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# assumptions of discriminant analysis

It consists of two closely … Discrimination is … Discriminant Analysis Data Considerations. As part of the computations involved in discriminant analysis, you will invert the variance/covariance matrix of the variables in the model. Assumes that the predictor variables (p) are normally distributed and the classes have identical variances (for univariate analysis, p = 1) or identical covariance matrices (for multivariate analysis, p > 1). Key words: assumptions, further reading, computations, validation of functions, interpretation, classification, links. A few … Discriminant analysis assumptions. Here, there is no … The basic assumption for discriminant analysis is to have appropriate dependent and independent variables. Quadratic Discriminant Analysis . The analysis is quite sensitive to outliers and the size of the smallest group must be larger than the number of predictor variables. (ii) Quadratic Discriminant Analysis (QDA) In Quadratic Discriminant Analysis, each class uses its own estimate of variance when there is a single input variable. This paper considers several alternatives when … In this blog post, we will be discussing how to check the assumptions behind linear and quadratic discriminant analysis for the Pima Indians data. The non-normality of data could be as a result of the … Fisher’s LDF has shown to be relatively robust to departure from normality. Discriminant analysis is a very popular tool used in statistics and helps companies improve decision making, processes, and solutions across diverse business lines. Understand how to examine this assumption. The Flexible Discriminant Analysis allows for non-linear combinations of inputs like splines. The code is available here. The assumptions of discriminant analysis are the same as those for MANOVA. (Avoiding these assumptions gives its relative, quadratic discriminant analysis, but more on that later). : 1-good student, 2-bad student; or 1-prominent student, 2-average, 3-bad student). The dependent variable should be categorized by m (at least 2) text values (e.g. We will be illustrating predictive … Understand how predict classifies observations using a discriminant analysis model. The basic idea behind Fisher’s LDA 10 is to have a 1-D projection that maximizes … The analysis is quite sensitive to outliers and the size of the smallest group must be larger than the number of predictor variables. Before we move further, let us look at the assumptions of discriminant analysis which are quite similar to MANOVA. Linear discriminant analysis is a classification algorithm which uses Bayes’ theorem to calculate the probability of a particular observation to fall into a labeled class. Discriminant function analysis makes the assumption that the sample is normally distributed for the trait. Linear discriminant analysis is a form of dimensionality reduction, but with a few extra assumptions, it can be turned into a classifier. This logistic curve can be interpreted as the probability associated with each outcome across independent variable values. Formulate the problem The first step in discriminant analysis is to formulate the problem by identifying the objectives, the criterion variable and the independent variables. Discriminant analysis (DA) is a pattern recognition technique that has been widely applied in medical studies. Logistic regression fits a logistic curve to binary data. Discriminant analysis assumes that the data comes from a Gaussian mixture model. With an assumption of an a priori probability of the individual class as p 1 and p 2 respectively (this can numerically be assumed to be 0.5), μ 3 can be calculated as: (2.14) μ 3 = p 1 * μ 1 + p 2 * μ 2. A distinction is sometimes made between descriptive discriminant analysis and predictive discriminant analysis. This Journal. Assumptions. Model Wilks' … Canonical Discriminant Analysis. Cases should be independent. The criterion … The K-NNs method assigns an object of unknown affiliation to the group to which the majority of its K nearest neighbours belongs. Quadratic discriminant analysis (QDA): More flexible than LDA. Linear discriminant analysis (LDA): Uses linear combinations of predictors to predict the class of a given observation. Regular Linear Discriminant Analysis uses only linear combinations of inputs. Steps in the discriminant analysis process. Unstandardized and standardized discriminant weights. It also evaluates the accuracy … … Steps for conducting Discriminant Analysis 1. Linear Discriminant Analysis is based on the following assumptions: The dependent variable Y is discrete. Discriminant function analysis is used to discriminate between two or more naturally occurring groups based on a suite of continuous or discriminating variables. [qda(); MASS] PCanonical Distance: Compute the canonical scores for each entity first, and then classify each entity into the group with the closest group mean canonical score (i.e., centroid). Logistic regression … The data vectors are transformed into a low … If the dependent variable is not categorized, but its scale of measurement is interval or ratio scale, then we should categorize it first. They have become very popular especially in the image processing area. A second critical assumption of classical linear discriminant analysis is that the group dispersion (variance-covariance) matrices are equal across all groups. Nonlinear Discriminant Analysis using Kernel Functions Volker Roth & Volker Steinhage University of Bonn, Institut of Computer Science III Romerstrasse 164, D-53117 Bonn, Germany {roth, steinhag}@cs.uni-bonn.de Abstract Fishers linear discriminant analysis (LDA) is a classical multivari ate technique both for dimension reduction and classification. When these assumptions hold, QDA approximates the Bayes classifier very closely and the discriminant function produces a quadratic decision boundary. Another assumption of discriminant function analysis is that the variables that are used to discriminate between groups are not completely redundant. Discriminant analysis is a group classification method similar to regression analysis, in which individual groups are classified by making predictions based on independent variables. However, in this, the squared distance will never be reduced to the linear functions. Little attention … The relationships between DA and other multivariate statistical techniques of interest in medical studies will be briefly discussed. Visualize Decision Surfaces of Different Classifiers. Discriminant function analysis (DFA) is a statistical procedure that classifies unknown individuals and the probability of their classification into a certain group (such as sex or ancestry group). Data. Linear vs. Quadratic … Stepwise method in discriminant analysis. This example shows how to visualize the decision … In this type of analysis, your observation will be classified in the forms of the group that has the least squared distance. Multivariate normality: Independent variables are normal for each level of the grouping variable. Normality: Correlation a ratio between +1 and −1 calculated so as to represent the linear … Linearity. Back; Journal Home; Online First; Current Issue; All Issues; Special Issues; About the journal; Journals. Eigenvalue. QDA assumes that each class has its own covariance matrix (different from LDA). Measures of goodness-of-fit. PQuadratic discriminant functions: Under the assumption of unequal multivariate normal distributions among