 7. Two consecutive numbers are removed from the progression 1, 2, 3, ...n. The arithmetic mean of the remaining numbers is 26 1/4. The value of n is a. 60 b. 81 c. 50 d. Cannot be determined Explanation: As the final average is 105/4, initial number of pages should be 2 more than a four multiple. So in the given options, we will check option C. Total = $\dfrac{{n(n + 1)}}{2} = \dfrac{{50 \times 51}}{2} = 1275$ Final total = $48 \times \dfrac{{105}}{4} = 1260$ So sum of the pages = 15. The page numbers are 7, 8 8. You need a 18% acid solution for a certain test, but your supplier only ships a 13% solution and a 43% solution. You need 120 lts of the 18% acid solution. the 13% solution costs Rs 82 per ltr for the first 67 ltrs, and Rs 66 per ltr for any amount in access of 67 ltrs. What is the cost of the 13% solution you should buy? a. 8002 b. 7012 c. 7672 d. 7342 Explanation: Let us assume we need "a" liters of 13% acid solution and "b" liters of 43% acid solution. Now $ \Rightarrow 18 = \dfrac{{a \times 13 + b \times 43}}{{a + b}}$ $ \Rightarrow \dfrac{a}{b} = \frac{5}{1}$ So we need 100 liters of 13% acid solution, and 20 liters of 18% acid solution. Final cost = 82 × 67 + 66 × 33 = 7672 9. A spherical solid ball of radius 58 mm is to be divided into eight equal parts by cutting it four times longitudinally along the same axis.Find the surface area of each of the final pieces thus obtained( in mm^2) ? (where pi= 22/7) a. 3365pi b. 5046pi c. 1682pi d. 3346pi Explanation: If a sphere is cut into 8 parts longitudinally, It something looks like below Now We have to find the surface area of one piece. This is ${\dfrac{1}{8}^{th}}$ of the initial sphere + 2 × area of the half circle = $\dfrac{1}{8}\left( {4\pi {r^2}} \right) + \pi \times {r^2}$= $\dfrac{1}{8}\left( {4 \times \pi \times {{58}^2}} \right) + \pi \times {58^2}$ = 5046pi 10. There is a lot of speculation that the economy of a country depends on how fast people spend their money in addition to how much they save. Auggie was very curious to test this theory.Auggie spent all of his money in 5 stores. In each store, he spent Rs.4 more than one-half of what he had when he went in. How many rupees did Auggie have when he entered the first store? a. 248 b. 120 c. 252 d. 250 Explanation: As he has spent all his money, He must spend Rs.8 in the final store. a simple equation works like this. Amount left = $\dfrac{1}{2}x - 4$ For fifth store this is zero. So x = 8. That means he entered fifth store with 8. Now for fourth store, amount left = 8 so $\dfrac{1}{2}x - 4 = 8 \Rightarrow $ x = 24 For third store, amount left = 24 so $\dfrac{1}{2}x - 4 = 24 \Rightarrow $ x = 56 For Second store, amount left = 56 so $\dfrac{1}{2}x - 4 = 56 \Rightarrow $ x = 120 For first store, amount left = 120 so $\dfrac{1}{2}x - 4 = 120 \Rightarrow $ x = 248 So he entered first store with 248. 11. A sudoku grid contains digits in such a manner that every row, every column, and every 3x3 box accommodates the digits 1 to 9, without repetition. In the following Sudoku grid, find the values at the cells denoted by x and y and determine the value of 6x + 15y. a. 87 b. 75 c. 66 d. 99 Explanation: 6x + 15y = 75 12. In how many ways can the letters of the english alphabet be arranged so that there are seven letter between the letters A and B, and no letter is repeated a. 24P7 * 2 * 18! b. 36 * 24! c. 24P7 * 2 * 20! d. 18 * 24! Explanation: We can fix A and B in two ways with 7 letters in between them. Now 7 letters can be selected and arranged in between A and B in ${}^{24}{P_7}$ ways. Now Consider these 9 letters as a string. So now we have 26 - 9 + 1 = 18 letters These 18 letters are arranged in 18! ways. So Answer is 2 x ${}^{24}{P_7}$ x 18! Infact, 2 x ${}^{24}{P_7}$ x 18! = 36 x 24!. So go for Option B as it was given as OA. 13. A certain function f satisfies the equation f(x)+2f(6-x)=x for all real numbers x. The value of f(1) is a. 1 b. 2 c. 3 d. Cannot be determined Explanation: Put x =1 ⇒ f(1)+2f(6-1) = 1 ⇒ f(1) + 2f(5) = 1 Put x = 5 ⇒ f(5)+2f(6-5) = 5 ⇒ f(5) + 2f(1) = 5 Put f(5) = 5 - 2f(1) in the first equation ⇒ f(1) + 2(5 - 2f(1)) = 1 ⇒ f(1) + 10 - 4f(1) = 1 ⇒ f(1) = 3 14. Professor absentminded has a very peculiar problem, in that he cannot remember numbers larger than 15. However, he tells his wife, I can remember any number up to 100 by remembering the three numbers obtained as remainders when the number is divided by 3, 5 and 7 respectively. For example (2,2,3) is 17. Professor remembers that he had (1,1,6) rupees in the purse, and he paid (2,0,6) rupees to the servant. How much money is left in the purse?A. 59 B. 61 C. 49 D. 56 Answer: D Explanation: Let the money with the professor = N Then N = 3a +1 = 5b + 1 = 7c + 6. Solving the above we get N = 181 (Explanation: See LCM formula 1 and 2: Click here) When a number is divided by several numbers and we got same remainder in each case, then the general format of the number is LCM (divisors).x + remainder. In this case 3, 5 are divisors. So N = 15x + 1. Now we will find the number which satisfies 15x + 1 and 7c + 6. ⇒ 15x + 1 = 7c + 6 ⇒ c = $\dfrac{{15x - 5}}{7}$ ⇒ c = $2x + \dfrac{{x - 5}}{7}$ Here x = 5 satisfies. So least number satisfies the condition is 5(15)+1 = 76. (x = 12 also satisfies condition. So substituting in 15x + 1 we get, 181 which satisfies all the three equations but this is greater than 100) Similarly Money given to servant = M = 3x + 2 = 5y = 7z + 6 Solving we get M = 25. (125 also satisfies but this is next number) Now N - M = 56 15. The sum of three from the four numbers A, B, C, D are 4024, 4087, 4524 and 4573. What is the largest of the numbers A, B, C, D?a. 1712 b. 1650 c. 1164 d. 1211 Answer: a Explanation: a+b+c=4024 b+c+d= 4087 a+c+d=4524 a+b+d=4573 Combining all we get 3(a+b+c+d) = 17208 ⇒ a + b + c +d = 3736 Now we find individual values. a = 1649, b = 1212, c = 1163, d = 1712. So maximum value is 1712. 16. Anand packs 304 marbles into packets of 9 or 11 so that no marble is left. Anand wants to maximize the number of bags with 9 marbles. How many bags does he need if there should be atleast one bag with 11 marblesa. 33 b. 32 c. 31 d. 30 Answer: B Explanation: Given 9x + 11y = 304. x = $\dfrac{{304 - 11y}}{9}$ = $33 + \dfrac{{7 - 2y}}{9} - y$ So y = - 1 satisfies. Now x = 35. But y cannot be negative. Now other solutions of this equation will be like this. Increase or decrease x by 11, decrease or increase y by 9. So we have to maximise x. next solution is x = 24 and y = 8. So bags required are 32. 17. When Usha was thrice as old as Nisha, her sister Asha was 25, When Nisha was half as old as Asha, then sister Usha was 34. their ages add to 100. How old is Usha?a. 37 b. 44 c. 45 d. 40 Answer: D Explanation: Let the age of Usha is 3x then Nisha is x and Asha is 25 Also Usha 34, Nisha y, and Asha 2y. We know that 3x - 34 = x - 2y = 25 - 2y Solving above three equations we get x = 9, y = 16 Their ages are 34, 16, 32. whose sum = 82. So after 18 years their ages will be equal to 100. So Usha age is 34 + 6 = 40 18. Find the number of zeroes in the expression 15322522407598112125a. 12 b. 9 c. 14 d. 7 Answer: B Explanation: Maximum power of 5 in the above expression can be calculated like this. Count all the powers of 5 in the above expression. So number of zeroes are 9. (Read this chapter) 19. Two vehicles A and B leaves from city Y to X. A overtakes B at 10:30 am and reaches city X at 12:00 pm. It waits for 2 hrs and return to city Y. On its way it meets B at 3:00 pm and reaches city Y at 5:00 pm. B reaches city X, waits for 1hr and returns to city Y. Afer how many hours will B reach city Y from the time A overtook him fro the first time?a. 50 hrs b. 49.5 hrs c. 41.5 hrs d. 37.5 hrs Answer: C Explanation: Let us understand the diagram. Vehicle A overtaken B at 10.30 am and reached X at 12 pm. It started at 2 pm and met B at 3 pm at Q. It means, Vehicle A took one hour to cover distance 'n', So it should be at Q at 11 pm. It is clear that Vehicle A takes 0.5 hour to cover distance 'm'. Now vehicle B travelled from 10.30 am to 3 pm to meet A. So it took 4.5 hours to cover m. So Speeds ratio = 4.5 : 0.5 = 9 : 1. Now Vehicle A took a total of 1.5 + 3 = 4.5 hours to travel fro P to Y. So It must take 4.5 × 9 + 1 = 41.5 hours 20. Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the twocircles is:a. (π/2) – 1 b. 4 c. √2 – 1 d. √5 Answer: Explanation: We have to find the area of the blue shaded one and double it to get the area common to the both. Now this can be calculated as Area of the sector OAB - Area of the Triangle OAB. As OA and OB are perpendicular, area of the sector OAB = $\dfrac{{90}}{{360}}\pi {(1)^2}$ = $\dfrac{\pi }{4}$ Area of the triangle OAB = $\dfrac{1}{2} \times 1 \times 1$ = $\dfrac{1}{2}$ Area common to both = $2\left( {\dfrac{\pi }{4} - \dfrac{1}{2}} \right) = \dfrac{\pi }{2} - 1$
Abstract The differential branching fraction of the decay Λ0b→Λμ+μ− is measured as a function of the square of the dimuon invariant mass, q2. A yield of 78±12 Λ0b→Λμ+μ− decays is observed using data, corresponding to an integrated luminosity of 1.0\,fb−1, collected by the LHCb experiment at a centre-of-mass energy of 7\,TeV. A significant signal is found in the q2 region above the square of the J/ψ mass, while at lower-q2 values upper limits are set on the differential branching fraction. Integrating the differential branching fraction over q2, while excluding the J/ψ and ψ(2S) regions, gives a branching fraction of $\BF($\Lambda_b^0\rightarrow\Lambda\mu^+\mu^-$)=(0.96\pm 0.16\stat\pm 0.13\syst\pm 0.21 (\mathrm{norm}))\times 10^{-6}$, where the uncertainties are statistical, systematic and due to the normalisation mode, $$\Lambda_b^0\rightarrow J/psi\Lambda$, respectively. Abstract The differential branching fraction of the decay Λ0b→Λμ+μ− is measured as a function of the square of the dimuon invariant mass, q2. A yield of 78±12 Λ0b→Λμ+μ− decays is observed using data, corresponding to an integrated luminosity of 1.0\,fb−1, collected by the LHCb experiment at a centre-of-mass energy of 7\,TeV. A significant signal is found in the q2 region above the square of the J/ψ mass, while at lower-q2 values upper limits are set on the differential branching fraction. Integrating the differential branching fraction over q2, while excluding the J/ψ and ψ(2S) regions, gives a branching fraction of $\BF($\Lambda_b^0\rightarrow\Lambda\mu^+\mu^-$)=(0.96\pm 0.16\stat\pm 0.13\syst\pm 0.21 (\mathrm{norm}))\times 10^{-6}$, where the uncertainties are statistical, systematic and due to the normalisation mode, $$\Lambda_b^0\rightarrow J/psi\Lambda$, respectively. Additional indexing
Thermalization Time Bounds for Pauli Stabilizer Hamiltonians 103 Downloads Citations Abstract We prove a general lower bound to the spectral gap of the Davies generator for Hamiltonians that can be written as the sum of commuting Pauli operators. These Hamiltonians, defined on the Hilbert space of N-qubits, serve as one of the most frequently considered candidates for a self-correcting quantum memory. A spectral gap bound on the Davies generator establishes an upper limit on the life time of such a quantum memory and can be used to estimate the time until the system relaxes to thermal equilibrium when brought into contact with a thermal heat bath. The bound can be shown to behave as ${\lambda \geq \mathcal{O}(N^{-1}\exp(-2\beta \, \overline{\epsilon}))}$, where ${\overline{\epsilon}}$ is a generalization of the well known energy barrier for logical operators. Particularly in the low temperature regime we expect this bound to provide the correct asymptotic scaling of the gap with the system size up to a factor of N −1. Furthermore, we discuss conditions and provide scenarios where this factor can be removed and a constant lower bound can be proven. Preview Unable to display preview. Download preview PDF. References 1. 2. 3. 4.Gottesman, D.: Stabilizer Codes and Quantum Error Correction. arXiv preprint arXiv:quant-ph/9705052 (1997) 5. 6.Xiao-Gang, W.: Quantum field theory of many-body systems from the origin of sound to an origin of light and electrons. Oxford University Press Inc., New York. ISBN 019853094, 1 (2004)Google Scholar 7. 8. 9. 10. 11. 12.Davies, E.B.: Quantum theory of open systems. IMA, London (1976)Google Scholar 13. 14. 15. 16. 17. 18. 19.Brown, B.J., Loss, D., Pachos, J.K., Self, C.N., Wootton, J.R.: Quantum Memories at Finite Temperature. arXiv preprint arXiv:1411.6643 (2014) 20. 21. 22.Yoshida, B.: Violation of the Arrhenius Law Below the Transition Temperature. arXiv preprint arXiv:1404.0457 (2014) 23. 24. 25. 26. 27. 28. 29.Weiss, U.: Quantum Dissipative Systems, vol. 10. World Scientific, Singapore (1999)Google Scholar 30. 31. 32. 33. 34.Fuchs, C.A., De Van Graaf, J: Cryptographic distinguishability measures for quantum-mechanical states. Inf. Theory IEEE Trans. 45(4), 1216–1227 (1999)Google Scholar 35. 36.Diaconis, P., Stroock, D.: Geometric bounds for eigenvalues of markov chains. Ann. Appl. Probab. pp. 36–61 (1991)Google Scholar 37.Fill, J.A.: Eigenvalue bounds on convergence to stationarity for nonreversible markov chains, with an application to the exclusion process. Ann. Appl. Probab. pp. 62–87 (1991)Google Scholar 38. 39. 40. 41. 42. 43. 44. 45. 46.Bhatia, R.: Matrix Analysis, vol. 169. Springer, New York (1997)Google Scholar 47. 48.Chung Fan, R.K.: Spectral Graph Theory, vol. 92. American Mathematical Soc, New York (1997)Google Scholar 49. 50. 51.Peierls, R.: On ising’s model of ferromagnetism. In: Mathematical Proceedings of the Cambridge Philosophical Society, 32:477–481, 10 (1936)Google Scholar 52. 53. 54.
To round off our series on round objects (see the first and second posts), let’s compute the sphere’s surface area. We can compute this in the same way we related the area and circumference of a circle two weeks ago. Approximate the surface of the sphere with lots of small triangles, and connect these to the center of the sphere to create lots of triangular pyramids. Each pyramid has volume $\frac{1}{3}(\text{area of base})(\text{height})$, where the heights are all nearly $r$ and the base areas add to approximately the surface area. By using more and smaller triangles these approximations get better and better, so the volume of the sphere is $$\frac{4}{3}\pi r^3 = \frac{1}{3}(\text{surface area})\cdot r,$$ meaning the surface area is $4\pi r^2$. (This and previous arguments can be made precise with the modern language of integral calculus.) Here’s an elegant way to rephrase this result: The surface area of a sphere is equal to the area of the curved portion of a cylinder that exactly encloses the sphere. In fact, something very surprising happens here!: Archimedes’ Hat-Box Theorem: If we draw any two horizontal planes as shown below, then the portions of the sphere and the cylinder between the two planes have the same surface area. We can prove this with (all!) the methods in the last few posts; here’s a quick sketch. To compute the area of the “spherical band” (usually called a spherical zone), first consider the solid spherical sector formed by joining the spherical zone to the center: By dividing this into lots of triangular pyramids as we did with the sphere above, we can compute the area of the spherical zone by instead computing the sector’s volume. This volume can be computed by breaking it into three parts: two cones and the spherical segment between the two planes (on the left of the next figure). Compute the volume of the spherical segment by comparing (via Cavalieri’s Principle) to the corresponding part of the vase (from the previous post), which can be expressed with just cylinders and cones. See if you can fill in the details!
Newform invariants Coefficients of the $q$-expansion are expressed in terms of a basis $1,\beta_1,\beta_2$ for the coefficient ring described below. We also show the integral $q$-expansion of the trace form. Basis of coefficient ring in terms of a root $\nu$ of $x^{3}\mathstrut -\mathstrut $ $x^{2}\mathstrut -\mathstrut $ $3$ $x\mathstrut +\mathstrut $ $1$: $\beta_{0}$ $=$ $ 1 $ $\beta_{1}$ $=$ $ \nu $ $\beta_{2}$ $=$ $ \nu^{2} - \nu - 2 $ $1$ $=$ $\beta_0$ $\nu$ $=$ $\beta_{1}$ $\nu^{2}$ $=$ $\beta_{2}\mathstrut +\mathstrut $ $\beta_{1}\mathstrut +\mathstrut $ $2$ For each embedding $\iota_m$ of the coefficient field, the values $\iota_m(a_n)$ are shown below. For more information on an embedded modular form you can click on its label. This newform does not admit any (nontrivial) inner twists. $ p $ Sign $2$ $1$ $3$ $-1$ $17$ $1$ $59$ $1$ This newform can be constructed as the intersection of the kernels of the following linear operators acting on $S_{2}^{\mathrm{new}}(\Gamma_0(6018))$: $T_{5}^{3} $ $\mathstrut -\mathstrut 4 T_{5}^{2} $ $\mathstrut -\mathstrut 4 T_{5} $ $\mathstrut +\mathstrut 20 $ $T_{7}^{3} $ $\mathstrut +\mathstrut T_{7}^{2} $ $\mathstrut -\mathstrut 13 T_{7} $ $\mathstrut -\mathstrut 23 $
What 'exactly' is power spectral density for discrete signal? I was always under the assumption that taking the Fourier transform of the signal, and then the ratio of desired freq range magnitude over entire freq range gives the power ratio for that freq range which is the same as power spectral density. Is that wrong? Reading a student paper got me confused as it says to computes PSD and then 'absolute and relative spectral powers in desired bands' as well. Are they different? If yes, how does one compute it? I have no idea what your calculation of power spectral density gives since I cannot understand it. If a signal $x(t)$ has Fourier transform $X(f)$, its power spectral densityis $\|X(f)\|^2 = S_X(f)$. The absolute spectral power in the band of frequencies from $f_0$ Hz to $f_1$ Hz is the total power in that band of frequencies, that is, the total power delivered at the output of an ideal (unit gain) bandpass filterthat passes all frequencies from $f_0$ Hz to $f_1$ Hz and stops everythingelse. Thus,$$\text{Absolute Spectral Power in Band} = \int_{-f_1}^{-f_0} S_X(f)\,\mathrm df + \int_{f_0}^{f_1} S_X(f)\,\mathrm df.$$The relative spectral power measures the ratio of the total power in the band (i.e., absolute spectral power)to the total power in the signal. Thus,$$\text{Relative Spectral Power in Band} = \frac{\displaystyle\int_{-f_1}^{-f_0} S_X(f)\,\mathrm df + \int_{f_0}^{f_1} S_X(f)\,\mathrm df}{\displaystyle\int_{-\infty}^{\infty} S_X(f)\,\mathrm df}.$$
I am trying to diagonalize hubbard model in real and K-space for spinless fermions. Hubbard model in real space is given as: $$H=-t\sum_{<i,j>}(c_i^\dagger c_j+h.c.)+U\sum (n_i n_j)$$ I solved this Hamiltonian using MATLAB. It was quite simple. t and U are hopping and interaction potentials. c, $c^\dagger$ and n are annihilation, creation and number operators in real space respectively. The first term is hopping and 2nd is two-body interaction term. is indicating that hopping is possible only to nearest neighbors. To solve this Hamiltonian I break it down as: (for M=# for sites=2 and N=# of particles=1) $$H=-t (c_1^\dagger c_2 + c_2^\dagger c_1)+U n_1 n_2 $$ The basis vectors that can be written in binary notation are: 01, 10 Using t=1, U=1 and above basis the Hamiltonian can be written as: H=[0 -1 -1 0] That is correct. I checked with different values of M,N,U and t this MATLAB program give correct results. In K-spaceTo diagonalize this Hamiltonian in K-space we can perform Fourier transform of operators that will results in:$$H(k)=\sum_k \epsilon_k n_k + U / L \sum_ {k,k,q} c_k^\dagger c_{k-q} c_{k'}^\dagger c_{k'+q}$$Where $\epsilon_k=-2tcos(k)$.To diagonalize this Hamiltonian I make basis by taking k-points between -pi and +pi (first brillion zone) i.e. for M=2 and N=1 allowed k-points are: [0,pi] Here first term is simple to solve and I have solved it already but I can't solve the 2nd term as it includes summation over three variables. To get in more details of my attempt you can see Hubbard model diagonalization in 1D K-space for spinless fermions My question:1. What is physical significance of 2nd term in H(k) given above? I mean what is it telling about which particles are hopping from where to where? What are limits on q, k and k'? 2. If you think any article can help me with this problem then please tell me about that. Thanks a lot.
[latex] \quicklatex{color=”#000000″ size=100} {A^{d}\alpha}\oint_r u_i \cdot T!} [/latex] Yesterday, I published a post about opamps here on the blog. This post utilized a new feature here at Adafruit: rendered LaTeX equations. For those that are unfamiliar, LaTeX is a markup language for the TeX system, originally developed by legendary computer scientist Donald Knuth. LaTeX (pronounced “Lay-Tek”), is used by scientists, educators and engineers around the world to format equations so that they look nice and neat, and are easy to read. About a month ago, in the course of originally drafting that opamp article, I started looking around at LaTeX plugins for WordPress. There are several of these available. All of them have their strengths and weaknesses, but eventually I settled on WP-QuickLaTeX by Pavel Holoborodko, Dmitriy Gubanov and Kim Kirkpatrick. WPQL supports automatic equation numbering, has built in tikz and pgfplots support, can render alpha-channel PNGs, and supports LaTeX markup in blog comments, which means that the conversation can go both ways. TeX and LaTeX have been around a long time, so there is information all over the place about how to use it, but here are a few tutorials (1, 2) What this means for you is that we can more easily do technical posts on the blog, and drop transfer functions like this: [latex] \begin{equation} \frac{V_o}{V_i} = \frac{(g_m R_{g})^2} { (s C R_f)^2 + 2 s C R_{g} g_m + (R_{g} g_m)^2} \end{equation} [/latex] Or plots like this: [latex] \begin{tikzpicture} [+preamble] \usepackage{pgfplots} \pgfplotsset{compat=newest} [/preamble] \begin{axis} \addplot3[surf,domain=0:360,samples=40] {sin(2x)cos(y)}; \end{axis} \end{tikzpicture} [/latex] We’re super-excited to have this new functionality here, and we hope you are too. If you want to take LaTeX for a test drive in the comments, you can use the ![latex] and [/latex] tags at the beginning and end of your LaTeX statements. If you want to make sure your code works before you post it, you can test it at quicklatex.com (include the preamble under “choose options” — thanks, zerth!) Try it out now!
1. Nope. For example, in Minkowski spacetime, consider the vector $(v^\mu)=(1,0,0,1)$, which is a null vector. The vector $ (u^\mu)=(0,1,0,0) $ is orthogonal to it, yet, it is not a null vector. 2. I don't understand this question. 3. Consider a null geodesic with tangent vector $u^\mu$ ($u^\mu u_\mu$=0). Let $\lambda$ be the parameter along the null geodesic. Let $\Sigma_p<T_pM$ be the orthogonal complement to $u^\mu$ at $p\in M$. Note that because $u^\mu$ is a null vector, it is orthogonal to itself, hence $u_p\in\Sigma_p$. Let us choose two additional vectors in $\Sigma_p$, $e^\mu_1$ and $e^\mu_2$. We can choose these vectors such that $e^\mu_Ae^\nu_Bg_{\mu\nu}=\delta_{AB}$ ($A,B=1,2$), and because $u^\mu$ is a normal vector, we have $u^\mu e^\nu_A g_{\mu\nu}=0$. The line element at $p$ can be expressed as $$ ds^2(p)=g_{\mu\nu}(p)u^\mu u^\nu d\lambda^2+2g_{\mu\nu}(p)e^\mu_A u^\nu d\sigma^Ad\lambda+g_{\mu\nu}(p)e^\mu_A e^\nu_B d\sigma^A d\sigma^B, $$ where $\sigma^A$ are some coordinates for which at $p$, $e^\mu_A$ are the coordinate basis vectors. Writing in the relations between the basis vectors gives $$ ds^2(p)=g_{\mu\nu}(p)e^\mu_A e^\nu_B d\sigma^A d\sigma^B\equiv \delta_{AB}d\sigma^A d\sigma^B, $$since the other contractions are all vanishing. We can then see that if you make an infinitesimal displacement $d\xi=(d\lambda,d\sigma^1,d\sigma^2)$ that is orthogonal to the null curve, the contribution from $d\lambda$ is zero, hence it doesn't matter. After all, $\lambda$ is a null parameter, along which there is vanishing arc length, and so the physical/geometric displacement corresponding to $d\xi$ depends only on $d\sigma^1$ and $d\sigma^2$, which is why the metric is effectively two dimensional.
Your confusion really just comes down to understanding the notation that is widely used for partial derivatives. For simplicity, I'll restrict the discussion to a system with one coordinate degree of freedom $x$. In this case, the Lagrangian is a real valued function of two real variables which we suggestively label by the symbols $x$ and $\dot x$. Mathematically, we would write $L:U\to\mathbb R$ where $U\subset \mathbb R^2$. Let's consider the simple example$$ L(x, \dot x) = ax^2+b\dot x^2$$When we write the expression$$ \frac{\partial L}{\partial \dot x}(x, \dot x)$$this is an instruction to differentiate the function $L$ with respect to its second argument (because we labeled the second argument $\dot x$) and then to evaluate the resulting function on the pair $(x, \dot x)$. But we just as well could have written$$ \partial_2L(x, \dot x)$$To represent the same expression. Both of these expressions simply mean that we imagine holding the first argument of the function constant, and we take the derivative of the resulting function with respect to what remains. In the case above, this therefore means that$$ \frac{\partial L}{\partial\dot x}(x, \dot x) = 2b\dot x$$because $x$ labels the first argument, and taking a partial derivative with respect to the second argument means that we treat $x$ like a constant whose derivative is therefore $0$. It it in this sense that the partial of $x^2$ with respect to $\dot x$ is zero. So to recap, when we are taking these derivatives, we just keep in mind that the symbols $x$ and $\dot x$ are just labels for the different arguments of the Lagrangian. You might ask, however, "if $x$ and $\dot x$ are just labels, then what relation do they have to position and velocity?" The answer is that after we have treated them as labels for the arguments of $L$ in order to take the appropriate derivatives, we then evaluate the resulting expressions on a $(x(t), \dot x(t))$, the position and velocity of a curve at time $t$, to obtain equations of motion. For example, if you take the example of $L$ that I started with, we get$$ \frac{\partial L}{\partial x}(x, \dot x) = 2 ax, \qquad \frac{\partial L}{\partial \dot x}(x, \dot x) = 2b\dot x$$now we evaluate these expressions on $(x(t), \dot x(t))$ to obtain$$ \frac{\partial L}{\partial x}(x(t), \dot x(t)) = 2 ax(t), \qquad \frac{\partial L}{\partial \dot x}(x(t), \dot x(t)) = 2b\dot x(t)$$so that the Euler-Lagrange equations become$$ 0=\frac{d}{dt}\left[\frac{\partial L}{\partial \dot x}(x(t), \dot x(t))\right] - \frac{\partial L}{\partial x}(x(t), \dot x(t))=\frac{d}{dt}(2b\dot x(t)) - 2ax(t)$$which gives$$ b\ddot x(t) = a x(t)$$Once you understand all of this, you can (and should) dispense with the long-winded notation I used here for illustrative purposes, and you should make no error in using the abbreviated notation in your original post.
A friend of mine asked me a question, which I considered trivial at first, but after a while gave rise to some doubts. For instance, we have a potential well in 1 dimension defined by $$ V(x)= \begin{cases} +\infty &\text{if}& x<0 \text{ and } x>L\\ 0 &\text{if} &0\leq x\leq L \end{cases} $$ We know the wave function that describes the particle in the potential at a given energy level $$ E_n=\frac{\hbar^2\pi^2n^2}{2mL^2} $$ Now if we take the state at the energy level $E_2$ we have a wavefunction that behaves like $\psi_2\sim\sin(\frac{2\pi x}{L})$. We are interested in the probability density, so we take the square modulus, which would be $0$ at $L/2$. According to this fact I would say that it's impossible to find the particle in the position $L/2$, which can be said as: the event: "find the particle at $L/2$" is impossible. The problem is that probability tells me that the fact that the probability is zero doesn't mean that the event is impossible. Of course to get the probability I should integrate over a length, but how can I say that the event IS impossible? Isn't it? Maybe it's a stupid question and I'm missing something, but I just can't fulfill my purpose.
I am trying to understand the solution of the optimization problem in geometric PCA. In my lecture notes I have the following: Given $$\{x_{1}, ..., x_n\}\in\mathbb{R}^{D}$$ we seek a low-dim representation $$\{y_1, ..., y_n\}\in\mathbb{R}^d$$ such that: $$x_i=Uy_i+\epsilon_i\ \ where\ \ U=[u_1, ..., u_d]\in\mathbb{R}^{D\times d}$$So we have the following optimization problem: $$\underset{U,\ \{y_i\}}{min}\underset{i}{\sum}\|\|x_i-Uy_i\|\|^2\ \ s.t.\ U^TU=I_d\ \ \ \ \ \ (1)$$And following Lagrangian:$$L=\underset{i}{\sum}\|\|x_i-Uy_i\|\|^2+Tr((I_d-U^TU)\Lambda)\ \ \ \ \ \ (2)\\ where\ \Lambda=\Lambda^T\ is\ a \ matrix\ of\ Langrangian\ multipliers$$ My first problem here is that I don't understand the trace representation. Since matrix U consists of orthonormal vectors I would write it like this:$$L=\underset{i}{\sum}\|\|x_i-Uy_i\|\|^2-\underset{i,\ j=1,\ i\neq\ j}{\sum}\lambda_{ij}u_i^Tu_j+\underset{i}{\sum}\lambda_{ii}(1-u_i^Tu_i)\ \ \ \ \ \ (3)$$ How do I get from (3) to (2)? I tried to write it with indices of each element in each matrix taking U to be 3x2, but is there an easier way to prove it? After differentiation with respect to y we get:$$y_i=U^Tx_i$$Then, in order to optimize over U we substitute y into (1) and we have:$$\underset{i}{\sum}\|\|x_i-UU^Tx_i\|\|^2\underset{()}{=}\underset{i}{\sum}x_i^T(I_D-UU^T)x_i\underset{()}{=}Tr((I_D-UU^T)XX^T)\ \ \ \ \ \ (4)$$ My second problem is that I don't understand these equalities in (4). For () I get:$$\underset{i}{\sum}\|\|x_i-UU^Tx_i\|\|^2=\underset{i}{\sum}x_i^T(I_D-UU^T)^2x_i$$And the (**) I don't understand completely. After substitution of y we obtain the optimization problem:$$\underset{U}{max}\ Tr(UU^TXX^T)\ s.t.\ U^TU=I_d\ \ \ \ \ \ (5)$$for which Lagrangian is:$$L=Tr(U^TXX^TU)+Tr((I_d-U^TU)\Lambda)\ \ \ \ \ \ (6)$$Differentiation with respect to U gives:$$\Lambda=U^TXX^TU$$ And then in my lecture notes I have the following: From (1) and (2) the reconstruction error can be computed as:$$\epsilon=Tr((I_D-UU^T)XX^T)=\overset{D}{\sum}\lambda_i-\overset{d}{\sum}\lambda_i=\sum_{i=d+1}^{D}\lambda_i$$So my third question(s)is why is this the error? Why eigenvalues of those eigenvectors that aren't included in a new low dimension represent an error? I would be glad to some intuitive explanation too. I tried to read Generalized Principal Component Analysis by Vidal, Ma, Sastry but explanation there is a bit more generalized and doesn't differ much. I also tried to use The Matrix Cookbook to understand those matrix formulas but couldn't find anything helpful. Thanks.
1) Off-shell vs. on-shell action. What may cause some confusion is that Noether's theorem in its original formulation only refers to the off-shell action functional $$\tag{1} I[q;t_i,t_f]~:=~ \int_{t_i}^{t_f}\! {\rm d}t \ L(q(t),\dot{q}(t),t), $$ while Feynman's proof [1]$^1$ mostly is referring to the Dirichlet on-shell action function $$\tag{2} S(q_f,t_f;q_i,t_i)~:=~I[q_{\rm cl};t_i,t_f], $$ where $q_{\rm cl}:[t_i,t_f] \to \mathbb{R}$ is the extremal/classical path, which satisfies the equation of motion (e.o.m.) $$\tag{3}\frac{\delta I}{\delta q}~:=~\frac{\partial L}{\partial q} - \frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot{q}}~\approx~ 0,$$ with the Dirichlet boundary conditions $$\tag{4} q(t_i)~=~q_i \qquad \text{and}\qquad q(t_f)~=~q_f.$$ See also e.g. this Phys.SE answer. [Here the $\approx$ symbol means equality modulo e.o.m. The words on-shell and off-shell refer to whether e.o.m. are satisfied or not.] 2) Noether's theorem. Let us recall the setting of Noether's theorem. The off-shell action is assumed to be invariant $$\tag{5} I[q^{\prime};t^{\prime}_i,t^{\prime}_f]~=~ I[q;t_i,t_f] $$ under an infinitesimal global variation $$\tag{6} t^{\prime}-t~=~\delta t~=~\varepsilon X(t) \qquad \text{and}\qquad q^{\prime}(t^{\prime})- q(t)~=~ \delta q(t) ~=~ \varepsilon Y(t).$$ Here $X$ is a horizontal$^2$ generator, $Y$ is a generator, and $\varepsilon$ is an infinitesimal parameter that is independent of $t$. Noether's theorem. The off-shell symmetry (5) implies that the Noether charge $$\tag{7} Q~:=~p Y - h X $$ is conserved in time $$\tag{8} \frac{\mathrm dQ}{\mathrm dt}~\approx~0$$ on-shell. Here $$ \tag{9} p~:=~\frac{\partial L}{\partial \dot{q}} \qquad \text{and}\qquad h~:=~p\dot{q}-L $$ are by definition the momentum and the energy function, respectively. 3) Assumptions. Let us assume$^3$: that the Lagrangian $L(q,v,t)$ is a smooth function of its arguments $q$, $v$, and $t$. that there exists a unique classical path $q_{\rm cl}:[t_i,t_f] \to \mathbb{R}$ for each set $(q_f,t_f;q_i,t_i)$ of boundary values. that the classical path $q_{\rm cl}$ depends smoothly on the boundary values $(q_f,t_f;q_i,t_i)$. 4) Differential ${\rm d}S$. Lemma. The Dirichlet on-shell action function $S(q_f,t_f;q_i,t_i)$ is a smooth function of its arguments $(q_f,t_f;q_i,t_i)$. The differential is $$ \tag{10} {\rm d}S(q_f,t_f;q_i,t_i) ~=~ (p_f {\rm d}q_f - h_f {\rm d}t_f) -(p_i {\rm d}q_i - h_i {\rm d}t_i), $$ or equivalently, $$ \tag{11} \frac{\partial S}{\partial q_f}~=~p_f , \qquad \frac{\partial S}{\partial q_i}~=~-p_i, $$ and $$ \tag{12} \frac{\partial S}{\partial t_f}~=~-h_f, \qquad \frac{\partial S}{\partial t_i}~=~h_i. $$ Proof of eq. (11): ^ q \| ____________________________ \| \| q_cl \| \| \| \| \| \|____________________________\| \| q_cl \| \| \|------\|----------------------------\|-----> t t_i t_f Fig. 1. Two neighbouring classical paths $q_{\rm cl}$ and $q^{}_{\rm cl}$. Consider a vertical infinitesimal variation $\delta q$ between two neighbouring classical paths $q_{\rm cl}$ and $q^{}_{\rm cl}=q_{\rm cl}+\delta q$, cf. Fig.1. The change in the Lagrangian is $$\tag{13} \delta L~=~ \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}} \delta \dot{q}~\stackrel{(3)+(9)}{=}~ \frac{\delta I}{\delta q} \delta q + \frac{\mathrm d}{\mathrm dt}(p~\delta q)~\stackrel{(3)}{\approx}~\frac{\mathrm d}{\mathrm dt}(p~\delta q),$$ so that $$ \tag{14} \delta S ~\stackrel{(2)}{\approx}~\delta I ~\stackrel{(1)}{=}~ \int_{t_i}^{t_f}\! {\rm d}t ~\delta L~\stackrel{(13)}{\approx}~[p~\delta q]_{t_i}^{t_f}~=~p_f~\delta q_f- p_i~\delta q_i. $$ This proves eq. (11). Proof of eq. (12): ^ q \| q_f\|-------------------/ \| /\| \| / \| \| / \| q_f\|---------------/ \| \| /\| \| \| / \| \| \| q_cl/ \| \| \| / \| \| q_i\|----------/ \| \| \| /\| \| \| \| / \| \| \| \| / \| \| \| q_i\|------/ \| \| \| \| \| \| \| \| \|------\|---\|----\|---\|-----> t t_i t_i t_f t_f Fig. 2. The classical path $q_{\rm cl}$. Next consider the classical path $q_{\rm cl}$ between $(t_i,q_i)$ and $(t_f,q_f)$, cf. Fig. 2. Imagine that we infinitesimally extend both ends of the time interval $[t_i,t_f]$ to $[t^{}_i,t^{}_f]$, where $$\tag{15}\delta t_i~:=~t^{}_i - t_i \qquad\text{and}\qquad \delta t_f~:=~t^{}_f - t_f$$ both are infinitesimally small. This induces a change of the boundary positions (4) of the fixed classical path $q_{\rm cl}$ as follows $$\tag{16} \delta q_i~:=~ q^{}_i - q_i~=~\dot{q}_i ~\delta t_i\qquad \text{and}\qquad \delta q_f~:=~ q^{}_f - q_f~=~\dot{q}_f ~\delta t_f,$$ which are dictated by the end point velocities. We would now like to calculate the variation $$ S(q^{}_f,t^{}_f;q^{}_i,t^{}_i) - S(q_f,t_f;q_i,t_i)~=~\delta S~\stackrel{(11)}{=}~p_f \delta q_f +\frac{\partial S}{\partial t_f} \delta t_f -p_i \delta q_i + \frac{\partial S}{\partial t_i}\delta t_i $$$$\tag{17} ~\stackrel{(16)}{=}~\left(p_f \dot{q}_f +\frac{\partial S}{\partial t_f}\right) \delta t_f -\left(p_i \dot{q}_i - \frac{\partial S}{\partial t_i}\right)\delta t_i. $$ Since the new classical path is just an infinitesimal extension of the same old classical path, we may also estimate the variation as $$ \tag{18} \delta S~=~S(q^{}_f,t^{}_f;q_f,t_f)+S(q_i,t_i;q^{}_i,t^{}_i)~=~ L_f \delta t_f - L_i \delta t_i.$$ Comparing eqs. (9), (17) and (18) yields eq. (12). Corollary. The Dirichlet on-shell action along an infinitesimal path segment generated by the infinitesimal symmetry transformation (6) is proportional to the Noether charge $$ \tag{19} S(q_i+\delta q,t_i+\delta t;q_i,t_i)~=~\varepsilon Q_i.$$ Proof of the Corollary: $$ \tag{20} S(q_i+\delta q,t_i+\delta t;q_i,t_i)~\stackrel{(10)}{=}~p_i\delta q -h_i \delta t ~\stackrel{(6)}{=}~\varepsilon(p_i Y -h_i X) ~\stackrel{(7)}{=}~\varepsilon Q_i.$$ 5) Feynman's four-point argument. We are finally ready to discuss Feynman's four-point argument. ^ q \| \| A' B' \| _______________________________ \| \| q' \| \| \| \| \| \| \| \| \|_______________________________\| \| A classical/on-shell q_cl B \| \| \|----------------------------------------------------> t Fig. 3. Feynman's four points. (Note that the two horizontal and the two vertical straight ASCII lines are in general an oversimplification of the actual paths.) We start with the on-shell action $$\tag{21} S(A\to B)~=~I(A\to B)$$ for some classical path $q_{\rm cl}$ between two spacetime events $A$ and $B$. We then apply the infinitesimal transformation (6) to produce a path $q^{\prime}$ between two infinitesimally shifted spacetime events $A^{\prime}$ and $B^{\prime}$. In turn, the path $q^{\prime}$ has an off-shell action $$\tag{22} I(A^{\prime}\to B^{\prime})~=~I(A\to B)$$ equal to the original action due to the off-shell symmetry (5). 6) Method 1: Using that $q^{\prime}$ is a classical path. The off-shell symmetry (5) implies that the path $q^{\prime}$ is in fact a classical path too, cf. e.g. this Phys.SE post. So we also have $$\tag{23} S(A^{\prime}\to B^{\prime})~=~I(A^{\prime}\to B^{\prime}),$$ cf. assumption 2. Since the Dirchlet on-shell action (2) is supposed to be differentiable in its arguments, cf. the Lemma, we have $$0~\stackrel{(22)}{=}~I(A^{\prime}\to B^{\prime})-I(A\to B) ~\stackrel{(21)+(23)}{=}~S(A^{\prime}\to B^{\prime})-S(A\to B)$$$$\tag{24}~\stackrel{\text{Lemma}}{=}~S(B\to B^{\prime}) -S(A\to A^{\prime})+{\cal O}(\varepsilon^2)~\stackrel{(19)}{=}~\varepsilon (Q_f-Q_i)+{\cal O}(\varepsilon^2).$$ We arrive at the main conclusion of Noether's theorem,namely that the Noether charge is conserved, $$\tag{25} Q_f~=~Q_i.$$ 7) Method 2: Not using that $q^{\prime}$ is a classical path. This method 2 tries to follow more faithfully Feynman's proof in the sense that we would like to make sense of the shifted path $A\to A^{\prime}\to B^{\prime}\to B$. Unfortunately, the two infinitesimal pieces $A\to A^{\prime}$ and $B^{\prime}\to B$ (which we will choose to be classical paths) may correspond to constant time. [The time-integration in the definition (1) of the off-shell action $I(A\to A^{\prime}\to B^{\prime}\to B)$ would not make sense in case of constant time.] In such cases we replace Feynman's four points with six points, i.e. we extend infinitesimally the original classical path $A\to B$ to a classical path $A^{}\to B^{}$, in such a way that the two new infinitesimal paths $A^{}\to A^{\prime}$ and $B^{\prime}\to B^{}$ (which we also will choose to be classical paths) do both not correspond to constant time. ^ q \| \| A' B' \| ________________________________ \| /\| \|\ \| / \| \| \ \| / \| \| \ \| A /___\|_______________________________\|___\ B* \| A classical/on-shell q_cl B \| \| \|----------------------------------------------------> t Fig. 4. Six points. We can now apply Feynman's argument to paths between $A^{}$ and $B^{}$. Since the virtual path $A^{}\to A^{\prime}\to B^{\prime}\to B^{}$ is an infinitesimal variation of the classical path $A^{}\to A\to B\to B^{}$, we conclude that the difference $$S(A^{}\to A^{\prime})+I(A^{\prime}\to B^{\prime})+S(B^{\prime}\to B^{})$$$$-S(A^{}\to A)-S(A\to B)-S(B\to B^{})$$$$\tag{26} ~=~I(A^\to A^{\prime}\to B^{\prime}\to B^)-S(A^\to A\to B\to B^)~=~{\cal O}(\varepsilon^2)$$ cannot contain contributions linear in $\varepsilon$. We next apply the Lemma and Corollary from Section 4. The six infinitesimal classical paths mentioned so far are all described by the differential (10), which is linear and hence obeys a (co-)vector addition rule. Therefore $$\tag{27} S(A^{}\to A^{\prime})-S(A^{}\to A) +{\cal O}(\varepsilon^2)~\stackrel{(10)}{=}~S(A\to A^{\prime})~\stackrel{(19)}{=}~\varepsilon Q_i,\qquad $$$$\tag{28} S(B\to B^{}) - S(B^{\prime}\to B^{})+{\cal O}(\varepsilon^2)~\stackrel{(10)}{=}~S(B\to B^{\prime})~\stackrel{(19)}{=}~\varepsilon Q_f.\qquad $$ Comparing eqs. (21)-(22), (26)-(28), we again arrive at the main conclusion (25) of Noether's theorem. References: R.P. Feynman, The Character of Physical Law, 1965, pp. 103 - 105. -- $^1$ For Feynman's proof, see approximately 50 minutes into this video. Noether's theorem is covered in 45:25-51:27. $^2$ Feynman uses the opposite convention for horizontal and vertical than this answer. $^3$ Noether's theorem works with less assumptions, but to avoid mathematical technicalities, we impose assumption 1, 2 and 3. Note that it is easy to find examples that satisfies assumption 1 and 2, but where the classical path $q_{\rm cl}$ may jump discontinuously for varying boundary values $(q_f,t_f;q_i,t_i)$, so that assumption 3 is not satisfied.
