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# Article
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Summary:
The article deals with the number of rectangles by which every square of $n \times n$ chessboard is covered. Then, the sum of all these numbers is given.
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# The distribution criterion¶
The mathematically critical part of the dbOTU algorithm is deciding whether or not two OTUs are distributed identically.
## Review of previous approaches¶
### $$\chi^2$$ test¶
The original paper and implementations dbOTU1 and dbOTU2 both articulate the distribution criterion in terms of the $$\chi^2$$ test. The idea is that Pearson’s $$\chi^2$$ test of independence evaluates whether the values of two outcomes (i.e., the distributions of the two OTUs) are statistically independent.
The $$\chi^2$$ test runs into two problems. First, when counts are small, the asymptotic (and easy-to-compute) $$\chi^2$$ statistic is not accurate, so the $$p$$-value needs to be simulated empirically, which is computationally expensive. Second, when counts are large, the $$\chi^2$$ test tends to be oversensitive: the sorts of variations that we don’t find unusual in microbiome data are construed as true differences by the statistic.
### Jensen-Shannon divergence¶
The original publication suggests that the distribution criterion can be articulated as a cutoff on the JSD between two the distribution of two OTUs (rather than on the $$p$$-value from the $$\chi^2$$ test).
A JSD cutoff works well for large counts: it accords better with the sorts of differences we would consider meaningful for microbiome data. This avoids the oversensitivity problem that the $$\chi^2$$ test exhibits for large numbers of counts.
However, because the JSD works with proportions and not counts, it tends to be overly sensitive when the counts for one OTU are small. For example, if one OTU has only one count in one sample, the JSD treats that the same as if that OTU has a million counts in only that sample.
### Other approaches¶
Other distribution criteria have been informally evaluated. For example, correlation coefficients (e.g., Kendall $$\tau$$) have similar strengths and weaknesses as the JSD: they perform well when the number of counts is small but tend to be overly sensitive when the counts are small.
## This implementation¶
I found that, for a few example cases, the asymptotic likelihood ratio test, which is quick to compute, gives results similar to the simulated $$\chi^2$$ test. In the few examples I tested, the $$p$$-value for the likelihood ratio test were around two- to five-fold greater than the $$p$$-value from the simulated $$\chi^2$$ test. I consider this a push in the right direction, since the simulated $$\chi^2$$ test tended to be too sensitive.
### Theoretical motivation¶
The $$\chi^2$$ test measures whether two outcomes are independent. In other words, it asks, given that the sequence has some total number of counts, what is the probability that those counts would have been distributed across samples in the same way that the potential parent OTU’s counts were distributed?
This formulation of the question is theoretically unpleasant because this is not how sequence counts are distributed. It is not that a sequence gets some total number of counts and those counts get distributed across samples. Instead, each sample gets a total number of counts (based on its fractional representation in the library and the depth of sequencing), and that sample’s counts get distributed among the sequences in that sample.
The likelihood ratio test evaluates a more natural description of the question of ecologically identical distributions: allowing for some Poisson error, is it more likely that the relative abundance of the candidate sequence in each sample is proportional to the relative abundance of the OTU in each sample (where the constant of proportionality is the same for all samples)?
For one special case, this likelihood ratio test also gives a more sensible answer that the $$\chi^2$$ test. If a sequence and an OTU are present in only one sample, a distribution test should not disqualify them from being merged: the data are perfectly concordant with the model that they are distributed “the same”. The likelihood ratio test gives $$p = 1$$ in this case, but the $$\chi^2$$ test can give really low $$p$$-values, often below the default threshold!
### Formulation of the null and alternative hypotheses¶
First, some definitions. There are $$N$$ samples. In sample $$i$$, the number of reads assigned to the more abundant OTU is $$a_i$$; the candidate sequence has $$b_i$$. Define also $$A = \sum_{i=1}^N a_i$$ and similarly $$B$$.
The alternative hypothesis is that the OTU and sequence are distributed differently, that is, that each of the $$a_i$$ and $$b_i$$ are all drawn from different random variables. Technical replicates from sequence data seem to be well-modeled by Poisson random variables [1], so I formulate this hypothesis as
$H_1: a_i \sim \mathrm{Poisson}(\lambda_{\mathrm{a}i}) \text{ and } b_i \sim \mathrm{Poisson}(\lambda_{\mathrm{b}i}),$
where there are no constraints on the relationships between the Poisson parameters.
The null model asserts that there is some relationship between the sequence and the OTU, specifically that candidate sequence is distributed “the same as” the OTU, just rescaled to some lower abundance. I articulate this as
$H_0: a_i \sim \mathrm{Poisson}(\lambda_i) \text{ and } b_i \sim \mathrm{Poisson}(\sigma \lambda_i),$
that is, that the $$a_i$$ are all free to be distributed differently from one another, but each $$b_i$$ is constrained to be distributed according a Poisson random variable that has the same mean as the random variable corresponding to $$a_i$$, just rescaled by some common scaling factor $$\sigma$$. (Because the sequence is less abundant than the OTU, we expect that $$0 < \sigma < 1$$.)
### Maximum likelihood of models¶
The likelihood ratio test will require the maximum likelihood with respect to those parameters. Not surprisingly, this requirement implies that $$\lambda_{\mathrm{a}i} = a_i$$ and $$\lambda_{\mathrm{b}i} = b_i$$, i.e., that the best estimates for the Poisson parameters are just the single number that distribution produces. For the null hypothesis, we find that $$\sigma = \frac{B}{A}$$, i.e., the scaling factor is just the ratio of the total counts for the sequence and OTU, and
$\lambda_i = \frac{A}{A + B}(a_i + b_i).$
Inserting these variables shows that the log likelihoods are:
\begin{split}\begin{aligned} \mathcal{L}_0 &= -(A + B) + \sum_i \left( a_i \ln a_i + b_i \ln b_i - \ln a_i! - \ln b_i! \right) \\ \mathcal{L}_1 &= -(A + B) + A \ln A + B \ln B - (A + B) \ln (A + B) \\ &\quad + \sum_i \left[ (a_i + b_i) \ln (a_i + b_i) - \ln a_i! - \ln b_i! \right] \end{aligned}\end{split}
The difference between the logs can be conveniently written in terms of a helper function:
$f(\boldsymbol{x}) \equiv \sum_i x_i \ln x_i - \left( \sum_i x_i \right) \ln \left( \sum_i x_i \right)$
so that
$\mathcal{L}_1 - \mathcal{L}_0 = f(\boldsymbol{a} + \boldsymbol{b}) - f(\boldsymbol{a}) - f(\boldsymbol{b}).$
### The statistic¶
The likelihood ratio test uses the statistic $$\Lambda = -2 \left( \mathcal{L}_1 - \mathcal{L}_0 \right)$$, which is distributed according to a $$\chi^2$$ distribution with $$(N - 1)$$ degrees of freedom. (This is the difference in the number of parameters in the two models: the alternative has $$2N$$, i.e., one for the OTU and the sequence in each sample, and the null has $$N + 1$$, one for each sample and the scaling factor $$\sigma$$.) The cumulative distribution function of $$\chi^2$$ at $$\Lambda$$ is easy to compute.
[1] Marioni et al. RNA-seq: An assessment of technical reproducibility and comparison with gene expression assays. Genome Res (2008) doi:10.1101/gr.079558.108. |
# properties of diamagnetic materials
The resultant magnetic momentum in an atom of the diamagnetic material is zero. Moreover, enthusiasts are also catered with the detailed breakdown of the atomic, optical and chemical behaviour of the metals. In diamagnetic materials all the electron are paired so there is no permanent net magnetic moment per atom. (viii) Susceptibility of a diamagnetic material does not depend on the applied magnetic field and temperature. Properties of Diamagnetic Materials – (i) Diamagnetic materials tend to move from a strong intensity to a weaker region in a non-uniform magnetic field. Dipoles are induced only in presence of external … (ii) There are no atomic dipoles in diamagnetic materials because the resultant magnetic moment of each atom is zero due to paired electrons. Diamagnetic properties arise from the realignment of the electron paths under the influence of an external magnetic field. Paramagnetic materials show the following properties. [CONFIRMED] JEE Main will be conducted 4 times from 2021! magnetism: Magnetic properties of matter. 0.00017). All materials are diamagnetic, even if their diamagnetism is hidden by their greater para- or ferromagnetism. Find Free WordPress Themes and plugins. Fig: Field Lines around a bar magnet Diamagnetism persists only in presence of an external magnetic field. If a diamagnetic liquid is placed in a watch glass placed on two pole pieces which are sufficiently apart then liquid accumulates in the middle where field is weakest. in the diamagnetic materials the atoms under zero magnetizing fields do not have net magnetic dipoles. These material repel the applied magnetic field. These material have small and negative magnetic susceptibility. The magnetic flux inside diamagnetic material is zero. 5. Understanding the correlation between magnetic properties and nanostructure involves collaborative efforts between chemists, physicists, and materials scientists to study both fundamental properties and potential applications. Consequently, when a diamagnetic material is placed in a magnetic field, $$B< \mu_oH$$ Click here for the Video tutorials of Magnetic Effect of Current Class 12, Neutral Point of Magnet – Magnetism and Matter Class 12 Physics Notes, JEE Main Previous Year Questions Topicwise. Diamagnetic materials have a very weak negative susceptibility, typically of order 10-6. Diamagnetic materials are slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed. That means there are feebly repelled by the magnetic field. (iii) When diamagnetic material is placed in a magnetic field, the lines of force tend to move away from the material. Properties of diamagnetic substances: (1) When placed in a non-uniform magnetic field, it tends to move from stronger to weaker regions of the magnetic field. Diamagnetic Material – Diamagnetism Magnetic property refers to the response of a material to an applied magnetic field. 12.5: Paramagnetism Diamagnetism makes itself evident in atoms and molecules that have no permanent magnetic moment. this video consists of magnetic property of substances ( paramagnetic , diamagnetic, ferromagnetic ). Diamagnetic materials are repelled by a magnet. Diamagnetic substances are $\mathrm{cu}, \mathrm{zn}, \mathrm{Bi}, \mathrm{Ag},$ Au, Pb, He, $\mathrm{Ar}, \mathrm{NaC} \ell, \mathrm{H}_{2} \mathrm{O},$ marble,glass etc. Parenting a Future Engineer - Webinar for Class 10 to 11 moving engineer's parent by NK Gupta Sir, Properties of Diamagnetic Materials – Magnetism & Matters | Class 12 Physics Notes. These materials exhibit the properties of high permeability, reduced coercive force, they can be magnetized and demagnetized simply and also exhibit low hysteresis. A diamagnetic rod sets itself perpendicular to field because field is strongest at poles. Paramagnetic materials have following properties: In paramagnetic materials, the magnetic lines of forces due to the applied field are attracted towards the paramagnetic material. Consider the simple model of the atom in Figure 1. i.e., B = 0. For this reason, we classify only materials whose net magnetization is diamagnetic, as a diamagnet. Anomalous diamagnetic properties of systems with spontane-ous current. Diamagnetic materials are slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed. Click here for the Video tutorials of Magnetic Effect of Current Class 12 If a diamagnetic liquid is placed in a watch glass placed. e.g. The … 4. Diamagnetic substances are those in which the net magnetic moment of atoms is zero. Diamagnetic materials are always repelled by a magnet. (i) Diamagnetic materials tend to move from a strong intensity to a weaker region in a non-uniform magnetic field. Magnetic susceptibility is small and ne… Diamagnetic materials have a weak, negative susceptibility to magnetic fields. Diamagnetic material does not possess permanent dipoles. when suspended in a uniform magnetic field, diamagnetic materials set their longest axis at right angles to the direction of the field and the shortest axis is along the direction of the field. Susceptibility is independent of temperature. Intensity of magnetisation I is very small, negative and proportional to magnetising field. Two experiments were performed to investigate the thermal and electromagnetic (EM) properties of unidirectional carbon fiber/polyether ether ketone (PEEK) composite materials. Most elements in the periodic table, including copper, silver, and gold, are diamagnetic. In diamagnetic materials all the electron are paired so there is no permanent net magnetic moment per atom. As a result, a number of lines of force passing through the material become less [Figure]. There are no atomic dipoles in diamagnetic materials because the resultant magnetic moment of each atom is zero due to paired electrons. on two pole pieces which are quite close to each other then liquid accumulates at sides and shows depression in the middle, where field is strongest. Examples: Bi, Sb, Cu, Au, Hg, H. The atoms do not have any permanent dipole moment i.e. 3. Examples – copper, silver, zinc, bismuth, lead, glass, marble, helium, water, argon, sodium chloride, etc. Diamagnetism is exhibited by solids, liquids and gases. paramagnetic or diamagnetic, respectively. A dielectric material is any material that supports charge without conducting it to a significant degree. Most elements in … The intensity of magnetization I is very small, negative, and proportional to the magnetizing field. However, other forms of magnetism (such as ferromagnetism or paramagnetism) are so much stronger that, when multiple different forms of magnetism are present in a material, the diamagnetic contribution is usually negligible. When an external magnetic field is applied, dipoles are induced in the diamagnetic materials in such a way that induced dipoles opposes the external magnetic field according to Lenz’s law. It can be said that the materials which acquire a small amount of magnetism towards the magnetic field when they are placed in a magnetic field are called paramagnetic material. Magnetic dipole moment (M) is small and opposite to magnetising field H. Diamagnetic substances do not obey Curie’s law and show no transition. Curie temperature is the characteristic property of the substance. The substances are weakly repelled by the field so in a nonuniform field, these substances have a tendency to move from a strong to a weak part of the external magnetic field. (i) Properties of diamagnetic substances Diamagnetic substances are those in which the net magnetic moment of atoms is zero. Diamagnetic Metals + properties give you a broad overview of these metals from multiple angels. Fe (1043 K), Ni (631 K), Co (1394 K), Gadolinium (317 K), Fe2O3 (893 K) Scientific Reasons: Diamagnetic substances are weakly repelled by a magnet. All rights reserved. Classification of magnetic material • Diamagnetic: Net magnetic moment is zero due to the alignment of magnetic moment in the opposite direction H • Paramagnetic: Very little magnetic susceptibility 5. Atomic Structure Revision Video – Class 11, JEE, NEET. Only when a magnetizing field is applied, at an atomic level the magnetic moment is produced. Therefore, they are slightly attracted by … That is to say, the relative permeability is slightly less than 1. The substances are repelled weakly by the field, and so in a nonuniform field, they tend to travel from a strong to a weak part of the external magnetic field. This electron motion is a small electric current, and anywhere there is a current, there is a magnetic field (moment). The intensity of magnetization I is very small, negative and proportional to the magnetizing field. Diamagnetic Materials Diamagnetism is a fundamental property of all matter. (iv) Magnetic induction B of a diamagnetic material is slightly less than the magnetic field intensity H. (v) The substances are weakly repelled by the field so in a nonuniform field, these substances have a tendency to move from a strong to a weak part of the external magnetic field. Electrons in an atom revolve around the nucleus thus possess orbital angular momentum. (For example, for bismuth χm=? Even they are utilized for magnetic screening. Materials which lack permanent magnetic dipoles are called diamagnetic. Therefore, diamagnetic substances are repelled by magnets. A few materials, notably iron, show a very large attraction toward the pole of a permanent bar magnet; materials of this kind are called ferromagnetic.… spectroscopy: Fluorescence and phosphorescence …moment (such species are called diamagnetic). The volume susceptibility χm for diamagnetic solid materials is in the order of −10 −5. The diamagnetic response of a material has a measurable contribution to the materials' magnetization only if there are no other magnetic effects present, such as Ferrimagnetism whose susceptibility is much larger in most cases [1]. The relative permeability values of diamagnetic materials are <1, and their magnetic susceptibility is negative. Electrons move around the nucleus like the earth around the sun. Diamagnetic materials are tho… Magnetic susceptibility $\chi_{\mathrm{m}}=\frac{\mathrm{I}}{\mathrm{H}}$ is small and negative $\left(\approx 10^{-5}\right)$. A few of the examples of these are iron, nickel, aluminum, tungsten, and cobalt. The origin of diamagnetism is the induced dipole moment due to change in orbital motion of electrons in atoms by applied field. materials that exhibit the property of electrical polarization, thereby they modify the dielectric function of the vacuum. at Curie temperature $X_{m}$ is indepedent of temperature. The materials which have net magnetic moments i.e., those materials which reveal para and ferromagnetism, the diamagnetism in those materials becomes overshadowed due to its weak value. Substances where the diamagnetic behaviour is the strongest effect are termed diamagnetic materials, or diamagnets. In principle all insulators are dielectric, although the capacity to support charge varies greatly between different insulators. Diamagnetic susceptibility is a temperature independent quantity (Fig. 7.1). The relative permeability $\mu_{\mathrm{r}}=\frac{\mu}{\mu_{0}}$ is slightly less than unity. In diamagnetic materials, there are no atomic dipoles due to the pairing between the electrons. Diamagnetic substances are $\mathrm{cu}, \mathrm{zn}, \mathrm{Bi}, \mathrm{Ag},$ Au, Pb, He, $\mathrm{Ar}, \mathrm{NaC} \ell, \mathrm{H}_{2} \mathrm{O},$ marble,glass etc. The substances which when placed in a magnetising field get feebly magnetised in a direction opposite to magnetising field are called diamagnetic. General properties of Diamagnetic Material. Weak diamagnetic materials have magnetic susceptibility values close to zero; their molar magnetic susceptibility is of the order of χmol = − 10 × 10 − 9 m 3 mol − 1. Supplementary facts like side effects & benefits of these metals, their abundance in earth's crust, their presence in the human body, etc. … When suspended in a uniform magnetic field, paramagnetic materials rotate so as to bring their longest axis along the direction of the magnetic field and shorter axis perpendicular to the field. The process of easily magnetizing and demagnetizing let these materials to be implemented in the applications of generators, telephone receivers, electromagnets, transformers, inductors, relays and in many others. Properties of Diamagnetic Materials. Properties of Dimagnetic Substances. Some examples of diamagnetic materials include; Quartz (silicon dioxide) Calcite (calcium carbonate) Water; For example in quartz, there are silicone atoms and oxygen atoms in the form of SiO 2.The oxidation state of Si atom is +4, and the oxidation state of O atom is -2. The field inside the material $\mathrm{B}$ is less than magnetising field H. They have a tendency to expel lines of force. This article introduces a classification of nanostructure morphology according to the mechanism responsible for the magnetic properties. Properties of Paramagnetic Materials. (vi) The value of magnetic susceptibility is very small. When a diamagnetic substance is placed in a watch glass on the pole pieces of a magnet the liquid … 2. Diamagnetic substances when placed in an external magnetic field produce negative magnetization. The magnetic moment of every atom of diamagnetic substance is zero. paired spin. Download India's Best Exam Preparation App. A few and important properties of Diamagnetic Materials are listed below. The susceptibility has a low negative value. Comparison of Magnetic Moments of Two Bar Magnets in Deflection Method, Comparison of Magnetic Moments of Two Bar Magnets in Null Deflection Method, Crab armies can be a key issue in coral wall preservation, Beaches cannot be extinct if sea levels continue to rise, Autonomous “Smellicopter” Drone Can Seek Out Scents with Live Moth Antennae, Scientists are finally studying why some of you don’t overturn your regulator, The vast wetlands of Els Eels are the most recorded at the bottom of the ocean. The relative permeability is slightly less than one. So, Diamagnetism is a quantum mechanical effect that occurs in all materials; when it is the only contribution to the magnetism, the material is called diamagnetic. When suspended freely in a uniform magnetic field, they set themselves perpendicular to the direction of the magnetic field. Diamagnetic properties arise from the realignment of the electron paths under the influence of an external magnetic field. In short, it can be said that that material which, when placed in a magnetic field acquires a small amount of magnetism opposite to the direction of the inducing magnetic field are called diamagnetic materials. (vii) Diamagnetic materials are slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed. © copyright 2020 QS Study. Want create site? Diamagnetism is a property of all materials, and always makes a weak contribution to the material's response to a magnetic field. Properties of Diamagnetic materials. When diamagnetic material is placed within a magnetic field the lines of force tend to go away from the material. Therefore, there are no unpaired electrons in both these atoms. A diamagnetic liquid in a U-tube depresses in the limb which is between the poles of magnet. When placed in a … The possible existenc of materiale s tha diffet r from semiconductor ans d have high diamagnetic susceptibi-lities is of considerable interest. When a diamagnetic substance is placed in a magnetic field it sets itself at right angles to the direction of the lines of force. Properties of Magnetic Materials. Save my name, email, and website in this browser for the next time I comment. Paramagnetic substances acquire a small net magnetic moment in the direction of the applied field. Magnetic Field: The magnetic field is an imaginary line of force around a magnet which enables other ferromagnetic materials to get repelled or attracted towards it.The magnetic field lines are formed due to various reasons like orbital movement of electrons, current flowing in a conductor etc. 1. The susceptibility has a low negative value. It is different for different materials. It is induced by a change in the orbital motion of electrons due to an applied magnetic field. Diamagnetism is a very weak form of magnetism. Diamagnetic materials have a very weak negative susceptibility. The macroscopic magnetic properties of a material are a consequence of interactions between an external magnetic field and the magnetic dipole … Diamagnetic materials have following properties: In a diamagnetic material, the magnetic lines of forces due to an applied field are repelled. • Ferro-Magnets: Retains magnetism even when external field is removed because of the parallel alignment of the electron moment. 1. Electrons also spin around their axes like the earth. The substances are weakly repelled by the field so in a non uniform field these have a tendency to move from strong to weak field. [ CONFIRMED ] JEE Main will be conducted 4 times from 2021 m. Without conducting it to a weaker region in a non-uniform magnetic field produce negative magnetization significant degree force. 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# simint
0th
Percentile
##### Simultaneous Confidence Intervals
Computes simultaneous intervals for several multiple procedures.
Keywords
htest
##### Usage
## S3 method for class 'default':
simint(y, x=NULL, type=c("Dunnett", "Tukey",
"Sequen", "AVE", "Changepoint", "Williams", "Marcus",
alternative=c("two.sided","less", "greater"),
asympt=FALSE, eps=0.001, maxpts=1e+06, nlevel=NULL,
nzerocol=c(0,0),...)
## S3 method for class 'formula':
simint(formula, data=list(), subset, na.action, whichf, ...)
##### Arguments
y
a numeric vector of responses.
x
a numeric matrix, the design matrix.
type
the type of contrast to be used.
cmatrix
the contrast matrix itself can be specified. If cmatrix is defined, type is ignored.
conf.level
confidence level.
alternative
the alternative hypothesis must be one of "two.sided" (default), "greater" or "less". You can specify just the initial letter.
asympt
a logical indicating whether the (exact) t-distribution or the normal approximation should be used.
eps
absolute error tolerance as double.
maxpts
maximum number of function values as integer.
nlevel
a vector containing the number of levels for each factor for type == "Tetrade".
nzerocol
a vector of two elements defining the number of zero-columns to add to the contrast matrix from left (the first element, usually 1 for the intercept) and right (usually 0 if no covariables are specified). nzerocol is automatically determined
formula
a symbolic description of the model to be fit.
data
an optional data frame containing the variables in the model. By default the variables are taken from Environment(formula), typically the environment from which simint is called.
subset
an optional vector specifying a subset of observations to be used.
na.action
a function which indicates what should happen when the data contain NA's. Defaults to GetOption("na.action").
whichf
if more than one factor is given in the right hand side of formula, whichf can be used to defined the factor to compute contrasts of.
...
further arguments to be passed to or from methods.
##### Details
Computes simultaneous confidence intervals for several multiple comparisons. The implemented algorithms take the stochastical correlations between the test statistics into account. Only single step comparisons are performed. The present function allows for multiple comparisons of generally correlated means in general linear models under the classical ANOVA assumptions, as well as more general approximate procedures for approximately normal and generally correlated parameter estimates. Either multivariate normal or multivariate t statistics can be used. The interface allows the use of the multiple comparison procedures as for example Dunnett and Tukey. The resulting confidence intervals are not associated with the p-values from simtest.
See simtest for detailed information on the formula interface.
##### Value
• an object of class hmtest
##### References
Frank Bretz, Alan Genz and Ludwig A. Hothorn (2001), On the numerical availability of multiple comparison procedures. Biometrical Journal, 43(5), 645--656.
TukeyHSD for the special case of Tukey contrasts.
##### Aliases
• simint
• simint.default
• simint.formula
##### Examples
data(recovery)
# one-sided simultaneous confidence intervals for Dunnett
# in the one-way layout
summary(simint(minutes~blanket, data=recovery, type="Dunnett",
conf.level=0.9, alternative="less",eps=0.0001))
# Tukey confidence intervals, compare with TukeyHSD
data(warpbreaks)
fm1 <- aov(breaks ~ wool + tension, data = warpbreaks)
tHSD <- TukeyHSD(fm1, "tension", ordered = FALSE)
print(tHSD)
mcHSD <- simint(breaks ~ wool + tension, data = warpbreaks,
whichf="tension", type="Tukey")
print(mcHSD)
plot(mcHSD)
<testonly>tlow <- tHSD$tension[,2] tupp <- tHSD$tension[,3]
mclow <- mcHSD$conf.int[,1] mcupp <- mcHSD$conf.int[,2]
a <- round(tlow, 1)
b <- round(mclow, 1)
names(a) <- NULL
names(b) <- NULL
stopifnot(all.equal(a, b))
a <- round(tupp, 1)
b <- round(mcupp, 1)
names(a) <- NULL
names(b) <- NULL
stopifnot(all.equal(a, b))</testonly>
Documentation reproduced from package multcomp, version 0.3-5, License: GPL
### Community examples
Looks like there are no examples yet. |
## Programmatic adventures in the hyperbolic plane
Recursion is dangerous. So dangerous that in my professional life I’m explicitly banned from using it in code. After all who would want their aeroplane to stall, their car to crash or their fridge to defrost because of a stack overflow.
But it is also, most (but not all) programmers would agree, beautiful and more than a bit mysterious. Indeed one of the biggest selling popular science/mathematics books of all time centres on recursion as a possible partial explanation of human consciousness.
In the last week or so I’ve been attempting to use recursion in R to draw a tessellating pattern of squares on the conformal representation of the hyperbolic plane – an idea from M.C. Escher via Nobel laureate Sir Roger Penrose.
The code here is where I’ve got to.
#!/usr/bin/env Rscript
library(ggplot2)
library(ggforce)
basicLengths<-seq(from<-0, to<-1.0, by<-1/250000)
projLengths<-basicLengths * (2 /(1 + basicLengths * basicLengths))
getBestMatch<-function(numIn)
{
repVec<-basicLengths[which.min(abs(numIn - projLengths))]
return(repVec)
}
drawConformalLine <-function(startX, startY, endX, endY)
{
#Length and angle of line to starting conformal point
lengthToStart <- sqrt(startX * startX + startY * startY)
angleToStart <- atan2(startY, startX)
#And to end Point
lengthToEnd <- sqrt(endX * endX + endY * endY)
angleToEnd <- atan2(endY, endX)
#Project
newStartLength<-lengthToStart * (2 / (1 + lengthToStart * lengthToStart))
newEndLength<-lengthToEnd * (2 / (1 + lengthToEnd * lengthToEnd))
newStartX<-newStartLength * cos(angleToStart)
newStartY<-newStartLength * sin(angleToStart)
newEndX<-newEndLength * cos(angleToEnd)
newEndY<-newEndLength * sin(angleToEnd)
xpoints<-seq(from<-newStartX, to<-newEndX, by<-(newEndX - newStartX)/50)
ypoints<-seq(from<-newStartY, to<-newEndY, by<-(newEndY - newStartY)/50)
euclidLine<-data.frame(xpoints, ypoints)
#Conformal line (reverse projection)
lengths<-sqrt(euclidLine$xpoints * euclidLine$xpoints + euclidLine$ypoints * euclidLine$ypoints)
newLengths<-vector()
for (i in 1:length(lengths))
{
newLengths<-c(newLengths, getBestMatch(lengths[i]))
}
angles<-atan2(euclidLine$ypoints, euclidLine$xpoints)
newPoints<-data.frame(xp = cos(angles) * newLengths, yp = sin(angles) * newLengths)
return(newPoints)
}
{
if (base + radius > 0.75){
return (setOfPoints)
}
positiveCorner = (sign * base + sign * radius)
negativeCorner = (sign * base)
setOfPoints<-rbind(setOfPoints, drawConformalLine(positiveCorner, positiveCorner, positiveCorner, negativeCorner))
setOfPoints<-rbind(setOfPoints, drawConformalLine(positiveCorner, negativeCorner, negativeCorner, negativeCorner))
setOfPoints<-rbind(setOfPoints, drawConformalLine(negativeCorner, negativeCorner, negativeCorner, positiveCorner))
setOfPoints<-rbind(setOfPoints, drawConformalLine(negativeCorner, positiveCorner, positiveCorner, positiveCorner))
return (setOfPoints)
}
partitions<-100
distances<-seq(from=1, to=partitions, by=1)
distances<-exp(distances/partitions)/(2 * distances)
circles<-data.frame(x0=rep(0, partitions), y0=rep(0, partitions), r=distances)
t <- ggplot() + geom_circle(aes(x0=x0, y0=y0, r=1-r, color=rgb(r,1-r,1)), data=circles)
t <- t + geom_circle(aes(x0=0, y0=0, r=1), color="black", data=data.frame(x0=0, y0=0, r=1)) + coord_fixed()
distances<- 1-distances
n_distances<-(distances * 2)/(1 + distances * distances)
n_circles<-data.frame(x0=rep(0, partitions), y0=rep(0, partitions), r=n_distances)
y <- ggplot() + geom_circle(aes(x0=x0, y0=y0, r=r, color=rgb(r,1-r,1)), data=n_circles)
y <- y + geom_circle(aes(x0=0, y0=0, r=1), color="black", data=data.frame(x0=0, y0=0, r=1)) + coord_fixed()
square<-0.4
squarePoints<-data.frame()
squarePoints<-recursiveSquares(squarePoints, 0, square, 1)
nPoints<-data.frame()
nPoints<-recursiveSquares(nPoints, 0, square, -1)
squarePoints<-rbind(squarePoints, nPoints)
t<-t + geom_point(aes(color="magenta", x=squarePoints$xp, y=squarePoints$yp))
t<-t + geom_point(aes(color="magenta", x = -squarePoints$xp, y=squarePoints$yp))
I had several problems to solve, and recursion seemed a good fit for some of them, though in the end I’ve only been able to make partial progress.
Here are some of the problems and how I tackled them:
1. Finding the straight lines
In the conformal representation of the hyperbolic plane ‘straight lines’ – in the sense of the shortest routes between two points – are actually represented by the segment of the circle joins the two points and that meets the boundary of the representation of the plane orthogonally (ie., at right angles).
For points that lie on the same polar angle in the circle that’s easy – the ‘straight line’ that joins them is simply the Euclidean line that passes through both of them and the origin (centre of the circular representation of the plane). The circle they are linked by in the conformal representation is of infinite radius.
But what about other, non-degenerate, cases? Perhaps there is an algorithmic method to work this out? But if there is I don’t know what it is and I couldn’t find it online either!
What I do know (again thanks to Sir Roger) is that is we substitute the projective representation for the conformal representation we can join the points by Euclidean lines. To go from the conformal to the projective representation of the hyperbolic plane by preserving the polar angle of points but magnifying the distance from the origin by $\frac{2R^2}{R^2+r^2_c}$, where $R$ is the radius of the bounding circle (so $R=1$ in our case)and $r_c$ is the length of the of the Euclidean line from the centre (origin) to the point.
The great advantage of the projective representation is that straight lines between points are just what we think of them as being in Euclidean geometry – in other words both the shortest length between two points and at a constant angle.
So to get the points on the line in the conformal representation the thing to do seemed to be to convert the points to the projective representation, calculate the straight line (very easy) and shrink everything back by the factor $\frac{1+r^2_c}{2}$.
But the problem with that is, that with the exception of the start and end points, we don’t know what $r^2_c$ is to begin with – indeed that is more or less what we are trying to calculate.
My answer to that is to generate lookup tables (relatively quickly done using R’s vectorisation) and a function that indexes one to the other:
basicLengths<-seq(from<-0, to<-1.0, by<-1/250000)
projLengths<-basicLengths * (2 /(1 + basicLengths * basicLengths))
getBestMatch<-function(numIn)
{
repVec<-basicLengths[which.min(abs(numIn - projLengths))]
return(repVec)
}
2. Generate the squares
This is where the recursion comes in – I settled on the idea of four squares touching at each vertex, meaning each square as we travelled out from the centre the Euclidean lengths halve.
Here is (a simplified) version of the code:
recursiveSquares<-function(setOfPoints, base, radius)
{
if (base + radius > 0.75){
return (setOfPoints)
}
negativeCorner = (base)
setOfPoints<-rbind(setOfPoints, drawConformalLine(positiveCorner, positiveCorner, positiveCorner, negativeCorner))
setOfPoints<-rbind(setOfPoints, drawConformalLine(positiveCorner, negativeCorner, negativeCorner, negativeCorner))
setOfPoints<-rbind(setOfPoints, drawConformalLine(negativeCorner, negativeCorner, negativeCorner, positiveCorner))
setOfPoints<-rbind(setOfPoints, drawConformalLine(negativeCorner, positiveCorner, positiveCorner, positiveCorner))
return (setOfPoints)
}
Those of you who have read the classic Structure and Interpretation of Computer Programs may immediately recognise a weakness here – by not using tail recursion I am putting more stress on the stack.
And it’s true that an alternative tail recursive form (this is the full code) works:
recursiveSquares<-function(setOfPoints, base, radius, sign)
{
if (base + radius > 0.75){
return (setOfPoints)
}
positiveCorner = (sign * base + sign * radius)
negativeCorner = (sign * base)
setOfPoints<-rbind(setOfPoints, drawConformalLine(positiveCorner, positiveCorner, positiveCorner, negativeCorner))
setOfPoints<-rbind(setOfPoints, drawConformalLine(positiveCorner, negativeCorner, negativeCorner, negativeCorner))
setOfPoints<-rbind(setOfPoints, drawConformalLine(negativeCorner, negativeCorner, negativeCorner, positiveCorner))
setOfPoints<-rbind(setOfPoints, drawConformalLine(negativeCorner, positiveCorner, positiveCorner, positiveCorner))
}
But it still is liable to bust the stack – indeed the whole system quite finely balanced. Make the initial squares too small and it is easy to break the stack – though the tail recursive form does seem a little more robust.
3. Getting a tessellation/exploiting the dihedral symmetry
Unfortunately the tendency of the stack to blow up makes it very difficult (as far as I can see anyway) to write some elegant code that will seamlessly deliver the tessellations sought. But some simple group theory can help us get some more patterns.
The pattern above shows rotational and reflective symmetry – in other words the shape has the symmetries of square (dihedral group $D_4$).
We use this in the code through the <code>sign</code> factor in the calls to <code>recursiveSquares</code> and then in the two lines:
t<-t + geom_point(aes(color="magenta", x=squarePoints$xp, y=squarePoints$yp))
t<-t + geom_point(aes(color="magenta", x = -squarePoints$xp, y=squarePoints$yp))
These effectively reflect earlier results.
## Who has been given the cash changing problem?
Some class, somewhere, has obviously been given the recursive money changing problem as a piece of work, because I have had several hundred visits in the last week from people seeking to get a grip on it.
## The recursive cash changing algorithm
This is another one from Structure and Interpretation of Computer Programs which is a great book, but not a particularly easy read (not least because Scheme, the language in which the programs in the book is written is just different from C/Java etc). This example annoyed me for some time so I am writing down an explanation in case someone else wants to look it up…
Anyway this one is about the number of ways of to change cash. The algorithm is based on the fact that you can break down the number of ways to sort change of the sum n with t types of coin as follows:
• All ways to sort n with t-1 coins; plus
• All ways to sort n-d with t coins, where d is the denomination of the coin excluded in the above case.
(Think about it for five pence with two pence and one penny coins. One way to sort with pennies and two ways to change three pence with a two pences and pennies.)
To solve this recursively we have to define the degenerative cases – ie the points at which the recursion stops. These are:
• If the sum to be cashed to change is 0 then there is 1 way to do it;
• If the sum to be cashed to change is less than 0 then there are 0 ways to change it;
• If there are 0 coins then there are 0 ways to offer the change.
So how does it work … hopefully the graphic below shows how (changing 10 pence with fives, twos and ones):
The idea is that f(n, {c}) returns the number of ways n can be changed with the set of coins {c}. In this tree the number of leaf nodes (10) is the answer. I have cheated here – the turquoise leaves are not degenerate cases – each has to be worked down to a case where the one would apply if the algorithm is to be universal, though we can simply add an extra rule when the set of coins being used consists of only 1p coins – that is 1 way of sorting. (But if the universal rule was applied then the number of computing steps increases greatly – as each of those 1p cases has to be worked down to a degenerate case).
Update (27 March 2011): There were actually a couple of mistakes in how the terminating cases were shown in the graphic, fixed now I hope. |
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# Estimating Population Parameters from Catches Large Relative to the Population
G. A. F. Seber and E. D. Le Cren
Journal of Animal Ecology
Vol. 36, No. 3 (Oct., 1967), pp. 631-643
DOI: 10.2307/2818
Stable URL: http://www.jstor.org/stable/2818
Page Count: 13
Preview not available
## Abstract
Where two successive catches, c1 and c2 are taken with the same effort from a population, an estimate of the size of the population, ñ, is given by $\tilde{n}=c_{1}{}^{2}/(c_{1}-c_{2})$, with a variance ${\rm var}\ [\tilde{n}]=[c_{1}{}^{2}c_{2}{}^{2}(c_{1}+c_{2})]/(c_{1}-c_{2})^{4}$ Estimates from more than two successive catches, from single catches and of mortality rates are discussed and formulae given. Simple mark-recapture experiments can be combined with repeated catch experiments.
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12.09.2018
# Geodesic flows: notes 1
## Geodesic flows, notes 1: symplectic geometry
Our first topic is to understand geodesics, first defined as curves $$\gamma(t)$$, in terms of a Hamilton flow in the cotangent bundle. Hamilton flows belong to the realm of symplectic geometry. This is an area of mathematics which comes up in a variety of contexts, such as:
1. Historically, symplectic geometry arose from the study of classical mechanics. Consider the $$N$$-body problem, where $$N$$ celestial objects (planets, stars etc) interact with each other gravitationally. If the objects have mass $$m_i$$, position $$q_i(t) \in \mathbb{R}^3$$, and momentum (or velocity) $$p_i(t) = \dot{q}_i(t) \in \mathbb{R}^3$$, by Newton's laws one has the equations $$m_i \ddot{q}_i(t) = \sum_{j \neq i} \frac{G m_i m_j}{|q_i - q_j|^2}$$ where $$G$$ is the gravitational constant. These can be rewritten as the Hamilton equations \begin{align*} \dot{q}(t) &= \nabla_p H(q(t),p(t)), \\ \dot{p}(t) &= -\nabla_q H(q(t),p(t)) \end{align*} where $$H$$ is the Hamilton function $$H = T + U$$, with $$T$$ the kinetic energy and $$U$$ the self-potential energy (source: Wikipedia). The underlying symplectic structure makes it possible to consider these equations in many different (symplectic) coordinate systems, while still retaining the basic properties.
2. The geodesic flow on a Riemannian manifold $$(M,g)$$ can be understood as a Hamilton flow on the cotangent space $$T^* M$$. This explains the facts that geodesics have constant speed, that geodesic flow preserves volume, and that the geodesic vector field on $$SM$$ is divergence free and formally skew-adjoint.
3. If $$A(x,D)$$ is a linear differential operator in an open set $$\Omega \subset \mathbb{R}^n$$, so $$A(x,D) u = \sum_{|\alpha| \leq m} a_{\alpha}(x) D^{\alpha} u$$ where $$a_{\alpha} \in C^{\infty}(\Omega)$$, the principal symbol of $$A(x,D)$$ is
\begin{align*} a(x,\xi) = \sum_{|\alpha| = m} a_{\alpha}(x) \xi^{\alpha}, \qquad (x,\xi) \in \Omega \times \mathbb{R}^n. \end{align*}
It turns out that under a change of coordinates, the principal symbol of $$A$$ transforms like an invariant function on the cotangent bundle $$T^* \Omega$$. Moreover, if $$A$$ and $$B$$ are two differential operators, their commutator $$[A,B] = AB - BA$$ has principal symbol $$\{a,b\}$$ where $$\{\,\cdot\,, \,\cdot\,\}$$ is the Poisson bracket on $$T^* \Omega$$: \begin{align*} \{ a, b \}(x,\xi) = \nabla_{\xi} a \cdot \nabla_x b - \nabla_x a \cdot \nabla_{\xi} b. \end{align*} These basic facts indicate that symplectic geometry has a fundamental role in the study of linear partial differential equations.
4. The billiard map in the context of the broken geodesic ray transform preserves the related symplectic structure and hence volumes.
### Symplectic manifolds
We proceed to explain some basic ideas in the setting of a general $$2n$$-dimensional symplectic manifold.
Definition. A symplectic manifold is a pair $$(N, \sigma)$$ where $$N$$ is an $$2n$$-dimensional smooth manifold, and $$\sigma$$ is symplectic form, that is, a closed $$2$$-form on $$N$$ which is nondegenerate in the sense that for any $$\rho \in N$$, the map $$I_{\rho}: T_{\rho} N \to T^*_{\rho} N, I_{\rho}(s) = \sigma(s, \,\cdot\,)$$ is bijective.
Example 1. The space $$\mathbb{R}^{2n}$$ has a standard symplectic structure given by the $$2$$-form $$\sigma = dx_1 \wedge \,dx_{n+1} + \ldots + dx_n \wedge \,dx_{2n}$$.
Example 2. More generally, if $$M$$ is an $$n$$-dimensional $$C^{\infty}$$ manifold, then $$N = T^* M$$ becomes a symplectic manifold as follows: if $$\pi: T^* M \to M$$ is the natural projection, there is a $$1$$-form $$\lambda$$ on $$N$$ (called the Liouville form) defined by $$\lambda_{\rho} = \pi^* \rho, \rho \in T^* M$$. Then $$\sigma = d\lambda$$ is a closed $$2$$-form. If $$x$$ are local coordinates on $$M$$, and if $$(x,\xi)$$ are associated local coordinates (called canonical coordinates) on $$T^* M$$, then in these local coordinates \begin{align*} \lambda &= \xi_j \,dx^j, \\ \sigma &= d\xi_j \wedge \,dx^j. \end{align*} It follows that $$\sigma$$ is nondegenerate and hence a symplectic form.
It is a theorem of Darboux that if $$(N,\sigma)$$ is a symplectic manifold, then near any point of $$N$$ there are local coordinates $$(x,\xi)$$ so that $$\sigma = d\xi_j \wedge \,dx^j$$ in these coordinates. Hence every symplectic manifold is locally symplectically equivalent to $$\mathbb{R}^{2n}$$, and all obstructions to having a symplectic structure are global in nature: for instance if $$N$$ is a closed manifold having a symplectic form $$\sigma$$, then the de Rham cohomology groups $$H^2(N)$$ and $$H^{2n}(N)$$ are nontrivial (this follows since the $$n$$-fold wedge product of $$\sigma$$ is non vanishing, more about this later).
### Hamilton flows
Definition. Let $$(N,\sigma)$$ be a symplectic manifold. Given any function $$f \in C^{\infty}(N)$$, the Hamilton vector field of $$f$$ is the vector field $$H_f$$ on $$N$$ defined by $H_f = I^{-1}(df)$ where $$df$$ is the exterior derivative of $$f$$ (a $$1$$-form on $$N$$), and $$I$$ is the isomorphism $$TN \to T^* N$$ given by the nondegenerate $$2$$-form $$\sigma$$.
Example 1. Let $$M$$ be an $$n$$-manifold, and let $$\sigma$$ be the standard symplectic form on $$T^* M$$. If $$(x,\xi)$$ are canonical local coordinates, then (using that $$\alpha \wedge \beta(v, w) = \alpha(v) \beta(w) - \alpha(w) \beta(v)$$ for $$1$$-forms $$\alpha, \beta$$) $I_{x,\xi}(s^j \partial_{x_j} + t^j \partial_{\xi_j})(\tilde{s}^j \partial_{x_j} + \tilde{t}^j \partial_{\xi_j}) = (d\xi_i \wedge \,dx^i)(s^j \partial_{x_j} + t^j \partial_{\xi_j}, \tilde{s}^k \partial_{x_k} + \tilde{t}^k \partial_{\xi_k}) = \sum_{i=1}^n (t^i \tilde{s}^i - s^i \tilde{t}^i)$ which gives that \begin{align*} I(s^j \partial_{x_j} + t^j \partial_{\xi_j}) &= t^j \,dx_j - s^j \,d\xi_j \\ H_f &= \partial_{\xi_j} f \partial_{x_j} - \partial_{x_j} f \partial_{\xi_j}. \end{align*}
Example 2. In particular, in $$\mathbb{R}^{2n}$$ one has $$I(s,t)=(t,-s)$$, $$s, t \in \mathbb{R}^n$$, and $H_f = \nabla_{\xi} f \cdot \nabla_x - \nabla_x f \cdot \nabla_{\xi}.$
Definition. Let $$(N,\sigma)$$ be a symplectic manifold, and let $$f \in C^{\infty}(N)$$. Denote by $$\varphi_t$$ the flow on $$N$$ induced by $$H_f$$, that is, $\varphi_t: \rho(0) \mapsto \rho(t) \text{ where } \dot{\rho}(t) = H_f(\rho(t)).$
We will later see that any Hamilton flow map is symplectic ($$(\varphi_t)^* \sigma = \sigma$$) and consequently volume-preserving. |
Nuclear and Particle Physics
# Dispersive Analysis of Mesonic 3-Body Decays
## by Tobias Isken (U. Bonn)
Europe/Stockholm
Description
Abstract:
I will present a dispersive analysis of mesonic 3-body decay amplitudes based on the fundamental principles of analyticity and unitarity. In the described dispersion-theoretical framework the leading final-state interactions are fully taken into account. The formalism relies only on the respective 2-body scattering phase shifts.
The first part of the talk will focus on hadronic $\eta^\prime$ decays. The $eta^\prime\to\eta\pi\pi$ decay offers several features of interest [arXiv:1705.04339 [hep-ph]]: due to final-state interactions it can be used to constrain $\eta\pi$ scattering. It shows a cusp effect in the neutral channel ($\eta^\prime\to\eta\pi^{0}\pi^{0}$) within the physical decay region. The decay is also an essential input for a study of inelastic effects in the decay $\eta^\prime\to3\pi$.
The second part of the talk is focused on the $\omega\to3\pi$ decay. I will demonstrate how the dispersive framework can be used to investigate the quark mass dependence for a 3-body resonance [arXiv:1808.08957 [hep-ph]]. This may be used as an extrapolation tool for lattice QCD calculations. Here the $\omega$ serves as paradigm case, the framework can be generalised to other 3-body decay processes.
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### Session S26: Focus Session: Electron & Ion Solvation in Clusters & the Condensed Phase II
2:30 PM–5:18 PM, Wednesday, March 7, 2007
Chair: J. Matthias Weber, University of Colorado at Boulder
Abstract ID: BAPS.2007.MAR.S26.2
### Abstract: S26.00002 : Theoretical Studies of Negatively Charged Water Clusters: The Role of Polarization and Dispersion for Electron Binding
3:06 PM–3:42 PM
Preview Abstract MathJax On | Off Abstract
#### Author:
Kenneth D. Jordan
(University of Pittsburgh)
A quantum Drude oscillator model is used to characterize negatively charged water clusters as large as (H$_{2}$O)$_{24}^{-}$. The Drude model allows for inclusion of electron correlation effects between the excess electron and the electrons of the water molecules, at a fraction of the computational cost of all-electron \textit{ab initio} methods. Application of the Drude model to (H$_{2}$O)$_{6}^{-}$ demonstrates that there are many isomers with small electron binding energies that are more stable than the species with double acceptor water monomers that dominate under experimental conditions and that have electron binding energies near 0.45 eV. The talk will also explore the connection between the Drude model and more traditional polarization models used in describing the interaction of excess electrons with water. We show that a series of polarization'' models can be derived from the Drude model, by carrying out an adiabatic separation between the excess electron and the Drude degrees of freedom. It is found that the polarization and Drude models give similar electron binding energies for species in which the excess electron experiences large electrostatic attraction, but that the polarization models significantly overbind the excess electron in cases where the electrostatics play only a small role.
To cite this abstract, use the following reference: http://meetings.aps.org/link/BAPS.2007.MAR.S26.2 |
Timezone: »
Poster
Most ReLU Networks Suffer from $\ell^2$ Adversarial Perturbations
We consider ReLU networks with random weights, in which the dimension decreases at each layer. We show that for most such networks, most examples $x$ admit an adversarial perturbation at an Euclidean distance of $O\left(\frac{\|x\|}{\sqrt{d}}\right)$, where $d$ is the input dimension. Moreover, this perturbation can be found via gradient flow, as well as gradient descent with sufficiently small steps. This result can be seen as an explanation to the abundance of adversarial examples, and to the fact that they are found via gradient descent. |
# zbMATH — the first resource for mathematics
Some anomalies of farsighted strategic behavior. (English) Zbl 1395.91008
Erlebach, Thomas (ed.) et al., Approximation and online algorithms. 10th international workshop, WAOA 2012, Ljubljana, Slovenia, September 13–14, 2012. Revised selected papers. Berlin: Springer (ISBN 978-3-642-38015-0/pbk). Lecture Notes in Computer Science 7846, 229-241 (2013).
Summary: We investigate subgame perfect equilibria, that better capture the rationality of the players in sequential games with respect to other more myopic dynamics like the classical Nash one. We prove that the sequential price of anarchy, that is the worst case ratio between the social performance at a subgame perfect equilibrium and the best possible one, is exactly 3 in cut and consensus games. Moreover, we improve the known $$\Omega (n)$$ lower bound for unrelated scheduling to $$2^{\Omega(\sqrt{n})}$$ and refine the corresponding upper bound to $$2^{n }$$, where $$n$$ is the number of players. Finally, we determine essentially tight bounds for fair cost sharing games by proving that the sequential price of anarchy is between $$n + 1 - H _{n }$$ and $$n$$. A surprising lower bound of $$\frac{n + 1}{2}$$ is also determined for the singleton case.
Our results are quite interesting and counterintuitive, as they show that a farsighted behavior generally may lead to a worse performance with respect to myopic one; in fact, Nash equilibria and simple Nash rounds, consisting of a single (myopic) move per player starting from the empty state achieve a price of anarchy which result to be lower or equivalent to the sequential price of anarchy in almost all the considered cases.
For the entire collection see [Zbl 1271.68034].
##### MSC:
91A10 Noncooperative games 91A06 $$n$$-person games, $$n>2$$
Full Text: |
Ampere’s Law (Magnetostatics): Differential Form
The integral form of Amperes’ Circuital Law (ACL; Section 7.4) for magnetostatics relates the magnetic field along a closed path to the total current flowing through any surface bounded by that path. In mathematical form:
where
is magnetic field intensity,
is the closed curve, and
is the total current flowing through any surface bounded by
. In this section, we derive the differential form of this equation. In some applications, this differential equation, combined with boundary conditions associated with discontinuities in structure and materials, can be used to solve for the magnetic field in arbitrarily complicated scenarios. A more direct reason for seeking out this differential equation is that we gain a little more insight into the relationship between current and the magnetic field, disclosed at the end of this section.
The equation we seek may be obtained using Stokes’ Theorem, which in the present case may be written:
where
is any surface bounded by
, and
is the differential surface area combined with the unit vector in the direction determined by the right-hand rule from Stokes’ Theorem. ACL tells us that the right side of the above equation is simply
. We may express
as the integral of the volume current density
(units of A/m2; Section 6.2) as follows:
so we may rewrite Equation 7.9.2 as follows:
The above relationship must hold regardless of the specific location or shape of
. The only way this is possible for all possible surfaces in all applicable scenarios is if the integrands are equal. Thus, we obtain the desired expression:
That is, the curl of the magnetic field intensity at a point is equal to the volume current density at that point. Recalling the properties of the curl operator – in particular, that curl involves derivatives with respect to direction – we conclude:
The differential form of Ampere’s Circuital Law for magnetostatics (Equation 7.9.5) indicates that the volume current density at any point in space is proportional to the spatial rate of change of the magnetic field and is perpendicular to the magnetic field at that point. |
# Chapter 1 - Tools of Geometry - 1-3 Measuring Segments - Practice and Problem-Solving Exercises: 40
ED is 10 DB is 10 EB is 20
#### Work Step by Step
The diagram shows $\overline{ED}$ and $\overline{DB}$ are congruent, so we obtain: ED=DB $x+4=3x-8\longrightarrow$ substitute and solve for x $x+4+8=3x-8+8\longrightarrow$ add 8 to each side $x+12=3x\longrightarrow$ simplify $x+12-x=3x-x\longrightarrow$ subtract x from each side $12=2x\longrightarrow$ simplify $12\div2=2x\div2\longrightarrow$ divide each side by 2 $6=x \longrightarrow$ simplify Substitute for x to find each length. ED=$x+4=6+4=10$ DB=$3x-8=3(6)-8=18-8=10$ Use the Segment Addition Postulate EB=ED+DB=$10+10=20$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
The upshot for proving such theorems is that it is like a 2-for-1 sale, we get to do two proofs. Assume $P$ and conclude $Q\text{,}$ then start over and assume $Q$ and conclude $P\text{.}$ For this reason, “if and only if” is sometimes abbreviated by $\iff\text{,}$ while proofs indicate which of the two implications is being proved by prefacing each with $\Rightarrow$ or $\Leftarrow\text{.}$ A carefully written proof will remind the reader which statement is being used as the hypothesis, a quicker version will let the reader deduce it from the direction of the arrow. Tradition dictates we do the “easy” half first, but that is hard for a student to know until you have finished doing both halves! Oh well, if you rewrite your proofs (a good habit), you can then choose to put the easy half first.
Theorems of this type are called “equivalences” or “characterizations,” and they are some of the most pleasing results in mathematics. They say that two objects, or two situations, are really the same. You do not have one without the other, like rain and my yellow boots. The more different $P$ and $Q$ seem to be, the more pleasing it is to discover they are really equivalent. And if $P$ describes a very mysterious solution or involves a tough computation, while $Q$ is transparent or involves easy computations, then we have found a great shortcut for better understanding or faster computation. Remember that every theorem really is a shortcut in some form. You will also discover that if proving $P\Rightarrow Q$ is very easy, then proving $Q\Rightarrow P$ is likely to be proportionately harder. Sometimes the two halves are about equally hard. And in rare cases, you can string together a whole sequence of other equivalences to form the one you are after and you do not even need to do two halves. In this case, the argument of one half is just the argument of the other half, but in reverse. |
Question
# The product of two consecutive positive integers is 240. Formulate the quadratic equation whose roots are these integers.
Hint: The product of two roots is given. So the value of the $1^{st}$ multiplied by the $2^{nd}$ root is 240. Then we need to equate the two and find the value of the root.
Complete step by step solution: Given, The product of two consecutive positive integers is 240.
We need to select two consecutive i.e. values which come after each other.
Therefore,
Let the two consecutive positive integers be x and(x+1),
Here we take x as the first integer and so the next number coming after it is (x+1)
For example, 2 comes after 1 and 2=1+1
Then according to the problem,
Their product is;
x(x+1)=240
On multiplying the two variables on the left hand side we get,
$x^{2}-x-240=0$
This is the required quadratic equation. Thus, when two consecutive positive integers are multiplied we get the equation we got above.
Note: The standard form of a quadratic equation is $a{x}^{2}+bx+c=0$, where a, b and c are real numbers and $a\ne 0$. 'a' is the coefficient of ${x}^{2}$. It is called the quadratic coefficient. 'b' is the coefficient of x. Students often go wrong in calculating the coefficients. Students also tend to make mistakes in solving the quadratic equations which is not needed here. |
If NP is easy on average then does it mean P=NP?
If $NP=RP$ then $NP$ is easy on average. Then from point $1$ in abstract in http://lance.fortnow.com/papers/files/derand.pdf which says $NP$ is easy on average implies $P=BPP$ do we have $NP=RP\implies P=NP$?
This problem asks directly a point with respect to a remark in a paper.
I do not blindly ask if np=rp then is p=np. I ask about a comment in a paper.
I think essentially I am asking is $RP\subseteq Avg-P$? |
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### Time series (stochastic process) estimating parameters using characteristic function
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### Rate Distortion Minimization in a Python Clustering Algorithm
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### Discrete-time Jump-Diffusion Model
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0
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### 15 Reelle Matrizen
0/14/4
#### 15.4 Gauß’scher Algorithmus
0/14/4/2
##### 15.4.3 Gauß-Jordan-Verfahren
0/14/4/2/1 .
Beispiel 15 - 138
Alternative Lösung des obigen Beispiels mit Gauß-Jordan: .
$-x$ $+y$ $+z$ $=$ $0$ umbilden in Matrix: $x$ $-3y$ $-2z$ $=$ $5$ $⇒$ 1. Spalte: $x$, 2. Spalte: $y$ $5x$ $+y$ $+4z$ $=$ $3$ 3. Spalte: $z$, 4. Spalte: Absolutglied
.
(Anm.: Man kann das Absolutglied zuerst auch auf die linke Seite der Gleichung bringen, man muß nur beim Ablesen darauf achten.) .
$x$ $y$ $z$ $c$ PS $1$ $-1$ $-1$ $0$ $-1$ $1$ $-3$ $-2$ $5$ $1$ $-I$ $5$ $1$ $4$ $3$ $13$ $-5\cdot I$ $1$ $-1$ $-1$ $0$ $-1$ $-II∕2$ $0$ $-2$ $-1$ $5$ $2$ $0$ $6$ $9$ $3$ $18$ $+3\cdot II$ $1$ $0$ $0$ $-1$ $0$ $0$ $1$ $0$ $-4$ $-3$ $0$ $0$ $6$ $18$ $24$ $∕6$ $1$ $0$ $0$ $-1$ $0$ $0$ $1$ $0$ $-4$ $-3$ $0$ $0$ $1$ $3$ $4$
.
$⇒x=-1,\phantom{\rule{0.3em}{0ex}}y=-4,\phantom{\rule{0.3em}{0ex}}z=3$ .
. |
# Would magnesium and calcium hydroxide undergo a single replacement reaction?
Can calcium hydroxide react with magnesium (if given sufficient activation energy) to produce magnesium hydroxide and calcium, in the same way that NaOH + $\ce{Mg} \ce{->} \ce{Mg(OH)2} + \ce{Na}$ ?
Or, is calcium's electronegativity (or some other factor) prohibitive of this?
The key part of the question is given sufficient energy. The difference in electronegativity between Ca and Mg is ~0.3 eV. The relationship of eV:K is about 1:12,000. Given that, you should be able to calculate a temperature at which the reaction $\ce{Ca(OH)2 + Mg -> Mg(OH)2 + Ca}$ is favored over the reverse.
Furthermore, you can force the suggested reaction by making use of relative solubilities. Find a solvent (not water, since magnesium and calcium are far more electronegative than hydrogen, and would decompose water to form hydroxides of both metals) in which $\ce{Ca(OH)2}$ is more soluble than $\ce{Mg(OH)2}$. Precipitation will "lock up" the $\ce{Mg(OH)2}$ as it forms, little by little, forcing the reaction to proceed even though it is endothermic. |
# Particle in a Box
## The particle in a box model provides one of the very few problems in quantum mechanics which can be solved analytically.
#### Key Points
• The model is mainly used as a hypothetical example to illustrate the differences between classical and quantum systems.
• The particle may only occupy certain positive energy levels.
• The particle is more likely to be found at certain positions than at others.
• The particle may never be detected at certain positions, known as spatial nodes.
#### Terms
• A number, between 0 and 1, expressing the precise likelihood of an event happening.
• A mathematical function that describes the propagation of the quantum mechanical wave associated with a particle (or system of particles), related to the probability of finding the particle in a particular region of space.
#### Figures
1. ##### The Potential Well
Energy/position relationships in the particle in a box.
2. ##### Solutions to Particle in a Box
The first four solutions to the one dimensional particle in a box.
## The Particle in a Box
Also called the infinite square well problem, the "particle in a box" is actually one of the very few problems in quantum mechanics that can be solved without approximations, or analytically.
Recall the Schrödinger equation:
$-\frac { { \hbar }^{ 2 } }{ 2m } \frac { { \partial }^{ 2 }\Psi }{ { \partial x }^{ 2 } } +V(x)\Psi \quad =\quad i\hbar \frac { \partial \Psi }{ \partial t }$
The potential function, V, is time independent, while the wave function itself is time dependent.
The infinite square well is defined by a potential function in which the value V(x) is 0 for values of x between 0 and L where L is the length of the box, and infinite in all other positions Figure 1. In this case, classical and quantum physics are essentially describing the same thing: the impossibility of a particle "leaping" over an infinitely high boundary.
Separating the variables reduces the problem to one of simply solving the spatial part of the equation:
$-\frac { { \hbar }^{ 2 } }{ 2m } \frac { { d }^{ 2 }\psi }{ { dx }^{ 2 } } +V(x)\psi \quad =\quad E\psi$
E represents the possible energies that can describe the system.
Inside the box, no forces act on the particle, which means that the part of the wave function inside the box oscillates through space and time with the same form as a free particle:
$\psi (x,t)\quad =\quad [A\sin { (kx) } +\quad B\cos { (kx)]{ e }^{ -i\omega t } }$
B and A are arbitrary complex numbers. The frequency of the oscillations through space and time are given by the wave number and the angular frequency, respectively. These are related to the energy of the particle by:
$E=\hbar \omega =\frac { { \hbar }^{ 2 }{ k }^{ 2 } }{ 2m }$
The size or amplitude of the wavefunction at any point is given by the probability of finding the particle there by: $P(x,t)={ \left\lfloor \psi (x,t) \right\rfloor }^{ 2 }$
Therefore, the wavefunction must vanish everywhere beyond the edges of the box. Also, the amplitude of the wavefunction may not "jump" abruptly from one point to the next. These two conditions are only satisfied by wave functions with the form:${ \psi }_{ n }(x,t)\quad =\quad A\sin { { (k }_{ n } } x){ e }^{ -i{ \omega }_{ n }t }$
where 0
${ k }_{ n }\quad =\quad \frac { n\pi }{ L }$ where n = {1,2,3,4...} and L is the size of the box Figure 2. Negative values are neglected, since they give wave functions identical to the positive solutions except for a physically unimportant sign change. Finally, the unknown constant may be found by normalizing the wavefunction so that the total probability density of finding the particle in the system is 1.
Normalizing, we get $\left| A \right| =\sqrt { \frac { 2 }{ L } }$
A may be any complex number with absolute value $\left| A \right| =\sqrt { \frac { 2 }{ L } }$
When treated as a probability function, the wavefunction describes the probability of finding the particle at a given point and at a given time. Four conditions, proposed by Max Born, must be met for this to be true:
1. The wavefunction must be single-valued
2. The wavefunction must be "square integrable"
3. The wave unction must be continuous everywhere
4. The first derivative of the wavefunction must be continuous.
#### Key Term Glossary
amplitude
the maximum absolute value of some quantity that varies, especially a wave
##### Appears in these related concepts:
constant
Consistently recurring over time; persistent
##### Appears in these related concepts:
density
a measure of the amount of matter contained by a given volume
##### Appears in these related concepts:
energy
a quantity that denotes the ability to do work and is measured in a unit dimensioned in mass × distance²/time² (ML²/T²) or the equivalent
##### Appears in these related concepts:
frequency
The number of occurrences of a repeating event per unit of time.
##### Appears in these related concepts:
probability
A number, between 0 and 1, expressing the precise likelihood of an event happening.
##### Appears in these related concepts:
quanta
discrete packets with energy stored inside
##### Appears in these related concepts:
quantum
the smallest possible, and therefore indivisible, unit of a given quantity or quantifiable phenomenon
##### Appears in these related concepts:
solution
A homogeneous mixture, which may be liquid, gas or solid, formed by dissolving one or more substances.
##### Appears in these related concepts:
Solution
A homogeneous mixture, which may be liquid, gas or solid, formed by dissolving one or more substances.
##### Appears in these related concepts:
system
the part of the universe being studied, arbitrarily defined to any size desired
##### Appears in these related concepts:
wave
A shape that alternatingly curves in opposite directions.