groups, dervie quadratic discriminant functions and classify each entity into the group with the highest score. As part of the computations involved in discriminant analysis, STATISTICA inverts the variance/covariance matrix of the variables in the model. The objective of discriminant analysis is to develop discriminant functions that are nothing but the linear combination of independent variables that will discriminate between the categories of the dependent variable in a perfect manner. Canonical correlation. The assumptions for Linear Discriminant Analysis include: Linearity; No Outliers; Independence; No Multicollinearity; Similar Spread Across Range; Normality; Let’s dive in to each one of these separately. It allows multivariate observations ("patterns" or points in multidimensional space) to be allocated to previously defined groups (diagnostic categories). #4. This also implies that the technique is susceptible to … The assumptions in discriminant analysis are that each of the groups is a sample from a multivariate normal population and that all the populations have the same covariance matrix. K-NNs Discriminant Analysis: Non-parametric (distribution-free) methods dispense with the need for assumptions regarding the probability density function. Pin and Pout criteria. Recall the discriminant function for the general case: $\delta_c(x) = -\frac{1}{2}(x - \mu_c)^\top \Sigma_c^{-1} (x - \mu_c) - \frac{1}{2}\log |\Sigma_c| + \log \pi_c$ Notice that this is a quadratic … In marketing, this technique is commonly used to predict … Predictor variables should have a multivariate normal distribution, and within-group variance-covariance matrices should be equal … We also built a Shiny app for this purpose. [9] [7] Homogeneity of variance/covariance (homoscedasticity): Variances among group … Since we are dealing with multiple features, one of the first assumptions that the technique makes is the assumption of multivariate normality that means the features are normally distributed when separated for each class. Assumptions – When classification is the goal than the analysis is highly influenced by violations because subjects will tend to be classified into groups with the largest dispersion (variance) – This can be assessed by plotting the discriminant function scores for at least the first two functions and comparing them to see if … Let’s start with the assumption checking of LDA vs. QDA. The posterior probability and typicality probability are applied to calculate the classification probabilities … The assumptions of discriminant analysis are the same as those for MANOVA. The main … There is no best discrimination method. Relax-ation of this assumption affects not only the significance test for the differences in group means but also the usefulness of the so-called "reduced-space transforma-tions" and the appropriate form of the classification rules. In practical cases, this assumption is even more important in assessing the performance of Fisher’s LDF in data which do not follow the multivariate normal distribution. Discriminant Function Analysis (DA) Julia Barfield, John Poulsen, and Aaron French . F-test to determine the effect of adding or deleting a variable from the model. Examine the Gaussian Mixture Assumption. It enables the researcher to examine whether significant differences exist among the groups, in terms of the predictor variables. The linear discriminant function is a projection onto the one-dimensional subspace such that the classes would be separated the most. Assumptions: Observation of each class is drawn from a normal distribution (same as LDA). Assumptions of Discriminant Analysis Assessing Group Membership Prediction Accuracy Importance of the Independent Variables Classiﬁcation functions of R.A. Fisher Discriminant Function Geometric Representation Modeling approach DA involves deriving a variate, the linear combination of two (or more) independent variables that will discriminate best between a-priori deﬁned groups. To perform the analysis, press Ctrl-m and select the Multivariate Analyses option from the main menu (or the Multi Var tab if using the MultiPage interface) and then … Independent variables that are nominal must be recoded to dummy or contrast variables. Quadratic Discriminant Analysis. However, the real difference in determining which one to use depends on the assumptions regarding the distribution and relationship among the independent variables and the distribution of the dependent variable.The logistic regression is much more relaxed and flexible in its assumptions than the discriminant analysis. Linear discriminant function analysis (i.e., discriminant analysis) performs a multivariate test of differences between groups. So so that we know what kinds of assumptions we can make about $$\Sigma_k$$, ... As mentioned, the former go by quadratic discriminant analysis and the latter by linear discriminant analysis. Wilks' lambda. The grouping variable must have a limited number of distinct categories, coded as integers. Real Statistics Data Analysis Tool: The Real Statistics Resource Pack provides the Discriminant Analysis data analysis tool which automates the steps described above. Abstract: “The conventional analysis of variance applied to designs in which each subject is measured repeatedly requires stringent assumptions regarding the variance-covariance (i. e., correlations among repeated measures) structure of the data. Most multivariate techniques, such as Linear Discriminant Analysis (LDA), Factor Analysis, MANOVA and Multivariate Regression are based on an assumption of multivariate normality. What we will be covering: Data checking and data cleaning Another assumption of discriminant function analysis is that the variables that are used to discriminate between groups are not completely redundant. Unlike the discriminant analysis, the logistic regression does not have the … In this type of analysis, dimension reduction occurs through the canonical correlation and Principal Component Analysis. Prediction Using Discriminant Analysis Models. … Introduction . Box's M test and its null hypothesis. In addition, discriminant analysis is used to determine the minimum number of dimensions needed to describe these differences. [7] Multivariate normality: Independent variables are normal for each level of the grouping variable. One of the basic assumptions in discriminant analysis is that observations are distributed multivariate normal. Violation of these assumptions results in too many rejections of the null hypothesis for the stated significance level. If any one of the variables is completely redundant with the other variables then the matrix is said to be ill … We now repeat Example 1 of Linear Discriminant Analysis using this tool. : the dependent variable should be categorized by m ( at least 2 ) text values e.g. Needed to describe these differences uses only linear combinations of predictors to predict the class of given! Of the variables that are nominal must be recoded to dummy or contrast variables are nominal be. Represent the linear discriminant function analysis makes the assumption checking of LDA vs. QDA the model of each has! Its own covariance matrix ( different from LDA ) more Flexible than LDA of... Will never be reduced to the group to which the majority of its K nearest neighbours belongs part of computations! Described above, 2-average, 3-bad student ), 2-bad student ; or 1-prominent student, 2-average, student! 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a hyperreal number?