Imagen de los τ-Productos Author Calderón Gómez, José Emilio AdvisorOrtiz Albino, Reyes M. CollegeCollege of Arts and Sciences - Art DepartmentDepartment of Mathematics TypeThesis Degree LevelM.S. Date2019-05-15 MetadataShow full item record Abstract The theory of $\tau$-factorizations, also known as theory of generalized factorizations, was developed by Anderson and Frazier in 2006. It was the result of a generalization of the comaximal factorizations by McAdams and Swam, replacing the condition of being comaximals to being related on the set of nonzero nonunits elements in the integral domain. Denote $D$ as an integral domain, $U(D)$ as the set of units of $D$ and $D^\#$ as the set of elements nonzero nonunits of $D$. The authors considered symmetric relations defined over the nonzero nonunits elements. The usual theory of factorizations came to be a particular case, where the relation used is $\tau=D^\#\times D^\#$. An expression of the form $a=\lambda a_1\cdot\cdot \cdot a_n$, where $\lambda\in U(D)$ and $a_i\tau a_j$ for all $1\leq i\neq j\leq n$, is called a $\tau$-factorizarion of $a$. Each $a_i$ is called a $\tau$-factor of $a$ and $a$ is a $\tau$-product of $a_i$. Furthermore, it is possible to obtain particular cases, such as factorizations in irreducibles elements, primals, and others, by taking $\tau=S\times S$, where $S$ is the set of irreducible elements or primals respectively. This work studied the relation $\tau_R$, where $R\subseteq D\times E$, $D$ and $E$ are integral domains, and $\tau$ is defined on $D^\#$. The relation $\tau_R$ is defined as $x\tau_R y$, if and only if there exist $a,b\in D^\#$ such that $a\tau b$, $aRx$, and $bRy$. That is, $\tau_R$ is ``the image of $\tau$ with respect to the relation $R$''. The properties of $\tau_R$ that can be inherited from $\tau$ in $\tau_R$ are analyzed . It must be clarified that although the definition is given with respect to the image of a relation, most of the work is focused in different types of functions, such as one to one and surjectives functions, homomorphisms, and others. The principal objective is to provide a way to study $\tau$-factorizations and structural properties using the images of the functions.
Let us now take a closer look at the hex-fractal we sliced last week. Chopping a level 0, 1, 2, and 3 Menger sponge through our slanted plane gives the following: This suggests an iterative recipe to generate the hex-fractal. Any time we see a hexagon, chop it into six smaller hexagons and six triangles as illustrated below. Similarly, any time we see a triangle, chop it into a hexagon and three triangles like this: In the limit, each triangle and hexagon in the above image becomes a hex-fractal or a tri-fractal, respectively. The final hex-fractal looks something like this (click for larger image): Now we are in a position to answer last week’s question: how can we compute the Hausdorff dimension of the hex-fractal? Let d be its dimension. Like last week, our computation will proceed by trying to compute the “ d-dimensional volume” of our shape. So, start with a “large” hex-fractal and tri-fractal, each of side-length 1, and let their d-dimensional volumes be h and t respectively. [1] Break these into “small” hex-fractals and tri-fractals of side-length 1/3, so these have volumes $h/3^d$ and $t/3^d$ respectively (this is how “ d-dimenional stuff” scales). Since $$\begin{gather}(\text{large hex}) = 6(\text{small hex})+6(\text{small tri}) \quad \text{and}\\ (\text{large tri}) = (\text{small hex})+3(\text{small tri}),\end{gather}$$ we find that $h=6h/3^d + 6t/3^d$ and $t=h/3^d+3t/3^d$. Surprisingly, this is enough information to solve for the value of $3^d$. [2] We find $3^d = \frac{1}{2}(9+\sqrt{33})$, so $$d=\log_3\left(\frac{9+\sqrt{33}}{2}\right) = 1.8184\ldots,$$ as claimed last week. As a final thought, why did we choose to slice the Menger sponge on this plane? Why not any of the (infinitely many) others? Even if we only look at planes parallel to our chosen plane, a mesmerizing pattern emerges: More Information It takes a bit more work to turn the above computation of the hex-fractal’s dimension into a full proof, but there are a few ways to do it. Possible methods include mass distributions [3] or similarity graphs [4]. This diagonal slice through the Menger sponge has been proposed as an exhibit at the Museum of Math. Sebastien Perez Duarte seems to have been the first to slice a Menger sponge in this way (see his rendering), and his animated cross section inspired my animation above. Thanks for reading! Notes We’re assuming that the hex-fractal and tri-fractal have the same Hausdorff dimension. This is true, and it follows from the fact that a scaled version of each lives inside the other. [↩] There are actually two solutions, but the fact that hand tare both positive rules one out. [↩] Proposition 4.9 in: Kenneth Falconer. Fractal Geometry: Mathematical Foundations and Applications.John Wiley & Sons: New York, 1990. [↩] Section 6.6 in: Gerald Edgar. Measure, Topology, and Fractal Geometry(Second Edition). Springer: 2008. [↩]
Can ordinary least squares estimation be considered an optimization technique? If so, how can I explain this? Note: From an AI perspective, supervised learning involves finding a hypothesis function $h_\vec{w}(\vec{x})$ that approximates the true nature between predictor variables and the predicted variable. Let some set of functions with the same model representation define the hypothesis space $\mathbb{H}$ (That is we hypothesise the true relationship to be a linear function of inputs or a quadratic function of inputs and so forth). The objective is to find the model $h\in\mathbb{H}$ that optimally maps inputs to outputs. This is done by application of some technique to finds optimal values for the adjustable parameters $\vec{w}$ that defines the function $h_w(\vec{x})$. In AI we call this parameter optimization. A parameter optimization technique/model inducer/learning algorithm would for example be the back propagation algorithm. OLS is used to find/estimate for $\beta$ parameters that defines the linear regression line that optimally maps predictor variables to output variables. This would be parameter optimization in the scenario above.
Definition:Path-Connected/Topology Contents Definition Let $T = \left({S, \tau}\right)$ be a topological space. Let $a, b \in S$ be such that there exists a path from $a$ to $b$. That is, there exists a continuous mapping $f: \left[{0 \,.\,.\, 1}\right] \to S$ such that $f \left({0}\right) = a$ and $f \left({1}\right) = b$. Then $a$ and $b$ are path-connected in $T$. Let $T = \left({S, \tau}\right)$ be a topological space. Let $U \subseteq S$ be a subset of $S$. Let $T' = \left({U, \tau_U}\right)$ be the subspace of $T$ induced by $U$. That is, $U$ is a path-connected set in $T$ if and only if: for every $x, y \in U$, there exists a continuous mapping $f: \left[{0 \,.\,.\, 1}\right] \to U$ such that $f \left({0}\right) = x$ and $f \left({1}\right) = y$. Let $T = \struct {S, \tau}$ be a topological space. That is, $T$ is a path-connected space if and only if: for every $x, y \in S$, there exists a continuous mapping $f: \closedint 0 1 \to S$ such that $\map f 0 = x$ and $\map f 1 = y$. Also known as Some sources do not hyphenate, but instead report this as path connected. Also see Results about path-connected spacescan be found here.
I saw this lovely tweet from PGS yesterday: Our basin studies team spotted this on fast-track imaging from Republic of Guinea. A 7.5 km diameter depression, with no salt or mobile shale, nor dissolution of fluid escape. We interpreted the structure as a complex meteorite impact crater. https://t.co/Z4TUOtsv54 #meteorite pic.twitter.com/hScJ31SoE3— PGS (@PGSNews) August 1, 2019 Kudos to them for sharing this. It’s always great to see seismic data and interpretations on Twitter — especially of weird things. And impact structures are just cool. I’ve interpreted them in seismic myself. Then uninterpreted them. I wish PGS were able to post a little more here, like a vertical profile, maybe a timeslice. I’m sure there would be tons of debate if we could see more. But not all things are possible when it comes to commercial seismic data. It’s crazy to say more about it without more data (one-line interpretation, yada yada). So here’s what I think. Impact craters are rare There are at least two important things to think about when considering an interpretation: How well does this match the model? (In this case, how much does it look like an impact structure?) How likely are we to see an instance of this model in this dataset? (What’s the base rate of impact structures here?) Interpreters often forget about the second part. (There’s another part too: How reliable are my interpretations? Let’s leave that for another day, but you can read Bond et al. 2007 as homework if you like.) The problem is that impact structures, or astroblemes, are pretty rare on Earth. The atmosphere takes care of most would-be meteorites, and then there’s the oceans, weather, tectonics and so on. The result is that the earth’s record of surface events is quite irregular compared to, say, the moon’s. But they certainly exist, and occasionally pop up in seismic data. In my 2011 post Reliable predictions of unlikely geology, I described how skeptical we have to be when predicting rare things (‘wotsits’). Bayes’ theorem tells us that we must modify our assigned probability (let’s say I’m 80% sure it’s a wotsit) with the prior probability (let’s pretend a 1% a priori chance of there being a wotsit in my dataset). Here’s the maths: $ \ \ \ P = \frac{0.8 \times 0.01}{0.8 \times 0.01\ +\ 0.2 \times 0.99} = 0.0388 $ In other words, the conditional probability of the feature being a rare wotsit, given my 80%-sure interpretation, is 0.0388 or just under 4%. As cool as it would be to find a rare wotsit, I probably need a back-up hypothesis. Now, what’s that base rate for astroblemes? (Spoiler: it’s much less than 1%.) Just how rare are astroblemes? First things first. If you’re interpreting circular structures in seismic, you need to read Simon Stewart’s paper on the subject (Stewart 1999), and his follow-up impact crater paper (Stewart 2003), which expands on the topic. Notwithstanding Stewart’s disputed interpretation of the Silverpit not-a-crater structure in the North Sea, these two papers are two of my favourites. According to Stewart, the probability P of encountering r craters of diameter d or more in an area A over a time period t years is given by: $ \ \ \ P(r) = \mathrm{e}^{-\lambda A}\frac{(\lambda A)^r}{r!} $ where $ \ \ \ \lambda = t n $ and $ \ \ \ \log n = - (11.67 \pm 0.21) - (2.01 \pm 0.13) \log d $ We can use these equations to compute the probability plot on the right. It shows the probability of encountering an astrobleme of a given diameter on a 2400 km² seismic survey spanning the Phanerozoic. (This doesn’t take into account anything to do with preservation or detection.) I’ve estimated that survey size from PGS’s tweet, and I’ve highlighted the 7.5 km diameter they mentioned. The probability is very small: about 0.00025. So Bayes tells us that an 80%-confident interpretation has a conditional probability of about 0.001. One in a thousand. So what? My point here isn’t to claim that this structure is not an astrobleme. I haven’t seen the data, I’ve no idea. The PGS team mentioned that they considered the possiblity of influence by salt or shale, and fluid escape, and rejected these based on the evidence. My point is to remind interpreters that when your conclusion is that something is rare, you need commensurately more and better evidence to support the claim. And it’s even more important than usual to have multiple working hypotheses. Last thing: if I were PGS and this was my data (i.e. not a client’s), I’d release a little cube (anonymized, time-shifted, bit-reduced, whatever) to the community and enjoy the engagement and publicity. With a proper license, obviously. References Hughes, D, 1998, The mass distribution of the crater-producing bodies. In Meteorites: Flux with time and impact effects, Geological Society of London Special Publication 140, 31–42. Davis, J, 1986, Statistics and data analysis in geology, John Wiley & Sons, New York. Stewart, SA (1999). Seismic interpretation of circular geological structures. Petroleum Geoscience 5, p 273–285. Stewart, SA (2003). How will we recognize buried impact craters in terrestrial sedimentary basins? Geology 31 (11), p 929–932.
Relation concatenation Set context $ R \in \text{Rel}(X,U) $ context $ S \in \text{Rel}(V,Y) $ definiendum $ \langle x,y \rangle \in S\circ R $ postulate $ \exists m.\ \langle x,m \rangle \in R \land \langle m,y \rangle \in S $ Discussion Concatenations/compositions are associative. A mayority of uses of the relation concatenation is when the relation is functional, i.e. one composes functions alla $g:X\to Y,\ \ f:Y\to Z$ then $f\circ g:X\to Z$ $(f\circ g)(x):=f(g(x))$ Notation If $f:X\to Y$ and $g:Y\to Z$ are functions, we'll often denote $f\circ g$ by $fg$. This convenient notation will also be used in more elaborate cases. For example, if by $f(x)$ we done the values of a function $f:X\to\mathbb R$ and $\|\cdot\|$ is the function which takes a real to its absolute value, then $\|f\|$ will denote the name of the function with values $\|f(x)\|$.
Question Let $R$ be the complex plane with the non positive real axis taken out. Find explicitly a conformal mapping $f$ of $R$ onto the unit disc $U$ such that $f(1)=0$ and $f'(1)\gt 0$ solution: We know that all automorphism of the unit disk are of the form: $$f(z)=e^{i\theta}\frac{z-\alpha}{1-\overline{\alpha}}$$ for $\|\alpha\|<1$ Let $g(z)$ be a conformal map. And let's define a conformal map $$f(z)=e^{i\theta}(\frac{g(z)-g(z_0)}{1-\overline{g(z_0)}g(z)})$$ Then $f(1)=0 \iff g(1)=g(z_0)$ And $$f'(z)=\frac{g'(1).e^{i\theta}}{1-\|g(1)\|^2}$$ From here, I dont know how to reach its result? Is this true way to solve this question? If it is true, how to continue? Thanks.
L # 1 Show that It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking. Offline Last edited by krassi_holmz (2006-03-09 02:44:53) IPBLE: Increasing Performance By Lowering Expectations. Offline It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking. Offline L # 2 If It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking. Offline Let log x = x' log y = y' log z = z'. Then: x'+y'+z'=0. Rewriting in terms of x' gives: IPBLE: Increasing Performance By Lowering Expectations. Offline Well done, krassi_holmz! It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking. Offline L # 3 If x²y³=a and log (x/y)=b, then what is the value of (logx)/(logy)? It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking. Offline loga=2logx+3logy b=logx-logy loga+3b=5logx loga-2b=3logy+2logy=5logy logx/logy=(loga+3b)/(loga-2b). Last edited by krassi_holmz (2006-03-10 20:06:29) IPBLE: Increasing Performance By Lowering Expectations. Offline Very well done, krassi_holmz! It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking. Offline L # 4 Offline It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking. Offline You are not supposed to use a calculator or log tables for L # 4. Try again! Last edited by JaneFairfax (2009-01-04 23:40:20) Offline No, I didn't I remember It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking. Offline You still used a calculator / log table in the past to get those figures (or someone else did and showed them to you). I say again: no calculators or log tables to be used (directly or indirectly) at all!! Last edited by JaneFairfax (2009-01-06 00:30:04) Offline Offline log a = 2log x + 3log y b = log x log y log a + 3 b = 5log x loga - 2b = 3logy + 2logy = 5logy logx / logy = (loga+3b) / (loga-2b) Offline Hi ganesh for L # 1 since log(a)= 1 / log(b), log(a)=1 b a a we have 1/log(abc)+1/log(abc)+1/log(abc)= a b c log(a)+log(b)+log(c)= log(abc)=1 abc abc abc abc Best Regards Riad Zaidan Offline Hi ganesh for L # 2 I think that the following proof is easier: Assume Log(x)/(b-c)=Log(y)/(c-a)=Log(z)/(a-b)=t So Log(x)=t(b-c),Log(y)=t(c-a) , Log(z)=t(a-b) So Log(x)+Log(y)+Log(z)=tb-tc+tc-ta+ta-tb=0 So Log(xyz)=0 so xyz=1 Q.E.D Best Regards Riad Zaidan Offline Gentleman, Thanks for the proofs. Regards. It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking. Offline log_2(16) = \log_2 \left ( \frac{64}{4} \right ) = \log_2(64) - \log_2(4) = 6 - 2 = 4, \, log_2(\sqrt[3]4) = \frac {1}{3} \log_2 (4) = \frac {2}{3}. \, Offline L # 4 I don't want a method that will rely on defining certain functions, taking derivatives, noting concavity, etc. Change of base: Each side is positive, and multiplying by the positive denominator keeps whatever direction of the alleged inequality the same direction: On the right-hand side, the first factor is equal to a positive number less than 1, while the second factor is equal to a positive number greater than 1. These facts are by inspection combined with the nature of exponents/logarithms. Because of (log A)B = B(log A) = log(A^B), I may turn this into: I need to show that Then Then 1 (on the left-hand side) will be greater than the value on the right-hand side, and the truth of the original inequality will be established. I want to show Raise a base of 3 to each side: Each side is positive, and I can square each side: ----------------------------------------------------------------------------------- Then I want to show that when 2 is raised to a number equal to (or less than) 1.5, then it is less than 3. Each side is positive, and I can square each side: Last edited by reconsideryouranswer (2011-05-27 20:05:01) Signature line: I wish a had a more interesting signature line. Offline Hi reconsideryouranswer, This problem was posted by JaneFairfax. I think it would be appropriate she verify the solution. It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking. Offline Hi all, I saw this post today and saw the probs on log. Well, they are not bad, they are good. But you can also try these problems here by me (Credit: to a book): http://www.mathisfunforum.com/viewtopic … 93#p399193 Practice makes a man perfect. There is no substitute to hard work All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam Offline JaneFairfax, here is a basic proof of L4: For all real a > 1, y = a^x is a strictly increasing function. log(base 2)3 versus log(base 3)5 2log(base 2)3 versus 2log(base 3)5 log(base 2)9 versus log(base 3)25 2^3 = 8 < 9 2^(> 3) = 9 3^3 = 27 < 25 3^(< 3) = 25 So, the left-hand side is greater than the right-hand side, because Its logarithm is a larger number. Offline
k-partite graph Set context $k\in\mathrm N$ context $V$ … set definiendum $\langle V,E\rangle \in \mathrm{it}(E,V) $ inclusion $ \langle V,E\rangle $ … undirected graph range $ i,j\in\{1,\dots,k\} $ range $ \bigcup_i X_i=V $ range $ \forall i,j.\ X_i\cap X_j=\emptyset $ range $ v,w\in V $ postulate $\exists X_1,\dots,X_k.\ \forall u,v.\ \{u,v\}\in E\implies \forall i.\ \neg(v\in X_i\land w\in X_i) $ Discussion The $X_i$ are the partitions of the graph and the condition says that there can be no edge within an $X_i$, i.e. there are only connection from one partition to another. One can also few the partitions as different coloring of their vertices. Theorems From any $v\in X_i$, there can be edges to only the other partitions, i.e. to at most $\|V\|-\|X_i\|$ different other vertices. If we sum up the edges for all partitions and divide the double-counting out, we find that for an $k$-partite graph $\|E\|\le \frac{1}{2}\sum_{i=1}^k \|X_i\|\cdot (\|V\|-\|X_i\|)$ Parents Subset of
It is "well known" that on the hand there is a "cubical line bundle" governing the fine structure of the Chern-Simons term in 11-dimensional supergravity/M-theory; on the other hand just such "cubical lines" on elliptic curves induce the elliptic cohomology refinement of the partition function of the heterotic string. I provide a review of that with pointers to the literature below, to be self-contained. First though my question: it is natural to speculate that these two "cubical structures" are in fact "the same", or at least closely related. In fact it seems to me that standard F-theory lore gives a way to relate them in some detail (this, too, I spell out below). But I am not really sure yet about the full story. My question is: has this or anything like this been considered/worked out anywhere? Here now more details on and pointers to what I have in mind here: Cubical Structure in M-Theory It is well known that when the higher Chern-Simons term in 11-dimensional supergravity is compactified on a 4-sphere to yield the 7-dimensional Chern-Simons theory which inside AdS7/CFT6 is dual to the M5-brane 6d (2,0)-superconformal QFT, the cup product square in ordinary differential cohomology that enters its definition is to receive a quadratic refinement. This was originally argued in (Witten 97) and then formalized and proven in (Hopkins-Singer 02). What though is the situation up in 11 dimensions before compactifying to 7-dimensions? In (DFM 03, section 9) it is claimed that the full 11-dimensional Chern-Simons term evaluated on the supergravity C-field (with its flux quantization correction, see there) indeed carries a cubic refinement. More precisely, and slightly paraphrasing, the transgression $\int_X \mathrm{CS}_{11}(\hat C)$ of the 11-dimensional Chern-Simons term of 11d SuGra to 10d spacetime X is a complex line bundle on the moduli space CField( X) of supergravity C-fields $\hat C$ is claimed to be such that its “cubical line” $\Theta^3\left(\int_X \mathrm{CS}_{11}(\hat C)\right)$ (in the notation at cubical structure on a line bundle) is the line bundle on the space of triples of C-field configurations which is given by the transgression of the three-fold cup product in ordinary differential cohomology, $\Theta^3\left(\int_X \mathrm{CS}_{11}(\hat -)\right) \simeq \int_X (\hat -)_1 \cup (\hat -)_2 \cup (\hat-)_3$ In the context of “F-theory compactifications” of M-theory, one considers C-fields on an elliptic fibration which are “factorizable fluxes”, in that their underlying cocycle $\hat C$ in ordinary differential cohomology is the cup product of a cocycle $\hat C_{fib }$ on the fiber with one $\hat C_{b}$ on the base $\hat C = \hat C_{b} \cup \hat C_{fib}$ In approaches like (GKP 12 (around p. 19), KMW 12) the C-field is factored as a cup product of a degree-2 cocycle on the elliptic fiber with a degree-2 class in the Calabi-Yau-base. This makes the component of the C-field on the elliptic fiber a complex line bundle (with connection). Notice that the space of complex line bundles on an elliptic curve is dual to the elliptic curve itself. On the other hand in e.g. (DFM 03, p.38) the factorization is taken to be that of two degree-3 cocycles in the base (which are then identified with the combined degree-3 RR-field/B-field flux coupled to the (p,q)-string) with, respectively, the two canonical degree-1 cocycles $\hat t_i$ on the elliptic fiber which are given by the two canonical coordinate functions $t_i$ (speaking of a framed elliptic curve). In this case the fiber-component of thesupergravity C-field “is” the elliptic curve-fiber, $\hat C = \hat B_{NS} \cup \hat t_1 + \hat B_{RR}\cup \hat t_2$ or equivalently: each point in the moduli space of H-flux in 10d induces an identification of the G-flux with the elliptic curve this way. This is maybe noteworthy in that when the C-field is identified with the compactification elliptic curve in this way, then the formula for $\Theta^3\left(\int_X \mathrm{CS}_{11}(\hat C)\right)$ as above is exactly that appearing in the definition of a cubical structure on a line bundle over an elliptic curve. But a “cubical” trivialization of $\Theta^3(\mathcal{O}(-\{0\}))$ over a given elliptic curve is what in (Hopkins 02, AHS01) is used to induce the sigma-orientation of the corresponding elliptic cohomology theory and in totality the string-orientation of tmf. But that is the refinement of the Witten genus, hence of the partition function of the heterotic string. Now, by the above M-theoretic equivalence, the cubical trivialization is also given by a trivialization of the topological class of the C-field. This is one way (or is at least closely related) to the trivialization of the anomaly line bundle which “sets the quantum integrand” of M-theory. So there is a curious coincidence of concepts here, which might want to become a precise identification: on the one hand there is naturally a cubical structure on a line bundle on the Chern-Simons line bundle over the moduli space of supergravity C-fields which for F-theory compactifications and factorizable flux configurations induces in particular a cubical structure on a line bundle over the compactification elliptic curve. On the other hand, the latter are the structures that enter the refined construction of the Witten genus via the string orientation of tmf. Has this been related further anywhere?
[This is the 6th post in the current series about Wythoff’s game: see posts #1, #2, #3, #4, and #5. Caveat lector: this post is a bit more difficult than usual. Let me know what you think in the comments!] Our only remaining task from last week was to prove the mysterious Covering Theorem: we must show that there is exactly one dot in each row and column of the grid (we already covered the diagonal case). Since the rows and columns are symmetric, let’s focus on columns. The columns really only care about the x-coordinates of the points, so let’s draw just these x-coordinates on the number-line. We’ve drawn $\phi,2\phi,3\phi,\ldots$ with small dots and $\phi^2,2\phi^2,3\phi^2,\ldots$ with large dots. We need to show that there’s exactly one dot between 1 and 2, precisely one dot between 2 and 3, just one between 3 and 4, and so on down the line. For terminology’s sake, break the number line into length-1 intervals [1,2], [2,3], [3,4], etc., so we must show that each interval has one and only one dot: Why is this true? One explanation hinges on a nice geometric observation: Take any small dot s and large dot t on our number-line above, and cut segment st into two parts in the ratio $1:\phi$ (with s on the shorter side). Then the point where we cut is always an integer! For example, the upper-left segment in the diagram below has endpoints at $s=2\cdot\phi$ and $t=1\cdot\phi^2$, and its cutting point is the integer 3: In general, if s is the jth small dot—i.e., $s=j\cdot\phi$—and $t=k\cdot\phi^2$ is the kth large dot, then the cutting point between s and t is $\frac{1}{\phi}\cdot s+\frac{1}{\phi^2}\cdot t = j+k$ (Why?! [1]). But more importantly, this observation shows that no interval has two or more dots: a small dot and a large dot can’t be in the same interval because they always have an integer between them! [2] So all we have to do now is prove that no interval is empty: for each integer n, some dot lies in the interval [ n, n+1]. We will prove this by contradiction. What happens if no dot hits this interval? Then the sequence $\phi,2\phi,3\phi,\ldots$ jumps over the interval, i.e., for some j, the jth dot in the sequence is less than n but the ( j+1)st is greater than n+1. Likewise, the sequence $\phi^2,2\phi^2,3\phi^2,\ldots$ jumps over the interval: its kth dot is less than n while its ( k+1)st dot is greater than n+1: By our observation above on segment $s=j\phi$ and $t=k\phi^2$, we find that the integer j+ k is less than n, so $j+k\le n-1$. Similarly, $j+k+2 > n+1$, so $j+k+2 \ge n+2$. But together these inequalities say that $n\le j+k\le n-1$, which is clearly absurd! This is the contradiction we were hoping for, so the interval [ n, n+1] is in fact not empty. This completes our proof of the Covering Theorem and the Wythoff formula! It was a long journey, but we’ve finally seen exactly why the Wythoff losing positions are arranged as they are. Thank you for following me through this! A Few Words on the Column Covering Theorem Using the floor function $\lfloor x\rfloor$ that rounds x down to the nearest integer, we can restate the Column Covering Theorem in perhaps a more natural context. The sequence of integers $$\lfloor\phi\rfloor = 1, \lfloor 2\phi\rfloor = 3, \lfloor 3\phi\rfloor = 4, \lfloor 4\phi\rfloor = 6, \ldots$$ is called the Beatty sequence for the number $\phi$, and similarly, $$\lfloor\phi^2\rfloor = 2, \lfloor 2\phi^2\rfloor = 5, \lfloor 3\phi^2\rfloor = 7, \lfloor 4\phi^2\rfloor = 8,\ldots$$ is the Beatty sequence for $\phi^2$. Today we proved that these two sequence are complementary, i.e., together they contain each positive integer exactly once. We seemed to use very specific properties of the numbers $\phi$ and $\phi^2$, but in fact, a much more general theorem is true: Beatty’s Theorem: If $\alpha$ and $\beta$ are any positive irrational numbers with $\frac{1}{\alpha}+\frac{1}{\beta}=1$, then their Beatty sequences $\lfloor\alpha\rfloor, \lfloor 2\alpha\rfloor, \lfloor 3\alpha\rfloor,\ldots$ and $\lfloor\beta\rfloor, \lfloor 2\beta\rfloor, \lfloor 3\beta\rfloor,\ldots$ are complementary sequences. Furthermore, our same argument—using $\alpha$ and $\beta$ instead of $\phi$ and $\phi^2$—can be used to prove the more general Beatty’s Theorem!
Atoms on the corners, edges, and faces of a unit cell are shared by more than one unit cell. An atom on a face is shared by two unit cells $\implies$ half an atom belongs to each unit cell. An atom on the edge is shared by four unit cells and an atom on the corner is shared by eight unit cells What does this have to do with the equations? Well lets say we have an sodium atom that is crystallized in a simple cubic unit cell, then there would be an sodium atom on each of the eight corners of the cell. However, only $\frac{1}{8}$ of these atoms can be assigned to a given cell. Hence the simple cubic structure is $$8 \, \text{(corners)} \times \frac{1}{8}=1 \, \text{atom}$$ if sodium formed a body centered cubic structure, there would be two atoms per unit cell (as seen in the top diagram). It isn't shared with an other cells. Therefore the body-centered cubic structure is given by $$8 \, \text{(corners)}\, \times \frac{1}{8} + 1 \, \text{(body)}= 2 \, \text{atoms}$$ now finally if sodium was a face-centered cubic structure the six atoms on the faces of the unit cell would contribute three net sodium atoms, for a total of four atoms per unit cell. Hence the face-centered cubic structure is $$8 \, \text{(corners)} \times \frac{1}{8} + \Big{(}6 \, \text{(faces)} \times \frac{1}{2}\Big{)}= 4 \, \text{atoms} $$ Hope this helps
GEOMETRY GEOMETRY Geometry is derived from two greek words "Geo" means "Earth" metron means "Measurement". That means measurement of Earth is called geometry. Basics terms of geometry There are three basics undefined terms of geometry. (i) Point (ii) Line (iii) Plane Point: Point is a mark of position, it is made by sharp tip of pen, pencil and nail. It is denoted by capital letter. It is represented by A point has no length, no breadth and no thickness. Line segment: The distance between two points in a same plane is called a line segment. It is denoted by\[\overline{AB}\]. It is measured in 'cm' or 'inch'. It can be measured. 1 inch\[=2.5\,\,cm,\] Rays: A line segment extended endlessly in one direction is called a ray. It is denoted by\[\overline{OA}\]. It can't be measured. Line: A line segment extended endlessly in both directions is called a line. It is denoted by\[\overline{AB}\] It can't be measured. Plane: A smooth flat surface which extended endlessly in all the directions is called a "plane". Example: (i) Surface of a blackboard in your class room. (ii) Floor of the classroom. Note: A plane has length and breadth. A plane has no thickens or boundary. Collmearity of points: Three points A, B, C in a plane are collinear if they lie on the same straight line. Non-ColIinear points: The points which do not lie on the same line are called non - collinear point. Note: Number of lines that can be drawn through 'n? non - collinear points is\[\frac{n\left( n-1 \right)}{2}\]. Properties of lines Passing through a point an infinite number of lines can be drawn. \[{{l}_{1}},\,\,{{l}_{2}}\]................... \[{{l}_{n}}\] all pass through 'p'. These lines are called concurrent line and the point P is called the point of concurrence. Two lines in a plane are either intersecting or parallel. ANGLE An angle is union of two different rays having the same initial point. Initial point is Q. Angle is denoted by\[\angle PQR=30{}^\circ .\] Unit of measurement of angle is degrees, which is represented as \[{}^\circ \] (e.g., \[60{}^\circ ,\] \[70{}^\circ ,\] \[80{}^\circ ,\]etc.) Types of Angle Zero Angle: Initially the terminal ray coincides with initial ray without any rotation then the angle formed is a zero angle and its measure is\[0{}^\circ .\] Acute Angle: An angle is measure greater than \[0{}^\circ \] but less than \[90{}^\circ \] is called an Acute Angle. Here, \[\angle AOB=30{}^\circ \]formed Acute Angle. Right Angle: An angle is measure \[90{}^\circ \] is called a right angle. Here, \[\angle PQR=90{}^\circ ,\]formed right angle. Obtuse Angle: An angle is measure greater than \[90{}^\circ \] but less than \[180{}^\circ \] is called an obtuse angle. Here, \[\angle PQR=120{}^\circ ,\]formed obtuse angle. Straight Angle: An angle is equal to \[180{}^\circ \]is called a straight angle. Here, \[\angle PQR=180{}^\circ ,\]formed straight angle. Reflex Angle: An angle is measure more than \[180{}^\circ \] but less than \[360{}^\circ \] is called a reflex angle. Here, \[\angle PQR=220{}^\circ ,\] formed reflex angle. Complete angle: An angle is equal to \[360{}^\circ \]is called a complete angle. Here, \[\angle AOB=360{}^\circ ,\]formed complete angle. Some more part of Angle Vertically opposite angle: It two lines are interested at a point then vertically opposite angles are equal. \[\angle POS=\angle QOR\]and \[\angle POR=\angle QOS\] \[\angle 1=\angle 2\]and\[\angle 3=\angle 4\] \[\angle POS=\angle QOR\] and \[\angle POR=\angle QOS\] are formed vertically opposite angle. Adjacent angle: Two angles in a plane are said to be adjacent angles if they have a' common vertex. In the adjoining figure x and y are called adjacent angles with common vertex A and common arm AC. Linear pair of angle: The pair of adjacent angles whose non common arms are two opposite rays is called a linear pair of angles. In the adjoining figure \[\angle PQS\] and \[\angle RQS\] form a linear pair of angles. \[\angle 1+\angle 2=180{}^\circ \] Complementary angle: if the sum of measures of any two angles is \[90{}^\circ ,\]then they are said to be complementary angles. \[\angle AOB+\angle PQR=30{}^\circ +60{}^\circ =90{}^\circ ~\] Supplementary Angle: If the sum of any two angles is \[180{}^\circ \]then they are said to be supplement angles. \[\angle AOB+\angle PQR=60{}^\circ +120{}^\circ =180{}^\circ \] Parallel lines: Two lines which never meet, even when they are extended infinitely, are known as parallel lines. \[l//m\] The distance between two parallel line be same. The angle between two parallel line is\[0{}^\circ \]. Transversal; A straight line which intersect two or more given lines at different points is called a transversal. Example: Classification of angles formed by a transversal If two parallel lines cut by a transversal then some angles are formed. 1. Corresponding Angles. \[(\angle 2,\,\,\angle 6),\,\,(\angle 1,\,\,\angle 5),\,\,(\angle 3,\,\,\angle 7)\]and\[(\angle 4,\,\,\angle 8)\]. Pair of corresponding angles are equal. 2. Alternate Interior Angles: \[(\angle 3,\,\,\angle 5);(\angle 4,\,\,\angle 6)\] Pair of alternate angles are equal. 3. Alternate exterior angles: \[(\angle 1,\,\,\angle 7);(\angle 2,\,\,\angle 8)\] Pair of alternate exterior angles are equal. 4. Co-interior angle or consecutive angle or allied angle. \[(\angle 4,\,\,\angle 5)\]and\[(\angle 3,\,\,\angle 6)\] Pair of co-interior angles are need not be equal. Sum of co-interior angles is \[180{}^\circ .\] \[\angle 4+\angle 5=180{}^\circ ,\]and\[\angle 3+\angle 6=180{}^\circ ,\] Example: 1. In the given figure \[AB\|\|CD,\] \[l\] is a transversal then\[x=\]? Solution: \[x=60{}^\circ \](by corresponding angle) Example: 2. In the given figure \[AB\|\|CD\|\|EF\] then \[x+y\] is Solution: \[AB\|\|CD\]and BC be the transversal then \[\angle ABC=\angle BCD\] \[85{}^\circ =y{}^\circ +35{}^\circ \] \[\therefore \]\[y=50{}^\circ \] Now, \[EF\|\|CD\]and CE be the transversal then \[\angle DCE+\angle CEF=180{}^\circ \](by co-interior Angle) \[35{}^\circ +x=180{}^\circ \] \[x=145{}^\circ \] \[\therefore x+y=50+145{}^\circ =195{}^\circ \] Triangle A closed figure bounded by three line segments is called a triangle. We read as 'triangle ABC" and it is denoted by\[\Delta ABC.\] Triangle have three sides and three angles. (These are called element of triangle) The sum of all angles of a \[\Delta \] is \[180{}^\circ \]\[\angle A+\angle B+\angle C=180{}^\circ \] Classification of triangles according to the side 1. Equilateral triangle: A triangle whose sides are equal in length is called an 'equilateral triangle'. \[AB=BC=AC\] All angle are equal. \[\angle A=\angle B=\angle C=60{}^\circ \] 2. Isosceles triangle: A triangle in which two sides are equal in length is called an isosceles triangle. \[AB=BC\]and \[\angle B=\angle C\] BC is called base and \[\angle B\] and \[\angle C\] are called base angles. 3. Scalene triangle: If no two sides of a triangle are equal in length, it is called a scalene triangle. \[AB\ne BC\ne AC\] Classification of triangles according to the angles 1. Acute angled triangle: If each angle of a triangle is an acute angle, men it is called an Acute' angled triangle. Measure of each angle is less than 90°. 2. Right angled triangle: A triangle in which one of its angles is a right angle is called a 'Right angled triangle'. \[\angle B=90{}^\circ \] The opposite side of the right angle is called Hypotenuse. \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\](Pythagoras theorem) 3. Obtuse angled triangle: A triangle containing an obtuse angle is called an obtuse angled triangle. Median A line segment in which joins a vertex of a triangle to the mid-point of the opposite side is called a median. AD, BE and FC are called medians. Point of intersection of the medians is called centroid. It is denoted by G. 'G' divide AD in the ratio \[2:1\]from the vertex side. Quadrilateral A closed figure bounded by four line segments is called a quadrilateral. AB, BC, CD and AD are four sides of ABCD quadrilateral. \[\angle A,\,\,\angle B,\,\,\angle C\]and \[\angle D\] are four angles of the quadrilateral. The quadrilateral has two diagonals (AC and BD). (AB, BC), (BC, CD), (CD, AD), (AD, AB) are four pairs of adjacent sides. \[(\angle A,\,\,\angle B),\] \[(\angle B,\,\,\angle C),\] \[(\angle C,\,\,\angle D),\] \[(\angle D,\,\,\angle A)\] are four pairs of adjacent angles. The sum of all angles of a quadrilateral is\[360{}^\circ \]. Types of Quadrilateral Trapezium: A quadrilateral having only one pair of parallel sides. Isosceles trapezium: A trapezium in which non - parallel sides are equal. Parallelogram: A quadrilateral having both pairs of opposite sides parallel and equal. \[AB\|\|CD\]and \[AD\|\|BC\] \[AB=CD\]and \[AD=BC\] \[\angle A=\angle C\]and \[\angle B=\angle D\] Rhombus: A parallelogram in which all sides are equal. \[AB\|\|CD\]and \[AD\|\|BC\] \[AB=BC=CD=AD\] Diagonals are perpendicular to each other. \[AO=CO\]and \[BO=DO\] (i.e. diagonals bisect each other) Rectangle: A parallelogram in which each angle is equal to\[90{}^\circ .\] \[AB\|\|CD\]and \[AD\|\|BC\] \[AB=CD\]and \[AD=BC\] \[\angle A=\angle B=\angle C=\angle D=90{}^\circ \] Diagonals are equal \[AC=BD\] \[~OA=OC=OD=OB\] Square: A rectangle in which whose all sides are equal. \[AB\|\|CD\]and \[AD\|\|BC\] \[AB=BC=CD=AD\] \[\angle A=\angle B=\angle C=\angle D=90{}^\circ \] Diagonals bisect each other at\[90{}^\circ \]. \[AC=BD\] Kite: A quadrilateral having two pairs of equal adjacent sides but unequal opposite sides is called a kite. \[AD=DB\]and \[AC=BC\] Diagonals are perpendicular to each other. \[\angle A=\angle B\] \[OA=OB\] Circle: A circle is a set of points in a plane whose distance from a fixed point is constant. O is called center. OA and OB are radius. AB is called diameter. Diameter \[=2\times \] Radius Diameter is the longest chord of circle. CD is called tangent of circle. PQ is called secant of a circle.