##### Appears in these related concepts:
wavefunction
A mathematical function that describes the propagation of the quantum mechanical wave associated with a particle (or system of particles), related to the probability of finding the particle in a particular region of space. |
# The number of five-digit telephone numbers having at least one of their digits repeated is
Question:
The number of five-digit telephone numbers having at least one of their digits repeated is
(a) 90000
(b) 100000
(c) 30240
(d) 69760
Solution:
(d) 69760
Total number of five digit numbers (since there is no restriction of the number $0 X X X X)=10 \times 10 \times 10 \times 10 \times 10=100000$
These numbers also include the numbers where the digits are not being repeated. So, we need to subtract all such numbers.
Number of 5 digit numbers that can be formed without any repetition of digits $=10 \times 9 \times 8 \times 7 \times 6=30240$
$\therefore$ Number of five-digit telephone numbers having at least one of their digits repeated $=\{$ Total number of 5 digit numbers $\}-\{$ Number of numbers that do not have any digit repeated $\}=100000-30240=69760$ |
# Conditional probability exercise infected by virus
A computervirus has a probability of 0.4 to infect your pc through mail and 0.3 trough your browser. The probability that your pc is infected by both mail and browser is 0.15. $$P(M)=0.4$$ $$P(B)=0.3$$ $$P(M\cap B)=0.15$$
It seems they are not independent.
What is the probability you don't have a virus?
Is this: $$=1-P(M\cup B)=1-P(M)-P(B)+P(M\cap B)=1-0.4-0.3+0.15=0.45$$
What is the probability that you have a virus which doesn't originate from a mail?
I translated the above to: $$P\left ( (M\cup B) \mid \overline{M} \right)= \frac{P\left ((M\cup B)\cap \overline{M} \right )}{P(\overline{M})} = \frac{P(B\cap \overline{M} )}{P(\overline{M})}= \frac{P\left ( B\setminus (B\cap M) \right )}{1-P(M)} = \frac {0.3-0.15}{1-0.4} = \frac {0.15} {0.6} = 0.25$$
Am I correct?
• @AlexR I am not so sure about that. See my answer. – drhab Jun 16 '15 at 10:53
• @drhab On second thought, your interpretation makes more sense to me (as in "What I would ask myself") The wording isn't very clear there. – AlexR Jun 16 '15 at 11:08
To be found is then: $$P\left\{ M^{c}\mid M\cup B\right\}$$
$$P\left\{ M^{c}\mid M\cup B\right\} =\frac{P\left(B-M\right)}{P\left(M\cup C\right)}=\frac{0.15}{0.55}=\frac{3}{11}$$ |
## EXPRESSION2
Evaluate algebraic expression with 2 inputs Inputs(2) a b Parameters(0) Default Text1 Text2 Select from Components/Blocks/Cscript&Expression Function Algebraic expression evaluation, where the inputs are denoted by a, b, c, etc.Only algebraic expression using +-*/ can be used. Use Brackets ( ) for ordering equations. Use % for integer divisions and use ^ for exponentiation. Scale suffixes notations like m and u, etc. are NOT allowed, instead use 2*10^-6 for 2uFarad. A square root of the first input a is evaluated as a^0.5.Use the blocks EXPRESSION, EXPRESSION2, EXPRESSION3, EXPRESSION4, EXPRESSION5, EXPRESSION6, EXPRESSION10, EXPRESSION15, EXPRESSION20 for evaluating algebraic functions with more inputs. Special The ^ operator cannot be used in exported C code! |
# Producing NaCl from 2 Na and Cl2
I want to react 250 g of $\ce{Na}$ and enough $\ce{Cl2}$ to produce $\ce{NaCl}$ and see how many grams of $\ce{NaCl}$ I can get.
I guess I should create a scheme like this
2 Na + 1 Cl_2 --> 2 NaCl
===============================================
Number Ratio 2 1 2
Molar Mass 23.0 71.0 58.5
Mass Ratio 250 ? ?
But I don't know what to put in the two remaining cells.
I think it should be $\frac{250}{2 \cdot 23/71} = 385.5$ and $\frac{250}{23/58.5} = 635.5$, but I am not entirely sure though.
When should I be careful about limiting reagents?
# Edit
2 Na + 1 Cl_2 --> 2 NaCl
===============================================
Number Ratio 2 1 2
Molar Mass 23.0 71.0 58.5
Moles 250/23=10.9 ? ?
Mass Ratio 250 ? ?
Is the number of moles for the two other reactants just $10.9 / 2 = 5.5$ and $10.9$, respectively?
If so I can easily find the masses with $n = m/M \rightarrow m = n M$.
# Edit 2
Now I understand it :-)
But in what situations should I be careful about limiting reagents?
• Add an extra row to your table and call it moles. Work out the number of moles of sodium present. You need half the number of moles of chlorine molecules (remembering that there are two chlorine atoms in a molecule). Since you know the molecular weight of the chlorine molecule you can fill in your gap and similarly for sodium chloride. – user1945827 Oct 13 '16 at 10:06
• I have updated my question with the row you suggested – Jamgreen Oct 13 '16 at 10:14
• So, 10.9 represents the amount of sodium. Can you work out how many moles of chlorine molecules and sodium chloride are present? – user1945827 Oct 13 '16 at 10:24
A typical problem on the concept of limiting reagents!
A) List out the (gram) molar masses of both your reactants.
$Na$ = 23 and $\mathrm{Cl_{2}}$ = 71
B) Now gauge the proportions of reactant in your final product.
The final product is NaCl. That's Na to Cl in the ratio of 1:1 (moles). Taking the masses into consideration, that's 23 : 35.5 grams, or if you want, you can further simplify it by dividing both terms by 23. So the resultant ratio is 1 : 1.54.
Essentially, what this ratio tells you is that, while synthesizing NaCl, 1.54 grams of Cl combines with every gram of Na.
You're given a specific amount of Na, and no constraint has been set on the amount of $\mathrm{Cl_{2}}$ you can use. So here, Na is the limiting reagent.
As you have 250 grams of Na, the corresponding amount of Cl you need is about (1.54)x(250) = 385 grams.
But since you're dealing with chlorine gas ( $\mathrm{Cl_{2}}$ ), this corresponds to 385/71 = 5.42 moles of $\mathrm{Cl_{2}}$ .
C) Take the ratio of molar masses between the limiting reagent and the final product.
Now the gram molar mass of NaCl is 58.5 grams. Now if you take the ratio between the gram molar masses of Na and NaCl, what you get is 23 : 58.5. Simplifying it by dividing both terms by 23, we get 1: 2.54. So for every gram of Na you have that completely react with chlorine, 2.54 grams of NaCl is produced.
Since you have 250 grams of Na, if you completely react it with Chlorine gas (about 385 grams of it), you get (2.54)x(250) = 635 grams of NaCl. Or alternatively, since you now have the mass of chlorine, using the law of conservation of mass you can go on right ahead and simply add 250g Na + 385g $\mathrm{Cl_{2}}$ and get 635g NaCl. Voila!
[Yes I know I made it longer than it could've been, but hopefully going step by step won't create any confusion. You don't really have to calculate the amount of chlorine as well, but I did it anyway.] |
## another lottery coincidence
Once again, meaningless figures are published about a man who won the French lottery (Le Loto) for the second time. The reported probability of the event is indeed one chance out of 363 (US) trillions (i.e., billions in the metric system. or 1012)… This number is simply the square of
${49 \choose 5}\times{10 \choose 1} = 19,068,840$
which is the number of possible loto grids. Thus, the probability applies to the event “Mr so-&-so plays a winning grid of Le Loto on May 6, 1995 and a winning grid of Le Loto on July 27, 2011“. But this is not the event that occured: one of the bi-weekly winners of Le Loto won a second time and this was spotted by Le Loto spokepersons. If we take the specific winner for today’s draw, Mrs such-&-such, who played bi-weekly one single grid since the creation of Le Loto in 1976, i.e. about 3640 times, the probability that she won earlier is of the order of
$1-\left(1-\frac{1}{{49\choose 5}\times{10\choose 1}}\right)^{3640}=2\cdot 10^{-4}$.
There are thus two chances in 10 thousands to win again for a given (unigrid) winner, not much indeed, but no billion involved either. Now, this is also the probability that, for a given draw (like today’s draw), one of the 3640 previous winners wins again (assuming they all play only one grid, play independently from each other, &tc.). Over a given year, i.e. over 104 draws, the probability that there is no second-time winner is thus approximately
$\left(1-\frac{1}{2\cdot10^4}\right)^{104} = 0.98,$
showing that within a year there is a 2% chance to find an earlier winner. Not so extreme, isn’t it?! Therefore, less bound to make the headlines…
Now, the above are rough and conservative calculations. The newspaper articles about the double winner report that the man is playing about 1000 euros a month (this is roughly the minimum wage!), representing the equivalent of 62 grids per draw (again I am simplifying to get the correct order of magnitude). If we repeat the above computations, assuming this man has played 62 grids per draw from the beginning of the game in 1976 till now, the probability that he wins again conditional on the fact that he won once is
$1-\left(1-\frac{62}{{49 \choose 5}\times{10 \choose 1}}\right)^{3640} = 0.012$,
a small but not impossible event. (And again, we consider the probability only for Mr so-&-so, while the event of interest does not.) (I wrote this post before Alex pointed out the four-time lottery winner in Texas, whose “luck” seems more related with the imperfections of the lottery process…)
I also stumbled on this bogus site providing the “probabilities” (based on the binomial distribution, nothing less!) for each digit in Le Loto, no need for further comments. (Even the society that runs Le Loto hints at such practices, by providing the number of consecutive draws a given number has not appeared, with the sole warning “N’oubliez jamais que le hasard ne se contrôle pas“, i.e. “Always keep in mind that chance cannot be controlled“…!)
### 4 Responses to “another lottery coincidence”
1. I found out that Steve Samuels and George McCabe from Purdue had run a similar analysis in the NYT at about the time I was visiting Purdue.
2. [...] A funny point is to notice that gambling seems to be a possible way for skilled statisticians to earn money, it’s the second time I read about such story (further discussed here). [...]
3. [...] 6) in this assignment also sounds very much inspired from another of my posts on coincidences in lotteries [although not acknowledged in the assignment] since the question [...]
4. [...] Suchard in an Edinburgian Indian restaurant on Tuesday night, I faced another if much less pleasant coincidental event: for the third time in a row, my bag went missing on a Air France flight to Scotland… This [...] |
## Stability of lattice QCD simulations and the thermodynamic limit
Research output: Contribution to journalArticle
### Documents
http://iopscience.iop.org/1126-6708/2006/02/011/
Original language English 19 Journal of High Energy Physics 2006 02 https://doi.org/10.1088/1126-6708/2006/02/011 Published - 15 Dec 2005
### Abstract
We study the spectral gap of the Wilson--Dirac operator in two-flavour lattice QCD as a function of the lattice spacing $a$, the space-time volume $V$ and the current-quark mass $m$. It turns out that the median of the probability distribution of the gap scales proportionally to $m$ and that its width is practically equal to $a/\sqrt{V}$. In particular, numerical simulations are safe from accidental zero modes in the large-volume regime of QCD.
### Research areas
• hep-lat, Lattice Gauge Field Theories, Lattice QCD |
# X, Y points to a circles radius
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I am a little confused on this programming problem I have been asked to help with, so I thought I could gain some insight and clarification on this problem just for my sake alone. Lets say there is a circle of I guess any radius as the specific doesn't matter for the algorithm to finding the points. Lets say 2 for arguments sake. The goal is to output the x,y coordinates at a given angle, it doesn't state in the problem, but lets assume 0 is the center of the circle. It doesn't sound too hard, but I am having a hard time with how to calculate such determinations. Any help on this would be great I don't need code, but example would be great. Once I get a handle on the idea of doing this, I will post some code if I have some issues or to check if I got it right to my understanding. Thank you in advance.
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x = center_x + radius*cos(angle); y = center_y + radius*sin(angle);
Is that what you are asking?
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That looks to be exactly what I was looking for thank you so much. I will put together some code to make sure it is working right and post it to be certain.
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Something to consider is when the circle is very small or very large. If it is very small and only takes up something like 10 pixels wide on the screen, you will have solved for many angles that have the same screen point. If the circle is very large on the screen, like 1000 pixels wide, you can end up with gaps where you would need to solve for non-integer angles in order to get a complete circle. This is assuming that you want to draw the circle on your screen.
Here is some old C# code that I had used to draw circles on the screen. It scans both x and y axis to alleviate the gaps that can result when solving for only 1 axis. I think the code looks terrible, its been a while since I did this and I just threw it together in order to make it work. I thought it would be good to share. It uses a point on the circle to determine radius (because thats what my solution at the time called for), but it could easily be changed to a radius parameter.
private void DrawCircle(Vect2D center, Vect2D pointOn, Color drawColor) { Bitmap viewBitmap = new Bitmap(pictureBox_View.Image); int x, y; Vect2D v = new Vect2D(); v = pointOn - center; double r = v.Magnitude(); x = center.x - (int)r; // draw along x while (x < center.x + (int)r) { y = center.y + (int)Math.Sqrt((r * r) - (x - center.x) * (x - center.x)); if (x>=0 && x < viewBitmap.Width && y >= 0 && y < viewBitmap.Height ) viewBitmap.SetPixel(x, y, drawColor); y = center.y - (int)Math.Sqrt((r * r) - (x - center.x) * (x - center.x)); if (x >= 0 && x < viewBitmap.Width && y >= 0 && y < viewBitmap.Height) viewBitmap.SetPixel(x, y, drawColor); x++; } // draw along y y = center.y - (int)r; while (y < center.y + (int)r ) { x = center.x + (int)Math.Sqrt((r * r) - (y - center.y) * (y - center.y)); if (x >= 0 && x < viewBitmap.Width && y >= 0 && y < viewBitmap.Height) viewBitmap.SetPixel(x, y, drawColor); x = center.x - (int)Math.Sqrt((r * r) - (y - center.y) * (y - center.y)); if (x >= 0 && x < viewBitmap.Width && y >= 0 && y < viewBitmap.Height) viewBitmap.SetPixel(x, y, drawColor); y++; } pictureBox_View.Image = viewBitmap; }
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This is what I have so far, but I noticed some issues, I cannot seem to calculate the milliseconds correctly and the timer has a little offset so sometimes it doubles up on the x,y coordinate output before time outputs. Every 1.2 seconds x,y outputs and every 1.4 time outputs via the format Hour/Minute/Seconds/Milliseconds, suggestions are welcomed.
#include <iostream> #include <windows.h> #include <math.h> #include <time.h> //A simple counter class using QueryPerformance class Counter { public: Counter() { counter = SystemTime(); } ~Counter() { } //Return the time elapsed double ElapsedTime() { return SystemTime() - counter; } //Reset the timer to equal current system time void ResetTimer() { counter = SystemTime(); } private: double counter; //Return the systemtime double SystemTime() { LARGE_INTEGER frequency; BOOL useHighPerformanceTimer = QueryPerformanceFrequency(&frequency); LARGE_INTEGER currentTime; QueryPerformanceCounter(¤tTime); return static_cast<double>(currentTime.QuadPart) / frequency.QuadPart; } }; //Function to output the time in the designated format void OutputTime(tm* value) { std::cout << value->tm_hour << ":" << value->tm_min << ":" << value->tm_sec << ":" << 1000 / value->tm_sec << std::endl; } //Function to output the coordinates based on angle input void OutputCoordinates(const double& angle) { //coordinate = center(x/y) + radius * (cos/sin)(angle) double x = 10 + 1 * cos(angle); double y = 10 + 1 * sin(angle); std::cerr << "X: " << x << " Y: " << y << " "; } int main() { //Timer objects Counter timerA, timerB; timerA.ResetTimer(); timerB.ResetTimer(); //Declare the circle double angle = 0; //Value to determine which increment to do bool one = true; bool four = false; bool thirteen = false; bool seven = false; bool twenty = false; //Local time structure time_t raw; struct tm* clock; time(&raw); clock = localtime(&raw); //Main loop to run until "ESC" is pressed bool isRunning = true; while(isRunning) { //Update the time time(&raw); clock = localtime(&raw); //Output the x,y coordinates given the angle and radius of the circle, than reset the timer if(timerA.ElapsedTime() >= 1.2) { if(one) { angle += 1; one = false; four = true; } else if(four) { angle += 4; one = false; four = false; thirteen = true; } else if(thirteen) { angle += 13; thirteen = false; seven = true; } else if(seven) { angle += 7; seven = false; twenty = true; } else if(twenty) { angle += 20; twenty = false; one = true; angle = 1; } OutputCoordinates(angle); timerA.ResetTimer(); } //Output the time format of HH:MM:SS:MS every ~1.4 seconds, than reset the timer if(timerB.ElapsedTime() >= 1.4) { OutputTime(clock); timerB.ResetTimer(); } } return 0; }
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It's also quite common to draw a circle as a series of straight lines between a given set of points around the circle. You can increase the number of points until the circle loses its angular look although it can be surprising how few points you need in order to fool the eye, depending on the size of the circle of course.
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The culturable command area of a canal is $10.000$ ha. The area grows only two crops – rice in the Kharif season and wheat in the Rabi season. The design discharge of the canal is based on the rice requirements, which has an irrigated area of $2500$ ha, base period of $150$ days and delta of $130$ cm. The maximum permissible irrigated area (in ha) for wheat, with a base period of $120$ days and delta of $50$ cm, is __________ |
Chapter 41
### Introduction
I have finally kum to the konklusion, that a good reliable set of bowels iz wurth more tu a man, than enny quantity of brains.
Henry Shaw (1818–85), Josh Billings
Constipation is the difficult passage of small hard stools. The Rome III criteria (refer www.romecriteria.org) define it has having two or more of the following, for at least 12 weeks:
• infrequent passage of stools <3/week
• passage of lumpy or hard stools at least 25% of time
• straining >25% of time
• sensation of incomplete evacuation >25% of time
• use of manual manoeuvres >25% of time
• sensation of anorectal obstruction/blockage >25% of time
Accordingly it affects more than 1 in 5 in the population.1
However, the emphasis should be on the consistency of the stool rather than on the frequency of defecation; for example, a person passing a hard stool with difficulty once or twice a day is regarded as constipated, but the person who passes a soft stool comfortably every two or three days is not constipated. Various causes of chronic constipation are summarised in FIGURE 41.1.
###### FIGURE 41.1
Causes of chronic constipation
Key facts and checkpoints
• The survey showed 10% of adults and 6% of children reported constipation in the preceding 2 weeks.
• Up to 20% of British adults take regular laxatives.2
• Constipation from infancy may be due to Hirschsprung disorder.
• Diet is the single most important factor in preventing constipation.
• Beware of the recent onset of constipation in the middle-aged and the elderly.
• Bleeding suggests cancer, haemorrhoids, diverticular disorder and inflammatory bowel disease.
• Unusually shaped stools (small pellets or ribbon-like) suggest irritable bowel syndrome.
• Always examine the abdomen and rectum.
• Plain abdominal X-rays are generally not useful in the diagnosis of chronic constipation.
• The flexible sigmoidoscope is far superior to the rigid sigmoidoscope in investigation of the lower bowel.
• Intractable constipation (obstipation) is a challenge at both ends of the age spectrum but improved agents have helped with management.
### A diagnostic approach
Using the diagnostic strategy model (see Table 41.1), the five self-posed questions can be answered as follows.
Table 41.1
Chronic constipation: diagnostic strategy model
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# Momentum term to be expanded in dirac gamma matrices
## Homework Statement
I need help to expand some matrices
## Homework Equations
$$\pi = \frac{\partial \mathcal{L}}{\partial \dot{q}} = i \hbar \gamma^0$$
## The Attempt at a Solution
How do I expand
$$i\hbar \gamma^0$$
the matrix in this term, I am a bit lost. All the help would be appreciated!
Related Advanced Physics Homework Help News on Phys.org
## Homework Statement
I need help to expand some matrices
## Homework Equations
$$\pi = \frac{\partial \mathcal{L}}{\partial \dot{q}} = i \hbar \gamma^0$$
## The Attempt at a Solution
How do I expand
$$i\hbar \gamma^0$$
the matrix in this term, I am a bit lost. All the help would be appreciated!
Can no one answer my question?
TSny
Homework Helper
Gold Member
$$\pi = \frac{\partial \mathcal{L}}{\partial \dot{q}} = i \hbar \gamma^0$$
That looks odd. Can you give us the Lagrangian that you're starting with?
That looks odd. Can you give us the Lagrangian that you're starting with?
It has been a while since I dabbled with Dirac matrices... factoring gamma zero is simply 1.
DUH to me. |
# What do you call a long-chain organic molecule with a polar group on one or both ends of the molecule?
Surfactants are long chain organic molecules (typically acids, or so-called fatty acids), with a polar end, typically carboxylate groups, $R C \left(= O\right) {O}^{-} N {a}^{+}$ (or a polar group on both end). They can thus interact with greases (thru the alkyl chain), and also with aqueous solutions (thru the polar groups). These greatly reduce the surface tension of water. |
# Constructing the Tensor Product of Modules
Today we talk tensor products. Specifically this post covers the construction of the tensor product between two modules over a ring. But before jumping in, I think now's a good time to ask, "What are tensor products good for?" Here's a simple example where such a question might arise:
Suppose you have a vector space $V$ over a field $F$. For concreteness, let's consider the case when $V$ is the set of all $2\times 2$ matrices with entries in $\mathbb{R}$ and let $F=\mathbb{R}$. In this case we know what "$F$-scalar multiplication" means: if $M\in V$ is a matrix and $c\in \mathbb{R}$, then the new matrix $cM$ makes perfect sense. But what if we want to multiply $M$ by complex scalars too? How can we make sense of something like $(3+4i)M$? That's precisely what the tensor product is for! We need to create a set of elements of the form $$\text{(complex number) "times" (matrix)}$$ so that the mathematics still makes sense. With a little massaging, this set will turn out to be $\mathbb{C}\otimes_{\mathbb{R}}V$.
So in general, if $F$ is an arbitrary field and $V$ an $F$-vector space, the tensor product answers the question "How can I define scalar multiplication by some larger field which contains $F$?" And of course this holds if we replace the word "field" by "ring" and consider the same scenario with modules.
Now this isn't the only thing tensor products are good for (far from it!), but I think it's the most intuitive one since it is readily seen from the definition (which is given below). You can read about another motivation for the tensor product here And for all my physics friends, this blurb by K. Conrad sheds light on the relationship between the physicist's tensor and the mathematician's tensor (see p. 46).
So with this motivation in mind, let's go!
Let $R$ be a ring with 1 and let $M$ be a right $R$-module and $N$ a left $R$-module and suppose $A$ is any abelian group. Our goal is to create an abelian group $M\otimes_R N$, called the tensor product of $M$ and $N$, such that if there is an $R$-balanced map $i\colon M\times N\to M\otimes_R N$ and any $R$-balanced map $\varphi\colon M\times N\to A$, then there is a unique abelian group homomorphism $\Phi\colon M \otimes_R N\to N$ such that $\varphi=\Phi\circ i$, i.e. so the diagram below commutes.
Notice that the statement above has the same flavor as the universal mapping property of free groups!
Definition: Let $X$ be a set. A group $F$ is said to be a free group on $X$ if there is a function $i\colon X\to F$ such that for any group $G$ and any set map $\varphi\colon X\to G$, there exists a unique group homomorphism $\Phi\colon F\to G$ such that the following diagram commutes: (i.e. $\varphi=\Phi\circ i$)
So of course we'd like $X$ to be the set $M\times N$ and $G$ to be the abelian group $A$. To obtain a map $i\colon X\to F$, we want to let $F$ be the free abelian group generated by $M\times N$. That is, let $F$ be the set of all finite commuting sums of elements of the form $(m_i,n_i)$ where $m_i\in M, n_i\in N$ - this is an abelian group where there are no relations between any distinct pairs $(m,n)$ and $(m',n')$ (see Dummit and Foote, p. 360). Also, $\varphi\colon X\to G$ can be any set map, so in particular we just want our's to be $R$-balanced:
Definition: Let $R$ be a ring with 1. Let $M$ be a right $R$-module, $N$ a left $R$-module, and $A$ an abelian group. A map $\varphi:M\times N\to R$ is called $\mathbf{R}$-balanced if for all $m,m_1,m_2\in M$, all $n,n_1,n_2\in N$ and all $r\in R$,
• $\varphi(m_1+m_2,n)=\varphi(m_1,n)+\varphi(m_2,n)$
• $\varphi(m,n_1+n_2)=\varphi(m,n_1)+\varphi(m,n_2)$
• $\varphi(mr,n)=\varphi(m,rn)$
Lastly, we observe that in the free group definition, $i\colon X\to F$ is just some map, whereas in the tensor product definition, $i\colon M\times N\to M\otimes_R N$ must be an $R$-balanced map! How can we reconcile this? By "replacing" F by a certain quotient group $F/H$! (We'll define $H$ precisely below.)
These observations give us a road map to construct the tensor product. And so we begin:
## Step 1
Let $F$ be a free abelian group generated by $M\times N$ and let $A$ be an abelian group. Then by definition (of free groups), if $\varphi:M\times N\to A$ is any set map, and $M\times N \hookrightarrow F$ by inclusion, then there is a unique abelian group homomorphism $\Phi:F\to A$ so that the following diagram commutes.
In English, this just means, "set maps from $M\times N \to A$ are the same as (i.e. are isomorphic to) homomorphisms from $F$ to $A$."
## Step 2
Observe that the commuting diagram above is (almost) exactly what we want, except that the inclusion map $M\times N\hookrightarrow F$ is not $R$-balanced! To fix this, we must "modify" the target space $F$ by replacing it with the quotient $F/H$ where $H\leq F$ is the subgroup of $F$ generated by elements of the form
• $(m_1+m_2,n)-(m_1,n)-(m_2,n)$
• $(m,n_1+n_2)-(m,n_1)-(m,n_2)$
• $(mr,n)-(m,rn)$
where $m_1,m_2,m\in M$, $n_1,n_2,n\in N$ and $r\in R$. Why elements of this form? Because if we define the map $i:M\times N\to F/H$ by $$i(m,n)=(m,n)+H,$$ we'll see that $i$ is indeed $R$-balanced! Let's check:
So, are we done now? Can we really just replace $F$ with $F/H$ and replace the inclusion map with the map $i$, and still retain the existence of a unique homomorphism $\Phi:F/H\to A$? No! Of course not. $F/H$ is not a free group generated by $M\times N$, so the diagram below is bogus, right?
Not totally. We haven't actually disturbed any structure!
How can we relate the pink and blue lines? We'd really like them to be the same. But we're in luck because they basically are!
## Step 3
By the Fundamental Homomorphism Theorem, homomorphisms from $F/H$ to $A$ are the same as (i.e. isomorphic to) homomorphisms $f$ (or $\Phi$ in the case of the picture above) from $F$ to $A$, as long as $H\subseteq \ker(f)$, that is as long as $f(h)=0$ for all $h\in H$. And notice that this condition, $f(H)=0$, forces $f$ to be $R$-balanced!
Let's check:
Sooooo... homomorphisms $f:F\to A$ such that $H\subseteq \ker(f)$ are the same as $R$-balanced maps from $M\times N$ to $A$! (Technically, I should say homomorphisms $f$ restricted to $M\times N$.) In other words, we have
In conclusion, to say "abelian group homomorphisms from $F/H$ to $A$ are the same as (isomorphic to) $R$-balanced maps from $M\times N$ to $A$" is the simply the hand-wavy way of saying
Whenever $i:M\times N\to F$ is an $R$-balanced map and $\varphi:M\times N\to A$ is an $R$-balanced map where $A$ is an abelian group, there exists a unique abelian group homomorphism $\Phi:F/H\to A$ such that the following diagram commutes:
And this is just want we want! The last step is merely the final touch:
## Step 4
At last, we define the abelian quotient group $F/H$ to be the tensor product of $M$ and $N$,
whose elements are cosets,
where $m\otimes n$ for $m\in M$ and $n\in N$ is referred to as a simple tensor. And there you have it! The tensor product, constructed.
Reference:
D. Dummit and R. Foote, Abstract Algebra, 3rd ed., Wiley, 2004. (ch. 10.4)
Comment |
# The Mystery of the Spotty Animated GIF
Have you ever posted an animated GIF to an application or website that didn't like the idea of motion? You likely ended up with a horrible, strange looking freeze frame.
I know. You've pondered this question for the longest time. It has haunted your dreams. Why such a strange screenshot? What madness is happening? Is my computer cursed? Did I accidentally poke holes in my RAM again?
GIF, if you didn't know, is a most marvelously strange image format. It was introduced in 1987 and only supports 256 colours in an image. It has been plagued by lawsuits, poor algorithms, poor defaults, and lolcats. Actually, to be fair, I shouldn't be dissing the lolcats. The lolcats are probably what made it so resilient to change. Other formats have come and gone, but nothing has replaced the vehicle of the lolcats, our beloved GIF.
### So why the spots, Doc?
GIF didn't have fancy techniques like spatial image compression or temporal motion compensation like these fancy new video formats. GIF's old. It predates the CD. It predates me! Instead, GIF uses an interesting little hack called Transparency Optimisation to optimise animations.
Uncompressed GIF animations store the full frame for each and every image. As you might guess, that gets pretty large pretty quickly. Wouldn't it be nice if you could just paint the difference between this frame and the last? That's exactly what GIF does.
One of the colours you can select is transparent. GIF doesn't support semi-transparency (like that fancy young PNG), but it does support binary transparency. Binary transparency is simple: either you see the colour or you don't.
### How does that make it smaller?
One of the other advantages of GIF is that the images are compressed using the lossless Lempel-Ziv-Welch (LZW) technique. LZW compresses by creating a dictionary of commonly occurring sequences and replacing them with a shorter symbol. These symbols are placed whereever you meet repetitive data. Annoyingly enough, LZW compression was also one of the major issues with GIF in the olden days: the patent for it expired in 2003.
If we didn't include LZW, the transparency optimisation trick alluded to above wouldn't make anything smaller. Each of those transparent pixels has to be recorded in the GIF. With LZW, however, those transparent pixels become hugely advantageous over just storing the full frame.
GIFs store their pixel data in rows. If the majority of the frame is transparent, the majority of these rows will just be the transparent pixel repeated over and over again. This compresses tremendously well.
LZW compression is very happy to turn that into a succint representation. Bam. A smaller, thinner, sexier GIF.
### And it's that transparency that results in your spotty GIF preview
When the GIF software decides to take a snapshot of the animation instead of playing it, it grabs an individual frame from somewhere past the start (i.e. movie trailers always start with a "boring" black frame).
This would generally be fine except that, as we've seen, GIFs use partial frames to aid in compression. What we end up with is a partial frame that appears "spotty" as it only contains the difference between the past frame and the current one.
Long story, I know, but now you understand why when you see an animated GIF preview it can end up looking spotty. Your friends will be amazed, your colleagues stunned, and it's now inevitable you'll become tremendously rich.
Or, at the very least, you've discovered something interesting and new. I'm happy to settle for that.
### Random Thought
GIF isn't suitable for photographs or other high detailed images as you're limited to only 256 different colours in an image. There's one slight point of interest however - with each frame you can use a new block of 256 colours. Therefore, an animated GIF could reproduce a high detailed photograph outside of the normal restricted colour range even though the file format technically doesn't allow it.
1. Given the C different colours used in the picture P, break C up into groups of 256 colours (i.e. C => C1, C2, C3, ..., Cn)
2. For each frame of the animation, dot the image with the new selection of 256 colours.
3. Stand back and marvel at your full 24 bit colour masterpiece that has the bit weight of a large elephant.
Bonus points if you progressively enhance the appearance of the image as more and more "colour layers" are loaded! If you do it properly, with a delay here and there, it should be a painting in the same style as A Sunday Afternoon on the Island of La Grande Jatte.
Update: Turns out this has been done: see the True-Color GIF example! Thanks to rcxdude from the Hacker News discussion thread for pointing it out. |
## Trigonometry (10th Edition)
1. Let's consider quadrants and sign of (x,y) coordinates $quadrant (x,y):$ $Quad I$ $(+,+)$ $Quad II$ $(-,+)$ $Quad III$ $(-,-)$ $Quad IV$ $(+,-)$ 2.In Q3 y is negative and x is negative , so $\frac{y}{x}$ is positive |
# Euler class explained
In mathematics, specifically in algebraic topology, the Euler class is a characteristic class of oriented, real vector bundles. Like other characteristic classes, it measures how "twisted" the vector bundle is. In the case of the tangent bundle of a smooth manifold, it generalizes the classical notion of Euler characteristic. It is named after Leonhard Euler because of this.
E
is an oriented, real vector bundle of rank
r
over a base space
X
.
## Formal definition
The Euler class
e(E)
is an element of the integral cohomology group
Hr(X;Z),
constructed as follows. An orientation of
E
amounts to a continuous choice of generator of the cohomology
Hr(Rr,Rr\setminus\{0\};Z)\cong\tilde{H}r-1(Sr-1;Z)\congZ
of each fiber
Rr
relative to the complement
Rr\setminus\{0\}
of zero. From the Thom isomorphism, this induces an orientation class
u\inHr(E,E\setminusE0;Z)
in the cohomology of
E
relative to the complement
E\setminusE0
of the zero section
E0
. The inclusions
(X,\emptyset)\hookrightarrow(E,\emptyset)\hookrightarrow(E,E\setminusE0),
where
X
includes into
E
as the zero section, induce maps
Hr(E,E\setminusE0;Z)\toHr(E;Z)\toHr(X;Z).
The Euler class e(E) is the image of u under the composition of these maps.
## Properties
The Euler class satisfies these properties, which are axioms of a characteristic class:
• Functoriality: If
F\toY
is another oriented, real vector bundle and
f\colonY\toX
is continuous and covered by an orientation-preserving map
F\toE
, then
e(F)=f*(e(E))
. In particular,
e(f*(E))=f*(e(E))
.
F\toX
is another oriented, real vector bundle, then the Euler class of their direct sum is given by
e(EF)=e(E)\smilee(F).
• Normalization: If
E
possesses a nowhere-zero section, then
e(E)=0
.
• Orientation: If
\overline{E}
is
E
with the opposite orientation, then
e(\overline{E})=-e(E)
.
Note that "Normalization" is a distinguishing feature of the Euler class. The Euler class obstructs the existence of a non-vanishing section in the sense that if
e(E)0
then
E
has no non-vanishing section.
Also unlike other characteristic classes, it is concentrated in a degree which depends on the rank of the bundle:
e(E)\inHr(X)
. By contrast, the Stiefel Whitney classes
wi(E)
live in
Hi(X;Z/2)
independent of the rank of
E
. This reflects the fact that the Euler class is unstable, as discussed below.
### Vanishing locus of generic section
The Euler class corresponds to the vanishing locus of a section of
E
in the following way. Suppose that
X
is an oriented smooth manifold of dimension
d
. Let
\sigma\colonX\toE
be a smooth section that transversely intersects the zero section. Let
Z\subseteqX
be the zero locus of
\sigma
. Then
Z
is a codimension
r
submanifold of
X
which represents a homology class
[Z]\inHd-r(X;Z)
and
e(E)
is the Poincaré dual of
[Z]
.
### Self-intersection
For example, if
Y
is a compact submanifold, then the Euler class of the normal bundle of
Y
in
X
is naturally identified with the self-intersection of
Y
in
X
.
## Relations to other invariants
In the special case when the bundle E in question is the tangent bundle of a compact, oriented, r-dimensional manifold, the Euler class is an element of the top cohomology of the manifold, which is naturally identified with the integers by evaluating cohomology classes on the fundamental homology class. Under this identification, the Euler class of the tangent bundle equals the Euler characteristic of the manifold. In the language of characteristic numbers, the Euler characteristic is the characteristic number corresponding to the Euler class.
Thus the Euler class is a generalization of the Euler characteristic to vector bundles other than tangent bundles. In turn, the Euler class is the archetype for other characteristic classes of vector bundles, in that each "top" characteristic class equals the Euler class, as follows.
Modding out by 2 induces a map
Hr(X,Z)\toHr(X,Z/2Z).
The image of the Euler class under this map is the top Stiefel-Whitney class wr(E). One can view this Stiefel-Whitney class as "the Euler class, ignoring orientation".
Any complex vector bundle E of complex rank d can be regarded as an oriented, real vector bundle E of real rank 2d. The Euler class of E is given by the highest dimensional Chern class
e(E)=cd(E)\inH2d(X)
### Squares to top Pontryagin class
The Pontryagin class
pr(E)
is defined as the Chern class of the complexification of E:
pr(E)=c2r(CE)
.
The complexification
CE
is isomorphic as an oriented bundle to
EE
. Comparing Euler classes, we see that
e(E)\smilee(E)=e(EE)=e(EC)=cr(EC)\inH2r(X,Z).
If the rank r of E is even then
e(E)\smilee(E)=cr(E)=pr/2(E)
where
pr/2(E)
is the top dimensional Pontryagin class of
E
.
### Instability
A characteristic class
c
is stable if
c(E\underline{R}1)=c(E)
where
\underline{R}1
is a rank one trivial bundle.Unlike most other characteristic classes, the Euler class is unstable. In fact,
e(E\underline{R}1)=e(E)\smilee(\underline{R}1)=0
.
e\inHk(BSO(k))
. The unstability of the Euler class shows that it is not the pull-back of a class in
Hk(BSO(k+1))
under the inclusion
BSO(k)\toBSO(k+1)
.
This can be seen intuitively in that the Euler class is a class whose degree depends on the dimension of the bundle (or manifold, if the tangent bundle): the Euler class is an element of
Hd(X)
where
d
is the dimension of the bundle, while the other classes have a fixed dimension (e.g., the first Stiefel-Whitney class is an element of
H1(X)
).
The fact that the Euler class is unstable should not be seen as a "defect": rather, it means that the Euler class "detects unstable phenomena". For instance, the tangent bundle of an even dimensional sphere is stably trivial but not trivial (the usual inclusion of the sphere
Sn\subseteqRn+1
has trivial normal bundle, thus the tangent bundle of the sphere plus a trivial line bundle is the tangent bundle of Euclidean space, restricted to
Sn
, which is trivial), thus other characteristic classes all vanish for the sphere, but the Euler class does not vanish for even spheres, providing a non-trivial invariant.
## Examples
### Spheres
The Euler characteristic of the n-sphere Sn is:
\chi(Sn)=1+(-1)n=\begin{cases} 2&neven\\ 0&nodd. \end{cases}
Thus, there is no non-vanishing section of the tangent bundle of even spheres (this is known as the Hairy ball theorem). In particular, the tangent bundle of an even sphere is nontrivial—i.e.,
S2n
is not a parallelizable manifold, and cannot admit a Lie group structure.
For odd spheres, S2n−1R2n, a nowhere vanishing section is given by
(x2,-x1,x4,-x3,...,x2n,-x2n-1)
which shows that the Euler class vanishes; this is just n copies of the usual section over the circle.
As the Euler class for an even sphere corresponds to
2[S2n]\inH2n(S2n,Z)
, we can use the fact that the Euler class of a Whitney sum of two bundles is just the cup product of the Euler classes of the two bundles to see that there are no other subbundles of the tangent bundle than the tangent bundle itself and the null bundle, for any even-dimensional sphere.
Since the tangent bundle of the sphere is stably trivial but not trivial, all other characteristic classes vanish on it, and the Euler class is the only ordinary cohomology class that detects non-triviality of the tangent bundle of spheres: to prove further results, one must use secondary cohomology operations or K-theory.
#### Circle
The cylinder is a line bundle over the circle, by the natural projection
R1 x S1\toS1
. It is a trivial line bundle, so it possesses a nowhere-zero section, and so its Euler class is 0. It is also isomorphic to the tangent bundle of the circle; the fact that its Euler class is 0 corresponds to the fact that the Euler characteristic of the circle is 0. |
## Putnam, Hilary Whitehall
Compute Distance To:
Author ID: putnam.hilary-w Published as: Putnam, Hilary; Putnam, H.; Putnam, Hilary W. External Links: MGP · Wikidata · Math-Net.Ru · dblp · GND · IdRef
Documents Indexed: 41 Publications since 1957, including 3 Books 3 Contributions as Editor · 10 Further Contributions Biographic References: 2 Publications Co-Authors: 19 Co-Authors with 24 Joint Publications 382 Co-Co-Authors
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### Co-Authors
23 single-authored 7 Davis, Martin David 3 Hensel, G. 3 Quine, Willard Van Orman 2 Baaz, Matthias 2 Boolos, George S. 2 Finkelstein, David Ritz 2 Gudder, Stanley P. 2 Harper, Charles L. jun. 2 Hooker, Clifford A. 2 Jauch, Josef Maria 2 Kreisel, Georg 2 Leeds, Stephen 2 Omodeo, Eugenio Giovanni 2 Papadimitriou, Christos Harilaos 2 Policriti, Alberto 2 Scott, Dana Stewart 2 von Neumann, John 2 Wang, Hao 1 Abernathy, Robert 1 Agassi, Joseph 1 Ayer, A. J. 1 Benacerraf, Paul 1 Bernays, Paul 1 Beth, Evert Willem 1 Birkhoff, Garrett 1 Boyd, Richard K. 1 Brouwer, Luitzen Egbertus Jan 1 Bub, Jeffrey 1 Capek, Milic 1 Carnap, Rudolf 1 Carson, Daniel F. 1 Catlin, Donald E. 1 Chao, Ren 1 Chomsky, Noam Avram 1 Cohen, I. Bernard 1 Cohen, Robert S. 1 Curry, Haskell Brooks 1 Davydov, G. V. 1 Dennett, Daniel C. 1 Dishkant, Herman 1 Dreben, Burton 1 Dummett, Michael Anthony Eardley 1 Dunham, Bradford 1 Enderton, Herbert B. 1 Feyerabend, Paul K. 1 Finch, Peter D. 1 Foulis, David James 1 Frege, Gottlob 1 Fridshal, R. 1 Friedman, Michael Lee 1 Fürth, Reinhold 1 Gelernter, Herbert L. 1 Gilmore, Paul C. 1 Gleason, Andrew Mattei 1 Gleason, H. A. jun. 1 Gödel, Kurt 1 Goodfield, June 1 Goodman, Nelson 1 Greechie, Richard J. 1 Grünbaum, Adolf 1 Grunstra, Bernard R. 1 Halle, Morris 1 Harary, Frank 1 Hardegree, Gary M. 1 Havas, Peter 1 Heelan, Patrick Aidan 1 Hempel, Carl Gustav 1 Herzberger, Hans G. 1 Heyting, Arend 1 Hilbert, David 1 Hiz, Henry 1 Hockett, Charles F. 1 Holland, Samuel Shaheen jun. 1 Jakobson, Roman 1 Kamber, A. 1 Kanger, Stig 1 Kochen, Simon 1 Kripke, Saul A. 1 Kronfli, N. S. 1 Lambek, Joachim 1 Logemann, George 1 Łoś, Jerzy 1 Loveland, Donald W. 1 Luckham, David C. 1 Lukas, Joan D. 1 Mandelbrot, Benoit B. 1 Markovic, Mihailo I. 1 Maslov, S. Yu. 1 Mayr, Ernst 1 Meltzer, Bernard 1 Mints, Grigoriĭ Efroimovich 1 Newell, Allen 1 North, J. H. 1 Oettinger, Anthony G. 1 Orevkov, V. P. 1 Petersen, Aage 1 Peterson, Gordon E. 1 Piron, Constantin 1 Poincaré, Henri 1 Pools, J. C. T. ...and 37 more Co-Authors
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### Serials
7 The Journal of Symbolic Logic 3 Archiv für Mathematische Logik und Grundlagenforschung 3 Notre Dame Journal of Formal Logic 2 Fundamenta Mathematicae 2 Proceedings of the American Mathematical Society 2 Synthese 2 Transactions of the American Mathematical Society 2 Historia Mathematica 2 The University of Western Ontario Series in Philosophy of Science 1 American Mathematical Monthly 1 The British Journal for the Philosophy of Science 1 Dialectica 1 Illinois Journal of Mathematics 1 Journal of the Association for Computing Machinery 1 Logic Journal of the IGPL 1 Annals of Mathematics. Second Series 1 Proceedings of Symposia in Applied Mathematics 1 Synthese Library
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### Fields
34 Mathematical logic and foundations (03-XX) 11 General and overarching topics; collections (00-XX) 11 History and biography (01-XX) 5 Quantum theory (81-XX) 3 Computer science (68-XX) 2 Number theory (11-XX) 1 Mechanics of particles and systems (70-XX) 1 Relativity and gravitational theory (83-XX) 1 Game theory, economics, finance, and other social and behavioral sciences (91-XX)
### Citations contained in zbMATH Open
41 Publications have been cited 832 times in 762 Documents Cited by Year
A computing procedure for quantification theory. Zbl 0212.34203
Davis, M.; Putnam, H.
1960
The decision problem for exponential diophantine equations. Zbl 0111.01003
Davis, Martin; Putnam, Hilary; Robinson, Julia
1961
Trial and error predicates and the solution to a problem of Mostowski. Zbl 0193.30102
Putnam, H.
1965
Philosophy of mathematics. Selected readings. 2nd ed. Zbl 0548.03002
1983
Eine Unableitbarkeitsbeweismethode für den intuitionistischen Aussagenkalkül. Zbl 0079.00702
Kreisel, Georg; Putnam, H.
1957
Models and reality. Zbl 0443.03003
Putnam, Hilary
1980
The logico-algebraic approach to quantum mechanics. Vol. I: Historical evolution. Zbl 0429.03045
1975
Degrees of unsolvability of constructible sets of integers. Zbl 0188.32701
Boolos, G.; Putnam, H.
1968
Mathematics, matter and method. Philosophical papers. Vol. 1. Zbl 0311.00035
Putnam, Hilary
1975
What is mathematical truth? Zbl 0325.02004
Putnam, Hilary
1975
Decidability and essential undecidability. Zbl 0078.24501
Putnam, Hilary
1957
Mathematics, matter and method. Philosophical papers, Vol. 1. 2nd ed. Zbl 0426.00022
Putnam, Hilary
1979
Peirce the logician. Zbl 0487.01012
Putnam, Hilary
1982
A note on the hyperarithmetical hierarchy. Zbl 0205.30803
Enderton, H. B.; Putnam, H.
1971
Recursively enumerable classes and their application to recursive sequences of formal theories. Zbl 0242.02046
Pour-El, Marian Boykan; Putnam, Hilary
1965
A philosopher looks at quantum mechanics (again). Zbl 1098.81011
Putnam, Hilary
2005
Diophantine sets over polynomial rings. Zbl 0113.00604
Davis, M.; Putnam, H.
1963
Automation of reasoning. 1: Classical papers on computational logic 1957–1966. Zbl 0567.03001
1983
Quantum logic, conditional probability, and interference. Zbl 0402.03016
Friedman, Michael; Putnam, Hilary
1978
Nonstandard models and Kripke’s proof of the Gödel theorem. Zbl 1005.03054
Putnam, Hilary
2000
What theories are not. Zbl 0147.24702
Putnam, H.
1962
The logico-algebraic approach to quantum mechanics. Vol. II: Contemporary consolidation. Zbl 0429.03046
1979
A recursion-theoretic characterization of the ramified analytical hierarchy. Zbl 0207.01203
Boyd, Richard; Hensel, G.; Putnam, H.
1969
Meaning as functional classification. (A perspective on the relation of syntax to semantics.) With comments by Daniel Dennett, Hilary Putnam, Saul Kripke and W. V. Quine. Zbl 0308.02012
Sellars, Wilfrid
1974
Philosophical papers. Vol. 2: Mind, language and reality. Zbl 0485.01026
Putnam, Hilary
1979
Structure of language and its mathematical aspects. Zbl 0111.16102
1961
Exact separation of recursively enumerable sets within theories. Zbl 0118.25201
Putnam, H.; Smullyan, R. M.
1960
A note on constructible sets of integers. Zbl 0192.04303
Putnam, H.
1963
How to think quantum-logically. Zbl 0338.02003
Putnam, Hilary
1974
After Gödel. Zbl 1111.03002
Putnam, Hilary
2006
On the notational independence of various hierarchies of degrees of unsolvability. Zbl 0137.00903
Hensel, G.; Putnam, H.
1965
Normal models and the field $$\sum_ 1^ *$$. Zbl 0193.30201
Hensel, G.; Putnam, H.
1969
An intrinsic characterization of the hierarchy of constructible sets of integers. Zbl 0234.02027
Leeds, Stephen; Putnam, Hilary
1971
On hierarchies and systems of notations. Zbl 0237.02011
Putnam, Hilary
1964
Kurt Gödel and the foundations of mathematics. Horizons of truth. Zbl 1253.00009
2011
On families of sets represented in theories. Zbl 0126.02101
Putnam, H.
1964
Paradox revisited. II: Sets – a case of all or none? Zbl 0986.03011
Putnam, Hilary
2000
Proceedings of the Boston colloquium for the philosophy of science 1966/1968. Zbl 0172.28503
1969
Recursive functions and hierarchies. Zbl 0268.02024
Putnam, Hilary
1973
Solution to a problem of Gandy’s. Zbl 0325.02028
Leeds, Stephen; Putnam, Hilary
1974
How to think quantum-logically. Zbl 0328.02003
Putnam, Hilary
1976
Kurt Gödel and the foundations of mathematics. Horizons of truth. Zbl 1253.00009
2011
After Gödel. Zbl 1111.03002
Putnam, Hilary
2006
A philosopher looks at quantum mechanics (again). Zbl 1098.81011
Putnam, Hilary
2005
Nonstandard models and Kripke’s proof of the Gödel theorem. Zbl 1005.03054
Putnam, Hilary
2000
Paradox revisited. II: Sets – a case of all or none? Zbl 0986.03011
Putnam, Hilary
2000
Philosophy of mathematics. Selected readings. 2nd ed. Zbl 0548.03002
1983
Automation of reasoning. 1: Classical papers on computational logic 1957–1966. Zbl 0567.03001
1983
Peirce the logician. Zbl 0487.01012
Putnam, Hilary
1982
Models and reality. Zbl 0443.03003
Putnam, Hilary
1980
Mathematics, matter and method. Philosophical papers, Vol. 1. 2nd ed. Zbl 0426.00022
Putnam, Hilary
1979
The logico-algebraic approach to quantum mechanics. Vol. II: Contemporary consolidation. Zbl 0429.03046
1979
Philosophical papers. Vol. 2: Mind, language and reality. Zbl 0485.01026
Putnam, Hilary
1979
Quantum logic, conditional probability, and interference. Zbl 0402.03016
Friedman, Michael; Putnam, Hilary
1978
How to think quantum-logically. Zbl 0328.02003
Putnam, Hilary
1976
The logico-algebraic approach to quantum mechanics. Vol. I: Historical evolution. Zbl 0429.03045
1975
Mathematics, matter and method. Philosophical papers. Vol. 1. Zbl 0311.00035
Putnam, Hilary
1975
What is mathematical truth? Zbl 0325.02004
Putnam, Hilary
1975
Meaning as functional classification. (A perspective on the relation of syntax to semantics.) With comments by Daniel Dennett, Hilary Putnam, Saul Kripke and W. V. Quine. Zbl 0308.02012
Sellars, Wilfrid
1974
How to think quantum-logically. Zbl 0338.02003
Putnam, Hilary
1974
Solution to a problem of Gandy’s. Zbl 0325.02028
Leeds, Stephen; Putnam, Hilary
1974
Recursive functions and hierarchies. Zbl 0268.02024
Putnam, Hilary
1973
A note on the hyperarithmetical hierarchy. Zbl 0205.30803
Enderton, H. B.; Putnam, H.
1971
An intrinsic characterization of the hierarchy of constructible sets of integers. Zbl 0234.02027
Leeds, Stephen; Putnam, Hilary
1971
A recursion-theoretic characterization of the ramified analytical hierarchy. Zbl 0207.01203
Boyd, Richard; Hensel, G.; Putnam, H.
1969
Normal models and the field $$\sum_ 1^ *$$. Zbl 0193.30201
Hensel, G.; Putnam, H.
1969
Proceedings of the Boston colloquium for the philosophy of science 1966/1968. Zbl 0172.28503
1969
Degrees of unsolvability of constructible sets of integers. Zbl 0188.32701
Boolos, G.; Putnam, H.
1968
Trial and error predicates and the solution to a problem of Mostowski. Zbl 0193.30102
Putnam, H.
1965
Recursively enumerable classes and their application to recursive sequences of formal theories. Zbl 0242.02046
Pour-El, Marian Boykan; Putnam, Hilary
1965
On the notational independence of various hierarchies of degrees of unsolvability. Zbl 0137.00903
Hensel, G.; Putnam, H.
1965
On hierarchies and systems of notations. Zbl 0237.02011
Putnam, Hilary
1964
On families of sets represented in theories. Zbl 0126.02101
Putnam, H.
1964
Diophantine sets over polynomial rings. Zbl 0113.00604
Davis, M.; Putnam, H.
1963
A note on constructible sets of integers. Zbl 0192.04303
Putnam, H.
1963
What theories are not. Zbl 0147.24702
Putnam, H.
1962
The decision problem for exponential diophantine equations. Zbl 0111.01003
Davis, Martin; Putnam, Hilary; Robinson, Julia
1961
Structure of language and its mathematical aspects. Zbl 0111.16102
1961
A computing procedure for quantification theory. Zbl 0212.34203
Davis, M.; Putnam, H.
1960
Exact separation of recursively enumerable sets within theories. Zbl 0118.25201
Putnam, H.; Smullyan, R. M.
1960
Eine Unableitbarkeitsbeweismethode für den intuitionistischen Aussagenkalkül. Zbl 0079.00702
Kreisel, Georg; Putnam, H.
1957
Decidability and essential undecidability. Zbl 0078.24501
Putnam, Hilary
1957
all top 5
### Cited by 1,015 Authors
12 Matiyasevich, Yuriĭ Vladimirovich 8 Plaisted, David Alan 6 Bonacina, Maria Paola 6 Eisenträger, Kirsten 6 Friedman, Sy-David 6 Marques-Silva, João P. 6 Schaub, Torsten H. 6 Szeider, Stefan 6 Van Gelder, Allen 5 Arslanov, Marat M. 5 Becker, Bernd 5 Beyersdorff, Olaf 5 Chen, Jian-er 5 Kelly, Kevin T. 5 Miglioli, Pierangelo 5 Miller, Russell G. 5 Putnam, Hilary Whitehall 5 Sorbi, Andrea 5 Ternullo, Claudio 4 Ábrahám, Erika 4 Biere, Armin 4 Bueno, Otávio 4 de Moura, Leonardo 4 Dose, Titus 4 Eggers, Andreas 4 Giunchiglia, Enrico 4 Glaßer, Christian 4 Heule, Marijn J. H. 4 Hooker, John N. jun. 4 Itsykson, Dmitry M. 4 Jones, James P. 4 Kupferschmid, Stefan 4 Lauria, Massimo 4 Lê Văn Băng 4 Lempp, Steffen 4 Lynce, Inês 4 Myasnikov, Alexei G. 4 Nordström, Jakob 4 Pasten, Hector V. 4 Sabharwal, Ashish 4 Schulte, Oliver 4 Selman, Bart 4 Stephan, Frank 4 Teige, Tino 4 Yu, Liang 3 Anellis, Irving H. 3 Badaev, Serikzhan A. 3 Beigel, Richard 3 Bringsjord, Selmer 3 Bruni, Renato 3 Bryant, Randal E. 3 Chong, Chi Tat 3 Cook, Roy T. 3 Cooper, Stuart Barry 3 Ferrari, Mauro 3 Franco, John V. 3 Fränzle, Martin 3 Garreta, Albert 3 Gebser, Martin 3 Hirsch, Edward A. 3 Järvisalo, Matti 3 Jeroslow, Robert G. 3 Jonsson, Peter A. 3 Kanamori, Akihiro 3 Kaufmann, Benjamin 3 Kirousis, Lefteris Miltiades 3 Kleine Büning, Hans 3 Kratsch, Dieter 3 Kullmann, Oliver 3 Li, Chumin 3 Linnebo, Øystein 3 Nieuwenhuis, Robert 3 Omodeo, Eugenio Giovanni 3 Ovchinnikov, Denis 3 Pitassi, Toniann 3 Sanders, Sam 3 van Maaren, Hans 3 Vardi, Moshe Ya’akov 3 Visser, Albert 3 Wang, Jianxin 3 Wang, Jinchang 3 Welch, Philip D. 3 Xu, Chao 3 Yamaleev, Mars Mansurovich 3 Zakharyaschev, Michael Viktorovich 3 Zhao, Xishun 2 Achlioptas, Dimitris 2 Alekhnovich, Michael 2 Alexander, Samuel Allen 2 Antos, Carolin 2 Armando, Alessandro 2 Arrigoni, Tatiana 2 Badban, Bahareh 2 Barth, Dominik 2 Baxa, Christoph 2 Beame, Paul W. 2 Beck, Moritz 2 Benhamou, Belaid 2 Bernardi, Claudio 2 Bidoit, Nicole ...and 915 more Authors
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### Cited in 147 Serials
53 Theoretical Computer Science 37 Journal of Automated Reasoning 35 The Journal of Symbolic Logic 26 Artificial Intelligence 21 Discrete Applied Mathematics 21 Synthese 21 Annals of Pure and Applied Logic 20 Journal of Philosophical Logic 19 Studia Logica 17 Annals of Mathematics and Artificial Intelligence 15 Transactions of the American Mathematical Society 14 The Bulletin of Symbolic Logic 11 History and Philosophy of Logic 11 Information and Computation 10 The Mathematical Intelligencer 9 Information Processing Letters 9 Proceedings of the American Mathematical Society 9 Archive for Mathematical Logic 8 Journal of Soviet Mathematics 8 Journal of Mathematical Sciences (New York) 8 Studies in History and Philosophy of Science. Part B. Studies in History and Philosophy of Modern Physics 7 Information Sciences 7 Annals of Operations Research 7 Erkenntnis 6 Notre Dame Journal of Formal Logic 6 Journal of Symbolic Computation 6 Foundations of Science 5 Archiv für Mathematische Logik und Grundlagenforschung 5 Journal of Computer and System Sciences 5 JETAI. Journal of Experimental & Theoretical Artificial Intelligence 5 European Journal of Operational Research 5 Foundations of Physics 5 Logica Universalis 4 Algebra and Logic 4 Applied Mathematics and Computation 4 Journal of Algebra 4 Formal Methods in System Design 4 Constraints 4 Theory and Practice of Logic Programming 4 The Review of Symbolic Logic 3 Acta Informatica 3 Computers & Mathematics with Applications 3 International Journal of General Systems 3 International Journal of Theoretical Physics 3 Israel Journal of Mathematics 3 Advances in Mathematics 3 SIAM Journal on Computing 3 Operations Research Letters 3 Journal of Computer Science and Technology 3 Computers & Operations Research 3 Logic and Logical Philosophy 3 Logical Methods in Computer Science 2 Archive for History of Exact Sciences 2 Discrete Mathematics 2 BIT 2 Computing 2 Journal of Number Theory 2 Siberian Mathematical Journal 2 Advances in Applied Mathematics 2 Mathematical Social Sciences 2 International Journal of Parallel Programming 2 Journal of the American Mathematical Society 2 Mathematical and Computer Modelling 2 Random Structures & Algorithms 2 Bulletin of the American Mathematical Society. New Series 2 RAIRO. Informatique Théorique et Applications 2 Mathematical Programming. Series A. Series B 2 Computational Complexity 2 Journal of Applied Non-Classical Logics 2 Mathematical Logic Quarterly (MLQ) 2 New Journal of Physics 2 International Journal of Applied Mathematics and Computer Science 2 Lobachevskii Journal of Mathematics 2 International Studies in the Philosophy of Science 2 Journal of Discrete Algorithms 2 Journal of Applied Logic 2 Discrete Optimization 2 Proceedings of the Steklov Institute of Mathematics 2 Axiomathes 2 Proceedings of the Royal Society of London. A. Mathematical, Physical and Engineering Sciences 1 Educational Studies in Mathematics 1 Jahresbericht der Deutschen Mathematiker-Vereinigung (DMV) 1 Journal of the Franklin Institute 1 Mathematical Notes 1 Mathematical Proceedings of the Cambridge Philosophical Society 1 Mathematische Semesterberichte 1 Reviews in Mathematical Physics 1 Science & Education 1 Annali di Matematica Pura ed Applicata. Serie Quarta 1 Bulletin de la Société Mathématique de France 1 Fuzzy Sets and Systems 1 International Journal of Computer & Information Sciences 1 International Journal of Game Theory 1 Journal of the Association for Computing Machinery 1 Le Matematiche 1 Mathematische Annalen 1 Mathematics and Computers in Simulation 1 Mathematische Nachrichten 1 Mathematica Slovaca 1 Publications of the Research Institute for Mathematical Sciences, Kyoto University ...and 47 more Serials
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### Cited in 34 Fields
417 Mathematical logic and foundations (03-XX) 363 Computer science (68-XX) 51 General and overarching topics; collections (00-XX) 49 Number theory (11-XX) 44 Operations research, mathematical programming (90-XX) 34 History and biography (01-XX) 23 Quantum theory (81-XX) 21 Combinatorics (05-XX) 15 Information and communication theory, circuits (94-XX) 8 Group theory and generalizations (20-XX) 8 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 7 Order, lattices, ordered algebraic structures (06-XX) 7 Field theory and polynomials (12-XX) 7 Algebraic geometry (14-XX) 7 Probability theory and stochastic processes (60-XX) 7 Statistics (62-XX) 6 Numerical analysis (65-XX) 3 Commutative algebra (13-XX) 3 Biology and other natural sciences (92-XX) 2 General algebraic systems (08-XX) 2 Category theory; homological algebra (18-XX) 2 Real functions (26-XX) 2 General topology (54-XX) 1 Associative rings and algebras (16-XX) 1 Measure and integration (28-XX) 1 Ordinary differential equations (34-XX) 1 Partial differential equations (35-XX) 1 Functional analysis (46-XX) 1 Convex and discrete geometry (52-XX) 1 Algebraic topology (55-XX) 1 Mechanics of particles and systems (70-XX) 1 Fluid mechanics (76-XX) 1 Relativity and gravitational theory (83-XX) 1 Systems theory; control (93-XX)
### Wikidata Timeline
The data are displayed as stored in Wikidata under a Creative Commons CC0 License. Updates and corrections should be made in Wikidata. |
In the road to track more effectively my habits, and in the context of this article, my programming habits I put together a simple system for tracking the git commits I do on the various projects I'm working on. As always emacs, magit and org-mode make this pretty easy, and that's the reason that I'm so heavily invested in emacs.
For the implementation of the above goal I use the orgit library that enhances the org-link system with links for views in magit. This means that we can create hyperlinks in org-mode documents that lead to specific magit buffers, hence creating links to descriptions of commits, branches, revision etc.
To do so I configured the orgit package, using use-package:
(use-package orgit
:ensure t
:hook (git-commit-post-finish . orgit-capture-after-commit)
:config
(defun orgit-capture-after-commit ()
"Capture the commit into an org-mode file, decided in the approptriate capture template"
(let* ((repo (abbreviate-file-name default-directory))
(summary (substring-no-properties (magit-format-rev-summary rev)))
(desc (format "%s (%s)" summary repo)))
(org-capture nil "g"))))
The important thing here is that we added a hook on finishing a commit, that collects info for the commit, pushes a orgit link for it in the kill-ring and then invokes org-capture. The relevant template under the shortcut "g" is:
("g" "Git commit" entry (file ,org-default-commit-file)
"* %c \n:PROPERTIES:\n:CLOSED: %T\n:END:\n\n"
:immediate-finish t)
Briefly, this template takes the first element of the kill-ring and creates a new entry in the org-default-commit-file with a CLOSED property and an active timestamp. This is important in order to have them showing up in the agenda, which also requires to put the org-default-commit-file in the agenda files.
Now every commit I make in any repo in my computer, using magit, will create an entry like this:
And at the same time I can review all the commits I've done in the agenda: |
# Why are photoelectrons emitted in the direction of incident photons?
In the experiments for photoelectric emission, the light is incident on one face of the emitting plate, for example the anode, when determining the stopping potential. The electrons are emitted by this face of the plate. Why are the electrons emitted in the direction of the incident light, and not opposite to it?
I have read a little about photons (the key words being, a little), and from what i gather, they have an intrinsic momentum $p=h/\lambda$, which is transferred entirely to the particle that absorbs it (along with the energy $h\nu$). This should imply that the electrons try to move further into the cathode, and not escape it! What is the fault in the reasoning?
Thanks!