Who?
Uğur Doğan (HU Berlin)
When?
2013/10/25, 16:00
Where?
HU Berlin, Rudower Chaussee 25, at room 1.023
The set of hyperreal numbers is a field which contains $\mathbb R$ (real numbers) and some "infinitesimal" and "infinite" numbers. It will be constructed using model theory, so some basic facts and theorems in model theory will also be mentioned, such as Łoś's theorem and compactness of first order logic (and some applications if time permits).

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# Chebyshev’s Theorem: Concepts, Formula & Examples

Chebyshev’s theorem is a fundamental concept in statistics that allows us to determine the probability of data values falling within a certain range. This theorem makes it possible to calculate the probability of a given dataset being within k standard deviations away from the mean. It is important for data scientists, statisticians, and analysts to understand this theorem as it can be used to assess the spread of data points around a mean value.

## What is Chebyshev’s Theorem?

Chebyshev’s Theorem is used to determine the approx. percentage of values that lie within a given number of standard deviations from the mean of a set of data whose shape of distribution is unknown or it is unknown whether the data is normally distributed. This theorem can be applied to all distributions regardless of their shape and can be used whenever the data distribution shape is unknown or is non normal. If the data distribution is known as normal distribution, one can apply the empirical rule (68-95-99.7) which looks like the following and states that given normal data distribution, 68% of the data falls within 1 standard deviation, 95% of data falls within two standard deviation and 99.7 % of data falls within 3 standard deviations.

Chebyshev’s theorem can be applied to data that are normally distributed as well as data that are non-normally distributed. However, for normal data distribution, empirical rule is widely used.

As per Chebyshev’s theorem, at least $$1 – \frac{1}{k^2}$$ values will fall within ±k standard deviations of the mean regardless of the shape of the distribution for values of k > 1. This looks like the following when plotted. The plot represents that 75% of values will fall under 2 standard deviations of mean and 88.88% of values will fall within 3 standard deviations of the mean.

75% is calculated as 1 − 1/k2 = 1 − 1/22 = 3/4 = .75. Note the value of k = 2. When comparing with the empirical rule if the data are normally distributed, 95% of all values are within μ ± 2σ (2 standard deviations). Similarly, the percentage of values within 3 standard deviations of the mean is at least 89%, in contrast to 99.7% for the empirical rule.

## Chebyshev’s Formula

The formula for Chebyshev’s theorem looks like the following:

## Chebyshev’s Example

Let’s look at an example to better understand how Chebyshev’s theorem works in practice. The Chebyshev theorem states that if the mean (μ) and standard deviation (σ) of a data set are known, then at least 75% of the data points should lie within two standard deviations of the mean (μ ± 2σ). This means that any two numbers that are two standard deviations away from the mean will contain at least 75% of the points in the data set. For example, if μ = 10 and σ = 2, then all points between 6 and 14 will contain at least 75% of our data points.

## Conclusion

In summary, Chebyshev’s theorem provides us with an easy way to calculate how many data points should fall within a certain range from their mean value based on their standard deviation and desired variance level if the data distribution is unknown or non-normal. This understanding can help us better assess where outliers may exist in our datasets and use this information for further analysis or predictions about our datasets’ behaviors over time. Whether you’re working with small or large datasets, having an understanding of Chebychev’s theorem can help you get more accurate insights into your data!