Search Now showing items 1-1 of 1 Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2014-09) Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, \|y\| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Measure Theory Notes by Anwar Khan Name Measure Theory: Notes Provider Mr. Anwar Khan Pages 169 pages Format PDF (see Software section for PDF Reader) Size 7.75 MB Partial contents Algebra on $X$ Sigma Algebra i.e. $\sigma-$algebra on $X$ Trivial $\sigma-$algebra; Largest $\sigma-$algebra Increasin & sequence of sets Decreasing sequence of sets Define $\lim\limits_{k\to \infty} \sup A_k$ and $\lim\limits_{k\to \infty} \inf A_k$ Smallest $\sigma-$algebra Borel set & Borel $\sigma-$algebra $G_\sigma-$set; $F_\sigma-$set Set of extended real numbers; Set function; Properties of set function Measure Finite measure; $\sigma-$finite measure Monotone convergence theorem Measurable space and measure space; Finite measure space; $\sigma-$finite measure space; $\mathcal{A}-$measurable set $\sigma-$finite set Null set Complete $\sigma-$algebra; Complete measure space; Outer measure $\mu^-$measurable set Lebesgue outer measure Lebesgue measurable set or $\mu^-$measurable set; Lebesgue $\sigma-$algebra; Lebesgue measurable space Lebesgue measure space Dense sub set of $X$ Translation of a set; Dielation of a set Translation invarient Addition modulo 1 Translation of $E$ mod 1 Measurable function Characteristic function Almost every where property; Equal almost every where Limit inferior and limit superior of real value sequence Sequence of $\mathcal{A}-$measurable functions & its limits & their properties Larger & smaller of two function; Positive part of $f$; Negative part of $f$; Absolute function of $f$ Limit existence almost every where Step function Riemann integral Simple function; Canonical representation of simple function Lebesgue integral of simple function Bounded function; Lower Lebesgue integral; Upper Lebesgue integral Lebesgue integral of bounded function Uniform convergence Almost uniform convergence; Egoroff's theorem; Bounded convergence theorem; Non-negative function; Lebesgue integralof non-negative function Monotone convergence theorem Fatou's lemma Please click on View Online to see inside the PDF. Download or view online Notes of other subjects Advanced Analysis: Handwritten Notes Algebra II by Syed Sheraz Asghar Complex Analysis (Easy Notes of Complex Analysis) Complex Analysis (Quick Review) Differential Geometry by M Usman Hamid Differential Geometry by Syed Hassan Waqas Differential Geometry: Handwritten Notes Fluid Mechanics I by Dr Rao Muzamal Hussain Fluid Mechanics II by Dr Rao Muzamal Hussain Functional Analysis by Mr. Tahir Hussain Jaffery Functional Analysis by Prof Mumtaz Ahmad Fundamental of Complex Analysis (Solutions of Some Exercises) General Topology by Raheel Ahmad Group Theory: Important Definitions and Results Groups (Handwritten notes) Linear Algebra: Important Definitions and Results Mathematical Method by Sir Muhammad Awais Aun Mathematical Statistics I by Muzammil Tanveer Mathematical Statistics II by Sir Haidar Ali Measure Theory Notes by Anwar Khan Mechanics (Easy Notes of Mechanics) Mechanics by Sir Nouman Siddique Metric Spaces (Notes) Number Theory by Dr Muhammad Umer Shuaib Number Theory: Handwritten Notes Number Theory: Notes by Anwar Khan Numerical Analysis by M Usman Hamid Numerical Analysis II Operation Research by Sir Haidar Ali Operation Research: Handwritten Notes Partial Differential Equations Real Analysis (Notes by Prof. Syed Gul Shah) Ring (Notes) by Prof. M. Dabeer Mughal Rings (Handwritten notes) Special Functions by Dr. Muhey-U-Din Theory of Optimization by Ma'am Iqra Razzaq Theory of Relativity & Analytic Dynamics: Handwritten Notes Topology Notes by Azhar Hussain Topology: Handwritten Notes Vector Spaces (Handwritten notes) msc/notes/measure_theory_by_anwar_khan Last modified: 19 months ago by Administrator
Previous abstract Next abstract Session 67 - Stars: Evolution, Atmospheres, Intrinsic. Display session, Thursday, June 11 Atlas Ballroom, We report on a rich set of observations of the K0 III star \beta Geminorum (Pollux) with the Goddard High-Resolution Spectrograph (GHRS). The dataset consists of low-dispersion spectra (\lambda/\Delta \lambda = 2,000) of the 1169 - 1671 Å\ region, moderate dispersion spectra (\lambda/\Delta \lambda = 20,000) of selected spectral intervals, and echelle observations (\lambda/\Delta \lambda = 90,000) of the O I resonance lines and Mg II h and k lines. We perform an analysis of intersystem lines and determine the emission measure distribution, from which we infer properties of the stellar transition region. The line profile shapes do not reveal evidence of broad components, similar to what is observed in the atmosphere of the inactive star Procyon and unlike what is observed in active stars. We find a trend of increasing redshift with line formation temperature up to a maximum log T of 5.2, and a decrease for temperatures greater than log T of 5.2, similar to what is seen on the Sun and dissimilar to what is seen on Procyon. We interpret these findings in light of differences in activity and atmospheric structure. Program listing for Thursday
Complex Networks Toolbox (NetLogo) Introduction Interface Scripts Generators Global Measures Utilities Dynamics Input/Output This NetLogo model is a toy tool to launch experiments for (small) Complex Networks. It provides some basic commands to generate and analyze small networks by using the most common and well-known graph models (random graphs, scale free networks, small world, etc). Also, it provides some methods to test dynamics on networks (spread processes, page rank, cellular automata,...). All the funtionalities have been designed to be used as extended NetLogo commands. In this way, it is possible to create small scripts to automate the generating and analyzing process in an easier way. Of course, they can be used in more complex and longer NetLogo procedures, but the main aim in their design is to be used by users with no previous experience on this language (although, if you know how to program in NetLogo, you can probably obtain stronger results in an easier way). You can find the last version of the tool in the associated Github poject. In the next sections you will learn some details about how to use it, but remember that the best way is to take a look at the sample scripts and playing around with the tool. Although the real power of the system is obtained via scripting, and it is its main goal, the interface has been designed to allow some interactions and facilitate to handle the creation and analysis of the networks. Indeed, before launching a batch of experiments is a good rule to test some behaviours by using the interface and trying to obtain a partial view and understanding of the networks to be analyzed... in this way, you will spend less time in experiments that will not work as you expect. The interface has 3 main areas: Network Representation: In the left side. It has a panel where the network is represented and it allows one only interaction: to inspect node information when the button Inspect Nodeis pressed. Under this panel some widgets to manage the visualization properties are located: select layouts and fix parameters for it. Measures Panel: In the middle. It contains a collection of plots where the several centralities and some monitors with global information about the current network are shown. The measures that are computed for every node are: Degree Clustering Betweenness Eigenvector Closeness Page-Rank Script Panel: In the right side. It has two input widgets: the first one, OneCommmand, to write one command to be executed (for example, the name of a script); and a multiline script widget, where you can test simple scripts to be run. Some functionalities are not available in this last widget, and we advise you to use the script file to write the more complex scripts you can need (also, the file editor is more comfortable than that from the widget). In the top of this panel you can find also some buttons to load, save and clear the current network. Use scripts.nls to write your customized scripts. In order to acces this file, you must go to Code Tab and then choose scripts.nls from Included Files chooser. You can add as many aditional files as you want if you need some order in your experiments and analysis (load them with the __includes command from main file). Remember that a script is only a NetLogo procedure, hence it must have the following structure: to script-name ... Commands ... end After defining your scripts, you can run them directly from the Command Center, from the OneCommand input in the interface, or as auxiliary procedure in other scripts. In the document you can find specific network commands that can be used to write scripts for creating and analyzing networks. In fact, these scripts can be written using any NetLogo command, but in this library you can find some shortcuts to make easier the process. Some of the useful NetLogo commands you will probably need are: let v val: Creates a new variable vand sets its value to val. set v val: Changes the value of the variable vto val. [ ]: Empty list, to store values in a repetition. range x0 xf incx: Returns an ordered list of numbers from x0to xfwith incxincrements. foreach [x1....xn] [ [x] -> P1...Pk ]: for each xin [x1 ... xn]it executes the comands P1to Pk. repeat N [P1...Pk]: Repeats the block of commands P1to Pk, Ntimes. store val L: Stores value valin list L. Returns the sum/max/min/mean of values of list sum / max / min / mean L: L. print v: Prints value of vin the Output. plotTable [x1...xn] [y1...yn]: Plots the points (x1,y1)...(xn,yn). In the next example we study how the diameter of a Scale Free Network (built with Barabasi-Albert algorithm) changes in function of the size of the network: foreach (range 10 1000 10) [ [N] -> clear BA-PA N 2 1 print diameter ] And the next one will perform a experiment moving a parameter from $0$ to $0.01$ with increments of $0.001$, and for every value of this parameter it will prepare $10$ networks to compute their diameters (these networks are obtained by adding random edges to a preferential attachtment ntework). to script5 clear foreach (range 0 .01 .001) [ [p] -> print p repeat 10 [ clear BA-PA 300 2 1 ER-RN 0 p print diameter ] ] end From the Output widget we can copy/export the printed values and then analyze them in any other analysis software (R, Python, NetLogo, Excel, etc.) The current generators of networks (following several algorithms to get different structures) are: In fact this is the Gilbert variant of the model introduced by Erdős and Rényi. Each edge has a fixed probability of being present ($p$) or absent ($1-p$), independently of the other edges. N - Number of nodes, p - link probability of wiring ER-RN N p The Watts–Strogatz model is a random graph generation model that produces graphs with small-world properties, including short average path lengths and high clustering. It was proposed by Duncan J. Watts and Steven Strogatz in their joint 1998 Nature paper. Given the desired number of nodes $N$, the mean degree $K$ (assumed to be an even integer), and a probability $p$, satisfying $N >> K >> ln(N) >> 1$, the model generates an undirected graph with $N$ nodes and $NK/2$ edges in the following way: Construct a regular ring lattice, a graph with $N$ nodes each connected to $K$ neighbors, $K/2$ on each side. Take every edge and rewire it with probability $p$. Rewiring is done by replacing $(u,v)$ with $(u,w)$ where $w$ is chosen with uniform probability from all possible values that avoid self-loops (not $u$) and link duplication (there is no edge $(u,w)$ at this point in the algorithm). N - Number of nodes, k - initial degree (even), p - rewiring probability WS N k p The Barabási–Albert (BA) model is an algorithm for generating random scale-free networks using a preferential attachment mechanism. Scale-free networks are widely observed in natural and human-made systems, including the Internet, the world wide web, citation networks, and some social networks. The algorithm is named for its inventors Albert-László Barabási and Réka Albert. The network begins with an initial connected network of $m_0$ nodes. New nodes are added to the network one at a time. Each new node is connected to $m \leq m_0$ existing nodes with a probability that is proportional to the number of links that the existing nodes already have. Formally, the probability $p_i$ that the new node is connected to node $i$ is \[p_i=\frac{k_i}{\sum_j k_j}\] where $k_i$ is the degree of node $i$ and the sum is made over all pre-existing nodes $j$ (i.e. the denominator results in twice the current number of edges in the network). Heavily linked nodes ("hubs") tend to quickly accumulate even more links, while nodes with only a few links are unlikely to be chosen as destination for new links. New nodes have a "preference" to attach themselves to already heavily linked nodes. N - Number of nodes, m0 - Initial complete graph, m - Number of links in new nodes BA-PA N m0 m The algorithm of Klemm and Eguílez manages to combine all three properties of many “real world” irregular networks – high clustering coefficient, short average path length (comparable with that of the Watts and Strogatz small-world network), and scale-free degree distribution. Indeed, average path length and clustering coefficient can be tuned through a “randomization” parameter, $\mu$, in a similar manner to the parameter $p$ in the Watts and Strogatz model. It begins with the creation of a fully connected network of size $m_0$. The remaining $N−m_0$ nodes in the network are introduced sequentially along with edges to/from $m_0$ existing nodes. The algorithm is very similar to the Barabási and Albert algorithm, but a list of $m_0$ “active nodes” is maintained. This list is biased toward containing nodes with higher degrees. The parameter $\mu$ is the probability of new edges to be connected to non-active nodes. When new nodes are added to the network, each new edge is connected from the new node to either a node in the list of active nodes or, with probability $\mu$, to a randomly selected “non-active” node. The new node is added to the list of active nodes, and one node is then randomly chosen, with probability proportional to its degree, for removal from the list, i.e., deactivation. This choice is biased toward nodes with a lower degree, so that nodes with the highest degree are less likely to be chosen for removal. N - Number of nodes, m0 - Initial complete graph, μ - Probability of connect with low degree nodes KE N m0 μ It is a simple algorithm to be used in metric spaces. It generates $N$ nodes that are randomly located in 2D, and after that two every nodes $u,v$ are linked if $d(u,v) < r$ (a prefixed radius). N - Number of nodes, r - Maximum radius of connection Geom N r This algorithm is similar to the geometric one, but we can prefix the desired mean degree of the network, $g$. It starts by creating $N$ randomly located nodes, and then create the number of links needed to reach the desired mean degree. This link creation is random in the nodes, but choosing the shortest links to be created from them. N - Number of nodes, g - Average node degree SCM N g A Grid of $N\times M$ nodes is created. It can be chosen to connect edges of the grid as a torus (to obtain a regular grid). M - Number of horizontal nodes, N - Number of vertical nodes, t? - torus? Grid N M t? Creates a Bipartite Graph with $N$ nodes (randomly typed to two families) and $M$ random links between nodes of different families. N - Number of nodes, M - Number of links BiP N M The model introduced by Kleinberg et al consists of a itearion of three steps: Node creation and deletion: In each iteration, nodes may be independently created and deleted under some probability distribution. All edges incident on the deleted nodes are also removed. $pncd$ - creation, $(1 - pncd)$ - deletion. Edge creation: In each iteration, we choose some node $v$ and some number of edges $k$ to add to node $v$. With probability $\beta$, these $k$ edges are linked to nodes chosen uniformly and independently at random. With probability $1 − \beta$, edges are copied from another node: we choose a node $u$ at random, choose $k$ of its edges $(u, w)$, and create edges $(v, w)$. If the chosen node $u$ does not have enough edges, all its edges are copied and the remaining edges are copied from another randomly chosen node. Edge deletion: Random edges can be picked and deleted according to some probability distribution. Iter - Number of Iterations, pncd - probability of creation/deletion random nodes, k - edges to add to the new node, beta - probability of new node to uniform connet/copy links, pecd - probability of creation/deletion random edges Edge-Copying Iter pncd k beta pecd Global measures that we can take from any network are: Number-Nodes Number-Links Density Average-Degree Average-Path-Length Diameter Average-Clustering Average-Betweenness, Average-Eigenvector, Average-Closeness, Average-Page-Rank All Some of them need to compute centralities before they can be used. If you choose All, you obtain a list with all the measures of the network. For example, you can print any of them in the Output Window: Print measure The system provides the following layouts: circle radial (centered in a max degree node) tutte spring (1000 iterations) bipartite To use it from a script you can write: layout type-string For example: layout "circle" Current Centralities of every node: Degree, Betweenness, Eigenvector, Closeness, Clustering, Page-Rank The way to compute all of them is by executing: compute-centralities Communities of the current network can be computed by using the Louvain method (maximizing the modularity measure of the network). Applies the Page Rank diffusion algorithm to the current Network a number of prefixed iterations. Iter - Number of iterations PRank Iter Rewires all the links of the current Network with a probability $p$. For every link, one of the nodes is fixed, while the other is rewired. p - probability of rewire every link Rewire p Applies a spread/infection algorithm on the current network a number of iterations. It starts with a initial number of $Nmi$ infected/informed nodes and in every step: The infected/informed nodes can spread the infection/message to its neighbors with probability $ps$ (independently for every neighbor). Every infected/informed node can recover/forgot with a probability of $pr$. Every recovered node can become inmunity with a probability of $pin$. In this case, it will never again get infected / receive the message, and it can't spread it. Nmi - Number of initial infected nodes, ps - Probability of spread of infection/message, pr - Probability od recovery / forgotten, pin - Probability of inmunity after recovering, Iter - Number of iterations Spread Nmi ps pr pin Iter Nodes have $2$ possible values: on/ off. In every step, every node changes its state according to the ratio of activated states. Iter - Number of iterations, pIn - Initial Probability of activation, p0 ac - ratio of activated neighbors to activate if the node is $0$, p1ac - ratio of activated neighbors to activate if the node is $1$ DiscCA Iter pIn p0_ac p1_ac Nodes have a continuous possitive value for state. In every step, every node changes its state according to the states of the neighbors: s'(u) = p * s(u) + (1 - p) * avg {s(v): v neighbor of u} Iter - Number of iterations, pIn - Initial Probability of activation, p - ratio of memory in the new state ContCA Iter pIn p Opens a Dialog windows asking for a file-name to save/load current network. Save Load You can export the several distributions of the measures for a single network: export view Where view can be any of the following: Degree Clustering Betweenness Eigenvector Closeness PageRank The program will automatically name the file with the distribution name and the date and time of exporting.
In some of Boundary value problems involving ordinary differential equations,, subsidiary conditions are imposed locally. In some other cases, nonlocal conditions are imposed. In this paper: Existence and uniqueness of a classical solution to a functional-differential abstract nonlocal Cauchy problem Byszewski studied this form of functional-differential nonlocal problem: $ (1)\left\{\begin{matrix} u'(t)=f(t,u(t),u(a(t))),\:\:t\in I \ u(t_0)+\sum_{k=1}^{p}c_ku(t_k)=x_0 \end{matrix}\right.$ With $ I:=[t_0,t_0+T], t_0<t_1<…<t_p\leq t_0+T, T>0$ and $ f:I\times E^2\rightarrow E \:$ and $ \:a:I\rightarrow I \:$ are given functions satisfying some assumptions; $ E$ is a Banach space with norm $ \:\left \\| . \right \\|; x_0\in E, c_k\neq 0 \:\:(k=1,…,p)\: p \in \mathbb N$ . And here, in the classical Robin problem: $ $ u”(t) + f(t,u(t),u'(t)) = 0$ $ With localconditions: $ u(0)= 0$ and $ u'(1) = 0.$ Or With nonlocalconditions: $ u(0)= 0$ and $ u(1) = u(\eta)\;\:\eta\in(0,1)$ My question is: -When we say that the boundary conditions are local or nonlocal? -In which situation we impose local or nonlocal conditions? Thank you!
Notes of Introduction to Statistical Learning ===================================== ## Statistical Learning - basic concepts - two main reasons to estimate f: prediction and inference - trade-off: complex models may be good for accurate prediction, but it may also be hard to interpret - reducible vs. irreducible error - how to estimate f: parametric vs. non-parametric approach - pros of non-parametric - flexibility, could possibly fit a wider range of f - cons - does not reduce the problem to a set of parameters, may be complicated and need a large dataset to get a nice result - accessing model accuracy - training/test error and complexity/flexibility of the model - note: error could be MSE for regression and error rate for classification - bias-variance trade-off - complex models typically have a higher variance and lower bias, see notes for deep learning for derivation ## Linear Regression - simple linear regression - estimate coefficients - note that the estimation is unbiased, which means that average of a large number of estimations for $\hat{\mu}$ will be very close to true $\mu$. - confidence interval, hypothesis testing for linear relationship using t-statistic, p-value - multiple linear regression - important questions - whether there exists a relationship between response and predictors - use F-statistic for all predictors, instead of use p-value for every single predictor, the reason is when the number of predictors is large, it is very likely to see at least one small p-values (which rejects hypothesis of corresponding coefficient is 0, indicating there is some linear relationship), instead F-statistic takes all predictors into consideration - deciding on important variables (variable selection) - methods - try all possible models with different combinations of variables - forward/backward/mixed selection - model fit - prediction - extensions of linear model - linear model is based on two assumptions: additive and linear - remove additive assumption: introduce interaction term - nonlinear relationships: polynomial regression, kernel trick, etc. - potential problems - nonlinearity issue - can be identified by residual plot - correlation of error terms - typically happens for time series data, need to check autocorrelation - non-constant variance of error terms - outliers - high leverage points - collinearity ## Classification - why not linear regression? - hard to convert multiple (more than 2) class labels to quantitative numerical values - even for binary responses, it does not make sense to use linear regression since the output may be outside [0,1] interval - logistic regression - determine coefficients - use maximum likelihood method - confounding - the results obtained from using one predictor may be quite different from those obtained using multiple predictors, especially when there is correlation among the predictors (see the student-balance example in the text) - linear discriminant analysis - basic idea: classify data points using Bayesian approach, that is to find $k$ such that $$p_k(x)=\frac{\pi_kf_k(x)}{\sum_l \pi_lf_l(x)}$$ is maximum, where $\pi_k$ is prior, $f_k(x)$ is likelihood function. We do further assumption that $f_k$ are Gaussian that share the same $\Sigma$ but have different $\mu_k$. $\mu_k$, $\Sigma$ and $\pi_k$ can be estimated from training set, and then the decision boundary is determined based on Bayesian method by $$\delta_k(x) = -\frac{1}{2}\log\|\Sigma_k\| - \frac{1}{2}(x-\mu_k)^T\Sigma_k^{-1}(x-\mu_k) + \log \pi_k$$ since $\Sigma_k$ are equal, we have $$\delta_k(x)=x^T \Sigma^{-1}\mu_k-\frac{1}{2}\mu_k^T \Sigma^{-1}\mu_k + \log \pi_k + const$$ This gives linear decision boundary. If $\Sigma_k$ are different, boundary may be quadratic - quadratic discriminant analysis - it assumes that classes have different covariance matrices. - LDA vs. QDA - this is a typical bias-variance trade-off issue. Since QDA contains more parameters, it is better when there are a large number of data points or it is clear that different classes have different covariance matrices. In contrast, LDA works well when number of data is small. - comparison of classification methods - when true boundary is linear, it is better to use LDA or logistic regression, if is nonlinear, it may be better to use QDA or KNN, but the level of smoothness of KNN should be carefully chosen. ## Resampling methods - cross validation - leave-one-out cross-validation - k-fold cross-validation - bootstrap ## Linear model selection and regression - subset selection - best subset selection - basic idea - fix number of predictors k, find the best model $M_k$ for each k - may use $R^2$, MSE, etc - for all $M_k$, select the best one - should use AIC, BIC, adjusted $R^2$, etc, since as number of predictors increases, MSE always decreases and $R^2$ always increases, AIC/BIC/adjusted $R^2$ includes penalization for the number of predictors - definition of AIC/BIC/adjusted $R^2$ - AIC = $\frac{1}{n\hat{\sigma}^2}(RSS+2d\hat{\sigma}^2)$ - BIC = $\frac{1}{n}(RSS+\log(n)d\hat{\sigma}^2)$ - adjusted $R^2 = 1- \frac{RSS/(n-d-1)}{TSS/(n-1)}$ - note that in addition to using these metrics above related to training set directly, we may also use cross validation (estimate test error directly) to select the optimal model, which requires fewer assumptions about the model and is more general - problem - need to list all possibilities, it is super computationally expensive - stepwise selection - forward stepwise selection: also generate best model $M_k$ for each k, using greedy approach to add predictors one at a time, then select the best among $M_k$ - backward stepwise selection - hybrid approaches - shrinkage methods - ridge regression - basic idea: apply a L-2 norm regularization term - since regularization term is affected by scaling of the training data, it is best to apply ridge regression after standardization of predictors - pros and cons - pros: very convenient for model selection, does not require to fit many different models - cons: the model still depends on all predictors, instead of depending on a subset of predictors. Lasso method can solve this issue. - lasso - basic idea: apply a L-1 norm regularization term - this method forces some of coefficients to be exactly zero, this performs variable selection (a diagram explaining this is shown in Fig. 6.7 in the text) - another formulation for ridge and lasso, which reveals a close connection between lasso, ridge and best subset selection, see text - comparing ridge and lasso (see text for details) - ridge works better for models related to many predictors with coefficients of roughly equal size, while lasso works better for models where only a relatively small number of variables are important - effects of two methods - roughly speaking, ridge shrinks every dimension of data by the same proportion, while lasso shrinks every dimension towards 0 by a similar amount. - Bayesian interpretation of ridge and lasso - Bayesian formulation says there is a prior distribution $p(\beta)$, and posterior distribution of $\beta$ given $X,Y$ is $p(\beta\|X,Y)\propto f(Y\|X,\beta)p(\beta)$, training process is to find the maximum $p(\beta\|X,Y)$, when prior is Gaussian with zero mean, we get ridge, when prior is double-exponential, we get lasso. - dimension reduction methods - previous methods use original predictors, dimension reduction methods use transformed predictors. - methods - PCA - note that standardization is typically needed - partial least squares (PLS) - motivation: in PCA, response Y is not used, only X is used in training in an unsupervised way. The principal directions extracted by PCA well explains predictors X, but may not be well correlated with Y. - basic idea: to find first PLS direction, set the weight of each predictor $X_j$ to be the coefficient of simple linear regression of $Y$ onto $X_j$. To find subsequent PLS directions, take orthogonalized data and do processing iteratively. - considerations in high dimensions ## Moving beyond linearity - polynomial regression - step functions - basis functions - regression splines - piecewise polynomials - constraints and splines - the spline basis representation - choosing the number and locations of the knots - see "notes of numerical analysis course" for more information on this part - comparison to polynomial regression - regression splines are typically better and more stable than polynomial regression, due to its lower degree - smoothing splines - definition: $$\sum_{i=1}^n (y_i-g(x_i))^2+\lambda \int g''(t)^2dt$$ it penalizes roughness of the function - property: it can be proved that $g(x)$ that minimizes error function is a piecewise cubic polynomial with knots at the unique values of $\\{x_i\\}$ - local regression - basic idea - for each $x_0$, find a neighbor data set $K_0$, and train a model with weighted error (weight associated to each training data is determined by distance between this point to $x_0$), then use this local model to predict $y$ for $x_0$ - generalized additive models - basic idea - apply a nonlinear transformation to each feature and then do linear fitting on transformed features ## tree-based models - basics of decision trees - regression decision trees - basic idea - given predictors $X_1, ..., X_p$, find the optimal predictor to split $j$ and optimal cutpoint $s$, such that for two defined regions $R_1(j,s)=\\{X\|X_j<s\\}$ and $R_2(j,s)=\\{X\|X_j\ge s\\}$, the error $$\sum_{k=1,2}\sum_{i:x_i\in R_k}(y_i-\hat{y}_{R_k})^2 $$ is minimized. Then pick one of these regions and repeat this process until stopping criterion is reached (e.g. each leaf node contains less than a specific number of data). This greedy approach is called recursive binary splitting - pruning - a complex tree may lead to overfitting, pruning the tree may solve this issue - cost complexity pruning: try to minimize error function $$\sum_{m=1}^{\|T\|} \sum_{i:x_i \in R_m} (y_i-\hat{y}_{R_m})^2+\alpha \|T\|$$ - classification decision trees - basic idea - most part is the same, except that error function is Gini index or cross entropy - pros and cons - pros - interpretability - easy to handle both quantitative and qualitative predictors - cons - may not have good predictive accuracy (can be improved by bagging, boosting, etc) - bagging, random forests, boosting - bagging - motivation: reduce prediction variance - methods: train several un-pruned trees (which has high variance but low bias) with different training sets, and make prediction based on average outputs/majority vote - how are different training sets generated: bootstrap - error estimation - in addition to cross-validation, we may use out-of-bag observations to estimate error. The basic idea is that when using bootstrap approach, each bagged tree contains about $(1-1/e)$ of all data, so the remaining $1/e$ of data (called "out-of-bag" data) can be used to estimate test error. - variable importance measures - in bagging model, we do not have the good interpretability of an individual decision tree, variance importance measures work as a metric to determine the relative importance of each feature - random forests - basic idea: similar to bagging, except that instead of using all predictors for splitting, we select a subset of predictors and request that the splitter can only be selected from this subset. This method reduces correlation of the trees and makes the prediction more reliable, especially when one of the predictors is particularly important compared with others. - boosting - basic idea: instead of building many trees in parallel, boosting builds trees sequentially: each tree fits the residue that has not been explained by previous trees. By controlling shrinkage parameter $\lambda$, we can control how fast the new tree learns the residue. ## support vector machines - see notes for course "advanced data science" ## unsupervised learning - challenges of unsupervised learning - PCA - clustering - K-means - hierarchical clustering - pros: does not require pre-specify number of clusters, one single dendrogram can be used to obtain any number of clusters.
I'm in trouble with the exercise problem iii.12.3 in Hartshorne's AG. Let $X_{1}$ be a rational normal curve in $\mathbb{P^4}$ ( = image of 4th veronese embedding of $\mathbb{P^1}$). $X_{0}$ be a rational quartic curve in $\mathbb{P^3}$ with parametrization $[t^4, t^3u, tu^3, u^4]$. Construct flat family $X$ using projection $\pi: \mathbb{P^4}\to \mathbb{P^3}$ parametrized by $\mathbb{A^1}$,with the given fibers $X_{1}$ and $X_{0}$ for $t=1$ and $t=0$. More precisely, $X_{a}$ is parametrized by $[t^4, t^3u, at^2u^2, tu^3, u^4]$. Let $\mathcal{I} \subset$ $\mathcal{O_{\mathbb{P^4}\times\mathbb{A^1}}}$ be a total idael sheaf of $X$. Show that the functions $h^0(t,\mathcal I_{t})$ and $h^1(t,\mathcal{I}_{t})$ are jump at $t=0$. Still, my calculation doesn't lead to such answer. Rather, it seems to be that they are constant functions. After some calculation, I found out that $\mathcal{I}\otimes k(t\ne 0)$ $\simeq$ idael sheaf of rational normal curve $\mathcal{I}\otimes k(0)$ $\simeq$ [$(x_{2}x_{0}, x_{2}x_{1}, x_{2}^2, x_{2}x_{3}, x_{2}x_{4})$ + ideals of quartic rational cuves in $\mathbb{P^3}$] In any cases, there zeroth and first cohomology group on $\mathbb{P^4}$ vanish, which comes from the exact sequence $0$ $\rightarrow$ $\mathcal{I}$ $\rightarrow$ $\mathcal{O_{P^4}}$ $\rightarrow$ $\mathcal{O}_{closed subscheme} \rightarrow 0$ (By the flat base change theorem, they do computes functions $h^0$ and $h^1$ above) I think above result is more acceptable because all fibers of $X$ are just $\mathbb{P^1}$. What's wrong with this calculation? I'll appriciate any comments. (edited) Thank you, Sandor-kovacs. I'm really appriciate about your kind explaination. But still, there are two things make me confuse. It seems to me that......according to your answer, calculation leads to $h^1=0$. (I forgot about flatness and just calculated it) Honestly, I still don't know why $x_{2}$ becomes independent. In my opinion, the relations $x_{2}x_{0} = x_{2}x_{1}=x_{2}^2= x_{2}x_{3}=x_{2}x_{4}=0$ are already in the construction of the structure sheaf. After sheafifying those relation, $x_{2}$ itself vanishes at any affine chart, because $\frac {x_{2}}{x_{i}}=\frac{x_{2}x_{i}}{x_{i}^2}=0$. I'ii appreciate to any feedbacks. (edited) I realize that my first question is foolish (I think that the geometric genus is 1), as Sander-Kovacs pointed out. But there are some problems remain. At first, I think Sander-Kovacs' answer is right. But if local sections $x_{2}$ were "renegades" and form a single global section, then it does not contribute to higher cohomology modules. This was a real reason of my confusing. Second question still remains. Precisely, $x_2 = \frac {x_2 x_0}{x_0}\in (x_2x_0, x_2x_1, x_2^2, x_2x_3, x_2x_4)_{(x_0)}\subset \mathscr {I} (U_{0})$ (I assumed $x_0$ has degree 0, as Sander-kovacs implicitly do. But my argument is same even if it has degree 1) There are some related problems. I'm continuously recalling that when one defines closed subscheme of $Proj S$ using its homogeneous ideal $I$, cutting off some lower degree part changes nothing, because saturation of ideal(not the ideal it self) determines scheme structure. (see Hartshorne's book ch ii, ex 3.12) Now, note that $(x_0x_2,x_1x_2,x_2^2,x_3x_2,x_4x_2)$ = $\bigoplus_{d \geq 2} (x_2)_d$ If I were wrong, then there are very serious gaps and errors of my whole undegraduate AG study. I'll waiting for any comments p.s Thank you Yemon Choi. I did not know about MO's policy, and I'm not good at English..... Apologize for those mistakes.
In the end, the choice of a single specific number comes from the necessity to standardize.However, we can make some physical observations to understand why that final choice had to fall in a certain range. Why a standard? First of all, why do we even need a standard?Can't individual appliances convert the incoming electricity to whatever frequency they want?Well, in principle it's possible, but it's rather difficult.Electromagnetism is fundamentally time invariant and linear; the differential equations we use to describe it Maxwells' equations are such that a system driven by a sinusoidal input at frequency $\omega$ responds only at that same frequency.In order to get out a frequency different from $\omega$ the electromagnetic fields have to interact with something else, notably charged matter.This can come in the form of a mechanical gear box or a nonlinear electrical elements such as transistors.Nonlinear elements such as the transistor can generate harmonics of the input, i.e. frequencies $2 \omega$, $3 \omega$, etc.However, in any case, frequency conversion introduces efficiency loss, cost, and bulkiness to the system. In summary, because of the time invariance and linearity of electromagnetism, it is considerably more practical to choose a single frequency and stick to it Light flicker In a historical note by E. L. Owen (see references), it is noted that the final decision between 50 and 60 Hz was somewhat arbitrary, but based partially on the consideration of light flicker. During the lecture, while Bibber recounted Steinmecz’s contributions to technical standards, he briefly repeated the story of the frequencies. By his account, “the choice was between 50- and 60-Hz, and both were equally suited to the needs. When all factors were considered, there was no compelling reason to select either frequency. Finally, the decision was made to standardize on 60-Hz as it was felt to be less likely to produce annoying light flicker.” The consideration of light flicker comes up elsewhere in historical accounts and explains why very low frequencies could not be used.When we drive a pure resistance with an ac current $I(t) = I_0 \cos(\omega t)$, the instantaneous power dissipation is proportional to $I(t)^2$.This signal oscillates in time at a frequency $2\omega$ (remember your trig identities).Therefore, if $\omega$ is lower than around $40 \, \text{Hz}$$^{[a]}$, the power dissipated varies slowly enough that as a visual stimulus you could perceive it.This sets a rough lower limit on the frequency you can use for driving a light source.Note that the arc lamps in use when electrical standards were developed may not have had purely resistive electrical response (see Schwern's answer where cooling on each cycle is mentioned) but the source frequency is always present in the output even in nonlinear and filtered systems. Reflections / impedance matching Alternating current signals travelling on a wire obey wave-like behavior.In a rough sense, the higher the frequency the more wavy the signal.A good rule of thumb is that if the length of wires is comparable to or much longer than the wavelength of the signal, then you have to worry about wave-like phenomena such as reflection.The wavelength $\lambda$ of an electrical signal is roughly$$\lambda = c / f$$where $c$ is the speed of light and $f$ is the frequency.Suppose we'd like to transmit the electricity from an electrical substation to a house and we want to keep the wavelength big enough to prevent reflection physics without having to deal with careful impedance matching.Let's put in a length of $1000 \, \text{m}$ to be conservative.Then we get$$f \leq c / 1000 \, \text{m} = 300 \, \text{KHz} \, .$$ Further reading Detailed document by E. L. Owen with references $[a]$: I'm not an expert in human flicker perception. This number is a rough guess based on personal experience and some literature. P.S. I consider this answer a work in progress and will add more as I learn more.