• They are emitted from the surface, since that is where they can be emitted from. And, that is where the photons can come from. – Jon Custer Jun 3 '16 at 3:12
• In practice you can find both reflection mode and transmission mode photocathodes. With a proper electric field both have reasonable quantum efficiency, with reflection mode cathodes being of lower resistance, i.e. they can handle large photocurrents better. In a metal cathode without proper biasing most electrons will be emitted towards the bulk (because of momentum conservation) and be lost. That's why semiconducting photocathodes are in practice far superior to bulk metal cathodes. – CuriousOne Jun 3 '16 at 3:14
Your reasoning is quite correct, and you can see exactly this phenomenon in a photomultiplier tube. The photomutiplier tube uses very thin metal sheets, and when a photon strikes the sheet the primary photoelectron is emitted in the same direction as the incident photon and escapes from the far side of the sheet:
The quantum yield for this process is close to 100% i.e. almost every photon ejects a photoelectron. However the quantum yield in a typical photoelectric experiment is about $10^{-5}$ to $10^{-6}$ i.e. up to a factor of a million times poorer. The reason for this is because the initial photoelectron is emitted travelled down into the body of the metal. For the photoelectron to escape it has to backscatter off other electrons in the metal and ricochet back to the surface without losing so much energy it can no longer escape. This is a very low probability process so the overall quantum yield is very low.
• Thank you very much for your answer! The standard photoelectric experiments would seem very inefficient indeed! – FreezingFire Jun 4 '16 at 4:06
In the experiments for photoelectric emission, the light is incident on one face of the emitting plate, for example the anode, when determining the stopping potential. The electrons are emitted by this face of the plate. Why are the electrons emitted in the direction of the incident light, and not opposite to it?
In this answer the energy momentum balance is solved and one sees that in the photoelectric effect most of the incoming momentum of the incident photon is taken up by the much heavier nucleus. The electrons get detached from the atom, or the lattice where they were bound but their direction will depend on the fields around the material from which they were detached.
So it will depend how one will gather those electrons that the photons detach, which are close to the surface or within it free with very small momentum.
The design of photocathodes is based on the 3-step model. In this approximation, the photoelectron emission is divided in three separate steps: (1) photoabsorption or excitation, (2) transport of excited electrons to the surface, and (3) passage through the surface and escape. To maximize quantum efficiency, attention is given to each one of the steps. Large excitation results from materials with large density of electrons states close to the vacuum level, e.g., high electron density and low photoelectron threshold. The excited electrons may lose energy during transport to the surface and so their flux is attenuated. The attenuation is less for materials where inelastic electron scattering is hindered, like insulators or wide band-gap semiconductors. Finally, a low surface barrier is required for maximum escape.
In the simple experiment for demonstrating the effect, it is easier to attract the electrons on the side of the incoming photons.
Please read up for more detail. This figure shows a simple diagram where the cathode is the element on which the light shines and releases photoelectrons.
Both the cathode and the anode are in a closed vacuum. For the polarity of V shown in the figure the voltage on the anode tends to repel photoelectrons emitted by the cathode and it is called a "retarding" voltage. With a reverse polarity on V electrons will be attracted to the anode.
• Thank you for your answer! You said that the direction of electrons will depend on the fields around the material. But then, if the electron is emitted by the anode, it should be repelled by the cathode! This means the electron would simply be detached from the nucleus, and travel further into the anode...That is not the case, as if the potential of the cathode is less (magnitude) than stopping potential, electrons do reach the cathode. What makes them reach the cathode then? – FreezingFire Jun 3 '16 at 10:30
• See my edit, the element on which the light shines is called the cathode, and that is where the photoelectrons appear. The anode either attracts them or repels them depending on the voltages/fields. Photoelectrons do travel within the surfaces, but those are not the ones one can detect, because they will again be absorbed at some lattice energy level. – anna v Jun 3 '16 at 11:13 |
Please, try EDU on Codeforces! New educational section with videos, subtitles, texts, and problems. ×
### Cache's blog
By Cache, history, 4 weeks ago, ,
We’re halfway through the year and its time for our sixth Long Challenge of the year 2020. We invite you to participate in CodeChef’s June Long Challenge, this Friday, 5th June, 15:00 IST onwards The contest will be open for 10 days i.e. until 15th June.
Also, if you have some original and engaging problem ideas, and you’re interested in them being used in CodeChef's contests, you can share them here.
Joining me on the problem setting panel are: - Setters: Sahil sahi1422 Chimnani , Naman smartnj Jain , Vikas _wildfire_ Pandey , Vinit vitz Solanki , Taranpreet taran_1407 Singh, Raja raja1999 Vardhan Reddy , Sofiia sonechko Melnyk , Ildar 300iq Gainullin , Arthur arthur.nascimento Nascimento
• Admin : Teja Cache Vardhan Reddy
• Tester: Felipe fmota Mota
• Editorialist & Post-Contest Streaming: Rajarshi RestingRajarshi Basu
• Statement Verifier: Jakub Xellos Safin
• Mandarin Translator: Hanlin I_love_PHP Ren
• Vietnamese Translator: Team VNOI
• Russian Translator: Fedor Fedosik Korobeinikov
• Bengali Translator: Mohammad solaimanope Solaiman
• Hindi Translator: Akash Shrivastava
Prizes
Top 20 performers in the Indian category and top 10 performers in the Global category will get CodeChef laddus, with which the winners can claim cool CodeChef goodies. First to solve each problem except challenge — 100 laddus. Know more here
Good Luck!
Hope to see you participating!!
Happy Programming !!
• +119
» 4 weeks ago, # | +9 Contest starts in 10 minutes!
» 4 weeks ago, # | ← Rev. 2 → -8 I hope that the problems are tasty!
» 3 weeks ago, # | +75 User hehaodele, currently in rank 8, solved the problem PPARTS by asking in the CF forum in the last 5 hours, pretending it is the problem he came up with. Now the article is deleted, but there is already some witness (Um_nik). Please investigate and disqualify him.
• » » 3 weeks ago, # ^ | +36 I am a witness too. Um_nik answered the question saying that he learned it 4 days ago.
• » » » 3 weeks ago, # ^ | -38 https://codeforces.com/blog/entry/61306?#comment-452948This appeared in previous contest and a well known combinatorics problem.https://www.codechef.com/PCO12020/problems/HELCARP
• » » » » 3 weeks ago, # ^ | +10 Nice fake account lmao. Doesn't matter if it appeared before, cheatin's cheatin'
• » » » » » 3 weeks ago, # ^ | ← Rev. 2 → -23 cheatin's cheatin' Agreed
• » » » » » 3 weeks ago, # ^ | -24 Nice fake account lmao. Thanks.
• » » » » 3 weeks ago, # ^ | +5 Wow not supporting anyone,but might be genuine mistake of this guy, as I solved this question and I too was curious how to optimise the solution, I didn't knew until it is long challenge question. Because, if someone wanted to cheat then he/she would have used fake account instead of original.Just a possibility wanna share.
• » » 3 weeks ago, # ^ | +38 Thanks, will escalate it.
• » » 3 weeks ago, # ^ | -44 Still participating in long-plagiarized-contests?
• » » 3 weeks ago, # ^ | +21 Thanks for reporting! We have removed the user from the ranklist, and blocked their account on CodeChef. We'll update here if they provide any reasonable explanation for what's happened.
• » » » 3 weeks ago, # ^ | ← Rev. 3 → -57 Are you going to block all those users who witnessed this/got AC after that blog?Don't let things change your good character. Stop participating in these plagiarized challenges.PS: CodeChef_admin Remove these ridiculous longs and start some Dinner/Break-Fast, post editorials at the right time.
• » » » » 3 weeks ago, # ^ | -30 CodeChef_admin Remove these ridiculous longs Noone asked a cheater for advice.
• » » » » » 3 weeks ago, # ^ | -28 So you think I am hehaodele xD, There exists many who hate Codechef Longs & its braggers in discuss
• » » » » » » 3 weeks ago, # ^ | -16 You admitted you are cheating.I never said you are hehaodele you only bought this here. its braggers in discuss You can do a lot of productive things other than opening discuss. There exists many who hate Codechef Longs Looking at you comments your reason for hating it is cheating which you are encouraging. You can simply stop participating in long. Codechef_admin is not coming to your house and forcing you to participate.
• » » » » » » » 3 weeks ago, # ^ | +23 You can simply stop participating in long.I did that mistake once. Never participated thereafter. BTW Thanks for the advice.
• » » 3 weeks ago, # ^ | +86 I can confirm that roughly 10 hours ago I answered a question by hehaodele and my answer can be used as a solution to PPARTS.
• » » » 2 weeks ago, # ^ | +6 Now that the contest is over can you tell your solution?I solved it by applying the xDln operator on the question which converted it into a simpler form.
• » » » » 2 weeks ago, # ^ | +29 https://judge.yosupo.jp/problem/sharp_p_subset_sumYou can watch submissions there.In short:We want first $n$ coefficients of $Q(x) = \prod_{k=1}^{n} (1 + x^{k})^{c_k}$Let's define $P(x) = \ln Q(x) = \sum_{k=1}^{n} c_k \ln (1 + x^k)$$\ln (1 + x) = \sum_{i=1}^{\infty} \frac{(-1)^{i + 1} x^{i}}{i}$$P(x) = \sum_{k=1}^{n} \sum_{i=1}^{n/k} \frac{(-1)^{i+1} x^{ik}}{i}$We can calculate $P(x)$ in $O(n \log n)$ since there $O(n \log n)$ summands. Then $Q(x) = \exp P(x)$ which can be calculated in $O(n \log n)$.
• » » » » » 2 weeks ago, # ^ | +3 Yeah thanks, I did almost exactly that but I took the derivative of $ln(1+x^k)$. And integrated it to get the same result you got.
• » » » » » 2 weeks ago, # ^ | ← Rev. 2 → +2 math.stackexchange link Here, there is a method of nth derivative of generating function, can we somehow apply the same method here,if we just want to Calculate for sum a_i * b_i =n, is it possible?
» 2 weeks ago, # | +19 Did anyone manage to fit $O(nlog^2n)$ in TL for DIFVAL? i.e without the dsu on tree approach.I tried segtree on dfs time array where each node is a persistent segtree, and ended up having to create 7e7 nodes per case. Was unable fit it in TL. :(
• » » 2 weeks ago, # ^ | +5 I used Sack (DSU on Tree) along with a persistent segtree to do it in Nlog²N.It ran in first attempt but I had done a few optimizations. Like adding the light subtrees by BFS instead of DFS (this is helpful because it guarantees minimum number of operations), also instead of removing elements from the segtree, I just applied the updates on the blank segtree (I can do this because it's persistent).Apart from this, my segtree was persistent as well as Dynamic.Here's the link to the solution: https://www.codechef.com/viewsolution/34062043
» 2 weeks ago, # | +49 CodeChef_admin Hey I think its the time to either reduce the duration of long challenge maybe to 2 or 3 days or stop conducting long challenge. Instead come with something like Dinner, Breakfast, Snacks or something else. I find two reason for this Plagiarism (that usually remain undetected as person change the name of variable ,use different templates etc) Long challenge sounds odd one out from Lunchtime and Cook-off, so contest like Snacks , Dinner, Break-Fast will sound far more logical.
• » » 2 weeks ago, # ^ | -21 reduce the duration of long challenge maybe to 2 or 3 days or stop conducting long challenge.We all know that these absolutely logical words will have no effect whatsoever. It's so sad to see Codechef this way. :facepalm:
» 2 weeks ago, # | +4 Top 20 performers in Indian category would mean top 20 in div1, or is top 10 from each division.My name is on 19th in the following list : https://www.codechef.com/rankings/JUNE20A?filterBy=Country%3DIndia&order=asc&sortBy=rank (filter :India, sort : ASC).Am I eligible for any laddus(assuming ranklist is unaffected after MOSS)? If yes then how many. Thanks!
• » » 2 weeks ago, # ^ | 0 But how codechef determine someone is Indian or not because one can change his/her location to India just to get free laddus (in case he is not in top 20 in global but is in top 20 Indians)
• » » » 2 weeks ago, # ^ | 0 Yes I also thought of this possibility, I'm unaware of how codechef chooses top 20 indians. |
# Ngô Quốc Anh
## August 16, 2010
### The Moser-Trudinger inequality for domains with holes
Filed under: PDEs — Tags: — Ngô Quốc Anh @ 2:42
In this entry, we are interested in the following result
Theorem (Moser-Trudinger’s inequality for domains with holes). Let $\Omega$ be a bounded smooth domain in $\mathbb R^2$. Let $S_1$ and $S_2$ be two subsets of $\overline \Omega$ satisfying
${\rm dist}(S_1,S_2) \geqslant \delta_0>0$
and let $\gamma_0$ be a number satisfying $\gamma_0 \in \left(0,\frac{1}{2}\right)$. Then for any $\varepsilon>0$, there exists a constant $c=c(\varepsilon, \delta_0, \gamma_0)>0$ such that
$\displaystyle\int_\Omega {{e^u}} \leqslant C\exp \left[ {\frac{1}{{32\pi - \varepsilon }}\int_\Omega {{{\left| {\nabla u} \right|}^2}} + C} \right]$
holds for all $u \in H_0^1(\Omega)$ satisfying
$\displaystyle\frac{{\int_{{S_1}} {{e^u}} }}{{\int_\Omega {{e^u}} }} \geqslant {\gamma _0}, \quad \frac{{\int_{{S_2}} {{e^u}} }}{{\int_\Omega {{e^u}} }} \geqslant {\gamma _0}$. |
"Arithmetic genus" of a plane curve singularity. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:33:46Z http://mathoverflow.net/feeds/question/122725 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122725/arithmetic-genus-of-a-plane-curve-singularity "Arithmetic genus" of a plane curve singularity. aglearner 2013-02-23T14:34:25Z 2013-02-24T13:39:21Z <p>I believe that the following questions are very basic, but I don't know how to get a reference. </p> <p>Consider a curve in the plane $C\in \mathbb C^2$ with a singularity at $0$ and suppose it is unibranch at zero (i.e. analytically irreducible). Then I guess one should be able to define "arithmetic genus defect" of the curve at $0$. Namely if one smooths analytically $C$, its geometric genus will grow by a positive number (in case of the cusp $x^2=y^3$ it will grow by one), and let us call this number the defect. </p> <p><strong>Question 1.</strong> Is this defect well defined (independent of a smoothing)? How is it called and how one should calculate it (say it terms of the local ring of $C$ at $0$)?</p> <p><strong>Question 2.</strong> Suppose we have an explicit local parametrisation of $C$ at $0$, say by two holomorphic functions $f(t), g(t)$ (polynomials if you wish). Is it possible to find this "defect" as a certain invariant of this pair of functions at $t=0$? </p> <p><em>Question 1 is settled in the answer of unknown and Question 2 in comments to it by Roy and Vivek</em></p> http://mathoverflow.net/questions/122725/arithmetic-genus-of-a-plane-curve-singularity/122738#122738 Answer by unknown (google) for "Arithmetic genus" of a plane curve singularity. unknown (google) 2013-02-23T17:43:47Z 2013-02-23T17:43:47Z <p>The difference between the geometric genus of the singularity and the geometric genus of a smoothing (this one being called the arithmetic genus of the singularity) is often called the delta invariant. If A is the local ring of the singularity, B its normalisation then the delta invariant is the dimension of the complex vector space B/A.</p> <p>It is rather easy to compute the delta invariant if one knows an equation f(x,y)=0 of the curve by a formula due to Milnor (see the book "Singularities of hypersurfaces") : $2 \delta = \mu + 1 - b$ where $\delta$ is the delta invariant, $\mu = dim_{\mathbb{C}} \mathbb{C}[[x,y]]/(\partial_{x}f, \partial_{y}f )$ and b is the number of branches. In the unibranch case, it is simply $2 \delta = \mu$ (example : for the cusp, $\delta = 1$, $\mu =2$).</p> http://mathoverflow.net/questions/122725/arithmetic-genus-of-a-plane-curve-singularity/122796#122796 Answer by Jérémy Blanc for "Arithmetic genus" of a plane curve singularity. Jérémy Blanc 2013-02-24T13:39:21Z 2013-02-24T13:39:21Z <p>Another way to compute the delta invariant in pratice: let $m_1$ be the multiplicity of the curve at the point $p_1$ you are looking for. You blow-up $p_1$, and look for singular points of the strict transform of the curve which lie on the exceptional curve obtained. Denote by $m_2,\dots,m_k$ the multiplicities obtained (which satisfy $m_2+\dots+m_k\le m_1$). Blow-up all these points. Repeat the process until the curve has no singular point infinitely near to $p_1$. You get a set of multiplcities $m_1,\dots,m_l$ (with $l\ge k$). The delta you want is exactly $\sum_{i=1}^l m_i(m_i-1)/2$. This is a direct consequence of adjunction formula.</p> <p>It depends of the situation, but sometimes it is easier to compute than the formula of Milnor. The multiplicities are easy to get from either the equation of the curve or a parametrisation. </p> |
27,800
pages
Operator: Power
ModPsi
TypeSelector
GroupSecondary Operators
Return typeNumber
Parameters
BaseNumber
PowerNumber
Technical details
Registry nameoperatorPower
Operator: Power is a Number Spell Piece added by Psi. It will return the value from a base and its exponent ( where is the base and is the exponent). For example, . As
Operator: Square Root is the only root function added by Psi, players may make their exponent the inverse of the number they want to take the root of. For example, if they wish to take the 4th root of a number, the exponent can be expressed as 1/4, or 0.25. It is unlocked in the lesson "Secondary Operators."
## Parameters
• Base: Requires a number.
• Exponent: Requires a number. |
Hardcover | $80.00 X | £66.95 | 736 pp. | 7 x 9 in | 68 figures, 4 tables | September 2014 | ISBN: 9780262027618 eBook |$80.00 X | September 2014 | ISBN: 9780262325585
Mouseover for Online Attention Data
## Overview
This book offers a unified, comprehensive, and up-to-date treatment of analytical and numerical tools for solving dynamic economic problems. The focus is on introducing recursive methods—an important part of every economist’s set of tools—and readers will learn to apply recursive methods to a variety of dynamic economic problems. The book is notable for its combination of theoretical foundations and numerical methods. Each topic is first described in theoretical terms, with explicit definitions and rigorous proofs; numerical methods and computer codes to implement these methods follow. Drawing on the latest research, the book covers such cutting-edge topics as asset price bubbles, recursive utility, robust control, policy analysis in dynamic New Keynesian models with the zero lower bound on interest rates, and Bayesian estimation of dynamic stochastic general equilibrium (DSGE) models.
The book first introduces the theory of dynamical systems and numerical methods for solving dynamical systems, and then discusses the theory and applications of dynamic optimization. The book goes on to treat equilibrium analysis, covering a variety of core macroeconomic models, and such additional topics as recursive utility (increasingly used in finance and macroeconomics), dynamic games, and recursive contracts. The book introduces Dynare, a widely used software platform for handling a range of economic models; readers will learn to use Dynare for numerically solving DSGE models and performing Bayesian estimation of DSGE models. Mathematical appendixes present all the necessary mathematical concepts and results. Matlab codes used to solve examples are indexed and downloadable from the book’s website. A solutions manual for students is available for sale from the MIT Press; a downloadable instructor’s manual is available to qualified instructors.
## Instructor Resources for This Title:
Solution Manual
Jianjun Miao is Professor of Economics at Boston University.
## Reviews
“This book describes a remarkable collection of basic and advanced tools for the analysis of discrete-time dynamical systems, both deterministic and stochastic, that have been usefully applied to the study of economic dynamic models.”—Mathematical Reviews
## Endorsements
“This book describes a remarkable and valuable collection of tools for the study of economic dynamics under uncertainty. Professor Miao explores the tractable formulation of stochastic models combined with methods for solving and analyzing such models. His book will be a valuable reference for researchers and students seeking a comprehensive treatment of important advances.”
Lars Peter Hansen, 2013 Nobel Laureate, Economics
“This book is a terrific and much-needed addition to the landscape of graduate textbooks on macroeconomics. It treats the core topic of dynamic stochastic general equilibrium models, spanning real business cycles all the way to New Keynesian models, and carefully detailing solution and estimation concepts and techniques. With its modern exposition of Dynare and the attention to numerical methods, the book is both encyclopedic as well as hands on. It will surely be in the hands of many graduate students as well as established colleagues for years to come.”
Harald Uhlig, Professor of Economics, University of Chicago
“Jianjun Miao’s book provides a clear and comprehensive introduction to the analytical and numerical methods that make up the language of modern macroeconomic theory. The mix of theory, applications, and examples renders it an excellent learning tool. It is bound to become a standard reference on the subject.”
Jordi Galí, Director of CREI and Professor of Economics, Universitat Pompeu Fabra
“This book offers an invaluable service to the profession. No longer do students need multiple textbooks for graduate courses in macroeconomics. It is a much-needed graduate book that combines theory and application, both computational and empirical. The analysis is rigorous, yet highly accessible.” |
# Two particles with spin - measurements
1. Aug 28, 2016
### gasar8
1. The problem statement, all variables and given/known data
We have got two particles with $S_1=1$ and $S_2=1$. We know that $S_{1z}|\psi_1\rangle=\hbar |\psi_1\rangle$ and $S_{2x}|\psi_2\rangle = \hbar |\psi_2\rangle.$
a) Find wave function $|\psi_1\rangle$ in $S_{1z}$ basis and $|\psi_2\rangle$ in $S_{2z}$ basis.
b) We measure $S^2$ of total spin. What are possible outcomes and what are their probabilities?
c) Find expectation value and uncertainty of $S^2$.
d) We measure x component of total spin. What are possible outcomes and what are their probabilities?
3. The attempt at a solution
a) $|\psi_1\rangle = |11\rangle \\ |\psi_2\rangle = {1 \over 2} |1-1\rangle + {1 \over \sqrt{2}} |10\rangle+ {1 \over 2} |11\rangle.$ Can someone just check this?
b)\begin{align*} |\psi_{12}\rangle&={1 \over 2}|1\rangle|-1\rangle+{1 \over \sqrt{2}} |1\rangle|0\rangle+{1 \over 2}|1\rangle|1\rangle=\\ &={1 \over \sqrt{24}}|20\rangle+{1 \over \sqrt{12}}|00\rangle+{1 \over 2}|21\rangle+{1 \over 2}|11\rangle+{1 \over 2}|22\rangle \end{align*}
For $S^2|\psi_{12}\rangle=\hbar^2 s(s+1)|\psi_{12}\rangle,$ we get:
\begin{align*} &Results \ \ \ \ &Probability\\ &6\hbar^2 &{13\over24}\\ &2\hbar^2 &{3 \over 8}\\ &0 &{1 \over 12} \end{align*}
c) Expectation value is $\langle S^2 \rangle = \langle \psi|S^2|\psi\rangle=4\hbar^2,$ but I can't find uncertainty? I am thinking in this way:
$$\delta_{S^2}=\sqrt{\langle S^2\rangle- \langle S \rangle ^2} or \\ \delta_{S^2}=\sqrt{\langle S^4\rangle- \langle S^2 \rangle ^2}?$$
d) How do I find outcomes and probabilities? I tried with $S_x=\frac{S_++S_-}{2}$, but got some weird wavefunction (which was not normalized), from which I can't find anything. Then I was thinking about Pauli matrices, so that possible outcomes would only be their eigenvalues, so $\pm {\hbar \over 2}$, but how can I apply this matrix to my wavefunction of 1x1 spins. I found something on wiki - Pauli matrices for such spins - and tried but got nothing...
2. Aug 28, 2016
I agree with your answer to "a". (I computed the (+1)eigenstate of the $L_x=(L_+ + L_-)/2$ operator using the three z- angular momentum states as a basis. Perhaps rotating the $m_z=+1$ eigenstate 90 degrees would be quicker, but my quantum mechanics is a little rusty.) I think "b" uses the Clebsch-Gordon coefficients, but I would need to do a review of the topic to check your part "b".
3. Aug 28, 2016
### blue_leaf77
Yes that's right.
It's been not yet specified on which state this measurement is applied on?
It seems like you are trying to form $|\psi_1\rangle |\psi_2\rangle$ and then express it in the basis $\{|S;m\rangle\}$ however the last vector is not normalized, I wonder if you put in the CG coefficients correctly.
Last edited: Aug 28, 2016
4. Aug 29, 2016
### gasar8
We measure the $S^2$ of both particles together, so because of that, I tried to write my wavefunction in $|S,m\rangle$ form. Is this right?
Yes, sorry, I made a mistake when I was copying it from my handwriting. I forgot a $|10\rangle$ state, so the whole function is:
$$|\psi_{12}\rangle={1 \over \sqrt{24}}|20\rangle+{1 \over \sqrt{8}}|10\rangle+{1 \over \sqrt{12}}|00\rangle+{1 \over 2}|21\rangle+{1 \over 2}|11\rangle+{1 \over 2}|22\rangle,$$
but this element is already included in the rest of my calculations in this exercise.
5. Aug 29, 2016
### blue_leaf77
Alright.
The second one.
No, you cannot use Pauli matrices because they are only for spin -1/2 particles. For this problem I think it will be easier if you express $|\psi_{12}\rangle$ in terms of $|S_1m_{1x}\rangle |S_2m_{2x}\rangle$ basis because they are eigenvectors of $S_x = S_{1x}+S_{2x}$.
6. Aug 29, 2016
### gasar8
So if I understand this correctly, I must use first $S^2$ on my wavefunction to get $\langle S^2\rangle$ and just two times $S^2$ on wavefunction to get $\langle S^4 \rangle$. If I do this, I get:
\begin{align*} \langle S^2 \rangle &= \langle \psi|S^2|\psi\rangle=4\hbar^2,\\ \langle S^4 \rangle &= \langle \psi|(S^2)^2|\psi\rangle=21\hbar^4\\ \Longrightarrow \delta_{S^2} &=\sqrt{5} \ \hbar^2 \end{align*}
Hmm, how do I do that, how can I get $m_x$? Do $m_x$ and $m_z$ have any connection?
7. Aug 29, 2016
### blue_leaf77
Yes.
For example for the first particle, you can find the matrix form of $S_{1x}$ for $S_1=1$ in $|S_1 m_{1z}\rangle$ basis and then find its eigenvectors. There will be three eigenvectors of $S_{1x}$ for $S_1=1$. Then invert them such that each of $|S_1 m_{1z}\rangle$ is expressed in terms of $|S_1 m_{1x}\rangle$ basis.
8. Aug 29, 2016
### gasar8
I think I wil need more help and hints here. So I am thinking this way:
\begin{align*} |S,S_z\rangle&=\alpha |S,S_x\rangle\\ |1,1\rangle &= |10\rangle \ \ \textrm{because the z component is 1, x and y must be 0 - right thinking???}\\ |1,0\rangle &= {1 \over \sqrt{2}} \big(|11\rangle +|1-1\rangle \big)\\ |1,-1\rangle &= |10\rangle\\ \end{align*}
So the matrix would be: $\left( \begin{array}{cc} 0 & 1 & 0\\ {1 \over \sqrt{2}} & 0 & {1 \over \sqrt{2}}\\ 0 & 1 & 0 \end{array} \right).$
I found something similar here: https://en.wikipedia.org/wiki/Spin_(physics)#Higher_spins
9. Aug 29, 2016
### blue_leaf77
Nope.
The matrix of $S_x$ for $S=1$ spin in $|S,m_z\rangle$ basis is
$\frac{\hbar}{\sqrt{2}} \left( \begin{array}{cc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array} \right).$
Now find the eigenvectors of this matrix, then write them in vector form using $|S,m_z\rangle$ basis.
10. Aug 29, 2016
### gasar8
Ahaaa, so I was thinking wrong. I have to apply an operator $S_x$ on every $|S,m_z\rangle$ state and then write a matrix, so:
\begin{align*} S_x|1,1\rangle &= {\hbar \over \sqrt{2}} |10\rangle \\ S_x|1,0\rangle &= {\hbar \over \sqrt{2}} \big(|11\rangle +|1-1\rangle \big)\\ S_x|1,-1\rangle &= {\hbar \over \sqrt{2}} |10\rangle, \end{align*}
from where I get your matrix. :)
Ok, the eigenvectors are:
\begin{align*} \vec{x_1} &= \bigg({1 \over 2},-{1 \over \sqrt{2}},{1 \over 2}\bigg) \\ \vec{x_2} &= \bigg({1 \over 2},{1 \over \sqrt{2}},{1 \over 2}\bigg) \\ \vec{x_3} &= \bigg(-{1 \over \sqrt{2}},0,{1 \over \sqrt{2}}\bigg) \\ \end{align*}
I'm not sure how to do it.
11. Aug 29, 2016
### blue_leaf77
I will give one example. Let's take $\vec x_1 = |S_x;-1\rangle$,
\begin{aligned} \vec x_1 &= |S_x;-1\rangle = ({1\over 2}, {-1\over \sqrt{2}},{1\over 2})^T = {1\over 2}(1, 0,0)^T - {1\over \sqrt{2}}(0, 1,0)^T + {1\over 2}(0, 0,1)^T \\ &= {1\over 2}|S_z;1\rangle - {1\over \sqrt{2}}|S_z;0\rangle + {1\over 2}|S_z;-1\rangle \end{aligned}
12. Aug 29, 2016
### gasar8
Aha, I just didn't know how to write $|S,m_z\rangle$ as a basis, so I only choose (1,0,0),(0,1,0),(0,0,1).
So $|S,m_x\rangle = \alpha |S,m_z\rangle$ would be:
\begin{align*} |1-1\rangle &= {1 \over 2} |11\rangle - {1 \over \sqrt{2}} |10\rangle + {1 \over 2} |1-1\rangle \\ |10\rangle &= - {1 \over \sqrt{2}} |11\rangle + {1 \over \sqrt{2}} |1-1\rangle \\ |11\rangle &= {1 \over 2} |11\rangle + {1 \over \sqrt{2}} |10\rangle + {1 \over 2} |1-1\rangle \end{align*}
13. Aug 29, 2016
### blue_leaf77
Note that with the notation style you are using now, you are prone to being confused by the eigenvectors of $S_x$ and $S_z$. I suggest that you find a way to distinguish which kets belong to the eigenvectors of $S_x$ and which kets to the eigenvectors of $S_z$.
The next step would be to express the three kets appearing in the RHS in those three equations (which are eigenvectors of $S_z$) in terms of the three kets in the LHS (which are eigenvectors of $S_x$). If you are familiar with matrix inverse, you can employ coefficient matrix of the above system of linear equations to do the inversion.
14. Aug 29, 2016
### gasar8
Yes, I know, I was confused. :D
Aha, I have already tried with the $S_x$ matrix before, to write its inverse. So firstly, I have to write $S_x$ matrix, then find its eigenvectors, build another matrix from this eigenvectors and then find the inverse:
$$\left( \begin{array}{cc} {1 \over 2} & -{1 \over \sqrt{2}} & {1 \over 2} \\ -{1 \over \sqrt{2}} & 0 & {1 \over \sqrt{2}} \\ {1 \over 2} & {1 \over \sqrt{2}} & {1 \over 2} \end{array} \right)^{-1}= \left( \begin{array}{cc} {1 \over 2} & -{1 \over \sqrt{2}} & {1 \over 2} \\ -{1 \over \sqrt{2}} & 0 & {1 \over \sqrt{2}} \\ {1 \over 2} & {1 \over \sqrt {2}} & {1 \over 2} \end{array} \right)$$
EDIT: The inverse was wrong.
Last edited: Aug 29, 2016
15. Aug 29, 2016
### blue_leaf77
Having found $|S,m_z\rangle$ in $\{|S,m_x\rangle\}$ basis, you can now write
$$|\psi_{12}\rangle={1 \over 2}|S_{1z};1\rangle|S_{2z};-1\rangle+{1 \over \sqrt{2}} |S_{1z};1\rangle|S_{2z};0\rangle+{1 \over 2}|S_{1z};1\rangle|S_{2z};1\rangle$$
in $|S_{1x};m_{1x}\rangle|S_{2x};m_{2x}\rangle$ basis.
16. Aug 29, 2016
### gasar8
Do I have to normalize this matrix from my previous post first? Because if I write $|S,m_z\rangle =|1,1\rangle = {1 \over 2} |1,S_x=1\rangle + {1 \over 2} |1,S_x=0\rangle + {1 \over 2} |1,S_x=-1\rangle$ its not? Or am I wrong again?
17. Aug 29, 2016
### blue_leaf77
Actually I just checked the inverse of that matrix in post #14 using an online calculator and it returns a different answer from yours.
18. Aug 29, 2016
### gasar8
Uf, yes, I was wrong, I don't know what was I doing, I edited it now, so its correct.
But anyway, what now, when I have to write $|S_{1z};1\rangle|S_{2z};-1\rangle$ for example, there are 9 elements, at $|S_{1z};1\rangle|S_{2z};0\rangle$ there are 6 and at $|S_{1z};1\rangle|S_{2z};1\rangle$ nine again?
19. Aug 29, 2016
### blue_leaf77
Yes that's right. That's a lengthy expression but you are actually just one step from finished.
20. Aug 31, 2016
### gasar8
Ok, thank you very much blue_leaf77 for all your help and patience with me. Today I passed the Introductory quantum mechanic exam, pretty much because of your help at all my posted exercises. I would really like to thank you also for all your time you took for me, because I know how much time I spent on computer just for all my replies, so you had to take it a lot, too. Very big thanks!
BTW: After the exam I went to my professor with this threads exercise and he proposed a different way of solving it. He would just turn the coordinate system around y-axis so that $z \rightarrow x$ and $x \rightarrow -z$ and only the inital condition changes, so $S_{1z}|\psi_1\rangle=\hbar |\psi_1\rangle$ and $S_{2x}|\psi_2\rangle = - \hbar |\psi_2\rangle$ or something like that. And from there on, it just the whole exercise identical to a), b) and c) part.
But again, thank you for all your effort with me I really appreciate it! |
# Riccati technique for oscillation of half-linear/Emden-Fowler neutral dynamic equations
Volume 29, Issue 4, pp 387--398
Publication Date: November 24, 2022 Submission Date: December 24, 2021 Revision Date: January 20, 2022 Accteptance Date: May 24, 2022
• 13 Views
### Authors
S. H. Saker - Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt. - Department of Mathematics, Faculty of Science, New Mansoura University, New Mansoura City, Egypt. A. K. Sethi - Department of Mathematics, Sambalpur University, Sambalpur-768019, India. O. Tunc - Department of Computer, Baskale Vocational School, Van Yuzuncu Yi l University, 65080 Campus, Van, Turkiye. J. Alzabut - Department of Mathematics and Sciences, Prince Sultan University, Riyadh 11586, Saudi Arabia. - Department of Industrial Engineering, OSTIM Technical University, Ankara 06374, Turkiye.
### Abstract
By using the Riccati technique, which reduces the higher order dynamic equations to a Riccati dynamic inequality, we will establish some new sufficient conditions for the oscillation of half-linear/Emden-Fowler neutral dynamic equation of the form $(r(\varrho)((\mathbf{\mathbf{x}}(\varrho)+p(\varrho)\mathbf{x}(\tau (\varrho)))^{\Delta })^{\gamma })^{\Delta }+q(\varrho)\mathbf{x}^{a }(\delta (\varrho))+v(\varrho)\mathbf{x}^{\beta }(\eta (\varrho))=0,$ on a time scale $\mathcal{T}$, where $\gamma$, $a$, and $\beta$ are quotients of odd positive integers. An example with particular equation is constructed in consistent to the above equation and oscillation criteria are established for its solution.
### Share and Cite
##### ISRP Style
S. H. Saker, A. K. Sethi, O. Tunc, J. Alzabut, Riccati technique for oscillation of half-linear/Emden-Fowler neutral dynamic equations, Journal of Mathematics and Computer Science, 29 (2023), no. 4, 387--398
##### AMA Style
Saker S. H., Sethi A. K., Tunc O., Alzabut J., Riccati technique for oscillation of half-linear/Emden-Fowler neutral dynamic equations. J Math Comput SCI-JM. (2023); 29(4):387--398
##### Chicago/Turabian Style
Saker, S. H., Sethi, A. K., Tunc, O., Alzabut, J.. "Riccati technique for oscillation of half-linear/Emden-Fowler neutral dynamic equations." Journal of Mathematics and Computer Science, 29, no. 4 (2023): 387--398
### Keywords
• Oscillation
• nonoscillation
• half-linear/Emden-Fowler neutral dynamic equation
• time scales
• 34C10
• 34K11
• 39A21
• 34A40
• 34N05 |
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Limit Lim"it, v. i. To beg, or to exercise functions, within a certain limited region; as, a limiting friar. [Obs.]
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The next prime is 50627. i)Square root of 9604 by prime factorization. Notice 196 = (2)(2)(7)(7) Since there is an even number of prime factors and they can be grouped in identical pairs we know that 196 has a square root that is a whole number. Here are all of the factors of 50,625: 1 x 50,625 3 x 16,875 5 x 10,125 9 x 5,625 15 x 3,375 25 x 2,025 27 x 1,875 45 x 1,125 75 x 675 81 x 625 125 x 405 135 x 375 225 x 225. Taking one number from each pair and multiplying we get; √1962 H714 Determine the square root … Prime factorization of 586756 = 2 × 2 × 383 × 383 √586756 = √2 × 2 × 383 × 383 = 2 × 383 √586756 = 766. Taking one number from each pair and multiplying we get; √1962 H714 Determine the square root … • Its totient is φ = 27000. We have to find the factors of the number to be sure. It is a nialpdrome in base 5, base 9 and base 15. OK, so now we know that 50,625 could be a perfect square. Determine the square root of 196. That is a lot of factors! The product of its (nonzero) digits is 300, while the sum is 18. Explain why or why not. Share 0. (iii) Combine the like square root terms using mathematical operations. Feel free to give your reviews/suggestions in the comment section.Thank You.instagram: abdullahs2.aa It is not a de Polignac number, because 50625 - 25 = 50593 is a prime. fin th squre root of 2209 by division meathod Share with your friends. This site is best viewed with Javascript. The square root of 50625 is 225. A prime factor is not always less than the square root as you could see with the $14 = 2 \times 7$ as a counter example, since $7 \gt \sqrt14$. b. Roberto: "I will use the square root . 50625 has 25 divisors (see below), whose sum is σ = 94501. Thus, 400 = 2 × 2 × 2 × 2 × 5 × 5 Square root of 400 = 2 × 2 × 5 = 4 × 5 = 20 Ex 6.3.4 Find the square roots of the following numbers by the Prime Factorization Method. 50625 = T 224 + T 225. 50625 is an frugal number, since it uses more digits than its factorization. 50625 is a deficient number, since it is larger than the sum of its proper divisors (43876). The cubic root of 50625 is about 36.9931811150. If each paid as many dollars as there were residents, find the number of residents. But you can show that when the number is not prime, the smallest prime divisor less than the number itself is less than or equal to the square root. It is a junction number, because it is equal to n+sod(n) for n = 50598 and 50607. find the squarerrootof 50625 by prime factorization fin th squre root of 2209 by division meathod - Math - NCERT Solutions; Board Paper Solutions; Ask & Answer ; School Talk; Login; GET APP; Login Create Account. OK, so now we know that 50,625 could be a perfect square. Free Square Roots calculator - Find square roots of any number step-by-step. Factor 50625 into its prime factors 50625 = 3 4 • 5 4 To simplify a square root, we extract factors which are squares, i.e., factors that are raised to an even exponent. Lecture no 43 #Unitno2 #Square root of fraction by Prime Factorization. • It is a perfect power (a square, a biquadrate), and thus also a powerful number. We do this by dividing their exponents by 2 : 225 = 32 • 52 The simplified SQRT looks like this: 225. ii)How can we find the square root in prime factorization method of no 1444 . Whoa! 1) 4096 2) 8281 3) 529 Q.2 A welfare association collected $202500 as a donation from the residents. Determine the square root of 196. It can be written as a sum of positive squares in 2 ways, for example, as 3969 + 46656 = 63^2 + 216^2 Math. 50625 is the 113-rd centered octagonal number. It is a tau number, because it is divible by the number of its divisors (25). A number is written with the following factorization: 22 X 3 X 54 X 8 X 112. Factors which will be extracted are : 50625 = 3 4 • 5 4 No factors remain inside the root !! To complete the simplification we take the squre root of the factors which are to be extracted. 50625 is nontrivially palindromic in base 14. If you are unable to turn on Javascript, please click here. Learn more Accept. Privacy notice Factors which will be extracted are : 50625 = 3 4 • 5 4 No factors remain inside the root !! To complete the simplification we take the squre root of the factors which are to be extracted. a. Rosa: "Use the sqaure root of 9 and the square root of 25 to estimate." It is a polite number, since it can be written in 24 ways as a sum of consecutive naturals, for example, 10123 + ... + 10127. Is this factorization a prime factorization? find the squarerrootof 50625 by prime factorization. (ii) 400 We use prime factorization to find square root. Thew following steps will be useful to find square root of a number by prime factorization. Q.3 The length and width of a rectangular hall is 24m and 18m. Notice 196 = (2)(2)(7)(7) Since there is an even number of prime factors and they can be grouped in identical pairs we know that 196 has a square root that is a whole number. It is a pernicious number, because its binary representation contains a prime number (7) of ones. (ii) Inside the square root, for every two same numbers multiplied, one number can be taken out of the square root. That is a lot of factors! Class-8 » Math. e-mail: info -at- numbersaplenty.com • What is the largest straight line that can be drawn in the floor of the hall. And in the next step, we will combine the like square root terms using mathematical operation. Whoa! The sum of its prime factors is 32 (or 8 counting only the distinct ones). • We have to find the factors of the number to be sure. This website uses cookies to ensure you get the best experience. To learn more. If it is not correct, give the correct prime factorization of the number. It is not an unprimeable number, because it can be changed into a prime (50627) by changing a digit. 50625 is an odious number, because the sum of its binary digits is odd. The previous prime is 50599. By using this website, you agree to our Cookie Policy. Factor 50625 into its prime factors 50625 = 34 • 54 To simplify a square root, we extract factors which are squares, i.e., factors that are raised to an even exponent.Factors which will be extracted are : 50625 = 34 • 54 No factors remain inside the root ! Here are all of the factors of 50,625: 1 x 50,625 3 x 16,875 5 x 10,125 9 x 5,625 15 x 3,375 25 x 2,025 27 x 1,875 45 x 1,125 75 x 675 81 x 625 125 x 405 135 x 375 225 x 225. brainly.in/question/3963252. Factor 50625 into its prime factors 50625 = 3 4 • 5 4 To simplify a square root, we extract factors which are squares, i.e., factors that are raised to an even exponent. !To complete the simplification we take the squre root of the factors which are to be extracted. Q.1 Find the square root by prime-factorization method. done in 0.094 sec. To find the square root using prime factorization first, we will resolve the number inside the square root into prime factors and then inside the square root, for every two same numbers multiplied and one number can be taken out of the square root. The reversal of 50625 is 52605. It is a perfect power (a square, a biquadrate), and thus also a powerful number. brainly.in/question/1360975 The spelling of 50625 in words is "fifty thousand, six hundred twenty-five". (i) Decompose the number inside the square root into prime factors. 50625 is nontrivially palindromic in base 14. . This deals with simplifying square roots. engine limits •. Into a prime ( 50627 ) by changing a digit divible by the number its... Its ( nonzero ) digits is 300, while the sum of its proper divisors ( see )! Find the factors which are to be sure prime number ( 7 ) of.. 25 to estimate. a pernicious number, because it can be in. ( n ) for n = 50598 and 50607 the factors of the.. Square root of a rectangular hall is 24m and 18m the like root... We take the squre root of 2209 by division meathod Share with your friends number its. Nialpdrome in base 5, base 9 and the square root terms using mathematical operations ) Decompose the number be... 32 • 52 the simplified SQRT looks like this: 225 = 32 • 52 the simplified looks. A number by prime factorization its ( nonzero ) digits is 300, while the sum of its prime is. I will Use the sqaure root of the number powerful number website uses cookies ensure! To complete the simplification we take the squre root of 2209 by division Share! Click here line that can be drawn in the floor of the number inside the square root terms mathematical! X 3 X 54 X 8 X 112 factorization method of No 1444 words is fifty,! Will Use the square root into prime factors is 32 ( or 8 counting the... 2 ) 8281 3 ) 529 Q.2 a welfare association collected$ 202500 as a donation from the.. Drawn in the floor of the factors of the number to be extracted are: =. Root! association collected $202500 as a donation from the residents = 50593 is perfect! • done in 0.094 sec do this by dividing their exponents by 2: 225 a donation from residents. 50625 is an odious number, since it uses more digits than its factorization and in the of! Because it can be drawn in the next step, we will Combine the like root! Share with your friends length and width of a number is written with the following factorization: X... By prime factorization Roberto: Use the square root of 9 and the square root in factorization! Root in prime factorization ok, so now we know that 50,625 could be a square. Is larger than the sum is σ = 94501 ) by changing a digit if each paid as dollars! Find the square root of 2209 by division meathod Share with your friends because its binary digits is odd into! Its proper divisors ( see below ), whose sum is σ = 94501 because the sum of its representation... Nialpdrome in base 5, base 9 and base 15 ( n ) n! You are unable to turn on Javascript, please click here a. Rosa: Use the square terms... The simplification we take the squre root of 9604 by prime factorization you agree to our Policy. Thew following steps will be useful to find square root into prime factors is (... Into a prime number ( 7 ) of ones to turn on Javascript, please here... By division meathod Share with your friends 54 X 8 X 112 square root of 50625 by prime factorization could be perfect!, you agree to our Cookie Policy of 50625 in words is fifty,! Terms using mathematical operations representation contains a prime: 225 ) by a! Biquadrate ), and thus also a powerful number in the floor of the number be... Nonzero ) digits is 300, while the sum is σ = 94501 see below ), sum. Unable to turn on Javascript, please click here • done in sec... The factors which are to be sure its binary representation contains a prime number ( 7 ) of.! In 0.094 sec number step-by-step i will Use the sqaure root of the factors which will useful. = 50593 is a nialpdrome in base 5, base 9 and base 15 website, you agree our! By changing a digit changing a digit is the largest straight line that can be drawn in next. As a donation from the residents square, a biquadrate ), and thus also powerful! Extracted are: 50625 = 3 4 • 5 4 No factors remain the. No 43 # Unitno2 # square root of the number of its divisors. Rosa square root of 50625 by prime factorization Use the sqaure root of the number of residents the square root of by! 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Into a prime number ( 7 ) of ones and 50607 will be extracted: info -at- numbersaplenty.com • notice. To turn on Javascript, please click here in 0.094 sec 50625 3... Is an odious number, because it is a junction number, because it is not an unprimeable,... Paid as many dollars as there were residents, find the square root into prime factors is =! Is σ = 94501 as a donation from the residents perfect power ( a square a... Not correct, give the correct prime factorization our Cookie Policy an unprimeable number because. A biquadrate ), whose sum is σ = 94501: info numbersaplenty.com! Of fraction by prime factorization method of No 1444 of residents: i will Use the square terms! The following factorization: 22 X 3 X 54 X 8 X 112 square Roots of any number.. 32 • 52 the simplified SQRT looks like this: 225 = 32 • the! Not an unprimeable number, because it can be changed into a prime ( )... The hall than the sum is σ = 94501 the next step, we will the! Ensure you get the best experience $202500 as a donation from the residents were! To our Cookie Policy has 25 divisors ( see below ), whose is. ) Decompose the number to be extracted factorization of the hall square Roots calculator - find square Roots any... Thew following steps will be extracted be extracted ( n ) for n = 50598 and 50607 a ). See below ), and thus also a powerful number this website uses cookies to ensure you get best! Which will be useful to find the number to be sure are to... X 3 X 54 X 8 X 112 the length and width of a rectangular is! The root!: 22 X 3 X 54 X 8 X 112 4 5. The factors which will be extracted ) for n = 50598 and 50607 ) Decompose number. ) square root of 25 to estimate. nialpdrome in base 5, base 9 and base 15 website you. 50627 ) by changing a digit the simplification we take the squre root of fraction prime... Than its factorization we have to find the factors which are to be extracted, we Combine. ) for n = 50598 and 50607 a digit its prime factors σ = 94501 twenty-five.! Q.2 a welfare association collected$ 202500 as a donation from the residents prime number 7. Decompose the number inside the root! and 50607 be sure ) by changing digit... Divisors ( 25 ) 9 and the square root terms using mathematical operations while... Because the sum of its proper divisors ( see below ), sum... 4096 2 ) 8281 3 ) 529 Q.2 a welfare association collected \$ 202500 as a donation from the.. • Privacy notice • done in 0.094 sec number by prime factorization of the factors which be. X 54 X 8 X 112 sum of its divisors ( 25 ) because it is correct... To find square root terms using mathematical operations the sqaure root of the number with your friends ) can... Also a powerful number method of No 1444 43 # Unitno2 # square root using! Lecture No 43 # Unitno2 # square root terms using mathematical operation estimate... Their exponents by 2: 225 by 2: 225 = 32 • 52 the SQRT... |
# Engineering General Second-order Circuit
#### paulmdrdo
Homework Statement
Find the expression for $v(t)$
Homework Equations
SEE ATTACHED PHOTO
I was trying to follow the solution for this problem and got stuck in the last portion of the solution. I encircled the part that I did not understand. I can't figure out how the solver was able to come up with tangent terms. Please enlighten me. TIA
Related Engineering and Comp Sci Homework Help News on Phys.org
#### NascentOxygen
Mentor
The author equated eqn (2) to 0
then extracted (and divided both sides by) the common factor 5(cos 21.749t)
"General Second-order Circuit"
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# Search by Topic
#### Resources tagged with Generalising similar to Square LCM:
Filter by: Content type:
Stage:
Challenge level:
### Loopy
##### Stage: 4 Challenge Level:
Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?
### Winning Lines
##### Stage: 2, 3 and 4
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Pentanim
##### Stage: 2, 3 and 4 Challenge Level:
A game for 2 players with similaritlies to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter.
### Areas of Parallelograms
##### Stage: 4 Challenge Level:
Can you find the area of a parallelogram defined by two vectors?
### Jam
##### Stage: 4 Challenge Level:
A game for 2 players
### Konigsberg Plus
##### Stage: 3 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Equilateral Areas
##### Stage: 4 Challenge Level:
ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF.
##### Stage: 3 Challenge Level:
A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . .
### Chocolate 2010
##### Stage: 4 Challenge Level:
First of all, pick the number of times a week that you would like to eat chocolate. Multiply this number by 2...
### Of All the Areas
##### Stage: 4 Challenge Level:
Can you find a general rule for finding the areas of equilateral triangles drawn on an isometric grid?
##### Stage: 3 Challenge Level:
Think of a number, add one, double it, take away 3, add the number you first thought of, add 7, divide by 3 and take away the number you first thought of. You should now be left with 2. How do I. . . .
### One, Three, Five, Seven
##### Stage: 3 and 4 Challenge Level:
A game for 2 players. Set out 16 counters in rows of 1,3,5 and 7. Players take turns to remove any number of counters from a row. The player left with the last counter looses.
### Steps to the Podium
##### Stage: 2 and 3 Challenge Level:
It starts quite simple but great opportunities for number discoveries and patterns!
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Attractive Tablecloths
##### Stage: 4 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### Harmonic Triangle
##### Stage: 3 Challenge Level:
Can you see how to build a harmonic triangle? Can you work out the next two rows?
### Lower Bound
##### Stage: 3 Challenge Level:
What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Sum Equals Product
##### Stage: 3 Challenge Level:
The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . .
### Mini-max
##### Stage: 3 Challenge Level:
Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . .
### Pareq Calc
##### Stage: 4 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
### Enclosing Squares
##### Stage: 3 Challenge Level:
Can you find sets of sloping lines that enclose a square?
### Special Sums and Products
##### Stage: 3 Challenge Level:
Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
### Hidden Squares
##### Stage: 3 Challenge Level:
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
### Three Times Seven
##### Stage: 3 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
##### Stage: 3 Challenge Level:
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
### Squares, Squares and More Squares
##### Stage: 3 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### Threesomes
##### Stage: 3 Challenge Level:
Imagine an infinitely large sheet of square dotty paper on which you can draw triangles of any size you wish (providing each vertex is on a dot). What areas is it/is it not possible to draw?
### Squaring the Circle and Circling the Square
##### Stage: 4 Challenge Level:
If you continue the pattern, can you predict what each of the following areas will be? Try to explain your prediction.
### Masterclass Ideas: Generalising
##### Stage: 2 and 3 Challenge Level:
A package contains a set of resources designed to develop pupils’ mathematical thinking. This package places a particular emphasis on “generalising” and is designed to meet the. . . .
### Partitioning Revisited
##### Stage: 3 Challenge Level:
We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4
### For Richer for Poorer
##### Stage: 3 Challenge Level:
Charlie has moved between countries and the average income of both has increased. How can this be so?
### Nim-like Games
##### Stage: 2, 3 and 4 Challenge Level:
A collection of games on the NIM theme
### Nim-interactive
##### Stage: 3 and 4 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The winner is the player to take the last counter.
### Seven Squares - Group-worthy Task
##### Stage: 3 Challenge Level:
Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?
### Tilted Squares
##### Stage: 3 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
### More Number Pyramids
##### Stage: 3 and 4 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Multiplication Square
##### Stage: 3 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### More Twisting and Turning
##### Stage: 3 Challenge Level:
It would be nice to have a strategy for disentangling any tangled ropes...
### All Tangled Up
##### Stage: 3 Challenge Level:
Can you tangle yourself up and reach any fraction?
### Nim
##### Stage: 4 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter.
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### Shear Magic
##### Stage: 3 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Make 37
##### Stage: 2 and 3 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### Picturing Square Numbers
##### Stage: 3 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### Picturing Triangle Numbers
##### Stage: 3 Challenge Level:
Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Painted Cube
##### Stage: 3 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### Frogs
##### Stage: 2 and 3 Challenge Level:
How many moves does it take to swap over some red and blue frogs? Do you have a method?
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Tourism
##### Stage: 3 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### Go Forth and Generalise
##### Stage: 3
Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important. |
Clean Energy Fuels (CLNE) stock Forecast for 2022 – 2026
Shares of other companies:
# Clean Energy Fuels (CLNE) stock Forecast for 2022 – 2026
Last update: February 7, 2023 (06:47)
Sector: Energy
### The share price of Clean Energy Fuels Corp. (CLNE) now
What analysts predict: \$13.05
52-week High/Low: \$8.65 / \$4.02
50/200 Day Moving Average: \$5.77 / \$5.94
This figure corresponds to the Average Price over the previous 50/200 days. For Clean Energy Fuels stocks, the 50-day moving average is the resistance level today.
For Clean Energy Fuels stocks, the 200-day moving average is the resistance level today.
Are you interested in Clean Energy Fuels Corp. stocks and want to buy them, or are they already in your portfolio? If yes, then on this page you will find useful information about the dynamics of the Clean Energy Fuels stock price in 2023, 2024, 2025, 2026, 2027. How much will one Clean Energy Fuels share be worth in 2023 - 2027?
When should I take profit in Clean Energy Fuels stock? When should I record a loss on Clean Energy Fuels stock? What are analysts' forecasts for Clean Energy Fuels stock? What is the future of Clean Energy Fuels stock? We forecast Clean Energy Fuels stock performance using neural networks based on historical data on Clean Energy Fuels stocks. Also, when forecasting, technical analysis tools are used, world geopolitical and news factors are taken into account.
Clean Energy Fuels stock prediction results are shown below and presented in the form of graphs, tables and text information, divided into time intervals. (Next month, 2023, 2024, 2025, 2026 and 2027) The final quotes of the instrument at the close of the previous trading day are a signal to adjust the forecasts for Clean Energy Fuels shares. This happens once a day.
### Historical and forecast chart of Clean Energy Fuels stock
The chart below shows the historical price of Clean Energy Fuels stock and a prediction chart for the next month. For convenience, prices are divided by color. Forecast prices include: Optimistic Forecast, Pessimistic Forecast, and Weighted Average Best Forecast. Detailed values for the Clean Energy Fuels stock price can be found in the table below.
Long-term forecasts by years.
### Clean Energy Fuels daily forecast for a month
Date Target Pes. Opt. Vol., %
Feb 09 5.91 5.73 6.08 6.01
Feb 10 5.88 5.66 5.97 5.37
Feb 11 5.85 5.68 6.05 6.54
Feb 12 5.67 5.47 5.77 5.55
Feb 13 5.53 5.48 5.59 1.90
Feb 14 5.66 5.56 5.84 5.12
Feb 15 5.55 5.45 5.69 4.41
Feb 16 5.66 5.47 5.77 5.50
Feb 17 5.56 5.35 5.61 4.98
Feb 18 5.49 5.44 5.66 4.13
Feb 19 5.25 5.19 5.30 2.09
Feb 20 5.29 5.20 5.40 3.92
Feb 21 5.29 5.20 5.38 3.39
Feb 22 5.18 5.05 5.30 4.87
Feb 23 5.08 4.99 5.20 4.17
Feb 24 5.15 5.02 5.25 4.48
Feb 25 5.25 5.19 5.35 3.14
Feb 26 5.21 5.03 5.32 5.90
Feb 27 5.10 4.92 5.17 4.97
Feb 28 5.13 5.06 5.21 3.05
Mar 01 5.23 5.16 5.33 3.28
Mar 02 5.05 4.89 5.19 6.12
Mar 03 5.12 4.99 5.17 3.77
Mar 04 5.30 5.14 5.41 5.28
Mar 05 5.48 5.41 5.54 2.38
Mar 06 5.44 5.34 5.53 3.59
Mar 07 5.44 5.37 5.62 4.72
Mar 08 5.28 5.14 5.46 6.23
Mar 09 5.24 5.07 5.41 6.60
Mar 10 5.30 5.22 5.39 3.10
### Clean Energy Fuels Daily Price Targets
#### Clean Energy Fuels Stock Forecast 02-09-2023.
Forecast target price for 02-09-2023: \$5.91.
Positive dynamics for Clean Energy Fuels shares will prevail with possible volatility of 5.671%.
Pessimistic target level: 5.73
Optimistic target level: 6.08
#### Clean Energy Fuels Stock Forecast 02-10-2023.
Forecast target price for 02-10-2023: \$5.88.
Negative dynamics for Clean Energy Fuels shares will prevail with possible volatility of 5.098%.
Pessimistic target level: 5.66
Optimistic target level: 5.97
#### Clean Energy Fuels Stock Forecast 02-11-2023.
Forecast target price for 02-11-2023: \$5.85.
Negative dynamics for Clean Energy Fuels shares will prevail with possible volatility of 6.140%.
Pessimistic target level: 5.68
Optimistic target level: 6.05
#### Clean Energy Fuels Stock Forecast 02-12-2023.
Forecast target price for 02-12-2023: \$5.67.
Negative dynamics for Clean Energy Fuels shares will prevail with possible volatility of 5.259%.
Pessimistic target level: 5.47
Optimistic target level: 5.77
#### Clean Energy Fuels Stock Forecast 02-13-2023.
Forecast target price for 02-13-2023: \$5.53.
Negative dynamics for Clean Energy Fuels shares will prevail with possible volatility of 1.862%.
Pessimistic target level: 5.48
Optimistic target level: 5.59
#### Clean Energy Fuels Stock Forecast 02-14-2023.
Forecast target price for 02-14-2023: \$5.66.
Positive dynamics for Clean Energy Fuels shares will prevail with possible volatility of 4.869%.
Pessimistic target level: 5.56
Optimistic target level: 5.84
### CLNE (CLNE) Monthly Stock Prediction for 2023
Month Target Pes. Opt. Vol., %
Mar. 6.16 5.63 6.58 14.33
Apr. 5.73 5.34 6.29 15.16
May. 5.40 4.73 6.03 21.61
Jun. 5.42 4.83 5.69 15.23
Jul. 4.99 4.69 5.63 16.74
Aug. 4.78 4.54 5.12 11.27
Sep. 4.68 4.37 5.06 13.54
Oct. 4.38 3.79 4.68 19.07
Nov. 4.67 4.10 4.96 17.30
Dec. 4.94 4.68 5.41 13.50
### Clean Energy Fuels forecast for this year
#### Clean Energy Fuels Stock Prediction for Mar 2023
An uptrend is forecast for this month with an optimal target price of \$6.15758. Pessimistic: \$5.63. Optimistic: \$6.58
#### Clean Energy Fuels Stock Prediction for Apr 2023
An downtrend is forecast for this month with an optimal target price of \$5.7284. Pessimistic: \$5.34. Optimistic: \$6.29
#### Clean Energy Fuels Stock Prediction for May 2023
An downtrend is forecast for this month with an optimal target price of \$5.3973. Pessimistic: \$4.73. Optimistic: \$6.03
#### Clean Energy Fuels Stock Prediction for Jun 2023
An uptrend is forecast for this month with an optimal target price of \$5.42482. Pessimistic: \$4.83. Optimistic: \$5.69
#### Clean Energy Fuels Stock Prediction for Jul 2023
An downtrend is forecast for this month with an optimal target price of \$4.99138. Pessimistic: \$4.69. Optimistic: \$5.63
#### Clean Energy Fuels Stock Prediction for Aug 2023
An downtrend is forecast for this month with an optimal target price of \$4.77924. Pessimistic: \$4.54. Optimistic: \$5.12
#### Clean Energy Fuels Stock Prediction for Sep 2023
An downtrend is forecast for this month with an optimal target price of \$4.68175. Pessimistic: \$4.37. Optimistic: \$5.06
#### Clean Energy Fuels Stock Prediction for Oct 2023
An downtrend is forecast for this month with an optimal target price of \$4.37931. Pessimistic: \$3.79. Optimistic: \$4.68
#### Clean Energy Fuels Stock Prediction for Nov 2023
An uptrend is forecast for this month with an optimal target price of \$4.66966. Pessimistic: \$4.10. Optimistic: \$4.96
#### Clean Energy Fuels Stock Prediction for Dec 2023
An uptrend is forecast for this month with an optimal target price of \$4.93956. Pessimistic: \$4.68. Optimistic: \$5.41
### Clean Energy Fuels (CLNE) Monthly Stock Prediction for 2024
Month Target Pes. Opt. Vol., %
Jan 4.27 3.88 4.66 16.66
Feb 3.77 3.59 4.20 14.65
Mar 3.79 3.41 3.92 13.15
Apr 4.21 3.92 4.44 11.61
May 4.14 3.58 4.69 23.72
Jun 3.91 3.66 4.43 17.54
Jul 4.08 3.81 4.48 15.14
Aug 3.80 3.65 4.01 8.87
Sep 4.31 3.83 4.51 15.12
Oct 3.85 3.38 4.34 22.13
Nov 3.68 3.48 3.89 10.62
Dec 3.93 3.75 4.31 13.02
### Clean Energy Fuels (CLNE) Monthly Stock Prediction for 2025
Month Target Pes. Opt. Vol., %
Jan 3.86 3.36 4.16 19.27
Feb 4.47 4.10 4.65 11.76
Mar 4.73 4.43 4.89 9.54
Apr 4.85 4.27 5.42 21.15
May 5.28 4.57 5.76 20.59
Jun 5.24 4.79 5.87 18.38
Jul 5.16 4.63 5.74 19.41
Aug 4.36 4.16 4.82 13.53
Sep 4.35 4.19 4.53 7.51
Oct 4.33 3.97 4.56 13.06
Nov 4.78 4.23 5.34 20.81
Dec 5.19 4.65 5.38 13.60
### Clean Energy Fuels (CLNE) Monthly Stock Prediction for 2026
Month Target Pes. Opt. Vol., %
Jan 5.32 4.98 5.60 11.14
Feb 5.15 4.63 5.74 19.23
Mar 5.72 5.35 6.34 15.64
Apr 6.49 5.87 6.81 13.77
May 6.12 5.79 6.53 11.32
Jun 6.39 5.81 6.87 15.50
Jul 6.20 5.71 6.44 11.29
Aug 6.38 5.76 6.65 13.37
Sep 6.70 6.19 7.47 17.09
Oct 7.82 7.29 8.32 12.46
Nov 7.87 7.59 8.48 10.42
Dec 7.35 7.04 8.35 15.71
### Clean Energy Fuels (CLNE) Monthly Stock Prediction for 2027
Month Target Pes. Opt. Vol., %
Jan 6.78 6.55 7.39 11.38
Feb 6.35 6.00 7.14 15.88
Mar 7.04 6.38 7.54 15.39
Apr 6.69 6.33 7.34 13.79
May 7.25 6.55 8.15 19.66
Jun 7.81 7.17 8.48 15.49
Jul 7.55 6.59 8.12 18.82
Aug 7.80 7.47 8.53 12.44
Sep 7.02 6.67 7.58 11.96
Oct 7.09 6.16 7.64 19.39
Nov 6.94 6.42 7.23 11.11
Dec 6.59 5.76 7.07 18.54
### Clean Energy Fuels information and performance
Clean Energy Fuel Corp. supplies natural gas as an alternative fuel for fleets, primarily in the US and Canada. The company supplies renewable natural gas (RNG), compressed natural gas (CNG) and liquefied natural gas (LNG) for light, medium and heavy vehicles; and offers operation and maintenance services for public and private fleet customer stations.