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## Equivalent conditions for a closed immersion of schemes

In Hartshorne, a closed immersion of schemes is defined to be a scheme morphism Φ:Y→X such that Φ is a homeomorphism onto Φ(Y), Φ(Y) is closed in |X| and (∗)Φ#:OX→Φ∗OY is surjective. Can I replace (∗) by the condition that Φ#U:OX(U):→OY(Φ−1(U)) is surjective for all affine open subsets U⊆X, or does this lead to another definition? … Read more

## Exercise 6.3.N. on Ravi Vakil’s notes (on morphism from SS scheme to PnB\mathbb{P}_B^n)

I am working on Exercise 6.3.N. on Ravi Vakil’s notes (on morphism from S scheme to PnB) and I would like some assistance. The exercise states: Let B be a ring. If X is a B-scheme, and f0,…,fn are n+1 functions on X with no common zeros, then show that [f0,…,fn] gives a morphism of … Read more

## canonical representation of three skew lines in P3\mathbb{P}^3

Consider three skew (non-intersecting) lines L,M,N in P3. Each line is given by two equations of the form α⊤i,jz=0,i=1,2,3,j=1,2, where z=(z0,z1,z2,z3) are the homogeneous coordinates of P3 and αi,j∈P3. What i want is to show that up to projective equivalence the equations for the three lines can be written as (this is possible according to … Read more

## How does an irreducible quadric in projective space look like?

I read the answer to the following question: Quadrics are birational to projective space Here it is stated that: Over a field k of characteristic ≠2 every irreducible quadric Q⊂Pnk has equation q(x)=x0x1+x22+…+x2m=0 in suitable coordinates . Can anyone tell me why an irreducible quadric would look like this (and would not have the terms … Read more

## If Y Y is irreducible set so is cl(Y)cl(Y). [duplicate]

This question already has answers here: The closure of an irreducible subset of an irreducible space is irreducible. (2 answers) Closed 6 years ago. If Y is irreducible set so is cl(Y). If cl(Y) is reducible then cl(Y)=A∪B where both A and B is closed in cl(Y). Now how do we proceed? Answer Suppose Y … Read more

## If C⊆P2C \subseteq \mathbb{P}^2 is a plane curve, then genus(C)=12(d−1)(d−2)genus(C)=\frac{1}{2}(d-1)(d-2). Compare with example in the notes

In my Algebraic Geometry notes (see http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2002/main.pdf) there is the following exercise: If C⊆P2 is a plane curve of degree d, then its arithmetic genus g(C) is equal to 12(d−1)(d−2). Compare this to example 0.1.3 (of the notes of course). In fact, the first part of the exercise is easy to solve using the Hilbert … Read more

## A curve of genus $g\geq 2$ has a closed point of degree at most $2g-2$ over base field.

I am working on the following problem [R. Vakil] Exercise 19.8.B: Suppose $C$ is a curve of genus $g>1$ over a field $k$ that is not algebraically closed. Show that $C$ has a closed point of degree at most $2g-2$ over the base field. I have no idea how to do this question. This is … Read more

## Morphisms induced by effective divisors on P1\mathbf P^1

This question is about the proof of Theorem V.2.17 in Hartshorne’s Algebraic Geometry. Here everything is defined over some algebraically closed field k. Define O=OP1. Let e≥0 be an integer. Then the following is what I do not understand: The proof says that when n≥e, we can have a surjective map O⊕O(−e)→O(n−e)→0 by arguing that … Read more

## Real Points of resolution of singularities of $\mathrm{Spec} \mathbb{R}[x,y]/(x^2+y^2)$

Consider the scheme $X = \mathrm{Spec} \mathbb{R}[x,y]/(x^2+y^2)$. Scheme-theoretically, it’s a one dimensional scheme with one real, singular point (there are, of course, other complex points). I’m interested in finding the structure of the real points of the resolution of $X$, but I’m having trouble visualizing exactly what I get. I think I want to do … Read more

## Does D(X)≅D(Y)\operatorname{D}(X)\cong \operatorname{D}(Y) imply X≅YX\cong Y?

Let X,Y be smooth projective varieties and by D(X) we mean the derived category of the category of bounded complexes of coherent sheaves on X. It is well-known that there might exist an equivalence D(X)≃D(Y) even if X≇. What if there is an isomorphism \operatorname{D}(X) \cong \operatorname{D}(Y)? Does it necessarily imply that X\cong Y? Answer … Read more


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# Conditional distribution on arrival time (Poisson process)

Suppose that $\{N_t: t\geq 0\}$ is a Poisson process of rate $\lambda$ and $T_1< T_2< \dotsb\$ are its arrival times (i.e. $T_i := \min \{t\geq 0 : N_t \geq i\}$). What is the conditional distribution of $N_2$, given $T_3=1$?

Bonus: I am wondering whether it is possible to generalize this to something like:

Let $0<t_0<t_1$ and let $n$ be a positive integer. What is the conditional distribution of $N_{t_1}$, given $T_n=t_0$?