Loading... Error There was an error loading the page. History and Feedback Elliott Fletcher 2 years, 2 months ago This is a really good question, i just have a couple of comments Main Parts a) and b) i would use a \displaystyle before the fractions in the questions to make the fractions look bigger. Advice a) in the very last calculation when you are factorising the quadratic, you write the constant term as a power of a different number like 4^3, i would just leave this as it's simplified number, e.g x^2-30x-4^3 = 0 could be x^2-30x-64 = 0 b) There is a random ' after the ln on the first line here I think you are just missing a few full stops after some equations. Specifically, the one before "We then just need to rearrange our equation" and the very last line where you give the final value of x. Stanislav Duris 2 years, 3 months ago I feel like some parts in advice need a bit more explanation: - In part a.iv)I think you should add a bit more explanation why the answer is always 0. Something along the lines that as $b^0 = 1$ then $\log_b(1) = 0$. - In part b)at the end, "the only value for x is..." I feel should be something along the lines of "the only possible value for x is ...". - In part c)"Laws for logarithms can also be applied to $\ln$." maybe you could add "as $\ln(x) = \log_e(x)$." - Also, the last two lines of the advice is only right when p = 2. The right answer in advice always takes a square root rather than the appropriate root. Fix this. There are some small mistakes in punctuation: - Some parts are missing a blank space between the = and the gap. - In advice, I feel like many of the commas you used are unnecesseary, for example "We need to use the rule," is followed by that rule so I would remove the comma so it flows better. You've done this correctly in part a.iii). Maybe if you want to keep the comma there, you should put comma after the rule/equation on the separate line as well, to make it stand out from the sentence. Similarly, "Subsituting in our values for x and y gives," I would remove these commas too for the same reason. - You're missing a few full stops when your sentences finish with an equation in advice. For example, in part a) after the last line in each part i/ii/iii or last line of the whole advice, I think there should be full stops. - In advice part b), near the end - "to write our equation as," is followed by two lines of equations. You put full stop after the first line by mistake, but it should be at the end of the second line. Just move the \text{.} to the end of that expression. Finally, part c)asks for the answer to be put in the form $\frac{e^{a}}{b}$, but accepts answers in different forms such as ($(\frac{e^{m}}{q})^{1/p}$ or ${e^{a}}{b^{-1}}$. I think this could be easily fixed by putting some required/forbidden strings in the string restriction for the question. I would advise trying "(e" and ")^" and "" as forbidden strings and add "e" as a required string just to be sure. Hopefully that will fix this tiny problem. No variables have been defined in this question. Name Type Generated Value Error in variable testing condition There's an error in the condition you specified in the Variable testing tab. Variable values can't be generated until it's fixed. No parts have been defined in this question. Select a part to edit. Pattern restriction Variables String restrictions Answers Choices Test that the marking algorithm works Check that the marking algorithm works with different sets of variables and student answers using the interface below. Create unit tests to save expected results and to document how the algorithm should work. There's an error which means the marking algorithm can't run: Name Value Note Value Feedback Click on a note's name to show or hide it. Only shown notes will be included when you create a unit test. 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The variable replacements you've chosen will cause the following variables to be regenerated each time the student submits an answer to this part: These variables have some random elements, which means they're not guaranteed to have the same value each time the student submits an answer. You should define new variables to store the random elements, so that they remain the same each time this part is marked. to This question is used in the following exams: Logarithms by Christian Lawson-Perfect in Transition to university. maths practice-revision by David Martin in David's workspace. Joshua's copy of Printable assignment using NUMBAS by Joshua Capel in Joshua's workspace. franco's copy of Joshua's copy of Printable assignment using NUMBAS by franco chooramen in franco's workspace. Logarithms by Angus Rosenburgh in Angus's workspace. Huuuuuge test by Angus Rosenburgh in Angus's workspace. Nick's copy of Logarithms by Nick Walker in Nick's workspace. Logarithms [L6 Randomised] by Matthew James Sykes in CHY1205. NUMBAS - Logarithms by Katy Dobson in Katy's workspace. Blathnaid's copy of NUMBAS - Logarithms by Blathnaid Sheridan in Blathnaid's workspace. Initial maths test by J. Richard Snape in J. Richard's workspace. Rules of Logarithms by Jo-Ann Lyons in Jo-Ann's workspace. MATH1601 Quiz 1 by Blathnaid Sheridan in Blathnaid's workspace. Printable assignment using NUMBAS by Joshua Capel in Daniel's workspace. Foundation Maths B-Portfolio 4 Logarithm by Jean jinhua Mathias in Core Foundation Maths. Miscellaneous by Mario Orsi in Mario's workspace. Maria's copy of Logarithms by Maria Aneiros in Maria's workspace. Logarithms by Leonardo Juliano in BiomedSkills.
Hask Note todo: objects are types arrows are extensionally identified Haskell functions identity morphism: for each object, the identity morphism is the instance of the polymorphic identity for that type The identity for a type 'a' is '(id :: a→a)' and concatenation is 'f . g', defined as '\x → f (g x)'. Discussion Hask is the syntactic subset of Haskell which permits a whole bunch of operations that are seen in basic category theory. With $\to$ and $\times$ etc., Hask is almost Cartesian closed. A particular problem is the polymorphic term 'undefined', which is defined to be term of every type. It prevents, for example, initial objects (i.e. there no analog to the “empty set”). Or when it comes to setting up the categorical product, the projections $\pi_1,\pi_2$ couldn't distinguish between $'(\text{undefined},\text{undefined})'$ and just $\text{undefined}$, spoiling uniqueness. See Hask for more examples. As far as some basic category theoretical concepts go, Hask is the largest collection of types in Haskell but by more restriction, you can find some actual and nicely behaved categories. See also the category-extras package. Another deficit is that when it comes to passing functions as arguments, Haskell sees more than just Hask morphisms. A nicer definition of the largest category of Haskell would be if arrows weren't Haskell functions identified by function extensionally ($\forall x. f(x)=g(x)\implies f=g$), but rather identified if equivalent when passed to other Haskell functions as arguments ($\forall h. h(f)=h(g)\implies f=g$, see indiscernibility of identicals. We can back the first definition if we consider only the eval-functions for $h$'s). However, this doesn't work: With the definition of concatenation given above, the category laws imply that '(undefined :: Int → Char).id' must be '\x →(undefined :: Int → Char) x'. But there are functions which can detect the non-extensive property of 'undefined :: Int → Char' being the 'undefined' term for the type 'Int → Char'. The function 'seq' (which is implemented to make enforcement of strict evaluation possible) will return a different result when passed the extensionally equal functions 'undefined :: Int → Char' and '\x →(undefined :: Int → Char) x'. Functors In the following I present some images elaborating on the connection between Haskell-features to category theory, just click the images to display a larger version. Functors and fmap: Polymorphism / Fibre bundles: Natural transformations: this has the right type. todo: check defining property of Natural transformation. The $\text{flatten}$ example given in this blog in the section “2.2. Meaning of polymorphism” is a non-academic example. Monads: These are explained in the entry Hom-set adjunction.
Event detail Probabilistic Operator Algebra Seminar: An introduction to monotonic independence Seminar \| October 2 \| 3:45-5:45 p.m. \| 748 Evans Hall Ian Charlesworth, NSF Postdoctoral Fellow UC Berkeley One may think of an "independence relation" as a prescription for building joint distributions of (non-commutative) random variables, satisfying some nice universality properties. Work of Muraki and of Ben Ghorbal and Schurmann has shown that there are very few such universal independences; even with the fewest required "nice properties", there are no more than five. In this talk I will give an introduction to monotonic independence of random variables which fits in only the broadest category as it is not symmetric: $X$ being monotonically independent from $Y$ is \emph {not} equivalent to $Y$ being monotonically independent from $X$. Our goal will be to investigate the behaviour of additive monotonic convolution: given probability measures µ and ν, and random variables $X \sim \mu $ and $Y \sim \nu $ in some algebra where $X$ is monotonically indepenedent from $Y$, what is the distribution of $X+Y$? I will cover the analytic techniques necessary to answer this question, and in the time remaining, begin an investigation into monotone infinite divisibility and semigroups of convolution. This talk will draw material variously from papers of Muraki, of Bercovici and of Hasebe.
Asymptotic expansion of the mean-field approximation 1. CMLS, Ecole polytechnique, CNRS, Université Paris-Saclay, 91128 Palaiseau Cedex, France 2. International Research Center on the Mathematics and Mechanics of Complex Systems, MeMoCS, University of L'Aquila, Italy We consider the $ N $-body quantum evolution of a particle system in the mean-field approximation. We show that the $ j $th order marginals $ F^N_j(t) $, for factorized initial data $ F(0)^{\otimes N} $, are explicitly expressed, modulo $ N^{-\infty} $, out of the solution $ F(t) $ of the corresponding non-linear mean-field equation and the solution of its linearization around $ F(t) $. The result is valid for all times $ t $, uniformly in $ j = O(N^{\frac12-\alpha}) $ for any $ \alpha>0 $. We establish and estimate the full asymptotic expansion in integer powers of $ \frac1N $ of $ F^N_j(t) $, $ j = O(\sqrt N) $, whose computation at order $ n $ involves a finite number of operations depending on $ j $ and $ n $ but not on $ N $. Our results are also valid for more general models including Kac models. As a by-product we get that the rate of convergence to the mean-field limit in $ \frac1N $ is optimal in the sense that the first correction to the mean-field limit does not vanish. Keywords:Mean-field limit, quantum mechanics, Kac model, asymptotic expansion, mathematical physics. Mathematics Subject Classification:35Q83, 35Q20, 35Q40, 34E05. Citation:Thierry Paul, Mario Pulvirenti. Asymptotic expansion of the mean-field approximation. Discrete & Continuous Dynamical Systems - A, 2019, 39 (4) : 1891-1921. doi: 10.3934/dcds.2019080 References: [1] [2] H. van Beijeren, O. E. Landford, J. L. Lebowitz and H. Spohn, Equilibrium time correlation functions in the low-density limit, [3] [4] C. Boldrighini, A. De Masi and A. Pellegrinotti, Non equilibrium fluctuations in particle systems modelling Reaction-Diffusion equations, [5] [6] [7] [8] S. Caprino, M. Pulvirenti and W. Wagner, A particle systems approximating stationary solutions to the Boltzmann equation, [9] [10] [11] A. De Masi, E. Orlandi, E. Presutti and L. Triolo, Glauber evolution with Kac potentials. Ⅰ.Mesoscopic and macroscopic limits, interface dynamics, [12] [13] A. De Masi, E. Presutti, D. Tsagkarogiannis and M. E. Vares, Truncated correlations in the stirring process with births and deaths, [14] [15] C. Graham and S. Méléard, Stochastic particle approximations for generalized Boltzmann models and convergence estimates, [16] K. Hepp and E. H. Lieb, Phase transitions in reservoir-driven open systems with applications to lasers and superconductors, [17] M. Kac, Foundations of kinetic theory, [18] M. Kac, [19] [20] [21] S. Lang, [22] [23] [24] [25] D. Mitrouskas, S. Petrat and P. Pickl, Bogoliubov corrections and trace norm convergence for the Hartree dynamics, preprint.Google Scholar [26] T. Paul, M. Pulvirenti and S. Simonella, On the size of kinetic chaos for mean field models, to appear in ARMA.Google Scholar [27] M. Pulvirenti and S. Simonella, The Boltzmann Grad limit of a hard sphere system: Analysis of the correlation error, [28] B. Schlein, Derivation of effective evolution equations from microscopic quantum dynamics, [29] [30] show all references References: [1] [2] H. van Beijeren, O. E. Landford, J. L. Lebowitz and H. Spohn, Equilibrium time correlation functions in the low-density limit, [3] [4] C. Boldrighini, A. De Masi and A. Pellegrinotti, Non equilibrium fluctuations in particle systems modelling Reaction-Diffusion equations, [5] [6] [7] [8] S. Caprino, M. Pulvirenti and W. Wagner, A particle systems approximating stationary solutions to the Boltzmann equation, [9] [10] [11] A. De Masi, E. Orlandi, E. Presutti and L. Triolo, Glauber evolution with Kac potentials. Ⅰ.Mesoscopic and macroscopic limits, interface dynamics, [12] [13] A. De Masi, E. Presutti, D. Tsagkarogiannis and M. E. Vares, Truncated correlations in the stirring process with births and deaths, [14] [15] C. Graham and S. Méléard, Stochastic particle approximations for generalized Boltzmann models and convergence estimates, [16] K. Hepp and E. H. Lieb, Phase transitions in reservoir-driven open systems with applications to lasers and superconductors, [17] M. Kac, Foundations of kinetic theory, [18] M. Kac, [19] [20] [21] S. Lang, [22] [23] [24] [25] D. Mitrouskas, S. Petrat and P. Pickl, Bogoliubov corrections and trace norm convergence for the Hartree dynamics, preprint.Google Scholar [26] T. Paul, M. Pulvirenti and S. Simonella, On the size of kinetic chaos for mean field models, to appear in ARMA.Google Scholar [27] M. Pulvirenti and S. Simonella, The Boltzmann Grad limit of a hard sphere system: Analysis of the correlation error, [28] B. Schlein, Derivation of effective evolution equations from microscopic quantum dynamics, [29] [30] [1] [2] Seung-Yeal Ha, Jeongho Kim, Jinyeong Park, Xiongtao Zhang. Uniform stability and mean-field limit for the augmented Kuramoto model. [3] Rong Yang, Li Chen. Mean-field limit for a collision-avoiding flocking system and the time-asymptotic flocking dynamics for the kinetic equation. [4] Seung-Yeal Ha, Jeongho Kim, Xiongtao Zhang. Uniform stability of the Cucker-Smale model and its application to the Mean-Field limit. [5] Seung-Yeal Ha, Jeongho Kim, Peter Pickl, Xiongtao Zhang. A probabilistic approach for the mean-field limit to the Cucker-Smale model with a singular communication. [6] [7] [8] Franco Flandoli, Enrico Priola, Giovanni Zanco. A mean-field model with discontinuous coefficients for neurons with spatial interaction. [9] Young-Pil Choi, Samir Salem. Cucker-Smale flocking particles with multiplicative noises: Stochastic mean-field limit and phase transition. [10] [11] [12] Yufeng Shi, Tianxiao Wang, Jiongmin Yong. Mean-field backward stochastic Volterra integral equations. [13] [14] Weinan E, Jianfeng Lu. Mathematical theory of solids: From quantum mechanics to continuum models. [15] Hayato Chiba, Georgi S. Medvedev. The mean field analysis of the kuramoto model on graphs Ⅱ. asymptotic stability of the incoherent state, center manifold reduction, and bifurcations. [16] Jianhui Huang, Xun Li, Jiongmin Yong. A linear-quadratic optimal control problem for mean-field stochastic differential equations in infinite horizon. [17] Haiyan Zhang. A necessary condition for mean-field type stochastic differential equations with correlated state and observation noises. [18] [19] Juan Li, Wenqiang Li. Controlled reflected mean-field backward stochastic differential equations coupled with value function and related PDEs. [20] Hancheng Guo, Jie Xiong. A second-order stochastic maximum principle for generalized mean-field singular control problem. 2018 Impact Factor: 1.143 Tools Metrics Other articles by authors [Back to Top]
Chandrasekhar Venkat Raman Chandrasekhar Venkat Raman, also known as Sir CV Raman, was a Physicist, Mathematician and a Nobel Laureate. Venkat (his first name) was a Tamil Brahmin and was the second of the eight children of his parents. He was born at Thiruvanaikaval, near Tiruchirappalli on 7th November 1888. He was the second of their eight children. His father was a lecturer in Mathematics and physics which helped him aspire careers in the same fields. Here is how his life progressed: At an early age, Raman moved to Visakhapatnam. Started studying in St. Aloysius Anglo-Indian High School. After his graduation he was selected into government services where he worked for years. In 1917, Raman resigned from his government service and took up the newly created Palit Professorship in Physics at the University of Calcutta at the age of 28. On February 28, 1928, (the reason National Science Day is celebrated in India) through his experiments on the scattering of light, he discovered the Raman effect. C. V. Raman was awarded the 1930 Physics Nobel Prize for this. Raman Effect In contrast to other conventional branches of spectroscopy, Raman spectroscopy deals with the scattering of light& not with its absorption. Chandrasekhar Venkat Raman discovered in 1928 that if light of a definite frequency is passed through any substance in gaseous, liquid or solid state, the light scattered at right angles contains radiations not only of the original frequency (Rayleigh Scattering) but also of some other frequencies which are generally lower but occasionally higher than the frequency of the incident light. The phenomenon of scattering of light by a substance when the frequencies of radiations scattered at right angles are different (generally lower and only occasionally higher) from the frequency of the incident light, is known as Raman Scattering or Raman effect. The lines of lower frequencies as known as Stokes lines while those of higher frequencies are called anti-stokes lines. If $f$ is the frequency of the incident light & $f’$ that of a particular line in the scattered spectrum, then the difference $f-f’$ is known as the Raman Frequency. This frequency is independent of the frequency of the incident light. It is constant and is characteristic of the substance exposed to the incident light. A striking feature of Raman Scattering is that , within the limits of experimental error, with those obtained from rotation-vibration (infrared) spectra of the substance. Raman Frequencies are identical Here is a home-made video explaining the Raman Scattering of Yellow light: And here is another video guide for Raman Scattering: Advantage of Raman Effects Raman Spectroscopy can be used not only for gases but also for liquids & solids for which the infrared spectra are so diffuse as to be of little quantitative value. Raman Effect is exhibited not only by polar molecules but also by non-polar molecules such as $O_2$, $N_2$, $Cl_2$ etc. The rotation-vibration changes in non-polar molecules can be observed only by Raman Spectroscopy. The most important advantage of Raman Spectra is that it involves measurement of frequencies of scattered radiations, which are only slightly different from the frequencies of incident radiations. Thus, by appropriate choice of the incident radiations, the scattered spectral lines are brought into a convenient region of the spectrum, generally in the visible region where they are easily observed. The measurement of the corresponding infrared spectra is much more difficult. It uses visible or ultraviolet radiation rather than infrared radiation. Uses Investigation of biological systems such as the polypeptides and the proteins in aqueous solution. Determination of the structures of molecules. Classical Theory of Raman Effect The classical theory of Raman effect, also called the polarizability theory, was developed by G. Placzek in 1934. I shall discuss it briefly here. It is known from electrostatics that the electric field $ E $ associated with the electromagnetic radiation induces a dipole moment $ \mu $ in the molecule, given by $ \mu = \alpha E $ …….(1) where $ \alpha $ is the polarizability of the molecule. The electric field vector $ E $ itself is given by $ E = E_0 \sin \omega t = E_0 \sin 2\pi \nu t $ ……(2) where $ E_0 $ is the amplitude of the vibrating electric field vector and $ \nu $ is the frequency of the incident light radiation. Thus, from equations (1) & (2), $ \mu= \alpha E_0 \sin 2\pi \nu t $ …..(3) Such an oscillating dipole emits radiation of its own oscillation with a frequency $ \nu $ , giving the Rayleigh scattered beam. If, however, the polarizability varies slightly with molecular vibration, we can write $ \alpha =\alpha_0 + \frac {d \alpha} {dq} q $ …..(4) where the coordinate q describes the molecular vibration. We can also write q as: $ q=q_0 \sin 2\pi \nu_m t $ …..(5) Where $ q_0$ is the amplitude of the molecular vibration and $ \nu_m $ is its (molecular) frequency. From equations. 4 & 5, we have $ \alpha =\alpha_0 + \frac {d\alpha} {dq} q_0 \sin 2\pi \nu_m t $ …..(6) Substituting for $ alpha $ in (3), we have $ \mu= \alpha_0 E_0 \sin 2\pi \nu t + \frac {d\alpha}{dq} q_0 E_0 \sin 2\pi \nu t \sin 2\pi \nu_m t $ …….(7) Making use of the trigonometric relation $ \sin x \sin y = \frac{1}{2} [\cos (x-y) -\cos (x+y) ] $ this equation reduces to: $ \mu= \alpha_0 E_0 \sin 2\pi \nu t + \frac {1}{2} \frac {d\alpha}{dq} q_0 E_0 [\cos 2\pi (\nu – \nu_m) t – \cos 2\pi (\nu+\nu_m) t] $ ……(8) Thus, we find that the oscillating dipole has three distinct frequency components: The exciting frequency $ \nu $ with amplitude $ \alpha_0 E_0 $ $ \nu – \nu_m $ $ \nu + \nu_m $ (2 & 3 with very small amplitudes of $ \frac {1}{2} \frac {d\alpha}{dq} q_0 E_0 $ Hence, the Raman spectrum of a vibrating molecule consists of a relatively intense band at the incident frequency and two very weak bands at frequencies slightly above and below that of the intense band. If, however, the molecular vibration does not change the polarizability of the molecule then $ (d\alpha / dq )=0$ so that the dipole oscillates only at the frequency of the incident (exciting) radiation. The same is true for the molecular rotation. We conclude that for a molecular vibration or rotation to be active in the Raman Spectrum, it must cause a change in the molecular polarizability, i.e., $ d\alpha/dq \ne 0$ …….(9) Homo-nuclear diatomic molecules such as $ \mathbf {H_2 , N_2 , O_2} $ which do not show IR Spectra since they don’t possess a permanent dipole moment, do show Raman spectra since their vibration is accompanied by a change in polarizability of the molecule. As a consequence of the change in polarizability, there occurs a change in the induced dipole moment at the vibrational frequency. REFERENCES:- Principles in Physical Chemistry, Puri, Sharma & Pathania Physics Chemistry, Atkins Spectroscopy, Raj Kumar Feel free to ask questions, send feedback and even point out mistakes. Great conversations start with just a single word. How to write better comments?
@egreg It does this "I just need to make use of the standard hyphenation function of LaTeX, except "behind the scenes", without actually typesetting anything." (if not typesetting includes typesetting in a hidden box) it doesn't address the use case that he said he wanted that for @JosephWright ah yes, unlike the hyphenation near box question, I guess that makes sense, basically can't just rely on lccode anymore. I suppose you don't want the hyphenation code in my last answer by default? @JosephWright anway if we rip out all the auto-testing (since mac/windows/linux come out the same anyway) but leave in the .cfg possibility, there is no actual loss of functionality if someone is still using a vms tex or whatever I want to change the tracking (space between the characters) for a sans serif font. I found that I can use the microtype package to change the tracking of the smallcaps font (\textsc{foo}), but I can't figure out how to make \textsc{} a sans serif font. @DavidCarlisle -- if you write it as "4 May 2016" you don't need a comma (or, in the u.s., want a comma). @egreg (even if you're not here at the moment) -- tomorrow is international archaeology day: twitter.com/ArchaeologyDay , so there must be someplace near you that you could visit to demonstrate your firsthand knowledge. @barbarabeeton I prefer May 4, 2016, for some reason (don't know why actually) @barbarabeeton but I have another question maybe better suited for you please: If a member of a conference scientific committee writes a preface for the special issue, can the signature say John Doe \\ for the scientific committee or is there a better wording? @barbarabeeton overrightarrow answer will have to wait, need time to debug \ialign :-) (it's not the \smash wat did it) on the other hand if we mention \ialign enough it may interest @egreg enough to debug it for us. @DavidCarlisle -- okay. are you sure the \smash isn't involved? i thought it might also be the reason that the arrow is too close to the "M". (\smash[t] might have been more appropriate.) i haven't yet had a chance to try it out at "normal" size; after all, \Huge is magnified from a larger base for the alphabet, but always from 10pt for symbols, and that's bound to have an effect, not necessarily positive. (and yes, that is the sort of thing that seems to fascinate @egreg.) @barbarabeeton yes I edited the arrow macros not to have relbar (ie just omit the extender entirely and just have a single arrowhead but it still overprinted when in the \ialign construct but I'd already spent too long on it at work so stopped, may try to look this weekend (but it's uktug tomorrow) if the expression is put into an \fbox, it is clear all around. even with the \smash. so something else is going on. put it into a text block, with \newline after the preceding text, and directly following before another text line. i think the intention is to treat the "M" as a large operator (like \sum or \prod, but the submitter wasn't very specific about the intent.) @egreg -- okay. i'll double check that with plain tex. but that doesn't explain why there's also an overlap of the arrow with the "M", at least in the output i got. personally, i think that that arrow is horrendously too large in that context, which is why i'd like to know what is intended. @barbarabeeton the overlap below is much smaller, see the righthand box with the arrow in egreg's image, it just extends below and catches the serifs on the M, but th eoverlap above is pretty bad really @DavidCarlisle -- i think other possible/probable contexts for the \over*arrows have to be looked at also. this example is way outside the contexts i would expect. and any change should work without adverse effect in the "normal" contexts. @DavidCarlisle -- maybe better take a look at the latin modern math arrowheads ... @DavidCarlisle I see no real way out. The CM arrows extend above the x-height, but the advertised height is 1ex (actually a bit less). If you add the strut, you end up with too big a space when using other fonts. MagSafe is a series of proprietary magnetically attached power connectors, originally introduced by Apple Inc. on January 10, 2006, in conjunction with the MacBook Pro at the Macworld Expo in San Francisco, California. The connector is held in place magnetically so that if it is tugged — for example, by someone tripping over the cord — it will pull out of the socket without damaging the connector or the computer power socket, and without pulling the computer off the surface on which it is located.The concept of MagSafe is copied from the magnetic power connectors that are part of many deep fryers... has anyone converted from LaTeX -> Word before? I have seen questions on the site but I'm wondering what the result is like... and whether the document is still completely editable etc after the conversion? I mean, if the doc is written in LaTeX, then converted to Word, is the word editable? I'm not familiar with word, so I'm not sure if there are things there that would just get goofed up or something. @baxx never use word (have a copy just because but I don't use it;-) but have helped enough people with things over the years, these days I'd probably convert to html latexml or tex4ht then import the html into word and see what come out You should be able to cut and paste mathematics from your web browser to Word (or any of the Micorsoft Office suite). Unfortunately at present you have to make a small edit but any text editor will do for that.Givenx=\frac{-b\pm\sqrt{b^2-4ac}}{2a}Make a small html file that looks like<!... @baxx all the convertors that I mention can deal with document \newcommand to a certain extent. if it is just \newcommand\z{\mathbb{Z}} that is no problem in any of them, if it's half a million lines of tex commands implementing tikz then it gets trickier. @baxx yes but they are extremes but the thing is you just never know, you may see a simple article class document that uses no hard looking packages then get half way through and find \makeatletter several hundred lines of trick tex macros copied from this site that are over-writing latex format internals.
Positive measurable numerical function Set context $ \langle X,\Sigma_X\rangle\in \mathrm{MeasurableSpace}(X) $ postulate $f\in \mathcal M^+$ context $f\in \mathrm{Measurable}(X,\overline{\mathbb R})$ $x\in X$ postulate $f(x)\ge 0$ Discussion For the definition of the integral, it's crucial to know that for every $f\in \mathcal M^+$, there is a sequence $u_n$ with elements in the step functions $\mathcal T^+$, with $u_n\uparrow f$.
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Search Now showing items 1-2 of 2 Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE (Elsevier, 2017-11) Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ... Investigations of anisotropic collectivity using multi-particle correlations in pp, p-Pb and Pb-Pb collisions (Elsevier, 2017-11) Two- and multi-particle azimuthal correlations have proven to be an excellent tool to probe the properties of the strongly interacting matter created in heavy-ion collisions. Recently, the results obtained for multi-particle ...
It looks like you're new here. If you want to get involved, click one of these buttons! One of the main lessons of category theory is that whenever you think about some kind of mathematical gadget, you should also think about maps between gadgets of this kind. For example, when you think about sets you should also think about functions. When you think about vector spaces you should also think about linear maps. And so on. We've been talking about various kinds of monoidal preorders. So, let's think about maps between monoidal preorders. As I explained in Lecture 22, a monoidal preorder is a crossbreed or hybrid of a preorder and a monoid. So let's think about maps between preorders, and maps between monoids, and try to hybridize those. We've already seen maps between preorders: they're called monotone functions: Definition. A monotone function from a preorder $(X,\le_X)$ to $(Y,\le_Y)$ is a function $f : X \to Y$ such that $$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ for all elements $x,x' \in X$. So, these functions preserve what a preorder has, namely the relation $\le$. A monoid, on the other hand, has an associative operation $\otimes$ and a unit element $I$. So, a map between monoids should preserve th0se! That's how this game works. Just to scare people, mathematicians call these maps "homomorphisms": Definition. A homomorphism from a monoid $ (X,\otimes_X,I_X) $ to a monoid $ (Y,\otimes_Y,I_Y) $ is a function $f : X \to Y$ such that: $$ f(x \otimes_X x') = f(x) \otimes_Y f(x') $$ for all elements $x,x' \in X$, and $$ f(I_X) = I_Y .$$ You've probably seen a lot of homomorphisms between monoids. Some of them you barely noticed. For example, the set of integers $\mathbb{Z}$ is a monoid with addition as $\otimes$ and the number $0$ as $I$. So is the set $\mathbb{R}$ of real numbers! There's a function that turns each integer into a real number: $$ i: \mathbb{Z} \to \mathbb{R} . $$It's such a bland function you may never have thought about it: it sends each integer to itself, but regarded as a real number. And this function is a homomorphism! What does that mean? Look at the definition. It means you can either add two natural numbers and then regard the result as a real number... or first regard each of them as a real number and then add them... and you get the same answer either way. It also says that integer $0$, regarded as a real number, is the real number we call $0$. Boring facts! But utterly crucial facts. Computer scientists need to worry about these things, because for them integers and real numbers (or floating-point numbers) are different data types, and $i$ is doing "type conversion". You've also seen a lot of other, more interesting homomorphisms between monoids. For example, the whole point of the logarithm function is that it's a homomorphism. It carries multiplication to addition: $$ \log(x \cdot x') = \log(x) + \log(x') $$ and it carries the identity for multiplication to the identity for addition: $$ \log(1) = 0. $$ People invented tables of logarithms, and later slide rules, precisely for this reason! They wanted to convert multiplication problems into easier addition problems. You may also have seen linear maps between vector spaces. A vector space gives a monoid with addition as $\otimes$ and the zero vector as $I$; any linear map between vector spaces then gives a homomorphism. Puzzle 80. Tell me a few more homomorphisms between monoids that you routinely use, or at least know. I hope I've convinced you: monotone functions between preorders are important, and so are homomorphisms between monoids. Thus, if we hybridize these concepts, we'll get a concept that's likely to be important. It turns out there are a few different ways! The most obvious way is simply to combine all the conditions. There are other ways, so this way is called "strict": Definition. A strict monoidal monotone from a monoidal preorder $ (X,\le_X,\otimes_X,I_X) $ to a monoidal preorder $ (Y,\le_Y,\otimes_Y,I_Y) $ is a function $f : X \to Y$ such that: $$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ and $$ f(x) \otimes_Y f(x') = f(x \otimes_X x') $$ for all elements $x,x' \in X$, and also $$ I_Y = f(I_X) . $$ For example, the homomorphism $$ i : \mathbb{Z} \to \mathbb{R} ,$$ is a strict monoidal monotone: if one integer is $\le$ another, then that's still true when we regard them as real numbers. So is the logarithm function. What other definition could we possibly use, and why would we care? It turns out sometimes we want to replace some of the equations in the above definition by inequalities! Definition. A lax monoidal monotone from a monoidal preorder $(X,\le_X,\otimes_X,I_X)$ to a monoidal preorder $(Y,\le_Y,\otimes_Y,I_Y)$ is a function $f : X \to Y$ such that: $$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ and $$ f(x) \otimes_Y f(x') \le_Y f(x \otimes_X x') $$ for all elements $x,x' \in X$, and also $$ I_Y \le_Y f(I_X). $$Fong and Spivak call this simply a monoidal monotone, since it's their favorite kind. But I will warn you that others call it "lax". We could also turn around those last two inequalities: Definition. An oplax monoidal monotone from a monoidal preorder $(X,\le_X,\otimes_X,I_X)$ to a monoidal preorder $(Y,\le_Y,\otimes_Y,I_Y)$ is a function $f : X \to Y$ such that: $$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ and $$ f(x) \otimes_Y f(x') \ge_Y f(x \otimes_X x') $$ for all elements $x,x' \in X$, and also $$ I_Y \ge_Y f(I_X). $$ You are probably drowning in definitions now, so let me give some examples to show that they're justified. The monotone function $$ i : \mathbb{Z} \to \mathbb{R} $$ has a right adjoint $$ \lfloor \cdot \rfloor : \mathbb{R} \to \mathbb{Z} $$which provides the approximation from below to the nonexistent inverse of $i$: that is, $ \lfloor x \rfloor $ is the greatest integer that's $\le x$. It also has a left adjoint $$ \lceil \cdot \rceil : \mathbb{R} \to \mathbb{Z} $$which is the best approximation from above to the nonexistent inverse of $i$: that is, $ \lceil x \rceil $ is the least integer that's $\ge x$. Puzzle 81. Show that one of the functions $ \lfloor \cdot \rfloor : \mathbb{R} \to \mathbb{Z} $, $ \lceil \cdot \rceil : \mathbb{R} \to \mathbb{Z} $ is a lax monoidal monotone and the other is an oplax monoidal monotone, where we make the integers and reals into monoids using addition. So, you should be sensing some relation between left and right adjoints, and lax and oplax monoidal monotones. We'll talk about this more! And we'll see why all this stuff is important for resource theories. Finally, for the bravest among you: Puzzle 82. Find a function between monoidal preorders that is both lax and oplax monoidal monotone but not strict monoidal monotone. In case you haven't had enough jargon for today: a function between monoidal preorders that's both lax and oplax monoidal monotone is called strong monoidal monotone.
It looks like you're new here. If you want to get involved, click one of these buttons! I've spent four lectures on the logic of partitions; you may be wondering why. One reason was to give you examples illustrating this important fact: Theorem. Left adjoints preserve joins and right adjoints preserve meets. Suppose $f : A \to B$ and $g : B \to A$ are monotone functions between posets. Suppose that $f$ is the left adjoint of $g$, or equivalently, $g$ is the right adjoint of $f$. If the join of $a,a' \in A$ exists then so does the join of $f(a), f(a') \in B$, and $$ f(a \vee a') = f(a) \vee f(a'). $$ If the meet of $b,b' \in B$ exists then so does the meet of $g(b), g(b') \in A$, and $$ g(b \wedge b') = g(b) \wedge g(b'). $$ The proof is very easy, so this deserves being called a "Theorem" only because it's so fundamental! I will prove it later, in more generality. Right now let's see how it's relevant to what we've been doing. In Lecture 9 we saw something interesting about the subsets. Given any set $X$ there's a poset $P(X)$ consisting of all subsets of $X$. Given any function $f : X \to Y$ there's a monotone map $$ f^* : P(Y) \to P(X) $$sending any subset of $Y$ to its preimage under $f$. And we saw that $ f^{\ast} $ has both a left adjoint and a right adjoint. This means that $ f^{\ast} $ is both a right adjoint and a left adjoint. (Remember: having a left adjoint means being a right adjoint, and vice versa.) So by our Theorem, we see that $f^* : P(Y) \to P(X)$ preserves both meets and joins! You can also see this directly - see Puzzle 41 in Lecture 13. But what matters here is the general pattern. In Lecture 13 we also saw something interesting about partitions. Given any set $X$ there's a poset $ \mathcal{E}(X)$ consisting of all partitions of $X$. Given any function $f : X \to Y$ there's a monotone map $$ f^* : \mathcal{E}(Y) \to \mathcal{E}(X) $$ sending any partition of $Y$ to its pullback along $f$. And we saw that while $ f^{\ast} $ preserves meets, it does not preserve joins! So by our Theorem, we see that $f^* : \mathcal{E}(Y) \to \mathcal{E}(X)$ cannot be a left adjoint. On the other hand, it might be a right adjoint. And indeed it is! So, this strange difference between the logic of subsets and the logic of partitions is really all about adjoints. Puzzle 42. Given a function $f : X \to Y$ there is a way to push forward any partition $P$ on $X$ and get a partition $f_{!} (P)$ on $Y$. In pictures it looks like this: although this is not the most exciting example, since here $f_{!}(P)$ has just one part. Figure out the precise general description of $f_{!} (P)$. If you get stuck read Section 1.5.2 of Seven Sketches. Puzzle 43. Show that for any function $f : X \to Y$, pushing forward partitions along $f$ gives a monotone map $$ f_{!} : \mathcal{E}(X) \to \mathcal{E}(Y) . $$ Puzzle 44. Show that for any function $f : X \to Y$, $f^* : \mathcal{E}(Y) \to \mathcal{E}(X)$ is the right adjoint of $f_{!}: \mathcal{E}(X) \to \mathcal{E}(Y)$.
On wikipedia I can find that a de Sitter-space has maximal symmetry and a constant curvature. Recently the interest of de Sitter spaces has increased as it could serve as a model for the universe, an universe which has (almost) no matter density, however a sensible non-zero value of the cosmological constant (large amount of dark energy compared to much smaller amound of dark & visible matter). So how can the de Sitter-space serve a model for the universe if the latter is supposed to be flat whereas the de Sitter space has constant curvature (I guess it is meant non-zero constant curvature)? Those are two different curvatures you are talking about. First, you can talk about curvature of the spacetime i.e. treating one temporal and three spatial coordinates on equal footing. Then de Sitter spacetime has constant spacetime curvature, it's basically 4d hyperboloid. Realistic cosmological solutions also all have some spacetime curvature that however is not constant. On the other hand, in cosmology it's common to consider a slice of constant time getting some 3d space. The time in question is chosen in such a way that everything on this 3d space is to a high degree homogeneous. The resulting 3d space can be of various topology and has some 3d curvature that is completely different from the spacetime curvature. Now the observed cosmology corresponds to zero 3d curvature but non-zero spacetime curvature. You may ask what would be the 3d curvature for the de Sitter spacetime? The curious thing is how do you define the slice of constant time. The de Sitter spacetime is highly symmetric and you can actually slice it in many ways obtaining homogeneous space. Those possibilities fall into three categories that can be illustrated by this picture (which I made from this)So while de Sitter spacetime has some positive spacetime curvature it can be viewed as having arbitrary constant 3d curvature. UPD: Let me be more technical now. The spacetime is characterized by its metric $g_{\mu\nu}$ determined by the Einstein equations. You can however consider various coordinates on the same spacetime. The definition of the space (and "now") as a slice of constant time $t\equiv x^0=\mathrm{const}$ is totally artificial. You may consider any coordinate system you like with all kinds of non-equivalent "spaces". However in cosmology we are interested in the particular form of the metric, namely the Friedmann-Robertson-Walker (FRW) metric, \begin{equation} ds^2=g_{\mu\nu}dx^\mu dx^\nu=dt^2-a^2(t)d\vec{\Sigma}_k^2 \end{equation} where $d\vec{\Sigma}_k^2$ is a metric of 3d homogeneous space, either Euclidean space ($k=0$), unit sphere ($k=1$) or unit hyperboloid ($k=-1$). The slice of constant time $t=\mathrm{const}$ defines the 3d space with the metric $a^2(t)d\vec{\Sigma}_k^2$. If you try to consider different coordinates $g_{\mu\nu}$ transforms as any tensor does. If the transformation is not a time shift, spatial rotation or spatial translation (or its analog in case of sphere and hyperboloid) you will lose the FRW form of the metric. That's why our desire to study the particular form of the metric prefers certain coordinate system and therefore certain notion of 3d space (which of course exists only to the extent that we can approximate actual spacetime with FRW metric) The de Sitter spacetime however is exception. First, as I've already mentioned you can find three classes of the coordinate systems that will give you all three versions of the FRW metrics. So all three metrics, \begin{aligned} &ds^2=dt^2-e^{2t}d\vec{\Sigma}_0^2,\\ &ds^2=dt^2-\cosh^2(t)d\vec{\Sigma}_{+1}^2,\\ &ds^2=dt^2-\sinh^2(t)d\vec{\Sigma}_{-1}^2 \end{aligned} all describe the same de Sitter spacetime in different coordinates while being in the FRW form. Second, the de Sitter spacetime has $SO(1,4)$ symmetry group coming from the Lorentz symmetry of its possible embedding as a hyperboloid into the $(1+4)$ Minkowski spacetime. Because of that all three metrics actually admit continuous symmetries that you can consider as generalizations of the Lorentz boosts.