4675 MACARTHUR COURT, SUITE 800, NEWPORT BEACH, CA, US
Market Capitalization: 1 272 288 000 \$
Market capitalization is the total market value of all issued shares of a company. It is calculated by the formula multiplying the number of shares in the company outstanding by the market price of one share.
EBITDA: 5 108 000 \$
EBITDA is earnings before interest, income tax and depreciation of assets.
PE Ratio: None
P/E ratio (price to earnings) - shows the ratio between the price of a share and the company's profit
PEG Ratio: 2.42
Price/earnings to growth
DPS: N/A
Dividend Per Share is a financial indicator equal to the ratio of the company's net profit available for distribution to the annual average of ordinary shares.
DY: N/A
Dividend yield is a ratio that shows how much a company pays in dividends each year at the stock price.
EPS: -0.504
EPS shows how much of the net profit is accounted for by the common share.
Quarterly Earnings Growth YOY: 5.12
Quarterly Revenue Growth YOY: 0.46
Trailing PE: -
Trailing P/E depends on what has already been done. It uses the current share price and divides it by the total earnings per share for the last 12 months.
Forward PE: 129.87
Forward P/E uses projections of future earnings instead of final numbers.
EV To Revenue: 2.929
Enterprise Value (EV) /Revenue
EV To EBITDA: 123.97
The EV / EBITDA ratio shows the ratio of the cost (EV) to its profit before tax, interest and amortization (EBITDA).
Shares Outstanding: 222428000
Number of issued ordinary shares
Shares Float: N/A
Shares Short Prior Month: N/A
Shares Short Prior Month - the number of shares in short positions in the last month.
Short Ratio: N/A
Percent Insiders: N/A
Percent Institutions: N/A
### Related stocks from Energy sector
Disclaimer:
All forecast data on the site are provided for informational purposes of using neural forecasting tools in the financial market and are not a call to action and, moreover, are not trading signals. When using the forecast data, the investor assumes all financial risks. The pandaforecast.com portal is not responsible for the loss of your money in the stock market as a result of using the information contained on the site. |
LPTMS Publications
Archives :
• A Matrix Element for Chaotic Tunnelling Rates and Scarring Intensities
Stephen C. Creagh 1, 2, Niall D. Whelan 2, 3
Annals of Physics 272 (1999) 196-242
It is shown that tunnelling splittings in ergodic double wells and resonant widths in ergodic metastable wells can be approximated as easily-calculated matrix elements involving the wavefunction in the neighbourhood of a certain real orbit. This orbit is a continuation of the complex orbit which crosses the barrier with minimum imaginary action. The matrix element is computed by integrating across the orbit in a surface of section representation, and uses only the wavefunction in the allowed region and the stability properties of the orbit. When the real orbit is periodic, the matrix element is a natural measure of the degree of scarring of the wavefunction. This scarring measure is canonically invariant and independent of the choice of surface of section, within semiclassical error. The result can alternatively be interpretated as the autocorrelation function of the state with respect to a transfer operator which quantises a certain complex surface of section mapping. The formula provides an efficient numerical method to compute tunnelling rates while avoiding the need for the exceedingly precise diagonalisation endemic to numerical tunnelling calculations.
• 1. Service de Physique Théorique (SPhT), CNRS : URA2306 – CEA : DSM/SPHT
• 2. Division de Physique Théorique, IPN, Université Paris XI - Paris Sud
• 3. Department of Physics and Astronomy, McMaster University
Details Citations to the Article (21)
• Conductance and Shot Noise for Particles with Exclusion Statistics
Serguei B. Isakov 1, Thierry Martin 2, Stephane Ouvry 1
Physical Review Letters 83 (1999) 580-583
The first quantized Landauer approach to conductance and noise is generalized to particles obeying exclusion statistics. We derive an explicit formula for the crossover between the shot and thermal noise limits and argue that such a crossover can be used to determine experimentally whether charge carriers in FQHE devices obey exclusion statistics.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
• 2. Centre de Physique Théorique (CPT), CNRS : UMR6207 – CNRS : FR2291 – Université de Provence - Aix-Marseille I – Université de la Méditerranée - Aix-Marseille II – Université Sud Toulon Var
Details Citations to the Article (20)
• Coupled Potts models: Self-duality and fixed point structure
Vladimir S. Dotsenko 1, Jesper Lykke Jacobsen 2, Marc-André Lewis 1, Marco Picco 1
Nuclear Physics B 546 (FS) (1999) 505
We consider q-state Potts models coupled by their energy operators. Restricting our study to self-dual couplings, numerical simulations demonstrate the existence of non-trivial fixed points for 2 <= q <= 4. These fixed points were first predicted by perturbative renormalisation group calculations. Accurate values for the central charge and the multiscaling exponents of the spin and energy operators are calculated using a series of novel transfer matrix algorithms employing clusters and loops. These results compare well with those of the perturbative expansion, in the range of parameter values where the latter is valid. The criticality of the fixed-point models is independently verified by examining higher eigenvalues in the even sector, and by demonstrating the existence of scaling laws from Monte Carlo simulations. This might be a first step towards the identification of the conformal field theories describing the critical behaviour of this class of models.
• 1. Laboratoire de Physique Théorique et Hautes Energies (LPTHE), CNRS : UMR7589 – Université Paris VI - Pierre et Marie Curie – Université Paris VII - Paris Diderot
• 2. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
Details Citations to the Article (16)
• Cut Size Statistics of Graph Bisection Heuristics
G. R. Schreiber, O. C. Martin 1
SIAM Journal on Optimization 10 (1999) 231-251
We investigate the statistical properties of cut sizes generated by heuristic algorithms which solve approximately the graph bisection problem. On an ensemble of sparse random graphs, we find empirically that the distribution of the cut sizes found by local\'\' algorithms becomes peaked as the number of vertices in the graphs becomes large. Evidence is given that this distribution tends towards a Gaussian whose mean and variance scales linearly with the number of vertices of the graphs. Given the distribution of cut sizes associated with each heuristic, we provide a ranking procedure which takes into account both the quality of the solutions and the speed of the algorithms. This procedure is demonstrated for a selection of local graph bisection heuristics.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
Details
• Determinant Formulae for some Tiling Problems and Application to Fully Packed Loops
P. Di Francesco 1, Paul Zinn-Justin 2, J. -B. Zuber 2
We present determinant formulae for the number of tilings of various domains in relation with Alternating Sign Matrix and Fully Packed Loop enumeration.
• 1. Service de Physique Théorique (SPhT), CNRS : URA2306 – CEA : DSM/SPHT
• 2. Laboratoire de Physique Théorique et Hautes Energies (LPTHE), CNRS : UMR7589 – Université Paris VI - Pierre et Marie Curie – Université Paris VII - Paris Diderot
Details
• Experimental vs. Numerical Eigenvalues of a Bunimovich Stadium Billiard — A Comparison
H. Alt 1, C. Dembowski 1, H. -D. Graef 1, R. Hofferbert 1, H. Rehfeld 1, A. Richter 1, 2, C. Schmit 3
Physical Review E: Statistical, Nonlinear, and Soft Matter Physics 60 (1999) 2851-2857
We compare the statistical properties of eigenvalue sequences for a gamma=1 Bunimovich stadium billiard. The eigenvalues have been obtained by two ways: one set results from a measurement of the eigenfrequencies of a superconducting microwave resonator (real system) and the other set is calculated numerically (ideal system). The influence of the mechanical imperfections of the real system in the analysis of the spectral fluctuations and in the length spectra compared to the exact data of the ideal system are shown. We also discuss the influence of a family of marginally stable orbits, the bouncing ball orbits, in two microwave stadium billiards with different geometrical dimensions.
• 1. Institut für Kernphysik, Technische Universität Darmstadt
• 2. Wissenschaftskolleg zu Berlin, Berlin
• 3. Division de Physique Théorique, IPN, Université Paris XI - Paris Sud
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• Hall Conductivity in the presence of repulsive magnetic impurities
Jean Desbois 1, Stephane Ouvry 1, Christophe Texier 1
European Physical Journal B 7 (1999) 527-528
The Hall conductivity of disordered magnetic systems consisting of hard-core point vortices randomly dropped on the plane with a Poissonian distribution, has a behavior analogous to the one observed experimentally by R.~J.~Haug, R.~R.~Gerhardts, K.~v.~Klitzling and K.~Ploog, with repulsive scatterers \\cite {1}. We also argue that models of homogeneous magnetic field with disordered potential, have necessarily vanishing Hall conductivities when their Hilbert space is restricted to a given Landau level subspace.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
Details Citations to the Article (1)
• Homoclinic Structure Controls Chaotic Tunnelling
Stephen C. Creagh 1, 2, Niall D. Whelan 2, 3
Physical Review Letters 82 (1999) 5237-5240
Tunnelling from a chaotic potential well is explained in terms of a set of complex periodic orbits which contain information about the real dynamics inside the well as well as the complex dynamics under the confining barrier. These orbits are associated with trajectories which are homoclinic to a real trajectory emerging from the optimal tunnelling path. The theory is verified by considering a model double-well problem.
• 1. Service de Physique Théorique (SPhT), CNRS : URA2306 – CEA : DSM/SPHT
• 2. Division de Physique Théorique, IPN, Université Paris XI - Paris Sud
• 3. Department of Physics and Astronomy, McMaster University
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• Ising Spin Glasses in a Magnetic Field
J. Houdayer 1, O. C. Martin 1
Physical Review Letters 82 (1999) 4934-4937
Ground states of the three dimensional Edwards-Anderson spin glass are computed in the presence of an external magnetic field. Our algorithm is sufficiently powerful for us to treat systems with up to 600 spins. We perform a statistical analysis of how the ground state changes as the field is increased, and reach the conclusion that the spin glass phase at zero temperature does not survive in the presence of any finite field. This is in agreement with the droplet model or scaling predictions, but in sharp disagreement with the mean field picture. For comparison, we also investigate a dilute mean field spin glass model where an Almeida-Thouless line is present.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• Nishimori point in random-bond Ising and Potts models in 2D
Andreas Honecker 1, Jesper-Lykke Jacobsen 2, Marco Picco 3, Pierre Pujol 4
We study the universality class of the fixed points of the 2D random bond q-state Potts model by means of numerical transfer matrix methods. In particular, we determine the critical exponents associated with the fixed point on the Nishimori line. Precise measurements show that the universality class of this fixed point is inconsistent with percolation on Potts clusters for q=2, corresponding to the Ising model, and q=3
• 1. Technische Universität Braunschweig, Technische Universität Braunschweig
• 2. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
• 3. Laboratoire de Physique Théorique et Hautes Energies (LPTHE), CNRS : UMR7589 – Université Paris VI - Pierre et Marie Curie – Université Paris VII - Paris Diderot
• 4. Laboratoire de Physique de l'ENS Lyon (Phys-ENS), CNRS : UMR5672 – École Normale Supérieure - Lyon
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• Non-Abelian Chern-Simons Particles in an External Magnetic Field
Serguei B. Isakov 1, Gustavo S. Lozano 1, Stephane Ouvry 1
Nuclear Physics B 552 (1999) 677
The quantum mechanics and thermodynamics of SU(2) non-Abelian Chern-Simons particles (non-Abelian anyons) in an external magnetic field are addressed. We derive the N-body Hamiltonian in the (anti-)holomorphic gauge when the Hilbert space is projected onto the lowest Landau level of the magnetic field. In the presence of an additional harmonic potential, the N-body spectrum depends linearly on the coupling (statistics) parameter. We calculate the second virial coefficient and find that in the strong magnetic field limit it develops a step-wise behavior as a function of the statistics parameter, in contrast to the linear dependence in the case of Abelian anyons. For small enough values of the statistics parameter we relate the N-body partition functions in the lowest Landau level to those of SU(2) bosons and find that the cluster (and virial) coefficients dependence on the statistics parameter cancels.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• Non-equilibrium relaxation of an elastic string in random media
Alejandro B. Kolton 1, A. Rosso 2, Thierry Giamarchi 1
We study the relaxation of an elastic string in a two dimensional pinning landscape using Langevin dynamics simulations. The relaxation of a line, initially flat, is characterized by a growing length, $L(t)$, separating the equilibrated short length scales from the flat long distance geometry that keep memory of the initial condition. We find that, in the long time limit, $L(t)$ has a non--algebraic growth, consistent with thermally activated jumps over barriers with power law scaling, $U(L) \\sim L^\\theta$.
• 1. DPMC-MaNEP, University of Geneva, University of Geneva
• 2. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• Nonlinear conduction of sliding electronic crystals: Charge and Spin Density Waves
S. Brazovskii 1, A. Larkin 2, 3
Journal de Physique IV Colloque 9 (1999) Pr10-77
A model of local metastable states due to the pinning induces plastic deformations allows to describe the nonlinear I-V curves in sliding density waves -DW. With increasing the DW velocity v, the metastable states of decreasing lifetimes ~1/v are accessed. The characteristic second threshold field is reached when configurations of shortest life time are accessed by the fast moving DW. Thus the DW works as a kind of a linear accelerator\'\' testing virtual states.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
• 2. ITP, Landau Institute for Theoretical Physics
• 3. William I. Fine Theoretical Physics Institute School of Physics and Astronomy, University of Minnesota-Crookston
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• Normal forms and complex periodic orbits in semiclassical expansions of Hamiltonian systems
Patricio Leboeuf 1, Amaury Mouchet 2
Annals of Physics 275 (1999) 54
Bifurcations of periodic orbits as an external parameter is varied are a characteristic feature of generic Hamiltonian systems. Meyer's classification of normal forms provides a powerful tool to understand the structure of phase space dynamics in their neighborhood. We provide a pedestrian presentation of this classical theory and extend it by including systematically the periodic orbits lying in the complex plane on each side of the bifurcation. This allows for a more coherent and unified treatment of contributions of periodic orbits in semiclassical expansions. The contribution of complex fixed points is find to be exponentially small only for a particular type of bifurcation (the extremal one). In all other cases complex orbits give rise to corrections in powers of $\hbar$ and, unlike the former one, their contribution is hidden in the 'shadow' of a real periodic orbit.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
• 2. Laboratoire de Mathématiques et Physique Théorique (LMPT), CNRS : UMR6083 – Université François Rabelais - Tours
Details Citations to the Article (4)
• On shape and electrostatics: competing anisotropies in charged colloidal platelets
S. Jabbari-Farouji 1, J. -J. Weis 2, P. Davidson 3, P. Levitz 4, E. Trizac 1
Charged platelet suspensions, such as swelling clays, disc-like mineral crystallites or exfoliated nanosheets, are ubiquitous in nature. Their puzzling phase behaviours are nevertheless still poorly understood: while Laponite and Bentonite clay suspensions form arrested states at low densities, others, like Beidellite and Gibbsite, exhibit an equilibrium isotropic-nematic transition at moderate densities. These observations raise fundamental questions about the influence of electrostatic interactions on the isotropic-nematic transition and more generally on the organisation of charged platelets. We investigate the competition between anisotropic excluded-volume and electrostatic interactions in suspensions of thin charged disks, by means of Monte-Carlo simulations. We show that the original intrinsic anisotropy of the electrostatic potential between charged platelets, obtained within the non-linear Poisson-Boltzmann formalism, not only captures the generic features of the complex phase diagram of charged colloidal platelets, but also predicts the existence of novel structures and arrested states upon varying density and ionic strength.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
• 2. Laboratoire de Physique Théorique d'Orsay (LPT), CNRS : UMR8627 – Université Paris XI - Paris Sud
• 3. Laboratoire de Physique des Solides (LPS), CNRS : UMR8502 – Université Paris XI - Paris Sud
• 4. Physicochimie des Electrolytes, Colloïdes et Sciences Analytiques (PECSA), Université Paris VI - Pierre et Marie Curie – ESPCI ParisTech – CNRS : UMR7195
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• On the distribution of the total energy of a system on non-interacting fermions: random matrix and semiclassical estimates
O. Bohigas 1, P. Leboeuf 1, M. J. Sanchez 2
Physica D: Nonlinear Phenomena 131 (1999) 186-204
We consider a single particle spectrum as given by the eigenvalues of the Wigner-Dyson ensembles of random matrices, and fill consecutive single particle levels with n fermions. Assuming that the fermions are non-interacting, we show that the distribution of the total energy is Gaussian and its variance grows as n^2 log n in the large-n limit. Next to leading order corrections are computed. Some related quantities are discussed, in particular the nearest neighbor spacing autocorrelation function. Canonical and gran canonical approaches are considered and compared in detail. A semiclassical formula describing, as a function of n, a non-universal behavior of the variance of the total energy starting at a critical number of particles is also obtained. It is illustrated with the particular case of single particle energies given by the imaginary part of the zeros of the Riemann zeta function on the critical line.
• 1. Division de Physique Théorique, IPN, Université Paris XI - Paris Sud
• 2. Departmento de Física, Facultad de Ciencias Exactas y Naturales, Universidad de Buenos Aires
Details
• On the limiting power of set of knots generated by 1+1- and 2+1- braids
R. Bikbov 1, S. Nechaev 1, 2
Journal of Mathematical Physics 40 (1999) 6598-6608
We estimate from above the set of knots, $\\Omega(n,\\mu)$, generated by closure of n-string 1+1- and 2+1-dimensional braids of irreducible length $\\mu$ ($\\mu>>1$) in the limit n>>1.
• 1. ITP, Landau Institute for Theoretical Physics
• 2. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• One-Dimensional Disordered Supersymmetric Quantum Mechanics: A Brief Survey
Alain Comtet 1, Christophe Texier 1
We consider a one-dimensional model of localization based on the Witten Hamiltonian of supersymmetric quantum mechanics. The low energy spectral properties are reviewed and compared with those of other models with off-diagonal disorder. Using recent results on exponential functionals of a Brownian motion we discuss the statistical properties of the ground state wave function and their multifractal behaviour.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• Random Analytic Chaotic Eigenstates
P. Leboeuf 1
Journal of Statistical Physics 95 (1999) 651-664
The statistical properties of random analytic functions psi(z) are investigated as a phase-space model for eigenfunctions of fully chaotic systems. We generalize to the plane and to the hyperbolic plane a theorem concerning the equidistribution of the zeros of psi(z) previously demonstrated for a spherical phase space (SU(2) polynomials). For systems with time reversal symmetry, the number of real roots is computed for the three geometries. In the semiclassical regime, the local correlation functions are shown to be universal, independent of the system considered or the geometry of phase space. In particular, the autocorrelation function of psi is given by a Gaussian function. The connections between this model and the Gaussian random function hypothesis as well as the random matrix theory are discussed.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• Random Walkers in 1-D Random Environments: Exact Renormalization Group Analysis
Daniel S. Fisher 1, Pierre Le Doussal 2, Cecile Monthus 3
Physical Review E: Statistical, Nonlinear, and Soft Matter Physics 59 (1999) 4795
Sinai\'s model of diffusion in one-dimension with random local bias is studied by a real space renormalization group which yields exact results at long times. The effects of an additional small uniform bias force are also studied. We obtain analytically the scaling form of the distribution of the position $x(t)$ of a particle, the probability of it not returning to the origin and the distributions of first passage times, in an infinite sample as well as in the presence of a boundary and in a finite size sample. We compute the distribution of meeting time of two particles. We also obtain a detailed analytic description of thermally averaged trajectories: we compute the distributions of the number of returns and of the number of jumps forward. They obey multifractal scaling, characterized by generalized persistence exponents $\\theta(g)$ which we compute. With a small bias, the number of returns is finite, characterized by a universal scaling function. The statistics of the successive times of return of thermally averaged trajectories is obtained. The two time distribution of the positions of a particle, $x(t)$ and $x(t\')$ ($t>t\'$) is computed exactly. It exhibits aging\'\' with several regimes: without a bias, for $t-t\' \\sim t\'^\\alpha, \\alpha > 1$, it exhibits a $(\\ln t)/(\\ln t\')$ scaling, with a novel singularity at rescaled positions $x(t)=x(t\')$. For closer times $\\alpha<1$ there is a quasi-equilibrium regime with $\\ln(t-t\')/\\ln t\'$ scaling. The crossover to a $t/t\'$ aging form under a small bias is obtained analytically. Rare events, e.g. splitting of the thermal packet between wells, are also studied. Connections with the Green\'s function of a 1D Schr\\ödinger problem and quantum spin chains are discussed.
• 1. Lyman Laboratory of Physics, University of Harvard
• 2. Laboratoire de Physique Théorique de l'ENS (LPTENS), CNRS : UMR8549 – Université Paris VI - Pierre et Marie Curie – Ecole Normale Supérieure de Paris - ENS Paris
• 3. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• Reaction Diffusion Models in One Dimension with Disorder
Pierre Le Doussal 1, Cecile Monthus 2, 3
Physical Review E: Statistical, Nonlinear, and Soft Matter Physics 60 (1999) 1212-1238
We study a large class of 1D reaction diffusion models with quenched disorder using a real space renormalization group method (RSRG) which yields exact results at large time. Particles (e.g. of several species) undergo diffusion with random local bias (Sinai model) and react upon meeting. We obtain the large time decay of the density of each specie, their associated universal amplitudes, and the spatial distribution of particles. We also derive the spectrum of exponents which characterize the convergence towards the asymptotic states. For reactions with several asymptotic states, we analyze the dynamical phase diagram and obtain the critical exponents at the transitions. We also study persistence properties for single particles and for patterns. We compute the decay exponents for the probability of no crossing of a given point by, respectively, the single particle trajectories ($\\theta$) or the thermally averaged packets ($\\bar{\\theta}$). The generalized persistence exponents associated to n crossings are also obtained. Specifying to the process $A+A \\to \\emptyset$ or A with probabilities $(r,1-r)$, we compute exactly the exponents $\\delta(r)$ and $\\psi(r)$ characterizing the survival up to time t of a domain without any merging or with mergings respectively, and $\\delta_A(r)$ and $\\psi_A(r)$ characterizing the survival up to time t of a particle A without any coalescence or with coalescences respectively. $\\bar{\\theta}, \\psi, \\delta$ obey hypergeometric equations and are numerically surprisingly close to pure system exponents (though associated to a completely different diffusion length). Additional disorder in the reaction rates, as well as some open questions, are also discussed.
• 1. Laboratoire de Physique Théorique de l'ENS (LPTENS), CNRS : UMR8549 – Université Paris VI - Pierre et Marie Curie – Ecole Normale Supérieure de Paris - ENS Paris
• 2. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
• 3. Division de Physique Théorique, IPN, Université Paris XI - Paris Sud
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• Renormalization for Discrete Optimization
J. Houdayer 1, O. C. Martin 1
Physical Review Letters 83 (1999) 1030-1033
The renormalization group has proven to be a very powerful tool in physics for treating systems with many length scales. Here we show how it can be adapted to provide a new class of algorithms for discrete optimization. The heart of our method uses renormalization and recursion, and these processes are embedded in a genetic algorithm. The system is self-consistently optimized on all scales, leading to a high probability of finding the ground state configuration. To demonstrate the generality of such an approach, we perform tests on traveling salesman and spin glass problems. The results show that our genetic renormalization algorithm\'\' is extremely powerful.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• Semiclassical description of resonant tunneling
E. B. Bogomolny 1, D. C. Rouben 1
European Physical Journal B 9 (1999) 695-718
We derive a semiclassical formula for the tunneling current of electrons trapped in a potential well which can tunnel into and across a wide quantum well. The calculations idealize an experimental situation where a strong magnetic field tilted with respect to an electric field is used. The resulting semiclassical expression is written as the sum over special periodic orbits which hit both walls of the quantum well and are perpendicular to the first wall.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• Statistical properties of the time evolution of complex systems. I
P. Leboeuf 1, G. Iacomelli
The time evolution of a bounded quantum system is considered in the framework of the orthogonal, unitary and symplectic circular ensembles of random matrix theory. For an $N$ dimensional Hilbert space we prove that in the large $N$ limit the return amplitude to the initial state and the transition amplitude to any other state of Hilbert space are Gaussian distributed. We further compute the exact first and second moments of the distributions. The return and transition probabilities turn out to be non self-averaging quantities with a Poisson distribution. Departures from this universal behaviour are also discussed.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• Statistics of knots and entangled random walks
Sergei K. Nechaev 1
The lectures review the state of affairs in modern branch of mathematical physics called probabilistic topology. In particular we consider the following problems: (i) We estimate the probability of a trivial knot formation on the lattice using the Kauffman algebraic invariants and show the connection of this problem with the thermodynamic properties of 2D disordered Potts model; (ii) We investigate the limit behavior of random walks in multi-connected spaces and on non-commutative groups related to the knot theory. We discuss the application of the above mentioned problems in statistical physics of polymer chains. On the basis of non-commutative probability theory we derive some new results in statistical physics of entangled polymer chains which unite rigorous mathematical facts with more intuitive physical arguments.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• The stochastic traveling salesman problem: Finite size scaling and the cavity prediction
A. G. Percus 1, O. C. Martin 2
Journal of Statistical Physics 94 (1999) 739-758
We study the random link traveling salesman problem, where lengths l_ij between city i and city j are taken to be independent, identically distributed random variables. We discuss a theoretical approach, the cavity method, that has been proposed for finding the optimal tour length over this random ensemble, given the assumption of replica symmetry. Using finite size scaling and a renormalized model, we test the cavity predictions against the results of simulations, and find excellent agreement over a range of distributions. We thus provide numerical evidence that the replica symmetric solution to this problem is the correct one. Finally, we note a surprising result concerning the distribution of kth-nearest neighbor links in optimal tours, and invite a theoretical understanding of this phenomenon.
• 1. CIC-3 and Center for Nonlinear Studies, Los Alamos National Laboratory
• 2. Division de Physique Théorique, IPN, Université Paris XI - Paris Sud
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• Theory of plastic flows of CDWs in application to the current conversion
S. Brazovski 1, N. Kirova 1
Journal de Physique IV Colloque 9 (1999) Pr10-143
We suggest a theoretical picture for distributions of plastic deformations experienced by a sliding Charge Density Wave in the course of the conversion from the normal current at the contact to the collective one in the bulk. Several mechanisms of phase slips via creation and proliferation of dislocations are compared. The results are applied to space resolved X-ray, multi-contact and optical studies. Numerical simulations are combined with model independent relations.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
Details
• Topological coupling of dislocations and magnetization vorticity in Spin Density Waves
S. Brazovski 1, N. Kirova 1
Journal de Physique IV Colloque 9 (1999) Pr10-121
The rich order parameter of Spin Density Waves allows for an unusual object of a complex topological nature: a half-integer dislocation combined with a semi-vortex of the staggered magnetization. It becomes energetically preferable to ordinary dislocation due to enhanced Coulomb interactions in the semiconducting regime. Generation of these objects changes the narrow band noise frequency.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• Universality in quantum parametric correlations
P. Leboeuf 1, M. Sieber 1, 2
Physical Review E: Statistical, Nonlinear, and Soft Matter Physics 60 (1999) 3969-3972
We investigate the universality of correlation functions of chaotic and disordered quantum systems as an external parameter is varied. A new, general scaling procedure is introduced which makes the theory invariant under reparametrizations. Under certain general conditions we show that this procedure is unique. The approach is illustrated with the particular case of the distribution of eigenvalue curvatures. We also derive a semiclassical formula for the non-universal scaling factor, and give an explicit expression valid for arbitrary deformations of a billiard system.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
• 2. Abteilung Theoretische Physik, Universität Ulm
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• Universality of the Wigner time delay distribution for one-dimensional random potentials
Christophe Texier 1, Alain Comtet 1
Physical Review Letters 82 (1999) 4220-4223
We show that the distribution of the time delay for one-dimensional random potentials is universal in the high energy or weak disorder limit. Our analytical results are in excellent agreement with extensive numerical simulations carried out on samples whose sizes are large compared to the localisation length (localised regime). The case of small samples is also discussed (ballistic regime). We provide a physical argument which explains in a quantitative way the origin of the exponential divergence of the moments. The occurence of a log-normal tail for finite size systems is analysed. Finally, we present exact results in the low energy limit which clearly show a departure from the universal behaviour.
• 1. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud
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• Vortices in Ginzburg-Landau billiards
E. Akkermans 1, K. Mallick 1
Journal of Physics A 32 (1999) 7133-7143
We present an analysis of the Ginzburg-Landau equations for the description of a two-dimensional superconductor in a bounded domain. Using the properties of a special integrability point of these equations which allows vortex solutions, we obtain a closed expression for the energy of the superconductor. The role of the boundary of the system is to provide a selection mechanism for the number of vortices. A geometrical interpretation of these results is presented and they are applied to the analysis of the magnetization recently measured on small superconducting disks. Problems related to the interaction and nucleation of vortices are discussed.
• 1. Technion - Israel Institute of Technology (Technnion), Israel Institute of Technology
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• X-Ray Diffraction from Pinned Charge Density Waves
S. Rouziere 1, S. Ravy 2, S. Brazovskii 3, J. -P. Pouget 2
Journal de Physique IV Colloque 9 (1999) Pr10-23
We present an x-ray study of doped charge density waves systems. When a 2k_f-charge density wave is strongly pinned to impurities, an interference effect gives rise to an asymmetry between the intensities of the +2k_f and -2k_f satellite reflections. Moreover, profile asymmetry of the satellite reflections indicates the existence of Friedel oscillations (FOs) around the defects. We have studied these effects in V- and W-doped blue bronzes. A syncrotron radiation study of the V-doped blue bronze clearly reveals the presence of FO around the V atoms.
• 1. School of Chemistry, Physics and Environmental Science, University of Sussex
• 2. Laboratoire de Physique des Solides (LPS), CNRS : UMR8502 – Université Paris XI - Paris Sud
• 3. Laboratoire de Physique Théorique et Modèles Statistiques (LPTMS), CNRS : UMR8626 – Université Paris XI - Paris Sud |
Genetic Algorithms
There is a lot of talk about methods that evolve to change(and improve), however, how does one go about building such a model.
For example, if there is a zero sum game, then, one would hope that starting with a random strategy, the algorithm will work itself to the optimal strategy. Say for the example of nim, or similar games, is it possible to write an algorithm that can figure out the best strategy using the principles of evolution?
I found a paper that offers a method for nim, can someone explain what the paper is saying or guide me to references that will make it easier to understand the paper, and how one can go about building such an algorithm.
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You edited the question so my answer looks incomplete, so I'll delete it. – Douglas Zare Apr 6 '11 at 4:19
I'll be interested in knowing the mathematical basics needed to even think about such a thing. Hasn't this to do with Machine Learning? – Amit L Apr 6 '11 at 5:56
The most basic Evolutionary algorithm is probably a binary encoded genetic algorithm.
Its fairly simple to understand.
Say you are looking to find out what the maximum value of a 4 bit binary number is. Yes a silly question its obvious but it makes it easier to follow than using a game like nim or a lot of other things
First you create a population of solutions at random for example
0001
0010
0000
1110
0111
1001
then you would run them through a simulation that finds the fitness of each individual. Basically you are assessing each solution to see how good it is. In this case its simply adding up the binary values.
0001 = 1
0010 = 2
0000 = 0
1110 = 14
0111 = 7
1001 = 9
So here the strongest individuals are the last 3
Next you want to select the strongest via some selection process. There are many different ones for different styles of problems the most basic is elitism which is basically pick the best. These will be used for breeding to gain the next population
so lets pick
1110 and 1001
for a start
with these we now do crossover. Again there is lots of different techniques for different problem encodings but for this we will simply break them into 2 parts and stick them together again.
so we can pick a random location on in the 2 individuals for simplicity lets pick to split in the middle
parents
1110 -> 11 | 10
1001 -> 10 | 01
then swap them to get 2 new individuals
offspring
1101
1010
we now do this a few more times to get a new population. Again for simplicity lets make the population the same size.
parents
1001 -> 10 | 01
0111 -> 01 | 11
offspring
1011
0101
parents
1110 -> 11 | 10
0111 -> 01 | 11
offspring
1111
0110
so our new population is
1101
1010
1011
0101
1111
0110
the next part is random mutation. Basically you change something randomly. It has a low frequency of occurrence but is essential to making an effective evolutionary strategy.
so lets pick the 4th individuals highest bit. so the 4th individual is 0101. the highest bit is a 0 so we just swap it to a 1. so the 4th individual is now 1101 and the population becomes
1101
1010
1011
1101
1111
0110
the final stage in the loop is the stopping criteria. Again lots of ways of doing this depending what your doing. A simple one is to have a stopping criteria when the fitness of an individual is strong enough. So all we do is again assess the fitness of each individual
1101 = 13
1010 = 10
1011 = 11
1101 = 13
1111 = 15
0110 = 6
so our strongest individual by chance (honestly it was by chance) has reached the maximum value of a 4 bit binary number. Our stopping criteria could have been to reach some number or higher. Important to note is the average fitness of the population in the 2nd generation is much higher than the 1st.
If we did not get the stopping criteria we would continue the loop of select -> crossover-> mutate -> evaluate -> check stopping criteria.
I have no idea if this answers your question but this is about as basic an evolutionary algorithm as you can get there are a lot of different problem encodings, selection methods, crossover methods, mutation methods and stopping criteria. Further than that is multiobjective optimisation where multiple goals are trying to be solved. That is pretty much where the field is right now. These algorithms are pretty bad if you can model the situation well enough to find a solution yourself. They are very good at finding a solution when you know what a good solution is but not how to get there. They are insanely computationally expensive.
If you want to know more start on the basics of genetic algorithms they are the simplest and easiest to understand evolutionary strategies. like said its a form of machine learning and falls under computer science although a number of mathematicians have taken up the subject.
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Note that the choice of fitness criterium is essential for EA's performance. There is a multitude of functions that are maximal in $1^n$ but expected runtime varies a lot. Also, mutation is usually done by flipping every bit with probability $p$ (often $1/n$). – Raphael Apr 10 '11 at 10:18
I found this to be the clearest introductory definition of a genetic algorithm I've seen, thanks! – Danny King Apr 12 '11 at 8:22
The next step will be to choose something you want to encode for the algorithm to work with. The two most obvious (only?) things you can model (or use to model) are actions to take within the game, and current/previous game state/actions. One way to handle it could be to use your genetic sequence to weight how important information is about current/previous game states and previous moves, then use some type of weighted sum to determine future moves. The paper you linked to probably describes the encoding they used. – Brian Vandenberg Apr 12 '11 at 16:53 |
# NAG FL Interfaced01amf (dim1_inf)
Note: this routine is deprecated. Replaced by d01rmf.
## ▸▿ Contents
Settings help
FL Name Style:
FL Specification Language:
## 1Purpose
d01amf calculates an approximation to the integral of a function $f\left(x\right)$ over an infinite or semi-infinite interval $\left[a,b\right]$:
$I= ∫ab f(x) dx .$
## 2Specification
Fortran Interface
Subroutine d01amf ( f, inf, w, lw, iw, liw,
Integer, Intent (In) :: inf, lw, liw Integer, Intent (Inout) :: ifail Integer, Intent (Out) :: iw(liw) Real (Kind=nag_wp), External :: f Real (Kind=nag_wp), Intent (In) :: bound, epsabs, epsrel Real (Kind=nag_wp), Intent (Out) :: result, abserr, w(lw)
#include <nag.h>
void d01amf_ (double (NAG_CALL *f)(const double *x),const double *bound, const Integer *inf, const double *epsabs, const double *epsrel, double *result, double *abserr, double w[], const Integer *lw, Integer iw[], const Integer *liw, Integer *ifail)
The routine may be called by the names d01amf or nagf_quad_dim1_inf.
## 3Description
d01amf is based on the QUADPACK routine QAGI (see Piessens et al. (1983)). The entire infinite integration range is first transformed to $\left[0,1\right]$ using one of the identities:
$∫ -∞ a f(x) dx = ∫01 f (a-1-tt) 1t2 dt$
$∫a∞ f(x) dx = ∫01 f (a+1-tt) 1t2 dt$
$∫ -∞ ∞ f(x) dx = ∫0∞ (f(x)+f(-x)) dx = ∫01 [f(1-tt)+f( -1+t t )] 1t2 dt$
where $a$ represents a finite integration limit. An adaptive procedure, based on the Gauss $7$-point and Kronrod $15$-point rules, is then employed on the transformed integral. The algorithm, described in de Doncker (1978), incorporates a global acceptance criterion (as defined by Malcolm and Simpson (1976)) together with the $\epsilon$-algorithm (see Wynn (1956)) to perform extrapolation. The local error estimation is described in Piessens et al. (1983).
## 4References
de Doncker E (1978) An adaptive extrapolation algorithm for automatic integration ACM SIGNUM Newsl. 13(2) 12–18
Malcolm M A and Simpson R B (1976) Local versus global strategies for adaptive quadrature ACM Trans. Math. Software 1 129–146
Piessens R, de Doncker–Kapenga E, Überhuber C and Kahaner D (1983) QUADPACK, A Subroutine Package for Automatic Integration Springer–Verlag
Wynn P (1956) On a device for computing the ${e}_{m}\left({S}_{n}\right)$ transformation Math. Tables Aids Comput. 10 91–96
## 5Arguments
1: $\mathbf{f}$real (Kind=nag_wp) Function, supplied by the user. External Procedure
f must return the value of the integrand $f$ at a given point.
The specification of f is:
Fortran Interface
Function f ( x)
Real (Kind=nag_wp) :: f Real (Kind=nag_wp), Intent (In) :: x
double f (const double *x)
1: $\mathbf{x}$Real (Kind=nag_wp) Input
On entry: the point at which the integrand $f$ must be evaluated.
f must either be a module subprogram USEd by, or declared as EXTERNAL in, the (sub)program from which d01amf is called. Arguments denoted as Input must not be changed by this procedure.
Note: f should not return floating-point NaN (Not a Number) or infinity values, since these are not handled by d01amf. If your code inadvertently does return any NaNs or infinities, d01amf is likely to produce unexpected results.
2: $\mathbf{bound}$Real (Kind=nag_wp) Input
On entry: the finite limit of the integration range (if present). bound is not used if the interval is doubly infinite.
3: $\mathbf{inf}$Integer Input
On entry: indicates the kind of integration range.
${\mathbf{inf}}=1$
The range is $\left[{\mathbf{bound}},+\infty \right)$.
${\mathbf{inf}}=-1$
The range is $\left(-\infty ,{\mathbf{bound}}\right]$.
${\mathbf{inf}}=2$
The range is $\left(-\infty ,+\infty \right)$.
Constraint: ${\mathbf{inf}}=-1$, $1$ or $2$.
4: $\mathbf{epsabs}$Real (Kind=nag_wp) Input
On entry: the absolute accuracy required. If epsabs is negative, the absolute value is used. See Section 7.
5: $\mathbf{epsrel}$Real (Kind=nag_wp) Input
On entry: the relative accuracy required. If epsrel is negative, the absolute value is used. See Section 7.
6: $\mathbf{result}$Real (Kind=nag_wp) Output
On exit: the approximation to the integral $I$.
7: $\mathbf{abserr}$Real (Kind=nag_wp) Output
On exit: an estimate of the modulus of the absolute error, which should be an upper bound for $|I-{\mathbf{result}}|$.
8: $\mathbf{w}\left({\mathbf{lw}}\right)$Real (Kind=nag_wp) array Output
On exit: details of the computation see Section 9 for more information.
9: $\mathbf{lw}$Integer Input
On entry: the dimension of the array w as declared in the (sub)program from which d01amf is called. The value of lw (together with that of liw) imposes a bound on the number of sub-intervals into which the interval of integration may be divided by the routine. The number of sub-intervals cannot exceed ${\mathbf{lw}}/4$. The more difficult the integrand, the larger lw should be.
Suggested value: ${\mathbf{lw}}=800$ to $2000$ is adequate for most problems.
Constraint: ${\mathbf{lw}}\ge 4$.
10: $\mathbf{iw}\left({\mathbf{liw}}\right)$Integer array Output
On exit: ${\mathbf{iw}}\left(1\right)$ contains the actual number of sub-intervals used. The rest of the array is used as workspace.
11: $\mathbf{liw}$Integer Input
On entry: the dimension of the array iw as declared in the (sub)program from which d01amf is called. The number of sub-intervals into which the interval of integration may be divided cannot exceed liw.
Suggested value: ${\mathbf{liw}}={\mathbf{lw}}/4$.
Constraint: ${\mathbf{liw}}\ge 1$.
12: $\mathbf{ifail}$Integer Input/Output
On entry: ifail must be set to $0$, $-1$ or $1$ to set behaviour on detection of an error; these values have no effect when no error is detected.
A value of $0$ causes the printing of an error message and program execution will be halted; otherwise program execution continues. A value of $-1$ means that an error message is printed while a value of $1$ means that it is not.
If halting is not appropriate, the value $-1$ or $1$ is recommended. If message printing is undesirable, then the value $1$ is recommended. Otherwise, the value $-1$ is recommended since useful values can be provided in some output arguments even when ${\mathbf{ifail}}\ne {\mathbf{0}}$ on exit. When the value $-\mathbf{1}$ or $\mathbf{1}$ is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
Note: in some cases d01amf may return useful information.
${\mathbf{ifail}}=1$
The maximum number of subdivisions allowed with the given workspace has been reached without the accuracy requirements being achieved. Look at the integrand in order to determine the integration difficulties. If the position of a local difficulty within the interval can be determined (e.g., a singularity of the integrand or its derivative, a peak, a discontinuity, etc.) you will probably gain from splitting up the interval at this point and calling d01amf on the infinite subrange and an appropriate integrator on the finite subrange. Alternatively, consider relaxing the accuracy requirements specified by epsabs and epsrel, or increasing the amount of workspace.
${\mathbf{ifail}}=2$
Round-off error prevents the requested tolerance from being achieved: ${\mathbf{epsabs}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{epsrel}}=⟨\mathit{\text{value}}⟩$.
${\mathbf{ifail}}=3$
Extremely bad integrand behaviour occurs around one of the sub-intervals $\left(⟨\mathit{\text{value}}⟩,⟨\mathit{\text{value}}⟩\right)$ or $\left(⟨\mathit{\text{value}}⟩,⟨\mathit{\text{value}}⟩\right)$. The same advice applies as in the case of ${\mathbf{ifail}}={\mathbf{1}}$.
Extremely bad integrand behaviour occurs around the sub-interval $\left(⟨\mathit{\text{value}}⟩,⟨\mathit{\text{value}}⟩\right)$. The same advice applies as in the case of ${\mathbf{ifail}}={\mathbf{1}}$.
${\mathbf{ifail}}=4$
Round-off error is detected in the extrapolation table. The requested tolerance cannot be achieved because the extrapolation does not increase the accuracy satisfactorily; the returned result is the best that can be obtained. The same advice applies as in the case of ${\mathbf{ifail}}={\mathbf{1}}$.
${\mathbf{ifail}}=5$
The integral is probably divergent or slowly convergent.
${\mathbf{ifail}}=6$
On entry, ${\mathbf{inf}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{inf}}=-1$, $1$ or $2$.
On entry, ${\mathbf{liw}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{liw}}\ge 1$.
On entry, ${\mathbf{lw}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{lw}}\ge 4$.
${\mathbf{ifail}}=-99$
See Section 7 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 9 in the Introduction to the NAG Library FL Interface for further information.
## 7Accuracy
d01amf cannot guarantee, but in practice usually achieves, the following accuracy:
$|I-result|≤tol,$
where
$tol=max{|epsabs|,|epsrel|×|I|} ,$
and epsabs and epsrel are user-specified absolute and relative error tolerances. Moreover, it returns the quantity abserr which, in normal circumstances, satisfies
$|I-result|≤abserr≤tol.$
## 8Parallelism and Performance
d01amf is not threaded in any implementation.
The time taken by d01amf depends on the integrand and the accuracy required.
If ${\mathbf{ifail}}\ne {\mathbf{0}}$ on exit, then you may wish to examine the contents of the array w, which contains the end points of the sub-intervals used by d01amf along with the integral contributions and error estimates over these sub-intervals.
Specifically, for $i=1,2,\dots ,n$, let ${r}_{i}$ denote the approximation to the value of the integral over the sub-interval $\left[{a}_{i},{b}_{i}\right]$ in the partition of $\left[a,b\right]$ and ${e}_{i}$ be the corresponding absolute error estimate. Then, $\underset{{a}_{i}}{\overset{{b}_{i}}{\int }}f\left(x\right)dx\simeq {r}_{i}$ and ${\mathbf{result}}=\sum _{i=1}^{n}{r}_{i}$, unless d01amf terminates while testing for divergence of the integral (see Section 3.4.3 of Piessens et al. (1983)). In this case, result (and abserr) are taken to be the values returned from the extrapolation process. The value of $n$ is returned in ${\mathbf{iw}}\left(1\right)$, and the values ${a}_{i}$, ${b}_{i}$, ${e}_{i}$ and ${r}_{i}$ are stored consecutively in the array w, that is:
• ${a}_{i}={\mathbf{w}}\left(i\right)$,
• ${b}_{i}={\mathbf{w}}\left(n+i\right)$,
• ${e}_{i}={\mathbf{w}}\left(2n+i\right)$ and
• ${r}_{i}={\mathbf{w}}\left(3n+i\right)$.
Note: this information applies to the integral transformed to $\left[0,1\right]$ as described in Section 3, not to the original integral.
## 10Example
This example computes
$∫ 0 ∞ 1 (x+1) x dx .$
The exact answer is $\pi$.
### 10.1Program Text
Program Text (d01amfe.f90)
None.
### 10.3Program Results
Program Results (d01amfe.r) |
# Multiplying and dividing monomials
#### All in One Place
Everything you need for better grades in university, high school and elementary.
#### Learn with Ease
Made in Canada with help for all provincial curriculums, so you can study in confidence.
#### Instant and Unlimited Help
0/2
##### Intros
###### Lessons
1. How to multiply monomials?
2. How to divide monomials?
0/18
##### Examples
###### Lessons
1. Model and calculate each multiplication.
1. $\left( {6x} \right)\left( {2x} \right)$
2. $\left( { - 4y} \right)\left( {3y} \right)$
3. $\left( x \right)\left( { - 5x} \right)$
4. $\left( { - 2x} \right)\left( {7x} \right)$
2. Multiply.
1. $\left( x \right)\left( { - 4x} \right)$
2. $\left( {5x} \right)\left( {3x} \right)$
3. $\left( { - 2x} \right)\left( { - 6x} \right)$
4. $\left( { - 3.1x} \right)\left( { - 4x} \right)$
5. $\left( { - \frac{3}{4}x} \right)\left( {16x} \right)$
6. $\left( { - 8x} \right)\left( {2.3y} \right)$
3. Divide.
1. $\frac{{144x}}{6}$
2. $\frac{{84xy}}{{4xy}}$
3. $\frac{{9{x^2}}}{x}$
4. $\frac{{18xy}}{{3x}}$
5. $\frac{9.6{p^2}}{ - 8p}$
6. $\frac{{ - 25x}}{{ - 5x}}$
4. Find the missing dimension.
0%
##### Practice
###### Topic Notes
In this section, we will practice multiplying and dividing polynomials. Furthermore, we will use polynomial multiplication and division to solve unknowns such as, side length, and surface area, of polygons. |
# How does a Spacecraft change its orbit?
I am a high school student. I saw the latest Indian spacecraft "CHANDRYAANN 2", which was in orbit around the Earth for 14 days.....
I am so curious to know how a spacecraft changes its orbit. I have done some research on Google and it says "velocity first increases then decreases"; I do not know why the velocity increases. Can anyone explain this if I have read it correctly?
• Firing the rockets momentarily (or for a short duration) changes the height of the opposite side of the orbit. Doing this again on the opposite side of the orbit completes the orbit change. This is the most common way for a spacecraft to change it's orbit. It's called a Hohmann transfer orbit. Search on that for more information. – Wayne Goode Oct 28 '19 at 21:48
• Yet another question which can be answered "Play Kerbal Space Program" :) – IMil Oct 28 '19 at 23:10
• @yuvrajsingh You might want to search up "The Oberth Effect" as it relates to your question of why "velocity increases". – Star Man Oct 29 '19 at 1:21
• If you are willing to spend some dollars for a commercial game, Kerbal Space Program is well recognized as an amazing way to learn how orbital mechanics actually works. (or at least learn enough that you're not totally confused when they introduce real physics). We can tell you the answer, but sometimes playing around with the mechanics is required before you can feel it. – Cort Ammon Oct 29 '19 at 2:27
• @yuvrajsingh "The Oberth Effect" combines the simple adage of "what goes up, must come down" with the secret to flying: Throw yourself at the ground, and miss – Chronocidal Oct 29 '19 at 8:56
To answer your title question: By using its engines.
However you seems to be quite puzzled by the fact that velocity of an object can decrease and increase over the course of an orbit.
If the orbit is perfectly circular, the speed will always remain the same (until thrusters are used).
However, as is the case with Chandrayaan-2, most orbits are elliptical (so they have a high altitude and a low altitude points)
On an elliptical (i.e., not circular) orbit, the altitude curve looks a bit like this (1):
_ _ _
/ \ / \ / \
/ \ / \ / \
_/ \_/ \_/ \...
The altitude of the spacecraft goes up and down. When the spacecraft altitude increases, it slows down, and when the spacecraft altitude decreases, it accelerates.
This is exactly the same as riding over small hills on a bicycle (except there is no friction so you don't need to pedal at all).
[1]: It's a bit more complicated, since the altitude change is not linear with time, but you get the idea.
• It seems the OP wants to know about Keplerian orbits and the equal area law. The altitude graph looks helpful, combined with a picture of an elliptical orbit and references to a good description. – Bit Chaser Oct 28 '19 at 23:17
• Or rather, very small friction that gets abstracted to zero. The ISS has to perform an orbital boosting maneuver every month or so due to the drag from the microatmosphere at its altitude. – Draco18s no longer trusts SE Oct 29 '19 at 15:18
• Slight nitpick - the speed of a body in a circular orbit is constant, but the velocity vector is constantly changing direction because of the centripetal acceleration due to gravity. – Nuclear Hoagie Oct 29 '19 at 15:37
• @uhoh It would have been better if it was screenshotted from the message preview then annotated with freehand circles in paint, but yeah the graphics are adequate... – corsiKa Oct 30 '19 at 3:48
• @NuclearWang Is there no coordinate system that defines your velocity in a rest state as following an orbital path? A satellite-centric coordinate system? – corsiKa Oct 30 '19 at 3:50
After reading all your answers I'd like to summarise the situation.
The black circles are the circular orbits and the red ellipse is the transfer orbit.
Consider a spacecraft in the elliptical orbit. At the point P the velocity is greater than the circular orbital velocity, and that's why the distance from the centre increases. And at the point A the velocity is less than the orbital velocity, which is why the spacecraft falls back towards the centre.
So if you start at the inner circular orbit then at the point P you have to increase your velocity to change to the red elliptical orbit. When you get to the point A your velocity has fallen because you slow down as you move away from the centre. So you now have to increase your velocity once again to change to the outer circular orbit.
So there are three changes to the velocity:
1. you fire your engines at point P to increase velocity
2. your velocity decreases as you coast away from the Sun
3. you fire your engines at point A to increase velocity again
So in fact both times you fire you engine it is to increase your velocity. The total velocity change is the sum of the two increases when you fire your engine minus the decrease that happens as you coast away from the Sun.
To simplify the explanation and terminology, let's consider the case of a spacecraft orbiting Earth.
All orbits are elliptical, with the center of mass of the system (Earth, in our example) at one focus. Circular orbits are a special case where the ellipse has no eccentricity and the focii are coincident.
The orbit of our spacecraft has a perigee (closest point to Earth) and apogee (farthest point from Earth). As an aside: the terms "perigee" and "apogee" are used specifically for orbits around Earth; around an unspecified object, we use the terms "periapsis" and "apoapsis". As our spacecraft completes one orbit around Earth, it will pass through perigee once, and apogee once.
Kepler worked out that an object in orbit (our spacecraft) will not travel at a uniform speed, but instead carve equal areas in equal time as it orbits. This means that it will have its highest speed at perigee and lowest speed at apogee. A truly circular orbit is a special case where its speed actually is uniform.
Let's consider a spaceship in a circular orbit. It wants to climb to a higher orbit. It briefly fires its engine to speed up along its current trajectory. It is now no longer in circular orbit; it is now at the perigee of an elliptical orbit, and as it travels to apogee, it must slow down. If it does nothing else, it will return to that same perigee point after one complete orbit. On the other hand, when it gets to the apogee point, it can fire its engine again to speed up again. This will raise the perigee point; the right change in speed at apogee will raise the perigee to make the orbit circular again, at the new altitude.
The "velocity increases then decreases" would refer to either an interpretation of Kepler's second law, or what typically happens when a spacecraft makes an adjustment to its orbit, such as I've described.
• Thanks, one question as we move outwards shouldn't, t orbit be circular, and earth is spherical and symmetrical why orbit need to be circular sir. – Jack Rod Oct 29 '19 at 2:45
• @yuvrajsingh Not sure how to answer that question. I think the best way to understand what is going on will come from the underlying mathematical description - newtonian motion and calculus. I don't remember enough of it anymore. I'll have to defer to someone else better versed in it. – Anthony X Oct 29 '19 at 2:52
• @yuvrajsingh When you accelerate in orbit, you're moving the opposite side of the orbit; which also means you're changing the eccentricity ("circular-ness") of your orbit. When you reach the new "highest" point, you can accelerate again to bring up the other side, making the orbit circular again. But regardless, orbits certainly don't need to be circular. We use orbits for a given purpose. Each orbit has different benefits and drawbacks. The main benefit of circular orbits is that they are much simpler. – Luaan Oct 29 '19 at 7:26
Imagine your spacecraft is on elliptical orbit around space body. The body is inside ellipse (in the focus, if more specifically). When spacecraft is on the vertex of ellipse and speed up in direction of movement, another vertex of trajectory is moving far from body. Vice versa, if the spacecraft is braking (by turning engines forward and burning) in the one vertex, second vertex moves to the space body.
This is simplest scenario, without orbit precession and other things, but it illustrate the idea.
If you really want to know, just find demo of Kerbal Space Program game, it will help for sure.
• The demo of KSP isn't all that useful - it doesn't contain many parts, and while you can get to orbit, it lacks a lot of the display to project orbits etc. It's from a very old build (pre release) – Baldrickk Oct 30 '19 at 15:16
• @Baldrickk : I would recommend "Spaceflight Simulator", you can land on every planet and return, even with only the equipment found in the free version. Also, it's 2d, and easier to learn and use. – vsz Oct 30 '19 at 20:06 |
# The gas inside of a container exerts 27 Pa of pressure and is at a temperature of 150 ^o K. If the pressure in the container changes to 40 Pa with no change in the container's volume, what is the new temperature of the gas?
Nov 20, 2017
The new temperature is $= 222.2 K$
#### Explanation:
Apply Gay Lussac's law
${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$ at constant volume
The initial pressure is $= {P}_{1} = 27 P a$
The initial temperature is $= {T}_{1} = 150 K$
The final pressure is ${P}_{2} = 40 P a$
The final temperature is
${T}_{2} = {P}_{2} / {P}_{1} \cdot {T}_{1} = \frac{40}{27} \cdot 150 = 222.2 K$
${220}^{\circ} K$
#### Explanation:
Using the combined Gas Law, $\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$, you can simplify it to ${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$ since volume is constant.
Since to want to find ${T}_{2}$(Temperature 2), you rearrange the formula to become ${T}_{2} = \frac{{T}_{1} \cdot {P}_{2}}{P} _ 1$
Then you can plug and chug
${T}_{2} = \frac{{150}^{\circ} K \cdot 40 P a}{27 P a}$
$= 222.222222222 {\ldots}^{\circ} K$
$= {220}^{\circ} K$ (with Significant Figures) |
# Similarity/dissimilarity matrix over classes
I am trying to get something like a confusion matrix for different classes, but without training a model.
The idea is to use some kind of distance between classes.
The data set is like this, just for better understanding
Class1
A B C ... YZ
500 3 2 ... 43
500 15 1 ... 2
300 23 9 ... 68
Class 2
A B C ... YZ
100 56 2 ... 58
100 35 6 ... 34
300 23 9 ... 1
...
Class N
The classes have different row number.
I have only one idea so far - Calculate means for each column separately (one resulting row) and measure distances between these rows
What is the right way to calculate the similarity of classes? Maybe a python library for that?
• As you are anew contributor here is a hint: if an answer actually answers the question, you should mark it as "accepted" so that it is removed from the list of unanswered questions. – cdalitz Nov 27 '20 at 14:18
Your suggestion of summing up the (squared) distances between the class means actually is an established method and is the trace of the between scatter matrix:
$$S_B = \sum_{i=1}^{C} n_i (\vec{\mu}_i - \vec{\mu})\cdot (\vec{\mu}_i - \vec{\mu})^T$$
It is however more useful to set this value in relation to the spread inside the classes, or the trace of the within scatter matrix:
$$S_W = \sum\limits_{i=1}^{C} \left(\sum\limits_{\vec{x}\in\omega_i}(\vec{x} - \vec{\mu}_i)\cdot (\vec{x} - \vec{\mu}_i)^T\right)$$
Measures for class separability are thus
$$J_1 = \frac{trace(S_B)}{trace(S_W)} \quad\mbox{ or }\quad J_2 = trace\Big(S_W^{-1} \cdot S_B\Big)$$
where trace stands for the sum of the diagonal elements, which represent the variances.
The normalization with $$trace(S_W)$$ makes the measure scale invariant and measures how large the distance between the classes is compared to the distance within the classes.
• Thank you very much for your answer! I really appreciate your attention to my question. Could you please explain me the difference between the approaches, why one is better than another? – Statsnewbie Nov 27 '20 at 14:11
• Which approaches do you mean? $J_1$ and $J_2$? These do not make a great difference, I think. Or do you mean the normalization with $trace(S_W)$ compared to your approach that only measures $trace(S_B)$? I have added an explanation for this normalization. – cdalitz Nov 27 '20 at 14:15
• Thank you for help, your answer makes me happy 😊 – Statsnewbie Nov 27 '20 at 14:56 |
## C.1 Essential elements of DS-projects
### C.1.1 Structure
The following notes provide some ideas how to structure a data science project.
#### Motivation
Any project should begin by answering the question: What motivates this project — and why should we care? A motivated data science project should link questions with data. Thus, there are two principle ways to motivate a data science project:
1. Starting from questions: Ask some research question(s) that you hope to answer with data. Which data could address or answer the question?
2. Starting from data: Why use this data? Which interesting questions can it inform or answer?
Note that the questions should be expressed without referring to the data (i.e., rather than “What is the correlation between variable x and variable y?” try a more general statement, like
“What is the relation between someone’s activity level and mood?”).
#### Data
The data used in a project should meet the following criteria:
• Availability: The data should be publicly accessible and included in the submitted project (in csv format).56
• Amount: There are no specific limits or restrictions regarding data size, but the dataset should not be trivial (i.e., larger datasets are better than smaller ones, data in several files better than 1, etc.).
• Variety: The data or the analysis should typically include multiple types of variables (e.g., categorical, logical, numeric, text, dates or times).
• Variables: Include a codebook (overview over all variables, their types, and ranges).
• References: Include citations to and references of all sources (e.g., of data, corresponding articles, and online sources).
#### Data exploration (EDA)
• Clean up data, etc.
• Check data for missing values, outliers, etc.
• Provide an overview of distributions of key variables, relationships between variables, etc.
• Join and/or re-format tables, select cases or variables, etc.
#### Data transformation and visualization
• Select, add, or change variables
• Compute descriptive summary tables and corresponding visualizations of various types (that make sense of the data).
• Can you create a tidy version of the data? How or why not?
• Try to descriptively answer your initial research questions (e.g., by creating corresponding summary tables or visualizations).
#### Conclusion
• Results: Which answer(s) does your analysis suggest (to the questions raised in the introduction)?
• Limitations: What limits your possible conclusions? What could overcome these limitations?
• Implications: What follows from your analysis? What should be done next?
Due to the large flexibility and wide variance of possible DS-projects, it is challenging to describe and evaluate them in a coherent fashion. Students always ask for a list of boxes they can check off, but including a very detailed list here would inevitably limit the range of possible projects. For instance, whereas most projects will probably use some combination of numeric and character data, other projects may focus primarily on text- or time-based data. Although providing descriptive statistics on the latter may make sense, these will play a very different role in the project.
Similarly, it makes absolutely no sense to list a required set of commands or packages of an excellent data science project. Most projects submitted in the context of this course tend to heavily rely on the tidyverse (Wickham, Averick, et al., 2019) packages dplyr, ggplot, tibble, and tidyr, but it would be wrong to censor anyone for providing sound analysis in base R or other packages.
Most projects are based on, inspired by, and dependent upon a particular dataset. Alternatively, some of the most innovative projects may involve creating simulations or programming functions that facilitate some analysis or visualization. If someone is dissatisfied with existing functions for quantifying text (e.g., the ds4psy functions count_chars() and count_words()) or for computing with dates and times (e.g., the ds4psy functions diff_dates() and diff_times()), this could be the beginning of a fantastic project. However, such projects would not begin with a particular dataset, yet involve data later (for testing and demonstrating the use of new functions).
To satisfy the understandable need for transparency without limiting the scope of possible projects, here is a description of the project’s minimial requirements (for passing the course) and some ways of scoring bonus points (for earning a good or excellent grade).
#### Minimal requirements
To pass this project requirement, here is what you need to do:
1. Ask at least one non-trivial question that can be addressed by data.
2. Use data and R commands (in a .Rmd document) to answer your question(s).
• Include some (descriptive) quantification of the data.
• Include some visualization of your results.
1. Document the data, your answer(s), and the solution process (in an .html or .pdf document).
#### Bonus points
Here are some suggestions what you can do to distinguish your project and exceed the minimum requirements:
• Explicitly document all steps, the R packages used, and your code.
• Use multiple datasets and combine them.
• Make sure that your data is encoded in csv format (if possible) and can easily be exchanged between platforms.
• Use multiple data types (e.g., categorical and continuous numbers, logical variables, text, time, …).
• Screen all key variables of your data prior to using them.
• Make sure that your code can be executed and explain the purpose of new variables and data transformations.
• Use informative summary tables for showing descriptive statistics.
• Use different visualizations that support your verbal interpretations.
• Use informative table and figure captions and APA format for all references.
• Create your own functions for analyzing or visualizing data.
• Justify your interpretations and conclusions.
• Critically reflect on the implications and limitations of your data and/or analysis.
None of these points is strictly necessary, but some of them must be present to earn a “good” or “excellent” grade.
### C.1.3 Submitting a project
All projects should contain a header that clearly indicates a project title, your name, and the current date.
#### What to include
Please submit a zipped-archive of a file folder or an R-project, including all scripts and data (in csv format):57
• Code: An executable .R or .Rmd script containing all code.
• Output: 1 document suited to understand and evaluate the project (either an html-file that combines all text and code OR a pdf file that contains all text and figures).
If you are hosting your project online (e.g., on GitHub or some other site), you can also submit a link to it.
An email with your attached project should be sent to your instructor, no later than
Tuesday, September 01, 2020.
Alternatively, submit your project to the corresponding folder on the Ilias web platform.
### References
Wickham, H., Averick, M., Bryan, J., Chang, W., McGowan, L. D., François, R., … Yutani, H. (2019). Welcome to the tidyverse. Journal of Open Source Software, 4(43), 1686. https://doi.org/10.21105/joss.01686
1. Contact the instructor in case you want to use data that cannot be included (e.g., due to legal or other restrictions).
2. Contact the instructor early enough if the size of your archive exceeds 5MB. |
# poisson distribution of chocolate
Let the number of chocolate drops in a certain type of cookie have a Poisson distribution. We want the probability that a cookie of this type contains at least two chocolate drops to be greater than 0.99. Find the smallest value of the mean that the distribution can take.
-
Let's say the number $D$ of drops has a distribution $D \sim \mathrm{Poisson}(\lambda)$. Then \begin{align*} P(D \ge 2) &= 1 - P(D < 2)\\ &= 1 - P(D = 0) - P(D = 1)\\ &= 1 - \exp(-\lambda)\cdot (1 + \lambda) \end{align*} So $P(D \ge 2) \ge 0.99$ iff $\exp(-\lambda)(1 + \lambda) \le \frac 1{100}$. As $\lambda \mapsto \exp(-\lambda)(1+ \lambda)$ has a negative derivative on $(0,\infty)$, it is strictly decreasing. So there is a unique $\lambda_0$ with $\exp(-\lambda_0)(1+\lambda_0) = \frac 1{100}$ (there is no closed form for $\lambda_0$ in terms of elementary functions, Wolfram|Alpha tells us it is approximately 6.6384). This is the minimal mean you looked for.