• I tried rewriting $T_3=1$ as $\{N[0,1)=2, N[1,1+\epsilon)=1\}$ for suitably small $\epsilon$ and using the fact that $N[1+\epsilon,2)$ is independent of $N[0,1)$ and $N[1,1+\epsilon)=1$, letting then $\epsilon$ tend to 0, but I am not sure if it is correct. – AdrianoMeis Mar 31 '15 at 17:04
• $P(N_2=n|T_3=1)$ means that the third jump was at $t=1$, so you're asking what is the probability of $N$ in $t=2$, given $N_1=3$. Therefore: $P(N_2=n|T_3=1)=P(N_2=n|N_1=3)=P(N_2-N_1=n-3|N_1=3)$ and by the increment independence, this is $P(N_2-N_1=n-3)$, and this is Poisson distribution with parameter $\lambda$ and argument $n-3$. – yoki Mar 31 '15 at 17:05
• @yoki This is what I obtained as a result proceeding as in the comment above, but I wasn't sure it was correct. Thank you! – AdrianoMeis Mar 31 '15 at 17:06
• Look up results on connections between Poisson, exponential, and order statistics of uniform. – BruceET Mar 31 '15 at 17:54


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# Surface Area Of A Sphere

A Sphere is a three-dimensional solid object having a round structure, just like a circle. The difference between a sphere and a circle is that, a circle is in 2-dimension, whereas, a sphere is a 3-dimensional object.

From a visual perspective, it has a three-dimensional structure that forms by rotating a disc that is circular with one of the diagonals. Let us consider an instance where spherical ball faces are painted. To paint the whole surface, the paint quantity required has to be known beforehand. Hence area of every face has to be known to calculate the paint quantity for painting the same. We define this term as the total surface area. Surface area of a sphere is equal to the areas of entire face surrounding it.

Area of a Sphere- It is the amount of region covered by the surface of a sphere.

Mathematical Formula for Surface Area of a Sphere-

 $4\pi r^{2}$

Total Surface Area- It is the sum of total area of all the surfaces be it curved surface or flat surface.

Curved Surface Area- It is the sum of the area of curved surface (leaving the flat bases).

In case of a Sphere, we only have curved surfaces (no flat surface).

Therefore the Total surface area = Curved surface area

 Example- Calculate the cost required to paint a football which is in shape of a sphere having radius of 7 cm. If the painting cost of football is INR 2.5/$cm^{2}$. (Take ? = 22/7)Solution- We know, Total surface area of a sphere = $4\pi r^{2}$ = 4 × (22/7) × 7 × 7 $= 616 \; cm^{2}$ Total cost of painting the container = 2.5 × 616 = Rs. 1540 Example- Calculate the curved surface area of a sphere having radius equals to 3.5 cm.(Take ? = 22/7) Solution- We know, Curved surface area = Total surface area = $4\pi r^{2}$ = 4 × (22/7) × 3.5 × 3.5 $= 616 \; cm^{2}$ $= 154 \; cm^{2}$<

This was just a brief introduction to the calculation of surface area of a sphere. To solve NCERT solutions related to calculation of surface area of a sphere, visit our site BYJU’S.

#### Practise This Question

When we rotate a semi-circle along the axis parallel to its plane. What shape do we obtain?


text:
# ContourPlot or DensityPlot ColorFunction with x, y and z arguments

Is it possible to apply a color function to ContourPlot or DensityPlot that is function of x, y and z, and not simply of z?

Example:

ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi},
ColorFunction -> (If[#1 > 1/2, Red, White] &)]

but colored only for x > 2 Pi (obviously, there are alternative methods for this example, but typical cases are more complex)

-
Not if the docs are correct. ContourPlot appears to pass only a single value. You could try to generate the contours and the background colour separately, then combine them together to achieve the effect. –  Szabolcs Feb 11 '14 at 16:53
@Szabolcs on this case it is easy, for others, only by doing a ContourPlot that outputs the color value on z, and then use it as backbroud, which is clearly a non optimum solution. –  P. Fonseca Feb 11 '14 at 16:55
This is not going to be of as good a quality as ContourPlot, and it's again just a hack: DensityPlot[(x - 2 Pi) Sign[Cos[x] + Cos[y]], {x, 0, 4 Pi}, {y, 0, 4 Pi}, MeshFunctions -> {Function[{x, y}, Cos[x] + Cos[y]]}, Mesh -> 5, PlotPoints -> 100, ColorFunction -> "RedBlueTones", Exclusions -> None] –  Szabolcs Feb 11 '14 at 16:57
The colouring and the contours are treated separately by specifying one as "the function to plot" and the other as "mesh functions". Instead of using mesh function, you could generate the contour lines with ContourPlot for better quality, then combine with Show. –  Szabolcs Feb 11 '14 at 17:00

The problem is that ColorFunction receives only one parameter, the value of the function, when called by ContourPlot. This is explained in the details of the documentation. A crude way to see that would be printing the argument received by ColorFunction :

ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi}
, ColorFunction ->
Function[{x, y, z}, Print[{x, y, z}]; If[x > 1/2, Red, White]]]

So I guess the answer is no, if you want to use ColorFunction on ContourPlot.