Stability of ground states for logarithmic Schrödinger equation with a $δ^{\prime}$-interaction Department of Mathematics, IME-USP, Cidade Universitária, CEP 05508-090, São Paulo, SP, Brazil $δ^{\prime}$ $i{\partial _t}u + \partial _x^2u + {\rm{ }}{\gamma ^\prime }(x)u + u{\mkern 1mu} {\rm{Log\|}}u\|2 = 0,(x,t) \in \mathbb{R} \times \mathbb{R} ,$ Mathematics Subject Classification:35Q51, 35Q55, 37K40, 34B37. Citation:Alex H. Ardila. Stability of ground states for logarithmic Schrödinger equation with a $δ^{\prime}$-interaction. Evolution Equations & Control Theory, 2017, 6 (2) : 155-175. doi: 10.3934/eect.2017009 References: [1] R. Adami and D. Noja, Existence of dynamics for a 1-d NLS equation perturbed with a generalized point defect [2] [3] R. Adami and D. Noja, Stability and symmetry-breaking bifurcation for the ground states of a NLS with a [4] R. Adami and D. Noja, Exactly solvable models and bifurcations: The case of the cubic NLS with a [5] R. Adami, D. Noja and N. Visciglia, Constrained energy minimization and ground states for NLS with point defects, [6] [7] J. Angulo and A. H. Ardila, Stability of standing waves for logarithmic Schrödinger equation with attractive delta potential, Indiana Univ. Math. J., to appear.Google Scholar [8] [9] [10] [11] [12] T. Cazenave, [13] [14] R. Fukuizumi and L. Jeanjean, Stability of standing waves for a nonlinear Schrödinger equation with a repulsive {D}irac delta potential, [15] R. Fukuizumi, M. Ohta and T. Ozawa, Nonlinear Schrödinger equation with a point defect, [16] R. Fukuizumi and A. Sacchetti, Bifurcation and stability for nonlinear Schrödinger equations with double well potential in the semiclassical limit, [17] A. Haraux, [18] [19] R.K. Jackson and M. Weinstein, Geometric analysis of bifurcation and symmetry breaking in a {G}ross-{P}itaevskii equation, [20] M. Kaminaga and M. Ohta, Stability of standing waves for nonlinear {S}chrödinger equation with attractive delta potential and repulsive nonlinearity, [21] [22] E.W. Kirr, P. Kevrekidis and D. Pelinovsky, Symmetry-breaking bifurcation in the nonlinear Schrödinger equation with symmetric potentials, [23] A. Kostenko and M. Malamud, Spectral theory of semibounded Schrödinger operators with $δ^{\prime}$-interactions, [24] S. Le Coz, R. Fukuizumi, G. Fibich, B. Ksherim and Y. Sivan, Instability of bound states of a nonlinear Schrödinger equation with a dirac potential, [25] [26] A. ~Sacchetti, Universal critical power for nonlinear Schrödinger equations with symmetric double well potential [27] [28] [29] K. Zloshchastiev, Logarithmic nonlinearity in theories of quantum gravity: {O}rigin of time and observational consequences, show all references References: [1] R. Adami and D. Noja, Existence of dynamics for a 1-d NLS equation perturbed with a generalized point defect [2] [3] R. Adami and D. Noja, Stability and symmetry-breaking bifurcation for the ground states of a NLS with a [4] R. Adami and D. Noja, Exactly solvable models and bifurcations: The case of the cubic NLS with a [5] R. Adami, D. Noja and N. Visciglia, Constrained energy minimization and ground states for NLS with point defects, [6] [7] J. Angulo and A. H. Ardila, Stability of standing waves for logarithmic Schrödinger equation with attractive delta potential, Indiana Univ. Math. J., to appear.Google Scholar [8] [9] [10] [11] [12] T. Cazenave, [13] [14] R. Fukuizumi and L. Jeanjean, Stability of standing waves for a nonlinear Schrödinger equation with a repulsive {D}irac delta potential, [15] R. Fukuizumi, M. Ohta and T. Ozawa, Nonlinear Schrödinger equation with a point defect, [16] R. Fukuizumi and A. Sacchetti, Bifurcation and stability for nonlinear Schrödinger equations with double well potential in the semiclassical limit, [17] A. Haraux, [18] [19] R.K. Jackson and M. Weinstein, Geometric analysis of bifurcation and symmetry breaking in a {G}ross-{P}itaevskii equation, [20] M. Kaminaga and M. Ohta, Stability of standing waves for nonlinear {S}chrödinger equation with attractive delta potential and repulsive nonlinearity, [21] [22] E.W. Kirr, P. Kevrekidis and D. Pelinovsky, Symmetry-breaking bifurcation in the nonlinear Schrödinger equation with symmetric potentials, [23] A. Kostenko and M. Malamud, Spectral theory of semibounded Schrödinger operators with $δ^{\prime}$-interactions, [24] S. Le Coz, R. Fukuizumi, G. Fibich, B. Ksherim and Y. Sivan, Instability of bound states of a nonlinear Schrödinger equation with a dirac potential, [25] [26] A. ~Sacchetti, Universal critical power for nonlinear Schrödinger equations with symmetric double well potential [27] [28] [29] K. Zloshchastiev, Logarithmic nonlinearity in theories of quantum gravity: {O}rigin of time and observational consequences, [1] Patricio Felmer, César Torres. Radial symmetry of ground states for a regional fractional Nonlinear Schrödinger Equation. [2] Wen Feng, Milena Stanislavova, Atanas Stefanov. On the spectral stability of ground states of semi-linear Schrödinger and Klein-Gordon equations with fractional dispersion. [3] Chuangye Liu, Zhi-Qiang Wang. A complete classification of ground-states for a coupled nonlinear Schrödinger system. [4] [5] Zupei Shen, Zhiqing Han, Qinqin Zhang. Ground states of nonlinear Schrödinger equations with fractional Laplacians. [6] Daniele Garrisi, Vladimir Georgiev. Orbital stability and uniqueness of the ground state for the non-linear Schrödinger equation in dimension one. [7] Jaime Angulo Pava, César A. Hernández Melo. On stability properties of the Cubic-Quintic Schródinger equation with $\delta$-point interaction. [8] Giuseppe Maria Coclite, Helge Holden. Ground states of the Schrödinger-Maxwell system with dirac mass: Existence and asymptotics. [9] Eugenio Montefusco, Benedetta Pellacci, Marco Squassina. Energy convexity estimates for non-degenerate ground states of nonlinear 1D Schrödinger systems. [10] Dongdong Qin, Xianhua Tang, Qingfang Wu. Ground states of nonlinear Schrödinger systems with periodic or non-periodic potentials. [11] Zhitao Zhang, Haijun Luo. Symmetry and asymptotic behavior of ground state solutions for schrödinger systems with linear interaction. [12] Alexander Komech, Elena Kopylova, David Stuart. On asymptotic stability of solitons in a nonlinear Schrödinger equation. [13] Kenji Nakanishi, Tristan Roy. Global dynamics above the ground state for the energy-critical Schrödinger equation with radial data. [14] Chenmin Sun, Hua Wang, Xiaohua Yao, Jiqiang Zheng. Scattering below ground state of focusing fractional nonlinear Schrödinger equation with radial data. [15] Xiaoyan Lin, Yubo He, Xianhua Tang. Existence and asymptotic behavior of ground state solutions for asymptotically linear Schrödinger equation with inverse square potential. [16] Silvia Cingolani, Mónica Clapp. Symmetric semiclassical states to a magnetic nonlinear Schrödinger equation via equivariant Morse theory. [17] Minbo Yang, Yanheng Ding. Existence and multiplicity of semiclassical states for a quasilinear Schrödinger equation in $\mathbb{R}^N$. [18] [19] Reika Fukuizumi. Stability and instability of standing waves for the nonlinear Schrödinger equation with harmonic potential. [20] François Genoud. Existence and stability of high frequency standing waves for a nonlinear Schrödinger equation. 2018 Impact Factor: 1.048 Tools Metrics Other articles by authors [Back to Top]
Maybe a little sketch maybe helpful here. Consider a one dimensional case first $$\ln y = \alpha_0 + \alpha_1 \ln x + \frac{1}{2}\beta_{11}\ln^2x \tag{1}$$ At first order (that is, ignore $\ln^2 x$), note that $$\frac{{\rm d}\ln y}{{\rm d}\ln x} = \alpha_1 \tag{2}$$ i.e. $\alpha_1$ is just the logarithmic slope of the output function (a.k.a elasticity), so it basically can help you identify the difference between the two plots below Now the second term, for that one note that $$\frac{{\rm d}^2\ln y}{{\rm d}\ln^2 x} = \beta_{11} \tag{3}$$ there's no mystery involved in the factor $2$ in Eq. (1), just there to make things simpler. Again, this is just the second derivative of the output function, so it can help you tell the difference between the two cases below Now comes the trick, How does this extend to higher dimensions? Well Eq.(1) still holds $$\alpha_k = \frac{\partial \ln y}{\partial \ln x_k}$$ so $\alpha_k$ is the elasticity keeping all other factors constant. Or in geometric terms $\boldsymbol{\alpha} = \nabla_{\ln {\bf x}} \ln y$. An the second derivatives (Hessian) are simply $\beta_{jk}$ $$\frac{\partial^2 \ln y}{\partial \ln x_k \partial \ln x_l} = \frac{1}{2}(\beta_{kl} + \beta_{lk})$$ If the matrix with entries $\beta_{kl}$ is diagonal, then the interpretation above is the same, it is just expressing the convexity of the production. If it is not, then you need to look for the eigenvalues of the matrix $$\boldsymbol{H} = \frac{1}{2}(\boldsymbol{\beta} + \boldsymbol{\beta}^T)$$ which always exist and are real, since $\boldsymbol{H} = \boldsymbol{H}^T$
Lipschitz function Let a function $f:[a,b]\to \mathbb R$ be such that for some constant M and for all $x,y\in [a,b]$ \begin{equation}\label{eq:1} \|f(x)-f(y)\| \leq M\|x-y\|. \end{equation} Then the function $f$ is called Lipschitz on $[a,b]$, and one writes $f\in \operatorname{Lip}_M[a,b]$. The concept can be readily extended to general maps $f$ between two metric spaces $(X,d)$ and $(Y, \delta)$: such maps are called Lipschitz if for some constant $M$ one has \begin{equation}\label{eq:2} \delta (f(x), f(y)) \leq M d (x,y) \qquad\qquad \forall x,y\in X\, . \end{equation} A mapping $f:X\to Y$ is called bi-Lipschitz if it is Lipschitz and has an inverse mapping $f^{-1}:f(X)\to X$ which is also Lipschitz. Properties If a mapping $f:U\to \mathbb R^k$ is Lipschitz (where open set $U\subset\mathbb R^n$), then $f$ is differentiable almost everywhere (Rademacher theorem). How to Cite This Entry: Lipschitz function. Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Lipschitz_function&oldid=28901
3GPP 5G has been focused on structured LDPC codes known as quasi-cyclic low-density parity-check (QC-LDPC) codes, which exhibit advantages over other types of LDPC codes with respect to the hardware implementations of encoding and decoding using simple shift registers and logic circuits.5G NR QC-LDPC Circulant Permutation Matrix A circular permutation matrix ${\bf I}(P_{i,j})$ of size $Z_c \times Z_c$ is obtained by circularly shifting the identity matrix $\bf I$ of... For any B-DMC $W$, the channels $\{W_N^{(i)}\}$ polarize in the sense that, for any fixed $\delta \in (0, 1)$, as $N$ goes to infinity through powers of two, the fraction of indices $i \in \{1, \dots, N\}$ for which $I(W_N^{(i)}) \in (1 − \delta, 1]$ goes to $I(W)$ and the fraction for which $I(W_N^{(i)}) \in [0, \delta)$ goes to $1−I(W)^{[1]}$. Mrs. Gerber’s Lemma Mrs. Gerber’s Lemma provides a lower bound on the entropy of the modulo-$2$ sum of two binary random... Channel Combining Channel combining is a step that combines copies of a given B-DMC $W$ in a recursive manner to produce a vector channel $W_N : {\cal X}^N \to {\cal Y}^N$, where $N$ can be any power of two, $N=2^n, n\le0^{[1]}$. The notation $u_1^N$ as shorthand for denoting a row vector $(u_1, \dots , u_N)$. The vector channel $W_N$ is the virtual channel between the input sequence $u_1^N$ to a linear encoder and the output sequence $y^N_1$ of $N$... There are many DSP48 Slices in most Xilinx® FPGAs, one DSP48 slice in Spartan6® FPGA is shown in Figure 1, the structure may different depending on the device, but broadly similar. Figure 1: A whole DSP48A1 Slice in Spartan6 (www.xilinx.com)2. Symmetric Systolic Half-band FIR Figure 2: Symmetric Systolic Half-band FIR Filter3. Two-channel Symmetric Systolic Half-band FIR Figure 3: 2-Channel... 3GPP 5G has been focused on structured LDPC codes known as quasi-cyclic low-density parity-check (QC-LDPC) codes, which exhibit advantages over other types of LDPC codes with respect to the hardware implementations of encoding and decoding using simple shift registers and logic circuits.5G NR QC-LDPC Circulant Permutation Matrix A circular permutation matrix ${\bf I}(P_{i,j})$ of size $Z_c \times Z_c$ is obtained by circularly shifting the identity matrix $\bf I$ of... There are many DSP48 Slices in most Xilinx® FPGAs, one DSP48 slice in Spartan6® FPGA is shown in Figure 1, the structure may different depending on the device, but broadly similar. Figure 1: A whole DSP48A1 Slice in Spartan6 (www.xilinx.com)2. Symmetric Systolic Half-band FIR Figure 2: Symmetric Systolic Half-band FIR Filter3. Two-channel Symmetric Systolic Half-band FIR Figure 3: 2-Channel... Channel Combining Channel combining is a step that combines copies of a given B-DMC $W$ in a recursive manner to produce a vector channel $W_N : {\cal X}^N \to {\cal Y}^N$, where $N$ can be any power of two, $N=2^n, n\le0^{[1]}$. The notation $u_1^N$ as shorthand for denoting a row vector $(u_1, \dots , u_N)$. The vector channel $W_N$ is the virtual channel between the input sequence $u_1^N$ to a linear encoder and the output sequence $y^N_1$ of $N$... For any B-DMC $W$, the channels $\{W_N^{(i)}\}$ polarize in the sense that, for any fixed $\delta \in (0, 1)$, as $N$ goes to infinity through powers of two, the fraction of indices $i \in \{1, \dots, N\}$ for which $I(W_N^{(i)}) \in (1 − \delta, 1]$ goes to $I(W)$ and the fraction for which $I(W_N^{(i)}) \in [0, \delta)$ goes to $1−I(W)^{[1]}$. Mrs. Gerber’s Lemma Mrs. Gerber’s Lemma provides a lower bound on the entropy of the modulo-$2$ sum of two binary random... There are many DSP48 Slices in most Xilinx® FPGAs, one DSP48 slice in Spartan6® FPGA is shown in Figure 1, the structure may different depending on the device, but broadly similar. Figure 1: A whole DSP48A1 Slice in Spartan6 (www.xilinx.com)2. Symmetric Systolic Half-band FIR Figure 2: Symmetric Systolic Half-band FIR Filter3. Two-channel Symmetric Systolic Half-band FIR Figure 3: 2-Channel... Channel Combining Channel combining is a step that combines copies of a given B-DMC $W$ in a recursive manner to produce a vector channel $W_N : {\cal X}^N \to {\cal Y}^N$, where $N$ can be any power of two, $N=2^n, n\le0^{[1]}$. The notation $u_1^N$ as shorthand for denoting a row vector $(u_1, \dots , u_N)$. The vector channel $W_N$ is the virtual channel between the input sequence $u_1^N$ to a linear encoder and the output sequence $y^N_1$ of $N$... For any B-DMC $W$, the channels $\{W_N^{(i)}\}$ polarize in the sense that, for any fixed $\delta \in (0, 1)$, as $N$ goes to infinity through powers of two, the fraction of indices $i \in \{1, \dots, N\}$ for which $I(W_N^{(i)}) \in (1 − \delta, 1]$ goes to $I(W)$ and the fraction for which $I(W_N^{(i)}) \in [0, \delta)$ goes to $1−I(W)^{[1]}$. Mrs. Gerber’s Lemma Mrs. Gerber’s Lemma provides a lower bound on the entropy of the modulo-$2$ sum of two binary random... 3GPP 5G has been focused on structured LDPC codes known as quasi-cyclic low-density parity-check (QC-LDPC) codes, which exhibit advantages over other types of LDPC codes with respect to the hardware implementations of encoding and decoding using simple shift registers and logic circuits.5G NR QC-LDPC Circulant Permutation Matrix A circular permutation matrix ${\bf I}(P_{i,j})$ of size $Z_c \times Z_c$ is obtained by circularly shifting the identity matrix $\bf I$ of...
If you are aware of elementary facts of geometry, then you might know that the area of a disk with radius $ R$ is $ \pi R^2$ . The radius is actually the measure(length) of a line joining the center of disk and any point on the circumference of the disk or any other circular lamina. Radius for a disk is always same, irrespective of the location of point at circumference to which you are joining the center of disk. The area of disk is defined as the ‘ measure of surface‘ surrounded by the round edge (circumference) of the disk. The area of a disk can be derived by breaking it into a number of identical parts of disk as units — calculating their areas and summing them up till disk is reformed. There are many ways to imagine a unit of disk. We can imagine the disk to be made up of several concentric very thin rings increasing in radius from zero to the radius of disc. In this method we can take an arbitrary ring, calculate its area and then in similar manner, induce areas of other rings — sum them till whole disk is obtained. Mathematically, we can imagine a ring of with radius $ x$ and thickness $ dx$ , anywhere in the disk having the same center as disk, calculate its area and then sum up (integrate) it from $ x=0$ to $ x=R$ . Area of a thin ring is since $ \pi x dx$ . And after integrating we get, area of disk $ A=2 \int_0^R \pi x dx$ or $ A=\pi R^2$ . There is another approach to achieve the area of a disk, A. Imagine a disk is made up of a number equal sections or arcs. If there are $ n$ number of arcs then interior angle of an arc is exactly $ \frac{2\pi}{n}$ , since $ 2 \pi$ is the total angle at the center of disk and we are dividing this angle into $ n$ equal parts. If we join two ends of each sections –we can get $ n$ identical triangles in which an angle with vertex O is $ \frac{2 \pi}{n}$ . Now, if we can calculate the area of one such section, we can approach to the area of the disk intuitively. This approach is called the method of exhaustion. Let, we draw two lines joining center O of the disk and points A & B at circumference. It is clear that OB and OA are the radius of the disk. We joined points A and B in order to form a triangle OAB. Now consider that the disk is made up of n-number of such triangles. We see that there is some area remaining outside the line AB and inside the circumference. If we had this triangle thinner, the remaining area must be lesser. So, if we increase the number of triangles in disk —-we decrease the remaining areas. We can achieve to a point where we can accurately calculate the area of disk when there are infinitely many such triangles or in other words area of one such triangle is very small. So our plan is to find the area of one triangle —sum it up to n — make $ n$ tending to infinity to get the area of disk. It is clear that the sum of areas of all identical triangles like OAB must be either less than or equal to area of the disk. We can call triangles like OAB as inscribed triangles. Now, if we draw a radius-line OT’, perpendicular to AB at point T and intersecting the circumference at point T’, we can easily draw another triangle OA’B’ as shown in figure. AOB and A’OB’ are inscribed and superscribed triangles of disk with same angle at vertex O. So, it is clear that the angle A’OB’ is equal to the angle AOB. Triangle A’OB’ is larger than the circular arc OAB and circular arc OAB is larger than the inscribed triangle AOB. Also, the sum of areas of triangles identical to OA’B’ is either greater than or equal to area of the disk. Let disk be divided into n- such inscribed and (thus) superscribed triangles. Since, total angle at point O is 360° or 2π —-the angles AOB and A’OB’ are of $ 2 \pi/n$ . And also since OT and OT’ are normals at chord AB and line A’B’ respectively, then they must divide the angles AOB and A’OB’ in two equal parts, each of $ \pi/n$ radians. In triangle AOB, the area of triangle AOB is the sum of the area of triangles AOT and BOT. But since both are equal to each other in area, area of AOB must be twice of the area of triangle AOB (or BOT). Our next target is to find, the area of AOT in order to find the area of AOB. From figure, area of $ \bigtriangleup{AOT}= \frac{1}{2} \times AT \times OT$ ….(1) $ OA=R$ And, $ \angle{AOT}= \frac{\pi}{n}$ . Thus, $ \frac{AT}{OA}=\sin {\frac{\pi}{n}}$ or, $ AT=OA \sin {(\pi/n)}$ or, $ AT=R \sin {(\pi/n)}$ …..(2) Similarly, $ OT=R \cos {(\pi/n)}$ Therefore, area of $ \bigtriangleup {AOT}=\frac{1}{2} \times R \sin {(\pi/n)} \times R \cos {(\pi/n)}=\frac{1}{2} R^2 \sin{(\pi/n)} \cos {(\pi/n)}$ And thus, area of $ \bigtriangleup{AOB}=2 \times \frac{1}{2} R^2 \sin{(\pi/n)} \cos{(\pi/n)}=R^2 \sin{(\pi/n)} \cos{(\pi/n)}$ Since there are $ n$ such triangles: sum of areas of such triangles $ S_1=n \times R^2 \sin{(\pi/n)} \cos{(\pi/n)}$ . In triangle A’OB’, the total area of triangle A’OB’ is the sum of areas two identical triangles A’OT’ and B’OT’. Therefore, area of $ \bigtriangleup{A’OB’}=2 \times \text{area of} \bigtriangleup{A’OT’}$ . And area of $ \bigtriangleup{A’OT’}=\frac{1}{2} \times AT’ \times OT’$ . We have $ OT’=R$ and angle A’OT’=$ \frac{\pi}{n}$ Thus, A’T’/OT’= $ \tan{\frac{\pi}{n}}$ or, $ A’T’=OT’ \tan{\frac{\pi}{n}} =R \tan{\frac{\pi}{n}}$ . Therefore, area of triangle A’OT’= $ \frac{1}{2} \times R \times R \tan{\frac{\pi}{n}}=\frac{1}{2} R^2 \tan{\frac{\pi}{n}}$ . Hence, area of $ \bigtriangleup A’OB’=2 \times \frac{1}{2}R^2 \tan{\frac{\pi}{n}}$ As, there are $ n$ such triangles: sum of areas of those triangles $ S_2=n \times R^2 \tan{\frac{\pi}{n}}$ . As it is clear that Sum of areas of triangles like $ \bigtriangleup AOB$ is an approximation for the area of disk from below, i.e., $ S_1 \le A$ when $ n \to \infty$ or, $ \displaystyle{\lim_{n \to \infty}} n \times R^2 \sin{(\pi/n)} \cos{(\pi/n)} \le A$ $ \displaystyle{\lim_{n \to \infty}} n \times R^2 \frac{\pi}{n} \dfrac{\sin{(\pi/n)}}{\pi/n} \cos{(\pi/n)} \le A$ $ \pi R^2 \le A \ldots (I)$ Similarly, $ \displaystyle{\lim_{n\to \infty}} n \times R^2 \tan{\frac{\pi}{n}} \ge A$ or, $ \pi R^2 \ge A \ldots (II)$ From (I) and (II), we have $ A=\pi R^2$ . So the area of a disk is $ \pi R^2$ . Want to catch up with studies or work and access your important Windows Application and software as you travel? It’s possible with a cloud desktop from CloudDesktopOnline.com . Also, for more hosted Microsoft applications such as Exchange, SharePoint, Dynamics CRM, Project Server and more, try Apps4Rent. Feel free to ask questions, send feedback and even point out mistakes. Great conversations start with just a single word. How to write better comments?
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 ...
If you only need to pick 5 out of 10 and want equal weights then just enumerate all 252 possibilities (as pointed out above) and compute the portfolio volatility $(\textbf{1}'K^{(i)}\textbf{1})^{1/2} = \left( \sum_{ij}K^{(i)}_{ij} \right)^{1/2}$, where $K^{(i)}$ is the covariance matrix for the $i$th subset. Then use whatever subset gives the lowest portfolio volatility. Here you are minimizing portfolio volatility so you will be biased towards lower volatility stocks. If you don't care about volatility per se and just want to minimize the contribution to portfolio risk related to correlation (somewhat loosely defined) then you can use the Most Diversified Portfolio (MDP) method. This method aims to minimize the diversification ratio $\frac{w'\sigma^{(i)}}{\left(w'K^{(i)}w\right)^{1/2}} =\frac{\sum_j\sigma_j}{ \left( \sum_{ij}K^{(i)}_{ij} \right)^{1/2}}$ Again, just plug in values for each subset and use whatever gives the largest value. Personally, I would argue that a few aspects of what your doing are inefficient. Why equal weights? If you have a covariance matrix then you can almost always find less risky portfolios. Each stock has a different volatility so equal weights tends to take too much risk in more volatile stocks. Why consider only your top 10? It is possible your best 5-stock portfolio includes stocks outside of your top-10 rankings due to correlations. Instead, consider attempting to generate expected returns for your stocks. You can do this by running a simple linear regression using your sorting metric. As has been pointed out, the full mean-variance optimization is hard to solve when you have a cardinality constraint and a large number of stocks to consider. A common approach is to employ $l_1$ norm based methods. The gist of it is instead of solving the standard mean-variance QP $\min_w \{ \lambda w'Kw - r'w \}, w \geq 0, \sum_i w_i = 1$, drop the budget constraint and add an $l_1$ penalty, i.e. $\min_w \{ \lambda w'Kw - r'w + \gamma \|\|w\|\|_1 \}, w \geq 0$. As you slowly increase $\gamma$ the $w$ vector will get sparser. Stop once you only have 5 non-zero values. Afterwards, re-scale the weights to sum to one. This version of the problem is a convex relaxation of the actual cardinality constrained problem. The $l_1$-norm penalty can also be motivated as the solution to a robust portfolio optimization problem where returns are uncertain, but satisfy a box constraint.
What is Variance in Statistics? Variance meaning – It is a measure of how data points differ from the mean. According to layman’s terms, it is a measure of how far a set of data( numbers) are spread out from their mean (average) value. For the purpose of solving questions, it is, Var (X) = E[ ( X – $\mu$)) 2] Put into words; this means that variance is the expectation of the deviation of a random set of data from its mean value, squared. Here, “µ” is equal to E(X) so the above equation may also be expressed as, Var(X) = E[(X – E(X) 2)] Var(X) = E[ X 2 -2X E(X) +(E(X)) 2] Var(X) = E(X 2) -2 E(X) E(X) + (E(X)) 2 Var(X) = E(X 2) – (E(X)) 2 Now let’s have a look at the relationship between Variance and Standard Deviation. Variance Standard Deviation Standard deviation is the positive square root of the variance. The symbols σ and S are used correspondingly to represent population and sample standard deviations. Standard Deviation is a measure of how spread out the data is. Its formula is simple; it is the square root of the variance for that data set. It’s represented by the Greek symbol sigma (σ ). Formulas The corresponding formulas are hence, Population standard deviation σ = $\sqrt{\frac{\sum (X-\mu )^{2}}{N}}$ and Sample standard deviation S = $\sqrt{\frac{\sum (X-\overline{X})^{2}}{n-1}}$ Variance Properties The variance, var(X) of a random variable X has the following properties. Var(X + C) = Var(X), where C is a constant. Var(CX) = C 2.Var(X), where C is a constant. Var(aX + b) = a 2.Var(X), where a and b are constants. If X 1, X 2,……., X nare n independent random variables, then Var(X 1 + X 2 +……+ X n) = Var(X 1) + Var(X 2) +……..+Var(X n). Learning Variance Concept with Example The concepts mentioned above sound a little boring don’t they? The concept of statistics often appears this way since it means dealing with large volumes of data; It is essential that you, as a student, understand that these are not just numbers; If read properly, it tells you a story. So let’s take a fun example of how to calculate variance in everyday life situation: Let’s say the heights at their shoulders (in mm) are 610, 450, 160, 420, 310. Mean and Variance are interrelated. The first step is finding the mean which is done as follows, Mean = ( 610+450+160+420+310)/ 5 = 390 So the mean average is 390 mm. Let’s plot this on the chart. To calculate the Variance, compute the difference of each from the mean, square it and find then find the average once again. So for this particular case the variance is : = (220 2 + 60 2 + (-230) 2 +30 2 + (-80) 2)/5 = (48,400 + 3,600 + 52,900 + 900 + 6,400)/5 Final answer : Variance = 23,700 Variance Example Example: Find the variance of the numbers 3, 8, 6, 10, 12, 9, 11, 10, 12, 7. Solution: Step 1: Compute the mean of the 10 values given. x = (3+8+6+10+12+9+11+10+12+7) / 10 = 88 / 10 = 8.8 Step 2: Make a table with three columns, one for the X values, the second for the deviations and the third for squared deviations. Value X X – $\bar{X}$ $(X-\bar{X})^2$ 3 -5.8 33.64 8 -0.8 0.64 6 -2.8 7.84 10 1.2 1.44 12 3.2 10.24 9 0.2 0.04 11 2.2 4.84 10 1.2 1.44 12 3.2 10.24 7 -1.8 3.24 Total 0 73.6 Step 3: As the data is not given as sample data we use the formula for population variance. σ2 = $\frac{\sum (X-\bar X )^{2}}{N}$ (Here $\mu = \bar X$) = 73.6 / 10 = 7.36 Stay tuned with Byju’s to learn more about Covariance Formula and other maths concepts with the help of interactive videos.
One can construct a Chebyshev series approximation to the integrand for an interval, such as -5 <= x <=5 mentioned in the comments, and integrate it to get a series expansion for the integral. It is well known that Chebyshev series representation have numerical advantages over power series. I saw a comment about a NPU-supported method, but I don't know what that is, this being a Mathematica site. Most numerical systems have access to an FFT, in one way or another, I think, and, of course, to trigonometric functions. That is all that is really needed. Some auxiliary functions used here (code below): chebSeries[f, {a, b}, n, precision] computes the Chebyshev expansion of order n for f[x] over a <= x <= b. iCheb[c, {a, b}, k] integrates the Chebyshev series c, plus k. f = chebFunc[c,{a,b}], computes the function represented by the Chebyshev coefficients c = {c0, c1,...} over the interval {a, b}. The first step is to compute the coefficients of the integrand as a function of x for a given value of c. The are saved (memoized) for the sake of speed, but it is not essential. An antiderivative of the integrand is computed with iCheb[coeff[c], {-5, 5}]. Depending on c, one needs a series of order 70 to 90+ to get a theoretical error of less than machine precision, so computing one of order 2^7 = 128 is sufficient for all c. (See Boyd, Chebyshev and FourierSpectral Methods, Dover, New York,2001, ch. 2., for a discussion of the convergence theory of Chebyshev series.) ClearAll[coeffs]; coeffs[c_?NumericQ] := coeffs[c] = Module[{ cc, (* Chebyshev coefficients ) pg = 16, ( Precision goal: ) sum, ( sum of tail = error bound ) max, ( max coefficient to measure rel. error ) len}, ( how many terms of the tail to drop ) cc = chebSeries[Function[x, Exp[-x^2] Erf[x + c]], {-5, 5}, 128]; max = Max@Abs@cc; sum = 0; ( trim tail of series of unneeded coefficients (smaller than desired precision) ) len = LengthWhile[Reverse[cc], (sum += Abs[#1]) < 10^-pg max &]; Drop[cc, -len] ]; Next we can define the user's sought-after function in terms of the antiderivative: func[a_, b_, c_] := With[{antiderivative = chebFunc[ N@iCheb[coeffs[SetPrecision[c, Infinity]], {-5, 5}, 0], {-5, 5}]}, antiderivative[b] - antiderivative[a] ]; The following computes the relative and absolute error of func on a hundred random inputs, using a high-precision NIntegrate[] to compute the "true" value. cmp[a_, b_, c_] := {(#1 - #2)/#2, #1 - #2} &[ func[a, b, c], NIntegrate[Exp[-x^2] Erf[x + SetPrecision[c, Infinity]], {x, a, b}, WorkingPrecision -> 40] ] ListLinePlot[ Transpose@RealExponent[cmp @@@ RandomReal[{-5, 5}, {100, 3}]], PlotRange -> {-20, 0}, PlotLabel -> "Error", PlotLegends -> {"Rel.", "Abs."}] The yellow line shows the absolute error is limited to a few ulps. The theoretical bound on the error does not take into account rounding error (in both the coefficients and the evaluation of the series). The lines that drop off the bottom are the result of an error of . zero Auxiliary functions ( Chebyshev extreme points ) chebExtrema::usage = "chebExtrema[n,precision] returns the Chebyshev extreme points of order n"; chebExtrema[n_, prec_: MachinePrecision] := N[Cos[Range[0, n]/n Pi], prec]; ( Chebyshev series approximation to f ) Clear[chebSeries]; chebSeries::usage = "chebSeries[f,{a,b},n,precision] computes the Chebyshev expansion \ of order n for f[x] over a <= x <= b."; chebSeries[f_, {a_, b_}, n_, prec_: MachinePrecision] := Module[{x, y, cc}, x = Rescale[chebExtrema[n, prec], {-1, 1}, {a, b}]; y = f /@ x; ( function values at Chebyshev points ) cc = Sqrt[2/n] FourierDCT[y, 1]; ( get coeffs from values ) cc[[{1, -1}]] /= 2; ( adjust first & last coeffs ) cc ]; ( Integrate a Chebyshev series -- cf. Clenshaw-Norton, Comp.J., 1963, p89, eq. (12) ) Clear[iCheb]; iCheb::usage = "iCheb[c, {a, b}, k] integrates the Chebyshev series c, plus k"; iCheb[c0_, {a_, b_}, k_: 0] := Module[{c, i, i0}, c[1] = 2 First[c0]; c[n_] /; 1 < n <= Length[c0] := c0[[n]]; c[_] := 0; i = 1/2 (b - a) Table[(c[n - 1] - c[n + 1])/(2 (n - 1)), {n, 2, Length[c0] + 1}]; i0 = i[[2 ;; All ;; 2]]; Prepend[i, k - Sum[(-1)^ni0[[n]], {n, Length[i0]}]]] (* chebFunc[c,...] computes the function represented by a Chebyshev series ) chebFunc::usage = "f = chebFunc[c,{a,b}], c = {c0,c1,...} Chebyshev coefficients, over the interval {a,b} y = chebFunc[c,{a,b}][x] evaluates the function"; chebFunc[c_, dom_][x_] := chebFunc[c, dom, x]; chebFunc[c_?VectorQ, {a_, b_}, x_] := Cos[Range[0, Length[c] - 1] ArcCos[(2 x - (a + b))/(b - a)]].c; Update: Comparison of Chebyshev and power series Perhaps it would be worth illustrating the difference between power series and Chebyshev series approximations for those who are not familiar with it. (One should become familiar with it, for Chebyshev expansions are to functions what decimal expansions are to numbers.) Key differences: Symbolic series expansion of the function Erf[x + c] grows extremely fast and takes a much longer time to evaluate than the DCT-I used to compute the Chebyshev coefficients. Attempting a degree 40 expansion hanged the computer and I had to kill the kernel. Aside from not being able to compute the series expansion to arbitrary order, it is probably impossible to get convergence for a fixed precision due to rounding error. At machine precision, you cannot even get 2 digits throughout the interval {c, -5, 5}, for a = -4, b = 4 up to order 25. OTOH, the Chebyshev series has an exponential order of convergence and can nearly achieve machine precision with machine precision coefficients. It is fairly easy to figure out when you have enough terms of a Chebyshev series $\sum a_j T_j$, because the error is bounded by the tail $\sum \|a_j\|$ and the $a_n \rightarrow 0$ roughly geometrically on average. If you don't have fast trigonometric functions, then instead of the code in chebFunc above, you can use Clenshaw's algorithm (see chebeval) to evaluate the series. Here's another implementation of Mariusz's power series idea. I speed up the integration with a "power rule" int[{n}] for$$ \int \exp\left(-x^2\right) x^n \; dx\,.$$Of course it turned out that Series was the bigger bottleneck. ClearAll[sol, int]; int[{0}] = Integrate[Exp[-x^2], x, Assumptions -> x ∈ Reals]; int[{n_}] = Integrate[Exp[-t^2] t^n, {t, 0, x}, Assumptions -> n > 0 && n ∈ Integers && t ∈ Reals]; $seriesCoefficientPart = 3; sol[n_] := sol[n] = Total@MapIndexed[ First@Differences[#1 int[#2 - 1] /. {{x -> a}, {x -> b}}] &, Series[Erf[x + c], {x, 0, n}][[$seriesCoefficientPart]] ]; ( Times ) First@AbsoluteTiming@sol /@ {2, 10, 20, 22, 23, 24, 25} ( {0.013267, 0.071065, 2.01908, 7.6296, 23.4752, 33.1553, 48.3542} ) ( Sizes ) LeafCount /@ sol /@ {2, 10, 20, 22, 23, 24, 25} ( {163, 693, 2413, 2765, 1100459, 1779935, 2879267} ) Chebyshev speed for c = 4, per evaluation of c: First@AbsoluteTiming@func[a, b, 4, #] & /@ {2, 10, 20, 25} ( {0.002047, 0.000184, 0.000248, 0.000276} *) To illustrate the issue with power series, the graphics below shows the error in approximating Erf[x + c] by its Taylor series (times Exp[-x^2] for c = -4, -2, 0, 2, 4 and various orders. It does pretty good for Abs[x] < 1 as the order increases, but it gets worse for Abs[x] > 4. GraphicsRow[ Table[ Plot[ Evaluate@Table[ Exp[-x^2] (Erf[x + c] - Normal@Series[Erf[x + c], {x, 0, n}]) // RealExponent, {c, -4, 4, 2}], {x, -5, 5}, PlotRange -> {-18, 0}, Frame -> True, Axes -> False, PlotLabel -> Row[{"Order ", n}], AspectRatio -> 1, FrameLabel -> {"x", "Log error"}], {n, {2, 10, 20, 25}}], PlotLabel -> "Error in approximating integrand by power series"] The two plots below compare the absolute error of approximating by power series and by Chebyshev series. The convergence of the Chebyshev series is remarkable by comparison. Plot[Evaluate@Table[ sol[n] - exact[a, b, c] /. {a -> -4, b -> 4} // RealExponent, {n, {2, 10, 20, 25}}], {c, -5, 5}, PlotRange -> {-18, 0}, Frame -> True, Axes -> False, GridLines -> {None, -Range[2, 16, 2]}, PlotLabel -> "Log error for power series of order n", FrameLabel -> {"c", "Log error"}, PlotLegends -> {2, 10, 20, 25}] Plot[Evaluate@Table[ func[a, b, c, n] - exact[a, b, c] /. {a -> -4, b -> 4} // RealExponent, {n, {2, 10, 20, 30, 40, 50, 60}}], {c, -5, 5}, PlotRange -> {-18, 0}, Frame -> True, Axes -> False, GridLines -> {None, -Range[2, 16, 2]}, PlotLabel -> "Log error for Chebyshev series of order n", FrameLabel -> {"c", "Log error"}, PlotLegends -> {2, 10, 20, 30, 40, 50, 60}] Determining the order of the approximation: trim[cc_, eps_] := Module[{sum, max, len}, max = Max@Abs@cc; sum = 0; len = LengthWhile[Reverse[cc], (sum += Abs[#1]) < eps max &]; Drop[cc, -len] ] Manipulate[ With[{cc = iCheb[chebSeries[Exp[-#^2] Erf[# + c] &, -5, 5, 128, 32], {-5, 5}]}, With[{order = Length@trim[cc, 10^-accuracy]}, ListPlot[ RealExponent@cc, PlotLabel -> Column[{ Row[{"Chebyshev coefficients a[n] for ", HoldForm["c"] -> Chop[N[c, {2, 1.5}], 0.05], ", "}], Row[{"For accuracy ", SetPrecision[10^-accuracy, 2], " use order ", order}] }, Alignment -> Center], Frame -> True, FrameLabel -> {"n", "exponent of a[n]"}, GridLines -> {{order}, {-accuracy}}, PlotRange -> {-31, 1}] ]], {{c, 4}, -5, 5, 1/10}, {{accuracy, 16.}, 2, 28}] Addendum: Failed ideas - Maybe someone can make them work... The Chebyshev series of Exp[-x^2] can be computed over {-x0, x0} exactly in terms of modified Bessel functions BesselI[]. Therefore, so can the series for Erf[x]. I was seduced into trying to come up with a way to compute the OP's function in this way but the +c in Erf[x + c] was too ornery. One thing that would be needed is a way to write ChebyshevT[n, x + c] as a Chebyshev series in ChebyshevT[n, x]. The coefficients would be polynomials in c (with integer coefficients), which themselves could be represented as Chebyshev expansions. This can be done, in fact, but it's a bit cumbersome and slow. Further the Chebyshev coefficients for n = 64 get bigger than 2^100, and I worried about numerical stability. For the moment, I have given up without testing it. The way above seems superior, in simplicity, as well as (probably) in speed and numerics.