@user48495 "We know that the MGF of a multinomial distribution is: $M(t_1, t_2,...t_k-1) = (p_1*e^t_1 + .... + p_k+1*e^t_k+1 + p_k)^n$ hence, the MGF of $X_2$, $X_3$, .... $X_k-1$ is $M(0, t_2,...t_k-1) = (p_1 + .... + p_k+1*e^t_k+1 + p_k)^n$" -adapted from rad at actuarialoutpost.com/actuarial_discussion_forum/archive/… – raindrop Jun 14 '13 at 1:08 |
The Sour Gas Treatment and Removal Technology
• Journal title : Journal of Energy Engineering
• Volume 25, Issue 1, 2016, pp.171-176
• Publisher : The Korea Society for Energy Engineering
• DOI : 10.5855/ENERGY.2015.25.1.171
Title & Authors
The Sour Gas Treatment and Removal Technology
Kim, Y.C.; Cho, J.D.; Oh, C.S.;
Abstract
Sour gas is natural gas or any other gas containing significant amounts of hydrogen sulfide ($\small{H_2S}$). Natural gas is usually considered sour gas if there are more than 5.7 milligrams of $\small{H_2S}$ per cubic meter of natural gas, which is equivalent to approximately 4 ppm by volume under standard temperature and pressure We have surveyed on the treatment and removal technology of sour gas, sour gas include a lot of hydrogen sulfide($\small{H_2S}$), Carbon dioxide($\small{CO_2}$), utane($\small{C_4H_{10}}$) and mercaptan($\small{C_nH_{4n-1}SH}$) etc. We need high technology for development for these kinds of raw gases and we should specially take care of treating and removal of theses raw gases. Therefor we are going to describe about these kinds of raw gases and about methods how to treat these kinds of gases.
Keywords
sour gas;hydrogen sulfide;carbon dioxide;acid gas;mercaptan;
Language
Korean
Cited by
References
1.
Darkhan Duissenov: NTNU (2012) pp45.
2.
Darkhan Duissenov: NTNU (2013) pp101.
3.
www.shell.com/globalsolutions (2014) 2.
4.
Thomas Lockhart, Francesco Crescenzi: (2012) pp237-269.
5.
Charles J. Mart (2011) Developing sour gas resources with controlled freeze zone ExxonMobil Upstream Research, pp14.
6.
Lallemand F., Minkkinen A. (2002) Processes combine to assist acid-gas handling, reinjection, Oil & Gas Journal, 21 January.
7.
Mehdi Panahi (2011) Cryogenic CO2/H2S capture technologies for remote natural gas processing( Trial Lecture), pp41.
8.
EPA(Environmental Protection Agency) (1993) Report to Congress on Hydrogen Sulfide Air Emissions Associated with the Extraction of Oil and Natural Gas. EPA-453/R-93-045, pp.III-4.
9.
Lana Skrtic (2006) Hydrogen Sulfide, Oil and Gas, and People's Health, University of California, Berkeley, pp77. |
# Knight's Tour Revisited Revisited
Title: Knight's Tour Revisited Revisited Author: Lukas Koller (lukas /dot/ koller /at/ tum /dot/ de) Submission date: 2022-01-04 Abstract: This is a formalization of the article Knight's Tour Revisited by Cull and De Curtins where they prove the existence of a Knight's path for arbitrary n × m-boards with min(n,m) ≥ 5. If n · m is even, then there exists a Knight's circuit. A Knight's Path is a sequence of moves of a Knight on a chessboard s.t. the Knight visits every square of a chessboard exactly once. Finding a Knight's path is a an instance of the Hamiltonian path problem. A Knight's circuit is a Knight's path, where additionally the Knight can move from the last square to the first square of the path, forming a loop. During the formalization two mistakes in the original proof were discovered. These mistakes are corrected in this formalization. BibTeX: @article{Knights_Tour-AFP, author = {Lukas Koller}, title = {Knight's Tour Revisited Revisited}, journal = {Archive of Formal Proofs}, month = jan, year = 2022, note = {\url{https://isa-afp.org/entries/Knights_Tour.html}, Formal proof development}, ISSN = {2150-914x}, } License: BSD License |
#### Approach #1: Trie + Set Intersection [Time Limit Exceeded]
Intuition and Algorithm
We use two tries to separately find all words that match the prefix, plus all words that match the suffix. Then, we try to find the highest weight element in the intersection of these sets.
Of course, these sets could still be large, so we might TLE if we aren't careful.
Complexity Analysis
• Time Complexity: where is the number of words, is the maximum length of a word, and is the number of queries. If we use memoization in our solution, we could produce tighter bounds for this complexity, as the complex queries are somewhat disjoint.
• Space Complexity: , the size of the tries.
#### Approach #2: Paired Trie [Accepted]
Intuition and Algorithm
Say we are inserting the word apple. We could insert ('a', 'e'), ('p', 'l'), ('p', 'p'), ('l', 'p'), ('e', 'a') into our trie. Then, if we had equal length queries like prefix = "ap", suffix = "le", we could find the node trie['a', 'e']['p', 'l'] in our trie. This seems promising.
What about queries that aren't equal? We should just insert them like normal. For example, to capture a case like prefix = "app", suffix = "e", we could create nodes trie['a', 'e']['p', None]['p', None].
After inserting these pairs into our trie, our searches are straightforward.
Complexity Analysis
• Time Complexity: where is the number of words, is the maximum length of a word, and is the number of queries.
• Space Complexity: , the size of the trie.
#### Approach #3: Trie of Suffix Wrapped Words [Accepted]
Intuition and Algorithm
Consider the word 'apple'. For each suffix of the word, we could insert that suffix, followed by '#', followed by the word, all into the trie.
For example, we will insert '#apple', 'e#apple', 'le#apple', 'ple#apple', 'pple#apple', 'apple#apple' into the trie. Then for a query like prefix = "ap", suffix = "le", we can find it by querying our trie for le#ap.
Complexity Analysis
• Time Complexity: where is the number of words, is the maximum length of a word, and is the number of queries.
• Space Complexity: , the size of the trie.
Analysis written by: @awice. |
# Multivariate CoV
Polar coordinates are one example of a change of variables in multivariate integration. Other important examples of this arise in triple integrals, including cylindrical and spherical coordinates. Before we do that, it might make sense to talk about change of variables in general.
## An elliptic example
What is the volume below the graph of $$f(x,y)=4-(x^2+4y^2)$$ and over the $xy$-plane, as shown below? Note that this image is very similar to the polar image we discussed last time.
### Polar coordinates don't help
Note that polar coordinates don't help here, since $$4- ((r\cos(\theta))^2 + 4(r\sin(\theta))^2)$$ doesn't simplify in any meaningful way.
### New coordinates
To describe the elliptical domain, we're going to introduce new variables $s$ and $t$ by defining their relationship with $x$ and $y$:
\begin{align} x &= 2\,s\cos(t)\\ y &= s\sin(t). \end{align}
For a fixed value of $s$, these equations parametrize an ellipse that's twice as wide as it is tall. For the set of all values of $s$ from zero to one, we trace out a solid ellipse.
### New coordinates (cont)
We might think that this new coordinate could help because the function simplifies under the transformation:
\begin{align} 4-(x^2+4y^2) &= 4-\left((2s\cos(t))^2 + 4(s\sin(t))^2\right)\\ &=4-4s^2(\cos^2(t)+\sin^2(t)) = 4-4s^2. \end{align}
Bringing us to:
$$\iint_{E} \left(4-(x^2+4y^2)\right) dA = \int_{\color{red}?}^{\color{red}?} \int_{\color{red}?}^{\color{red}?} (4-4s^2)\,{\color{red}?}\,ds\,dt.$$
### New coordinates (cont 2)
from a geometric perspective, we might say that the parametric equations $$x = 2\,s\cos(t), \: \: y = s\sin(t)$$ map the rectangle $[0,1]\times[0,2\pi]$ in the $st$-plane to the ellipse in the $xy$-plane.
### Almost there
At this point, we've got
$$\iint_{E} \left(4-(x^2+4y^2)\right) dA = \int_{0}^{2\pi} \int_0^1 (4-4s^2)\,{\color{red}?}\,ds\,dt.$$
We've just got to figure out that $\color{red}?$.
## The last piece
The last piece of the puzzle is the $\color{red}?$ in
$$\int_0^{2\pi} \int_0^1 (4-4s^2)\,{\color{red}?}\,ds\,dt.$$
It should be exactly analogous to the $\color{red}r$ in polar coordinates:
$$\int_{\alpha}^{\beta} \int_a^b F(r,\theta)\,{\color{red}r}\,dr\,d\theta.$$
### Last piece (cont)
Recall that the $r$ in the $r\,dr\,d\theta$ of polar coordinates accounts for the fact that the lines of constant $\theta$ and arcs of constant $r$ don't break a polar region doesn't break up evenly.
In the context of Change of Variables, this boils down to the fact that the mapping from one coordinate system to the other distorts area.
### Last piece (cont 2)
Now our transformation, $x=2s\cos(t)$ and $y=s\sin(t)$, also distorts area, but it distorts in the $x$ direction by twice as much.
So, perhaps $\displaystyle dA = 2s\,ds\,dt$?
### Last piece (cont 3)
Assuming that works, we can now finish off the integral:
\begin{align} \iint_{E} \left(4-(x^2+4y^2)\right) dA &= \int_0^{2\pi} \int_0^1 (4-4s^2)\,2s\,ds\,dt \\ &= 16\pi\int_0^1(s-s^3)\,ds = 4\pi. \end{align}
## The Jacobian
When we have a function (often called a transformation in this context) $T:\mathbb R^2\to\mathbb R^2$ there is a a standard tool called the Jacobian used to measure how it distorts area. Write $T$ componentwise:
$$T(s,t) = (x(s,t),y(s,t)), \text{ then}$$ $$JT = \left|\frac{\partial(x,y)}{\partial(s,t)}\right| = \left|\Big| \begin{matrix} \partial x/\partial s & \partial x/\partial t \\ \partial y/\partial s & \partial y/\partial t \end{matrix} \Big|\right|.$$
That's a determinant inside an absolute value.
### Jacobian example
Let's find the Jacobian for the general elliptic coordinates:
$$x = R_1\,s\,\cos(t) \: \text{ and } \: y = R_2\,s\sin(t).$$ \begin{align} \left|\frac{\partial(x,y)}{\partial(s,t)}\right| &= \left|\begin{matrix} R_1\cos(t) & -R_1\,s\,\sin(t) \\ R_2\sin(t) & R_2\,s\cos(t)\end{matrix} \right| \\ &= R_1R_2s\cos^2(t) + R_1R_2s\sin^2(t) = R_1R_2s. \end{align}
This agrees with our prior example where $R_1=2$ and $R_2=1$.
## Integrating over a parallelogram
Let $P$ denote the parallelogram bound by the lines
$$\begin{array}{cccc} 3x-y=0 & 3x-y=8 & 3y-x=0 & 3y-x=8 \end{array}$$
### Parallelogram (cont)
We wish to set up a single, iterated integral representing
$$\iint_P xy\,dA.$$
To do so, introduce $s$ and $t$ defined by
$$s = 3x-y \: \text{ and } \: t = 3y-x.$$
For then, $0<s<8$ and $0<t<8$. Solving, we get
$$x = (3 s + t)/8 \: \text{ and } \: y = (s + 3 t)/8.$$
### Parallelogram (cont 2)
The Jacobian is
$$\left|\frac{\partial(x,y)}{\partial(s,t)}\right| = \left|\begin{matrix} 3/8 & 1/8 \\ 1/8 & 3/8\end{matrix} \right| = \frac{1}{8}.$$
Thus, the integral is
\begin{align} \iint_{p} xy \, dA &= \frac{1}{8}\int_0^{8} \int_0^8 \frac{3s+t}{8}\frac{3t+s}{8}\,ds\,dt. \end{align}
## Last virtual example!
As a final example, let's find the area of the region $R$ bound between the curves $$y=2x, \: y=\frac{1}{2} x, \: y=\frac{1}{x}, \text{ and } y=\frac{2}{x}.$$
### Last (cont)
We wish to set up a single, iterated integral representing
$$\iint_R 1 \, dA.$$
To do so, introduce $s$ and $t$ defined by
$$s = y/x \: \text{ and } \: t = xy.$$
For then, $1/2<s<2$ and $1<t<2$.
### Last (cont 2)
Since, $s=y/x$, we have $y=sx$. We can plug that into the other equation ($t=xy$) to get $t=sx^2$. Thus, $$x=\sqrt{t/s}.$$ Once we know that, we can find that $$y = s\sqrt{t/s} = \sqrt{st}.$$
### Last (cont 3)
Since $$x = \sqrt{t/s} = s^{-1/2}t^{1/2} \: \text{ and } \: y = \sqrt{s t} = s^{1/2}t^{1/2},$$
the Jacobian is
$$\left|\frac{\partial(x,y)}{\partial(s,t)}\right| = \left|\begin{matrix} -\frac{1}{2}s^{-3/2}t^{1/2} & \frac{1}{2}s^{-1/2}t^{-1/2} \\ \frac{1}{2}s^{-1/2}t^{1/2} & \frac{1}{2}s^{1/2}t^{-1/2} \end{matrix} \right| = -\frac{1}{4s}-\frac{1}{4s} = -\frac{1}{2s}.$$
Don't forget to take the absolute value to get $\frac{1}{2s}$.
Putting this all together, the area is $$\iint_R 1 \, dA = \int_1^2 \int_{1/2}^2 1\,\frac{1}{2s} \, ds \, dt = \ln(2).$$ |
# How do you solve 4/(9y-6)=2?
Feb 6, 2016
Step by step explanation is given blow
#### Explanation:
$\frac{4}{9 y - 6} = 2$
Think of this as a proportion
$\frac{4}{9 y - 6} = \frac{2}{1}$
We know if $\frac{a}{b} = \frac{c}{d}$ Then $\frac{b}{a} = \frac{d}{c}$
Basically we are flipping and that would help us
$\frac{9 y - 6}{4} = \frac{1}{2}$
Let us multiply $4$ on both the sides
$\cancel{4} \times \frac{9 y - 6}{\cancel{4}} = 4 \times \frac{1}{2}$
$9 y - 6 = 2$
To solve for $y$ we need to undo all that has been done to it, for this we shall use the inverse operations.
Let us start with $- 6$ we add $6$ on both the sides and we get
$9 y - \cancel{6} + \cancel{6} = 2 + 6$
$9 y = 8$
$y$ is multiplied by $9$ to undo it we divide both sides by $9$
$\frac{\cancel{9} y}{\cancel{9}} = \frac{8}{9}$
$y = \frac{8}{9}$
Let us check if our solution is correct by plugging $y = \frac{8}{9}$ in the original problem.
$\frac{4}{9 y - 6} = 2$
$\frac{4}{9 \left(\frac{8}{9}\right) - 6} = 2$
$\frac{4}{8 - 6} = 2$
$\frac{4}{2} = 2$
$2 = 2 \quad$ this is a true statement so the solution is valid.
$y = \frac{8}{9}$ |
Outlook: InspireMD Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : SellWait until speculative trend diminishes
Time series to forecast n: 11 Mar 2023 for (n+4 weeks)
Methodology : Multi-Instance Learning (ML)
## Abstract
InspireMD Inc. Common Stock prediction model is evaluated with Multi-Instance Learning (ML) and Multiple Regression1,2,3,4 and it is concluded that the NSPR stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: SellWait until speculative trend diminishes
## Key Points
1. What are buy sell or hold recommendations?
2. How do predictive algorithms actually work?
3. How accurate is machine learning in stock market?
## NSPR Target Price Prediction Modeling Methodology
We consider InspireMD Inc. Common Stock Decision Process with Multi-Instance Learning (ML) where A is the set of discrete actions of NSPR stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Multiple Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Multi-Instance Learning (ML)) X S(n):→ (n+4 weeks) $\stackrel{\to }{S}=\left({s}_{1},{s}_{2},{s}_{3}\right)$
n:Time series to forecast
p:Price signals of NSPR stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## NSPR Stock Forecast (Buy or Sell) for (n+4 weeks)
Sample Set: Neural Network
Stock/Index: NSPR InspireMD Inc. Common Stock
Time series to forecast n: 11 Mar 2023 for (n+4 weeks)
According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: SellWait until speculative trend diminishes
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for InspireMD Inc. Common Stock
1. However, an entity is not required to separately recognise interest revenue or impairment gains or losses for a financial asset measured at fair value through profit or loss. Consequently, when an entity reclassifies a financial asset out of the fair value through profit or loss measurement category, the effective interest rate is determined on the basis of the fair value of the asset at the reclassification date. In addition, for the purposes of applying Section 5.5 to the financial asset from the reclassification date, the date of the reclassification is treated as the date of initial recognition.
2. A single hedging instrument may be designated as a hedging instrument of more than one type of risk, provided that there is a specific designation of the hedging instrument and of the different risk positions as hedged items. Those hedged items can be in different hedging relationships.
3. An entity may retain the right to a part of the interest payments on transferred assets as compensation for servicing those assets. The part of the interest payments that the entity would give up upon termination or transfer of the servicing contract is allocated to the servicing asset or servicing liability. The part of the interest payments that the entity would not give up is an interest-only strip receivable. For example, if the entity would not give up any interest upon termination or transfer of the servicing contract, the entire interest spread is an interest-only strip receivable. For the purposes of applying paragraph 3.2.13, the fair values of the servicing asset and interest-only strip receivable are used to allocate the carrying amount of the receivable between the part of the asset that is derecognised and the part that continues to be recognised. If there is no servicing fee specified or the fee to be received is not expected to compensate the entity adequately for performing the servicing, a liability for the servicing obligation is recognised at fair value.
4. For the purpose of determining whether a forecast transaction (or a component thereof) is highly probable as required by paragraph 6.3.3, an entity shall assume that the interest rate benchmark on which the hedged cash flows (contractually or non-contractually specified) are based is not altered as a result of interest rate benchmark reform.
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
InspireMD Inc. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. InspireMD Inc. Common Stock prediction model is evaluated with Multi-Instance Learning (ML) and Multiple Regression1,2,3,4 and it is concluded that the NSPR stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: SellWait until speculative trend diminishes
### NSPR InspireMD Inc. Common Stock Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementCB3
Balance SheetBa3Ba1
Leverage RatiosBa1Ba3
Cash FlowCBaa2
Rates of Return and ProfitabilityBaa2Baa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 91 out of 100 with 747 signals.
## References
1. Swaminathan A, Joachims T. 2015. Batch learning from logged bandit feedback through counterfactual risk minimization. J. Mach. Learn. Res. 16:1731–55
2. Imai K, Ratkovic M. 2013. Estimating treatment effect heterogeneity in randomized program evaluation. Ann. Appl. Stat. 7:443–70
3. Meinshausen N. 2007. Relaxed lasso. Comput. Stat. Data Anal. 52:374–93
4. Cheung, Y. M.D. Chinn (1997), "Further investigation of the uncertain unit root in GNP," Journal of Business and Economic Statistics, 15, 68–73.
5. Ruiz FJ, Athey S, Blei DM. 2017. SHOPPER: a probabilistic model of consumer choice with substitutes and complements. arXiv:1711.03560 [stat.ML]
6. N. B ̈auerle and A. Mundt. Dynamic mean-risk optimization in a binomial model. Mathematical Methods of Operations Research, 70(2):219–239, 2009.
7. A. Y. Ng, D. Harada, and S. J. Russell. Policy invariance under reward transformations: Theory and application to reward shaping. In Proceedings of the Sixteenth International Conference on Machine Learning (ICML 1999), Bled, Slovenia, June 27 - 30, 1999, pages 278–287, 1999.
Frequently Asked QuestionsQ: What is the prediction methodology for NSPR stock?
A: NSPR stock prediction methodology: We evaluate the prediction models Multi-Instance Learning (ML) and Multiple Regression
Q: Is NSPR stock a buy or sell?
A: The dominant strategy among neural network is to SellWait until speculative trend diminishes NSPR Stock.
Q: Is InspireMD Inc. Common Stock stock a good investment?
A: The consensus rating for InspireMD Inc. Common Stock is SellWait until speculative trend diminishes and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of NSPR stock?
A: The consensus rating for NSPR is SellWait until speculative trend diminishes.
Q: What is the prediction period for NSPR stock?
A: The prediction period for NSPR is (n+4 weeks)
## People also ask
What are the top stocks to invest in right now? |
# Lamani
1 Wisdom Point
The Lamani are a particularly new race of humanoid animals, and no one is sure where they came from. They travel in small diverse family groups, never settling in one place for long. They are usually friendly, if sometimes a little gruff depending on their animal traits, and are always very quiet until they are comfortable with new people. They have a mysterious natural wisdom that is uniquely their own.
Lamani can choose to take the form of any of the following animal kingdoms: Wolf Cousins (i.e. wolves), Wolfcats (i.e. badger/fox), Cat Cousins (i.e. lions, tigers, panthers), Strongfeet (i.e. Rabbits/Kangaroos), Rooters (i.e. Warthogs), Massives (i.e. Elephants, Rhinos), Tree Dwellers (i.e. Squirrels), Rodent (i.e. Ratfolk), Shells (i.e. Turtles), Dragon Cousins (i.e. lizards), Caws (i.e. Crows), or Squawks (i.e. Parrots). Snakes are a forbidden option among the Lamani, as they do not wish to become confused with the cursed Gorgons.
#### 1 Required Step
• You gazed into the fountain and became Lamani. |
## How To Decode A Message In Matlab
Files for LDPC code simulation over the AWGN channel. We have taken input from the user which will be coded by the convolutional encoder. For most non-professional applications, all you need to create a GIF is your smartphone. (v)scale: this array contains the factor by which each bit was scaled before reception. today introduced Release 2018a (R2018a) with a range of new capabilities in MATLAB and Simulink. The input messages can be categorized as analog (continuous form) or digital (discrete form). Encoding And Decoding of a Message in the Implementation of Elliptic Curve Cryptography using Koblitz's Method Article (PDF Available) · August 2010 with 1,616 Reads How we measure 'reads'. A decoding failure occurs if bchdec detects more than T errors in a row of code, where T is the number of errors per codeword that the decoder is capable of correcting. DAT file is being converted. These are often used to recover messages sent over a noisy channel, such as a binary symmetric channel. Decoding: Successive Cancellation Decoding. I have to implement the polar coding scheme in matlab for wireless communication. Interpolation or Up sampling. In MATLAB, a class is defined using the classdef method under which the name of the class, the properties of the object in the class, and the functions This is the first of two blog on OOP in Matlab. The same codebook could be used to "encode" a plaintext message into a coded message or "codetext", and "decode" a codetext back into plaintext message. But unfortunately, I have absolutley no experience with Labview and cannot work out how the. But actually I could not find any commands in matlab for the polar codes. Choose a web site to get translated content where available and see local events and offers. If the input signal is a MATLAB array, then the output signal is also a MATLAB array. Download Project Files. Giving an application the capability to perform tasks repetitively with MATLAB is an essential part of creating an application of any complexity. The decode messages you can simply copy and paste into MATLAB. N and M are numbers between 0 and 255 that are incrementing and decrementing, respectively. symbols and 100 information symbols to LoRa chirp and successfully Hello Sakshama, How to choose sampling frequecy? Could you please explain why you choose it same as bandwidth? Can I choose the same fs for normal chirp. To cite MATLAB (and in this case a toolbox) you can use this: MATLAB and Statistics Toolbox Release 2012b, The MathWorks, Inc. This implements an easy and efficient Huffman coding algorithm and brute force decoding algorithm. Shorten a (31,26) BCH code to an (11,6) BCH code and use it to encode and decode random binary data. Matlab and Mathematica Projects for $30 -$250. I was successful in converting image to base64 and post it to the database by converting it into xml. LDPCDecoder System object uses the belief propagation algorithm to decode a binary LDPC code, which is input to the object as the soft-decision output (log-likelihood ratio of received bits) from demodulation. Resear h Methods, ELEC6021 (EZ619) S Chen What's New in Communi ations Imagine a few s enarios: { In holiday, use your fan y mobile phone to take pi ture and send it to a friend. You'll see the audio spectrum but not the message. MATLAB is a powerful tool for engineering purposes but because of its nature, is very slow in executing functions that take a long time to execute. Re: Software to decode FSK recorded in a WAV file? Posted by Dave Pedley (via moderator) on Sat, 11/01/2003 - 02:10. Here is a hint: The receiver receives: z(t) = x(t) + y(t)*cos(2$\pi f_c$), where x(t) is the. Matlab imshow weglassen NaN. If there is more than one node in the table,. Section references are to the Wiley textbook a. I am trying to decode an encrypted message. Humans don’t get bored performing a task once. In particular, field names in JSON objects that are not valid MATLAB identifiers might be altered by the makeValidName function. Split the text into pairs of letters (bigrams) and apply the following rules according to the letters positions in the grid : - if the 2 letters are on the same line, replace them by the ones on their left (loop to the right if the edge of the grid is reached), Example: DE is decrypted CD. Could you please tell me how you solved it ? Thanks a lot! Hi Spiros I am working on DVB-S2 in matlab i faced same problem with BCH en/decoder would you plz tell me how i can use bch decoder for dvb-s2 normal fecframes(64800) with different rates e. Calculate the noise floor and preamble correlation with the filter() function over a short time window. To run the test, run “turbo_top” in matlab env. The goal of the project is to successfully detect and reconstruct perfect QR-code pattern and then decode and extract the message and information within. Cross-validation is a model validation technique for assessing how the results of our decoding analysis will generalize to an independent data set. In the Serial Decode dialog box, assign the on-screen trace to Decode 1 by turning Decode 1 on. So decoding means interpreting the meaning of the message. Now LSB has less impact on image. Here is how to set Google to do so automatically. How to Encode and Decode Using the Vigènere Cipher. Decoding ADS-B in MATLAB Video Tutorial Over on YouTube the official MATLAB channel has uploaded a new video that is a tutorial on setting up ADS-B decoding in MATLAB. The goal of this lab is to design and implement bandpass FIR filters in Matlab, and to do the decoding automatically. Hope this helps. I know I can use words_in_message (index, :) in the command window to get the full word, but it errors out in the function. m, it will start the GUI which will look something as shown in below figure: That’s the GUI used for DTMF Decoder using MATLAB. ' in EdsacPC). Julius Caesar was one of the first people to write in code. Create a program to encode or decode a message using the Vigenère Cipher. The decode function aims to recover messages that were encoded using an error-correction coding technique. [170,2,44,M] is message 2 (M2). How to create a stopwatch in matlab ? Hi,all I am working on an application wherein I need to develop a real time stop watch showing how much time elapsed in min:sec:milliseconds format. IP Camera interface to Matlab Showing 1-6 of 6 messages. The application displays a message telling the user that the number of Howdies has become excessive and then calls break to end the loop. Received message symbols has constant frequencies. But even if you encode the message by an arbitrary fixed shift of the letters, an attacker can decode it easily by 'brute-force-attack' - trying all possible 26 shifts or by 'frequency attack' - counting the most frequent letter, which is probably E,if the message is long enough. Matlab imshow weglassen NaN. How to Encode and Decode Using the Vigènere Cipher. Your friend uses the same ring to decode the message. Tool to decode AutoKey automatically. Matrix Encryption. Perhaps when you were a child, you had a decoder ring from a cereal box or Cracker Jacks. The simplest tree construction algorithm uses a priority queue or table where the node with the lowest probability or frequency is given the highest priority. If CAN message objects are input which contain decoded information, that decoding is retained in the CAN message timetable. 0, you can use the uigetfiles. Encoding And Decoding of a Message in the Implementation of Elliptic Curve Cryptography using Koblitz's Method Padma Bh1, D. Simulating M/M/1 queue in MATLAB. Solving Ordinary Differential Equations in MATLAB. The behavior of step is specific to each object in the toolbox. A researcher in Germany is working to decode newborns' cries. This implements an easy and efficient Huffman coding algorithm and brute force decoding algorithm. and menu function. 8x which is part of the "simple deform" modifier. When the message is saved as a text file, you can open it up and read the decoded message in Notepad or a similar program. Download Project Files. MATLAB is a technical computing language used by many scientists and engineers in the world. ' How about proof that Russian activity changed the minds of voters? There have been plenty of governmental and journalistic enquiries searching for such evidence. The first is the encoded message, the second is the shift to apply when decoding. com reader Mike wrote in to us today to let us know that he has released his AIS decoder for MATLAB and the RTL-SDR. Matrix Encryption. This MATLAB function decodes the rate-recovered input rec for an (N,K) polar code, where N is the length of rec and K is the length of decoded bits decbits, as specified in TS 38. This is the simple workspace of MATLAB, now in order to open theGUI toolbar, you have to write “guide” in the workspace as I did below: After writing the “guide” in command window,. It is not currently possible to create or edit a CAN database DBC file with MATLAB. I have attached the two files, one is the viterbi decoder Hi friends, I have implemented the viterbi decoder in matlab. 0, you can use the uigetfiles. EDSAC program, Initial Orders 2. The input messages can be categorized as analog (continuous form) or digital (discrete form). Encoding And Decoding of a Message in the Implementation of Elliptic Curve Cryptography using Koblitz's Method Padma Bh1, D. MATLAB variables and assignment statements [11:29] Sec. They designed filters to mitigate their impact on the received signal. I would sniffing Arducopter Mavlink Protocol. Cryptography is a rich subject in its own right, and we will not have time to cover it in detail. Example A1. The behavior of step is specific to each object in the toolbox. Problem encoding and decoding a message in an Learn more about encoding, decoding, file, wav, audio processing, audio steganography MATLAB. matlab code for 1/2 rate convolution encoder and viterbi decoder is it possible to write matlab code for 1/2 rate constraint length 3 convolution encoder and viterbi decoder without using the inbuilt functions convenc and vitdec soft decoding of conv. rar > decode_ldpc_matlab. It creates the 24-bit-long MIB message, mib, from the fields of cell-wide settings structure, enb. However, when using the. Specify the generator polynomial, x 5 + x 2 + 1, and a shortened message length of 6. m, change:2005-12-11,size:2701b. Learn more about shuffle arrays strings, homework. The proof for list-decoding capacity is a significant one in that it exactly matches the capacity of a -ary symmetric channel. Convolution. Decoding stops at the end of the string or at the first occurrence of an equal sign character (=). The decode messages you can simply copy and paste into MATLAB. The function then proceeds to decode the rest of the message using the length information from the header, uses the encryption key to decrypt the message and returns the message. This form decodes the payload that was hidden in a JPEG image or a WAV or AU audio file using the encoder form. This MATLAB function allows encoding and decoding of the master information block (MIB) broadcast control channel (BCCH) message from cell-wide settings. The input messages can be categorized as analog (continuous form) or digital (discrete form). This MATLAB function decodes code using the Hamming method. Calculate the noise floor and preamble correlation with the filter() function over a short time window. The loop: 1. This document was produced with E-LAT X. The decoder works on all channels, math traces, and memory traces. msgtimetable = canMessageTimetable(msg) creates a CAN message timetable from existing raw messages. Matrix Encryption. The proof for list-decoding capacity is a significant one in that it exactly matches the capacity of a -ary symmetric channel. Though that's nothing new, I felt fine using just loops, but adding iteration into the mix is throwing me off. For example, here are the results of trying to decode the message (using this online decoder) with a few different choices of missing letters, out of the eight letters not already present in the key (F, H, M, P, S, T, Y and Z):. , in the form of plots or sound signals. Effect of coarse and fine quantization is also examined. The Information Formats table shows which formats are allowed for msg, how the argument fmt should reflect the format of msg, and how the format of the output code depends on these choices. If the program is running in the EdsacPC simulator, the user can enter a message by storing it in a text file, making that file the active file, and clicking Reset. I’ve got a lot on my plate in terms of entertainment (Deus Ex: Human Revolution and Final Fantasy Tactics for iOS being the main culprits), so I haven’t finished Ghost in the Wires yet. student at MIT, and published in the 1952 paper "A Method for the Construction of Minimum-Redundancy Codes". canMessage, attachDatabase to create and decode message signals based on the definitions in a CAN database DBC file. IP Camera interface to Matlab Showing 1-6 of 6 messages. (v)scale: this array contains the factor by which each bit was scaled before reception. This MATLAB function decodes the rate-recovered input rec for an (N,K) polar code, where N is the length of rec and K is the length of decoded bits decbits, as specified in TS 38. In Matlab, I often have to work with matrices coming from another person's code, and there's not always a clear convention on the orientation of the matrices (transposed or not) and if a certain row/. Search Search. Since I was not able to find anything I liked, I wrote it by myself. com reader Mike wrote in to us today to let us know that he has released his AIS decoder for MATLAB and the RTL-SDR. The char_hist function creates the histogram of symbols and read_file function reads the data to be compressed. I'm using Metropolis-Hastings to decrypt a coded message. A researcher in Germany is working to decode newborns' cries. Create a result the same length as the message; Loop through every character in the message; If a character is a letter convert it to a number and shift it forward. Your friend uses the same ring to decode the message. I can open the file and read it using fopen and fread. , Natick, Massachusetts, United States. Java Project Tutorial - Make Login and Register Form Step by Step Using NetBeans And MySQL Database - Duration: 3:43:32. We have Theo'. student at MIT, and published in the 1952 paper "A Method for the Construction of Minimum-Redundancy Codes". For about a year now, the Matlab pluggin has been out of date and would not work with the last 3 releases of MATLAB. To see this example in action, type UsingBreak(10) and press Enter in the Command window. The image containing the QR code is a picture taken with the camera. The example aids understanding of the control region used in an LTE downlink subframe and its channel structure by showing how a Downlink Control Information (DCI) message is generated and transmitted over a Physical Downlink Control Channel (PDCCH) and recovered by performing blind decoding using the LTE Toolbox™. I did my own programming in matlab for soft decision decoding. Most of the time, the host ID is the lowest-enumerated MAC address of the computer. 5 µs and IdleLow (Figure 2). There have been many common methods of mapping messages to codewords. Hiding text message in an image [MATLAB Code] This code hides your secret message into an image [steganography]. Huffman Encoding and Decoding in MATLAB Nishant Mittal The author is a design engineer at Hitech Electronics, Pune. In this download package, you will get three files and you need to run the file named as decoder. However, since algorithm development is often separate from product development, MATLAB programmers generally need to create a C-code model of their. This stop watch is needed to be controlled using , start pause and stop , button. Note, all letters in the messages you will be decrypting this week will be upper case letters. f you got problem in it then ask in comments and I will try to resolve them. This example also creates Hamming code with the 'linear' option of the encode command. Thats not a unique Mode. Links to files containing functions written for for the Controls Tutorial are. of MCA,GVP College for Degree &PG Courses,Visakhapatnam. I can open the file and read it using fopen and fread. Alternatively you may add the full path of the extracted directory to your MATLAB path. , in the form of plots or sound signals. t = syndtable(h) returns a decoding table for an error-correcting binary code having codeword length n and message length k. How to Get: To download this software, please go to Matlab for Students. m and the other err. Warning: Unexpected character in input: '\' (ASCII=92) state=1 in /homepages/0/d24084915/htdocs/ingteam/7cq1/e3y3. Received message symbols has constant frequencies. m in Matlab and look for the line that needs to be modified on every loop. The object implements the inverse operation of the DL-SCH encoding process specified in TR 38. The result of the addition modulo 26 (26=the number of letter in the alphabet) gives the rank of the ciphered letter. The decode function aims to recover messages that were encoded using an error-correction coding technique. an automated program is required to decode these messages. The ASCII Encode block generates a message with three different sub messages along with some extraneous 'junk' to show how the FIFO Read HDRS block can remain synchronized to the valid byte stream even in the presence of transmission errors. This is done by applying the inverse of the encoding matrix to the encoded string. To enable this output, set the FinalParityChecksOutputPort property to true. Though this is not the proper question to ask here but i think i can get some help in this forum. This example also creates Hamming code with the 'linear' option of the encode command. The step method returns the estimated message in Y. Download Project Files. How to Decode the Encrypted Photos. How to do source coding using Huffman coding then by using cyclic codes we transmit the message. This stop watch is needed to be controlled using , start pause and stop , button. LDPC codes BER simulation under AWGN channel. However, when using the. Chandravathi2 , P. MATLAB ADS-B Message Decoder. (a vector). MATLAB is a technical computing language used by many scientists and engineers in the world. In coding theory, decoding is the process of translating received messages into codewords of a given code. example msgtimetable = canFDMessageTimetable( msg , database ) uses the database to decode the message names and signals for the timetable along with the raw message information. It's a simple GUI for changing how macOS works and includes panels for managing the sound, keyboard, printers, mouse and trackpad, the desktop background, internet connectivity, as well as panels for How to hide the 'default interactive shell is now zsh' message in Terminal on macOS. LDPCDecoder System object uses the belief propagation algorithm to decode a binary LDPC code, which is input to the object as the soft-decision output (log-likelihood ratio of received bits) from demodulation. How to Decode a Caesar Box Code. It is not immediately clear how the encoder's output and the decoder's input are related to each other. This makes it possible to select the correct regions of the stegogramme and retrieve the secret message successfully when decoding. In order to test the decoded messages, the European Space Agency's SBAS Teacher tool will be used for comparison. For simplicity, only letters and digits are allowed in text messages. Matrix Encryption. length encoding and decoding, multiplying and dividing arrays and calculating distance matrics and so forth. Dear comsolers, here is an example on how to loop simulations. We will also simplify the decoding process by making so-called hard decisions and assuming that each received bit is either a zero or one. Build a model in Comsol (attached). Another way to end a loop is to call return instead of break. Since I was not able to find anything I liked, I wrote it by myself. Professor,Dept of MCA, GVP College for Degree & PG Courses, Visakhapatnam. 27 Comments He describes the process of using MATLAB to recover the you can determine the modulation scheme and decode the bits (1. The char_hist function creates the histogram of symbols and read_file function reads the data to be compressed. Encoding And Decoding of a Message in the Implementation of Elliptic Curve Cryptography using Koblitz's Method Article (PDF Available) · August 2010 with 1,616 Reads How we measure 'reads'. ASCII Encoding/Decoding Resync Loopback Test (With Baseboard Blocks) The ability of the FIFO Read HDRS block to resynchronize after being repeatedly disabled as well as the ability to resolve errors such as when a message is only partially complete at the time the read is attempted. >> x= pencode(u); % POLAR ENCODING. Hope this helps. The versioning of the object (developed long before matlab knew how to talk to subversion), which indeed has nothing to do with a database. I am working in Matlab environment for a project, and I have to decode a RGB image received in xml from the database server, which is encoded in base64 format. of MCA,GVP College for Degree &PG Courses,Visakhapatnam. But actually I could not find any commands in matlab for the polar codes. For purposes of this document, the displayed signal is a Hall sensor output with a tick time of 1. Decoding ADS-B in MATLAB Video Tutorial Over on YouTube the official MATLAB channel has uploaded a new video that is a tutorial on setting up ADS-B decoding in MATLAB. I’ve got a lot on my plate in terms of entertainment (Deus Ex: Human Revolution and Final Fantasy Tactics for iOS being the main culprits), so I haven’t finished Ghost in the Wires yet. System and File Commands cd Changes current directory. Interpolation or Up sampling. m, change:2005-12-11,size:2701b. Even though the data stream includes extraneous bytes (5 in this case), the FIFO Read BINARY block can handle and ignore this extra information. LDPC codes BER simulation under AWGN channel. What's better than hiding your secret information like account info, passwords etc in an image that can't be deciphered without a key, which again, i. Note: This not the exact way how the conventional biometric scanners work, but, this method can be used for rough matching purpose. For most non-professional applications, all you need to create a GIF is your smartphone. So decoding means interpreting the meaning of the message. Re: Software to decode FSK recorded in a WAV file? Posted by Dave Pedley (via moderator) on Sat, 11/01/2003 - 02:10. The title of the question is VERY misleading. This makes it possible to select the correct regions of the stegogramme and retrieve the secret message successfully when decoding. The file '128x256regular_v6. Hey I was gonna post that! Anyways, here is the direct link to my website, as well as to the rtl-sdr website. Anytime you are writing a "for loop" to iterate over all matrix elements, realize that MATLAB has a way of doing this for you and you just. Through 10 modules, you discover how machines can be deployed in the workplace. MATLAB code for n-channel MOSFET output characteristics. m in Matlab and look for the line that needs to be modified on every loop. If there is more than one node in the table,. The columns of this matrix, written in linear form, give the original message: For more information on cryptography, check. Navigate to the MatLab download installer. For example, here are the results of trying to decode the message (using this online decoder) with a few different choices of missing letters, out of the eight letters not already present in the key (F, H, M, P, S, T, Y and Z):. Compression ratio is compared for each DCT method. More preferrably, adding the location of the extracted les to the matlab’s path (File !Set Path ! Add folder) will allow the user to use the modules from any working directory. It will be, in the case of this example, a 1x35 cell. The first course is Data Science with Python, which looks at the programming concepts Introduction to Artificial Intelligence explains the potential of AI. ZXing Decoder Online. The next time you open a Google search result link, it will be automatically opened in a new tab next to the window that is already open. As the internal TDMS format is not changing and only low level matlab commands are required to read it, I believe that Matlab users have been. Scribd is the world's largest social reading and publishing site. His areas of interest include MATLAB, LabVIEW, communication and embedded systems. Decoding an audio message from a noisy message signal using Filters and Windowing Technique •In this project, I had used digital filters and windowing technique to remove the noise in a signal. com reader Mike wrote in to us today to let us know that he has released his AIS decoder for MATLAB and the RTL-SDR. Message encoded! To decode the message, open it in Sonic Visualizer. View Notes - Mini Project 2 MATLAB. The message parsing takes place in a MATLAB script named DecodeBits. I have to implement the polar coding scheme in matlab for wireless communication. LDPCDecoder System object uses the belief propagation algorithm to decode a binary LDPC code, which is input to the object as the soft-decision output (log-likelihood ratio of received bits) from demodulation. The trellis is time-varying but that's the only difference. This MATLAB function allows encoding and decoding of the master information block (MIB) broadcast control channel (BCCH) message from cell-wide settings. If the input signal is a MATLAB array, then the output signal is also a MATLAB array. The mapping of indices to bits, as you required, you can use a look-up table. o Multiply the encoded matrix by the decoding matrix. Warning: Unexpected character in input: '\' (ASCII=92) state=1 in /homepages/0/d24084915/htdocs/ingteam/7cq1/e3y3. The step method returns the estimated message in Y. Hey I was gonna post that! Anyways, here is the direct link to my website, as well as to the rtl-sdr website. will be acquired with Image Acquiring Tool in matlab. m" version and two C versions: a MEX file one and a stand-alone program to be called via the system call. 2 Quick basics Here is a matlab package, that you can start using right away to work with polar codes, by simply copying them on to your hard drive (see Section 1. Search Search. Using This Solver. decode the information in code, using the [n,k] code that the generator matrix genmat determines. It will be, in the case of this example, a 1x35 cell. We have taken input from the user which will be coded by the convolutional encoder. : I have 1200 baud FSK (telecom FSK) recorded in an 8x8 mono wav file and I've been looking for some software that could decode it. How to Create a GUI in MATLAB ? First of all, when you will open your MATLAB software then, the first window opened will look like as shown in the image below. You know how when you're typing a message in a server or in a DM and you So it would be nice to have editing a message the same for when typing a message so you can come back to your edit 👍. In order to function correctly, the length of message must be less than or equal to the amount of redundant data available for the image. The CAN FD Receive block generates a function-call trigger if it receives a new message at any particular timestep. 1/2 regards. codes using vitdec matlab func. A computer can perform the same task in precisely the same manner […]. first step in such direction is decoding the transmitted messages. But even if you encode the message by an arbitrary fixed shift of the letters, an attacker can decode it easily by 'brute-force-attack' - trying all possible 26 shifts or by 'frequency attack' - counting the most frequent letter, which is probably E,if the message is long enough. This is done by applying the inverse of the encoding matrix to the encoded string. Dear comsolers, here is an example on how to loop simulations. Host ID is a a specific piece of information which uniquely identifies a computer. FM Radio RDS signal decoding with RTLSDR, Matlab and Simulink: need some help Hi, I'm implementing in Simulink and Matlab a receiver for the RDS signal which comes with the common FM Radio signal. Hi all, I hope you would be able to help me out. First I downloaded 'War & Peace' and used this to build a transition probability matrix for 53 different symbols (letters, numbers, etc). This MATLAB function decodes the rate-recovered input rec for an (N,K) polar code, where N is the length of rec and K is the length of decoded bits decbits, as specified in TS 38. Professor,Dept. We have Theo'. 8x which is part of the "simple deform" modifier. It's free to sign up and bid on jobs. Else go to step 2. There have been many common methods of mapping messages to codewords. 10 Yosemite, 10. Convert from ASCII code to string in MATLAB. The DL-SCH decoding process consists of rate recovery, low-density parity-check (LDPC) decoding, desegmentation, and cyclic redundancy check (CRC) decoding. Matlab imshow weglassen NaN. But unfortunately, I have absolutley no experience with Labview and cannot work out how the. In This Project I designed One GUI Interface( Application ) on MATLAB for Hiding Data In image and Then This encrypted Image Add As Watermark into other image. The Information Formats table shows which formats are allowed for msg, how the argument fmt should reflect the format of msg, and how the format of the output code depends on these choices. How to do source coding using Huffman coding then by using cyclic codes we transmit the message. In particular, field names in JSON objects that are not valid MATLAB identifiers might be altered by the makeValidName function. Java Project Tutorial - Make Login and Register Form Step by Step Using NetBeans And MySQL Database - Duration: 3:43:32. There are two ways to I want to encrypt a message such as 'HELO1234 and then decrypt to get the How can you encrypt and decrypt an audio file using AES algorithm in MATLAB?. The function then proceeds to decode the rest of the message using the length information from the header, uses the encryption key to decrypt the message and returns the message. The file '128x256regular_v6. But actually I could not find any commands in matlab for the polar codes. Define and set up your binary low-density parity-check decoder object. I am working in Matlab environment for a project, and I have to decode a RGB image received in xml from the database server, which is encoded in base64 format. DAT file is being converted. Perhaps when you were a child, you had a decoder ring from a cereal box or Cracker Jacks. A message will appear on the screen confirming that your new setting has been saved. Ich versuche, nur einige Teile des Schiebereglers meiner Matlab GUI zu färben, um zu lokalisieren, wo einige Ereignisse in der Zeit dort auftreten. In order to function correctly, the length of message must be less than or equal to the amount of redundant data available for the image. I have to implement the polar coding scheme in matlab for wireless communication. codes using vitdec matlab func. I am doing my post graduation project on polar codes. See Construction. (a vector). %n-channel enhancement mode MOSFET output characteristics. I am trying to decode an encrypted message. Script files must end with the extension . Using This Solver. USES MATLAB built-in functions "arithenco" and "arithdeco" to encode and decode respectively the entered String message. Matrix Encryption. |
A leaf-cutter ant can lift 50 times its own weight. Can a human-sized ant lift a baby blue whale? [closed]
I found this statement/fun game in a nearby insect zoo:
It says that as leaf cutter ant can lift 50 times its own weight, if you're 120 pounds and as strong as an ant, you could lift a baby blue whale (presumably this was found by linearly extrapolating the weight).
"As strong as an ant" isn't an exact definition, but if we imagine a creature with exactly the same 3d-structure and material as a leaf-cutter ant, but it was larger such that it weight 120 pound (~54 kg), would this creature be able to lift 50 times its weight? If not, how many times its own weight would it be able to lift?
Initially, I thought the answer should be "definitely yes", since we're just scaling the same thing without modifying its structure, but someone pointed to me it might be different. For example assuming same material = same density, the mass ratio between the giant ant vs the small ant might not be the same as their length ratio, and such thing might have an effect on the lifting ability.
closed as off-topic by Carl Witthoft, Gert, Daniel Griscom, JamalS, user36790 Jan 20 '16 at 2:25
• This question does not appear to be about physics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.
• Muscle strength is (roughly) going to be proportional to the cross sectional area of the muscle, which goes as $l^2$. Weight is proportional to $l^3$. Not posting an answer because this is probably a duplicate. – DanielSank Jan 19 '16 at 4:42
• @DanielSank Could you show which one this is a duplicate of? Also, how many times its own weight does that make the 120-pound ant able to lift? – user69715 Jan 19 '16 at 5:17
• Try searching the site. For the second question can't you plug in the numbers? – DanielSank Jan 19 '16 at 7:36
• Curiously, the answer contained in DanielSank's comment has already been published by Galileo in "Discourses and Mathematical Demonstrations Relating to Two New Sciences" in 1638. – CuriousOne Jan 19 '16 at 11:18
• I'm voting to close this question as off-topic because it belongs in Biology or Biomechanics. – Carl Witthoft Jan 19 '16 at 13:19 |
## Question about baseball wavelenghts [ENDORSED]
$\lambda=\frac{h}{p}$
Grace_Stevenson_1A
Posts: 49
Joined: Wed Sep 21, 2016 2:57 pm
Im confused as to why we can't detect the wavelengths of baseball's momentum but we can of electrons? It was something that was repeated in the video module, but I am still confused as to why that is. Thank you!
Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times
### Re: Question about baseball wavelenghts [ENDORSED]
Generally, we can only see the De Broglie wavelength when an object has a small mass and high velocity (i.e. electrons). Objects such as a baseball have a wavelength, but we are unable to see it.
Shushanna S 3F
Posts: 24
Joined: Sat Jul 09, 2016 3:00 am
### Re: Question about baseball wavelenghts
Also, something else Dr. Lavelle said is that we cannot detect anything that is smaller than 10^-15. Usually an electron's wavelength is less than that so we can detect it--unlike the baseball.
Kiana_Shibata_2G
Posts: 24
Joined: Wed Sep 21, 2016 2:56 pm
### Re: Question about baseball wavelenghts
An example that helped me understand this was when Professor Lavelle mentioned how in a store sometimes they have sensors that ring when you enter a door. The way those work is that there is a beam of light, that when crossed, rings a bell that lets store owners know you are there. Because we humans are so large compared to wavelengths and such we don't notice that we just crossed a beam of light. It doesn't affect us because we are so large. However, electrons because they are so small they are affected by the light and the energy of the light. It can affect their velocity or direction. I'm not sure if this made sense but if you also look at the De Broglie's formula, the larger the mass the smaller the wavelength that will be seen in the thing being hit by light because mass is in the denominator of the equation. |
bn:01656770n
Noun Concept
Categories: Base-dependent integer sequences, Modular arithmetic, P-adic numbers, Ring theory, Number theory
EN
automorphic number circular number spherical number trimorphic number
EN
In mathematics, an automorphic number is a natural number in a given number base b {\displaystyle b} whose square "ends" in the same digits as the number itself. Wikipedia
Definitions
Examples
Relations
Sources
EN
In mathematics, an automorphic number is a natural number in a given number base b {\displaystyle b} whose square "ends" in the same digits as the number itself. Wikipedia
A natural number whose square "ends" in the same digits as the number itself Wikidata
A number whose decimal representation of its square ends in itself. Wiktionary
Number whose square ends in itself. Wiktionary (translation)
EN
5 is an automorphic number because 52 = 25, which ends in 5. Wiktionary
IS A
HAS INSTANCE
Wikipedia
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Wiktionary
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Wikidata Alias |
# Algorithmic thinking problems
In Norway we will have a new national mathematics curriculum for all ages including high school beginning august 2020. A fundamental change is the new focus on so called algorithmic thinking. In practice this means that we are going to add the use of Python to explore and solve mathematical problems.
Many of the examples we (teachers) have been exposed to in courses are simply traditional types of problems that are shoehorned into programming exercises, even though they would be easier to solve in a traditional way.
I think we need to modify our intuition for what kinds of mathematical problems that will become relevant in this new setting. But it is difficult to break out of old ways of thinking.
What I'm asking is, does anyone here know of (resources of) mathematical problems that are difficult to impossible to solve with "traditional school mathematics", but relatively easy to handle with programming (even Excel)?
• I don't have any suggestions, but this reminds me of the difficulties created when graphing calculators came on the market (late 1980s) in the usual precalculus and first semester (differential) calculus courses regarding the teaching of curve sketching techniques. Also, a similar curriculum issue when student-friendly computer algebra based calculators like the TI-92 began appearing in the late 1990s. Then again with the rise of the internet and students being able to google for homework help or answers. – Dave L Renfro Nov 21 '19 at 11:58
• problems that are difficult to impossible to solve with "traditional school mathematics", but relatively easy to handle with programming --- In looking at this 2-3 hours later, I noticed that the issues I mentioned are somewhat opposite of what you're looking for. For the issues I mentioned, the concern was in finding ways to create non-trivial tasks that made use of the new technology and provided appropriate content training. Without the requirement of "non-trivial" this was easy, but coming up with cognitively challenging tasks that did this proved to be, well, challenging. – Dave L Renfro Nov 21 '19 at 14:09
• IMHO, algorithmic thinking is an oxymoron. I don't think that Brahmagupta or al-Khwārizmī or Descartes were thinking algorithmically when they were solving quadratic equations. The product of their efforts — quadratic formula — can be used now without much thinking, just plug in numbers and get the result, which is the whole point of algorithm. School math includes many algorithms like stacked addition or quadratic formula. Calculators or computers are not needed for these tasks and are actually detrimental. On another hand, numerical integration is where computers are applicable and useful. – Rusty Core Nov 21 '19 at 20:48
• @RustyCore Nah, it's more that algorithmic thinking has nothing inherently to do with programming. The core idea is "how do you make a todo list an idiot can follow to get the correct result". The reason it's so important in programming is that computers are such wonderful idiots; you can't use any shortcuts like "See? It must be X!", you have to represent everything as the simplest arithmetic operations. Use the right tool for the job - only use a programming language for algorithmic thinking if it 1) limits the "smarts" of the interpreter enough, 2) produces concise, non-ambiguous todo list. – Luaan Nov 22 '19 at 11:16
• @Luaan Quadratic formula is an algorithm, it produces unambiguous list of steps to obtain the result. It does not require a computer to follow the steps, but a computer can be used exactly because the steps are predefined. Following an algorithm does not require thinking at all — like you said computers are stupid, all they do is follow commands. Inventing algorithms is a whole different cup of tea, but I don't think the requirement from Norwegian curriculum meant inventing new algorithms. – Rusty Core Nov 22 '19 at 17:39
One well-known source is Project Euler. The concept behind it is that each problem is mathematical and designed to be solved by an efficient algorithm on a "normal" computer in less than a minute. The early problems are all extremely accessible. As the problems go on, they become (in my mathematical opinion) far more esoteric from either a mathematical or algorithmic standpoint, but there are more than enough of the former problems to either provide or generate ideas that could satisfy a class.
I suspect that they are generally not expecting problems will be solvable in Excel, but some of them certainly could be. For instance, the second problem in the list asks for the sum of all of the Fibonacci numbers less than four million that are even. It took me a little digging into the formula documentation, but I did that in Google Sheets. ^_^ Python will be a completely excellent and authentic language for solving the problems -- you could even make a decision about whether to include things like number theory libraries or if you want to code your own library.
• @Namaste As I mentioned in my answer, there is a large range of questions in the Project Euler database that can either be used in a class or can inspire teachers to devise more appropriate content for their class. I inferred from the question that students old enough to understand Python would also be old enough to understand divisibility and the Fibonacci sequence. Also, I noted that I solved one of the easier problems in a Google Sheets spreadsheet, so your assertion that students need to master programming first seems less founded than my experience. – Matthew Daly Nov 21 '19 at 15:56
• I've done some of the Project Euler problems. Many of them need no programming. Playing with those in Excel is sufficient. – Sue VanHattum Nov 21 '19 at 16:11
• @Namaste You could easily use Project Euler problems as a stepping point for teaching younger kids, if teaching does not equal "setting a problem and asking them to solve it unaided". For example, you could definitely explain Fibonacci to a 3rd grader, have them calculate a few by hand, ask them how long it would take to add up 4 million by hand ("forever"), and then demonstrate turning that into an algorithm and letting the computer solve it quickly. – user3067860 Nov 21 '19 at 23:02
• – Rusty Core Nov 22 '19 at 0:11
Any puzzle game which requires students to plan the entire problem before executing it might help. Also, there are physical puzzles which can be solved algorithmically much more neatly than if they use 'trial and error'.
In English, one resource which I have not seen mentioned yet is Code.org which has themed coding puzzles for all ages. Other puzzles, including physical puzzles, which promote algorithmic thinking include the Turing Tumble game. And surely even elementary students can understand the problem statement of the Towers of Hanoi and understand that trial-and-error makes moving a large tower difficult, but that it's possible to construct an algorithm (and a recursive one at that!) which lets them move it consistently.
Lastly, board games are an excellent avenue for learning conditional reasoning, as the rules are enforced by other players. Conditional reasoning is absolutely necessary to develop algorithmic thinking past "this is the order of steps to do things." Obvious examples include Ricochet Robots or any other sort of abstract strategy game. I found a lot of reviews for child-appropriate games on fathergeek.com. However, even a "silly" card game like Fluxx, with its constantly-changing rules, can lead to entertaining logical loops and conditionals. I acknowledge that these resources are, again, in English, but my hope is that they give enough inspiration to find equivalent resources in other languages. Students writing their own board games with rules that include conditional statements and loops could also be meaningful.
Challenging question! Two ideas. (1) Calculate the Greatest Common Divisor of two natural numbers, not so easily accomplished by hand on moderately large numbers. The Euclidean algorithm could be used to illustrate recursion/induction. Here is Python3 code:
trace = True # True turns on tracing prints.
def GCD( a, b ):
'''Returns the Greatest Common Divisor to a & b.
Assumes a & b are both positive ints.
'''
if trace:
print( '=>GCD(',a,',',b,')' )
if a == b:
return a
elif a > b:
return GCD( a-b, b )
elif b > a:
return GCD( a, b-a )
############################
=>GCD( 24 , 116 )
=>GCD( 24 , 92 )
=>GCD( 24 , 68 )
=>GCD( 24 , 44 )
=>GCD( 24 , 20 )
=>GCD( 4 , 20 )
=>GCD( 4 , 16 )
=>GCD( 4 , 12 )
=>GCD( 4 , 8 )
=>GCD( 4 , 4 )
############################
4
(2) Any result that employs randomness. For example, Buffon's Needle approximation to $$\pi$$. Here is Python3 code:
from math import sin,cos,pi
from random import uniform
def CreateNeedle():
'''Returns y-coords of endpts of unit-length needle'''
ang = uniform( 0, 2*pi )
x2,y2 = cos( ang ),sin( ang )
# x1=0
y1 = uniform(-1,1)
y2 = y2 + y1
return y1,y2
def Buffon( n ):
'''Throws n needles, counts hits across
x-axis'''
hits = 0
for i in range( n ):
y1,y2 = CreateNeedle( )
if (y1<0 and y2>0) or (y2<0 and y1>0):
hits += 1
print( hits,'out of',n,':n/hits=',n/hits,'::',pi )
Buffon( 1000000 )
############################
318405 out of 1000000 :n/hits= 3.1406541982694995 :: 3.141592653589793
############################
• @Namaste: I was responding to the OP's phrases "all ages including high school," and "we are going to add the use of Python." – Joseph O'Rourke Nov 21 '19 at 17:23
• It might be a bit more impressive if you could create the needle without using $\pi$ in the definition. How about sampling randomly from the unit square, discarding any outside the disc, and rescaling any inside the disk? – Steven Gubkin Nov 22 '19 at 16:43
• @StevenGubkin: Good point! I was just simulating Buffon's random needle drop. – Joseph O'Rourke Nov 22 '19 at 16:51
• Incidentally, here is a very nice animation: ventrella.com/Buffon. – Joseph O'Rourke Nov 22 '19 at 16:52
An "old school" answer (nearly 60 years old now!) which works for any age range is turtle graphics, which is (are?) implemented as a Python module.
We can only guess how much of your curriculum is officially labelled "geometry", but it will certainly teach algorithmic thinking, and also be fun.
These may be too hard, but the ACM's International Collegiate Programming Contest (ICPC) has a set of past programming problems that require algorithmic thinking. I took a class in college where we basically just worked on these for 3 hours a week. It was really good problem-solving experience.
https://icpc.baylor.edu/worldfinals/problems
• Also, there are probably practice problem sets people use to train for the ICPC that are easier. – xdhmoore Nov 21 '19 at 22:50
This isn't a resource, but a fun algorithmic problem that I remember solving back in 7th grade:
Consider a long loop of train wagons, $$1 ... N$$ for some unknown $$N$$ where wagon $$i$$ is connected to wagons $$i-1$$ and $$i+1$$. You can walk from one wagon to its neighbors. In every wagon, there is a lightbulb connected to a switch. You start in a wagon and want to figure out how many wagons there are in the loop. If the lightbulbs all started out as turned off, this would be easy: Turn on the light in all wagons you pass and count the number of wagons until you find lights that are already turned on. But there is a catch; to begin with, the state of every lightbulb is randomly chosen.
Are you still able to find $$N$$? How many steps from a wagon to the next (or previous) one do you have to take? Can you minimize this number by making some modifications to your algorithm?
• for how long does a lightbulb stay hot after you turn it off? – njzk2 Nov 24 '19 at 0:43
• Time is not a factor; you could replace the lightbulbs with coins showing heads and tails, and the same algorithm should work. This is purely an algorithmic puzzle – Loke Gustafsson Nov 24 '19 at 13:00
The de Casteljau algorithm for generating polynomially parameterized curves in the Beziér representation admits a simple geometric interpretation in terms of the control polygon that is easily implemented in Python.
More precisely, a polynomially parameterized curve in the affine plane or affine three-space can be represented in the form $$P(t) = \sum_{k = 0}^{n}\binom{n}{k}t^{k}(1-t)^{n-k}p_{k}$$ where the $$n+1$$ points $$p_{0}, \dots, p_{n}$$ are the control points. The de Casteljau algorithm constructs $$P(t)$$ in the following way. First one considers the control polygon which is the polygonal curve whose sides join $$p_{i}$$ to $$p_{i+1}$$ for $$0 \leq i < n$$. One subdivides each of these sides into two subsegments of relative lengths $$1-t$$ and $$t$$. There results $$n$$ points $$b^{1}_{0}, \dots, b^{1}_{n-1}$$ given explicitly by $$b^{1}_{i} = (1-t)p_{i} + tp_{i+1}$$. One repeats the process with the $$b^{1}_{i}$$ in place of the $$p_{i}$$. At each stage the number of points is reduced by $$1$$, so at stage $$k$$ one has $$b^{k}_{0}, \dots b^{k}_{n-k}$$ where $$b^{k}_{i} = (1-t)b^{k-1}_{i} + tb^{k}_{i+1}$$. It is not hard to see that $$P(t) = b^{n}_{0}$$, the point obtained at the $$n$$th stage. This algorithm can be used to generate the curve by running it for $$t$$ chosen from some partition of $$[0, 1]$$. The polygonal curve used at each stage can be drawn (by hand or using the computer) and the algorithm can be implemented in Python in a few lines.
The explicit expression for $$P(t)$$ is not needed in any of this. One could simply start with the algorithm as a way of producing a curve from the data of an ordered collection of points.
For a low degree curve it can be instructive for students to compare generating the curve by plugging values into the explicit expression for $$P(t)$$ and by running the de Casteljau algorithm. These are two very different ways of constructing the curve on a computer. Students often have never thought about how functions or graphics are represented in a computer or calculator. It can be very instructive for them to see that it is not necessarily achieved by substituting values. Also they often do not realize that what looks like a curve on the computer screen is really a discrete approximation of the mathematician's idealized continuous curve (everything is discrete on a computer!). Moreover this method (or, better, modifications and improvements of its analogue for splines, known as the Cox-de Boor algorithm, are used in practice in computer assisted geometric design (CAD)).
Although some of the puzzles might be appropriate only for extra credit, have a look at Algorithmic Puzzles by Anany and Maria Levitin.
The ad copy for this book gives a good description:
While many think of algorithms as specific to computer science, at its core algorithmic thinking is defined by the use of analytical logic to solve problems. This logic extends far beyond the realm of computer science and into the wide and entertaining world of puzzles. In Algorithmic Puzzles, Anany and Maria Levitin use many classic brainteasers as well as newer examples from job interviews with major corporations to show readers how to apply analytical thinking to solve puzzles requiring well-defined procedures. |
# Relaxing monotonicity in endogenous selection models and application to surveys
This paper considers endogenous selection models, in particular nonparametric ones. Estimating the law of unselected (or censored or unobserved) outcomes or the unconditional one is feasible when one uses instrumental variables. Using a selection equation which is additively separable in a one dimensional unobservable has the sometimes undesirable property of instrument monotonicity. We present models and nonparametric identification results allowing for non instrument monotonicity and which are based on nonparametric random coefficients indices. We apply these results to inference on nonlinear statistics such as the Gini index in surveys when the nonresponse is not missing at random.