Another solution could be to superimpose your ContourPlot into a RegionPlot or another graphic function that would provide {x,y} to the ColorFunction:

Show[
RegionPlot[
Cos[x] + Cos[y] > 0 && 2 x > y, {x, 0, 4 Pi}, {y, 0, 4 Pi},
ColorFunction -> Function[{x, y}, Hue[y]]],
ContourPlot[
Cos[x] + Cos[y] == Range[-2, 2, 0.25], {x, 0, 4 Pi}, {y, 0, 4 Pi}]
]

-


text:
## Traces and Pedersen ideals of tensor products of non-unital C*-algebras

##### Author
Ivanescu, Cristian
Kucerovski, Dan
2019
##### Keywords
C ∗ -algebra , tensor product
##### Abstract (summary)
We show that positive elements of a Pedersen ideal of a tensor product can be approximated in a particularly strong sense by sums of tensor products of positive elements. This has a range of applications to the structure of tracial cones and related topics, such as the Cuntz-Pedersen space or the Cuntz semigroup. For example, we determine the cone of lower semicontinuous traces of a tensor product in terms of the traces of the tensor factors, in an arbitrary C*-tensor norm. We show that the positive elements of a Pedersen ideal are sometimes stable under Cuntz equivalence. We generalize a result of Pedersen’s by showing that certain classes of completely positive maps take a Pedersen ideal into a Pedersen ideal. We provide theorems that in many cases compute the Cuntz semigroup of a tensor product.
##### Publication Information
Ivanescu, C. and Kucerovski, D. (2019). Traces and Pedersen ideals of tensor products of non-unital C*-algebras. New York Journal of Mathematics, 25, 423-450. https://arxiv.org/abs/1811.04430v2
Article
English


text:
# nLab Ek-Algebras

### Context

#### Higher algebra

higher algebra

universal algebra

# Contents

## Idea

This are notes on

• Jacob Lurie, $𝔼\left[k\right]$-Algebras (pdf)

Using the definition of the notion (∞,1)-operad in terms of a vertical categorification of the notion of category of operators, the article discusses the ${𝔼}_{k}$- or little cubes operads and its En-algebras.

A major application in the second part of the article is the study of topological chiral homology.

## Definitions and results

### Grouplike monoid objects

Let $𝒳$ be an ∞-stack (∞,1)-topos and $X:$ Assoc $\to 𝒳$ be a monoid object in $𝒳$. Say that $X$ is grouplike if the composite

${\Delta }^{\mathrm{op}}\to \mathrm{Ass}\to 𝒳$\Delta^{op} \to Ass \to \mathcal{X}

(see 1.1.13 of Commutative Algebra)

is a groupoid object in $𝒳$.

Say an $𝔼\left[1\right]$-algebra object is grouplike if it is grouplike as an Assoc-monoid. Say that an $𝔼\left[k\right]$-algebra object in $𝒳$ is grouplike is the restriction along $𝔼\left[1\right]↪𝔼\left[k\right]$ is. Write

${\mathrm{Mon}}_{𝔼\left[k\right]}^{\mathrm{gp}}\left(𝒳\right)\subset {\mathrm{Mon}}_{𝔼\left[k\right]}\left(𝒳\right)$Mon^{gp}_{\mathbb{E}[k]}(\mathcal{X}) \subset Mon_{\mathbb{E}[k]}(\mathcal{X})

for the (∞,1)-category of grouplike $𝔼\left[k\right]$-monoid objects.

### Main result: $k$-fold delooping, monoidalness and $𝔼\left[k\right]$-action

The following result makes precise for parameterized ∞-groupoids – for ∞-stacks – the general statement that $k$-fold delooping provides a correspondence betwen n-categories that have trivial r-morphisms for $r and k-tuply monoidal n-categories.

###### Theorem (k-tuply monoidal $\infty$-stacks)

Let $k>0$, let $𝒳$ be an ∞-stack (∞,1)-topos and let ${𝒳}_{*}^{\ge k}$ denote the full subcategory of the category ${𝒳}_{*}$ of pointed objects, spanned by those pointed objects thar are $k-1$-connected (i.e. their first $k$ ∞-stack homotopy groups) vanish. Then there is a canonical equivalence of (∞,1)-categories

${𝒳}_{*}^{\ge k}\simeq {\mathrm{Mon}}_{𝔼\left[k\right]}^{\mathrm{gp}}\left(𝒳\right)\phantom{\rule{thinmathspace}{0ex}}.$\mathcal{X}_*^{\geq k} \simeq Mon^{gp}_{\mathbb{E}[k]}(\mathcal{X}) \,.
###### Proof

This is EKAlg, theorem 1.3.6..

Specifically for $𝒳=\mathrm{Top}$, this reduces to the classical theorem by Peter May

###### Theorem (May recognition theorem)

Let $Y$ be a topological space equipped with an action of the little cubes operad ${𝒞}_{k}$ and suppose that $X$ is grouplike. Then $Y$ is homotopy equivalent to a $k$-fold loop space ${\Omega }^{k}X$ for some pointed topological space $X$.

###### Proof

This is EkAlg, theorem 1.3.16.