I've been reading up on the internet as to why the sky is blue. The answer usually cites Rayleigh scattering that I've checked on wikipedia: https://en.wikipedia.org/wiki/Rayleigh_scattering: $$ I=I_0 \frac{1+\cos^2\theta}{2R^2}\left(\frac{2\pi}{\lambda}\right)^4\left(\frac{n^2-1}{n^2+2}\right)^2\left(\frac{d}{2}\right)^6 $$ This answer raises more questions in my mind, that I hope some people can help to answer. First of all, I can't understand the $\lambda^{-4}$ dependence in that equation. It means that the scattered intensity goes to infinity as $\lambda\to 0$. It also means that the observed intensity $I$ can be greater than the incident intensity $I_0$. On several web pages, the $\lambda^{-4}$ dependence has been cited to explain why the sky is blue, but that doesn't make sense either. According to this reasoning the sky should be purple or indigo which has a higher frequency than blue. I've seen another explanation online that says that the sunlight impinging on our atmosphere has less indigo frequency than blue. However I can see the indigo part of a rainbow; it doesn't seem significantly dimmer than the blue part, so sunlight must have a decent indigo frequency content and as mentioned, the 4th power is a very strong dependency. This argument says that the sky ought to be indigo. A second question I have is concerning the angle dependency. $1+\cos^2 \theta$ has a maximum at zero and at 180, and a minimum at 90 degrees. According to this dependency the sky should look brightest when looking at 0 degrees (towards the sun), and 180 degrees (with the sun to your back), but it should have half the intensity at 90 degrees. This doesn't match with our experience of the sky. Given the hand-wavy nature of the explanations, I wonder if Rayleigh scattering truly is the explanation for why the sky is blue.
№ 9 All Issues Volume 60, № 9, 2008 Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1155-1156 Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1157–1167 We study the relationship between the (min, max)-equivalence of posets and properties of their quadratic Tits form related to nonnegative definiteness. In particular, we prove that the Tits form of a poset S is nonnegative definite if and only if the Tits form of any poset $(\min, \max)$-equivalent to S is weakly nonnegative. Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1168–1188 We study the inverse problem for the Strum-Liouville equation on a graph that consists of two quasione-dimensional loops of the same length having a common vertex. As spectral data, we consider the set of eigenvalues of the entire system together with the sets of eigenvalues of two Dirichlet problems for the Sturm-Liouville equations with the condition of total reflection at the vertex of the graph. We obtain conditions for three sequences of real numbers that enable one to reconstruct a pair of real potentials from L 2 corresponding to each loop. We give an algorithm for the construction of the entire set of potentials corresponding to this triple of spectra. Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1189–1195 Let X and Y be topological spaces such that every mapping f : X → Y for which the set f - 1( G) is an f σ -set in X for any set G open in Y can be represented as a pointwise limit of continuous mappings f n : X→ Y. The question of subspaces Zof the space Yfor which mappings f: X→ Zhave the same property is investigated. Best M-term trigonometric approximations of the classes of periodic functions of many variables in the space L q q Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1196 – 1214 We obtain order estimates for the best M-term trigonometric approximations of classes B Ω p,θ of periodic multivariable functions in the space L q for some values of the parameters pand q. Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1215 – 1233 Necessary and sufficient conditions for a scalar-type spectral operator in a Banach space to be a generator of a Carleman ultradifferentiable C 0-semigroup are found. The conditions are formulated exclusively in terms of the spectrum of the operator. Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1234–1242 According to the well-known Skitovich-Darmois theorem, the independence of two linear forms of independent random variables with nonzero coefficients implies that the random variables are Gaussian variables. This result was generalized by Krakowiak for random variables with values in a Banach space in the case where the coefficients of forms are continuous invertible operators. In the first part of the paper, we give a new proof of the Skitovich-Darmois theorem in a Banach space. Heyde proved another characterization theorem similar to the Skitovich-Darmois theorem, in which, instead of the independence of linear forms, it is supposed that the conditional distribution of one linear form is symmetric if the other form is fixed. In the second part of the paper, we prove an analog of the Heyde theorem in a Banach space. Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1243–1269 We consider the BCS Hamiltonian with sources, as proposed by Bogolyubov and Bogolyubov, Jr. We prove that the eigenvectors and eigenvalues of the BCS Hamiltonian with sources can be exactly determined in the thermodynamic limit. Earlier, Bogolyubov proved that the energies per volume of the BCS Hamiltonian with sources and the approximating Hamiltonian coincide in the thermodynamic limit. Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1270–1274 We present the equivalent definition for spaces of functions analytic in the half-plane ${\mathbb C}_+ = \{z: Re z > 0 \}$ for which $$\sup_{\|\varphi\| < \frac{\pi}2} \left\{\int\limits_0^{+\infty}\left\| f(r e^{i\varphi})\right\|^p e^{-p\sigma r\|\sin \varphi\|} dr \right\} < +\infty.$$ Regularization inertial proximal point algorithm for unconstrained vector convex optimization problems Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1275–1281 The purpose of this paper is to investigate an iterative regularization method of proximal point type for solving ill posed vector convex optimization problems in Hilbert spaces. Applications to the convex feasibility problems and the problem of common fixed points for nonexpansive potential mappings are also given. Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1282–1286 We investigate an impulsive storage process switched by a jump process. The switching process is, in turn, averaged. We prove the weak convergence of the storage process in the scheme of series where a small parameter ε tends to zero. Ukr. Mat. Zh. - 2008. - 60, № 9. - pp. 1287–1296 We establish sufficient conditions for the existence of periodic R-solutions of linear differential inclusions with impulses at fixed times.
Previous abstract Next abstract Session 12 - Cosmology, Large-Scale Structure and Distance Scales. Display session, Monday, June 10 Great Hall, The production of LiH in the postrecombination epoch was previously investigated by Stancil et al. (1996, ApJ, 458, 401) who incorporated quantal rate coefficients for the (spontaneous) radiative association reaction Li + H \to LiH + \nu (Dalgarno et al. 1996, ApJ, 458, 397) into a comprehensive lithium chemistry. The fractional abundance [LiH]/[H] was found to approach \sim 10^-16 for a redshift z\sim 10; an abundance so small as to limit the utility of LiH in erasing cosmic background radiation (CBR) anisotropies through Thompson scattering (Maoli et al. 1994, ApJ, 425, 374) or to discriminate between various big bang nucleosynthesis models (Signore et al. 1994, ApJS, 92, 535). It was suggested by Dubrovich (1995, private communication) that the production of LiH could be enhanced by stimulated radiative association due to the microwave CBR field. Using a fully quantal approach (cf. Dalgarno et al. 1996), we have determined stimulated radiative association rate coefficients for the formation of LiH over a range of matter temperatures T_m and black-body radiation temperatures T_r. We find, with a slight T_m dependence, the total rate coefficients to be enhanced by factors of about 1.1, 1.3, and 4 for T_r = 500, 1000, and 5000 K, respectively. Below T_r = 100 K there is no significant enhancement. Inclusion of stimulated radiative association into the early universe chemistry results in a slight increase of the LiH fractional abundance by a maximum of \sim25% for \sim 350 > z > 150. For higher z, the enhancement is negligible since the fractional abundance is small due to a high dissociation efficiency. For z < 100, the enhancement is also insignificant since T_r<550 K and LiH is predominantly formed through the associative detachment reaction Li^- + H \to LiH + e^-. The conclusions in Stancil et al. (1996) regarding the attenuation of CBR anisotropies and constraints on the lithium primordial abundance remain unchanged. Funding provided by NSF grants OSR-9353227 and AST 93-01099. Program listing for Monday
The integral formula for the Moyal/Groenewold $$ product in Ref. 1 reads $$\tag{1} (f g)(y)~=~\int_{\mathbb{R}^4} \! d^2u~d^2v~ \exp \left[i u_{\alpha} \epsilon^{\alpha\beta} v_{\beta}\right]f(y+u)g(y+v).$$ Remark: The argument $u_{\alpha} \epsilon^{\alpha\beta} v_{\beta}$ inside the exponential of the integral formula (1) has an interesting geometric interpretation as (twice) a signed area of a triangle in phase space $\mathbb{R}^2$, cf. Ref. 2. Usually the Moyal/Groenewold $$ product is defined as $$\tag{2} (f g)(y)~:=~ f(y) \exp \left[\stackrel{\leftarrow}{\frac{\partial}{\partial y_{\alpha}}} i\epsilon_{\alpha\beta} \stackrel{\rightarrow}{\frac{\partial}{\partial y_{\beta}}} \right] g(y). $$ Let us for completeness prove$^1$ the integral formula (1) from the definition (2) for a pair of sufficiently well-behaved functions $f$ and $g$: $$(f * g)(y)~\stackrel{(2)}{:=}~ f(y) \exp \left[\stackrel{\leftarrow}{\frac{\partial}{\partial y_{\alpha}}} i\epsilon_{\alpha\beta} \stackrel{\rightarrow}{\frac{\partial}{\partial y_{\beta}}} \right] g(y) $$$$ ~=~\int_{\mathbb{R}^4} \! d^2u~d^2v ~\delta^2(u)~\delta^2(v) \exp \left[\frac{\partial}{\partial u_{\alpha}} i\epsilon_{\alpha\beta} \frac{\partial}{\partial v_{\beta}}\right]f(y+u)g(y+v)$$$$ ~=~\int_{\mathbb{R}^8} \! d^2u~d^2v~\frac{d^2p}{(2\pi)^2}\frac{d^2q}{(2\pi)^2} \exp i\left[p\cdot u+q\cdot v+\frac{\partial}{\partial u_{\alpha}} \epsilon_{\alpha\beta} \frac{\partial}{\partial v_{\beta}}\right]f(y+u)g(y+v)$$$$ ~\stackrel{\text{int. by part}}{=}~\int_{\mathbb{R}^8} \! d^2u~d^2v~\frac{d^2p}{(2\pi)^2} \frac{d^2q}{(2\pi)^2} \exp i\left[p\cdot u+q\cdot v-p^{\alpha} \epsilon_{\alpha\beta} q^{\beta}\right]f(y+u)g(y+v)$$$$ ~=~\int_{\mathbb{R}^6} \! d^2u~d^2v~\frac{d^2q}{(2\pi)^2} ~\delta^2(u_{\alpha}-\epsilon_{\alpha\beta} q^{\beta}) ~e^{i q\cdot v}f(y+u)g(y+v)$$$$ ~\stackrel{(6)}{=}~\int_{\mathbb{R}^6} \! d^2u~d^2v~\frac{d^2q}{(2\pi)^2} ~\delta^2( q^{\beta}-u_{\alpha}\epsilon^{\alpha\beta} ) ~e^{i q\cdot v}f(y+u)g(y+v)$$$$\tag{3} ~=~\int_{\mathbb{R}^4} \! d^2u~d^2v~ \exp \left[i u_{\alpha} \epsilon^{\alpha\beta} v_{\beta}\right]f(y+u)g(y+v).$$ The functions $f$ and $g$ should be sufficiently well-behaved in order for the above integrals and manipulations (3) to make mathematical sense. Let us here give a necessary and sufficient condition for the integrand $$\tag{4} h(u,v) ~:=~\exp \left[i u_{\alpha} \epsilon^{\alpha\beta} v_{\beta}\right]f(y+u)g(y+v), \qquad u,v,y ~\in ~\mathbb{R}^2, $$ of the integral formula (1) to be integrable. In general an integrand $h$ is integrable (i.e belongs to $L^1$) if and only if (i) the integrand $h$ is Lebesgue measurable, and (ii) the absolute value $\|h\|$ of the integrand has a finite integral $\int \|h\| dm<\infty$. Note that $\|h(u,v)\|=\|f(y+u)\|~\|g(y+v)\|$ factorizes in a $u$- and a $v$-dependent factor. Assuming that both functions $f$, $g$ (and therefore $h$) are Lebesgue measurable, this means (via Tonelli's and Fubini's theorems) that the integrand (4) is integrable $h\in L^1(\mathbb{R}^4)$ if and only if (i) both $f,g\in L^1(\mathbb{R}^2)$ are integrable, or (ii) at least one of the two functions $f$ and $g$ vanishes almost everywhere. For instance, inserting the two non-integrable first-order polynomials $f(y)=y_{\alpha}$ and $g(y)=y_{\beta}$ inside the integral formula (1) is ill-defined as OP also mentions. On the other hand, the definition (2) does not involve integrals. So we can put $f(y)=y_{\alpha}$ and $g(y)=y_{\beta}$ in eq. (2) to derive the sought-for formula $$\tag{5} y_{\alpha} * y_{\beta}~=~y_{\alpha} y_{\beta}+ i\epsilon_{\alpha\beta}. $$ References: M.A. Vasiliev, Unfolded representation for relativistic equations in 2 + 1 anti-de Sitter space, Class. Quant. Grav. 11, (1994) 649. Note that there is an imaginary unit $i$ missing in the published version of formula (1). A preprint version from KEK Preprint Library has the $i$ factor so it seems that the $i$ was lost during the publishing phase. C. Zachos, Geometrical Evaluation of Star Products, J. Math. Phys. 41 (2000) 5129, arXiv:hep-th/9912238. -- $^1$ Note that Ref. 1 uses the convention $$\tag{6} \epsilon_{\alpha\beta} \epsilon^{\beta\gamma}~=~-\delta_{\alpha}^{\gamma}. $$
Analytic Signal In communication theory and modulation theory we always deal with two phases: In-phase (I) and Quadrature-phase (Q). The question that I will discuss in this blog is that why we use two phases and not more. Any real band-limited signal along with its Hilbert transformed pair form an analytic signal. We normally use the analytic signal for modulation. A modulated signal is actually a carrier or the sine signal that one attribute of it is changing with time which is our signal. These attributes might be amplitude, phase or frequency. A sine signal has not any other parameter to change. Albeit, we can discuss about modulating the phase second, third,... derivative or changing the amplitude derivatives. But current modulations rely on amplitude and phase and phase first derivative (frequency). The analytic signal in the pass-band is the summation of the pass-band signal and its Hilbert transform. The envelope of this modulated signal will be the real signal. $$ x(t)={ I cos(\omega t) -Q sin(\omega t)}.\tag {1} $$ Now the question is that why in current communication we are limited to this representation? The answer lies in the definition of Fourier transform. As we know the Euler equation $$ \exp(j \theta) ={cos(\theta) + j sin( \theta)}.\tag {2} $$ since there are only two basis functions that are orthogonal to each other, we can only have two phases in the communications. If we extend the idea to more than two dimensions we will have the Quaternions and the hyper-analytic signals [1]. The reason that why we use just use $sin$ and $cos$ lies in their simplicity and orthogonality. Any other set of orthogonal pairs in two dimensions, based on the Fourier theorem can be described in terms of these two functions. To extend the idea of analytic signal we have to follow the 4-D Hamilton space or higher dimensional spaces to have more phases in the modulation theory. There is a question: Why we are limited to circular trigonometry? Does non-circular trigonometry lead to more efficient and faster communications? One answer to this question is that our electromagnetic theory has been developed for sine waves. And if the theory changes for non-circular waves, these waves may change our communications. [1] Sangwine, S.J.; Le Bihan, N., "Hypercomplex analytic signals : Extension of the analytic signal concept to complex signals," in Signal Processing Conference, 2007 15th European , vol., no., pp.621-624, 3-7 Sept. 2007 Previous post by Mehdi : Maximum Likelihood Estimation Your definition of Equation (1) is a bit confusing. You say it's "the complex envelope of this modulated signal is the real signal". However, from my understanding the complex envelope is complex not real. Equation (1) uses I and Q, which I think is the baseband complex envelope defined in (2.1-10) and (2.1-11) of Dig. Comm. 5th ed., J. Proakis and Salehi. Can you help clear this for me? Thanks. Hello, The definition of the complex envelope can be found in [1]. The idea of the envelope of a modulated carrier which is real in analytic signals is extended to the complex envelope of hyper analytic signal which is complex. Please refer to [1] for more details. Regards, Mehdi This is clearer but your statement in the article is then imprecise. Shouldn't it be changed to: "the envelope of this modulated signal is the real signal"? Thanks for the comment. I will fix that. Regards, Mehdi To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments. Registering will allow you to participate to the forums on ALL the related sites and give you access to all pdf downloads.
Ten Little Algorithms, Part 2: The Single-Pole Low-Pass Filter Other articles in this series: Part 1: Russian Peasant Multiplication Part 3: Welford's Method (And Friends) Part 4: Topological Sort Part 5: Quadratic Extremum Interpolation and Chandrupatla's Method Part 6: Green’s Theorem and Swept-Area Detection I’m writing this article in a room with a bunch of other people talking, and while sometimes I wish they would just SHUT UP, it would be better if I could just filter everything out. Filtering is one of those things that comes up a lot in signal processing. It’s either ridiculously easy, or ridiculously difficult, depending on what it is that you’re trying to filter. I’m going to show you a one-line algorithm that can help in those cases where it’s ridiculously easy. Interested? Here’s a practical example of a filtering problem: you have a signal you care about, but it has some noise that snuck in through a little hole in the wall, and darnit, you’re stuck with the noise. And the question is, can we get rid of the noise? Here we’ve got a lot of noise: relative to a square-wave signal with peak-to-peak amplitude of 2.0, the peak-to-peak noise waveform is in the 6-7 range with even larger spikes. Yuck. The signal processing approach is to look at the frequency spectrum. Now this is some good news for signal processing. The signal spectrum content really stands out above the noise. Square waves have harmonics that are odd multiples of the fundamental frequency: here 4Hz, 12Hz, 20Hz, 28Hz, etc. The harmonic content decays as 1/f. White noise, on the other hand, has a roughly flat spectrum that is spread out across all harmonics, so it is lower in amplitude for the same amount of energy. Kind of like making a peanut butter sandwich — if you put a glob of peanut butter on a slice of bread, it’s pretty thick, but if you spread it over the bread evenly, it’s much thinner, even though it’s the same amount of peanut butter. Mmmm. I hope you like peanut butter; this is making me hungry. This effect is easier to see on a log scale: What we’d really like is to keep the low frequencies only; most of the signal energy is only in the lowest harmonics, and if we want to get rid of the noise, there’s no point in keeping the higher frequencies where the noise energy is much higher. Looks like the crossover here is roughly around 500Hz; below 500Hz the signal dominates, and above that point the noise dominates. So let’s do it! We’ll use a low-pass filter to let the low frequencies pass through and block the high frequencies out. In an ideal world, we’d use a low-pass filter with a very sharp cutoff, in other words one that lets everything through below 500Hz and nothing through above 500Hz. But in practice, sharp-cutoff filters are challenging to implement. It’s much easier to create a gradual-cutoff filter, and the simplest is a single-pole infinite impulse response (IIR) low-pass filter, sometimes called a exponential moving average filter. We’re going to use a filter which has a transfer function of $ H(s) = \frac{1}{\tau s + 1} $. This is a first-order system, which I’ve talked about in a previous article, in case you’re interested in more information. But here we’re just going to use it. We’ll start with the differential equation of this filter, which is $ \tau \frac{dy}{dt} + y = x $. We can take this continuous-time equation and convert to a discrete-time system with timestep $ \Delta t $, namely $ \tau \frac{\Delta y}{\Delta t} + y = x $, and if we solve for $ \Delta y $ we get $ \Delta y = (x-y)\frac{\Delta t}{\tau} $. Blah blah blah, math math math, yippee skip. Here’s the upshot, if all you care about is some computer code that does something useful: // // SHORTEST USEFUL ALGORITHM **EVER** // y += alpha * (x-y); // // # # # ## ## ##### # // # # # # # # # # # // # # # # # # # # // # # ## ## # # // That’s it! One line of C code. The input is x. The output is y. The filter constant is alpha. Every timestep we add alpha times the difference between input and output. So simple. Now we just have to calculate what alpha is. The approximate answer is $ \alpha = \frac{\Delta t}{\tau} $: we just divide $ \Delta t $ by the time constant $ \tau $. The more accurate answer is $ \alpha = 1 - e^{-\Delta t/\tau} \approx \frac{\Delta t}{\tau} $, with the approximation being more accurate for time constants much longer than the timestep, namely $ \tau \gg \Delta t $. Let’s see it in action. If we want to have a filter with a cutoff of 500Hz (in other words, pass through frequencies much lower than 500Hz, block frequencies much higher than 500Hz), we use a value of $ \tau $ = 1/500 = 2msec. Hey, that’s not bad, especially considering that the original signal including noise had a huge amount of noise on it. There is always a tradeoff in filtering: if you want less noise, you can use a lower cutoff frequency, at the cost of more phase lag. Phase lag is a frequency-dependent delay that rounds off sharp edges and plays havoc with feedback control systems, if they aren’t designed to take that phase lag into account. We could use a lower cutoff frequency here, say 100Hz → $ \tau $ = 10 msec: There you go: less noise, more rounding / distortion / phase lag / whatever you want to call it. Here are the signal spectra for signals shown in the two previous graphs: The higher time constant filter has a lower cutoff frequency, so the remaining amplitude of the output signal is lower. Now in practice, this example we’ve used contains a huge amount of noise on the input signal. We always hope for signals that are less noisy than that. Here’s a more reasonable example, with only 10% of the noise we showed earlier: Here we can go all the way out to a cutoff frequency of 2000 Hz (time constant of 0.5 msec), get fairly nice edges in our signal, but still remove most of the noise. In general the noise content gets reduced in amplitude by a factor of about $ \sqrt{\alpha} $: here we have $ \alpha = \Delta t/\tau = 1.0/f\tau $ where f is the sampling frequency, and for f = 65536Hz we have $ \alpha = 1/32.768 $, so should expect about a reduction to about 1/5.7 of the original noise level. We might have had a peak-to-peak noise of about 0.5 in the original signal, and we’re down to maybe 0.08 now, which is around 1/6 of the original noise level. Measuring peak noise is kind of subjective, but we’re in the ballpark of what theory predicts. The interesting thing here is the value $ \alpha $ drops if the timestep $ \Delta t $ gets smaller: we can improve the noise filtering if we sample more often. This is called oversampling and it can be used, up to a point, to reduce the impact of noise on signals of interest, by sampling them more than normal and then using low-pass filters to get rid of the noise. Oversampling has diminishing returns, because if we want half the noise amplitude, we have to sample four times as often, and compute the filter four times as often. It also only works on noise. If our unwanted signal is, say, a Rick Astley song, with the same general type of frequency spectrum content of our signal of interest, then we’re totally out of luck if we try to filter it out, and we just have to hope the person playing it has some sense of decency and will turn it off. The one line of code posted above: // curtains part, angels sing.... y += alpha * (x-y); // now back to normal is applicable for floating-point math. Most of the lower-end processors don’t have floating-point math, and have to emulate it, which takes time and CPU power. If you want to work in fixed-point instead, the code to implement a low-pass filter looks like this: y_state += (alpha_scaled * (x-y)) >> (Q-N); y = y_state >> N; where alpha_scaled = $ 2^Q \alpha $ for some fixed-point scaling Q, and you have to keep N additional state bits around to accurately represent the behavior of the low-pass filter. The value of N is more or less logarithmic in alpha, e.g. $ N \approx - \log_2 \alpha $, so if you want to use very long time constants, you have to keep around more bits in your filter state. I worked with a software contractor around 2002 or 2003, who was convinced I was wrong about this; if I wanted to filter a 16-bit ADC reading, what was wrong about storing everything in 16 bits? Why would I have to use 32-bit numbers? But you do; try it without adding extra bits: you’ll find that there is a minimum difference in input that is needed in order to get the output to change, and small changes in the input will make the output look stair-steppy. This is an unwanted quantization effect. Don’t believe me? Let’s try it, first with floating-point, then with fixed-point, on a pair of signals. Signal x1 will be a square wave plus some unwanted high-frequency stuff; signal x2 will be the plain square wave. Here we go: Now we’ll do a low-pass filter in fixed-point. We’ll plot the filter output (scaled back from integer counts to a floating-point equivalent), along with the error between the fixed-point and floating-point implementations. The proper number of extra state bits is $ N = -\log_2 \frac{1}{1024} = 10 $, but we used 6, 4, 2, and 0 in these examples. You’ll notice a couple of things: the lower N is, the worse the error gets for N = 6 the error is small but still present for N = 4, 2 and 0 the error is visually noticeable there is a DC bias: the output is always lower than the input. (This is true because a shift right represents a rounding down in the integer math.) the error for the y2 signal is always much higher; this is the output for the plain square wave with no harmonic content. This last point is kind of interesting. The error caused by not having enough state bits is quantization error, and when high-frequency signals are present, it helps alleviate quantization error, in part because these signals are helping the filter state “bounce” around. This effect is called dithering, and it’s a trick that can be used to reduce the number of state bits somewhat. But to use it, you really need to know what you’re doing, and it also takes extra work (both up-front design time, and also CPU time) on an embedded system to generate the right dithering signal. Aside from the dithering trick, you’re either stuck with using more bits in your low-pass filters, or you can use a larger timestep $ \Delta t $ so the filter constant $ \alpha $ doesn’t get so small. Well, aside from the fixed-point quirks, the basic one-pole low-pass filter algorithm is pretty simple. If you need more filtering than a one-pole low-pass filter can provide, for example you have lots of 1kHz noise on a 3Hz signal, another thing you can do is to cascade two of these one-pole low-pass filters (in other words, filter twice). It’s not as good as an optimally-designed two-pole low-pass filter, but it’s stable and easy to implement. That’s it for today – happy filtering! © 2015 Jason M. Sachs, all rights reserved. Previous post by Jason Sachs: Understanding and Preventing Overflow (I Had Too Much to Add Last Night) Next post by Jason Sachs: Padé Delay is Okay Today Alternatively, I could try to strip out text cells and publish the code as IPython notebooks; that's something that might be quicker. It would be Mercurial, not Git -- while I use both (Mercurial for personal projects, Git for some things at work), and I definitely like Github best for code hosting, I just hate that Git occasionally shoves more complexity in your face when you try to do common tasks, and that the major Powers That Be (Github, and Atlassian via their Stash platform) don't support Mercurial as a simpler alternative that handles 99% of the cases you need in a DVCS. >:P >:P >:P I like your peanut butter analogy. Jason, I'm unable to read (understand) the code you posted here. But I'd still like to learn the material in your blog. Will you please translate the 'y += alpha * (x-y);' line of code into a standard difference equation using 'y(n)', 'y(n-1)', 'x(n)', etc. terms? I'd like to model that line of code using Matlab software. Thanks, [-Rick-] y[n] = alpha * x[n] + (1-alpha) * y[n-1] Ah, Thanks. An "exponential averager." I should have figured that out on my own. Maybe your readers would be interested in the exponential averager material at: http://www.dsprelated.com/showarticle/72.php [-Rick-] I thought that fft() converts from the time domain to the frequency domain, but that doesn't seem to be the case here. I'm trying to understand the delta T term, in context of working with a sample buffer from the ADC, Is that the interval between samples. i.e., if I am sampling at 100Hz, delta T is simply x[n] - x[n-1] ? Thanks! Gary Yes. If sampling at 100Hz, delta T would be 10ms. A very nice write-up on single-pole IIR filter. I guess many people will continue to read this post, so I would like to point out one thing-the cutoff frequency didn't get right. 1/tau is not the cutoff frequency in Hertz, but in rad/s, that is, omega=2pif=1/tau, where omega is the frequency in rad/s. So when tau was selected to be 2msec, the cutoff frequency should be 80Hz instead of 500Hz. This would be clear if frequency responses of the low pass filters had been plotted. That is why I suggest, when designing filters, both of time-domain and frequency-domain responses should be simulated and analyzed. Thanks for the nice post. To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments. Registering will allow you to participate to the forums on ALL the related sites and give you access to all pdf downloads.
AI News, MachineLearning MachineLearning So for example, it may have 25 inputs which are gray-levels of a 5x5 image and a single output that should, for example, tell whether a cat is in the image or not. If you are familiar with linear algebra you will notice that this can be done with the dot product of the normal vector with the data point. This operation is exactly the same as weighting the inputs, summing them and determine if the sum is positive or negative. If a data set with labels is available, a learning rule can be applied to find the correct line. If you want to understand SVMs make sure to understand Perceptrons correctly, what a dot product is, how the figure with the artificial neuron corresponds to the figure with the 2-dimensional input space and why the learning rule actually works. In this form the weight vector is expressed as a sum over dot products between all data points and the input vector. The crucial point here is that that a dot-product of the input data is used to define the weight vector. Different kernels have different feature spaces and so choosing a kernel function means choosing a feature space. A mathematical framework called statistical learning theory was developed in order to give a precise definition what "best" Select a Web Site A 2-input hard limit neuron is trained to classify 5 input vectors into two categories. Each of the five column vectors in X defines a 2-element input vectors and a row vector T defines the vector's target categories. Here the input and target data are converted to sequential data (cell array where each column indicates a timestep) and copied three times to form the series XX and TT. ADAPT updates the network for each timestep in the series and returns a new network object that performs as a better classifier. The perceptron correctly classified our new point (in red) as category 'zero' (represented by a circle) and not a 'one' (represented by a plus). Perceptron In machine learning, the perceptron is an algorithm for supervised learning of binary classifiers (functions that can decide whether an input, represented by a vector of numbers, belongs to some specific class or not).[1] a classification algorithm that makes its predictions based on a linear predictor function combining a set of weights with the feature vector. perceptron was intended to be a machine, rather than a program, and while its first implementation was in software for the IBM 704, it was subsequently implemented in custom-built hardware as the 'Mark 1 perceptron'. This machine was designed for image recognition: it had an array of 400 photocells, randomly connected to the 'neurons'. In a 1958 press conference organized by the US Navy, Rosenblatt made statements about the perceptron that caused a heated controversy among the fledgling AI community; based on Rosenblatt's statements, The New York Times reported the perceptron to be 'the embryo of an electronic computer that [the Navy] expects will be able to walk, talk, see, write, reproduce itself and be conscious of its existence.'[4] Although the perceptron initially seemed promising, it was quickly proved that perceptrons could not be trained to recognise many classes of patterns. This caused the field of neural network research to stagnate for many years, before it was recognised that a feedforward neural network with two or more layers (also called a multilayer perceptron) had far greater processing power than perceptrons with one layer (also called a single layer perceptron).[dubious in 1969 a famous book entitled Perceptrons by Marvin Minsky and Seymour Papert showed that it was impossible for these classes of network to learn an XOR function. (See the page on Perceptrons (book) for more information.) Three years later Stephen Grossberg published a series of papers introducing networks capable of modelling differential, contrast-enhancing and XOR functions. While the complexity of biological neuron models is often required to fully understand neural behavior, research suggests a perceptron-like linear model can produce some behavior seen in real neurons [7][8]. In the modern sense, the perceptron is an algorithm for learning a binary classifier: a function that maps its input x (a real-valued vector) to an output value i = 1 m i i (0 or 1) is used to classify x as either a positive or a negative instance, in the case of a binary classification problem. The most famous example of the perceptron's inability to solve problems with linearly nonseparable vectors is the Boolean exclusive-or problem. In the context of neural networks, a perceptron is an artificial neuron using the Heaviside step function as the activation function. The perceptron algorithm is also termed the single-layer perceptron, to distinguish it from a multilayer perceptron, which is a misnomer for a more complicated neural network. Alternatively, methods such as the delta rule can be used if the function is non-linear and differentiable, although the one below will work as well. When multiple perceptrons are combined in an artificial neural network, each output neuron operates independently of all the others; These weights are immediately applied to a pair in the training set, and subsequently updated, rather than waiting until all pairs in the training set have undergone these steps. The perceptron is a linear classifier, therefore it will never get to the state with all the input vectors classified correctly if the training set D is not linearly separable, i.e. In this case, no 'approximate' solution will be gradually approached under the standard learning algorithm, but instead learning will fail completely. x j {\displaystyle \mathbf {w} \cdot \mathbf {x} _{j}>\gamma } j x j {\displaystyle \mathbf {w} \cdot \mathbf {x} _{j}<-\gamma } j 2 2 The idea of the proof is that the weight vector is always adjusted by a bounded amount in a direction with which it has a negative dot product, and thus can be bounded above by O(√t), where t is the number of changes to the weight vector. However, it can also be bounded below by O(t) because if there exists an (unknown) satisfactory weight vector, then every change makes progress in this (unknown) direction by a positive amount that depends only on the input vector. While the perceptron algorithm is guaranteed to converge on some solution in the case of a linearly separable training set, it may still pick any solution and problems may admit many solutions of varying quality.[11] The perceptron of optimal stability, nowadays better known as the linear support vector machine, was designed to solve this problem (Krauth and Mezard, 1987)[12]. The pocket algorithm with ratchet (Gallant, 1990) solves the stability problem of perceptron learning by keeping the best solution seen so far 'in its pocket'. However, these solutions appear purely stochastically and hence the pocket algorithm neither approaches them gradually in the course of learning, nor are they guaranteed to show up within a given number of learning steps. In the linearly separable case, it will solve the training problem – if desired, even with optimal stability (maximum margin between the classes). In all cases, the algorithm gradually approaches the solution in the course of learning, without memorizing previous states and without stochastic jumps. The algorithm starts a new perceptron every time an example is wrongly classified, initializing the weights vector with the final weights of the last perceptron. Each perceptron will also be given another weight corresponding to how many examples do they correctly classify before wrongly classifying one, and at the end the output will be a weighted vote on all perceptron. The so-called perceptron of optimal stability can be determined by means of iterative training and optimization schemes, such as the Min-Over algorithm (Krauth and Mezard, 1987)[12] The perceptron of optimal stability, together with the kernel trick, are the conceptual foundations of the support vector machine. -perceptron further used a pre-processing layer of fixed random weights, with thresholded output units. Another way to solve nonlinear problems without using multiple layers is to use higher order networks (sigma-pi unit). In this type of network, each element in the input vector is extended with each pairwise combination of multiplied inputs (second order). Indeed, if we had the prior constraint that the data come from equi-variant Gaussian distributions, the linear separation in the input space is optimal, and the nonlinear solution is overfitted. ≈Learning again iterates over the examples, predicting an output for each, leaving the weights unchanged when the predicted output matches the target, and changing them when it does not. a r g m a x y In recent years, perceptron training has become popular in the field of natural language processing for such tasks as part-of-speech tagging and syntactic parsing (Collins, 2002). On Monday, September 23, 2019 Perceptron Training Watch on Udacity: Check out the full Advanced Operating Systems course for free .. Unit 5 48 Perceptron Unit 5 48 Perceptron. 10.2: Neural Networks: Perceptron Part 1 - The Nature of Code In this video, I continue my machine learning series and build a simple Perceptron in Processing (Java). Perceptron Part 2: This .. 3. A Geometrical View of Perceptrons Video from Coursera - University of Toronto - Course: Neural Networks for Machine Learning: How SVM (Support Vector Machine) algorithm works In this video I explain how SVM (Support Vector Machine) algorithm works to classify a linearly separable binary data set. The original presentation is available ... Support Vector Machine Algorithm Support Vector Machines are one of the most popular and talked about machine learning algorithms. This algorithm is used for classification. It is done through ... 3. Decision Boundary Video from Coursera - Standford University - Course: Machine Learning: Lecture 03 -The Linear Model I The Linear Model I - Linear classification and linear regression. Extending linear models through nonlinear transforms. Lecture 3 of 18 of Caltech's Machine ... How Powerful is a Perceptron Unit Watch on Udacity: Check out the full Advanced Operating Systems course for free ..
I would like to solve the following nonlinear PDE: $$ \frac{\partial^2 \phi}{\partial x^2} - \frac{\partial^2 \phi}{\partial t^2} = \lambda \|\phi\|^2 \phi $$ I was trying: NDSolve[{D[f[x, t], x, x] - D[f[x, t], t, t] == f[x, t]^3, f[x, 0] == Sin[2Pix], f[0, t] == 0, f[1, t] == 0}, f, {x, 0, 1}, {t, 0, 1}] but, I am consistenly getting NDSolve::femnonlinear: Nonlinear coefficients are not supported in this version of NDSolve. Is there any solver for non-linear PDEs?
Search Now showing items 1-1 of 1 Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2014-09) Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, \|y\| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
What is Fluid Dynamics? What is computational fluid dynamics? Applications of fluid dynamics Fluid Dynamics can be applied in the following ways: Fluid dynamics is used to calculate the forces acting upon the aeroplane. It is used to find the flow rates of material such as petroleum from pipelines. It can also be used in traffic engineering (traffic treated as continuous liquid flow). Equations in Fluid Dynamics : Bernoulli’s Equation $\large \frac{P}{\rho }+g\;z+\frac{v^{2}}{2}=k$ $\large \frac{P}{\rho g}+z+\frac{v^{2}}{2g}=k$ $\large \frac{P}{\rho g}+\frac{v^{2}}{2g}+z=k$ Here, $\frac{P}{\rho g}$ is the pressure head or pressure energy per unit weight fluid $\frac{v^{2}}{2g}$ is the kinetic head or kinetic energy per unit weight z is the potential head or potential energy per unit weight P is the Pressure ρ is the Density K is the Constant The Bernoulli equation is different for isothermal as well as adiabatic processes. $\large \frac{dP}{\rho } + V \; dV + g \; dZ = 0$ $\large \int \left ( \frac{dP}{\rho }+ V\; dV + g\; dZ \right )=K$ $\large \int \frac{dP}{\rho}+\frac{V^{2}}{2}+g\;Z=K$ Where, Z is the elevation point ρ is the density of fluid The equation can also be written as, $\large q + P = P_{o}$ Where, q is the dynamic pressure P O is the total pressure P is the static pressure Continue learning about isothermal processes with engaging video lectures and diagrams at BYJU’S.
Intersection Set context $X,Y\in\mathfrak U$ definiendum $ x\in X \cap Y $ postulate $ x\in X \cap Y \Leftrightarrow (x\in X\land x\in Y) $ Discussion $ X \cap Y $ is commutative and idempotent. The intersection and union are associative and distributive with respect to another. Reference Wikipedia: Intersection
Many of us are familiar with the word problem, but are we aware of the fact and problems related to variables and constants? When we say 5 it means a number but what if we say x=5 or 5y or something like that? This is where algebra came into existence algebra is that branch of mathematics which not only deals with numbers but also variable and alphabets. The versatility of Algebra is very deep and very conceptual, all the non-numeric character represents variable and numeric as constants For example. 1. $-5y+3=2\left(4y+12\right)$ 2. $\frac{4}{x^{2}-2x}-\frac{2}{x-2}=-\frac{1}{2}$ 3. $x\sqrt{x}=-x$ 4. $\begin{vmatrix} x-a \end{vmatrix}=a^{2}-x^{2}$ 5. $4^{x^{2}+1}-2^{x^{2}+2}=8$ 6. $log_{2}(2^{x}-1)+x=log_{4}(144)$ 7. $\left\{\begin{matrix}x^{2}+y^{2}=17+2x & \\ (x-1)^{2}+(y-8)^{2}=34 & \end{matrix}\right.$ Algebra Word Problems deal with real time situations and solutions which can be solved using algebra for example: Linear Equation in One Variable There are various methods For Solving the Linear Equations Cross multiplication method Replacement method Hit and trial method There is another category of problems as such Quadratic Equations which is of the form $ax^{2}+bx+c$ Algebra problems along with their solutions Basic Algebra Problems $(x-1)^{2}=(4\sqrt{x-4})^{2}$ $x^{2}-2x+1=16(x-4)$ $x^{2}-2x+1=16x-64$ $x^{2}-18x+65=0$ $(x-13)(x-5)=0$ There are Variety of different Algebra problem present, and are solved depending upon their functionality and state, for example, a linear equation problem can’t be solved using a quadratic equation formula and vice verse for, e.g., x+x/2=7 then solve for x is an equation in one variable for x which can be satisfied by only one value of x. whereas x 2+5x+6 is a quadratic equation which is satisfied for two values of x the domain of algebra is huge and vast so for more information; please visit byjus.com where different techniques are explained different algebra problem Currently, 12 lakh students are using byjus to overcome their phobia against algebra and much more topics.