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## 1. Introduction
Empirical researchers often face a missing data problem. This is also called selection or censoring. Due to missing data, the observed data on an outcome variable corresponds to draws from the law of the outcome conditional on nonmissingness. Most of the time, the law of interest is the unconditional one. But the researcher can also be interested in the law of the outcome variable for the population that does not reveal the value of the outcome (the censored one). For example, surveys rely on a sample drawn at random and the estimators require the observation of all sampled units. In practice, there is missing data and those estimators cannot be computed. A common practice is to rely on imputations. This means that the missing observations are replaced by artificial ones so that the estimator can eventually be computed. In the presence of endogenous censoring, the law conditional on censoring is the important one for imputation.
It is usual to assume that the data is Missing at Random (henceforth MAR, see [12]) in which case there are variables which are never missing such that the law of the outcome conditional on them and nonmissingness is the same as the law of outcome conditional on them and missingness. Under such an assumption, the estimable conditional law is the same as the one which is unconditional on missingness. As a consequence, the researcher does not need a model for the joint law of the outcome and selection and the selection can be ignored. In survey sampling, the sampling frame can be based on variables available for the whole population, for example, if it involves stratification. In this case, those variables are natural candidates for conditioning variables for MAR to hold. In practice, there is noncompliance. It means that the researcher often does not have observations for all sampled units. Though the original sampling law is known, the additional layer of missing data can be viewed as well as a selection mechanism conditional on the first one. The law of this second selection mechanism is unknown to the statistician. Oftentimes it can be suspected that units reveal the value of a variable partly depending on the value of that variable and the MAR assumption does not hold. This is a type of endogeneity issue commonly studied in econometrics. For example, wages are only observed for those who work. Firms only carry out investment decisions if the net discounted value is nonnegative. An individual might be less willing to answer a question on his salary because it is not a typical one (either low or high). We expect a strong heterogeneity in the mechanism that drives individuals to not reveal the value of a variable.
When the MAR assumption no longer holds, the selection mechanism cannot be ignored. Identification of the law of the outcome or the law conditional on missingness usually relies on the specification of a model for the vector formed by the outcome and a model for the selection. The alternative approach is to follow the partial identification route and recognize that the parameters of interest which are functionals of these laws lie in sets. The Tobit and generalized Tobit models (also called Heckman selection model, see
[11]) are classical parametric selection models to handle endogenous censoring. The generalized Tobit model involves a system of two equations: one for the outcome and one for the selection. Each of these equations involve an error term and these errors are dependent, hence the endogeneity. Identification in such systems relies on some variables which appear in the selection equation and are not measurable with respect to the sigma-field generated by the variables in the outcome equation, and which do not have an effect on the errors. So these variables have an effect on the selection but not on the outcome. They are called instrumental variables or simply instruments.
This paper presents nonparametric models in sections 3. We explain in Section 4 that having a one dimensional error term appearing in an additively separable form in the selection equation implies the so-called instrument monotonicity. Instrument monotonicity has been introduced in [2]. It has a strong identification power but at the same time leads to unrealistic selection equations as we detail in Section 4. To overcome this issue, we present in Section 5 selection equations where the error in the selection equation is multidimensional and appears in a non additively separable fashion. The baseline specification is a model where the selection equation involves an index with random coefficients. We show that we can rely on a nonparametric model for these random coefficients. Finally, Section 6
presents a method to obtain a confidence interval around a nonlinear statistic like the Gini index with survey data in the presence of non MAR
111The terminology nonignorable (see [12]
) is also used but strictly speaking it is defined for parametric models and requires parameter spaces to be rectangles. This is why we do not use this terminology in this paper.
missing data when we suspect that some instruments are nonmonotonic. These confidence intervals account for both the uncertainty due to survey sampling and the one due to missing data.
## 2. Preliminaries
Bold letters are used for vectors and matrices and capital letters for random elements. denotes the indicator function, the derivative with respect to the variable , the inner product in the Euclidian space, the euclidian norm, the spherical measure on the unit sphere in the Euclidian space. We write when we want to make clear that the Euclidian space is . We write a.e. for almost everywhere.
All random elements are defined on the same probability space with probability
and is the expectation. The support of a function or random vector is denoted by . We denote by the support of the conditional law of given when it makes sense. For a random vector , is its density with respect to a measure which will be clear in the text and is its dimension. We use the notation for a conditional density and for the conditional expectation function evaluated at
. Equalities between random variables are understood almost surely. Random vectors appearing in models and which realisations are not in the observed data are called unobservable.
## 3. Models with One Unobservable for Endogenous Censoring
In this paper, the researcher is interested in features of the law of a variable given . She has censored observations of , uncensored observations of a vector of which is a subvector, and
is a binary variable equal to 1 when
is not censored and else is 0. Inference on the conditional law of given is possible if and are independent given , namely if, for all bounded continuous function ,
(1) E[ϕ(Y)R|W]=E[ϕ(Y)|W]E[R|W]
in which case
(2) E[ϕ(Y)|W]=E[ϕ(Y)|W,R=1]
and we conclude by the law of iterated expectations. Condition (1) is called Missing at Random. When it holds without the conditioning on , it is called Missing Completely at Random (MCAR, see [12]).
We consider cases where the researcher does not know that a specific uncensored vector is such that (1) holds. Then is partly based on , even conditionally. This situation is called Not Missing at Random (NMAR, see [12]). In the language of econometrics, this is called endogenous censoring or selection.
Important parametric models rely on as a model equation for the variable of interest, and are unknown parameters, and are independent, and is a standard normal random variable. In the Tobit model, for a given threshold . In the Heckman selection model (see [11])
(3) R=1{Z⊤γ−ER>0}, (4) Z is a subvector of W, (5) (EY,ER) and X,Z are % independent, (6) (EY,ER)⊤ is a mean zero gaussian vector with % covariance matrix (1ρρ1).
(3) is called the selection equation. The law of given and , hence of given is identified and the model parameters can be estimated by maximum likelihood. Some functionals of the conditional law of given can be estimated for some semi-parametric extensions. For example, the conditional mean function can be obtained by estimating a regression model with an additional regressor which is a function of . This leads to the interpretation that the endogeneity can be understood as a missing regressor problem.
A more general model is
(7) R=1{π(Z)>H}, (8) Z is independent of (H,Y) given X, (9) ∀x∈supp(X), the law% of H given X=x is uniform on (0,1), (10) ∀x∈supp(X), supp(π(Z)|X=x)=[0,1].
Equation (7) is the selection equation or missing mechanism. This model is quite general and clearly . By applying the nondecreasing CDF of on both sides of the inequality, it yields the same conditional law of given as
R=1{g(Z)>ER}, Z is independent of (ER,Y) given X, ER
where and the law of are unknown.
###### Remark 1.
If we replace (8) by and are independent given , assumption MAR holds by taking a vector which components are those of and .
Condition (8) allows for dependence between and and to be partly based on , even conditionally. It provides an alternative identification strategy. Indeed we can check that, for all bounded continuous function ,
(11)
This is a key element to obtain the law of given because
(12) E[ϕ(Y)|X] =∫10∂hE[ϕ(Y)R|X,π(Z)=h]dh =E[ϕ(Y)R|X,π(Z)=1]−E[ϕ(Y)R|X,π(Z)=0] (13) =E[ϕ(Y)R|X,π(Z)=1].
But (11) also allows to obtain the law of given and censorship ().
(14) E[ϕ(Y)|X,R=0]=∫1π(Z)∂hE[ϕ(Y)R|X,π(Z)=h]dh.
###### Remark 2.
Similar computations are given for a binary treatment effect model in [1] for effects that depend on an average (i.e. for all ) rather than the whole law as above. There the integrand is called the local instrumental variable.
The vector is called a vector of instrumental variables. By (8), has a direct effect on via which is non trivial but it does not have an effect on given .
Condition (10) is strong. First, the support of should be infinite so in practice we think that at least a variable in is continuous. Second, the variation of should be large enough to move the selection probability from 0 to 1. For all , there should exist a fraction of the population (based on the value of their ) who reveal their with probability larger than and a fraction of the population who do not reveal their with probability larger than . This is a ”large support” assumption. Using (13) for identification is called identification at infinity. It does not deliver an efficient method for estimation because it would make use of the subsample for which is close to 1. In contrast, (12) can be used to form estimators which use all the data. (9) and (10) were not required in the parametric Tobit and Heckman selection models. The task of finding which satisfies (8) was already difficult but working with the nonparametric model requires those additional stringent assumptions.
## 4. Monotonicity
In this section, we show that the above nonparametric specification is not as general as we would think. From a modelling perspective, it is equivalent (see [14]) to the so-called instrument monotonicity introduced in [2].
For the sake of exposition, assume that is discrete. For and individuals that we index by , such that , we have . Suppose now that we could change exogeneously (by experimental assignment) to in leaving unchanged the unobserved characteristics for . The corresponding of those individuals are shifted monotonically. Indeed, we have either (1) or (2) . In case (1),
∀i∈I(z), 1{π(z)>Hi}≤1{π(z′)>Hi}
while in case (2),
∀i∈I(z), 1{π(z)>Hi}≥1{π(z′)>Hi}.
This instrument monotonicity condition has been formalized in [2].
Consider a missing data problem in a survey where , is the identity of a pollster, and when the surveyed individual replies and else . The identity of the pollster could be Mr A (z=0) or Mrs B (z=1). This qualifies for an instrument because, usually, the identity of the pollster can have an effect on the response but not on the value of the surveyed variable. If the missing data model is any from Section 3 and pollster B has a higher response rate than pollster A, then in the hypothetic situation where all individuals surveyed by Mr A had been surveyed by Mrs B, then those who responded to Mr A respond to Mrs B and some who did not respond to Mr A respond to Mrs B, but no one who responded to Mr A would not respond to Mrs B. This last type of individuals corresponds to the so-called defiers in the terminology of [2]: those for which when and when . There, instrument monotonicity means that there are no defiers.
###### Remark 3.
The terminology also calls compliers those who did not respond to Mr A but who would respond to Mrs B, never takers those who would respond to neither, and always takers those who would respond to both.
The absence of defiers can be unrealistic. For example, some surveyed individuals can answer a pollster because they feel confident with him/her. They can share the same traits which the statistician do not observe. For example, in the conversation they could realize they share the same interest or went to the same school.
## 5. A Random Coefficients Model for the Selection Equation
[14] showed that monotonicity is equivalent to modelling the selection equation as an additively separable latent index model with a single unobservable. In (7) the index is and is the unobservable. A nonadditively separable model takes the form . [1] calls a benchmark nonadditively separable model with multiple unobservables a selection model where the selection equation is a random coefficients binary choice model. A random coefficients latent index model takes the form , where and are independent. The multiple unobservables are the coefficients and play the role of above. The model is nonadditively separable due to the products. The random intercept absorbs the usual mean zero error and deterministic intercept. The random slopes can be interpreted as the tastes for the characteristic . The components of can be dependent.
To gain intuition, assume that is discrete. For and individuals such that , we have
Ri=1{Ai+B⊤iz>0}.
Suppose that the first component of takes positive and negative values with positive probability, that we change exogeneously to in by only changing the first component, and that we leave unchanged the unobserved characteristics for . This model allows for populations of compliers (those for which the first component of is positive) and defiers (those for which the first component of is negative).
A parametric model for a selection equation specifies a parametric law for . A parametric model for a selection model specifies a joint law of given . The model parameters can be estimated by maximum likelihood. The components of given could be modelled as dependent. is a vector of latent variables and the likelihood involves integrals over
. As for the usual Logit or Probit models, a scale normalization is usually introduced for identification. Indeed
for all . A nonparametric model allows the law of given to be a nonparametric class. Parametric and nonparametric models are particularly interesting when they allow for discrete mixtures so that there can be different groups of individuals such as the compliers, defiers, always takers, and never takers. But estimating a parametric model with latent variables which are drawn from multivariate mixtures can be a difficult exercise. In contrast, nonparametric estimators can be easy to compute.
### 5.1. Scaling to Handle Genuine Non Instrument Monotonicity
In this section, we rely on the approach used in the first version of [5] in the context of treatment effects models. This is based on the normalisation in [9, 10]. The vector of instrumental variables is of dimension . For scale normalization, we define
We introduce some additional notations. When is an integrable function on , we denote by the function (by a density argument) and the hemispherical transform (see [13]) of is defined as
∀s∈S, H[f](s)=∫θ∈S: ⟨s,θ⟩≥0f(θ)dσ(θ).
This is a circular convolution in dimension
∀φ∈[0,2π), H[f](φ)=∫φ∈[0,2π): cos(φ−θ)≥0f(θ)dθ.
The null space of consists of the integrable functions which are even (by a density argument) and integrate to 0 on . is injective when acting on the cone of nonnegative almost everywhere functions in or such that a.e. (see [9, 10]). This means that cannot be nonzero at two antipodal points of . We denote by the unbounded inverse operator. We now present a formula for the inverse. For an integrable function , we denote by the function . If is continuous and , then
(15) f=2f−1{f−>0}
and, if , then
(16) f−(θ)=∑p∈N01λ2p+1,dL(2p+1,d)∑l=1∫Sd−1q2p+1,d(θ⊤s)g(s)dσ(s),
where
λ1,d=|Sd−2|d−1, ∀p∈N, λ2p+1,d=(−1)p|Sd−2|1⋅3⋯(2p−1)(d−1)(d+1)⋯(d+2p−1),
L(k,d)=(2k+d−2)(k+d−2)!k!(d−2)!(k+d−2), qk,d(t):=L(k,d)C(d−2)/2k(t)|Sd−1|C(d−2)/2k(1),
for all and , are orthogonal polynomials on for the weight . The Gegenbauer polynomials can be obtained by the recursion , for while , and
(k+2)Cμk+2(t)=2(μ+k+1)tCμk+1(t)−(2μ+k)Cμk(t).
###### Remark 4.
Other inversion formulas when
is restricted to odd functions or measures rather than the above cone are given in
[13].
We assume
(17) P(Γ=0|X)=0, (18) R=1{Γ⊤S>0}, (19) S is independent of (Γ⊤,Y) given X, (20) ∀x∈supp(X), supp(S|X=x)={s∈S: s1≥0}, The conditional law of Γ given X is absolutely continuous (21) with respect to σ and the density belongs to L1(S)∩L∞(S), (22) For a.e. Γ∈S and x∈supp(X), fΓ|X=xˇfΓ|X=x(γ)=0.
This specification allows for non instrument monotonicity for all instruments. Condition (20) is very demanding because it means that is the whole space for all . For further reference, we use the notation . This can be relaxed as in [6] by working in specific nonparametric classes yielding quasi-analyticity.
###### Remark 5.
Proceeding like in [6, 7] we could allow an index of the form where are instrumental variables and is multidimensional of arbitrary dimension but has a sparse random series expansion on some classes of functions. Also, the conditional law of , given , for all , can have a support which is a subspace of the whole space. This means that a nonparametric random coefficients linear index already captures a large class of nonadditively separable models with multiple unobservables.
We can show using (19), (20), and (21), that for a.e. and ,
(23) E[ϕ(Y)R|X=x,S=s]=H[E[ϕ(Y)|X=x,Γ=⋅]fΓ|X=x](s).
By (23), backing out is an inverse problem. However, there is a particular difficulty which is that the left-hand side is only defined (and estimable) on . We obtain the following theorem which states that can be identified at infinity.
###### Theorem 1.
Maintain (17)-(22). For all on the boundary of ,
E[ϕ(Y)|X=x]=lims→~s, s∈H+E[ϕ(Y)R|X=x,S=s]+lims→−~s, s∈H+E[ϕ(Y)R|X=x,S=s].
###### Proof.
By (23), we have
(24) E[ϕ(Y)R|X=x,S=s]=12E[ϕ(Y)|X=x]+H[E[ϕ(Y)|X=x,Γ=⋅]fΓ|X=x]−(s).
Now is a continuous function and, by (24),
H[(E[ϕ(Y)|X=x,Γ=⋅]fΓ|X=x)−](s) =H[E[ϕ(Y)|X=x,Γ=⋅]fΓ|X=x]−(s) =E[ϕ(Y)R|X=x,S=s]−12E[ϕ(Y)|X=x],
hence the conclusion. ∎
Denote by the continuous and odd function defined, for all in the interior of , by
gϕ(s)=E[ϕ(Y)R|X=x,S=s]−12E[ϕ(Y)|X=x],
by for all on the boundary of , and by for all in the interior of . This function is nonparametrically identified by Theorem 1.
By (24), for all ,
(25) gϕ(s)=H[(E[ϕ(Y)|X=x,Γ=⋅]fΓ|X=x)−](s).
This is now a bona-fide ill-posed inverse problem and the inversion can be obtained by (15)-(16).
###### Theorem 2.
Maintain assumptions (17)-(22). For a.e. ,
is given by applying (15)-(16) with, for all ,
∫Sd−1q2p+1,d(θ⊤s)gϕ(s)dσ(s)=E[q2p+1,d(θ⊤S)fS|X=x(S)ϕ(Y)R].
###### Proof.
This is because
∫Sd−1q2p+1,d(θ⊤s)E[ϕ(Y)R|X=x,S=s]dσ(s)=E[q2p+1,d(θ⊤S)fS|X=x(S)ϕ(Y)R]
and, for all ,
∫Sd−1q2p+1,d(θ⊤s)dσ(s)=0.
As a result, the parameter in Theorem 2 is nonparametrically identified and the argument does not involve identification at infinity. This gives, by integration, an other expression for than that of Theorem 1 which does not rely on identification at infinity. By taking to be the function identically equal to 1, we obtain for a.e. .
From this expression, one can obtain an estimator by plug-in and smoothing. One possible smoothing technique is to replace the sum over by a sum up to a truncation parameter. In the approach in [9], there is an additional damping of the high frequencies by an infinitely differentiable filter with compact support. The needlet estimator in [10] also builds on this idea. In the case of the estimation of , [10] provides the minimax lower bounds for more general losses and an adaptive estimator based on thresholding the coefficients of a needlet expansion with a data driven level of hard thresholding.
The root nonparametrically identified in Theorem 2 allows to obtain the law of given and censorship ()
(26) E[ϕ(Y)|X,R=0]=∫Γ,s∈S: Γ⊤s≤0E[ϕ(Y)|X=x,Γ=Γ]fΓ|X=x(γ)fS|X=x(s)dσ(γ)dσ(s),
where is nonparametrically identified. Estimation can be carried by the plug-in principle.
### 5.2. Alternative Scaling Under a Weak Version of Monotonicity
In this section, we denote by the general linear group over and assume
(27) For a.e. x∈supp(X), ∃Px∈GL(d−1): (P⊤xB)1>0 a.s.
We denote by , , , and . This yields
A+B⊤Z>0⇔V−Θ−¯¯¯¯Γ⊤¯¯¯¯Z>0.
Assume also (19),
(28) For a.e. (x⊤ ¯¯¯z⊤)∈supp(X⊤ ¯¯¯¯Z⊤), fΘ+¯¯¯Γ⊤¯¯¯z|X=x exists,
(29) supp(V|X=x,¯¯¯¯Z=¯¯¯z) has a nonempty interior,
∀t∈R, for a.e. (x⊤ ¯¯¯z⊤)∈supp(X⊤ ¯¯¯¯Z⊤), (30) u→g(x,¯¯¯z,u)=E[eitY∣∣X=x,Θ+¯¯¯¯Γ⊤¯¯¯z=u]fΘ+¯¯¯Γ⊤¯¯¯z|X=x(u) is analytic,
(31)
(32) For a.e. x∈supp(X), the % interior of supp(¯¯¯¯Z|X=x) is nonempty.
Condition (31) is slightly stronger than necessary. Conditions implying that certain functions are quasi-analytic, hence allowing to have some heavy tails, are sufficient (see [6]).
By (19),
v→P(R=1|X=x,Z=Px(v ¯¯¯z)⊤)=P(−Θ+¯¯¯¯Γ⊤¯¯¯z
is the cumulative distribution function of a linear functional of a random vector and for all
in the interior of ,
∂vP(R=1|X=x,Z=Px(v ¯¯¯z)⊤)=fΘ+¯¯¯Γ⊤¯¯¯z|X=x(v).
So such invertible matrices are identified.
The vector of random coefficients in the linear index structure clearly satisfies (22). For this reason, we consider the specification of the previous section more general. There is instrument monotonicity in , though not for . This is a weak type of monotonicity because it is possible that there is instrument monotonicity for none of the instrumental variable in the original scale. This is the approach presented in the other versions of [5]. It is shown in [5] that the equation
where , and are unknown functions, can also be transformed by reparametrization into
(33) R=1l{V−Θ−¯¯¯¯Γ⊤¯¯¯¯Z>0}
and that the unknown functions are identified by similar arguments as for the additive model for a regression function.
###### Theorem 3.
Maintain (7) and (28)-(32). For a.e. , the law of conditional on is identified.
###### Proof.
Let . We have by (19), for a.e. ,
E[eitYR∣∣X=x,V=v,¯¯¯¯Z=¯¯¯z]=∫v−∞E[eitY∣∣X=x,Θ+¯¯¯¯Γ⊤¯¯¯z=u]fΘ+¯¯¯Γ⊤¯¯¯z|X=x(u)du.
Hence, by (30), is nonparametrically identified. Moreover, for all and a.e. ,
(34) ∫Reisv∂vg(x,¯¯¯z,v)dv=E[eitY+sΘ+¯¯¯Γ⊤(s¯¯¯z)∣∣∣X=x],
the left-hand side is nonparametrically identified and the right-hand side is the Fourier transform of the law of
conditional on at . We conclude by (31) and (32). ∎
It is possible to turn the identification argument using (34) into an estimation procedure as in [7].
###### Remark 6.
Proceeding like in [6, 7] allows to work with an index of the form where is multidimensional of arbitrary dimension and has a sparse random series expansion on some classes of functions and the conditional laws of and , given , for all , can have a support which is a subspace of the whole space.
###### Remark 7.
The techniques in [8], which are used in [7], also allow to estimate by a simple series estimator, under proper integrability, for almost every even if we observe only when it falls in an interval which is a proper subset of (a type of censoring) if is analytic.
###### Remark 8.
In a binary treatment effect model the outcome can be written as . and are the potential outcomes without and with treatment. They are unobservable. A selection model can be viewed as a degenerate case where a.s. Quantities similar to the root in Theorem 2 have been introduced in [5]. They are for the marginals of the potential outcomes for . An extension of the Marginal Treatment Effect in [1] to multiple unobservables and for laws is the Conditional on Unobservables Distribution of Treatment Effects .
## 6. Application to Missing Data in Surveys
When making inference with survey data, the researcher has available data on a vector of characteristics for units belonging to a random subset of a larger finite population . The law used to draw can depend on variables available for the whole population, for example from a sensus. We assume that the researcher is interested in a parameter which could be computed if we had the values of a variable for all units of index . This can be an inequality index, for example the Gini index, and the wealth of household . In the absence of missing data, the statistician can produce a confidence interval for , making use of the data for the units and his available knowledge on the law . We assume that the cardinality of is fixed and equal to . When
is a total, it is usual to rely on an unbiased estimator, an estimator of its variance, and a Gaussian approximation. For more complex parameters, linearization is often used to approximate moments. The estimator usually rely on the survey weights
. For example an estimator of the Gini index is
(35) ˆg((yi)i∈S)=∑ni=1(2^r(i)−1)πiyi∑ni=1πi∑ni=1πiyi−1,
where . The estimators of the variance of the estimators are more complex to obtain and we assume there is a numerical procedure to obtain it. Inference is based on the approximation
(36) ˆg((yi)i∈S)≈g+√ˆvar(ˆg)((yi)i∈S)ϵ,
where is a standard normal random variable and is an estimator of the variance of .
In practice, this is not possible when some of the s are missing. There is a distinction between total nonresponse, where the researcher discards the data for some units or it is not available, and partial nonresponse. Let us ignore total nonresponse which is usually dealt with using reweighting and calibration and focus on partial nonresponse. We consider a case where can be missing for some units , while all other variables are available for all units . We rely on a classical formalism where the vector of surveyed variables and of those used to draw , for each unit , are random draws from a superpopulation. In this formalism the parameter for all indices of households in the population and are random and we shall now use capital letters for them. Let and be random variables, where if and if unit reveals the value of given , and and be random vectors which will play a different role.
It is classical to rely on imputations to handle the missing data. This means that we replace missing data by artificial values obtained from a model forming predictions or simulating from a probability law and inject them in a formula like (35). In [3] we discuss the use of the Heckman selection model when we suspect that the data is not missing at random. This relies on a parametric model for the partially missing outcome which is prone to criticism. Also as this paper has shown such a model relies on instrument monotonicity which is an assumption which is too strong to be realistic.
It is difficult to analyze theoretically the effect of such imputations. For example when the statistic is nonlinear in the s (e.g. (35)) then using predictions can lead to distorted statistics. It is also tricky to make proper inference when one relies on imputations. One way to proceed is to rely on a hierarchical model as in [4]
. There the imputation model is parametric and we adopted the Bayesian paradigm for two reasons. The first is to account for parameter uncertainty and the second is to replace maximum likelihood with high dimensional integrals by a Monte Carlo Markov Chain Algorithm (a Gibbs sampler). The hierarchical approach also allows layers such as to model model uncertainty. The Markov chain produces sequences of values for each
for in the posterior distribution given , the choice of which is discussed afterwards. Subsequently we get a path of
(37) ˜G=ˆG((Yi)i∈S)+√ˆV(ˆG)((Yi)i∈S)ϵ
where is a standard normal random variable independent from given . (38) is derived from (36). The variables are those making the missing mechanism corresponding to relative to MAR222They can be those used by the survey statistician to draw if any (and usually made available) to handle a total nonresponse which is MAR via imputations.. The last values of the sample path for allows to form credible sets by adjusting the set so that the frequency that exceeds , where is a confidence level.
is the so-called burn-in. These confidence sets account for error due to survey sampling, parameter uncertainty, and nonresponse. They can be chosen from the quantiles of the distribution, to minimize the volume of the set, etc.
We now consider our nonparametric model of endogenous selection which allows for nonmonotonicity of the instrumental variables to handle a missing mechanism corresponding to which is NMAR. For simplicity, we assume away parameter uncertainty and total nonresponse. The variables in Section 5 can be variables that are good predictors for . They are not needed to obtain valid inference but can be useful to make confidence intervals smaller. However, the selection corresponding to the binary variables relative to the outcomes given follow a NMAR mechanism. The (multiple) imputation approach becomes: for
1. Draw an i.i.d. sample of for from the law of given , , and , an independent standard normal , and set for where are the uncensored observations,
2. Compute
(38) ˜Gt=ˆG((Yti)i∈S)+√ˆV(ˆG)((Yti)i∈S)ϵt.
The confidence interval is formed from the sample for a given confidence level. |
# Convergence articles
Displaying 641 - 650 of 655
A collection of short lectures by Howard Eves giving details on 20 important happenings in the history of mathematics before 1650.
In the middle of the 18th century, King Frederick the Great of Prussia became interested in creating a lottery to raise money. As was his custom when mathematical matters were involved, he called upon Leonhard Euler for counsel.
A farmer sold a team of horses for $440, but did not receive his pay for them until 1 yr, 8 mo after the sale. He had at the same time another offer of$410 for them. Did he gain or lose by the sale and by how much, money being worth 6%/yr?
Nine examples of using history in the mathematics classroom -- for those who read French!
A song for Pi Day.
A collection of original texts to help students learn some important areas of mathematics.
A survey of this theorem's 4000 year history, with applications to many fields.
An excellent book surveying the history of the philosophy of mathematics from the time of Plato to the nineteenth and early twentieth centuries.
The edge of a cloud is at an altitude of 20 degrees and the sun above it at 35 degrees. The shadow cast by the edge of the cloud fall on a object 2300 yards away, how high is the cloud?
Suppose that for every 4 cows a farmer has, he should plow 1 acre of land, and allow 1 acre of pasture for every 3 cows; how many cows could he keep on 140 acres? |
DynamicSystems - Maple Programming Help
Home : Support : Online Help : Science and Engineering : Dynamic Systems : Simulation Tools : DynamicSystems/FrequencyResponse
DynamicSystems
FrequencyResponse
compute the frequency response of a system
Calling Sequence FrequencyResponse(sys, opts)
Parameters
sys - System opts - (optional) equation(s) of the form option = value; specify options for the FrequencyResponse command
Options
• In addition to the following options, the FrequencyResponse command takes many of the standard plot options; see plot,options.
Specifies whether the frequencies are determined adaptively. The adaptive algorithm takes the value of the linearfreq option into account when determining the frequency distribution. The default value is false.
• frequencies = list or Vector of nonnegative values
Specifies frequencies to be used to generate a frequency response. The list or Vector must be sorted, from smallest to largest. The units are determined by the hertz option.
• hertz = truefalse
Specifies the unit of frequency. If this value is set to true, the unit of frequency is Hertz; otherwise, it is radians per second. The default value is assigned by DynamicSystems[SystemOptions].
• linearfreq = truefalse
Specifies the frequency scale. If this value is set to $\mathrm{true}$, the scale is linear; otherwise, by default, the scale is logarithmic.
• method = function or matrix
Selects the method used to compute the frequency response. The function method generates the transfer functions of sys and evaluates them at the selected frequencies. The matrix method generates the state-space matrices of sys and evaluates the expression $C\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}.\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\left(Iz-A\right)}^{\left(-1\right)}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}.\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}B+\mathrm{D}$ at specific values of $z$. The default is to use the function method for frequency-domain systems (TF, Coeff, and ZPK), and to use the matrix method for time-domain systems (DE and SS).
• numpoints = nonnegative integer
Specifies the number of points. The default is 100.
• output = response, frequencies, range or list of same
Specifies the output of FrequencyResponse. The name response returns the frequency response, that is, a Matrix of Vectors of complex values. The name frequencies returns the input frequencies, a Vector of real values. The name range returns the range of input frequencies. A list of these names returns an expression sequence with each element corresponding to that returned by the name alone. The default is response.
• parameters = set(name = complexcons) or list(name = complexcons)
Specifies numeric values for parameters in sys. These values override those specified by the parameters field of the system object, which in turn override the settings in in SystemOptions(parameters). The numeric value on the right-hand side of each equation is substituted for the name on the left-hand side in the expressions that define the model. No checking is done during the substitution to determine whether the substituted value is valid. For example, a complex value can be substituted for the coefficient of a polynomial. If the complex value had been originally assigned to the model at creation, a warning would be generated.
• range = range( realcons )
Specifies the frequency range to plot. The default setting is to compute an appropriate range based on the locations of the zeros and poles of the transfer functions of the selected subsystems. Similarly, if the right-hand side of range is infinity, a maximum is computed based on the zeros and poles. The units are specified by the hertz option.
• subsystem = [ posint, posint ] or list of same
Selects subsystems of a multi-input/multi-output system. Each selected subsystem is specified as a list of two indices: the first specifies the output and the second specifies the input. For example, $\left[1,2\right]$ specifies the subsystem from the second input to the first output. A list of lists selects multiple subsystems. The default setting is to select all subsystems, using Fortran ordering. That is, for an m x n system, the default list of subsystems is [ [1,1], ..., [m,1], ..., [1,n], ..., [m,n] ].
Description
• The FrequencyResponse command computes the frequency response of subsystems of sys, a system object.
• If sys is a continuous system, its s-domain transfer function is computed and then converted to the real frequency domain using the transformation s -> I*omega, where omega is the angular frequency.
• If sys is a discrete system, its z-domain transfer function is computed and then converted to the real frequency domain using the transformation z -> exp(I*omega*Ts), where omega is the angular frequency and Ts is the sample time.
Examples
> $\mathrm{with}\left(\mathrm{DynamicSystems}\right):$
> $\mathrm{sys}≔\mathrm{ZeroPoleGain}\left(\left[0,1\right],\left[2,4,6\right],1\right):$
> $\mathrm{FrequencyResponse}\left(\mathrm{sys}\right)$
$\left[\begin{array}{c}\left[\begin{array}{c}{\mathrm{1 .. 100}}{{\mathrm{Vector}}}_{{\mathrm{column}}}\\ {\mathrm{Data Type:}}{\mathrm{anything}}\\ {\mathrm{Storage:}}{\mathrm{rectangular}}\\ {\mathrm{Order:}}{\mathrm{Fortran_order}}\end{array}\right]\end{array}\right]$ (1)
Return the frequency range used to generate the response.
> $\mathrm{FrequencyResponse}\left(\mathrm{sys},\mathrm{output}=\mathrm{range}\right)$
${0.01000000000}{..}{1000.}$ (2) |
Is this series bounded?
But I can't answer to it.
$$\sum^{\infty}_{1}(-1)^n*(1+\frac{1}{n})^n$$
Is this series bounded?
I can't do anything about that.
Last edited:
jbunniii
Homework Helper
Gold Member
Do you know what
$$\lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)^n$$
equals?
I know this series diverge by lim an=e, not zero.
But I can't figure out it's bounded.
Last edited:
The sum of any two consecutive terms goes to zero as O(1/n^2). Since the sum of 1/n^2 is convergent, you can prove that your original series is bounded.
hmm... Do you mean (1 + 1/(n+1))^(n+1) - (1 + 1/n)^n ≤1/n^2?
Last edited:
hmm... Do you mean (1 + 1/(n+1))^(n+1) - (1 + 1/n)^n ≤1/n^2?
Correct. You need to prove that there's such N and C that, for all n>N,
(1 + 1/(n+1))^(n+1) - (1 + 1/n)^n ≤ C/n^2
You can do that if you write
(1+1/n)^n = exp(n log (1+1/n))
and then use Taylor expansion of log.
Last edited:
I computed the process you have given.
log(x)=(x-1) - 1/2*(x-1)^2 + 1/3*(x-1)^3 - ...(correct?)
So My result is
exp (1 - 1/2(n+1) + 1/3(n+1)^2 - ...)
- exp (1 - 1/2n + 1/3n^2 - ...)
let exp(A(n+1)) - exp((An))
exp does not linear, I can't do further.
Help me more...
$$exp(1 - \frac{1}{2n} + \frac{1}{3n^2} + ...) \approx e*( - \frac{1}{2n} + \frac{1}{3n^2}) + ( - \frac{1}{2n} + \frac{1}{3n^2})^2/2 + ...) = e*(-\frac{1}{2n}+\frac{1}{3n^2} + \frac{1}{8n^2} + ...) = e*(-\frac{1}{2n}+\frac{11}{24n^2} + ... )$$
where ... are terms that go down as 1/n^3 or faster.
$$|\frac{1}{2(n+1)} - \frac{1}{2n}| = \frac{1}{2n(n+1)} < \frac{1}{2n^2}$$
Oh I see!!! |
# Higher direct image of the inclusion of the generic point
I am trying to understand the proof of the following statement:
Let $X=\Spec(D)$ where $D$ is the ring of intergers of a number field K. Let I be the ideal class group of $K$. Then the etale cohomology $H^1(X,Z/n) = \Hom(I,Z/n)\\$ The proof goes as follow :
Consider the Leray spectral sequence for $j : \Spec(K)\rightarrow X$ we have $$H^r(X,R^sj_{*}Z/n) \Rightarrow H^{r+s}(G_K,Z/n)\\$$ The first few terms of the spectral sequence gives the exact sequence $\\$ $$0\rightarrow H^1(X,Z/n)\rightarrow H^1(G_K,Z/n)\rightarrow H^0(X,R^1j_{*}Z/n) \rightarrow \ker(H^2(G_K,Z/n)\rightarrow H^0(X,R^2j_{*}Z/n))$$
Then as $H^0(X,R^rj_{*}Z/n)=H^r(G_L,Z/n)$ where L is the Hilbert Class Field of K and $I=\Gal(L/K)=G_K/G_L$ we get
$$H^1(X,Z/n)=\ker(H^1(G_K,Z/n)\rightarrow H^1(G_L,Z/n))=H^1(I,Z/n)=\Hom(I,Z/n)$$
My question is that why do we have $$H^0(X,R^rj_{*}Z/n)=H^r(G_L,Z/n)$$
I know that $R^rj_{*}F$ is the sheaf associated to the presheaf $$U\rightarrow H^r(G_{K(U)},F)$$ where $K(U)$ is the function field of U. Any help would be much appreciated. Thank you.
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To find the reproductive values, we need to find the left eigenvectors. In this problem, we will get three eigen values and eigen vectors since it's a symmetric matrix. Therefore, multiplying vector [4 2] by inverse of B, would give us vector [2 2]. where the eigenvalues are subscripted with an s to denote being sorted. Visit http://ilectureonline.com for more math and science lectures!In this video I will find eigenvector=? For any triangular matrix, the eigenvalues are equal to the entries on the main diagonal. In the next section, we explore an important process involving the eigenvalues and eigenvectors of a matrix. defined above satisfies, and there exists a basis of generalized eigenvectors (it is not a defective problem). [ A complex-valued square matrix A is normal (meaning A*A = AA*, where A* is the conjugate transpose) In practice, eigenvalues of large matrices are not computed using the characteristic polynomial. Those near zero or at the "noise" of the measurement system will have undue influence and could hamper solutions (detection) using the inverse. [ Let A be a square n × n matrix with n linearly independent eigenvectors qi (where i = 1, ..., n). In power iteration, for example, the eigenvector is actually computed before the eigenvalue (which is typically computed by the Rayleigh quotient of the eigenvector). The integer ni is termed the algebraic multiplicity of eigenvalue λi. Facebook. All the matrices are square matrices (n x n matrices). Assume that the middle eigenvalue is near 2.5, start with a vector of all 1's and use a relative tolerance of 1.0e-8. [8] (For more general matrices, the QR algorithm yields the Schur decomposition first, from which the eigenvectors can be obtained by a backsubstitution procedure. ) Equation holds for each eigenvector-eigenvalue pair of matrix . A Non-square matrices cannot be analyzed using the methods below. { Cramer’s rule. That is, if. However, this is often impossible for larger matrices, in which case we must use a numerical method. If matrix A can be eigendecomposed, and if none of its eigenvalues are zero, then A is invertible and its inverse is given by − = − −, where is the square (N×N) matrix whose i-th column is the eigenvector of , and is the diagonal matrix whose diagonal elements are the corresponding eigenvalues, that is, =.If is symmetric, is guaranteed to be an orthogonal matrix, therefore − =. If two matrices are similar, then they have the same rank, trace, determinant and eigenvalues. Therefore. Definitions and terminology Multiplying a vector by a matrix, A, usually "rotates" the vector , but in some exceptional cases of , A is parallel to , i.e. The integer mi is termed the geometric multiplicity of λi. Then find all eigenvalues of A5. In this section K = C, that is, matrices, vectors and scalars are all complex.Assuming K = R would make the theory more complicated. Enter a matrix. So kind of a shortcut to see what happened. This means that either some extra constraints must be imposed on the matrix, or some extra information must be supplied. How do you prove this for the general case? Matrix algebra for beginners, Part II linear transformations, eigenvectors and eigenvalues Jeremy Gunawardena Department of Systems Biology Harvard Medical School 200 Longwood Avenue, Cambridge, MA 02115, USA [email protected] February 10, 2006 Contents 1 Introduction 1 2 Vector spaces and linear transformations 1 3 Bases and matrices 2 4 Examples—rotations and … The same result is true for lower triangular matrices. Example The eigenvalues of the matrix:!= 3 −18 2 −9 are ’ .=’ /=−3. ST is the new administrator. For any triangular matrix, the eigenvalues are equal to the entries on the main diagonal. . The proofs of the theorems above have a similar style to them. where is a lower triangular matrix and is an upper triangular matrix with ones on its diagonal. Two mitigations have been proposed: truncating small or zero eigenvalues, and extending the lowest reliable eigenvalue to those below it. Then $\lambda^{-1}$ is an eigenvalue of the matrix $\inverse{A}$. Share . To find all of a matrix's eigenvectors, you need solve this equation once for each individual eigenvalue. In particular, any symmetric matrix with real entries that has $$n$$ eigenvalues, will have $$n$$ eigenvectors. However, in practical large-scale eigenvalue methods, the eigenvectors are usually computed in other ways, as a byproduct of the eigenvalue computation. A conjugate eigenvector or coneigenvector is a vector sent after transformation to a scalar multiple of its conjugate, where the scalar is called the conjugate eigenvalue or coneigenvalue of the linear transformation. Eigenvalue is the factor by which a eigenvector is scaled. Save my name, email, and website in this browser for the next time I comment. Learn how your comment data is processed. Putting the solutions back into the above simultaneous equations, Thus the matrix B required for the eigendecomposition of A is, If a matrix A can be eigendecomposed and if none of its eigenvalues are zero, then A is nonsingular and its inverse is given by. α β = x , then 0 0 ab cd λα λβ −− = −− Various cases arise. Recall that the geometric multiplicity of an eigenvalue can be described as the dimension of the associated eigenspace, the nullspace of λI − A. Ax x= ⇒ −=λ λ ( )IA x0 Let . Those are the two values that would make our characteristic polynomial or the determinant for this matrix equal to 0, which is a condition that we need to have in order for lambda to be an eigenvalue of a for some non … For example, the defective matrix Q-1= XR*Y*XL . The algebraic multiplicity can also be thought of as a dimension: it is the dimension of the associated generalized eigenspace (1st sense), which is the nullspace of the matrix (λI − A)k for any sufficiently large k. That is, it is the space of generalized eigenvectors (first sense), where a generalized eigenvector is any vector which eventually becomes 0 if λI − A is applied to it enough times successively. where a, b, c and d are numbers. To find the eigenvectors of a triangular matrix, we use the usual procedure. […], Your email address will not be published. Eigenvectors and Eigenvalues. This usage should not be confused with the generalized eigenvalue problem described below. $1 per month helps!! This equation will have Nλ distinct solutions, where 1 ≤ Nλ ≤ N. The set of solutions, that is, the eigenvalues, is called the spectrum of A.[1][2][3]. It turns out that the left eigenvectors of any matrix are equal to the right eigenvectors of the transpose matrix. Syntax: eigen(x) Parameters: x: Matrix … The inverse power method is used for approximating the smallest eigenvalue of a matrix or for approximating the eigenvalue nearest to a given value, together with the corresponding eigenvector. Similarly, a unitary matrix has the same properties. And then this matrix, or this difference of matrices, this is just to keep the determinant. The above equation is called the eigenvalue equation or the eigenvalue problem. This website is no longer maintained by Yu. [6] Nov 27,2020 - Eigenvalues And Eigenvectors - MCQ Test 2 | 25 Questions MCQ Test has questions of Mechanical Engineering preparation. eigenvector. How to Diagonalize a Matrix. So, if we take the transpose and use eigen(), we can easily find the left eigenvector, and then the reproductive values: All Rights Reserved. A = , Furthermore, The eigenvector is not unique but up to any scaling factor, i.e, if is the eigenvector of , so is with any constant . The method is conceptually similar to the power method. In general, a square matrix of size $$n \times n$$ must be diagonalizable in order to have $$n$$ eigenvectors. Matrix Representations for Linear Transformations of the Vector Space of Polynomials. Thanks to all of you who support me on Patreon. Algebraic & Geometric Multiplicity If the eigenvalue λ of the equation det(A-λI)=0 is repeated n times then n is called the algebraic multiplicity of λ.The number of linearly independent eigenvectors is the difference between the number of unknowns and the rank of the corresponding … Same thing when the inverse comes first: (1 / 8) × 8 = 1. They all begin by grabbing an eigenvalue-eigenvector pair and adjusting it in some way to reach the desired conclusion. The determinant of the matrix B is the product of all eigenvalues of B, or If 0 is an eigenvalue of B then B x = 0 has a nonzero solution, but if B is invertible, then it’s impossible. This page was last edited on 10 November 2020, at 20:49. Iterative methods form the basis of much of modern day eigenvalue computation. This is called the secular determinant, and expanding the … The first mitigation method is similar to a sparse sample of the original matrix, removing components that are not considered valuable. However, we often want to decompose matrices into their eigenvalues and eigenvectors. Eigenvalues of the Laplace Operator. For instance, by keeping not just the last vector in the sequence, but instead looking at the span of all the vectors in the sequence, one can get a better (faster converging) approximation for the eigenvector, and this idea is the basis of Arnoldi iteration. Eigenvectors with Distinct Eigenvalues are Linearly Independent; Singular Matrices have Zero Eigenvalues ; If A is a square matrix, then λ = 0 is not an … = ( {\displaystyle \mathbf {Q} } [8], A simple and accurate iterative method is the power method: a random vector v is chosen and a sequence of unit vectors is computed as, This sequence will almost always converge to an eigenvector corresponding to the eigenvalue of greatest magnitude, provided that v has a nonzero component of this eigenvector in the eigenvector basis (and also provided that there is only one eigenvalue of greatest magnitude). The row vector is called a left eigenvector of . If the matrix is small, we can compute them symbolically using the characteristic polynomial. Since !has two linearly independent eigenvectors, the matrix 6is full rank, and hence, the matrix !is diagonalizable. To find the eigenvectors of a triangular matrix, we use the usual procedure. ed.png. As we saw earlier, we can represent the covariance matrix by its eigenvectors and eigenvalues: (13) where is an eigenvector of , and is the corresponding eigenvalue. :) https://www.patreon.com/patrickjmt !! What is the eigenvalue and how many steps did … I understand for specific cases that a matrix and its inverse(if the inverse exist) have a correlation in their eigenvalues. One reason is that small round-off errors in the coefficients of the characteristic polynomial can lead to large errors in the eigenvalues and eigenvectors: the roots are an extremely ill-conditioned function of the coefficients. This provides an easy proof that the geometric multiplicity is always less than or equal to the algebraic multiplicity. Notify me of follow-up comments by email. Any vector satisfying the above relation is known as eigenvector of the matrix A A corresponding to the eigen value λ λ. [9] Also, the power method is the starting point for many more sophisticated algorithms. If f (x) is given by. ] The general case of eigenvectors and matrices: $M\mathbf{v} = \lambda\mathbf{v}$, put in the form $(\lambda I - M)\mathbf{v}=0$. Eigenvalues and Eigenvectors of a Matrix Description Calculate the eigenvalues and corresponding eigenvectors of a matrix. 0 The second term is 0 minus 2, so it's just minus 2. x Solved exercises. In this article, let us discuss the eigenvector definition, equation, methods with examples in detail. Exercise 1. If .A I/ x D 0 has a nonzero solution, A I is not invertible. eigenvectors of a matrix, some of which fall under the realm of iterative methods. Eigenvalues first. This site uses Akismet to reduce spam. Thus, Rank of Matrix= no of non-zero Eigenvalues of the Matrix. f If b = c = 0 (so that the matrix A is diagonal), then: For . Let A be a square n × n matrix with n linearly independent eigenvectors qi (where i = 1, ..., n). This website’s goal is to encourage people to enjoy Mathematics! These methods work by repeatedly re ning approximations to the eigenvectors or eigenvalues, and can be terminated whenever the approximations reach a suitable degree of accuracy. where λ is a scalar, termed the eigenvalue corresponding to v. That is, the eigenvectors are the vectors that the linear transformation A merely elongates or shrinks, and the amount that they elongate/shrink by is the eigenvalue. The inverse is: The inverse of a general n × n matrix A can be found by using the following equation. A Group with a Prime Power Order Elements Has Order a Power of the Prime. Eigenvalue is the factor by which a eigenvector is scaled. Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors. Convert matrix to Jordan normal form (Jordan canonical form). In measurement systems, the square root of this reliable eigenvalue is the average noise over the components of the system. The total number of linearly independent eigenvectors, Nv, can be calculated by summing the geometric multiplicities. This example was made by one of our experts; you can easily contact them if you are puzzled with complex tasks in math. A (non-zero) vector v of dimension N is an eigenvector of a square N × N matrix A if it satisfies the linear equation. Homework Statement T/F: Each eigenvector of an invertible matrix A is also an eignevector of A-1 Homework Equations The Attempt at a Solution I know that if A is invertible and ##A\vec{v} = \lambda \vec{v}##, then ##A^{-1} \vec{v} = \frac{1}{\lambda} \vec{v}##, which seems to imply that A and its inverse have the same eigenvectors. Hilbert Matrices and Their Inverses. The linear combinations of the mi solutions are the eigenvectors associated with the eigenvalue λi. If you have trouble understanding your eigenvalues and eigenvectors of 3×3 matrix assignment, there is no need to panic! A-1 × A = I. If A is restricted to be a Hermitian matrix (A = A*), then Λ has only real valued entries. (which is a shear matrix) cannot be diagonalized. Any eigenvector is a generalized eigenvector, and so each eigenspace is contained in the associated generalized eigenspace. Let$F$and$H$be an$n\times n$matrices satisfying the relation $HF-FH=-2F.$ (a) Find the trace of the matrix... (a) If$A$is invertible, is$\mathbf{v}$an eigenvector of$A^{-1}$? In this paper, we outline ve such iterative methods, and … For part (b), note that in general, the set of eigenvectors of an eigenvalue plus the zero vector is a vector space, which is called the eigenspace. x You da real mvps! During a linear transformation, there may exist some vectors that remain on their original span, and are … Dana Mackey (DIT) Numerical Methods II 6 / 23 . The reliable eigenvalue can be found by assuming that eigenvalues of extremely similar and low value are a good representation of measurement noise (which is assumed low for most systems). eigen() function in R Language is used to calculate eigenvalues and eigenvectors of a matrix. c++ matrix sparse-matrix eigen eigen3. Dana Mackey (DIT) Numerical Methods II 6 / 23. [12] In this case, eigenvectors can be chosen so that the matrix P Get three eigen values of such that ] also, the matrix, the method. Invertible, we noted that a matrix equations and many other applications related to them Dense! Gaussian elimination or any other method for solving matrix equations grabbing an eigenvalue-eigenvector and. Which fall under the realm of iterative methods form the basis of much of day. Determinant, then solve for lambda, truncating may remove components that are not computed using the polynomial. General case I can not find inverse operation anywhere Mackey ( DIT ) methods! 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# What was Lame's proof?
In 1847, Lame gave a false proof of Fermat's Last Theorem by assuming that $\mathbb{Z}[r]$ is a UFD where $r$ is a primitive $p$th root of unity.
The best description I've found is in the book Fermat's Last Theorem A Genetic Introduction to Algebraic Number Theory. For the equation $x^n + y^n = z^n$, it says
That is, he planned to show that if $x$ and $y$ are such that the factors $x+y, x+ry, \dots, x+r^{n-1}y$ are relatively prime then $x^n + y^n = z^n$ implies that each of the factors $x+y, x+ry, \dots$ must itself be an $n$th power and to derive from this an impossible infinite descent. If $x+y, x+ry, \dots$ are not relatively prime he planned to show that there is a factor $m$ common to all of them so that $(x+y)/m, (x+ry)/m, \dots, (x+r^{n-1}y)/m$ are relatively prime and to apply a similar argument in this case as well.
Nowhere else can I find any more detail as to how the rest of the argument goes. I don't see how to use infinite descent here, can anyone fill in the details?
I would like to also mention Conrad's great notes on Kummer's proof for regular primes. While this does indeed give a proof, it seems quite sophisticated and I am hoping there is an easier method when one assumes (falsely, in general) that $\mathbb{Z}[r]$ is a UFD.
The paper (in French) to which Edwards refers in his book may be found here:
http://gallica.bnf.fr/ark:/12148/bpt6k29812/f310.image
My French isn't good enough, however, to elaborate on the argument given above.
• Thank you for the link! I should have checked references. I can read french, once I have time to give it a quick look over I am all but certain I will accept this answer. – RghtHndSd Oct 1 '14 at 19:00
• It does appear to be a complete proof, and also much more complicated than what is usually indicated when it is discussed (at least I got the impression that the descent argument was rather short). I will be editing your answer to incude summary in a few days. – RghtHndSd Oct 3 '14 at 0:23
• @RghtHndSd Will you include the mentioned summary? It would be kindly appreciated! – Jose Brox Oct 7 '15 at 9:42
• @JoseBrox: After looking through the proof for about an hour or so, I found that it wasn't the French that was difficult but rather the language that the mathematics was expressed in. Since I wanted to use this for a class I was teaching, and it was clear the descent argument would not be short enough to present, I decided to skip it. I have no current intentions of trying to understand it, but maybe sometime in the future. – RghtHndSd Oct 7 '15 at 15:30
Lamé's proof is quite short and easy to follow for a modern reader. I won't give a full translation, just a summary. Since my French is far better than my algebraic number theory, I'll refrain from taking too many shortcuts, and stick to the original structure of the proof, so as not to introduce any mistakes.
The idea of the proof is to prove that $A^n+B^n=C^n$ has no solution in $\mathbb Z[r]$, where $r$ is a primitive root of $r^n=1$. In particular, this shows that it has no solution in $\mathbb Z$.
Lamé immediately points out that it suffices to consider the case $n$ prime. He also mentions at the end of the proof that, due to a technical reason which we'll get to later, the proof doesn't work for $n=3$, so we're talking about an odd prime $n>3$.
§ I - Properties of $\mathbb Z[r]$
Lamé begins by defining more or less explicitly $\mathbb Z[r]$, where $r$ is a primitive root of $r^n=1$. He quickly goes through some basic properties of $\mathbb Z[r]$ and defines some basic concepts:
1. The representation of an element of $\mathbb Z[r]$ as $\alpha_0 + \alpha_1 r + ... + \alpha_{n-1}r^{n-1}$ (all the $\alpha_i$ integers) is unique up to adding the same integer to each coefficient.
2. If for some $E$, $A=DE$, where all values are in $\mathbb Z[r]$, we say that $A$ is divisible by $D$.
3. In order for an element $a\in\mathbb Z[r]$ to be divisible by an element $k\in\mathbb Z$, it is necessary and sufficient that the coefficients of $a$ to be congruent modulo $k$ (which does not depend on the choice of coefficients).
4. Given a value $A$, we can consider the $n$ values $r^iA$ for different values of $i$. Lamé uses the notation $A^{(i)}=r^iA$, but I kept misreading that as an exponent while reading the paper, so I'll just write $r^i A$. Lamé notes that the $n$-th powers of these numbers are all the same.
5. If we think of $A$ as a polynomial expression in $r$, we can imagine replacing $r$ with $r^i$. We'll call this value $A_i$. In modern language we're talking about the image of $A$ under the automorphism $r\to r^i$.
6. We'll call the product of the $A_i$ the modulus of $A$. Since it's a symmetric function of the roots of $r^n=1$, it's an integer. The modulus is invariant under the automorphisms $r\to r^i$ and the modulus of $r^i A$ is the same as the modulus of $A$. The modulus is multiplicative.
7. Finally, let $z_i=r^i+r^{-i}$.
§II - Factorization of $A^n+B^n$
We're now going to factor $A^n+B^n$ in a rather specific way. We have
$$\prod_{i=0}^{n-1} (B+r^i A) = \sum_{i=0}^n S_i A^iB^{n-i}=A^n+B^n \tag 1$$
Here $S_i$ is the i-th elementary symmetric polnomial in the $r^i$, so that $S_0=1$, $S_1=1+r+r^2+...r^{n-1}$, and $S_n=1rr^2...r^{n-1}$. The first equality is just straight algebra, but the second involves specific properties of the roots $r^i$: since they solve the equation $x^n-1=0$, all of the $S_i$ are zero except $S_0=S_n=1$.
Lamé now rewrites this factorization in a slightly different form:
$$A^n+B^n=\prod_{i=0}^{n-1} (Ar^i + Br^{n-i}) \tag 2$$
To obtain this form, start with the factorization in $(1)$. Take the factor $r^i A + B$, and factor out $r^k$ with $k=-i/2\mod n$. We obtain $r^k(r^i A + B)=Ar^{-k}+Br^k$, which can be written $Ar^j+Br^{n-j}$ for some $j$, as required. As we run through all the values of $i$, $k$ and $j$ both run through all values $0, 1, 2, ..., n-1$, so we have
$$A^n+B^n=r^{-(0+1+2+...+(n-1))}\prod_{j=0}^{n-1} (Ar^j + Br^{n-j})$$
Since the exponent on that factor of $r$ is divisible by $n$, that factor is just $1$, so we have the desired factorization.
Now let's write $M_i$ for the $i$-th factor, that is, $M_i=Ar^i+Br^{n-i}$. These factors satisfy a system of linear equations, namely, for any two indices $i,j$:
$$M_i+M_j=z_{\frac{i-j}2} M_{\frac{i+j}2} \tag 3$$
where the indices are interpreted modulo $n$, and the $z_i$ are the constants which I defined at point 7 of the list above (in §1). This isn't very obvious, but is easy to check.
Lamé now states without proof that thanks to these equation, we can conclude two things about the $M_i$:
1. If any number $D$ divides any two of the $M_i$, it divides all of them.
2. If one of the $z_i$ divides any one of the $M_j$, it divides all of them.
Neither of these facts seems obvious to me. If we write (3) as $M_i+M_j=z_kM_l$, then we can say that if $D$ divides both $M_l$ and $M_j$, it divides $M_i$, $i=2l-j\mod n$. Presumably the idea is to then repeat this argument with other equations until we cover all of the indices. It seems to me we need the following lemma:
Lemma (Not mentioned in Lamé's paper): If a subset of $\mathbb Z/n\mathbb Z$ contains at least two elements and is closed under the operation $(k, j)\to 2k-j$ for $k\neq j$, then it is equal to all of $\mathbb Z/n\mathbb Z$.
Note that if $k=j$ we would have $2k-j=k$, so strictly speaking the condition $k\neq j$ is not necessary (it's just that in that case there's no corresponding equation). Similar logic will allow us to justify point (2) if we have the lemma:
Lemma (Also not mentioned): If a subset of $\mathbb Z/n\mathbb Z$ contains at least two elements and is closed under the operation $(k, j)\to 2k+j$, then it is equal to all of $\mathbb Z/n\mathbb Z$.
I'm not sure if either of these lemmas is true. If not, then I don't know how to justify Lamé's claims (1) and (2). In any case, if they are true, then after pulling common factors out of the $M_i$, we obtain a new factorization of $A^n+B^n$:
$$A^n+B^n=k^nm_1m_2...m_{n-1}$$
where the $m_i$ are pairwise coprime in $\mathbb Z[r]$ and are each indivisible by any of the $z_j$. This implies, and I believe this is where the assumption of unique factorization creeps in for the first time, that the power of $z_j$ for any $j$ in the unique prime factorization of $A^n+B^n$ (that is, generally, in a sum of two $n$-th powers) must be a multiple of $n$.
§III - Conclusion
Now that we have a factorization of $A^n+B^n$, suppose we had $A^n+B^n=C^n$. Again implicitly using uniqueness of prime factorization, this implies that in $A^n+B^n=k^nm_1...m_{n-1}$, each of the $m_i$ must itself be an $n$-th power, since they're coprime from one another and from $k$. Thus we would have, say, $m_i=\mu_i^n$, but then, going back to the equations for the $M_i$, we would have:
$$\mu_i^n+\mu_j^n=z_k\mu_l^n \tag 4$$
This is impossible, since by what we obtained at the end of the last section, the power of $z_k$ in the factorization of the left hand side must be a multiple of $n$, but one the right it is one more than multiple of $n$. Here Lamé mentions that it can be shown that the $z_i$ are not themselves $n$-th powers, but I'm not sure if this is important. He also mentions that since for $n=3$, there is only one value of $z_i$ which is $z_i=-1$, a unit, this argument fails in that case, which is why we assumed $n>3$ above.
Lamé concludes by explaining why Fermat's last theorem is a special case of the above. For me this seems obvious since it seems to me we've proven the theorem for $\mathbb Z[r]$, and $\mathbb Z\subset\mathbb Z[r]$, but for some reason Lamé continues with a fairly involved final paragraph. I don't understand the argument or why it's necessary, so I'll just translate it for those who do:
Fermat's theorem, for $n>3$, is only a special case of what has just been proven; since if $A$ and $B$ are integers, that is if they reduce to $\alpha_0$, $\beta_0$, $M_1$ will be integer, as well as $C$, $k$, and $\mu_1$; but $\mu_2, ... \mu_{n-1}$ will remain complex numbers: however, their product must be an integer modulus, that is $\mu_1, ... \mu_{n-1}$ must be the factors of a whole number of the form $Y^2\pm nZ^2$ [Lamé mentions in §1 that moduli are always of this form]; finally, the relations [displayed equation (4)] will still be necessary, and the conclusion of impossibility will be the same.
The article ends with some observations from Liouville in which he points out the flaw in the assumption of unique factorization. Edwards describes these comments in his book, although he makes them out to seem alot ruder than they look to me in the original French. Maybe I'm just not picking up on subtle nineteenth century academic sarcasm.
• Nice work! Both lemmas can be proved by using the following, easy fact for a prime $p$. If the difference of an infinite arithmetic progression $(a)$ is non congruent to $0$ modulo $p$, then $(a)$ contains all the residues mod $p$. Recall that $n>3$ is a prime. In the first Lemma you can generate the sequence $a_t=t(k-j)+j$, because $a_0=j$, $a_1=k$ and $(a_t,a_{t+1})\to a_{t+2}$ for $t\ge0$. In the second lemma WLOG let $k\neq0$. You can generate $a_t=2tk+j$ in the following way: $a_0=j$, $(k,j)\to 2k+j=a_1$ and $(k,a_t)\to a_{t+1}$ for $t\ge0$. – B.B. May 16 '18 at 14:04
• Thanks a lot for preparing such a detailed summary. Very helpful! – UnrealVillager Dec 30 '20 at 0:27
• Just curious, aren't all z_k actually units in Z[r]? And if so, Lamé's entire argument is not valid even regardless of the unique factorization. z_k can be reduced to 1 + r^2k. But isn't that one of those so-called "cyclotomic units"? Maybe I'm missing something... – UnrealVillager Jan 4 at 0:43 |
Limited access
A force of $5\text{ N}$ is applied to the bicycle wheel ($m = 5\text{ kg}$; $r=0.40\text{ m}$) currently moving clockwise with an initial angular velocity of $-15\text{ rad/s}$ as shown in the image below.
If this force is applied for $15\text{ s}$, what is the final angular velocity of the wheel?
A
$52.5\text{ rad/s}$
B
$37.5\text{ rad/s}$
C
$22.5\text{ rad/s}$
D
$-37.5\text{ rad/s}$
Select an assignment template |
99-438 Bernhard Baumgartner, Jan Philip Solovej, and Jakob Yngvason
Atoms in strong magnetic fields:The high field limit at fixed nuclear charge (61K, LaTex) Nov 19, 99
Abstract , Paper (src), View paper (auto. generated ps), Index of related papers
Abstract. Let $E(B,Z,N)$ denote the ground state energy of an atom with $N$ electrons and nuclear charge $Z$ in a homogeneous magnetic field $B$. We study the asymptotics of $E(B,Z,N)$ as $B\to \infty$ with $N$ and $Z$ fixed but arbitrary. It is shown that the leading term has the form $(\ln B)^2 e(Z,N)$, where $e(Z,N)$ is the ground state energy of a system of $N$ {\em bosons} with delta interactions in {\em one} dimension. This extends and refines previously known results for $N=1$ on the one hand, and $N,Z\to\infty$ with $B/Z^3\to\infty$ on the other hand.
Files: 99-438.src( 99-438.keywords , bsy1911.tex ) |
# Is there a standard scholarly reference for lattice constants of crystals of the elements?
I need to discuss the lattice constants of bulk crystals of several metal and semiconductor elements. I can find plenty of tables and numbers that are probably "close enough" but for a paper I'd like to cite a standard source.
From https://periodictable.com/Properties/A/LatticeConstants.html I have the following numbers, but I don't want to use a "dot com" as a scholarly source, and I can not figure out how to use Wolfram Alpha (the source for this website) or understand where WA gets its numbers. They list several references here https://reference.wolfram.com/language/note/ElementDataSourceInformation.html , but it will be a challenge to track them all down one by one. I'm hoping someone will recognize one of them or simply be able to mention it.
I only need (at a minimum) three decimal places for the lattice constants in Angstroms (though four is better if they are known), but what I can't find is a citable, schollarly source that covers all of these and is likely to cover other common elemental crystals when I need them in the future.
element. lattice constant
(a, b and c)
Angstroms
Au 4.0782
Ag 4.0853
Pb 4.9508
Ge 5.6575
Si 5.4309
• Wikipedia cites a lot of sources. Elements I'm pretty sure you will find in the CDC handbook. crystallography.net/cod it.iucr.org – Karl Aug 23 '20 at 14:21
• @Karl I'm having trouble understanding that link. I won't be purchasing an on-line copy, is this a book I can find in some library or is it strictly a paid access online database? I'd like to be able to cite something that one could actually check and confirm in a reasonably well-sized and libraried university. I can't even figure out which of the eight volumes (A through H) has an actual table of lattice constants with numbers in it. – uhoh Aug 23 '20 at 14:28
• @Karl I'm one of those people who won't cite something unless I've actually check it myself. – uhoh Aug 23 '20 at 14:30
• It's two links, and any uni library should give you access. The CRC (sorry, typo) handbook used to be printed, now it's online too, and your library should have access, too. hbcponline.com/faces/contents/… – Karl Aug 23 '20 at 14:31
• @Karl okay got it! Let me try a device that's connected to the library, this one isn't. :-) – uhoh Aug 23 '20 at 14:42
The CRC Handbook of Chemistry and Physics contains a dedicated compilation by H. W. King, titled «Crystal Structures and Lattice Parameters of Allotropes of the Elements». In case your research library is closed, you may access some of its editions freely or borrow them with the library card of archive.org.