Lurie’s proof of the equivalence of $n+1$-connected objects with grouplike $E\left[k\right]$-objects is entirely at the level of (∞,1)-categories. One would hope that in addition there is a model for this equivalence at the level of model categories.

There is a model category structure on the category ${\mathrm{Top}}_{*}$ of pointed topological spaces, such that the cofibrant objects are $n$-connected CW-complexes, described in

• J. I. Extremiana Aldana, Luis Javier Hernández Paricio? and
Maria Teresa Rivas Rodriguez, A closed model category for $\left(n-1\right)$-connected spaces , Proc. AMS, 124, Number 11, 1996 (pdf)

### Stabilization hypothesis

A proof of the stabilization hypothesis for k-tuply monoidal n-categories is a byproduct of corollary 1.1.10, stated as example 1.2.3

It has been long conjectured that it should be true that when suitably defined, there is a tensor product of $\infty$-operads such that

${𝔼}_{k}\otimes {𝔼}_{k\prime }\simeq {𝔼}_{k+k\prime }\phantom{\rule{thinmathspace}{0ex}}.$\mathbb{E}_k \otimes \mathbb{E}_{k'} \simeq \mathbb{E}_{k + k'} \,.

This is discussed and realized in section 1.2. The tensor product is defined in appendix B.7.

### Deligne conjecture

Section 2.5 gives a proof of a generalization of the Deligne conjecture.

category: reference

Revised on March 22, 2012 07:55:49 by Tim_Porter (95.147.237.122)


text:
# Determine the number of paths to move from the top-left cell to the bottom-right cell such that there are an even number of direction changes

Determine the number of paths to move from the top-left cell to the bottom-right cell in the $8 \times 6$ (so the height is $6$ units and the length $8$ units) cell grid using a sequence of downwards moves and rightwards moves such that there are an even number of direction changes.

I'm a little lost on this. I know that there are $(8+6)!/(8!\cdot 6!)$ ways to get from the top left corner to the bottom left corner, but otherwise I'm not sure what to do. Of course, I did try to "get my hands dirty" and worked out possible paths that have an even and an odd number of direction changes, but I'm not sure if they have helped me much so far.

Observe that if you first move right, you must end with a move to the right, and if you first move down, you must end with a move down. In the former case, you are left with a grid of $6 \times 6,$ which has $12 \choose 6$ valid moves. In the latter case, you are left with a grid of $8 \times 4,$ which has $12 \choose 8$ valid moves. The number of ways to achieve an even number of direction changes thus equals:

$${12 \choose 6} + {12 \choose 8} = 924 + 495 = 1419$$

Let's generate all lattice paths with $h$ horizontal moves, $v$ vertical moves and $c$ changes in direction. The number of such lattice paths will be the $x^hy^vz^c$ coefficient of some generating function series $f(x,y,z)$.

This function can be further partitioned into a function of all such paths ending with a horizontal move $f_x(x,y,z)$, a function of all such paths ending with a vertical move $f_y(x,y,z)$ and the empty path $1$.

$$f=1+f_x+f_y\, .\tag{1}$$

Now, paths ending in a horizontal move $f_x$ are either

• a single move $x$,
• a path ending in a horizontal move $f_x$ followed by another horizontal move $x$,
• a path ending in a verical move $f_y$ followed by a horizontal move $x$ and therfore also a change in direction $z$.

\begin{align} && f_x&=x+f_xx +f_yxz\\[1ex] &\implies & f_x&=\frac{x+f_yxz}{1-x}\, .\tag{2}\end{align}

Similarly we can argue for $f_y$:

\begin{align} &&f_y&=y+f_yy+f_xyz\\[1ex] &\implies & f_y&=\frac{y+f_xyz}{1-y}\, .\tag{3}\end{align}

We can substitute $(3)$ in $(2)$ eliminating $f_y$ to give:

\begin{align} &&f_x&=\frac{x}{1-x}+\frac{xyz}{(1-x)(1-y)}+\frac{f_xxyz^2}{(1-x)(1-y)}\\[1ex] &\implies &f_x&=\frac{x(1-y)+xyz}{(1-x)(1-y)-xyz^2}\, .\tag{4}\end{align}

Similarly by eliminating $f_x$ we get

$$f_y=\frac{y(1-x)+xyz}{(1-x)(1-y)-xyz^2}\, ,\tag{5}$$

then inputting $(4)$ and $(5)$ into $(1)$ and simplifying gives:

$$f(x,y,z)=\frac{1-xy(1-z)^2}{1-x-y+xy-xyz^2}\, .\tag{6}$$

This is our generating function for lattice paths with direction changes.

We could, at this point, use a computer algebra system such a sage to find the coefficients of all $x^8y^6z^{\text{even}}$ in $(6)$ then add them to get our answer. However, there is a useful trick we can use called the Discrete Fourier Transform (DFT) which will make our answer easy to calculate.