We have discussed how to calculate the probability that an event will happen. Sometimes, we are interested in finding the probability that an event will not happen. The complement of an event [latex]E[/latex], denoted [latex]{E}^{\prime }[/latex], is the set of outcomes in the sample space that are not in [latex]E[/latex]. For example, suppose we are interested in the probability that a horse will lose a race. If event [latex]W[/latex] is the horse winning the race, then the complement of event [latex]W[/latex] is the horse losing the race. To find the probability that the horse loses the race, we need to use the fact that the sum of all probabilities in a probability model must be 1. The probability of the horse winning added to the probability of the horse losing must be equal to 1. Therefore, if the probability of the horse winning the race is [latex]\frac{1}{9}[/latex], the probability of the horse losing the race is simply A General Note: The Complement Rule The probability that the complement of an event will occur is given by Example 5: Using the Complement Rule to Calculate Probabilities Two six-sided number cubes are rolled. Find the probability that the sum of the numbers rolled is less than or equal to 3. Find the probability that the sum of the numbers rolled is greater than 3. Solution The first step is to identify the sample space, which consists of all the possible outcomes. There are two number cubes, and each number cube has six possible outcomes. Using the Multiplication Principle, we find that there are [latex]6\times 6[/latex], or [latex]\text{ 36 }[/latex] total possible outcomes. So, for example, 1-1 represents a 1 rolled on each number cube. [latex]\text{1 - 1}[/latex] [latex]\text{1 - 2}[/latex] [latex]\text{1 - 3}[/latex] [latex]\text{1 - 4}[/latex] [latex]\text{1 - 5}[/latex] [latex]\text{1 - 6}[/latex] [latex]\text{2 - 1}[/latex] [latex]\text{2 - 2}[/latex] [latex]\text{2 - 3}[/latex] [latex]\text{}[/latex] [latex]\text{2 - 4}[/latex] [latex]\text{2 - 5}[/latex] [latex]\text{2 - 6}[/latex] [latex]\text{3 - 1}[/latex] [latex]\text{3 - 2}[/latex] [latex]\text{3 - 3}[/latex] [latex]\text{3 - 4}[/latex] [latex]\text{3 - 5}[/latex] [latex]\text{3 - 6}[/latex] [latex]\text{4 - 1}[/latex] [latex]\text{4 - 2}[/latex] [latex]\text{4 - 3}[/latex] [latex]\text{4 - 4}[/latex] [latex]\text{4 - 5}[/latex] [latex]\text{4 - 6}[/latex] [latex]\text{5 - 1}[/latex] [latex]\text{5 - 2}[/latex] [latex]\text{5 - 3}[/latex] [latex]\text{5 - 4}[/latex] [latex]\text{5 - 5}[/latex] [latex]\text{5 - 6}[/latex] [latex]\text{6 - 1}[/latex] [latex]\text{6 - 2}[/latex] [latex]\text{6 - 3}[/latex] [latex]\text{6 - 4}[/latex] [latex]\text{6 - 5}[/latex] [latex]\text{6 - 6}[/latex] We need to count the number of ways to roll a sum of 3 or less. These would include the following outcomes: 1-1, 1-2, and 2-1. So there are only three ways to roll a sum of 3 or less. The probability is[latex]\frac{3}{36}=\frac{1}{12}[/latex] Rather than listing all the possibilities, we can use the Complement Rule. Because we have already found the probability of the complement of this event, we can simply subtract that probability from 1 to find the probability that the sum of the numbers rolled is greater than 3.[latex]\begin{array}{l}P\left({E}^{\prime }\right)=1-P\left(E\right)\hfill \\ \text{ }=1-\frac{1}{12}\hfill \\ \text{ }=\frac{11}{12}\hfill \end{array}[/latex] Try It 5 Two number cubes are rolled. Use the Complement Rule to find the probability that the sum is less than 10.
Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees Fahrenheit to degrees Celsius. She finds the formula and substitutes 75 for [latex]F[/latex] to calculate Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather forecast for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit. At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for [latex]F[/latex] after substituting a value for [latex]C[/latex]. For example, to convert 26 degrees Celsius, she could write After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature. The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function. Given a function [latex]f\left(x\right)[/latex], we represent its inverse as [latex]{f}^{-1}\left(x\right)[/latex], read as [latex]``f[/latex] inverse of [latex]x.\text{``}[/latex] The raised [latex]-1[/latex] is part of the notation. It is not an exponent; it does not imply a power of [latex]-1[/latex] . In other words, [latex]{f}^{-1}\left(x\right)[/latex] does not mean [latex]\frac{1}{f\left(x\right)}[/latex] because [latex]\frac{1}{f\left(x\right)}[/latex] is the reciprocal of [latex]f[/latex] and not the inverse. The “exponent-like” notation comes from an analogy between function composition and multiplication: just as [latex]{a}^{-1}a=1[/latex] (1 is the identity element for multiplication) for any nonzero number [latex]a[/latex], so [latex]{f}^{-1}\circ f\\[/latex] equals the identity function, that is, This holds for all [latex]x[/latex] in the domain of [latex]f[/latex]. Informally, this means that inverse functions “undo” each other. However, just as zero does not have a reciprocal, some functions do not have inverses. Given a function [latex]f\left(x\right)[/latex], we can verify whether some other function [latex]g\left(x\right)[/latex] is the inverse of [latex]f\left(x\right)[/latex] by checking whether either [latex]g\left(f\left(x\right)\right)=x[/latex] or [latex]f\left(g\left(x\right)\right)=x[/latex] is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.) For example, [latex]y=4x[/latex] and [latex]y=\frac{1}{4}x\\[/latex] are inverse functions. and A few coordinate pairs from the graph of the function [latex]y=4x[/latex] are (−2, −8), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function [latex]y=\frac{1}{4}x\\[/latex] are (−8, −2), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function. A General Note: Inverse Function For any one-to-one function [latex]f\left(x\right)=y[/latex], a function [latex]{f}^{-1}\left(x\right)[/latex] is an inverse function of [latex]f[/latex] if [latex]{f}^{-1}\left(y\right)=x[/latex]. This can also be written as [latex]{f}^{-1}\left(f\left(x\right)\right)=x[/latex] for all [latex]x[/latex] in the domain of [latex]f[/latex]. It also follows that [latex]f\left({f}^{-1}\left(x\right)\right)=x[/latex] for all [latex]x[/latex] in the domain of [latex]{f}^{-1}[/latex] if [latex]{f}^{-1}[/latex] is the inverse of [latex]f[/latex]. The notation [latex]{f}^{-1}[/latex] is read [latex]\text{``}f[/latex] inverse.” Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[/latex], so we will often write [latex]{f}^{-1}\left(x\right)[/latex], which we read as [latex]``f[/latex] inverse of [latex]x.''[/latex] Keep in mind that and not all functions have inverses. Example 1: Identifying an Inverse Function for a Given Input-Output Pair If for a particular one-to-one function [latex]f\left(2\right)=4[/latex] and [latex]f\left(5\right)=12[/latex], what are the corresponding input and output values for the inverse function? Solution The inverse function reverses the input and output quantities, so if Alternatively, if we want to name the inverse function [latex]g[/latex], then [latex]g\left(4\right)=2[/latex] and [latex]g\left(12\right)=5[/latex]. Try It 1 Given that [latex]{h}^{-1}\left(6\right)=2[/latex], what are the corresponding input and output values of the original function [latex]h?[/latex] How To: Given two functions [latex]f\left(x\right)\\[/latex] and [latex]g\left(x\right)\\[/latex], test whether the functions are inverses of each other. Determine whether [latex]f\left(g\left(x\right)\right)=x[/latex] or [latex]g\left(f\left(x\right)\right)=x[/latex]. If either statement is true, then both are true, and [latex]g={f}^{-1}[/latex] and [latex]f={g}^{-1}[/latex]. If either statement is false, then both are false, and [latex]g\ne {f}^{-1}[/latex] and [latex]f\ne {g}^{-1}[/latex]. Example 2: Testing Inverse Relationships Algebraically If [latex]f\left(x\right)=\frac{1}{x+2}[/latex] and [latex]g\left(x\right)=\frac{1}{x}-2[/latex], is [latex]g={f}^{-1}?[/latex] Solution [latex]\begin{cases} g\left(f\left(x\right)\right)=\frac{1}{\left(\frac{1}{x+2}\right)}{-2 }\hfill\\={ x }+{ 2 } -{ 2 }\hfill\\={ x }\hfill \end{cases}\\[/latex] so This is enough to answer yes to the question, but we can also verify the other formula. Try It 2 If [latex]f\left(x\right)={x}^{3}-4[/latex] and [latex]g\left(x\right)=\sqrt[3]{x+4}[/latex], is [latex]g={f}^{-1}?[/latex] Example 3: Determining Inverse Relationships for Power Functions If [latex]f\left(x\right)={x}^{3}[/latex] (the cube function) and [latex]g\left(x\right)=\frac{1}{3}x[/latex], is [latex]g={f}^{-1}?\\[/latex] Solution [latex]f\left(g\left(x\right)\right)=\frac{{x}^{3}}{27}\ne x\\[/latex] No, the functions are not inverses. Try It 3 If [latex]f\left(x\right)={\left(x - 1\right)}^{3}\text{and}g\left(x\right)=\sqrt[3]{x}+1\\[/latex], is [latex]g={f}^{-1}?\\[/latex]
It looks like you're new here. If you want to get involved, click one of these buttons! I've spent four lectures on the logic of partitions; you may be wondering why. One reason was to give you examples illustrating this important fact: Theorem. Left adjoints preserve joins and right adjoints preserve meets. Suppose $f : A \to B$ and $g : B \to A$ are monotone functions between posets. Suppose that $f$ is the left adjoint of $g$, or equivalently, $g$ is the right adjoint of $f$. If the join of $a,a' \in A$ exists then so does the join of $f(a), f(a') \in B$, and $$ f(a \vee a') = f(a) \vee f(a'). $$ If the meet of $b,b' \in B$ exists then so does the meet of $g(b), g(b') \in A$, and $$ g(b \wedge b') = g(b) \wedge g(b'). $$ The proof is very easy, so this deserves being called a "Theorem" only because it's so fundamental! I will prove it later, in more generality. Right now let's see how it's relevant to what we've been doing. In Lecture 9 we saw something interesting about the subsets. Given any set $X$ there's a poset $P(X)$ consisting of all subsets of $X$. Given any function $f : X \to Y$ there's a monotone map $$ f^* : P(Y) \to P(X) $$sending any subset of $Y$ to its preimage under $f$. And we saw that $ f^{\ast} $ has both a left adjoint and a right adjoint. This means that $ f^{\ast} $ is both a right adjoint and a left adjoint. (Remember: having a left adjoint means being a right adjoint, and vice versa.) So by our Theorem, we see that $f^* : P(Y) \to P(X)$ preserves both meets and joins! You can also see this directly - see Puzzle 41 in Lecture 13. But what matters here is the general pattern. In Lecture 13 we also saw something interesting about partitions. Given any set $X$ there's a poset $ \mathcal{E}(X)$ consisting of all partitions of $X$. Given any function $f : X \to Y$ there's a monotone map $$ f^* : \mathcal{E}(Y) \to \mathcal{E}(X) $$ sending any partition of $Y$ to its pullback along $f$. And we saw that while $ f^{\ast} $ preserves meets, it does not preserve joins! So by our Theorem, we see that $f^* : \mathcal{E}(Y) \to \mathcal{E}(X)$ cannot be a left adjoint. On the other hand, it might be a right adjoint. And indeed it is! So, this strange difference between the logic of subsets and the logic of partitions is really all about adjoints. Puzzle 42. Given a function $f : X \to Y$ there is a way to push forward any partition $P$ on $X$ and get a partition $f_{!} (P)$ on $Y$. In pictures it looks like this: although this is not the most exciting example, since here $f_{!}(P)$ has just one part. Figure out the precise general description of $f_{!} (P)$. If you get stuck read Section 1.5.2 of Seven Sketches. Puzzle 43. Show that for any function $f : X \to Y$, pushing forward partitions along $f$ gives a monotone map $$ f_{!} : \mathcal{E}(X) \to \mathcal{E}(Y) . $$ Puzzle 44. Show that for any function $f : X \to Y$, $f^* : \mathcal{E}(Y) \to \mathcal{E}(X)$ is the right adjoint of $f_{!}: \mathcal{E}(X) \to \mathcal{E}(Y)$.
I'm wondering is that possible to get insignificant beta estimates in the time-series context, but highly significant risk premium associated with that beta in the cross-sectional regression? Any help would be greatly appreciated! Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It only takes a minute to sign up.Sign up to join this community I'm wondering is that possible to get insignificant beta estimates in the time-series context, but highly significant risk premium associated with that beta in the cross-sectional regression? Any help would be greatly appreciated! I prefer thinking in terms of well measured vs. poorly measured rather than significant vs. insignificant: arbitrary p-value cutoffs and ignoring sensible priors can both be problematic. On the question, "can poorly measured betas from time-series regressions give rise to well measured factor premiums from cross-sectional regression?" The abstract answer is yes, but you have several problems working against you. Let $F_t$ denote some factor. You can estimate the beta for a return $R_i$ on the factor with a time-series regression: $$ R_{it} -R^f_t = \alpha_i + \beta_i F_t + \epsilon_{it}$$ The key idea behind all these factor models is that expected returns should be linearly increasing in the regression beta on the factor. To estimate factor premium $\gamma_F$, you'd like to run the regression: $$R_{it} - R^f_t = \gamma_0 + \gamma_F \beta_i + u_{it} $$ To do this sensibly, you need to confront several problems:
The scalar QED Lagrangian in your question is for a complex scalar field $\psi(x)$ interacting with an electromagnetic field given by potential $A_{\mu}(x)$. At any point $x$, the scalar field is a complex number. We model this situation by constructing a space - a vector bundle $V$ - which is isomorphic to $M \ X \ \mathbb{C}$. In general a bundle is only locally isomorphic to the product space, since it might have twists in it, but we'll ignore this here. The bundle has a projection map onto the spacetime manifold $\pi: V \rightarrow M$. The set of points which are projected onto $x \in M$ is called the fibre over $x$, and denoted $F_x$. Each $F_x$ is isomorphic to the complex numbers $\mathbb{C}$. Now to make explicit such an isomorphism, we effectively choose a coordinate $z$ for each fibre. So our bundle $V$ then has coordinates $\{x^{\mu}, z \}$. Our spacetime field $\psi(x)$ as a map from $M \rightarrow V$ is called a section of $V$. If we compose a section with the projection $\pi$ we get back the spacetime point we started with. We can think of the choice of fibre coordinates as a gauge choice. A gauge transformation is the choice of a new fibre coordinate, related to the old by $z \rightarrow g.z$ where $g \in U(1)$. In the case of a local gauge transformation, this new choice of coordinate becomes $z \rightarrow g(x).z$ where $g$ is now a function of $x$. Now, given the interpretation of $\psi(x)$ as a section of $V$, in order to construct the Lagrangian, we need to be able to differentiate it i.e. we need to be able to compute a derivative which is a limit $$ \lim_{\Delta x \to 0}\frac{\psi(x+\Delta x)-\psi(x)}{\Delta x} $$ The problem is: $\psi(x)$ lives in the fibre $F_x$ over $x$, and $\psi(x+dx)$ lives in the fibre $F_{x+dx}$ over $x+dx$. These are different spaces, so we can't perform the subtraction unless we can map points in $F_{x+dx}$ to points in $F_x$. If we've chosen a gauge, this is no problem - we have an explicit mapping of both fibres to the complex numbers, so we can perform the subtraction, but we want something that makes sense when we make changes of gauge, in particular local changes of gauge. The recipe to do this is to introduce a connection. If we start at a point $p$ in the fibre $F_x$ and infinitesimally perturb the point $x$ to $x+dx$, to specify where $p$ moves to, we need to give it in general a horizontal component (in the M coordinate direction), and a vertical component (in the fibre directions). Given a gauge, describing an infinitesimal fibrewise displacement is easy - we just apply an infinitesimal element of the gauge group. Such an infinitesimal element belongs to the Lie algebra of the group. For $U(1)$, this Lie algebra is just the real numbers, so the vertical displacements corresponding to movement of $p$ in the 4 spacetime coordinate directions are just given by four real numbers. As a function of spacetime coordinates, they become four functions $A_{\mu}(x)$. The gauge covariant derivative is then just $$ D_{\mu}\psi(x) = \partial_{\mu}\psi(x) + A_{\mu}(x)\psi(x)$$ If we perform a local gauge transformation $$\psi(x) \rightarrow \psi'(x) = g(x)\psi(x)$$ then, provided we make a corresponding transformation $$ A_{\mu}(x) \rightarrow A'_{\mu}(x) = A_{\mu}(x) + g(x)\partial_{\mu}g^{-1}(x) \ \ (1) $$ the gauge covariant derivative transforms like $$ D_{\mu}\psi'(x) = D_{\mu}(g(x)\psi(x)$$ $$ = \partial_{\mu}(g(x)\psi(x)) + A'_{\mu}(x)g(x)\psi(x)$$ $$ = \partial_{\mu}(g(x)\psi(x)) + [A_{\mu}(x)+g(x)\partial_{\mu}g^{-1}(x)]g(x)\psi(x)$$ $$ = g(x)\partial_{\mu}\psi(x) + (\partial_{\mu}g)\psi(x) + A_{\mu}(x)g(x)\psi(x) + g(x)(\partial_{\mu}g^{-1}(x))g(x)\psi(x)$$ $$ = g(x)(\partial_{\mu}\psi(x) + A_{\mu}(x)\psi(x) = g(x)D_{\mu}\psi(x)$$ Where for the last step we used $$ 0 = \partial_{\mu}1 = \partial_{\mu}(g(x)g^{-1}(x)) = (\partial_{\mu}g(x))g^{-1}(x)+g(x)(\partial_{\mu}g^{-1}(x)) $$ So $D_{\mu}\psi(x)$ transforms covariantly, in a way that ensures the Lagrangian is gauge invariant. If we write $g(x) = e^{i\alpha(x)}$ then the transformation law (1) becomes $$ A_{\mu}(x) \rightarrow A'_{\mu}(x) = A_{\mu}(x) - i\partial_{\mu}\alpha(x) $$This post imported from StackExchange Physics at 2014-03-22 17:12 (UCT), posted by SE-user twistor59
I like to think of QFT Weinberg-style: particles come first, and fields come later; and the latter are constructed so as to describe the former. Fields are not to be thought of as fundamental, but only as convenient tools to study particles. In this setting, what justifies the postulate that a fermion particle must be described by a field transforming according to a representation of the Clifford algebra? More details: Particles are classified according to the irreducible representations of the orthogonal group, $\mathrm{SO}(d-1)$ (where $d$ is the number of spacetime dimensions; I'm taking all particles to be massive for simplicity). Once you have a particle described by a representation $R$ of $\mathrm{SO}(d-1)$, you introduce a field $\psi$ which, by definition, lives in a representation $R'$ of the Lorentz group, $\mathrm{SO}(1,d-1)$. The non-trivial question is which representation $R'$ corresponds to a representation $R$; that is, what field is to be used to describe a certain particle. In general, the answer is non-unique, so we should actually look for "the simplest" $R'$ for a given $R$. In his book, Weinberg addresses this question in $d=4$. For example, the trivial representation of the orthogonal group corresponds to the trivial representation of the Lorentz group (i.e., scalar particles correspond to scalar fields). Similarly, the spin $s=1/2$ representation of the orthogonal group corresponds to the $(0,\frac12),(\frac12,0)$ representations of the Lorentz group. To conserve parity, these are to appear together, so $$ \text{spin $s=1/2$}\quad \Longleftrightarrow \quad (0,\tfrac12)\oplus(\tfrac12,0) $$ Now comes the key point: as it turns out, the right hand side of this equation actually corresponds to the an irreducible representation of the Clifford algebra $$ \gamma^{(\mu}\gamma^{\nu)}=\eta^{\mu\nu} $$ but this is only an a-posteriori realisation. There was no reason to expect Clifford to be relevant from the beginning. It just happens to be so. In higher dimensions, the logic is inverted. One declares that in higher dimensions the Clifford algebra is fundamental, and fermions are whatever particle is described by such fields. In the spirit of Weinberg, this is pretty unconvincing: particles should come first. Given higher dimensional fermions, one must ask which fields are to be used to describe them. And it may very well be the case that the answer is again Clifford, but this must be an a-posteriori conclusion, not a postulate. Thus, my question: what justifies the use of the Clifford algebra to describe higher dimensional fermions? How can we prove that such a representation of the Lorentz group is indeed the simplest representation that is able to describe fermions, when we take the latter as fundamental instead of the former? To be specific, let us define higher dimensional fermions as the first non-trivial projective representation $R$ of $\mathrm{SO}(d-1)$ (or a direct sum thereof, if necessary, in order to conserve parity).
This is a question that I have thought about myself, but I'm struggling a bit regarding how to answer it. This is my question: Imagine we have a universe with 3 bodies. Out of these 3, 2 are identical and massive, and 1 is small and insignificant. The 2 identical celestial bodies are literally identical in every way possible -- density, mass, volume, etc. They are also fixed in space and unable to move. If their centers are perfectly aligned vertically (although this doesn't really matter since we're in space) and separated by distance $d$, then their Lagrangian point will be at $\frac{d}{2}$ since the bodies are identical. Now, let's make another axis from this Lagrangian point that is orthogonal to the vertical axis (which $d$ falls on), and call it the horizontal axis. In other words, let's just say this produces some $xy$-plane. If our third body, which we may assume to be a perfect sphere and insignificant mass, is placed on this horizontal axis $x$ units of distance away from the origin of this $xy$-plane (which is the Lagrangian point), how will it's motion behave? Will it be harmonic and form some sort of space-pendulum or will it eventually fall into the Lagrangian point? And in either case, what will be its equation of motion? If it falls into the Lagrangian point, depending on $x$ and other parameters, how long will it take to fall in? I have included a diagram below. EDITED: So basically, you can start out with vector summation of the forces and get to here: $$2\mathbf{F_{G}}\cos(\theta)=m \cdot \mathbf{a}$$ $\theta$ is basically the angle above the horizontal axis. $$ \frac{2GMm}{\frac{d^2}{4}+\Big(x(t)\Big)^2} \cdot \frac{x(t)}{\sqrt{\frac{d^2}{4}+\Big(x(t)\Big)^2}} = m \cdot a(t)$$ $$\frac{2GM\cdot x(t)}{\bigg (\frac{d^2}{4}+\Big(x(t)\Big)^2 \bigg) \sqrt{\frac{d^2}{4}+\Big(x(t)\Big)^2}} = a(t)$$ But now I get kind of stuck.
There is a saying “a good compromise makes no one happy”. As distressing as that may be, I guess most people still think it is a good idea to seek the best agreement. The problem though, is to define the measure in which the word “best” gets a meaning. Let us consider $n$ people, each with a firm belief on $m$ different yes/no issues. We may think of the people’s beliefs as binary strings $b_ i$ of length $m$ for $0<i\leq n$, i.e. one string per person, with entries reflecting a person’s belief in each of the $m$ issues. An agreement is also a binary string $a$ of length $m$, indicating the agreed outcome of each of the $m$ issues. Some would argue that the agreement $a$ that minimises the maximum of $H(b_ i-a)$ over all $i$, where $H(x)$ is the Hamming function counting the number of non-zero entries of the vector $x$, is the best compromise. Unfortunately, it is widely believed that finding this agreement is computationally infeasible for $n$ and $m$ large. Another reasonable suggestion is the agreement $a$ that minimises $\left(\sum _{i=1}^ n H(b_ i-a)^ p\right)^{1/p}$ for some positive integer $p$. Note in particular that when $p\rightarrow \infty $, this measure coincides with the former. I for one am most content with the choice of $p=1$ though, don’t you agree? On the first line of input is a single positive integer $t$, telling the number of test scenarios to follow. Each test scenario is described by two positive integers $n\leq 100$ and $m\leq 100$, on a line of their own. Then follow $n$ lines, each containing a binary string $b_ i$ on $m$ ‘0’ and ‘1’ characters. For each test scenario, output the assignment $a$ that minimises the measure above with $p = 1$, on a row of its own. If there are several solutions, output anyone. Sample Input 1 Sample Output 1 2 5 5 00000 00100 01001 01101 00101 1 7 0110010 00101 0110010
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 exclusive γγ→ℓ+ℓ− production in proton–proton collisions at s=7 TeV with the ATLAS detector Physics Letters B, ISSN 0370-2693, 10/2015, Volume 749, Issue C, pp. 242 - 261 Journal Article 2. Search for heavy ZZ resonances in the l(+) l(-) l(+) l(-) and l(+) l(-) nu(nu)over-bar final states using proton-proton collisions at root s=13 TeV with the ATLAS detector EUROPEAN PHYSICAL JOURNAL C, ISSN 1434-6044, 04/2018, Volume 78, Issue 4 A search for heavy resonances decaying into a pair of Z bosons leading to l(+) l(-) l(+) l(-) and l(+) l(-) nu(nu) over bar final states, where l stands for... DISTRIBUTIONS \| BOSON \| DECAY \| MASS \| TAUOLA \| TOOL \| PHYSICS, PARTICLES & FIELDS \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences DISTRIBUTIONS \| BOSON \| DECAY \| MASS \| TAUOLA \| TOOL \| PHYSICS, PARTICLES & FIELDS \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences Journal Article 3. Search for heavy ZZ resonances in the ℓ + ℓ - ℓ + ℓ - and ℓ + ℓ - ν ν ¯ final states using proton-proton collisions at s = 13 TeV with the ATLAS detector The European physical journal. C, Particles and fields, ISSN 1434-6044, 2018, Volume 78, Issue 4, pp. 293 - 34 Journal Article 4. Observation of a new particle in the search for the Standard Model Higgs boson with the ATLAS detector at the LHC Physics Letters B, ISSN 0370-2693, 09/2012, Volume 716, Issue 1, pp. 1 - 29 A search for the Standard Model Higgs boson in proton–proton collisions with the ATLAS detector at the LHC is presented. The datasets used correspond to... TRANSVERSE-MOMENTUM \| BROKEN SYMMETRIES \| PARTON DISTRIBUTIONS \| MASSES \| DECAY \| ASTRONOMY & ASTROPHYSICS \| QCD CORRECTIONS \| TAU \| PHYSICS, NUCLEAR \| COLLIDERS \| PHYSICS, PARTICLES & FIELDS \| Collisions (Nuclear physics) \| Analysis \| Standards \| Detectors \| Searching \| Decay \| Elementary particles \| Higgs bosons \| Standard deviation \| Channels \| Physics - High Energy Physics - Experiment \| Physics \| High Energy Physics - Experiment \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS \| Qcd Corrections \| Hadron Colliders \| Transverse-Momentum \| Massless Particles \| Gauge Fields \| Proton-Proton Collisions \| Fysik \| Broken Symmetries \| Physical Sciences \| Naturvetenskap \| Parton Distributions \| Root-S=7 Tev \| Natural Sciences \| Cross-Sections TRANSVERSE-MOMENTUM \| BROKEN SYMMETRIES \| PARTON DISTRIBUTIONS \| MASSES \| DECAY \| ASTRONOMY & ASTROPHYSICS \| QCD CORRECTIONS \| TAU \| PHYSICS, NUCLEAR \| COLLIDERS \| PHYSICS, PARTICLES & FIELDS \| Collisions (Nuclear physics) \| Analysis \| Standards \| Detectors \| Searching \| Decay \| Elementary particles \| Higgs bosons \| Standard deviation \| Channels \| Physics - High Energy Physics - Experiment \| Physics \| High Energy Physics - Experiment \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS \| Qcd Corrections \| Hadron Colliders \| Transverse-Momentum \| Massless Particles \| Gauge Fields \| Proton-Proton Collisions \| Fysik \| Broken Symmetries \| Physical Sciences \| Naturvetenskap \| Parton Distributions \| Root-S=7 Tev \| Natural Sciences \| Cross-Sections Journal Article 5. Search for heavy ZZ resonances in the $$\ell ^+\ell ^-\ell ^+\ell ^-$$ ℓ+ℓ-ℓ+ℓ- and $$\ell ^+\ell ^-\nu \bar{\nu }$$ ℓ+ℓ-νν¯ final states using proton–proton collisions at $$\sqrt{s}= 13$$ s=13 $$\text {TeV}$$ TeV with the ATLAS detector The European Physical Journal C, ISSN 1434-6044, 4/2018, Volume 78, Issue 4, pp. 1 - 34 A search for heavy resonances decaying into a pair of $$Z$$ Z bosons leading to $$\ell ^+\ell ^-\ell ^+\ell ^-$$ ℓ+ℓ-ℓ+ℓ- and $$\ell ^+\ell ^-\nu \bar{\nu }$$... Nuclear Physics, Heavy Ions, Hadrons \| Measurement Science and Instrumentation \| Nuclear Energy \| Quantum Field Theories, String Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Astronomy, Astrophysics and Cosmology Nuclear Physics, Heavy Ions, Hadrons \| Measurement Science and Instrumentation \| Nuclear Energy \| Quantum Field Theories, String Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Astronomy, Astrophysics and Cosmology Journal Article 6. Measurement of the ZZ production cross section in proton-proton collisions at s = 8 $$ \sqrt{s}=8 $$ TeV using the ZZ → ℓ−ℓ+ℓ′−ℓ′+ and Z Z → ℓ − ℓ + ν ν ¯ $$ ZZ\to {\ell}^{-}{\ell}^{+}\nu \overline{\nu} $$ decay channels with the ATLAS detector Journal of High Energy Physics, ISSN 1029-8479, 1/2017, Volume 2017, Issue 1, pp. 1 - 53 A measurement of the ZZ production cross section in the ℓ−ℓ+ℓ′ −ℓ′ + and ℓ − ℓ + ν ν ¯ $$ {\ell}^{-}{\ell}^{+}\nu \overline{\nu} $$ channels (ℓ = e, μ) in... Quantum Physics \| Quantum Field Theories, String Theory \| Hadron-Hadron scattering (experiments) \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Nuclear Experiment Quantum Physics \| Quantum Field Theories, String Theory \| Hadron-Hadron scattering (experiments) \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Nuclear Experiment Journal Article The European Physical Journal C: Particles and Fields, ISSN 1434-6052, 2017, Volume 77, Issue 5, pp. 1 - 53 During 2015 the ATLAS experiment recorded $$3.8\,{\mathrm{fb}}^{-1}$$ 3.8 fb - 1 of proton–proton collision data at a centre-of-mass energy of... PHYSICS, PARTICLES & FIELDS \| Collisions (Nuclear physics) \| Protons \| Data acquisition systems \| Physics - High Energy Physics - Experiment \| High Energy Physics - Phenomenology \| Physics \| PARTICLE ACCELERATORS \| Regular - Experimental Physics \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences PHYSICS, PARTICLES & FIELDS \| Collisions (Nuclear physics) \| Protons \| Data acquisition systems \| Physics - High Energy Physics - Experiment \| High Energy Physics - Phenomenology \| Physics \| PARTICLE ACCELERATORS \| Regular - Experimental Physics \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences Journal Article 8. Combined Measurement of the Higgs Boson Mass in pp Collisions at √s=7 and 8 TeV with the ATLAS and CMS Experiments Physical Review Letters, ISSN 0031-9007, 2015, Volume 114, Issue 19 A measurement of the Higgs boson mass is presented based on the combined data samples of the ATLAS and CMS experiments at the CERN LHC in the H arrow right... CERN \| Large Hadron Collider \| Collisions \| Higgs bosons \| Decomposition \| Solenoids \| Channels \| Invariants \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS CERN \| Large Hadron Collider \| Collisions \| Higgs bosons \| Decomposition \| Solenoids \| Channels \| Invariants \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS Journal Article 9. Search for W W/W Z resonance production in ℓνqq final states in pp collisions at √s=13 TeV with the ATLAS detector Journal of High Energy Physics, ISSN 1126-6708, 03/2018, Volume 2018, Issue 3, pp. 1 - 45 A search is conducted for new resonances decaying into a W W or W Z boson pair, where one W boson decays leptonically and the other W or Z boson decays... Hadron-Hadron scattering (experiments) \| Luminosity \| Quarks \| Large Hadron Collider \| Particle collisions \| Searching \| Decay \| Fysik \| Physical Sciences \| Subatomic Physics \| Naturvetenskap \| Subatomär fysik \| Natural Sciences Hadron-Hadron scattering (experiments) \| Luminosity \| Quarks \| Large Hadron Collider \| Particle collisions \| Searching \| Decay \| Fysik \| Physical Sciences \| Subatomic Physics \| Naturvetenskap \| Subatomär fysik \| Natural Sciences Journal Article
I notice if I put heat under a laser beam it will cause the beam to shimmer. The heated air is causing a disturbance in the air which affects the light. Yet, I can blast a stream of air from a compressor across the beam and there is no effect on the light. What’s going on? Heat ripples the light beam because the temperature of the flame and its surroundings is sufficiently nonuniform that it creates significant differences in the index of refraction of the air that vary quickly with time. The blast of air coming out of the nozzle is close enough to ambient that it doesn't create anisotropies in the index of refraction of the air near the light beam, and so you don't see any ripples. Wind, and generally atmosphere, causes considerable effect on light propagation. However usually the effects are too small for human eye to detect on short length-scales and regular environment. A good example where this effect is big and important (that is, actively funded and researched) is astronomy, and a whole branch of optics called Adaptive optics tries to compensate the phenomena to increase the resolution of telescopes located down on earth. As @niels nielsen mentioned, the thermal effects are more prominent, since the average velocity of the air molecules is larger in comparison to the velocity of air out of the compressor, thus the compressor is just a slight perturbation to regular state. You have to compare the velocity of the air out of compressor to the the RMS velocity of "air particles" -the speed of sound, which is about $340 [\frac{m}{s}]$. Only when the velocity out of the compressor becomes comparable to the speed of sound you will start to notice effects on light trajectories. On the other hand, even the use of a crude estimation found in Wikipedia speed of sound shows that for oven at $T\approx 200[C^\circ]$ the speed of sound changes by more than $30\%$, thus disturbing more the light trajectories.
I am trying to compute the dBFS value of a group of samples (stereo wave file), according to this formula: $$p_{RMS} = \sqrt{\frac{ x_1^2 + x_2^2 + \ldots}n } $$ $$dbFS = 20\log_{10}\frac{ p_{RMS}}{p_{max}}$$ The value I get is wrong, and not stable (I am using a stereo constant -18dbFS 1000H sinewave audio file). I know the input/output of the program is working, because I did an fft on a music file and the output is fine. float get_db(Header header, unsigned int buffer_size, int32_t left[], int32_t right[]) { unsigned int i = 0; float dbfs = 0; //From -144 to 0dbfs double pmoy = 0; double pmax = 0; pmax = maxint(header->bits_per_sample); //Max positive range of sample, here it is 8388608 for (i=0; i < buffer_size; i++){ //Buffer size of 4096 here pmoy =+ pow(left[i], 2); //Just the left channel for now } printf("sum= %f\n", pmoy); //temporary sum pmoy = sqrt( pmoy / buffer_size); printf("pmoy= %f\n", pmoy); //pmoy rms of 4096 samples dbfs = 20 * log10( pmoy / pmax); return dbfs; } I really don't understand why I don't get a constant value of -18dBFS. I am still learning C, there should be some obvious reasons and I hope you could help me. Is there a normalized definition of a dBFS scale for discrete values, and a particular integration time that should be used ?
One thing you should learn, is that analysts like to think of functions as power series (or at worst Laurent series) In this sense, L'Hopital's rules is essentially saying that "When we have a function $ \frac {(x-7)f(x)}{(x-7)g(x)}$, then we can 'fill in' the hole and carry along our own merry way". So, if we don't have L'Hopital, and we know we want to use it, we simply force it out. For example, notice that $$(\sqrt[4]{x+9} - 2)(\sqrt[4]{x+9} + 2)= \sqrt[2]{x+9} -4,$$ which I'm sure you did. Does this help us? No, not yet, because we haven't forced out the troublesome $x-7$. So let's try again, and we use $$(\sqrt{x+9}-4)(\sqrt{x+9}+4) = x+9 - 16 = x-7.$$ Are we done with the denominator? You bet! How about the numerator? It is likely giving us problems with $x-7$, so let's force it out. Try $$(\sqrt{x+2} - \sqrt[3]{x+20})(\sqrt{x+2} + \sqrt[3]{x+20}) = x+2 - (x+20)^{2/3}.$$ Are we done? No not yet, I don't see a $x-7$. So let's use $$ [(x+2) - (x+20)^{2/3} ][(x+2)^2 + (x+2)(x+20)^{2/3} + (x+20^{4/3} ] = (x+2)^3 - (x+20)^2.$$ Are we done? I most certainly hope so, and you can check that we can factor out an $(x-7)$, since $(7+2)^3 - (7+20)^2 = 0$. What's the moral of the story? $$\frac {\sqrt{x+2} - \sqrt[3]{x+20}} {\sqrt[4]{x+9} - 2} \times \frac {\mbox{stuff}} {\mbox{same stuff}} = \frac {(x-7) \times \mbox {something}}{(x-7) \times \mbox {more something}}.$$ And now we rejoice and wait for the cows to come home.