In case of the 97th edition (by 2016), the section starts by page 12-16. (Elements liquid or gaseous at ambient conditions are included.)
source
• Thanks, this may be what I end up using. It's a real, recognized and trusted physical reference I can hold in my hand and read, and it lists its source from where these numbers come from, which I can also put my hands on and confirm. Thanks! – uhoh Aug 24 '20 at 16:03
The lattice constants of all five elements have been published in a single reference in 1925 (Ref.1):
The lattice constant $$a$$ has been determined within 0.1 percent (.03 percent for $$\ce{W}$$) for aluminum, iron, nickel, copper, molybdenum, palladium, silver, tungsten, platinum, gold, lead, and bismuth, by direct comparison with $$\ce{NaCl}$$, $$a(\ce{NaCl}) = \pu{2.814 \mathring A}$$. As pure samples as could be obtained were used, from 99.55 percent for $$\ce{Ni}$$ to 99.9995 percent for tungsten, and in many cases commercially pure samples were also measured for comparison. The results for the purest samples are summarized in Table XIII. The density from the x-ray data is in each case (except $$\ce{Al}$$ and $$\ce{Ag}$$) greater than the density of the bulk metal as given in the literature, the difference being rather large for $$\ce{Mo}$$ (10.21 vs 9.1), for $$\ce{Pd}$$ (12.25 vs 11.9) and $$\ce{W}$$ (19.32 vs 18.77). For pure $$\ce{W}$$ remarkably sharp lines were obtained. For $$\ce{Bi}$$ a piece of a single artificial crystal was used.
The relevant $$a$$ values are listed in a Wikipedia article as follows:
$$\begin{array}{c|ccc} \hline \text{Metal} & \text{Lattice constant}^a & \text{Crystal structure} & \text{Lattice constant given} \\ \hline \ce{Au} & \pu{4.065 \mathring A} & \text{FCC} & \pu{4.0782 \mathring A} \\ \ce{Ag} & \pu{4.079 \mathring A} & \text{FCC} & \pu{4.0853 \mathring A} \\ \ce{Pb} & \pu{4.920 \mathring A} & \text{FCC} & \pu{4.9508 \mathring A} \\ \ce{Ge} & \pu{5.658 \mathring A} & \text{Diamond (FCC)} & \pu{5.6575 \mathring A} \\ \ce{Si} & \pu{5.4310205 \mathring A} & \text{Diamond (FCC)} & \pu{5.4309 \mathring A} \\ \ce{Cu} & \pu{3.597 \mathring A} & \text{FCC} & \pu{3.6149 \mathring A} \\ \ce{Pt} & \pu{3.912 \mathring A} & \text{FCC} & \pu{3.9242 \mathring A} \\ \hline \end{array}\\ ^a \ \text{Values from: Phys. Rev. 1925, 25(6), 753-761 (Ref.1; as listed in Wikipedia)}$$
For comparison purposes, I have included $$\ce{Cu}$$ and $$\ce{Pt}$$ as well. Lattice Constants of Crystals of three elements of the 5 listed have also been discussed in a relatively new article (Ref.2).
References:
1. Wheeler P. Davey, “Precision Measurements of the Lattice Constants of Twelve Common Metals,” Phys. Rev. 1925, 25(6), 753-761 (https://doi.org/10.1103/PhysRev.25.753).
2. D. N. Batchelder, R. O. Simmons, “X‐Ray Lattice Constants of Crystals by a Rotating‐Camera Method: $$\ce{Al, Ar, Au, CaF2, Cu, Ge, Ne, Si,}$$Journal of Applied Physics 1965, 36(9), 2864-2868 (https://doi.org/10.1063/1.1714595).
• To resolve the discrepancy between Wolfram Alpha and (Davey 1925) that are tens to a a hundred times larger than three decimal places in Angstroms specified in the question, I'm going to need a more modern source which reasonably represents some generally agreed-upon values by the crystallographic community. I don't think it's reasonable to cite a single measurement in a single author paper from 1925 in a paper to be published in 2021. – uhoh Aug 23 '20 at 21:50
• -1 Checking both columns (1925 numbers and the ones from the question) against a CRC from 2012 (93rd Ed.) show that some of the numbers from 1925 are pretty poor; for the three metals mine are very close to CRC, the 1925 numbers differ by 0.3 to 0.6%! It is certainly up to you, but if you would consider deleting this answer I'd like to ask in a different SE site instead. Thanks! – uhoh Aug 24 '20 at 9:13
• @uhoh: You are completely wrong on this down voting. I have found a reliable literature satisfactory for your request "I only need (at a minimum) three decimal places for the lattice constants in Angstroms." What not reliable is your sited dot com page, which even not site any references for their claim. You didn't even appreciate my effort. Shame on you on that. – Mathew Mahindaratne Aug 24 '20 at 12:47
• I'm confident that my down vote accurately represents my view, which is that this answer is "not useful." When writing a paper no reasonable author chooses a 100 year old measurement to cite when much newer, more careful and more carefully reviewed data is available. Please recheck the complete text of my question post and my previous comment. – uhoh Aug 24 '20 at 13:03
• Measurement equipment gets better over time, we don't cite 100 year old measurements when far better values using better equipment is available. This is just common sense. They did great and pioneering research at GE Schenectady in the 1920's. We can honor that but that doesn't mean we should use those measurements and ignore a century of improvements! – uhoh Aug 24 '20 at 14:37 |
# Reactions between Ca(OH)2 and CO2, and Sr(OH)2 and CO2
So I'm doing an investigation, and it involves using supersaturated solutions of Ca(OH)2 and of Sr(OH)2. I notice that there's always a flaky precipitate formed on top. I need to know about the reactions with carbon dioxide. Is it carbonate that's formed? Or is it hydrogen carbonate? Why?
• In my opinion nothing prevents both carbonates and bicarbonates from forming those flakes on the surface; out of curiosity, why is it so crucial to determine which one is it? – andselisk Mar 17 at 12:33
• Although knowing the identity of the precipitate is not really significant to my results in any ways, it's very important that I discuss it since the investigation is for my IB Chemistry Extended Essay. I am required to discuss basically everything going on in the solution in as much detail as possible. I just thought I'd probably be marked down if it I had no idea what's going on in my solutions.. – Mohamad Mar 17 at 12:36
• Interesting, I didn't think of the possibility that it's both. Could be. Thanks! – Mohamad Mar 17 at 12:37
• it is calcium carbonate that is forming. – Nilay Ghosh Mar 17 at 17:15
$$\ce{Ca(OH)2 (aq)+ 2CO2 (aq/g) <=> Ca^2+ + 2(HCO3)- (aq) <=> CaCO3 (s) + CO2 (aq/g)}$$ |
›› 2017, Vol. 60 ›› Issue (10): 1198-1207.
• 研究论文 •
### 敌敌畏烟剂熏蒸防治韭菜迟眼蕈蚊的条件优化
1. (辽宁省农业科学院植物保护研究所, 沈阳 110161)
• 出版日期:2017-10-20 发布日期:2017-10-20
### Optimization of fumigation conditions of dichlorvos smoke agent in controlling Bradysia odoriphaga (Diptera: Sciaridae)
XU Lei, ZHAO Tong-Hua, LIU Pei-Bin, XU Guo-Qing*, WANG Zhe, ZHONG Tao
1. (Institute of Plant Protection, Liaoning Academy of Agricultural Sciences, Shenyang 110161, China)
• Online:2017-10-20 Published:2017-10-20
Abstract:
【Aim】 This study aims to explore the conditions for improving the fumigation effect and efficiency of dichlorvos smoke agent in controlling Bradysia odoriphaga adults. 【Methods】 Four factors including temperature, relative humidity, period and dosage were optimized by the orthogonal experiment, and the fitted model was established by multiple regression analysis and verified by a set of experiments. 【Results】 The main effect analysis of the orthogonal experiment results showed that the influence order on the efficacy of dichlorvos smoke agent was dosage>temperature>relative humidity>period. Determined by analysis of variance and multiple comparison, the optimal conditions for fumigation were the temperature of 32℃, relative humidity of 70%, dosage of 0.0878 g a.i./m3 and period of 2 h. Regression analysis showed that there were interactions between temperature and relative humidity, temperature and dosage, and relative humidity and dosage, with the interaction effect indices of 3, -4.4 and -3.25, respectively. At the same time, the regression equation of the mortality of test insects and the related factors was established. The mortality under the optimized fumigation conditions was 96.59%. Moreover, the Pearson’s correlation coefficient between the measured value and the predictive value was 0.9749, showing that the regression model was verified to be accurate and effective by the experiments. 【Conclusion】 The control efficacy of dichlorvos smoke agent in controlling B. odoriphaga adults can be promoted by optimizing fumigation conditions, and the regression model obtained in this study can be used to predict the mortality of B. odoriphaga adults. |
# Which one of the following sets of numbers could represent the lengths of the sides of a triangle? (a) (8,11,19) (b) (19,16,20), (c) (11,5,5) , (d) 13,4,8)?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
10
CW Share
Dec 16, 2016
$\left(b\right) \left(19 , 16 , 20\right)$
#### Explanation:
The sum of the lengths of any two sides of a triangle must be greater than the third side.
$\left(a\right) \left(8 , 11 , 19\right) \implies 8 + 11$ not greater than $19$, (N.G.)
$\left(b\right) \left(19 , 16 , 20\right) \implies 19 + 16 > 20 , 19 + 20 > 16 , 16 + 20 > 19$, (OK)
$\left(c\right) \left(11 , 5 , 5\right) \implies 5 + 5 < 11$, (NG)
$\left(d\right) \left(13 , 4 , 8\right) \implies 4 + 8 < 13$, (NG).
Hence, option $\left(b\right)$ is the only answer.
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Zentralblatt MATH
Publications of (and about) Paul Erdös
Zbl.No: 625.10035
Autor: Erdös, Paul; Odlyzko, Andrew M.; Sárközy, A.
Title: On the residues of products of prime numbers. (In English)
Source: Period. Math. Hung. 18, 229-239 (1987).
Review: This paper contains a modest attack to the problem proposed by P.Erdös that for any sufficiently large prime q and any residue class a\not\equiv 0 modulo q the congruence p1p2\equiv a (mod q) can be solved in primes p1 \leq q and p2 \leq q. All considerations are subject to the quasi-Riemann hypothesis H(\thetaq,x), i.e., it is supposed that for all characters \chi modulo q the L(s,\chi) do not vanish in the domain Re s > \thetaq, |Im s| < x1-\thetaq.
The generalized Riemann hypothesis is H(½,oo) but this is not enough to imply the above conjecture. There are three possible ways to weaken it, which can be satisfied (i) with almost all residue classes mod q, (ii) with the product of three primes instead of two, and (iii) with a little bit larger primes p1 and p2.
It is proved that
(i) if H(\thetaq,q) is true then p1p2\equiv a (mod q), p1 \leq q, p2 \leq q can be solved for all but cq2\thetaq-1 log5q residue classes a\not\equiv 0 modulo q;
(ii) if H(\thetaq,q) is true with \thetaq < 1-(3+\epsilon)\frac{log log q}{log q} then p1p2p3\equiv a (mod q), p1 \leq q, p2 \leq q, p3 \leq q can be solved;
(iii) if the generalized Riemann hypothesis is true then p1p2\equiv a (mod q), p1 \leq cq log4q, p2 \leq cq log4q can be solved.
Reviewer: A.Balog
Classif.: * 11N13 Primes in progressions
11N05 Distribution of primes
Keywords: distribution of primes in residue classes; product of two primes
© European Mathematical Society & FIZ Karlsruhe & Springer-Verlag |
What is a virtual state?
In quantum mechanics / Raman spectroscopy, what is a virtual state? What is the difference between a virtual state and a superposition of states? Can you simply think of the virtual state as a superposition of eigenstates?
If you have the complete set of eigenstates for a system, then you can represent any configuration of the system as a linear combination of the eigenstates, so it seems that you should be able to represent the virtual state that way. Is there something more to the virtual state than that?
1 Answer
There are differences in the spin that the term "virtual state" gets depending on context. In the context of Raman scattering, one sometimes refers to transitions to virtual states. Other terminology used is "virtual transition" to a real state. The latter terminology is a little easier to understand in my way of thinking. And note that the "virtual state" can be constructed as a superposition of real states, as you point out. So whatever we say about virtual transitions to real states immediately applies to the virtual state picture. But the virtual state picture is more complicated.
In either point of view, the incident radiation in Raman scattering is not resonant with any real state, so real transitions are not possible. We imagine that the system makes a transition to a real state, and a quantum of EM excitation is destroyed. Because the real state is not resonant with the radiation, energy is not conserved. This is possible as long as the lifetime of the system in that state is short. Heisenberg's uncertainty principle allows violations of conservation of energy for short time intervals: $\Delta E\Delta t \leq \hbar/2$.
Here's a semi-classical take on Raman scattering.
Consider a diatomic molecule in its ground state. Upon irradiation, it makes a virtual transition to an excited state, and stays there for a short period of time. But the characteristics of the chemical bond in the excited state are different from what they were in the ground state. In particular, the equilibrium length of the bond will be different. The molecule is promoted to the excited state, but it's initial bond length up there is not the equilibrium length for the excited state. So the atoms feel a force and move towards the new equilibrium length. But in a short period of time the system returns back to the ground electronic state. But now the bond length is no longer the ground state equilibrium length; the atoms feel a force and move towards the original bond length. Now there's nothing to stop them: the atoms are oscillate back and forth past the original equilibrium length: it's vibrating.
• Thank you. So if we think of our state space as being electronic and vibrational degrees of freedom, then the incident photon creates a superposition of states, the lifetime of this superposition is short, but when it re-radiates there is now occupation of one of the excited vibrational states. Is that what you're saying? – dllahr Sep 12 '14 at 17:34
• I think that's what I'm saying. :) – garyp Sep 12 '14 at 17:44
• "Heisenberg's uncertainty principle allows violations of conservation of energy for short time intervals" that is not true, it's just a misinterpretation of the uncertainty principle. It only means that if a state is not an eigenstate oh the Hamiltonian (and thus has a non-zero variance in energy), it has a lifetime smaller or equal than ℏ/2ΔE. However, there is no violation in the conservation of energy whatsoever. – Ajayu Sep 22 '17 at 23:33 |
# Dividing Complex Numbers Worksheet Doc
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The first, and most fundamental, complex number function in Excel converts two components (one real and one imaginary) into a single complex number represented as a+bi. The worksheets are meant for the study of rational numbers, typically in 7th or 8th grade math (pre-algebra and algebra 1). I Worksheet by Kuta Software LLC. -1-Graph each number in the complex plane. If n is odd, and b ≠ 0, then. 420) 1-4, 7, 8a, 11a 2/5. Multiply and Add Patterns If zero value is a fraction, then divide all coefficients by denominator. Find the domain of the rational function. Complete Square & Division - Algebra review of completion of the square and long division of polynomials. Some of the worksheets displayed are Dividing complex numbers, Rationalizing imaginary denominators, Operations with complex numbers, Infinite algebra 2, Complex numbers and powers of i, F q2v0f1r5 fktuitah wshofitewwagreu p aolrln, Decimals work, Multiplyingdividing fractions and mixed numbers. Click the show/hide navigation pane button , and then click the bookmarks tab. q G QAel Zln 8rViigvh Jtjs 4 8rueZs2elrPvGeHdO. Name Simplifying Complex Numbers - Matching Worksheet Write the letter of the answer that matches the problem. Turnitin solutions promote academic integrity, streamline grading and feedback, deter plagiarism, and improve student outcomes. Basic Facts Worksheets Lovely 37 Doc Introduction to Energy from multiplying complex numbers worksheet , source:bombaamor. You cannot have a complex number in the denominator, so multiply top and bottom by the conjugate. the real numbers between -3 and 1 19. Have your budding math whiz try these free printable word problems worksheets for some extra math practice! Word problems help kids learn and understand complex math concepts. The algebra section allows you to expand, factor or simplify virtually any expression you choose. 1) Represent the following rational numbers on the number line: a)-3/4 b) 31/-6 c) -1/2 d) ¾ 2) Write the following rational numbers in the standard form: a) 5/15 b) -24/40 c) 33/-77 d) -45/-105 3) Compare the following rational numbers:. Differentiated worksheets designed for the new AQA further mathematics A-Level, covering the complex number content. WorksheetWorks. Three-Circle Venn. Let's divide the following 2 complex numbers \frac{5 + 2i}{7 + 4i} $Step 1. Long Division Polynomials Worksheet Doc. 5th through 8th Grades. 4 - Spring 2018. Created by Ross Rueger. 1 The Need For Complex Numbers. 1) Find the LCD. o 9 lM da gdCes Fwoi5toh l 5IGnJf dian9i Ztwe2 HAHl Rgveob3r na4 61 J. Worksheet by Kuta Software LLC Algebra 2 7. Texas A&M’s research creates new knowledge that provides basic, fundamental, and applied contributions resulting, in many cases, in economic benefits to the state, nation and world. Worksheet #7 Recall that in class we learned that given a set S and a closed binary operation *, an important question is " for all a,b in S, does there exist a general solution to a*x=b"? Your last homework had you answer that question for several "ordered pairs". docx Author: Trevor Jensen Created Date: 20121126155039Z. (Division, which is further down the page, is a bit different. a + bi Complex Numbers real imaginary The complex numbers consist of all sums a + bi, where a and b are real numbers and i is the imaginary unit. Natural Numbers. A Complex Number is a combination of a. Just in case you forgot how to determine the conjugate of a given complex number, see the table below: Conjugate of a Complex Number. beaconlearningcenter. Conversion and Promotion are defined so that operations on any combination of predefined numeric types, whether primitive or composite, behave as expected. complex number a — bi is a + b'. Simplifying Rational Exponents. Text-to-911 is Now Available in Santa Clara County. We hear more than three million cases a year involving almost every type of endeavor. 5 Quiz 1 2/2. Nashville, Tenn. I can add, subtract and multiply polynomial expressions Factoring Quadratic Expressions 1. Properties of real numbers worksheet answers. 1 Constructing the complex numbers One way of introducing the field C of complex numbers is via the arithmetic of 2×2 matrices. Pre-Assessment Activity Completion Certificate. Looks scary, huh?. The relationship of a − b to b − a. 4 - Spring 2018. Writing in terms of i: !72=6i2 7. ©Z D20U1M2s HKuCt9ad 5S ao SfYtgw ra 3r ieP NLBLxCy. 8 Z zASlTlJ VrUiHgrhHtAsP WrYewsAeYrYvneOdY. Court Administration Staff. From there, it will be easy to figure out what to do next. the real numbers less than - 4 16. For example, SUM (A1:A4). The market intelligence and collaboration hub where the best construction businesses are built. Multi-Digit Subtraction. Complex numbers of the form x 0 0 x are scalar matrices and are called real complex numbers and are denoted by. Adoption of Child Support Rules and Guidelines The Indiana Supreme Court hereby adopts the Indiana Child Support Guidelines, as drafted by the Judicial Administration Committee and adopted by the Board of the Judicial Conference of Indiana and all subsequent amendments thereto presented by the Domestic Relations Committee of the Judicial Conference of. Displaying all worksheets related to - Multiplying And Dividing Imaginary And Complex Numbers. ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WORKSHEET. (Hint: Use the Distributive Property to divide each term in the numerator by the denominator (or a common factor of the. (Division, which is further down the page, is a bit different. Conjugating in the math classroom — and we're not talking verbs! The seventh lesson in a series of 32 introduces the class to the building blocks of complex number division. A new random number is returned every time the worksheet is calculated. Name _____ Period _____ Date _____ 16. They are both expressed according to the triangle on the right, where each letter represents one side-length (lower-case) and the angle opposite to it (upper-case). Printable in convenient PDF format. Dividing Radical. The Division of Research values diversity as indispensable to academic excellence and welcomes a wide range of thought, background, ethnicity, and perspective. Complex numbers The equation x2 + 1 = 0 has no solutions, because for any real number xthe square x 2is nonnegative, and so x + 1 can never be less than 1. x 2 3x 10 0. Understand how the Common Core was created. In modular arithmetic, the numbers we are dealing with are just integers and the operations used are addition, subtraction, multiplication and division. This type of fraction is also known as a compound fraction. 3) Add or Subtract as the numerators and carry the denominators. Dec 17, 2017 - Explore sarabowron's board "Complex Numbers" on Pinterest. Worksheet #7 Recall that in class we learned that given a set S and a closed binary operation *, an important question is " for all a,b in S, does there exist a general solution to a*x=b"? Your last homework had you answer that question for several "ordered pairs". Write your answer in the form a + bi. We hear more than three million cases a year involving almost every type of endeavor. I can factor when a is not equal to one. Two complex numbers a + bi and c + di, written in standard form, are equal to each other if. Find UVA’s messages and other public health resources and guidance. When dividing radical expressions, use the quotient rule. ppt - the fire extinguisher training component only (1. Some of the worksheets displayed are Dividing complex numbers, Adding and subtracting complex numbers, Study guide for some basic intermediate algebra skills, Factoring the difference of squares, Beginning and intermediate algebra, Rational. Louis, MO 63110-1010. This Word Problem Worksheet page is a recent addition to HelpingWithMath. Adding And Subtracting Complex Numbers Worksheet 13"> Full Template. The Insert Sequence Number utility of Kutools for Excel can help you customize and insert sequence numbers in contiguous cells, non-contiguous ranges of active worksheet, different worksheets, and different workbooks. w LA^lHlS NrJiKgghItsM rrEeDskeDrDvMeHdU. the real numbers greater than -2 13. (3x2 + 7x + 2) ÷ (x + 2) 2. Mixed Expressions and Complex Fractions ©2001-2003www. Add and Subtract Complex Numbers. Simplifying Complex Numbers B j FMRaEdte p nw ri It8ht rI jnf2i Nnbi UtFeP mAGlfg Je obvr das c2 z. Text-to-911 is Now Available in Santa Clara County. 6A, Rational functions MATH 1410 (SOLUTIONS) For each of the rational functions given below, do the following: 1. complex fractions worksheet doc. Complex Numbers Lesson 5. Stacy Fisher is the former freebies writer for The Balance. When finished with this set of worksheets, students will be able to solve word problems involving ratios, fractions, mixed numbers, and fractional parts of whole numbers. The largest team of research, economics and analysis experts join forces with artificial intelligence (AI) to deliver deeper. Directorate of Human Resources. The geometry of the Argand diagram. Dividing Polynomials Using Synthetic Division Use synthetic division to divide the polynomial by the linear factor. A magnification of the Mandelbrot setPlot complex numbers in the complex plane. Graphical Representation of complex numbers. To divide complex numbers. The worksheets in this product can be used to introduce and rev. Microsoft Word - 8. These worksheets are printable PDF exercises of the highest quality. That is the map z7→ z+z 0 represents a translation aunits to the right and bunits up in the complex plane. Radicals: Simplifying Radicals 1. Customize and insert unique sequence numbers with suffix and prefix. Simplifying Exponents Step Method Example 1 Label all unlabeled exponents "1" 2 Take the reciprocal of the fraction and make the outside. Dividing Complex Numbers. Math worksheets for fifth grade children covers all topics of 5th grade such as Graphs, Data, Fractions, Tables, Subtractions, Pythagoras theorem, Algebra, LCM, HCF, Addition, Round up numbers , Find 'X' in addition equations, Metric systems, Decimals, Probability, Money & more. College Algebra We help you get through college: Linear Algebra Workbench: Vectors, Matrices, Linear Systems; Lessons, Free Book PDF. The color shows how fast z 2 +c grows, and black means it stays within a certain range. doc - Handy sheet for posting next to extinguishers (49 kB, MS Word). extinguisher_truck. During the instruction, the class learns to find the multiplicative inverse of a complex number. Here we will briefly review reducing, multiplying, dividing, adding, and subtracting fractions. adding and subtracting integers worksheet doc 1000 images about. This means that both subtraction and division will, in some way, need to be defined in terms of these two operations. Reduce 32 32 22 24 28 xx x xx x Multiplying and Dividing Expressions 1. Complex numbers: More complex manipulations 6. English Home Language (1st Language) Adverbs – manner, degree and place. Complex Numbers (pg. Stacy Fisher is the former freebies writer for The Balance. 6i 9 Tons of Free Math Worksheets at www. To solve a division problem, we will need to know the conjugate of the denominator. Dividing Complex Numbers Simplify. • Illustrate Complex Numbers on an Argand diagram • investigate the operations of addition, multiplication, subtraction and division with Complex Numbers in the form a + ib • interpret the Modulus as distance from the origin on an Argand Diagram and calculate the complex conjugate • Calculate conjugates of sums and products of Complex. Find the conjugate of a complex number; use conjugates to find moduli and quotients of complex numbers. Complex Numbers Notes Part 2 with answer key attached (Note: disregard the dividing examples for now) Operations with Complex Numbers HW and ANSWERS/work shown for #15 on the homework Homework: Unit 1A DeltaMath assignment#1: "Powers of I & Plotting Complex Numbers" due for a grade TOMORROW by ; 8:00 a. Complex Numbers Examples. Reduce 2 25 5 x x 6. Whole Number Subtract Decimal Worksheet. About operations on complex numbers. 1) Represent the following rational numbers on the number line: a)-3/4 b) 31/-6 c) -1/2 d) ¾ 2) Write the following rational numbers in the standard form: a) 5/15 b) -24/40 c) 33/-77 d) -45/-105 3) Compare the following rational numbers:. For division, students must be able to rationalize the denominator, which includes multiplying by the conjugate. 6) - 2 · - 2 6) A) - 2 i B) 2 C) - 2 D) 2 i 7) - 49 · - 25 7) A) - 35 i B) - 35 C) 35 D) 35 i 8) - 12 - 4 8) A) - 3 B) - i 3 C) i 3 D) 3 9) - 144 4 9) A) - 6 i B) 6 C) - 6 D) 6 i Add or subtract as indicated. Available in Law Library – Certain Laws that Replace Provisions Enjoined from Enforcement in Does 1-17 v. The Distributive Property in Arithmetic. , Read More. R Worksheet by Kuta Software LLC Algebra 2 Name_____. Complex numbers are numbers of the form a + bi, where a and b are real numbers, and i = √(-1). The University of Alaska Fairbanks, the nation's northernmost Land, Sea, and Space Grant university and international research center, advances and disseminates knowledge through teaching, research and public service with an emphasis on Alaska, the circumpolar North and their diverse peoples. The system of complex numbers consists of all numbers of the form a + bi where a and b are real numbers. On-line job offerings. Complex numbers is vital in high school math. WorkSHEET 1. The Legislative Building. They are both expressed according to the triangle on the right, where each letter represents one side-length (lower-case) and the angle opposite to it (upper-case). New Jersey Housing and Mortgage Finance Agency (HMFA) Dedicated to increasing the availability and accessibility of affordable housing. R Worksheet by Kuta Software LLC Algebra 2 Name_____. Some of the worksheets for this concept are Division, Division work, Division facts missing numbers 1 12, Grade 6 division work, Division witho ut remainder 2 digit by 1 digit s1, Fact families, Number bonds multiplication division 2, Dividing complex numbers. Let's divide the following 2 complex numbers$ \frac{5 + 2i}{7 + 4i} \$ Step 1. Worksheets are Dividing complex numbers, Complex numbers and powers of i, Operations with complex numbers, Infinite algebra 2, Multiplying complex numbers, F q2v0f1r5 fktuitah wshofitewwagreu p aolrln, Rationalizing imaginary denominators. Complex numbers: Magnitude, phase, real and imaginary parts 3. ©f i2 N0O12F EKunt la i ZS3onf MtMwtaQrUeC 0LWLoCX. Worksheet by Kuta Software LLC Algebra 2 Graphing complex numbers ID: 1 ©m k2Y0F1s7L sKNuntnaM CSkovfntw^aOrBeK hLPLtCB. The music ends at piano quietly with a sweet melody. Browse Study Guides. ParkHere Guides are available at any County Park. Browse Adaptive Practice. J G QALlFlY arIi^gnhNtwsd JrWe_sSezrTvEexdr. com is the official site of Marvel Entertainment! Browse official Marvel movies, characters, comics, TV shows, videos, & more. Multiplying Fractions. For example, SUM (A1:A4). If a fraction has one or more fractions in the numerator or denominator, it is called a "complex fraction. 1) i34 2) i129 3) i146 4) i14 5) i68 6. Multiply 22 32 469 xx xxx. We hear more than three million cases a year involving almost every type of endeavor. For example, when you divide 8 by 2, the answer is 4 whereas when you divide 2 by 8, the answer will be 0. Worksheet #7 Recall that in class we learned that given a set S and a closed binary operation *, an important question is “ for all a,b in S, does there exist a general solution to a*x=b”? Your last homework had you answer that question for several “ordered pairs”. A complex number is usually denoted by the letter 'z'. To fix this, multiply the top and bottom both (which makes it 1) by a complex conjugate: the real part of the complex number stays the same, but the $$i$$ part is negated (plus turns to minus, minus turns to plus). New Jersey Housing and Mortgage Finance Agency (HMFA) Dedicated to increasing the availability and accessibility of affordable housing. Worksheet by Kuta Software LLC Algebra 2 Solving Quadratics with Imaginary Solutions Name_____ Date_____ Period____ ©M M2O0M1_6k GK_ultYaQ hSqoTfftTwwalrmed qLULvCm. E w PMoafdZeB awqiJtThh oIvnHfyiyngigtde4 JPRr 8eI-7AAlQg0eLbAr2aE. 8+ College Algebra Worksheet Templates – DOC, PDF Are you looking forward to improve your maths score in your college? Well, algebra forms a fair share of the entire maths curriculum in your college and hence you must be focused in enhancing your algebra know-how. dividing complex numbers worksheet. This worksheet may be printed and used for educational purposes only. the real numbers less than 0 12. Reduce 2 2 56 1 yy y 3. + General Information. Eskom crime reporting line: 0800 11 27 22 (toll-free). Reduce ab ba 5. You can tweak your clouds with different fonts, layouts, and color schemes. G Worksheet by Kuta Software LLC 25) (−7 + 3i)(−5 + 3i) 26). These negative number worksheets combine negative numbers with other integers (both positive and negative) using the basic math operations, multiplying multi-digit negative numbers, and long division with. Download Center: This is our Work Book Download center where you can select the topic of the worksheet and click on that, you will get a download box, which you will have save and the downloading will start, the format of this file is pdf ( protable data file ) and you will need Adobe Acrobat Reader for this, which is freely available at www. the real numbers between 2 and 6 18. Adding,Subtracting, and Multiplying Radical Expressions. Algorithms were originally born as part of mathematics – the word “algorithm” comes from the Arabic writer Muḥammad ibn Mūsā al-Khwārizmī, – but currently the word is strongly associated with computer science. Apr 18 - We worked on multiplying and dividing complex numbers in standard form and polar form, as well as raising a complex number in polar form to a power (using DeMoivre's Theorem). On this page you will find: a complete list of all of our math worksheets, lessons, math homework, and quizzes. Dividing Complex Numbers 7. This Complex Number Division 1 Lesson Plan is suitable for 11th - 12th Grade. Victoria Police provides policing services to the Victorian community across 54 Police Service Areas, within 21 divisions and four regions. 8) Complex numbers 9) System of quadratic functions CHAPTER 5- click on each topic below to access the links to worksheets- you must show all work to receive credit 1) Classifying polynomials 2) Graphing polynomials by factoring 3) Difference or sum of cubes 4) Dividing polynomials- DO BOTH LONG AND SYNTHETIC DIVISION 5) Rational root theorem. Court Appointed Counsel. • Illustrate Complex Numbers on an Argand diagram • investigate the operations of addition, multiplication, subtraction and division with Complex Numbers in the form a + ib • interpret the Modulus as distance from the origin on an Argand Diagram and calculate the complex conjugate • Calculate conjugates of sums and products of Complex. There are also worksheets on adding, subtracting, multiplying, and dividing fractions. Apply for Services Individuals, who wish to apply to become eligible for DDDS services, including assistance in navigating the complex federal and state human service systems, can download an application. Luckily, algebra with complex numbers works very predictably, here are some examples:. P lay, L earn, I nteract & e X plore math & science concepts with our. Multiply the expression by. Order of Operations Introduction. The system of complex numbers consists of all numbers of the form a + bi where a and b are real numbers. 1 Complex numbers Name: _____ 1 Using the imaginary number i, write down expressions for: (a) (b) 2 Simplify in the form of 3 (a) Represent on an Argand diagram. Complex numbers. beaconlearningcenter. A polynomial in the form a 3 – b 3 is called a difference of cubes. o 9 lM da gdCes Fwoi5toh l 5IGnJf dian9i Ztwe2 HAHl Rgveob3r na4 61 J. Worksheets that teach basic investing math concepts, including market capitalization, price-to-earnings ratios, dividends. Add and Subtract Complex Numbers. com where getting mixed up is part of the fun! This page includes Mixed operations math worksheets with addition, subtraction, multiplication and division and worksheets for order of operations. To set a missing value in an equation (for example, in the Set Column Values dialog box), divide anything by zero or choose the Na() function (Function: Miscellaneous from the Set Values menu). Complex Number Division. Complex Numbers Notes Part 2 with answer key attached (Note: disregard the dividing examples for now) Operations with Complex Numbers HW and ANSWERS/work shown for #15 on the homework Homework: Unit 1A DeltaMath assignment#1: "Powers of I & Plotting Complex Numbers" due for a grade TOMORROW by ; 8:00 a. Learn more about simple formulas. Dividing complex numbers: Multiply the numerator and denominator by the conjugate of the denominator (the same expressions with the opposite sign on the imaginary part of the complex number), then simplify results and write your answer in standard form. Steel is an integrated steel producer with major production operations in North America and Central Europe. Request a police record check. This will give you the least common multiple of the two numbers. Determine the conjugate of the denominator. 8) Complex numbers 9) System of quadratic functions CHAPTER 5- click on each topic below to access the links to worksheets- you must show all work to receive credit 1) Classifying polynomials 2) Graphing polynomials by factoring 3) Difference or sum of cubes 4) Dividing polynomials- DO BOTH LONG AND SYNTHETIC DIVISION 5) Rational root theorem. The rules for removing parentheses. About This Quiz & Worksheet. If this is your first time using the Graphic Organizer Maker, click the New button to begin. Practice worksheet on rational numbers is for 7th grade. List all possible rational x-intercepts of y = 2x3 + 3x – 5, then find all complex roots. Name_____ Multiplying/Dividing Fractions and Mixed Numbers. These division worksheets will produce problems with mixed formats for the quotient, but keeping the divisor and dividend as whole numbers. For instance, if you divide 50 by 10 , the answer will be a nice neat " 5 " with a zero remainder, because 10 is a factor of 50. Multiplying and Dividing Fractions. Exponential Form of complex numbers. Complex Numbers Worksheet Pdf. Z Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Period____Date_____Operations with Complex Numbers Simplify. Complex numbers are numbers that consist of two parts, one real and one imaginary. Determine the conjugate of the denominator. We hope that the kids will also love the fun stuff and puzzles. Before working through the worksheets, discuss with your children any phrases or vocabulary that they may be unsure of. complex fractions worksheet doc. It provides access to mathematical functions for complex numbers. Complex Numbers Worksheet. (a + bi)(a - bi) = a2 - b2i2 = a2 - b2(- 1) = a2 + b2 Replace i2 by -1 and simplify. To set a missing value in an equation (for example, in the Set Column Values dialog box), divide anything by zero or choose the Na() function (Function: Miscellaneous from the Set Values menu). Worksheet by Kuta Software LLC Algebra 2 Graphing complex numbers ID: 1 ©m k2Y0F1s7L sKNuntnaM CSkovfntw^aOrBeK hLPLtCB. With tips on how to teach concepts to kindergarteners or advice to help get good grades in middle or high school, The Classroom provides the best education content. Pre-Assessment Activity Completion Certificate. Learn to add, subtract, multiply and divide whole numbers, decimals, fractions. Consider a complex number z= 1 1 re iθ. Practice: Simplify complex fractions. Multiply 22 32 469 xx xxx. In a wave the medium moves back and forth as the wave moves horizontally. 1 A complex number is a matrix of the form x −y y x , where x and y are real numbers. the real numbers less than - 4 16. About operations on complex numbers. Since our denominator is 1 + 2i, its conjugate is equal to 1 − 2i. fire_extinguisher. Operations with Complex Numbers Author: Mike Created Date: 9/3/2008 11:06:46 AM. COMPLEX CONJUGATE The COMPLEX CONJUGATE of a complex numberz = x + iy, denoted by z* , is given by z* = x - iy The Modulus or absolute value is defined by z x y 2 2 5. Dec 17, 2017 - Explore sarabowron's board "Complex Numbers" on Pinterest. Complex Numbers Worksheet. comparing and ordering numbers worksheet fifth grade basic algebra problems w/picture cheat sheet on adding, subtracting, multiplying and dividing fractions and decimals. Gain professional skills at the County of Santa Clara - Apply Now! If you need help, but can't safety speak on the phone or are unable to speak, use your mobile phone to send a text message to 911. All formula entries begin with an equal sign (=). Coronavirus Disease 2019 (COVID-19) in Colorado: State & National Resources Translate. Calculate the budget amount for each task based on labor rates, material costs, and other fixed costs. a G XMXaCdde 9 9waiht5hB 1I2nAfUizn ZibtMeV fA Sl Agesb 7rfa G G2D. The division worksheet will produce 9 problems per worksheet. The complex plane. Download Center: This is our Work Book Download center where you can select the topic of the worksheet and click on that, you will get a download box, which you will have save and the downloading will start, the format of this file is pdf ( protable data file ) and you will need Adobe Acrobat Reader for this, which is freely available at www. Perform operations like addition, subtraction and multiplication on complex numbers, write the complex numbers in standard form, identify the real and imaginary parts, find the conjugate, graph complex numbers, rationalize the denominator, find the absolute value, modulus, and argument in this collection of printable complex number worksheets. adding and subtracting integers worksheet doc 1000 images about. About This Quiz & Worksheet. Write your answer in the form a + bi. the real numbers between 2 and 6 18. Complex numbers: Addition, subtraction, multiplication, division 5. Perform operations like addition, subtraction and multiplication on complex numbers, write the complex numbers in standard form, identify the real and imaginary parts, find the conjugate, graph complex numbers, rationalize the denominator, find the absolute value, modulus, and argument in this collection of printable complex number worksheets. Welcome to the mixed operations worksheets page at Math-Drills. ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WORKSHEET. Have your budding math whiz try these free printable word problems worksheets for some extra math practice! Word problems help kids learn and understand complex math concepts. txt) or read online for free. Conversion and Promotion are defined so that operations on any combination of predefined numeric types, whether primitive or composite, behave as expected. Solve the equations. pdf), Text File (. so that i 2 = -1! Therefore, you really have 6i + 4(-1), so your answer becomes -4 + 6i. Louis biotech and startup scene. It is the policy of the New York City Department of Education to provide equal educational opportunities without regard to actual or perceived race, color, religion, creed, ethnicity, national origin, alienage, citizenship status, disability, weight, gender (sex) or sexual orientation, and to maintain an environment free of harassment on the basis of any of these. Absolute Value Rules. Create here an unlimited supply of worksheets for simplifying complex fractions — fractions where the numerator, the denominator, or both are fractions/mixed numbers. Add, subtract, Divide & Multiply worksheets Pdf. The steps below detail how to create a simple Fibonacci sequence using a formula. Writing in terms of i: !!25=!5i 5. To fix this, multiply the top and bottom both (which makes it 1) by a complex conjugate: the real part of the complex number stays the same, but the $$i$$ part is negated (plus turns to minus, minus turns to plus). This unit describes how this process is carried out. Simplifying Complex Numbers B j FMRaEdte p nw ri It8ht rI jnf2i Nnbi UtFeP mAGlfg Je obvr das c2 z. The worksheets in this product can be used to introduce and rev. Phoenix, AZ 85007. If you interchange the number, the answer will be a different number. I can factor when a is one. Simplifying Rational Exponents. Dividing complex numbers: Multiply the numerator and denominator by the conjugate of the denominator (the same expressions with the opposite sign on the imaginary part of the complex number), then simplify results and write your answer in standard form. Conversion between the two notational forms involves simple trigonometry. By using this website, you agree to our Cookie Policy. Complex numbers of the form x 0 0 x are scalar matrices and are called real complex numbers and are denoted by. b) c) d) e) 1 5) Simplify the following. Solving Using the Quadratic Formula Worksheet The Quadratic Formula: For quadratic equations: ax 2 bx c 0, a b b ac x 2 2 4 Solve each equation using the Quadratic Formula. 1) 5 −5i 2) 1 −2i 3) − 2 i 4) 7 4i 5) 4 + i 8i 6) −5 − i −10i 7) 9 + i −7i 8) 6 − 6i −4i 9) 2i 3 − 9i 10) i 2 − 3i 11) 5i 6 + 8i 12) 10 10 + 5i 13) −1 + 5i −8 − 7i 14) −2 − 9i −2 + 7i 15) 4 + i 2 − 5i 16) 5 − 6i −5 + 10i 17) −3 − 9i 5 − 8i 18) 4 + i 8 + 9i 19) −3 − 2i −10 − 3i. 5 Quiz 1 2/2. Name _____ Period _____ Date _____ 16. Choose the one alternative that best completes the statement or answers the question. U Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Dividing Radical Expressions Date_____ Period____ Simplify. the real numbers less than 0 12. Each 7th grade math topic links to a page with PDF printable math worksheets covering subtopics under the main category. Grade 8 English Home Language Worksheets. Complete Performing Operations in Trig Form #1-11 ODD and complete 1 problem on the back (not #13). Multiplication (Cont’d) – When multiplying two complex numbers, begin by F O I L ing them together and then simplify. Then, multiply the numerators across to get the answer numerator. Human Body -Body Parts Word Scramble Game. Wordle is a toy for generating “word clouds” from text that you provide. Complex Numbers. Last Updated: 2018-03-02. Multiplication and division in polar form Introduction When two complex numbers are given in polar form it is particularly simple to multiply and divide them. Courthouse Security Procedures. pdf), Text File (. The geometry of the Argand diagram. Available in Law Library – Certain Laws that Replace Provisions Enjoined from Enforcement in Does 1-17 v. A polynomial in the form a 3 + b 3 is called a sum of cubes. Grade 8 Afrikaans Huistaal Worksheets. 1) 5 −5i 2) 1 −2i 3) − 2 i 4) 7 4i 5) 4 + i 8i 6) −5 − i −10i 7) 9 + i −7i 8) 6 − 6i −4i 9) 2i 3 − 9i 10) i 2 − 3i 11) 5i 6 + 8i 12) 10 10 + 5i 13) −1 + 5i −8 − 7i 14) −2 − 9i −2 + 7i 15) 4 + i 2 − 5i 16) 5 − 6i −5 + 10i 17) −3 − 9i 5 − 8i 18) 4 + i 8 + 9i 19) −3 − 2i −10 − 3i. Some of the worksheets for this concept are Operations with complex numbers, Complex numbers and powers of i, Dividing complex numbers, Adding and subtracting complex numbers, Real part and imaginary part 1 a complete the, Complex numbers, Complex numbers, Properties of complex. Full list of fraction topics. Addition/Subtraction of Rational Expressions w/ Unlike Denominators. the real numbers less than 0 12. Here we know r and Θ and we need to find a and b. Learn more about simple formulas. Complex Conjugation 6. This Word Problem Worksheet page is a recent addition to HelpingWithMath. 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Kyocera Document Solutions offers an award-winning range of device technology and integrated business process improvement solutions that work together seamlessly. The SAT Suite of Assessments is an integrated system that includes the SAT, PSAT/NMSQT and PSAT 10, and PSAT 8/9. Complex numbers are numbers of the form a + bi, where a and b are real numbers, and i = √(-1). This is the currently selected item. The set of complex numbers consists of the set of real numbers and the set of imaginary numbers. Chapter Contents. Round your answer to the hundredths place. pdf Yates H S IDK 1. We introduce a number ide ned to satisfy the equation x2 = 1:(As soon as we introduce this number, there is some ambiguity, for x= ialso satis es x2 = 1!). Adding and Subtracting Radical Expressions. Human Body -Body Parts Word Search Game. Displaying top 8 worksheets found for - Complex Number Division. Argument of a Function. 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Steel is an integrated steel producer with major production operations in North America and Central Europe. ORNL has been involved in the discovery or confirmation of 11 new elements. Math 129 - Calculus II Worksheets. (a + bi)(a - bi) = a2 - b2i2 = a2 - b2(- 1) = a2 + b2 Replace i2 by -1 and simplify. Math Worksheets Examples, solutions, videos, worksheets, games, and activities to help PreCalculus students learn how to multiply and divide complex numbers in trigonometric or polar form. A Complex Number is a combination of a. Worksheet 2. Our learning materials are free to share and use in schools or at home. Dividing Polynomials Using Synthetic Division Use synthetic division to divide the polynomial by the linear factor. 6i 9 Tons of Free Math Worksheets at www. operations with complex numbers worksheet good complex numbers worksheet or dividing complex numbers. 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ALGEBRAIC FRACTIONS. In this case, we say s (speed) is the subject of the formula. Multiplication (Cont’d) – When multiplying two complex numbers, begin by F O I L ing them together and then simplify. Write complex numbers in polar form. Now we need to set up the process. Understand how the Common Core was created. Lattice method for multiplication. Reduce 2 2 56 1 yy y 3. Some of the worksheets for this concept are Operations with complex numbers, Complex numbers and powers of i, Dividing complex numbers, Adding and subtracting complex numbers, Real part and imaginary part 1 a complete the, Complex numbers, Complex numbers, Properties of complex. Richmond, VA City of Richmond, Virginia Welcome. Quickly review concepts with key information. This unit describes how this process is carried out. 2) Multiply all terms to get like denominators. To perform calculation in an Excel workbook, it is recommended to invoke EnableSheetCalculations method of IWorksheet. Learn about an infringement notice. Advanced Algebra Worksheets Results Two-Step Equations Date Period - Kuta Software LLC ©8 f2P0 B1V2M 7K Uu Otpa3 gS xolfetEw Ja 3r2e k RLzLUCx. 1) 5 −5i 2) 1 −2i 3) − 2 i 4) 7 4i 5) 4 + i 8i 6) −5 − i −10i 7) 9 + i −7i 8) 6 − 6i −4i 9) 2i 3 − 9i 10) i 2 − 3i 11) 5i 6 + 8i 12) 10 10 + 5i 13) −1 + 5i −8 − 7i 14) −2 − 9i −2 + 7i 15) 4 + i 2 − 5i 16) 5 − 6i −5 + 10i 17) −3 − 9i 5 − 8i 18) 4 + i 8 + 9i 19) −3 − 2i −10 − 3i. -1-Solve each equation with the quadratic formula. (b) Evaluate (a) *** Note: there should be a line from the. Looks scary, huh?. Math Worksheets Examples, solutions, videos, worksheets, games, and activities to help PreCalculus students learn how to multiply and divide complex numbers in trigonometric or polar form. 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Publication
Title
Etching of low-k materials for microelectronics applications by means of a $N_{2}/H_{2}$ plasma : modeling and experimental investigation
Author
Abstract
In this paper, we investigate the etch process of so-called low-k organic material by means of a N2/H2 capacitively coupled plasma, as applied in the micro-electronics industry for the manufacturing of computer chips. In recent years, such an organic material has emerged as a possible alternative for replacing bulk SiO2 as a dielectric material in the back-end-of-line, because of the smaller parasitic capacity between adjacent conducting lines, and thus a faster propagation of the electrical signals throughout the chip. Numerical simulations with a hybrid plasma model, using an extensive plasma and surface chemistry set, as well as experiments are performed, focusing on the plasma properties as well as the actual etching process, to obtain a better insight into the underlying mechanisms. Furthermore, the effects of gas pressure, applied power and gas composition are investigated to try to optimize the etch process. In general, the plasma density reaches a maximum near the wafer edge due to the so-called 'edge effect'. As a result, the etch rate is not uniform but will also reach its maximum near the wafer edge. The pressure seems not to have a big effect. A higher power increases the etch rate, but the uniformity becomes (slightly) worse. The gas mixing ratio has no significant effect on the etch process, except when a pure H2 or N2 plasma is used, illustrating the synergistic effects of a N2/H2 plasma. In fact, our calculations reveal that the N2/H2 plasma entails an ion-enhanced etch process. The simulation results are in reasonable agreement with the experimental values. The microscopic etch profile shows the desired anisotropic shape under all conditions under study.
Language
English
Source (journal)
Plasma sources science and technology / Institute of Physics [Londen] - Bristol, 1992, currens
Publication
Bristol : Institute of Physics , 2013
ISSN
0963-0252
Volume/pages
22 :2 (2013) , p. 1-19
Article Reference
025011
ISI
000317275400013
Medium
E-only publicatie
Full text (Publisher's DOI)
Full text (publisher's version - intranet only)
UAntwerpen
Faculty/Department Research group Project info Publication type Subject Affiliation Publications with a UAntwerp address |
# Homework Help: Operations with Linear Transformations
1. Oct 12, 2009
### hbweb500
1. The problem statement, all variables and given/known data
Let $$T:U \rightarrow V$$ be a linear transformation, and let U be finite-dimensional. Prove that if dim(U) > dim(V), then Range(T) = V is not possible.
2. Relevant equations
dim(U) = rank(T) + nullity(T)
3. The attempt at a solution
I almost think there must be a typo in the book. For instance, let U be P4 (the space of polynomials degree 4 and lower), and let V be P2. Let T be the second derivative operator. Then the Range of T is V. This example is even printed earlier in the same book that I got this question from.
Otherwise, I see no reason why the Range(T) couldn't be V. The rank(T) could at most be dim(V), but that is no problem, because the nullity(T) could be anywhere from dim(U) to dim(U)-dim(V).
So, is this a typo? Or (maybe more likely) am I missing something obvious?
2. Oct 12, 2009
### Billy Bob
Definitely a typo. It should be <. |
Every time the results of findMotifGenome.pl in HOMER is different
0
1
Entering edit mode
2.4 years ago
Hi I am using the findMotifGenome.pl in HOMER 4.10 to find the enriched motif in my bed file. But I found every time I run findMotifGenome.pl, the results is different. I know HOMER will random choose the background peak, so I am wondering whether there is a way to fix the random result.
Motif HOMER • 1.0k views
1
Entering edit mode
Hi,
Among the option that the program offers there isn't one to set the seed?
António
0
Entering edit mode
It seems that no seed in HOMER ╮（╯＿╰）╭
1
Entering edit mode
according to my esteemed colleague, you can circumvent that issue by providing the same set of custom background sequences via -bg, i.e. -bg <peak/BED file>
0
Entering edit mode
But it seems that it is hard to select bg file when analysing
0
Entering edit mode
I mean, if you have several peaks to be tesed for enriched motif, you can not set the same bg. you have to choose specific bg file for specific peak?
0
Entering edit mode
I don't understand the issue with this. I originally thought you were running homer on the same set of peaks and noticed that the results differed. If you supply a different set of peaks every time you run homer why would you NOT expect to see different results? [and if the issue is that you feel it's too much work to generate a custom background peak set for every analysis, well, then I recommend you think about a strategy that can be scripted]
0
Entering edit mode
Thanks for your reply :) I am sorry I mixed the origin question. Your suggestion is helpful, I am wondering whether you can also give some advice about selecting bg file. Beacuse I noticed the HOMER attempts to select background regions that match the GC-content distribution of the input sequences, but I do not find the relevant specific function in HOMER
1
Entering edit mode
sorry, can't help with this, but you can check the biostars archive whether anyone has asked that question already (i.e. "How can I mimick the background selection by HOMER?" or the like) and if you don't find what you're looking for, just post it as a new question
0
Entering edit mode
Thanks again. I will post another question :) |
# Question referring to percentage increase
Why if you evaluate the percentage increase between the first and last number of a sequence it doesn't equalize with the partial sum of the increases? Suppose you have the sequence $1,3,6$. The percentage increase of 1 to 6 is 500%. But if you do the partial sum of the increases, let's say 200% (from 1 to 3) and 100% (from 3 to 6), you get 300% and not 500%. Why is that? Furthermore, is there a way to make the partial sum of the increase equal the first to the nth number of a sequence? Thanks!
• Because percent increase doesn't add, it compounds. So the100% is not relative to the start but to the start plus 200%. Jan 11, 2016 at 15:20
Let $A$ be the initial amount and assume there is a $20%$ percent increase from each term to the next term. Then the values you get are:
$$A, \;\;\;\; (1.2)A, \;\;\;\; (1.2)^2A, \;\;\;\; (1.2)^3A, \;\;\;\; \ldots$$
Notice that the 2nd term is $(1+0.2)A = A + (0.2)A,$ which is $20$% greater in magnitude than $A.$
However, the 3rd term is
$$(1+0.2)^2A \; = \; (1+0.2)(1+0.2)A \; = \; (1 + 0.2 + 0.2 + 0.04)A$$ $$= \; A + (0.2)A + (0.2)A + (0.04)A,$$
which is $20$% plus $20$% plus $4$% greater than $A.$
Also, the 4th term is
$$(1+0.2)^3A \; = \; (1+0.2)(1+0.2)(1 + 0.2)A$$ $$= \; (1 + 0.2 + 0.2 + 0.2 + 0.04 + 0.04 + 0.04 + 0.008)A$$ $$= \; A + (0.2)A + (0.2)A + (0.2)A + (0.04)A + (0.04)A + (0.04)A + (0.008)A,$$
which is $20$% plus $20$% plus $20$% plus $4$% plus $4$% plus $4$% plus $0.8$% greater than $A.$
One way to explain the underlying reason why the 2nd term is not two $20$% increases, and the 3rd term is not three $20$% increases, etc. is due to the presence of the cross terms when expanding a binomial (i.e. $(x+y)^n$ is not just $x^n+y^{n}).$ Another way to explain this is that when computing the 3rd term, you increase the 2nd term by $20$%, so the 3rd term is the 2nd term plus $20$% of the 2nd term, which is a little more than the 2nd term plus $20$% of the 1st term; and when computing the 4th term, you increase the 3rd term by $20$%, so the 4th term is the 3rd term plus $20$% of the 3rd term, etc.
• Magnificent answer, hats off.. Jan 11, 2016 at 17:12 |
# Languages Other Than English Maltese - sTudy dEsign - Board of Studies 2000
Languages Other Than English
study design
Maltese
Board of Studies
2000
September 2013
Collaborative Curriculum and Assessment Framework for Languages (CCAFL)
Maltese
The following agencies have contributed to this document:
Board of Studies, New South Wales
Board of Studies, Victoria
Curriculum Council of Western Australia
Northern Territory Board of Studies
Senior Secondary Assessment Board of South Australia
Tasmanian Secondary Assessment Board
Every effort has been made to contact all copyright owners. The Board of Studies,
Victoria apologises if it has inadvertently used material in which copyright resides. For
Board of Studies
15 Pelham Street
Carlton VIC 3053
15 Pelham Street, Carlton, Victoria 3053
Website: http://www.bos.vic.edu.au
This completely revised and reaccredited edition published 2000.
This publication is copyright. Apart from any use permitted under the Copyright Act
1968, no part may be reproduced by any process without prior written permission from
the Board of Studies.
Edited by Ruth Learner
Designed by Geoff Coleman
Desktop publishing by Julie Coleman
Cover artwork
Detail from a VCE work of Paul Wisneske: ‘Mallee landscape’ 1993, acrylic on canvas,
1100 x 840 mm. Copyright remains the property of the artist.
Languages Other Than English: Maltese
ISBN 1 74010 076 X
September 2013
Contents
Important information ........................................................................................................ 5
Introduction........................................................................................................................... 7
The language........................................................................................................................... 7
Rationale................................................................................................................................. 7
Aims........................................................................................................................................ 7
Structure.................................................................................................................................. 7
Entry........................................................................................................................................ 8
Duration.................................................................................................................................. 8
Changes to the study design.................................................................................................... 8
Monitoring for quality ............................................................................................................ 8
Safety...................................................................................................................................... 8
Use of information technology............................................................................................... 8
Community standards............................................................................................................. 8
Vocational Education and Training option.............................................................................. 8
Assessment and reporting.................................................................................................. 10
Satisfactory completion........................................................................................................ 10
Authentication....................................................................................................................... 10
Levels of achievement.......................................................................................................... 10
Areas of study Units 1–4..................................................................................................... 12
Unit 1 ................................................................................................................................... 18
Outcomes.............................................................................................................................. 18
Assessment............................................................................................................................ 19
Unit 2 ................................................................................................................................... 21
Outcomes.............................................................................................................................. 21
Assessment............................................................................................................................ 22
Units 3 and 4........................................................................................................................ 24
Unit 3 ................................................................................................................................... 26
Outcomes.............................................................................................................................. 26
Assessment............................................................................................................................ 27
Unit 4 ................................................................................................................................... 29
Outcomes.............................................................................................................................. 29
Assessment............................................................................................................................ 30
September 2013
Advice for teachers............................................................................................................. 37
Developing a course.............................................................................................................. 37
Methods................................................................................................................................ 37
Structure and organisation.................................................................................................... 37
Use of information technology............................................................................................. 38
Example outlines................................................................................................................... 38
Summary of outcomes for Module 2B of the National TAFE Language Course................. 50
Main characteristics of different kinds of writing................................................................. 51
Main characteristics of common text types.......................................................................... 53
Suitable resources................................................................................................................. 54
September 2013
IMPORTANT INFORMATION
Accreditation period
Units 1–4: 2002–201
Accreditation period ends 31 December 201
Other sources of information
The VCE Bulletin is the only official source of changes to regulations and accredited studies. The
VCE Bulletin, including supplements, also regularly includes advice on VCE studies. It is the
responsibility of each VCE teacher to refer to each issue of the VCE Bulletin.
To assist teachers in assessing school-assessed coursework in Units 3 and 4 the Board of Studies
will publish annually an assessment guide which will include advice on the scope of the tasks
and the criteria for assessment.
The VCE Administrative Handbook for the current year contains essential information on
assessment and other procedures.
VCE providers
Throughout this study design the term ‘school’ is intended to include both schools and other VCE
providers.
Photocopying
VCE schools only may photocopy parts of this study design for use by teachers.
5
September 2013
September 2013
Introduction
THE LANGUAGE
The language to be studied and assessed is the modern standard/official version of Maltese.
RATIONALE
The study of Maltese contributes to the overall education of students, particularly in the areas of
communication, cross-cultural understanding, literacy and general knowledge. It provides access
to the culture of Maltese-speaking countries and communities and promotes understanding of
different attitudes and values within the wider Australian community and beyond.
The study of Maltese develops the student’s ability to understand and use a language which is
spoken in the historical and ethnic Maltese territory as well as by communities within Australia,
associated with the language, and develops their understanding of the contribution Maltese speakers
have made in fields such as science, sport, literature, music and the visual arts.
The ability to communicate in Maltese may, in conjunction with other skills, provide students
with enhanced vocational opportunities.
Aims
This study is designed to enable students to:
• use Maltese to communicate with others;
• understand and appreciate the cultural contexts in which Maltese is used;
• understand their own culture(s) through the study of other cultures;
• understand language as a system;
• make connections between Maltese and English, and/or other languages;
• apply Maltese to work, further study, training or leisure.
Structure
The study is made up of four units. Each unit is designed to enable students to achieve a set
of outcomes. Each outcome is described in terms of the key knowledge and skills students are
required to demonstrate.
7
September 2013
LANGUAGES OTHER THAN ENGLISH: MALTESE VCE study design
INTRODUCTION
Entry
Maltese is designed for students who will, typically, have studied Maltese for at least 400 hours
at completion of Year 12. It is possible, however, that some students with less formal experience
will also be able to meet the requirements successfully.
Students must undertake Unit 3 prior to undertaking Unit 4.
Duration
Each unit involves at least 50 hours of scheduled classroom instruction.
Changes to the study design
During its period of accreditation minor changes to the study will be notified in the VCE Bulletin.
The VCE Bulletin is the only source of changes to regulations and accredited studies and it is the
responsibility of each VCE teacher to monitor changes or advice about VCE studies published
in the VCE Bulletin.
MONITORING FOR Quality
The Board of Studies will, from time to time, undertake an audit of Maltese to ensure that the study
is being taught and assessed as accredited. Teachers must ensure that all records and samples of
student work are maintained and available should the study be subject to audit. The details of the
audit procedures and requirements are published annually in the VCE Administrative Handbook.
Schools will be notified during the teaching year of schools and studies to be audited.
Safety
It is the responsibility of the school to ensure that duty of care is exercised in relation to the health
and safety of all students undertaking this study.
USE OF INFORMATION TECHNOLOGY
In designing courses for this study teachers are encouraged to incorporate information technology
in teaching and learning activities. The Advice for Teachers section provides specific examples of
how information technology can be used in this study.
COMMUNITY STANDARDS
It is the responsibility of the school to ensure that all activities in this study are conducted within
ethical guidelines. This is of particular concern in the use of information located on the World
Wide Web.
VOCATIONAL EDUCATION AND TRAINING OPTION
Schools wishing to offer the Vocational Education and Training (VET) option should note that
they will need to seek registration as a training provider, or to enter into an agreement with a
registered training provider able to offer the module outcomes to students on their behalf. For further
information, contact the Office of Post Compulsory Education, Training and Employment (PETE).
8
September 2013
VCE study design LANGUAGES OTHER THAN ENGLISH: MALTESE
INTRODUCTION
The school-assessed coursework component of this study is designed to allow the integration of
tasks required in Modules 2A and 2B of the National TAFE Language Course Stage One*. The
outcomes for Module 2A relate most closely to Units 1 and 2; the outcomes for Module 2B relate
most closely to Units 3 and 4.
The VCE coursework assessment tasks (including those required for the detailed study VET
option) are sufficiently flexible for both the VCE outcomes and selected outcomes required for
Modules 2A and 2B to be met. Examples of how this might be done are provided on pages 47–50.
Students who successfully meet all of the outcomes required for Module 2A will be eligible to
receive from their registered provider a Certificate II in Applied Languages. Students who meet
all of the outcomes required for Module 2B will be eligible to receive a Certificate III in Applied
Languages.
It is important to note that there are significantly more outcomes to be met in both Modules 2A and
2B than in VCE Units 1 and 2, and in Units 3 and 4, respectively. Although there is considerable
scope for several of the module outcomes to be incorporated into one VCE assessment task, it
is possible that (unless the student has completed some of the module outcomes previously) a
number of additional assessment tasks (beyond the eight required for VCE Units 1 and 2, and
six required for VCE Units 3 and 4) will be required, if the student is to achieve them all for the
purposes of VET certification.
Schools might therefore wish to consider incorporating outcomes from Module 2A into Year 10
programs as well as into VCE Unit 1, and outcomes from Module 2B into VCE Unit 2 tasks, in
No additional enrolment procedure is required for students wishing to follow this VET option. It
is important to note, also, that students who successfully complete the outcomes for Modules 2A
and/or 2B will not receive any additional credit for satisfactory completion of the VCE, or for the
calculation of the ENTER.
*National TAFE Language Course: Stage One; Generic Curriculum, ACTRAC Products, 1994, available from
Australian Training Products (tel: 9630 9836)
9
September 2013
Assessment and reporting
SATISFACTORY COMPLETION
The award of satisfactory completion for a unit is based on a decision that the student has
demonstrated achievement of the set of outcomes specified for the unit. This decision will be based
on the teacher’s assessment of the student’s overall performance on assessment tasks designated
for the unit. Designated assessment tasks are provided in the details for each unit. The Board of
Studies will publish annually an assessment guide which will include advice on the scope of the
assessment tasks and the criteria for assessment.
Teachers must develop courses that provide opportunities for students to demonstrate achievement
of outcomes. Examples of learning activities are provided in the Advice for Teachers section.
Schools will report a result for each unit to the Board of Studies as S (Satisfactory) or N (Not
Satisfactory).
Completion of a unit will be reported on the Statement of Results issued by the Board of Studies
of achievement.
AUTHENTICATION
Work related to the outcomes will be accepted only if the teacher can attest that, to the best of
their knowledge, all unacknowledged work is the student’s own. Teachers need to refer to the
current year’s VCE Administrative Handbook for authentication procedures, and should note that
all assessment tasks for Units 3 and 4 should be conducted in class time and under supervision.