SolidsWW Flash Applet Sample Problem 1 Line 6: Line 6: </p> </p> <p style="background-color:#93BED2;border:black solid 1px;padding:3px;">This applet and WeBWorK problem are based upon work supported by the National Science Foundation under Grant Number DUE-0941388.</p> <p style="background-color:#93BED2;border:black solid 1px;padding:3px;">This applet and WeBWorK problem are based upon work supported by the National Science Foundation under Grant Number DUE-0941388.</p> + <p> <p> A standard WeBWorK PG file with an embedded applet has six sections: A standard WeBWorK PG file with an embedded applet has six sections: Latest revision as of 14:24, 31 July 2013 Flash Applets embedded in WeBWorK questions solidsWW Example Sample Problem with solidsWW.swf embedded A standard WeBWorK PG file with an embedded applet has six sections: A tagging and description section, that describes the problem for future users and authors, An initialization section, that loads required macros for the problem, A problem set-up sectionthat sets variables specific to the problem, An Applet link sectionthat inserts the applet and configures it, (this section is not present in WeBWorK problems without an embedded applet) A text section, that gives the text that is shown to the student, and An answer and solution section, that specifies how the answer(s) to the problem is(are) marked for correctness, and gives a solution that may be shown to the student after the problem set is complete. The sample file attached to this page shows this; below the file is shown to the left, with a second column on its right that explains the different parts of the problem that are indicated above. A screenshot of the applet embedded in this WeBWorK problem is shown below: There are other example problems using this applet: solidsWW Flash Applet Sample Problem 2 solidsWW Flash Applet Sample Problem 3 And other problems using applets: Derivative Graph Matching Flash Applet Sample Problem USub Applet Sample Problem trigwidget Applet Sample Problem solidsWW Flash Applet Sample Problem 1 GraphLimit Flash Applet Sample Problem 2 Other useful links: Flash Applets Tutorial Things to consider in developing WeBWorK problems with embedded Flash applets PG problem file Explanation ##DESCRIPTION ## Solids of Revolution ##ENDDESCRIPTION ##KEYWORDS('Solids of Revolution') ## DBsubject('Calculus') ## DBchapter('Applications of Integration') ## DBsection('Solids of Revolution') ## Date('7/31/2011') ## Author('Barbara Margolius') ## Institution('Cleveland State University') ## TitleText1('') ## EditionText1('2011') ## AuthorText1('') ## Section1('') ## Problem1('') ########################################## # This work is supported in part by the # National Science Foundation # under the grant DUE-0941388. ########################################## This is the The description is provided to give a quick summary of the problem so that someone reading it later knows what it does without having to read through all of the problem code. All of the tagging information exists to allow the problem to be easily indexed. Because this is a sample problem there isn't a textbook per se, and we've used some default tagging values. There is an on-line list of current chapter and section names and a similar list of keywords. The list of keywords should be comma separated and quoted (e.g., KEYWORDS('calculus','derivatives')). DOCUMENT(); loadMacros( "PGstandard.pl", "AppletObjects.pl", "MathObjects.pl", ); This is the The TEXT(beginproblem()); $showPartialCorrectAnswers = 1; Context("Numeric"); $a = random(2,10,1); $b = random(2,10,1); $xy = 'y'; $func1 = "$asin(piy/8)+2"; $func2 = "$bsin(piy/2)+2"; $xmax = max(Compute("$a+2"), Compute("$b+2"),9); $shapeType = 'circle'; $correctAnswer = Compute("64$a+4pi$a^2+32pi"); This is the The solidsWW.swf applet will accept a piecewise defined function either in terms of x or in terms of y. We set ######################################### # How to use the solidWW applet. # Purpose: The purpose of this applet # is to help with visualization of # solids # Use of applet: The applet state # consists of the following fields: # xmax - the maximum x-value. # ymax is 6/5ths of xmax. the minima # are both zero. # captiontxt - the initial text in # the info box in the applet # shapeType - circle, ellipse, # poly, rectangle # piece: consisting of func and cut # this is a function defined piecewise. # func is a string for the function # and cut is the right endpoint # of the interval over which it is # defined # there can be any number of pieces # ######################################### # What does the applet do? # The applet draws three graphs: # a solid in 3d that the student can # rotate with the mouse # the cross-section of the solid # (you'll probably want this to # be a circle # the radius of the solid which # varies with the height ######################################### <p> This is the Those portions of the code that begin the line with ################################### # Create link to applet ################################### $appletName = "solidsWW"; $applet = FlashApplet( codebase => findAppletCodebase ("$appletName.swf"), appletName => $appletName, appletId => $appletName, setStateAlias => 'setXML', getStateAlias => 'getXML', setConfigAlias => 'setConfig', maxInitializationAttempts => 10, #answerBoxAlias => 'answerBox', height => '550', width => '595', bgcolor => '#e8e8e8', debugMode => 0, submitActionScript => '' ); <p>You must include the section that follows ################################### # Configure applet ################################### $applet->configuration(qq{<xml><plot> <xy>$xy</xy> <captiontxt>'Compute the volume of the figure shown.' </captiontxt> <shape shapeType='$shapeType' sides='3' ratio='1.5'/> <xmax>$xmax</xmax> <theColor>0x0000ff</theColor> <profile> <piece func='$func1' cut='8'/> <piece func='$func2' cut='10'/> </profile> </plot></xml>}); $applet->initialState(qq{<xml><plot> <xy>$xy</xy> <captiontxt>'Compute the volume of the figure shown.' </captiontxt> <shape shapeType='$shapeType' sides='3' ratio='1.5'/> <xmax>$xmax</xmax> <theColor>0x0000ff</theColor> <profile> <piece func='$func1' cut='8'/> <piece func='$func2' cut='10'/> </profile> </plot></xml>}); TEXT( MODES(TeX=>'object code', HTML=>$applet->insertAll( debug=>0, includeAnswerBox=>0, ))); The lines $applet->initialState(qq{<xml><plot> <xy>$xy</xy> <captiontxt>'Compute the volume of the figure shown.'</captiontxt> <shape shapeType='$shapeType' sides='3' ratio='1.5'/> <xmax>$xmax</xmax> <theColor>0x0000ff</theColor> <profile> <piece func='$func1' cut='8'/> <piece func='$func2' cut='10'/> </profile> </plot></xml>}); configure the applet. The configuration of the applet is done in xml. The argument of the function is set to the value held in the variable The code Answer submission and checking is done within WeBWorK. The applet is intended to aid with visualization and is not used to evaluate the student submission. TEXT(MODES(TeX=>"", HTML=><<'END_TEXT')); <script> if (navigator.appVersion.indexOf("MSIE") > 0) { document.write("<div width='3in' align='center' style='background:yellow'> You seem to be using Internet Explorer. <br/>It is recommended that another browser be used to view this page.</div>"); } </script> END_TEXT The text between the BEGIN_TEXT $BR $BR Find the volume of the solid of revolution formed by rotating the curve \[x=\begin{cases} $a\sin\left(\frac{\pi y}{8}\right)+2 &y\le 8\\ $b\sin\left(\frac{\pi y}{2}\right)+2 &8<y\le 10\end{cases}\] about the $y$-axis. \{ans_rule(35) \} $BR END_TEXT Context()->normalStrings; This is the ################################ # # Answers # ## answer evaluators ANS( $correctAnswer->cmp() ); ENDDOCUMENT(); This is the The License The Flash applets developed under DUE-0941388 are protected under the following license: Creative Commons Attribution-NonCommercial 3.0 Unported License.
Given $q = e^{2\pi i \tau}$ and the Eisenstein series $E_{2k}(\tau)$, i.e., $$E_2(\tau) = 1-24\sum_{n=1}^\infty \frac{n q^n}{1-q^n}$$ $$E_4(\tau) = 1+240\sum_{n=1}^\infty \frac{n^3 q^n}{1-q^n}$$ and so on. Define the function, $$F_{2k}(\tau) = \frac{E_{2k}(\tau)}{\left(E_2(\tau)-\frac{3}{\pi\; \Im(\tau)}\right)^k}$$ for $k \geq 2$, where $\tau = \frac{1+\sqrt{-d}}{2}$, $\Im(\tau)$ is the imaginary part of $\tau$, and $d$ has class number $h(-d) = m$. For example, we have, $$F_4\left(\tfrac{1+\sqrt{-163}}{2}\right) = \frac{5\cdot23\cdot29\cdot163}{2^2\cdot3\cdot181^2}$$ $$F_6\left(\tfrac{1+\sqrt{-163}}{2}\right) = \frac{7\cdot11\cdot19\cdot127\cdot163^2}{2^9\cdot181^3}$$ $$F_8(\tau) = F_4^2(\tau)$$ and so on. Question: In general, is it true that for $k \geq 2$ the function $F_{2k}$, ? (I've tested it with like the j-function, is an algebraic number of degree m = h(-d) dwith higher class numbers, and it seems to be true.)
1.A watch is sold for Rs.440 cash or for Rs.200 cash down payment together with Rs.244 to be paid after one month. Find the rate of interest charged in the installment scheme. a. 10% b. 15% c. 20% d. 25% Answer: C Explanation: Principal for the next month = 440 - 200 = 240 Amount paid after next month = 244 Therefore interest charged at Rs.240 = $\displaystyle\frac{{240 \times 1 \times R}}{{12 \times 100}} = 4$ Rate of interest (R) = 20% per annum Principal for the next month = 440 - 200 = 240 Amount paid after next month = 244 Therefore interest charged at Rs.240 = $\displaystyle\frac{{240 \times 1 \times R}}{{12 \times 100}} = 4$ Rate of interest (R) = 20% per annum 2.A cell phone is available for Rs. 600 or for Rs.300 cash down payment together with Rs.360 to be paid after two months. Find the rate of interest charged under this scheme. a. 20% b. 50% c. 120% d. None Answer: C Explanation: Amount as a principal for first and second month = 600 - 300 = Rs.300 Now, Interest = 360 - 300 = Rs.60 $60 = \displaystyle\frac{{300 \times 2 \times r}}{{100 \times 12}}$ r = 120% Amount as a principal for first and second month = 600 - 300 = Rs.300 Now, Interest = 360 - 300 = Rs.60 $60 = \displaystyle\frac{{300 \times 2 \times r}}{{100 \times 12}}$ r = 120% 3.Samsung mobile phone is available for Rs.2500 cash or Rs.520 cash down payments followed by 4 equal installments. If the rate of interest charged is 25% per annum Simple interest, calculate the monthly installment a. 520 b. 480 c. 550 d. None of these Answer: C Explanation: Balance price to be paid in installments = Rs.1980 At the rate of r% per annum after 4 months, Rs.1980 will amount to Rs.$P + \dfrac{{P \times T \times R}}{{100 \times 12}}$ = 1980 + $\dfrac{{1980 \times 4 \times 25}}{{12 \times 100}}$ = 2145 - - - - - (2) Now, the total amount for the 4 installments at the end of fourth month will be ${\left( {x + \dfrac{{25x \times 3}}{{12 \times 100}}} \right) + \left( {x + \dfrac{{25x \times 2}}{{12 \times 100}}} \right)}$ + ${\left( {x + \dfrac{{25x \times 1}}{{12 \times 100}}} \right) + x}$ = $4x + \displaystyle\frac{{25x}}{{1200}}\left( {1 + 2 + 3} \right) = \displaystyle\frac{{33x}}{8}$ - - - - - (2) From (1) and (2), Rs.2145 = $\displaystyle\frac{{33x}}{8}$ x = Rs.520 Balance price to be paid in installments = Rs.1980 At the rate of r% per annum after 4 months, Rs.1980 will amount to Rs.$P + \dfrac{{P \times T \times R}}{{100 \times 12}}$ = 1980 + $\dfrac{{1980 \times 4 \times 25}}{{12 \times 100}}$ = 2145 - - - - - (2) Now, the total amount for the 4 installments at the end of fourth month will be ${\left( {x + \dfrac{{25x \times 3}}{{12 \times 100}}} \right) + \left( {x + \dfrac{{25x \times 2}}{{12 \times 100}}} \right)}$ + ${\left( {x + \dfrac{{25x \times 1}}{{12 \times 100}}} \right) + x}$ = $4x + \displaystyle\frac{{25x}}{{1200}}\left( {1 + 2 + 3} \right) = \displaystyle\frac{{33x}}{8}$ - - - - - (2) From (1) and (2), Rs.2145 = $\displaystyle\frac{{33x}}{8}$ x = Rs.520 4.Kishore purchases a track suit for Rs.2400 cash or for Rs.1000 cash down payments and two monthly installments of Rs.800 each. Find the rate of interest. a. 75% b. 120% c. 50% d. None of these Answer: B Explanation: Amount as a principal for 2 month = 2400 - 1000 = 1400 At the rate of r% per annum after 2 months, Rs.1400 will amount to Rs. $1400 + \displaystyle\frac{{1400 \times r \times 2}}{{100 \times 12}}$ ......(1) Again total amount for the 2 installments at the end of second month will be Rs.$\left[ {800 + \left( {800 + \displaystyle\frac{{800 \times r \times 1}}{{100 \times 12}}} \right)} \right]$ - - - (2) From (1) and (2), we get $1400 + \displaystyle\frac{{2800 \times r}}{{1200}} = 1600 + \displaystyle\frac{{800r}}{{1200}}$ R = 120% Amount as a principal for 2 month = 2400 - 1000 = 1400 At the rate of r% per annum after 2 months, Rs.1400 will amount to Rs. $1400 + \displaystyle\frac{{1400 \times r \times 2}}{{100 \times 12}}$ ......(1) Again total amount for the 2 installments at the end of second month will be Rs.$\left[ {800 + \left( {800 + \displaystyle\frac{{800 \times r \times 1}}{{100 \times 12}}} \right)} \right]$ - - - (2) From (1) and (2), we get $1400 + \displaystyle\frac{{2800 \times r}}{{1200}} = 1600 + \displaystyle\frac{{800r}}{{1200}}$ R = 120% 5.What annual installment will discharge a debt of Rs 2,360 due in four years at 12% p.a. simple interest? a. 400 b. 500 c. 300 d. 600 Answer: B Explanation: Installments paid at the end of 1st, 2nd, 3rd and 4th years earn a simple interest at 12% p.a. for 3, 2, 1 and 0 years respectively. Hence the respective installments amount to, (100 + 3 x 12), (100 + 2 x 12), (100 + 1 x 12) and 100, when annual installment is Rs 100. Hence amount paid is Rs. 136 + Rs 124 + Rs 112 + Rs 100 i.e., Rs 472, when the annual installment is Rs.100 For an amount of Rs 2,360, annual installment = $\displaystyle\frac{{2,360 \times 100}}{{472}} = Rs.{\rm{ }}500$ Installments paid at the end of 1st, 2nd, 3rd and 4th years earn a simple interest at 12% p.a. for 3, 2, 1 and 0 years respectively. Hence the respective installments amount to, (100 + 3 x 12), (100 + 2 x 12), (100 + 1 x 12) and 100, when annual installment is Rs 100. Hence amount paid is Rs. 136 + Rs 124 + Rs 112 + Rs 100 i.e., Rs 472, when the annual installment is Rs.100 For an amount of Rs 2,360, annual installment = $\displaystyle\frac{{2,360 \times 100}}{{472}} = Rs.{\rm{ }}500$ 6.Hiralal gave a loan of Rs. 20 to Ramlal and recovered it at the rate of Rs 3.50 for eight months, commencing from the end of 1st month. What is the effective rate of simple interest? a. 60% b. 80% c. 20% d. 90% Answer: A Explanation: Principal = Rs.20 Amount = Rs.3.50 × 8 = Rs.28 Interest = Rs.28 - Rs.20 = Rs.8 Time = 8 months = 8/12 years 8 = $\displaystyle\frac{{20 \times 8 \times r}}{{12 \times 100}}$ r = $\displaystyle\frac{{12 \times 100}}{{20}}$ = 60% p.a. Principal = Rs.20 Amount = Rs.3.50 × 8 = Rs.28 Interest = Rs.28 - Rs.20 = Rs.8 Time = 8 months = 8/12 years 8 = $\displaystyle\frac{{20 \times 8 \times r}}{{12 \times 100}}$ r = $\displaystyle\frac{{12 \times 100}}{{20}}$ = 60% p.a. 7.Find the amount of equal installment, annual payment of which will discharge a debt of Rs. 848 due in 4 years at 4% p.a. of Simple interest. a. Rs.200 b. Rs.225 c. Rs.240 d. Rs.255 Answer: A Explanation: Amount of each installment =$\displaystyle\frac{{100P}}{{100n + \displaystyle\frac{{n(n - 1)r}}{2}}}$ = $\displaystyle\frac{{{\rm{100 \times 848}}}}{{{\rm{100 \times 4 + }}\dfrac{{{\rm{4 \times 4 \times 3}}}}{{\rm{2}}}}}$ = $\displaystyle\frac{{{\rm{848 \times 100}}}}{{{\rm{424}}}}$ = Rs. 200 Amount of each installment =$\displaystyle\frac{{100P}}{{100n + \displaystyle\frac{{n(n - 1)r}}{2}}}$ = $\displaystyle\frac{{{\rm{100 \times 848}}}}{{{\rm{100 \times 4 + }}\dfrac{{{\rm{4 \times 4 \times 3}}}}{{\rm{2}}}}}$ = $\displaystyle\frac{{{\rm{848 \times 100}}}}{{{\rm{424}}}}$ = Rs. 200 8.Find the amount of debt that will be discharged by 5 equal installments of Rs. 200 each, if the debt is due in 5 year at 5% p.a. a. Rs.1100 b. Rs.1200 c. Rs.1255 d. Rs.1400 Answer: A Explanation: We know that $x = \displaystyle\frac{{100P}}{{100n + \displaystyle\frac{{n(n - 1)r}}{2}}}$ $ \Rightarrow 200 = \displaystyle\frac{{100P}}{{100 \times 5 + \displaystyle\frac{{5(5 - 1)5}}{2}}} = \displaystyle\frac{{100P}}{{500 + \displaystyle\frac{{5.4.5}}{2}}}$ $ \Rightarrow P = \displaystyle\frac{{200}}{{100}} \times \left( {500 + 50} \right) = 1100$ We know that $x = \displaystyle\frac{{100P}}{{100n + \displaystyle\frac{{n(n - 1)r}}{2}}}$ $ \Rightarrow 200 = \displaystyle\frac{{100P}}{{100 \times 5 + \displaystyle\frac{{5(5 - 1)5}}{2}}} = \displaystyle\frac{{100P}}{{500 + \displaystyle\frac{{5.4.5}}{2}}}$ $ \Rightarrow P = \displaystyle\frac{{200}}{{100}} \times \left( {500 + 50} \right) = 1100$ 9.Gopal borrows Rs 1,00,000 from a bank at 10% p.a. simple interest and clears the debt in five years. If the installments paid at the end of the first, second, third and fourth years to clear the debt are Rs 10,000, Rs 20,000, Rs 30,000 and Rs 40,000 respectively, what amount should be paid at the end of the fifth year to clear the debt? a. Rs.20,000 b. Rs.24,500 c. Rs.30,000 d. Rs.35,900 Answer: B Explanation: Please remember, In the case of simple interest, installment amount will always be reduced from principal and the interest will be calculated on the remaining principal. Simple interest for the first year = $\displaystyle\frac{{P \times T \times R}}{{100}} = \displaystyle\frac{{100000 \times 1 \times 10}}{{100}}$ = Rs.10000 Amount after first instalment is paid = 1,00,000 - 10,000 = Rs 90,000 Simple interest for the second year = $\displaystyle\frac{{P \times T \times R}}{{100}} = \displaystyle\frac{{90000 \times 1 \times 10}}{{100}}$ = Rs.90,000 Amount after second instalment is paid = 90,000 - 20,000 = Rs 70,000 Simple interest for the third year = $\displaystyle\frac{{P \times T \times R}}{{100}} = \displaystyle\frac{{70000 \times 1 \times 10}}{{100}}$ = Rs.7000 Amount after third instalment is paid = 70,000 - 30,000 = Rs 40,000 Simple interest for the fourth year = $\displaystyle\frac{{P \times T \times R}}{{100}} = \displaystyle\frac{{40000 \times 1 \times 10}}{{100}}$ = Rs.4000 Amount after fourth instalment is paid = 40,000 - 40,000 = 0. Balance of debt for the fifth year = simple interest for four years = 10,000 + 9,000 + 7,000 + 4,000 = Rs 30,000. Please remember, In the case of simple interest, installment amount will always be reduced from principal and the interest will be calculated on the remaining principal. Simple interest for the first year = $\displaystyle\frac{{P \times T \times R}}{{100}} = \displaystyle\frac{{100000 \times 1 \times 10}}{{100}}$ = Rs.10000 Amount after first instalment is paid = 1,00,000 - 10,000 = Rs 90,000 Simple interest for the second year = $\displaystyle\frac{{P \times T \times R}}{{100}} = \displaystyle\frac{{90000 \times 1 \times 10}}{{100}}$ = Rs.90,000 Amount after second instalment is paid = 90,000 - 20,000 = Rs 70,000 Simple interest for the third year = $\displaystyle\frac{{P \times T \times R}}{{100}} = \displaystyle\frac{{70000 \times 1 \times 10}}{{100}}$ = Rs.7000 Amount after third instalment is paid = 70,000 - 30,000 = Rs 40,000 Simple interest for the fourth year = $\displaystyle\frac{{P \times T \times R}}{{100}} = \displaystyle\frac{{40000 \times 1 \times 10}}{{100}}$ = Rs.4000 Amount after fourth instalment is paid = 40,000 - 40,000 = 0. Balance of debt for the fifth year = simple interest for four years = 10,000 + 9,000 + 7,000 + 4,000 = Rs 30,000.
> Puzzle 71. Can you make the complex numbers, ℂ, into a commutative monoidal poset with the usual + and 0 and some concept of ≤? If so, how many ways can you do this? We can always just do \$x \preceq_1 y \iff \mathfrak{Re}(x) \leq \mathfrak{Re}(y)\$ where \$\mathfrak{Re}(x)\$ is the real component of \$x\$. But this is just a special case of looking at the magnitudes of the [orthogonal projection](https://en.wikibooks.org/wiki/Linear_Algebra/Orthogonal_Projection_Onto_a_Line) of two points in \$\mathcal{C}\$ onto a common point. Specifically, \$\mathfrak{Re}(x)\$ is the magnitude of \$x\$ projected onto \$1 + i0\$. We could pick any point in \$\mathbb{C}\$ and do this. The magnitude of the orthogonal projection of \$p + iq\$ onto \$x + i y\$ is \$p x + y q\$, or the usual [dot-product \$(\cdot)\$](https://en.wikipedia.org/wiki/Dot_product) from linear algebra. So we have \$2^{\aleph_0}\$ other examples, each one corresponding to $$ x \preceq_p y \iff p \cdot x \leq p \cdot y $$ for each point \$p \in \mathbb{C}\$. [wikipedia_norm]: https://en.wikipedia.org/wiki/Norm_(mathematics)
Denote $[r]\triangleq\{1,2,\ldots,r\}$. Consider a game with $n$ players, $[n]$, each has $m$ strategies, $[m]$. Each player $i$ has an associated payoff function, which considers only his selected strategy, and the number of players selected the same strategy: $$U_i:[m]\times[n]\to[0,1]$$ Furthermore, the utility function is monotonically decreasing in the number of players which picked the same strategy, i.e. $$\forall i\in[n],j\in[m],k\in[n-1]:U_i(j,k)\geq U_i(j,k+1)$$ Does this game always have a pure Nash equilibrium? Can we (computationally) find it efficiently? Notice that the special case, where all players are symmetric ($\forall i,j\in[n]: U_i\equiv U_j\equiv U$), the game reduces to an exact potential game and therefore is guaranteed to have a pure Nash equilibrium. The potential function for the symmetric case would be, given a strategy profile $s$: $$\phi(s) = \sum_{j\in[m]}\sum_{k=1}^{\#_j(s)} U(j,k)$$ Where $\#_j(s)$ is the number of players in $s$ playing strategy $j$.
broall wrote: EgmatQuantExpert wrote: What is the product of all values of x that satisfies the equation: $\sqrt{5x} + 1= \sqrt{(7x - 3)}$ ? A. 9 B. 11 C. 49 D. 99 E. None of the above First, make sure that $\sqrt{5x}$ and $\sqrt{7x-3}$ have real value so $x \geq 0$ and $x \geq \frac{3}{7}$. Hence $x \geq \frac{3}{7}$. Now, solve that equation $\begin{align} \quad \sqrt{5x}+1 &= \sqrt{7x-3} \\ 5x + 2\sqrt{5x}+1 &= 7x-3 \\ 2\sqrt{5x} &= 2x-4 \\ \sqrt{5x} &= x-2 \\ x - \sqrt{5x} - 2 &=0 \end{align}$ Set $t = \sqrt{x} \geq \sqrt{3/7}$ we have $t^2 -t\sqrt{5} -2 =0$ $\delta = 5 + 42 = 13 \implies t_{12}=\frac{\sqrt{5} \pm \sqrt{13}}{2}$ We have $t_1 = \frac{\sqrt{5}+ \sqrt{13}}{2} \approx 2.9 > 1 > \sqrt{\frac{3}{7}}$. Choose this root. $t_2 = \frac{\sqrt{5}- \sqrt{13}}{2} \approx -0.7 < 0 < \sqrt{\frac{3}{7}}$. Eliminate this root. Hence $x=t_1^2=\frac{9+\sqrt{65}}{2}$. The answer E. It's actually x=(9+√65)/2 and x=(9-√65)/2 There are 2 solutions the best way to check it is b^2-4ac 4 possible scenarios, one of them if the result is positive non-perfect square number then there are 2 irrational solutions but the product of them will be rational number, in that case 4 (9+√65)/2(9-√65)/2=(81-65)/4=4 the question wasn't formulated right I've spent like 3 or 4 minutes double and cross checking myself, 'cause i got the solution here it's but not among the available answers...
Search Now showing items 1-10 of 108 Measurement of electrons from beauty hadron decays in pp collisions at root √s=7 TeV (Elsevier, 2013-04-10) The production cross section of electrons from semileptonic decays of beauty hadrons was measured at mid-rapidity (\|y\| < 0.8) in the transverse momentum range 1 < pT <8 GeV/c with the ALICE experiment at the CERN LHC in ... Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC (Springer, 2014-10) Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ... 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 ... 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 ... Multiplicity dependence of the average transverse momentum in pp, p-Pb, and Pb-Pb collisions at the LHC (Elsevier, 2013-12) The average transverse momentum <$p_T$> versus the charged-particle multiplicity $N_{ch}$ was measured in p-Pb collisions at a collision energy per nucleon-nucleon pair $\sqrt{s_{NN}}$ = 5.02 TeV and in pp collisions at ... Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider (American Physical Society, 2016-02) The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ... Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV (Elsevier, 2016-02) Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < \|\eta\| < 4.0$) and associated particles in the central range ($\|\eta\| < 1.0$) are measured with the ALICE detector in ... 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 ... Directed flow of charged particles at mid-rapidity relative to the spectator plane in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV (American Physical Society, 2013-12) The directed flow of charged particles at midrapidity is measured in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV relative to the collision plane defined by the spectator nucleons. Both, the rapidity odd ($v_1^{odd}$) and ... Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV (Springer, 2016-08) The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ...
Geophysicists often assume that the earth is isotropic. This word comes from 'iso', meaning same, and 'tropikos', meaning something to do with turning. The idea is that isotropic materials look the same in all directions — they have no orientation, and we can make measurements in any direction and get the same result. Note that this is different from homogeneous, which is the quality of uniformity of composition. You can think of anisotropy as a directional (not just spatial) variation in homogeneity. In the illustration, I may have cheated a bit. The lower-left image shows a material that is homogeneous but anisotropic. The thin lines are supposed to indicate microfractures, say, or the alignment of clay flakes, or even just stress. So although the material has uniform composition, at least at this scale, it has an orientation. The recognition of the earth's anisotropy is a dominant theme among papers in our forthcoming 52 Things book on rock physics. It's not exactly a new thing — it was an emerging trend 10 years ago when Larry Lines at U of C reviewed Milo Backus's famous 'challenges' (Lines 2005). And even then, the spread of anisotropic processing and analysis had been underway for almost 20 years since Leon Thomsen's classic 1986 paper, Weak elastic anisotropy. This paper introduced three parameters that we need—alongside the usual $V_\text{P}$, $V_\text{S}$, and $\rho$—to describe anisotropy. They are $\delta$ (delta), $\epsilon$ (epsilon), and $\gamma$ (gamma), collectively referred to as Thomsen's parameters. $\delta$ or delta— the short offset effect — captures the relationship between the velocity required to flatten gathers (the NMO velocity) and the zero-offset average velocity as recorded by checkshots. It's easy to measure, but perhaps hard to understand in physical terms. $\epsilon$ or epsilon— the long offset effect — is, according to Thomsen himself: "the fractional difference between vertical and horizontal P velocities; i.e., it is the parameter usually referred to as 'the' anisotropy of a rock". Unfortunately, the horizontal velocity is rather hard to measure. $\gamma$ or gamma— the shear wave effect — relates, as rock physics meister Colin Sayers put it on Twitter, a horizontal shear wave with horizontal polarization to a vertical shear wave. He added, "$\gamma$ can be determined in a single well using sonic. So the correlation with $\epsilon$ and $\delta$ is of great interest." Sidenote to aspiring authors: Thomsen's seminal paper, which has been cited over 2800 times, is barely 13 pages long. Three and a half of those pages are taken up by... data! A huge table containing the elastic parameters of almost 60 samples. And this is from a corporate scientist at Amoco. So no more excuses: publish you data! </rant> Vertical transverse what now? The other bit of jargon you will come across is the concept of transverse isotropy, which is a slightly perverse (to me) way of expressing the orientation of the anisotropy effect. In vertical transverse isotropy, the horizontal velocity is different from the vertical velocity. Think of flat-lying shales with gravity dominating the stress field. Usually, the velocity is faster along the beds than it is across the beds. This manifests as nonhyperbolic moveout in the far offsets, in particular a pull-up or 'hockey stick' effect in the gathers — the arrivals are unexpectedly early at long offsets. Clearly, this will also affect AVO analysis. There's more jargon. If the rocks are dipping, we call it tilted transverse isotropy, or TTI. But if the anisotropies, so to speak, are oriented vertically — as with fractures, for example, or simply horizontal stress — then it's horizontal transverse isotropy, or HTI. This causes azimuthal (compass directional) travel-time variations. We can even venture into situations where we encounter orthorhombic anisotropy, as in the combined VTI/HTI model shown above. It's easy to imagine how these effects, if not accounted for in processing, can (and do!) result in suboptimal seismic images. Accounting for them is not easy though, and trying can do more harm than good. If you have handy rules of thumb of ways of conceptualizing anisotropy, I'd love to hear about them. Some time soon I want to write about thin-layer anisotropy, which is where this post was going until I got sidetracked... References Lines, L (2005). Addressing Milo's challenges with 25 years of seismic advances. The Leading Edge 24 (1), 32–35. DOI 10.1190/1.2112389. Thomsen, L (1986). Weak elastic anisotropy. Geophysics 51 (10), 1954–1966. DOI 10.1190/1.1442051.
Specifically, we offer a choice of FHE schemes based on the learning with error (LWE) or ring-LWE (RLWE) problems that have $2^\secparam$ security against known attacks. For RLWE, we have: 1. A leveled FHE scheme that can evaluate $L$-level arithmetic circuits with $\tilde{O}(\secparam \cdot L^3)$ per-gate computation -- i.e., computation {\em quasi-linear} in the security parameter. Security is based on RLWE for an approximation factor exponential in $L$. This construction does not use the bootstrapping procedure. 2. A leveled FHE scheme that uses bootstrapping {\em as an optimization}, where the per-gate computation (which includes the bootstrapping procedure) is $\tilde{O}(\secparam^2)$, {\em independent of $L$}. Security is based on the hardness of RLWE for {\em quasi-polynomial} factors (as opposed to the sub-exponential factors needed in previous schemes). We obtain similar results for LWE, but with worse performance. We introduce a number of further optimizations to our schemes. As an example, for circuits of large width -- e.g., where a constant fraction of levels have width at least $\secparam$ -- we can reduce the per-gate computation of the bootstrapped version to $\tilde{O}(\secparam)$, independent of $L$, by {\em batching the bootstrapping operation}. Previous FHE schemes all required $\tilde{\Omega}(\secparam^{3.5})$ computation per gate. At the core of our construction is a much more effective approach for managing the noise level of lattice-based ciphertexts as homomorphic operations are performed, using some new techniques recently introduced by Brakerski and Vaikuntanathan (FOCS 2011).
Difference between revisions of "Stiffened equation of state" (fixed typo in las formula) Line 20: Line 20: It is useful to notice that, given this equation of state, the [[adiabatic]] law is modified from its ideal form: It is useful to notice that, given this equation of state, the [[adiabatic]] law is modified from its ideal form: − :<math> \gamma p+p^* = (\gamma p_0 +p^* ) \left(\frac{\rho}{\rho_0}\right)^\gamma </math> + :<math> \gamma p+p^* = (\gamma p_0 +p^* ) \left(\frac{\rho}{\rho_0}\right)^\gamma </math> + ==References== ==References== Revision as of 13:34, 5 March 2015 The Stiffened equation of state is a simplified form of the Grüneisen equation of state [1].When considering water under very high pressures (typical applications are underwater explosions, extracorporeal shock wave lithotripsy, and sonoluminescence) the stiffened equation of state is often used: where is the internal energy per unit mass, given by (Eq. 15 in [2]): where is the heat capacity at constant pressure. is an empirically determined constant typically taken to be about 6.1, and is another constant, representing the molecular attraction between water molecules. The magnitude of the later correction is about 2 gigapascals (20,000 atmospheres). , from which the value of $p^*$ may be computed given all the other variables. Thus water behaves as though it is an ideal gas that is already under about 20,000 atmospheres (2 GPa) pressure, and explains why water is commonly assumed to be incompressible: when the external pressure changes from 1 atmosphere to 2 atmospheres (100 kPa to 200 kPa), the water behaves as an ideal gas would when changing from 20,001 to 20,002 atmospheres (2000.1 MPa to 2000.2 MPa). This equation mispredicts the heat capacity of water but few simple alternatives are available for severely nonisentropic processes such as strong shocks. It is useful to notice that, given this equation of state, the adiabatic law is modified from its ideal form: which, rearranged is known as the Cole equation of state.
Equivalence Class of Element is Subset Jump to navigation Jump to search Theorem $\forall x \in S: \eqclass x {\mathcal R} \subseteq S$ Proof $\displaystyle y$ $\in$ $\displaystyle \eqclass x {\mathcal R}$ $\displaystyle \leadsto \ \ $ $\displaystyle \tuple {x, y}$ $\in$ $\displaystyle \mathcal R$ Definition of Equivalence Class $\displaystyle \leadsto \ \ $ $\displaystyle x$ $\in$ $\displaystyle S \land y \in S$ Definition of Relation $\displaystyle \leadsto \ \ $ $\displaystyle \eqclass x {\mathcal R}$ $\subseteq$ $\displaystyle S$ Definition of Subset $\blacksquare$
In last week’s discussion of proofs by contradiction and nonconstructive proofs, we showed: Theorem: There exist irrational numbers $x$ and $y$ with the property that $x^y$ is rational. However, our proof was nonconstructive: it did not pinpoint explicit values for $x$ and $y$ that satisfy the condition, instead proving only that such numbers must exist. Would a more constructive proof be more satisfying? Let’s see! I claim $x=\sqrt{2}$ and $y=\log_2 9$ work, because $\sqrt{2}$ we already know to be irrational, $y=\log_2 9$ can be similarly proved to be irrational (try this!), and $$x^y = \sqrt{2}^{\log_2 9} = \sqrt{2}^{\log_{\sqrt{2}}3}=3,$$ which is rational. Let’s further discuss why last week’s proof was less satisfying. The following rephrasing of this proof may help shed some light on the situation: Proof: Assume the theorem were false, so that any time $x$ and $y$ were irrational, $x^y$ would also be irrational. This would imply that $\sqrt{2}^{\sqrt{2}}$ would be irrational, and by applying our assumption again, $\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}$ would also be irrational. But this last number equals 2, which is rational. This contradiction disproves our assumption and thereby proves the theorem, QED. So perhaps this argument seems less satisfactory simply because it is, at its core, a proof by contradiction. It does not give us evidence for the positive statement “$x$ and $y$ exist”, but instead only for the negative statement “$x$ and $y$ don’t not exist.” (Note the double negative.) This distinction is subtle, but a similar phenomenon can be found in the English language: the double negative “not bad” does not mean “good” but instead occupies a hazy middle-ground between the two extremes. And even though we don’t usually think of such a middle-ground existing between logic’s “true” and “false”, proofs by contradiction fit naturally into this haze. In fact, these ideas motivate a whole branch of mathematical logic called Constructive logic that disallows double negatives and proofs by contradiction, instead requiring concrete, constructive justifications for all statements. But wait; last week’s proof that $\sqrt{2}$ is irrational used contradiction, and therefore is not acceptable in constructive logic. Can we prove this statement constructively? We must show that $\sqrt{2}$ is not equal to any rational number; what does it even mean to do this constructively? First, we turn it into a positive statement: we must show that $\sqrt{2}$ is unequal to every rational number. And how do we constructively prove that two numbers are unequal? By showing that they are measurably far apart. So, here is a sketch of a constructive proof: $\sqrt{2}$ is unequal to every rational number $a/b$ because $$\left\|\sqrt{2} – \frac{a}{b}\right\| \ge \frac{1}{3b^2}.$$ See if you can verify this inequality! [1] PS. In case you are still wondering whether $\sqrt{2}^{\sqrt{2}}$ is rational or irrational: It is irrational (moreover, transcendental), but the only proof that I know uses a very difficult theorem of Gelfond and Schneider.
I would like to monitor step by step running of ECDSA. Parameters I am working on: The elliptic curve satisfies the equation: $y^2 = x^3 + x + 1$. I picked up the base point $G$ as: $(0, 1)$ Finally, the modulo $p$ is $977$ My private key is $19$. Thereby, my public key is equal to $19\times G$ and its coordinates are $(396, 650)$. Suppose that my hash is equal to $14$. Then, I picked up a random key. It would be $17$. So, random point would be $17\times G$ and it is equal to $(699, 739)$. My signature is pair of $r$ and $s$, where $r$ is $x$ coordinate of my random point, and $s$ can be calculated as: $$\begin{align} r &= 699\\ s &= ((\text{random key})^{-1} \bmod p)\cdot(\text{hash} + r \cdot (\text{my private key}))\\ &= (17^{-1}\bmod 977)\cdot(14 + 699\cdot19)\bmod 977\\ &= (115\cdot13295)\bmod977\\ &= 1528925\bmod977\\ &= 897 \end{align}$$ So, my signature is $(699, 897)$. Let's verify the signature $$\begin{align} w &= s^{-1}\bmod p\\ &= 897^{-1}\bmod 977\\ &= 403\\ u_1 &= \text{hash} \cdot w \bmod p\\ &= 14 \cdot 403 \bmod 977\\ &= 5642 \bmod 977\\ &= 757\\ u_2 &= r \cdot w \bmod p\\ &= 7699 \cdot 403 \bmod 977\\ &= 281697 \bmod977\\ &= 321 \end{align}$$ Now, I calculate the checkpoint $$\begin{align} \text{checkpoint}&=u_1\times G + u_2\times(\text{public key})\\ &= 757\times(0, 1) + 321\times(396, 650)\\ &=(707, 48) \end{align}$$ Finally, I need to compare the $x$ coordinate of checkpoint and $r$ value of signature. But $x$ coordinate of checkpoint is $707$ whereas $r$ value was $699$. They are not same. This means that signature is invalid. But they should be equal! Anyone can help where I am wrong?
Use Inclusion-Exclusion to determine the number of integer solutions to the equation $$x_1+x_2+x_3+x_4=14$$ Where $0{\leq}x_1{\leq}8; 0{\leq}x_2{\leq}5; x_3, x_4{\geq}0$. My thought process: I can disregard the last two restrictions because those two are given since the combination gives the number of nonnegative integer solutions. Have A= # solutions given $0{\leq}x_1{\leq}8$ B= # solutions given $0{\leq}x_2{\leq}5$ Which means $A^c$ is the number of solutions for $9{\leq}x_1{\leq}14$. Would the possible solutions for $A^c$ then be partitioning 9+1+1+1+1+1=14? Or rather, give us ${6+4-1\choose 4-1}$? And $B^c$ is the number of solutions for $6{\leq}x_2{\leq}14$, so we would have $6+1+1+1+1+1+1+1+1=14$ or ${9+4-1\choose 4-1}$? If $x_1{\gt8}$ and $x_2{\gt}5$ then for $A^c{\cap}B^c$ we have $$9+6+0+0=15$$ or similar and we see there are no possible solutions for $x_1+x_2+x_3+x_4=14$. Then $$A{\cap}B = A{\cup}B-A^c-B^c+A^c\cap B^c$$ $$={14+4-1\choose 4-1} - {6+4-1\choose 4-1}- {9+4-1\choose 4-1} +0$$ $$=376$$ *I'm a little nervous about my reasoning for $A^c$ and $B^c$. I always end up missing some crucial step in these kinds of problems. - resolved in comments
Single mode Suppose I have two $LC$ oscillators, one with $L_1$ and $C_1$, and the other with $L_2$ and $C_2$. If uncoupled, each oscillator has resonant frequency $\omega \equiv 1/\sqrt{LC}$. Using the flux in the inductor as the coordinate, the equation of motion for each oscillator is $$\ddot{\Phi} = -\omega^2 \Phi .$$ We can rewrite this in Hamiltonian form like this $$ \frac{d}{dt} \left[ \begin{array}{c} \Phi \\ Q \end{array} \right] = \left[ \begin{array}{cc} 0 & 1/C \\ -1/L & 0\end{array} \right] \left[ \begin{array}{c} \Phi \\ Q \end{array} \right]. $$ We can clean this up by defining $X \equiv (C/L)^{1/4} \Phi$ and $Y \equiv (L/C)^{1/4} Q$, which leads to $$ \frac{d}{dt} \left[ \begin{array}{c} X \\ Y \end{array} \right] = \omega^2 \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0\end{array} \right] \left[ \begin{array}{c} X \\ Y \end{array} \right]. $$ Coupled modes Now suppose the oscillators are coupled through an inductor $L_g$. The coupled equations of motion are $$ \left[ \begin{array}{c} \ddot{\Phi}_1 \\ \ddot{\Phi}_2 \end{array} \right] = \left[ \begin{array}{cc} -\omega_1^2(1+L_1/L_g) & \omega_1^2(L_1/L_g) \\ \omega_2^2 (L_2/L_g) & -\omega_2^2(1+L_2/L_g) \end{array} \right] \left[ \begin{array}{c} \Phi_1 \\ \Phi_2 \end{array} \right] $$ Note that the weak coupling limit is that where $L_g \gg L_1,L_2$. From here, one can find the normal frequencies by the usual means of finding eigenvalues of a matrix. One can also work with the Hamiltonian form: $$ \frac{d}{dt} \left[ \begin{array}{c} X_1 \\ Y_1 \\ X_2 \\ Y_2 \end{array} \right] = \left[ \begin{array}{cccc} 0 & \omega_1 & 0 & 0 \\ -\omega_1' & 0 & -g & 0 \\ 0 & 0 & 0 & \omega_2 \\ -g & 0 & -\omega_2' & 0 \end{array} \right] \left[ \begin{array}{c} X_1 \\ Y_1 \\ X_2 \\ Y_2 \end{array} \right] $$ where $\omega_1' \equiv \omega_1 (1-L_1/L_g)$, $\omega_2' \equiv \omega_2 (1-L_2/L_g)$, and $g\equiv (\sqrt{L_1 L_2}/L_g)\sqrt{\omega_1 \omega_2}$. Is there any advantage in terms of intuition for the physics or mathematical elegance/simplicity to extract information from the Hamiltonian matrix as opposed to the one which came from the 2$^{\text{nd}}$ order equations?

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