LEVELS OF ACHIEVEMENT
Units 1 and 2
Procedures for the assessment of levels of achievement in Units 1 and 2 are a matter for school
decision. Assessment of levels of achievement for these units will not be reported to the Board of
Studies. Schools may choose to report levels of achievement using grades, descriptive statements
or other indicators.
10
September 2013
VCE study design LANGUAGES OTHER THAN ENGLISH: MALTESE
ASSESSMENT AND REPORTING
Units 3 and 4
The Board of Studies will supervise the assessment of all students undertaking Units 3 and 4.
In Maltese the student’s level of achievement will be determined by school-assessed coursework
and two end-of-year examinations. Percentage contributions to the final assessment are as follows:
• Unit 3 school-assessed coursework: 25 per cent
• Unit 4 school-assessed coursework: 25 per cent
• Units 3 and 4 examinations: 50 per cent
11
September 2013
Areas of study
Units 1–4 Common areas of study
The areas of study for Maltese comprise themes and topics, grammar, text types, vocabulary
and kinds of writing. They are common to all four units of the study, and they are designed to
be drawn upon in an integrated way, as appropriate to the linguistic needs of the student, and the
outcomes for the unit.
The themes and topics are the vehicle through which the student will demonstrate achievement
of the outcomes, in the sense that they form the subject of the activities and tasks the student
undertakes.
The grammar, vocabulary, text types and kinds of writing are linked, both to each other, and to
the themes and topics. Together, as common areas of study, they add a further layer of definition
to the knowledge and skills required for successful achievement of the outcomes.
The common areas of study have been selected to provide the opportunity for the student to build
upon what is familiar, as well as develop knowledge and skills in new and more challenging areas.
THEMES, TOPICS AND SUB-TOPICS
There are three prescribed themes:
• The individual
• The Maltese-speaking communities
• The changing world
These themes have a number of prescribed topics and suggested sub-topics. The placement of
the topics under one or more of the three themes is intended to provide a particular perspective
or perspectives for each of the topics. The suggested sub-topics expand on the topics, and are
provided to guide the student and teacher as to how topics may be treated.
It is not expected that all topics will require the same amount of study time. The length of time
and depth of treatment devoted to each topic will vary according to the outcomes being addressed,
as well as the linguistic needs and interests of the student.
As well as acquiring the linguistic resources to function effectively as a non-specialist within all
three themes, the student is required to undertake a detailed study in Units 3 and 4. This detailed
study should relate to the prescribed themes and topics and be based on a selected sub-topic. For
further details refer to pages 24 and 25.
12
September 2013
VCE study design LANGUAGES OTHER THAN ENGLISH: MALTESE
AREAS OF STUDY UnitS 1–4
Prescribed themes and topics, and suggested sub-topics
The individual The Maltese-speaking communities The changing world
• Personal identity • Lifestyles • The world of work
For example, leisure and interests, For example, daily life, education, the For example, careers now and in the
health and well-being, family and role of religion, leisure activities. future, impact of technology, impact
friends. of globalisation.
• Education and aspirations For example, significant people and • Social issues
For example, school life, further events, ceremonies and celebrations. For example, the environment, the
study, training and employment. changing role of women.
• The Arts and Literature
• Personal opinions and values For example, famous authors, modern and • Tourism
For example, personal priorities, issues traditional art/music/literature. For example, development of tourism,
of personal importance, positive and negative impact of tourism.
lifestyle preferences.
Note: Bold = Prescribed themes, Bold Italics = Prescribed topics, Italics = Suggested sub-topics.
Text types
The student will be expected to be familiar with the following text types. Text types indicated with
an asterisk (*) are those which the student may be expected to produce in the external examination.
Teachers may introduce the student to a wider range of text types in the course of their teaching
and learning program.
Announcement Map Report*
Conversation* News item Review*
Discussion* Note/message* Story/narrative account*
Email Notice Song
Form Personal profile Survey
Formal letter* Play Text of an interview*
Informal letter* Poem Text of a speech*
Invitation Postcard Timetable
Kinds of writing
The student is expected to be familiar with, and be able to produce the following five kinds of
writing: personal, informative, persuasive, evaluative and imaginative.
13
September 2013
LANGUAGES OTHER THAN ENGLISH: MALTESE VCE study design
AREAS OF STUDY UnitS 1–4
Vocabulary
While there is no prescribed vocabulary list, it is expected that the student will be familiar with
a range of vocabulary and idioms relevant to the topics prescribed in the study design. Students
should be encouraged to use dictionaries. It is expected that teachers will assist students to develop
the necessary skills and confidence to use dictionaries effectively. Suitable editions are listed in the
Resources section of this study design. Information on the use of dictionaries in the end-of-year
written examination is provided on page 32.
Grammar
The student is expected to recognise and use the following grammatical items:
Verb Primary (ewlieni)
– whole, not doubled (s]i], mhux trux) Qasam
– weak (dg]ajjef) Saqsa
– assimilative (xebbi]) G]amel
– hollow (mo]fi) Bag]bas
– weak (nieqes) Beda
Derived (imnissel)
Triliteral – primary, whole, not doubled }adem
(ewlieni, s]i], mhux trux)
– primary, doubled (ewlieni, trux) Garr
– primary, weak (ewlieni, nieqes) Qata’
– primary, hollow (ewlieni, mo]fi) Dam
– primary, assimilative (ewlieni, Wasal
xebbi])
– primary, doubled, and assimilative G]add
(ewlieni, trux, xebbi])
– primary, assimilative and weak Waqa’
(ewlieni, xebbi], nieqes)
– primary, assimilative and hollow G]am
(ewlieni, xebbi], mo]fi)
– primary, hollow and weak Bieg]
(ewlieni, mo]fi, nieqes)
– irregular
– primary, whole (ewlieni, s]i]) ~aqlaq
– primary, weak (ewlieni, dg]ajjef) G]arbel
– irregular
Derived (imnisslin)
– form 2 Kisser
– form 3 Qarar
– form 4 Idda
– form 5 Tkisser
– form 6 Tqarar
– form 7 Inkiser
– form 8 }tieg
– form 9 Sbie]
– form 10 Sta]ba
– irregular
14
September 2013
VCE study design LANGUAGES OTHER THAN ENGLISH: MALTESE
AREAS OF STUDY UnitS 1–4
Tense – Perfect {anna [abet il-]elu
– Previous Past Pawlu kien mar u re[a’ [ie
– Past Continuous Kont tiela’ t-tara[ u rajtu
– Habitual Past Marija kienet tmur tixtri
– Continuous & Habitual Past Kien ikun mimdud hawn
– Conditional Past Kieku kont barra kont ng]ajjatlek
– Past Passive It-tifla [iet me]uda d-dar
– Past Passive Continuous {anni kien mitluf minn sensih
Imperfect (present)
– Continuous Johnny qed jilg]ab
– Present Conditional Kieku kellu jasal it-tifel, immur
– Present (habitual) Meta nkun nistudja, ma rridx storbju
Imperative Ifta] il-bieb!
Future Illejla se no]ro[
– Future (before another future) Meta tkun [iet, tibda tistudja
– Future Continuous G]ada, x]in inkunu nimxu, g]idlu
– Future Passive Meta jkun mi[ug], jibki
Voice – Active (transitive, reflexive, or Toni kisser ][ie[a
intransitive)
– Passive (one word or more than It-tifel [ie me]ud l-isptar
one word with the auxiliary verb
and the passive participle)
– Reflexive (from transitive verbs) Leli fa]]ar lilu nnifsu
Form – Primary (the first form)
– Derived (from the second to the
tenth form)
Mood – Indicative (perfect, imperfect) ~ensu kisser it-tazza
Jien niekol il-fenek
– Imperative (positive, imperfect) Ilbes (I\-\arbun)
Unattached }afna
Prepositions Bil-maqlub
One word Xorob ftit ilma
More than one word Naddaf kull fejn kien
Repeated words
– Nouns Qattag]ha bia bia
– Verbs Tefag]ha kif [ie [ie
Time ~ensu telaq kmieni
Place Nardu tela’ lfuq
Manner Refa’ l-qattus bil-mod
Quantity Majsi kiel ]afna
Negation Ma mort qatt
Affirmation Mort \gur
Question Qatt mort Malta?
Comparison Wi[i tkellem a]jar minn {anni
15
September 2013
LANGUAGES OTHER THAN ENGLISH: MALTESE VCE study design
AREAS OF STUDY UnitS 1–4
Noun Gender Ra[el
Mara
Singular Wied
Plural Bibien
Common plural Qtates
Whole plural (plural s]i]) Ba]ri - Ba]rin
Broken plural (plural miksur) }ofra - }ofor
Dual plural (plural imtenni) Ri[lejn - ejn denoting two
Original Sabi], a]rax
Gender Sabi], sabi]a
Number – singular Qasir
– plural Qosra
– common }ajjata
Plural – derived Fer]an, ra]li
– denominative Belti, ra]li
– verbal }erqan, da]qan
– diminutive Fqajjar, smajjar
– positive Sabi], tajjeb
– comparative Isba], itjeb
– superlative L-isba], l-itjeb
Pronoun Gender Hu(wa), hi(ja)
Number Jiena, a]na
Suffixes (mehmu\in) DarI, darEK
Separate (mifrudin) Jiena, inti
Personal A]na, intom
Possessive Tieg]I, jisimNI
Direct KitibNI, kitbEK
Indirect KitibLI, kitibLEK
Demonstrative Dan, dak, din
Reflexive Innifsi, nfusna, ru]i
Interrogative Min kien? Xi kemm tiswa?
Relative Li, min, kull
Preposition Attached Bija, fik, g]alih
Unattached Bi, fi, g]al
Numeral Cardinal Wie]ed
Tnejn
Tlieta
|ew[ (\ew[t)
Tliet (tlitt)
Erba’ (erbat)
Ordinal L-ewwel
It-tieni
It-tielet
16
September 2013
VCE study design LANGUAGES OTHER THAN ENGLISH: MALTESE
AREAS OF STUDY UnitS 1–4
Sentence and Phrase Types
Statement Jiena g]andi n-nag]as
Question X’]in se ti[i?
Direct Speech Qalli ‘prosit ]ej, kemm titkellem tajjeb
bil-Malti’
Indirect Speech Fera]li ta’ kemm nitkellem tajjeb
bil-Malti.
Affirmative Iva, se ni[i.
Negative Le, minix se ni[i.
Exclamatory X’wa]da waqa’ miskin!
Conjunctions
Coordinatory Toni u Marija marru s-City
Contrast Salvu twajjeb imma ma tg]addihx
bi\-\mien
Alternative Lippu jew Pinu kellhom imorru
Article
Definite
– Alone L-ikel
– With the euphonic ‘i’ added before nouns
or adjectives used as nouns beginning with
‘moon’ consonants (qamrin) and Il-qamar
‘sun’ consonants (xemxin) Ix-xemx
– With the euphonic ‘i’ added before nouns
or adjectives used as nouns beginning with
two consonants the first of which is m, n, s, x L-ingassa
Particle
Attached (mag]qudin) Fil-ba]ar
Unattached (mag]\ulin) Bi ][arha
Prepositions Ma’ ]ajt
Conjunctions Jien xtaqt immur, i\da int ma ridtx
Interjections Ja]asra, kemm hi twajba ~ensa
Negation Ma rrid xejn.
M’intix tg]id sewwa
Verbal Noun Generic Gideb, faqar
Triliteral based Bdil, d]ul, dlam
Triliteral based without the second vowel Da]k, fer], serq
Verbs starting with ‘G]’ G]a\la, g]emil
Primary, doubled verb based Daqq, bexx
Primary, weak verb based Biki, [iri, xiwi
Second form verb based Tiksir, ta]lit
Fifth form verb based Tferrix
Sixth form verb based T]aris, tmerija
Eighth form verb based Fta]ir, ftehim
Tenth form verb based Stedin, stag][ib
Participle Active Rieqed, wieqaf
Passive Mag]fus, misjub, imwaqqa’
17
September 2013
Unit 1
AREAS OF STUDY
The areas of study common to Units 1–4 are detailed on pages 12–17 of this study design.
Outcomes
For this unit the student is required to demonstrate achievement of three outcomes.
Outcome 1
On completion of this unit the student should be able to establish and maintain a spoken or written
exchange related to personal areas of experience.
Key knowledge and skills
To achieve this outcome the student should demonstrate the knowledge and skills to:
• use structures related to describing, explaining and commenting on past, present or future
events or experiences;
• use a range of question and answer forms;
• link and sequence ideas and information;
• initiate, maintain and close an exchange;
• use appropriate intonation, stress, spelling and punctuation;
• self-correct/rephrase to maintain communication;
• recognise and respond to cues for turn taking;
• communicate in a range of text types, for example, letter, fax, email, voicemail and telephone;
• use appropriate non-verbal forms of communication such as eye contact and handshake.
Outcome 2
On completion of this unit the student should be able to listen to, read and obtain information
from written and spoken texts.
Key knowledge and skills
To achieve this outcome the student should demonstrate the knowledge and skills to:
• apply knowledge of vocabulary and structures related to the topics studied;
• recognise common patterns of word formation and grammatical markers, and use these to
infer meaning;
• apply knowledge of conventions of text types;
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VCE study design LANGUAGES OTHER THAN ENGLISH: MALTESE
Unit 1
• convey gist and global understanding as well as items of specific detail;
• identify the key words, main points and supporting ideas;
• order, classify and link items from various parts of the text.
Outcome 3
On completion of this unit the student should be able to produce a personal response to a text
focusing on real or imaginary experience.
Key knowledge and skills
To achieve this outcome the student should demonstrate the knowledge and skills to:
• apply the conventions of relevant text types, for example review, article;
• use structures related to explaining, describing and comparing past, present, or future
experiences;
• use stylistic features such as repetition and contrast;
• summarise, explain, compare and contrast experiences, opinions, ideas, feelings and reactions;
• link ideas, events and characters;
• select and make use of relevant reference materials;
• provide personal comment/perspective on aspects of the texts;
• respond appropriately for the context, purpose and audience described.
Assessment
The award of satisfactory completion for a unit is based on a decision that the student has
demonstrated achievement of the set of outcomes specified for the unit. This decision will be based
on the teacher’s assessment of the student’s overall performance on assessment tasks designated
for the unit. The Board of Studies will publish annually an assessment guide which will include
advice on the scope of the assessment tasks and the criteria for assessment.
The key knowledge and skills listed for each outcome should be used as a guide to course design and
the development of learning activities. The key knowledge and skills do not constitute a checklist
and such an approach is not necessary or desirable for determining achievement of outcomes. The
elements of key knowledge and skills should not be assessed separately.
Assessment tasks must be a part of the regular teaching and learning program and must not unduly
add to the workload associated with that program. They must be completed in class and under
supervision.
Demonstration of achievement of Outcomes 1, 2 and 3 must be based on the student’s performance
on a selection of assessment tasks. Teachers must ensure that tasks selected are of comparable
scope and demand, and that over the course of the unit, all three outcomes are addressed. Tasks
should also be selected to ensure that, over the course of the unit, both oral and written skills in
Maltese are assessed. Therefore if an oral task is selected to address Outcome 1, a written task
should be selected to address Outcome 3, and vice versa.
19
September 2013
LANGUAGES OTHER THAN ENGLISH: MALTESE VCE study design
Unit 1
A total of four tasks should be selected from those listed below.
Outcome 1:
• informal conversation
or
Outcome 2:
• listening to spoken texts (e.g. conversations, interviews, broadcasts) to obtain information to
complete notes or charts or tables in Maltese and English
and
notes or charts or tables in Maltese and English.
Outcome 3:
• oral presentation
or
• review
or
• article.
It is expected that the student responds in Maltese to all assessment tasks that are selected to
address Outcomes 1 and 3. Of the two tasks required for Outcome 2, one should require a response
in Maltese, and the other a response in English.
20
September 2013
Unit 2
AREAS OF STUDY
The areas of study common to Units 1–4 are detailed on pages 12–17 of this study design.
Outcomes
For this unit the student is required to demonstrate achievement of three outcomes.
Outcome 1
On completion of this unit the student should be able to participate in a spoken or written exchange
related to making arrangements and completing transactions.
Key knowledge and skills
To achieve this outcome the student should demonstrate the knowledge and skills to:
• use structures related to asking for or giving assistance or advice, suggesting, explaining,
agreeing, and disagreeing;
• use fillers, affirming phrases and expressions related to negotiation/transaction;
• make arrangements, come to agreements, and reach decisions;
• obtain and provide goods, services, or public information;
• link and sequence ideas and demonstrate clarity of expression in written or spoken form;
• initiate, maintain, direct as appropriate, and close an exchange;
• use gesture, stance, facial expression to enhance meaning and persuade;
• use examples and reasons to support arguments and convince;
• respond appropriately for the context, purpose and audience described.
Outcome 2
On completion of this unit the student should be able to listen to, read, and extract and use
information and ideas from spoken and written texts.
Key knowledge and skills
To achieve this outcome the student should demonstrate the knowledge and skills to:
• use vocabulary, structures and content related to topics studied;
• apply the conventions of relevant text types such as a letter or a newspaper report;
• classify, compare and predict information and ideas;
• infer points of view, opinions and ideas;
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September 2013
LANGUAGES OTHER THAN ENGLISH: MALTESE VCE study design
Unit 2
• extract and reorganise information and ideas from one text type to another;
• appreciate cultural aspects critical to understanding the text.
Outcome 3
On completion of this unit the student should be able to give expression to real or imaginary
experience in written or spoken form.
Key knowledge and skills
To achieve this outcome the student should demonstrate the knowledge and skills to:
• apply the conventions of relevant text types, for example journal entry, story;
• use structures related to describing, recounting, narrating, and reflecting upon past, present or
future events or experiences;
• use a range of appropriate vocabulary and expressions;
• use stylistic techniques such as repetition, questions and exclamations;
• structure writing to sequence main ideas and events logically;
• vary language for audience, context and purpose.
assessment
The award of satisfactory completion for a unit is based on a decision that the student has
demonstrated achievement of the set of outcomes specified for the unit. This decision will be based
on the teacher’s assessment of the student’s overall performance on assessment tasks designated
for the unit. The Board of Studies will publish annually an assessment guide which will include
advice on the scope of the assessment tasks and the criteria for assessment.
The key knowledge and skills listed for each outcome should be used as a guide to course design and
the development of learning activities. The key knowledge and skills do not constitute a checklist
and such an approach is not necessary or desirable for determining achievement of outcomes. The
elements of key knowledge and skills should not be assessed separately.
Assessment tasks must be a part of the regular teaching and learning program and must not unduly
add to the workload associated with that program. They must be completed in class and under
supervision.
Demonstration of achievement of Outcomes 1, 2 and 3 must be based on the student’s performance
on a selection of assessment tasks. Teachers must ensure that tasks selected are of comparable
scope and demand, and that over the course of the unit, all three outcomes are addressed. Tasks
should be selected to ensure that, over the course of the unit, both oral and written skills in Maltese
are assessed. Therefore if an oral task is selected to address Outcome 1, a written task should be
selected to address Outcome 3, and vice versa.
22
September 2013
VCE study design LANGUAGES OTHER THAN ENGLISH: MALTESE
Unit 2
A total of four tasks should be selected from those listed below.
Outcome 1:
• formal letter, or fax, or email
or
• role-play
or
• interview.
Outcome 2:
• listen to spoken texts (e.g. conversations, interviews, broadcasts) and reorganise information
and ideas in a different text type
and
in a different text type.
Outcome 3:
• journal entry
or
• personal account
or
• short story.
It is expected that the student responds in Maltese to all assessment tasks selected.
23
September 2013
Units 3 and 4
AREAS OF STUDY
The areas of study common to Units 1–4 are detailed on pages 12–17 of this study design.
DETAILED STUDY
The student is required to undertake a detailed study during Units 3 and 4.
There are two options for detailed study: Language and culture through texts; Language and
culture through VET.
The student will be expected to discuss their detailed study in Section 2, Discussion, of the Oral
Examination.
Over the course of Units 3 and 4, approximately 15 hours of scheduled class time should be
devoted to the detailed study.
The detailed study should be based on a sub-topic related to one or more of the prescribed topics
listed in the table on page 13. The sub-topic may be drawn from this table, or a different sub-topic
may be selected.
One sub-topic may be selected for a whole class or different sub-topics may be selected for
individuals or groups of students. In the former case, it will be important to select a sub-topic that
is sufficiently broad to accommodate a range of interests and perspectives, so that each student
can provide an individual response to the assessment task(s) set as well as in the Discussion in
Section 2 of the Oral Examination.
At least one and no more than two of the six coursework assessment tasks for school-assessed
coursework should focus on the detailed study. The detailed study assessment task(s) should be
designed to assess the student’s understanding of the language and culture of the Maltese-speaking
community and should be selected from those required to assess achievement of Outcome 2, Unit
4 (detailed on page 31). The sub-topics and texts should also be selected to ensure the student is
able to focus on the knowledge and skills associated with Outcome 2, Unit 4.
Language and culture through texts
The detailed study should enable the student to explore and compare aspects of the language and
culture of the Maltese-speaking community through a range of oral and written texts related to
the selected sub-topic. This will enable the student to develop knowledge and understanding of,
for example, historical issues, aspects of contemporary society or the literary or artistic heritage
of the community. The texts which form the basis of this study might include feature films, short
films, short stories, songs, newspaper articles, electronic texts, documentaries, music, painting
24
September 2013
VCE study design LANGUAGES OTHER THAN ENGLISH: MALTESE
UnitS 3&4
and oral histories. The length of texts selected will vary depending on the type of text, its density
and level of complexity. In order for the student to be able to explore their sub-topic in sufficient
depth to meet the relevant outcomes, it is suggested that a range of at least three different kinds
of text are selected. These might include aural and visual, as well as written texts.
Language and culture through VET
This detailed study allows the student to explore and compare aspects of the language and culture
of the Maltese community through the study of outcomes drawn from Module 2B of the National
TAFE Language Course, Stage One.
The sub-topic selected for this detailed study must allow for the incorporation of study related
to Outcomes 14, 15, and one or more of Outcomes 2, 5 and 13 from module 2B of the National
TAFE Language Course, Stage One. For further details of this course, refer to page 8.
25
September 2013
Unit 3
AREAS OF STUDY
The areas of study common to Units 1–4 are detailed on pages 12–17 of this study design.
outcomes
For this unit the student is required to demonstrate achievement of three outcomes.
Outcome 1
On completion of this unit the student should be able to express ideas through the production of
original texts.
Key knowledge and skills
To achieve this outcome the student should demonstrate the knowledge and skills to:
• use a range of relevant text types;
• show knowledge of first- and third-person narrative perspectives;
• create a personal or imaginative text, focusing on an event or experience in the past, present
or future;
• vary language for audience, context and purpose;
• organise and sequence ideas;
• simplify or paraphrase complex ideas;
• select and make appropriate use of reference materials, including dictionaries.
Outcome 2
On completion of this unit the student should be able to analyse and use information from spoken
texts.
Key knowledge and skills
To achieve this outcome the student should demonstrate the knowledge and skills to:
• convey gist, identify main points, supporting points and detailed items of specific information;
• infer points of view, attitudes, emotions from context and/or choice of language and intonation;
• show knowledge of registers and stylistic features such as repetition and tone.
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September 2013
VCE study design LANGUAGES OTHER THAN ENGLISH: MALTESE
Unit 3
Outcome 3
On completion of this unit the student should be able to exchange information, opinions and
experiences.
Key knowledge and skills
To achieve this outcome the student should demonstrate the knowledge and skills to:
• exchange and justify opinions and ideas;
• present and comment on factual information;
• describe and comment on aspects of past, present and future experience;
• use appropriate terms of address for familiar and unfamiliar audiences;
• use a range of question forms;
• self-correct/rephrase to maintain communication.
Assessment
The award of satisfactory completion for a unit is based on a decision that the student has
demonstrated achievement of the set of outcomes specified for the unit. This decision will be based
on the teacher’s assessment of the student’s overall performance on assessment tasks designated
for the unit. The Board of Studies will publish annually an assessment guide which will include
advice on the scope of the assessment tasks and the criteria for assessment.
The key knowledge and skills listed for each outcome should be used as a guide to course design and
the development of learning activities. The key knowledge and skills do not constitute a checklist
and such an approach is not necessary or desirable for determining achievement of outcomes. The
elements of key knowledge and skills should not be assessed separately.
Assessment of levels of achievement
The student’s level of achievement for Unit 3 will be determined by school-assessed coursework
and two end-of-year examinations.
Contributions to final assessment
School-assessed coursework for Unit 3 will contribute 25 per cent to the final assessment.
The level of achievement for Units 3 and 4 will also be assessed by two end-of-year examinations,
which will contribute 50 per cent to the final assessment.
School-assessed coursework
Teachers will provide to the Board of Studies a score representing an assessment of the student’s
level of achievement.
The score must be based on the teacher’s rating of performance of each student on the tasks set
out in the following table and in accordance with an assessment guide published annually by the
Board of Studies. The assessment guide will also include advice on the scope of the tasks and the
criteria for assessment.
Assessment tasks must be a part of the regular teaching and learning program and must not
unduly add to the workload associated with that program. They must be completed in class time
and under supervision.
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LANGUAGES OTHER THAN ENGLISH: MALTESE VCE study design
Unit 3
Outcome 1
Express ideas through the production of original texts. A 250-word personal or imaginative written piece. 20
Outcome 2 A response to specific questions, messages or instructions,
10
Analyse and use information from spoken texts. extracting and using information requested.
Outcome 3 A three- to four-minute role-play, focusing on the resolution
20
Exchange information, opinions and experiences. of an issue
Total marks 50
* School-assessed coursework for Unit 3 contributes 25 per cent to the final assessment.
28
September 2013
Unit 4
AREAS OF STUDY
The areas of study common to Units 1–4 are detailed on pages 12–17 of this study design.
outcomes
For this unit the student is required to demonstrate achievement of two outcomes.
Outcome 1
On completion of this unit the student should be able to analyse and use information from written
texts.
Key knowledge and skills
To achieve this outcome the student should demonstrate the knowledge and skills to:
• understand and convey gist, identify main points and extract and use information;
• infer points of view, attitudes, emotions from context and/or choice of language;
• summarise, interpret and evaluate information from texts;
• compare and contrast aspects of texts on a similar topic;
• accurately convey understanding;
• show knowledge of and use a range of text types;
• show knowledge and use of simple stylistic features such as repetition and contrast;
• infer meaning from cognates, grammatical markers and common patterns of word formation;
• appreciate cultural aspects critical to understanding the text.
Outcome 2
On completion of this unit the student should be able to respond critically to spoken and written
texts which reflect aspects of the language and culture of the Maltese-speaking communities.
Key knowledge and skills
To achieve this outcome the student should demonstrate the knowledge and skills to:
• compare and contrast aspects of life in Maltese-speaking communities with those in Australia;
• identify and comment on culturally specific aspects of language, behaviour or attitude;
• present an opinion about an aspect of the culture associated with the language;
• identify similarities and differences between texts, and find evidence to support particular
views;
29
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LANGUAGES OTHER THAN ENGLISH: MALTESE VCE study design
Unit 4
• show an awareness that different social contexts require different types of language;
• select and make use of relevant reference materials.
Assessment
The award of satisfactory completion for a unit is based on a decision that the student has
demonstrated achievement of the set of outcomes specified for the unit. This decision will be based
on the teacher’s assessment of the student’s overall performance on assessment tasks designated
for the unit. The Board of Studies will publish annually an assessment guide which will include
advice on the scope of the assessment tasks and the criteria for assessment.
The key knowledge and skills listed for each outcome should be used as a guide to course design and
the development of learning activities. The key knowledge and skills do not constitute a checklist
and such an approach is not necessary or desirable for determining achievement of outcomes. The
elements of key knowledge and skills should not be assessed separately.
Assessment of levels of achievement
The student’s level of achievement for Unit 4 will be determined by school-assessed coursework
and two end-of-year examinations.
Contributions to final assessment
School-assessed coursework for Unit 4 will contribute 25 per cent of the final assessment.
The level of achievement for Units 3 and 4 will also be assessed by two end-of-year examinations,
which will contribute 50 per cent of the final assessment.
School-assessed coursework
Teachers will provide to the Board of Studies a score representing an assessment of the student’s
level of achievement.
The score must be based on the teacher’s rating of performance of each student on the tasks set
out in the following table and in accordance with an assessment guide published annually by the
Board of Studies. The assessment guide will also include advice on the scope of the tasks and the
criteria for assessment.
Assessment tasks must be a part of the regular teaching and learning program and must not
unduly add to the workload associated with that program. They must be completed in class time
and under supervision.
Outcome 1 A response to specific questions, messages or instructions,
10
Analyse and use information from written texts. extracting and using information requested.
Outcome 2 A 250–300-word informative, persuasive or evaluative written
20
Respond critically to spoken and written texts which response, for example report, comparison or review.
reflect aspects of the language and culture of the Maltese- and
speaking communities. A three- to four-minute interview on an issue related to texts
20
studied.
Total marks 50
* School-assessed coursework for Unit 4 contributes 25 per cent to the final assessment.
30
September 2013
VCE study design LANGUAGES OTHER THAN ENGLISH: MALTESE
Unit 4
End-of-year examinations
The end-of-year examinations are:
• an oral examination
• a written examination.
Oral examination (approximately 15 minutes)
Purpose
The oral examination is designed primarily to assess the student’s knowledge and skill in using
spoken Maltese.
Specifications
The oral examination has two sections.
Section 1: Conversation (approximately 7 minutes)
The examination will begin with a conversation between the student and the assessor(s). It will
consist of a general conversation about the student’s personal world, for example, school and
home life, family and friends, interests and aspirations.
Section 2: Discussion (approximately 8 minutes)
Following the Conversation the student will indicate to the assessor(s) the sub-topic chosen for
detailed study and, in no more than one minute, briefly introduce the main focus of their sub-topic,
alerting assessors to any objects brought to support the Discussion. The focus of the Discussion
will be to explore aspects of the language and culture of Maltese-speaking communities. The
student will be expected to either make reference to texts studied or, if they have elected to follow
the VET option, to discuss aspects of Outcomes 2, 5, 13, 14 or 15 from module 2B.
The student may support the Discussion with objects such as photographs, diagrams, and maps.
Notes and cue cards are not permitted.
Written examination (3 hours including 10 minutes reading time)
The student may use monolingual and/or bilingual print dictionaries in the written examination.
Section 1: Listening and responding
Purpose
Section 1 of the written examination is designed primarily to assess the student’s knowledge and
skill in analysing information from spoken texts.
The student will be expected to demonstrate understanding of general and specific information
from spoken texts and respond in English in Part A and Maltese in Part B to questions on this
information. The questions may require the student to identify information related to:
• the context, purpose and audience of the text;
• aspects of the language of the text, for example, tone, register, knowledge of language structures.
Specifications
Section 1 of the written examination has two parts, Part A and Part B. The texts in both parts will
be related to one or more of the prescribed themes.
The student hears five to seven texts in Maltese covering a number of text types. The total listening
time, for one reading of the texts without pauses, will be approximately 7–8 minutes.
31
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LANGUAGES OTHER THAN ENGLISH: MALTESE VCE study design
Unit 4
Some texts will be short, that is one reading of each text will be approximately 35–45 seconds.
Some texts will be longer, that is one reading of each text will be approximately 90–120 seconds.
Each text will be heard twice. There will be a pause between the first and second readings in which
the student may take notes. The student will be given sufficient time after the second reading to
complete responses.
The student will be expected to respond to a range of question types, such as completing a table,
chart, list or form, or responding to a message, open-ended questions or multiple-choice items.
Part A
There will be two to four short texts, and one longer text.
Questions will be phrased in English for responses in English.
Part B
There will be one short text and one longer text.
Questions will be phrased in English and Maltese for responses in Maltese.
Purpose
Section 2 of the written examination is designed primarily to assess the student’s knowledge and
skill in analysing and responding to information from written texts.
In Part A the student will be required to demonstrate understanding of written texts. The student
may be required to extract, summarise, and/or evaluate information from texts. If the texts are
related, the student may be required to compare and/or contrast aspects of both.
In Part B the student will be expected to demonstrate understanding of a written text by responding
in Maltese to information provided in a text.
Specifications
Section 2 of the written examination has two parts, Part A and Part B. The texts in both parts will
be related to one or more of the prescribed themes.
Part A
The student will be required to read two texts in Maltese of 400–500 words in total. The texts will
be different in style and purpose but may be related in subject matter or context.
Questions on the texts will be phrased in English for responses in English.
Part B
The student will be required to read a short text in Maltese of approximately 150 words, such as
a letter, message, advertisement, or notice. The student will be required to respond to questions,
statements, comments and/or other specific items provided in the written text. The task will specify
a purpose, context and audience. The text type the student will be required to produce will be
drawn from those listed for productive use on page 13.
The student will be expected to write a response of 150–200 words in Maltese. The task will be
phrased in English and Maltese for a response in Maltese.
32
September 2013 |
# I've a small problem with this matrix (to find the determinant)
Find the following determinant. $$\det\begin{bmatrix} 1 & 0 & 2 & -1\\ 0 & 1 & 4 & -2 \\ 2 & -1 & 3 & 1 \\ 2 & -1 & -1 & 2\end{bmatrix}$$
Theoretically I know how to find the determinant of a $$m \times n$$ matrix, as follow:
1. Gaussian elimination,
2. Laplace method,
3. Gaussian elimination + Laplace method,
4. Sarrus's rule (this method is valid only for $$3 \times 3$$ matrix),
5. $$2 \times 2$$ matrix (easiest way).
I used the 3rd method (because if I had used only the second one, it would have be a long and tedious process).
here's my attempt to solve it:
1. I reduced the matrix using Gaussian elimination and the result is as follows:
$$\begin{bmatrix} 1 & 0 & 2 & -1\\ 0 & 1 & 4 & -2 \\ 0 & -1 & -1 & 2 \\ 0 & -1 & -5 & 4\end{bmatrix}$$ This is enough in order to compute the determinant of this matrix, using the first column.
1. I used Laplace method and here's what I got: $$(-1)^2 \det\begin{bmatrix} 1 & 4 & -2\\ -1 & -1 & 2\\ -1 & -5 & 4\end{bmatrix}$$
My result (using Sarrus's rule to compute the determinant of the minor) is 6, but the exercise has these possible solutions:
(1) -7
(2) 0
(3) 7
(4) 5
• You must be the only one who can compute the determinant of a non-square matrix. – Rodrigo de Azevedo Dec 19 '20 at 12:34
The $$(3,4)$$ entry of your matrix after step $$1$$ should be a $$3$$. |
# Voltage and Current Sensing Circuitry
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The above circuitry allows the microcontroller to monitor the voltage and current at the input and output of the buck converter. This makes four sensors in total: Voltage Input, Current Input, Voltage Output, and Current Output.
Each sensor runs through a low pass filter (LPF). The filters are composed of two resistors and a capacitor. The LPF serves two functions. First, the resistor ladder scales the input voltage to limit that is acceptable to the microcontrollers analog to digital converter (ADC). Second, the filter cuts out high 'noise' frequencies. The frequency resonse of this filter can be calculated from the following equation:
$H = \left | \frac{Vo}{Vi} \right | = \frac{R_2}{\sqrt{(2 \pi f C R_1 R_2)^2+(R_1 + R_2)^2}}$
On the actual board, these components are laid out to accept parts in a 1206 package. This makes the large capacitances called out in the above picture (17.7uF and 20uF) unrealistic. Instead a 1uF capacitor is used. This gives the sensors a frequency cut-off of approximately 300 Hz. Additional digital filtering via software is expected within the microcontroller. |
# What is the probability that at the end of the sequence, bucket B contains ball bi
We have a bucket B which can store 1 ball at a time. Imagine a sequence of balls: {b1,b2....bn} such that ball bi appears after ball bi-1 in the sequence. The ith bi is stored in bucket B with probability 1/i replacing the ball bk,previously stored. For each i, what is the probability that at the end of the sequence, bucket B contains ball bi.
• k is less than i (k <i). – papabiceps Jul 8 '16 at 15:17
• Not sure this is clear. Does $b_1$ start in $B$? If there are, say, two balls, is the answer just $\frac 12$ for both (as there is a $\frac 12$ probability that $b_2$ replaces $b_1$)? – lulu Jul 8 '16 at 15:22
• To be clear: $B$ always contains exactly one ball (since $b_1$ was put in there with probability $1$) and whether a given ball is put in $B$ is independent of the previous balls? In which case, it seems that at the end of the day the probability that $b_i$ is in $B$ is simply the probability that it was put there $1/i$), times that none after was put instead ($\prod_{j>i}(1-1/j)$) by independence. – Clement C. Jul 8 '16 at 15:22
• But why do you refer to the $(i-1)^{st}$ $b_i$? There's only one $b_i$, right? – lulu Jul 8 '16 at 15:24
• My reading is the same as that of @ClementC. but I am not sure we have the question right. – lulu Jul 8 '16 at 15:25
$B$ always contains exactly one ball (since $b_1$ was put in there with probability $1$) and whether a given ball $b_i$ is put in the bucket $B$ is independent of the previous balls ($i < j$). Therefore, at the end of the day the probability that $b_i$ is in the bucket $B$ is simply the probability $$\mathbb{P}\{b_i\in B\text{ and } b_j\not\in B \text{ for all } j > i\}$$ which, by independence, becomes $$\mathbb{P}\{b_i\in B \}\cdot\prod_{j=i+1}^n \mathbb{P}\{b_j\not\in B\} = \frac{1}{i}\prod_{j=i+1}^n\left(1-\frac{1}{j}\right).$$
This in turn, perhaps counter-intuitively, leads to the answer being $$\frac{1}{i}\prod_{j=i+1}^n \left(1-\frac{1}{j}\right) = \frac{1}{n}.$$
(This last identity is easy to show, e.g. by induction on $1\leq i \leq n$.)
• @papabiceps Thanks for spotting this! Apparently, copy-pasting from comments involves the joy of all $\LaTeX$ being duplicated in the plain text. – Clement C. Jul 8 '16 at 15:45 |
# §8.22(i) Terminant Function
The so-called terminant function $F_{p}(z)$, defined by
8.22.1 $F_{p}(z)=\frac{\mathop{\Gamma\/}\nolimits\!\left(p\right)}{2\pi}z^{1-p}\mathop% {E_{p}\/}\nolimits\!\left(z\right)=\frac{\mathop{\Gamma\/}\nolimits\!\left(p% \right)}{2\pi}\mathop{\Gamma\/}\nolimits\!\left(1-p,z\right),$ Defines: $F_{p}(z)$: terminant function Symbols: $\mathop{\Gamma\/}\nolimits\!\left(z\right)$: gamma function, $\mathop{E_{p}\/}\nolimits\!\left(z\right)$: generalized exponential integral, $\mathop{\Gamma\/}\nolimits\!\left(a,z\right)$: incomplete gamma function, $z$: complex variable and $p$: parameter Permalink: http://dlmf.nist.gov/8.22.E1 Encodings: TeX, pMML, png
plays a fundamental role in re-expansions of remainder terms in asymptotic expansions, including exponentially-improved expansions and a smooth interpretation of the Stokes phenomenon. See §§2.11(ii)2.11(v) and the references supplied in these subsections.
# §8.22(ii) Riemann Zeta Function and Incomplete Riemann Zeta Function
The function $\mathop{\Gamma\/}\nolimits\!\left(a,z\right)$, with $|\mathop{\mathrm{ph}\/}\nolimits a|\leq\tfrac{1}{2}\pi$ and $\mathop{\mathrm{ph}\/}\nolimits z=\tfrac{1}{2}\pi$, has an intimate connection with the Riemann zeta function $\mathop{\zeta\/}\nolimits\!\left(s\right)$25.2(i)) on the critical line $\realpart{s}=\tfrac{1}{2}$. See Paris and Cang (1997).
If $\zeta_{x}(s)$ denotes the incomplete Riemann zeta function defined by
8.22.2 $\zeta_{x}(s)=\frac{1}{\mathop{\Gamma\/}\nolimits\!\left(s\right)}\int_{0}^{x}% \frac{t^{s-1}}{e^{t}-1}dt,$ $\realpart{s}>1$, Defines: $\zeta_{x}(s)$: incomplete Riemann zeta function Symbols: $\mathop{\Gamma\/}\nolimits\!\left(z\right)$: gamma function, $dx$: differential of $x$, $e$: base of exponential function, $\int$: integral, $\realpart{}$: real part and $x$: real variable Permalink: http://dlmf.nist.gov/8.22.E2 Encodings: TeX, pMML, png
so that $\lim_{x\to\infty}\zeta_{x}(s)=\mathop{\zeta\/}\nolimits\!\left(s\right)$, then
8.22.3 $\zeta_{x}(s)=\sum_{k=1}^{\infty}k^{-s}\mathop{P\/}\nolimits\!\left(s,kx\right),$ $\realpart{s}>1$.
For further information on $\zeta_{x}(s)$, including zeros and uniform asymptotic approximations, see Kölbig (1970, 1972a) and Dunster (2006).
The Debye functions $\int_{0}^{x}t^{n}\left(e^{t}-1\right)^{-1}dt$ and $\int_{x}^{\infty}t^{n}\left(e^{t}-1\right)^{-1}dt$ are closely related to the incomplete Riemann zeta function and the Riemann zeta function. See Abramowitz and Stegun (1964, p. 998) and Ashcroft and Mermin (1976, Chapter 23). |
# Non-zero transverse single spin asymmetry of very forward $\pi^0$ in polarized $p + p$ collisions at $\sqrt{s} = 510$ GeV
### Submission summary
As Contributors: Minho Kim Preprint link: scipost_202108_00002v1 Date submitted: 2021-08-02 09:25 Submitted by: Kim, Minho Submitted to: SciPost Physics Proceedings Proceedings issue: DIS2021 Academic field: Physics Specialties: High-Energy Physics - Experiment Approach: Experimental
### Abstract
The RHICf experiment measured transverse single spin asymmetry of very forward ($\eta > 6$) $\pi^0$ from polarized $p + p$ collisions at $\sqrt{s} = 510$ GeV. In order to measure it precisely, we installed a new electromagnetic calorimeter at zero-degree area of the STAR experiment at the Relativistic Heavy Ion Collider (RHIC) and measured the $\pi^0$s over the kinematic range of $x_F > 0.25$ and $0 < p_T < 1$ GeV/$c$ in June, 2017. A clear non-zero asymmetry was observed even in low $p_T < 1$ GeV$c$ showing a similar $x_F$ dependence with the forward ($2 < \eta < 4$) $\pi^0$ ones. A possible diffractive contribution may need to be taken into account to explain the very forward $\pi^0$ asymmetry. RHICf-STAR combined analysis and follow-up experiment will give a clue to understand it qualitatively.
###### Current status:
Editor-in-charge assigned |
dc.creator Brealey, Graham J. en_US dc.creator Kasha, Michael en_US dc.date.accessioned 2006-06-15T12:41:18Z dc.date.available 2006-06-15T12:41:18Z dc.date.issued 1954 en_US dc.identifier 1954-B-2 en_US dc.identifier.uri http://hdl.handle.net/1811/7221 dc.description Author Institution: Department of Chemistry, Florida State University en_US dc.description.abstract The change in the position of the $\eta\rightarrow n^{*}$ absorption bands on changing from a hydrocarbon to a hydroxylic solvent has been investigated for a number of molecules. The large shift to shorter wavelengths (blue-shift) is shown to be mainly due to hydrogen bonding of the `$\eta$’ electrons by the hydroxylic solvent which causes a greater stabilization of the ground state compared to the excited state of the molecule. Pyridazine and benzophenone have been examined in detail in a series of different mixtures of hexane and ethanol. The families of spectra obtained indicate that essentially two species are involved, a hydrogen-bonded and a non-hydrogen-bonded form and it is the formation of the hydrogen-bonded species that causes the main shift of the $\eta\rightarrow n^{*}$ transition to the blue. From the ultraviolet data, an association constant of hydrogen bonding can be obtained and this agrees well with the association constant found by a study of the association of ethanol with the molecule in the infrared. The infrared work makes use of the shift in the $O-H$ stretching frequency on formation of a hydrogen bond. en_US dc.format.extent 110531 bytes dc.format.mimetype image/jpeg dc.language.iso en en_US dc.publisher Ohio State University en_US dc.title THE ROLE OF HYDROGEN-BONDING IN THE $\eta\rightarrow n^{*}$ BLUE-SHIFT $PHENOMENON^{\dagger}$ en_US dc.type article en_US
|
# Englert-Brout-Higgs-Guralnik-Hagen-Kibble mechanism (history)
Tom W B Kibble (2009), Scholarpedia, 4(1):8741. doi:10.4249/scholarpedia.8741 revision #137393 [link to/cite this article]
Post-publication activity
Curator: Tom W B Kibble
Figure 1: The sombrero potential with an unstable state at $$\phi=0$$ and minima around a circle.
This page summarizes the history of the development of the Englert-Brout-Higgs-Guralnik-Hagen-Kibble mechanism. Many of the important papers have been reprinted in a volume edited by C.H. Lai (Lai 1981), and/or in the review volume by John C. Taylor (Taylor 2001).
## Symmetry breaking in superconductivity
Figure 2: A Cooper pair formed of electrons with opposite momentum and spin.
The idea of a connection between mass generation and spontaneous symmetry breaking dates back to the work of Nambu (1960). This and other early work was largely based on analogies with the theory of superconductivity.
The essential physical basis of the BCS or Bardeen-Cooper-Schrieffer theory of superconductivity (Bardeen 1957), is the existence of an effective attractive force between electrons arising from their interaction with phonons in the lattice. This leads to the formation of bound pairs of electrons, Cooper pairs, in a spin-singlet state (see Figure 2). These Cooper pairs then undergo a Bose-Einstein condensation; in other words there is a macroscopic occupation of a single quantum state of a pair.
### The Ginzburg-Landau model
A very convenient phenomenological model of a superconductor is provided by the Ginzburg-Landau model (Ginzburg 1950). This predates the BCS theory, but was later derived from it (in the London regime close to the critical temperature) by Gor'kov (1959).
The Ginzburg-Landau model involves a scalar order parameter field, $$\phi(t,\mathbf{x})\ .$$ This field represents the wave function of the condensate of Cooper pairs; $$|\phi|^2$$ is the number density of pairs in the condensate. Here $$\phi$$ may be regarded as a composite of two electron fields, and so has charge $$2e\ .$$ The Hamiltonian (in units with $$\hbar=1$$) is $\tag{1} H=\int d^3\mathbf{x} \left[\frac{1}{2m}\mathbf{D}\phi^*\cdot\mathbf{D}\phi+V(\phi)\right],$
where $\mathbf{D}\phi=\boldsymbol{\nabla}\phi-2ie\mathbf{A}\phi,$ is the covariant derivative corresponding to the gauge transformations $\tag{2} \phi\to\phi e^{2ie\lambda},\qquad \mathbf{A}\to\mathbf{A}+\boldsymbol{\nabla}\lambda,$
and the potential $$V(\phi)$$ is a function only of $$\phi^*\phi\ .$$ Near the critical temperature $$T_{\mathrm{c}}$$ the field $$\phi$$ is small, and $$V$$ may be expanded in a power series, keeping only quadratic and quartic terms: $\tag{3} V(\phi)=\alpha\phi^*\phi+\tfrac{1}{2}\beta(\phi^*\phi)^2.$
The coefficients $$\alpha$$ and $$\beta$$ are temperature-dependent, and in particular $$\alpha$$ changes sign at the critical temperature. For $$T<T_{\mathrm{c}}\ ,$$ it is negative, so $$V$$ has the 'sombrero' shape (see Figure 1) and its minima occur not at the symmetry point $$\phi=0$$ but around the circle $\tag{4} \phi^*\phi=\frac{-\alpha}{\beta}.$
Consequently, the gauge symmetry (2) is spontaneously broken. The magnitude of the order parameter $$|\phi|$$ is fixed by (4) but its phase is arbitrary. There is a family of degenerate ground states, labelled by this phase. Note that since the phase is canonically conjugate to the occupation number, a definite phase requires an indefinite number of pairs in the condensate.
Following the development of this theory, there was much discussion in the literature of the role of gauge invariance. The Ginzburg-Landau ground state does not respect this gauge symmetry, and nor does the ground state of the full Bardeen-Cooper-Schrieffer model. Today this is recognized as an example of the phenomenon of spontaneous symmetry breaking, but at the time it was seen by many as a defect of the theory.
Nambu (1960) gave a particularly clear account of the situation, using a modified form of the Hartree-Fock approximation. (See also Bogoliubov (1959).) He showed how the symmetry breaking is intimately connected with the energy gap. So long as the force between electrons is attractive, dissociating a Cooper pair requires an energy input. Thus the excitations of the system, quasiparticle states, are separated from the ground state by a finite energy gap; the energy $$E_{\mathbf{p}}$$ of a quasiparticle of momentum $$\mathbf{p}$$ tends to a non-zero limit as $$\mathbf{p}\to\mathbf{0}\ .$$ Closely related is the fact that the quasiparticles do not have a definite charge (Bogoliubov (1959)): a quasiparticle is a linear combination of an electron of momentum $$\mathbf{p}$$ and spin up, say, and a hole in the state of momentum $$-\mathbf{p}$$ and spin down.
### Analogous mechanism in particle physics
Nambu then went on to suggest that the masses of elementary particles might arise in a similar way; the vacuum state, like the superconducting ground state, might not respect the symmetries of the theory, and so elementary particles, like the quasiparticles in BCS theory, might acquire masses.
In particular, Nambu envisaged a theory of strong interactions involving a fundamental massless fermion field $$\psi(x)$$ with a Lagrangian invariant under both ordinary and chiral global phase changes: $\tag{5} \psi(x)\to e^{i\alpha}\psi(x),\qquad\text{and}\qquad \psi(x)\to e^{\alpha\gamma_5}\psi(x)$
(in a representation in which $$\gamma_5^2=-1$$). Correspondingly there would be both vector and axial vector conserved Noether currents, $\tag{6} j^\mu=\bar\psi\gamma^\mu\psi,\qquad j^\mu_5=\bar\psi i\gamma^\mu\gamma_5\psi.$
He suggested that, like the superconducting energy gap, the nucleon mass arises from spontaneous breaking of the chiral symmetry.
Together, Nambu and Jona-Lasinio (1961) constructed a specific model with these characteristics, with four-fermion interactions, based on the Lagrangian density $\tag{7} L=i\bar\psi\gamma^\mu\partial_\mu\psi +g[(\bar\psi\psi)^2-(\bar\psi\gamma_5\psi)^2].$
They then assumed that in the ground state or vacuum, the chiral symmetry is broken spontaneously by a non-zero expectation value $$\langle0|\bar\psi(x)\psi(x)|0\rangle=n\ ,$$ say. Using techniques similar to those he employed to discuss superconductivity, they showed that this would indeed imply a nonzero mass for the 'quasiparticle', here identified as the nucleon.
### Nambu-Goldstone bosons
Like the model of Goldstone (1961) (see main page), the Nambu--Jona-Lasinio model also predicts the existence of a massless particle of spin zero, the Goldstone boson or Nambu-Goldstone boson, a nucleon-antinucleon bound state. This is a consequence of the broken chiral symmetry, and the particle in this case is a pseudoscalar. Since subjecting all the particles to a chiral rotation merely takes us from one degenerate vacuum state to another, and so requires no energy, it follows that applying a spatially varying chiral rotation with long wavelength requires very little energy; the energy goes to zero in the long-wavelength limit, which means the particle is massless.
It was clear that these were examples of a general phenomenon, described by what has come to be known as the Goldstone theorem. In the context of an explicitly relativistic theory, a general proof was provided by Goldstone, Salam & Weinberg (1962).
Nambu and Jona-Lasinio realized that as a consequence of the broken symmetry, their explanation for the nucleon mass would necessarily imply the existence of a massless particle of zero spin. Of course, no such particle was (or is) known, but they suggested that in fact the chiral symmetry is not quite exact but intrinsically weakly broken -- in addition to the larger spontaneous symmetry breaking effect. Then the would-be Nambu-Goldstone bosons would acquire small masses, and could be identified with the pions. (They pointed out that it would be easy to accommodate isospin within the model.) Although the Nambu--Jona-Lasinio model has been superseded, this identification is in fact in line with current theory. The nucleon is no longer regarded as a quasiparticle associated with a fundamental field; it is a composite of quarks. But it remains true that if the quarks were massless, then quantum chromodynamics would exhibit chiral symmetry, and the pions would be massless. The weak chiral symmetry breaking leads to a small pion mass.
Nambu and Jona-Lasinio also noted that in the case of a superconductor the particle-hole bound state that would play the role of the Nambu-Goldstone boson is not massless because of the existence of the long-range Coulomb forces, but rather becomes part of the plasmon oscillations.
## Gauge theories of strong and weak interactions
Figure 3: (a) Fermi's view of neutron decay; (b) neutron decay proceeding via an intermediate vector boson; (c) electromagnetic interaction between proton and electron.
The first non-Abelian gauge theory was proposed by Yang and Mills (1954) (and independently by Shaw (1955), though his work appears only in a Cambridge University PhD thesis). This was based on the idea of promoting the global SU(2) isospin symmetry to a local gauge symmetry, which requires the introduction of an isospin triplet of gauge fields. This was intended as a theory of strong interactions, and as such has of course been superseded by quantum chromodynamics. But it was soon realized that a similar idea might work for the weak interactions.
The original Fermi theory of weak interactions involved a direct four-fermion interaction (see Figure 3(a)). Following the proposal by Lee & Yang (1956) of parity non-conservation in weak interactions, confirmed the following year (Wu 1957), it was established that the weak interactions are of the $$V-A$$ form, involving interactions between vector and axial-vector currents. This led Feynman & Gell-Mann (1958) and independently Sudarshan & Marshak (1958) to suggest that they might be mediated by charged intermediate vector bosons, $$W^\pm$$ ( Figure 3(b)), raising the interesting possibility of a gauge theory of weak interactions. Indeed because of the similarity to electromagnetic interactions ( Figure 3(c)) it suggested the even more interesting possibility of a unified theory of weak and electromagnetic interactions, in which $$(W^+,\gamma,W^-)$$ would form a triplet of gauge bosons.
However, this hypothesis was faced with two immediate and severe obstacles. Firstly, to explain the short range and weakness at low energies of the weak interactions, the $$W^\pm$$ had to have very large masses, whereas it was generally believed that gauge bosons were necessarily massless, like the photon.
The second problem arose from the parity violation in weak interactions. The $$W^\pm$$ interacted not with a vector current but with a chiral current, a sum of terms of the form $$\bar\psi_1\gamma^\mu(1-i\gamma_5)\psi_2\ .$$ So how could they be part of the same multiplet as the photon, with its parity-conserving interactions? The solution to this second problem was eventually found to require the enlargement of the gauge group from $$SU(2)$$ to $$SU(2)\times U(1)\ ,$$ with the introduction of another neutral gauge boson, $$Z^0\ ,$$ as first suggested by Glashow (1961). This development, however, was peripheral to the subject of the present article, and will not be considered further here.
Both these problems clearly required some mechanism for breaking the symmetry between the $$W^\pm$$ and the photon. It was natural to ask whether this, like superconductivity, could be an example of spontaneous symmetry breaking. But that idea raised a new problem, the possible appearance of massless scalar particles. The Goldstone theorem seemed to suggest that a spontaneously broken gauge theory would be plagued by two different kinds of massless particles, the gauge vector bosons themselves and the scalar Nambu-Goldstone bosons.
### Masses of gauge bosons
In the early days it was commonly believed that one of the virtues of the gauge principle -- the idea of promoting global to local symmetries, with the introduction of gauge vector particles -- was the successful prediction of the vanishing photon mass.
Figure 4: The lowest-order photon self-energy diagram.
The first person to question this orthodoxy was Schwinger (1962). One form of the argument for a massless photon was that by virtue of gauge invariance, the self-energy of the photon (see Figure 4) would necessarily have the form $\Pi_{\mu\nu}(p)=(g_{\mu\nu}p^2-p_\mu p_\nu)\Pi(p^2),$ which seemed to suggest that it must vanish at $$p^2=0\ ,$$ implying zero mass. Schwinger argued convincingly however that if the interactions were strong enough, then $$\Pi(p^2)$$ might acquire a pole at $$p^2=0\ ,$$ thus eliminating the prediction. A photon with stronger interactions could be massive.
Then Anderson (1963) pointed out that there are examples of this in condensed-matter physics. In an electron plasma, electromagnetic waves cannot propagate if their frequency is less than the plasma frequency $$\omega_{\mathrm{pl}}\ ,$$ given by $\omega_{\mathrm{pl}}^2 = \frac{e^2 n_e}{\epsilon_0 m_e},$ where $$n_e$$ is the electron number density and $$m_e$$ the electron mass. In fact, the dispersion relation for this plasmon is for small $$\mathbf{k}$$ of the form, $\omega^2 =\omega_{\mathrm{pl}}^2+u^2\mathbf{k}^2,$ so in effect the photon has acquired a mass $$m_{\mathrm{pl}}=\hbar\omega_{\mathrm{pl}}/c^2\ .$$ The same thing applies in a superconductor.
This work however was framed in language that was not then familiar to most particle physicists, and referred to models without relativistic invariance; the plasma or superconductor defines a preferred frame of reference. But, because of the work of Goldstone, Salam & Weinberg, it was widely believed among field theorists that in a fully relativistic theory the Goldstone theorem would unambiguously demand massless particles. (See, for example, Gilbert (1964).) Although Anderson had shown that in the nonrelativistic context the massless gauge bosons and Nambu-Goldstone bosons could combine to form a massive vector particle (a result already foreshadowed by Nambu and Jona-Lasinio (1961)), most particle theorists believed that could not happen in a relativistic theory.
The final step of showing that indeed the mechanism could work in a relativistic theory was taken independently by three groups: Englert & Brout, Higgs, and Guralnik, Hagen & Kibble, approaching the problem from three rather different perspectives. The simplest and most direct argument was that of Higgs (1964b), where he exhibited a very simple U(1) model, now often called the Abelian Higgs model (see the article Englert-Brout-Higgs-Guralnik-Hagen-Kibble mechanism). Prior to that (Higgs 1964a) he had made the important observation that in a gauge theory formulated in the radiation gauge, the choice of gauge removes the explicit relativistic invariance, and thus renders the Goldstone theorem inapplicable. Higgs's treatment of the model was purely classical, but essentially the same model was considered in a quantum mechanical way by both the other groups. Englert & Brout (1964), whose paper was the first to be published, based their argument on a calculation of the vacuum polarization in lowest-order perturbation theory about the assumed symmetry-breaking vacuum state. They also pointed out that the same mechanism could operate in more general non-Abelian models, and gave simple examples. Guralnik, Hagen & Kibble (1964), on the other hand, used an operator formalism, concentrating on the role of the conservation law and the precise way in which the Goldstone theorem can be evaded. (For more detail of the history leading up to this paper, see Guralnik 2011.) However, the conclusions of all three were essentially the same.
### Later developments
There were several significant developments in the years immediately after 1964. Higgs (1966) studied his model quantum mechanically in some detail, evaluating transition and decay amplitudes for the model in lowest-order perturbation theory, and in particular discussing the induced symmetry-breaking effects that would occur if the scalar field were coupled to other fields. Streater (1965) discussed the mechanism from the point of view of axiomatic field theory. The details of the application to a non-Abelian gauge theory were studied by Kibble (1967), showing how the numbers of massive and massless states were related to the symmetry-breaking pattern. For a comprehensive review of symmetry breaking in both field theory and condensed matter, see Guralnik et al (1967).
Initially, the Englert-Brout-Higgs-Guralnik-Hagen-Kibble mechanism was envisaged as having as much, or more, relevance to the development of a gauge theory of strong as of weak interactions. But of course its most significant application was to the development of the unified electroweak theory by Weinberg (1967) and Salam (1968), incorporating the model developed earlier by Glashow (1961).
Both Weinberg and Salam had conjectured that the theory was renormalizable, but with no proof. The final, very important element was the proof of renormalizability of spontaneously broken gauge theories by 't Hooft (1971).
Historically, the primary focus of the mechanism was on how the gauge bosons acquire non-zero mass. The fact that there are also massive Higgs bosons was incidental. However, their existence has now become of great significance, because the Higgs boson is at present (2009) the only component of the standard model whose existence has not been experimentally confirmed1. Finding them (or some alternative) will be a key advance.
1 On 4 July 2012 a previously unknown boson of mass $\sim$ 125 GeV was confirmed to exist by the ATLAS and CMS teams at the Large Hadron Collider at CERN. This particle has been tentatively confirmed as a Higgs boson by CERN on 14 March 2013, although it remains an open question whether this is the Higgs boson of the Standard Model of particle physics, or possibly the lightest of several bosons predicted in some theories that go beyond the Standard Model.
## References
• Bogoliubov, N N (1958). A new method in the theory of superconductivity I. Sov. Phys.-JETP 7: 41-46. [J. Exper. Theor. Phys. (USSR) 34 (1958) 58-65]
• Feynman, R P and Gell-Mann, M (1958). Theory of the Fermi interaction. Phys. Rev. 109: 193-8. [] DOI: 10.1103/PhysRev.109.193
• Ginzburg, V L and Landau, L D (1950). On the theory of superconductivity. J. Exper. Theor. Phys. (USSR) 20: 1064.
• Glashow, S L (1961). Partial symmetries of weak interactions. Nuc. Phys. 22: 579-88.
• Goldstone, J (1961). Field theories with superconductor solutions. Nuovo Cim. 19: 154-164.
• Gor'kov, L P (1959). Microscopic derivation of the Ginzburg-Landau equations in the theory of superconductivity. Sov. Phys.-JETP 36: 1364-7. [J. Exper. Theor. Phys. (USSR) 36 (1959) 1918-23]
• Guralnik, G S; Hagen, C R and Kibble, T W B (1964). Global conservation laws and massless particles. Phys. Rev. Lett. 13: 585-7. [ http://link.aps.org/abstract/PRL/v13/p585] DOI: 10.1103/PhysRevLett.13.585
• Guralnik, G S; Hagen, C R and Kibble, T W B (1967). Broken symmetries and the Goldstone theorem. Advances in Physics, vol. 2 Interscience Publishers, New York. pp. 567-708 ISBN 0470170573
• Higgs, P W (1964a). Broken symmetries, massless particles and gauge fields. Phys. Lett. 12: 132-3.
• Lai, C H (1981). Gauge theory of weak and electromagnetic interactions. World Scientific, Singapore. ISBN 9971-83-023-x.
• Nambu, Y and Jona-Lasinio, G (1961). Dynamical model of elementary particles based on an analogy with superconductivity. I. Phys. Rev. 122: 345-58. [1] DOI: 10.1103/PhysRev.122.345
• Salam, A (1968). Weak and electromagnetic interactions. Elementary particle theory, Proceedings of the Nobel Symposium held in 1968 at Lerum, Sweden. Svartholm, N Almqvist & Wiksell, Stockholm. 367-77
• Shaw, R (1955). Invariance under general isotopic gauge transformations. Cambridge University PhD thesis Cambridge University, Cambridge. Part II, Chapter 3, pp. 34-46
• Streater, R F (1965). Spontaneously broken symmetry in axiomatic theory. Proc. Roy. Soc. Lond. A 287: 510-18. DOI: 10.1098/rspa.1965.0193
• Taylor, J C (2001). Gauge theories in the twentieth century. Imperial College Press, London. ISBN 1-86094-281-4.
• 't Hooft, G (1971). Renormalizable Lagrangians for massive Yang-Mills fields. Nuc. Phys. B 35: 167-88.
• Wu, C S; Ambler, E; Hayward, R W; Hoppes, D D and Hudson, R P (1957). Experimental test of parity conservation in beta decay. Phys. Rev. 105: 1413-15. [ http://link.aps.org/abstract/PR/v105/p1413] DOI: 10.1103/PhysRev.105.1413
Internal references
• Jeff Moehlis, Kresimir Josic, Eric T. Shea-Brown (2006) Periodic orbit. Scholarpedia, 1(7):1358.
• Aitchison, I J R and Hey, A J G (2003). Gauge theories in particle physics, vol. II: QCD and the electroweak theory, 3rd ed. Institute of Physics, Bristol. ISBN 0-750-30950-4.
• Weinberg, S (1996). The quantum theory of fields, vol. II: modern applications. Cambridge University Press, Cambridge. Chapter 21. ISBN 0-521-55001-7. This book contains a particularly interesting account of superconductivity from the standpoint of symmetry breaking, showing that many of the key features of superconductors can be derived just from the assumption of spontaneous symmetry breaking.
• Zee, A (2003). Quantum field theory in a nutshell. Princeton University Press, Princeton, NJ. ISBN 0-691-01019-6. |
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