We note that the desired solution would be the $x^8y^6$ coefficient of

\begin{align}\tfrac{1}{2}(f(x,y,1)+f(x,y,-1))&=\frac{1-2xy}{1-(x+y)}\\[1ex] &=\sum_{k\ge 0}(x+y)^k-2xy\sum_{k\ge 0}(x+y)^k\end{align}\tag{7}

since we are using the square-roots of unity $\{1,-1\}$ odd powers of $z$ cancel and we sum the even coefficients.

So, taking the required $x^8y^6$ coefficient with the operator $[x^8y^6]$ on $(7)$:

\begin{align}[x^8y^6]\tfrac{1}{2}(f(x,y,1)+f(x,y,-1))&=[x^8y^6]\sum_{k\ge 0}(x+y)^k-2[x^7y^5]\sum_{k\ge 0}(x+y)^k\\[1ex] &=\binom{8+6}{6}-2\binom{7+5}{5}\\[1ex] &=\binom{14}{6}-2\binom{12}{5}\\[1ex]& =1419\tag{Answer}\end{align}

• A very complete answer (+1). I would like to point out, however, that the result in the second to last line can intuitively be reasoned as well: from all possible paths, subtract those in which you start with a downward movement and end with a move towards the right, and those in which you start with a move towards the right and end with a downward movement. – jvdhooft Jun 21 '18 at 19:39
• A worthwhile contribution, thank you! Buried in my method as I was, I didn't take a step back to make that observation, but it's as clear as day now you mention it. Thanks for the upvote too! I had already done so for your beautifully succinct answer. :) – N. Shales Jun 21 '18 at 19:48


text:
## Chapter Chosen

Mechanical Properties of Solids

## Book Store

Currently only available for.
CBSE Gujarat Board Haryana Board

## Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12
The stress-strain graph for a metal wire is shown in the figure. Up to the point E, the wire returns to its original state O along the curve EPO, when it is gradually unloaded. Point B corresponds to the fracture of the wire.

(a) Upto what point on the curve is the Hooke’s law obeyed? (This point is sometimes called ‘Proportional limit’).

(b) Which point on the curve corresponds to ‘elastic limit’ or yield point of the wire?

(c) Indicate the elastic and plastic regions of the stress-strain graph.

(d) Describe what happens when the wire is loaded upto a stress corresponding to the point A on the graph, and then unloaded gradually. In particular, explain the dotted curve.

(e) What is peculiar(odd) about the portion of the stress-strain graph from C to B? Upto what stress can the wire be subjected without causing fracture?

(a) Hooke's law states that that stress is directly proportional to strain. From the graph, we can say that Hooke’s law is obeyed from point O to P.

(b) E point on the curve corresponds to elastic limit or the yield point.

(c) Elastic region is the region in which the body regains its state after removal of stress and is from O to E. Plastic region is the region in which there is residue strain after removing the stress and that region is E to B.

(d) From O to P, the strain is directly proportional to stress, beyond P increase in strain is more for same increase in stress. Beyond E, when stress is increased, it does not retrace its path on decreasing the stress, but returns to point O' along path AO'. Therefore there is residue strain of 00' when stress is increased corresponding to the point A.

(e) From C to B, strain increases even if the wire is being unloaded and at B, it fractures. Till point C, stress can be applied without causing the fracture.

476 Views

What do you understand by intermolecular forces?

Intermolecular force is the force between the molecules of the substance is called intermolecular force.

There is a dipole-dipole interaction between the molecules.
630 Views

Define interatomic forces and give an example?

The force between atoms of the molecule is called interatomic force.

E.g.The nucleus and electron cloud of a molecule is binded by electrostatic force.
1032 Views

What is the nature of inter-atomic or inter-molecular forces?

Inter-atomic or inter-molecular forces are electrostatic in nature.

637 Views

What is the force of adhesion?

Adhesive force is the attractive force between two unlike molecules. They are caused by forces acting between two substances like electrostatic force (attraction due to opposing charges), mechanical force etc.

E.g., Force of attraction between gum and paper.

587 Views

What is the force of cohesion?

Cohesive attraction or cohesive force is the force of attraction between same types of molecules sticking together.
651 Views


text:
I'm working in Visual Basic.NET

My questions are

1. Is there a better place / way to show the MessageBox.Show?

2. I'm being forced to use a single Oracle Connection, in my sample cnn is declared with Friend scope so the entire solution can see it. Is there a better way to handle this Connection Object, perhaps I should pass it all about my app as a parameter?

3. The Sub SaveJobInfo I put this in a Sub so it wouldn't clutter up the ValidateLoginInfo code, was wondering if there is a better way to check for Nulls other than using the .Size property?

4. Any other ideas that can improve this code?
Public Class LoginValidator

Friend Shared Function ValidateLoginInfo(jobNum As String, userID As String, ByRef djnInfo As JobNumInfo) As Boolean
Try

Dim cmd As New OracleCommand() With {.Connection = cnn, .CommandType = CommandType.StoredProcedure, .CommandText = "owner.packageName.procedureName"}

cmd.Parameters.Add(New OracleParameter("param4", OracleDbType.Varchar2, 20, String.Empty, ParameterDirection.Output))
cmd.Parameters.Add(New OracleParameter("param5", OracleDbType.Varchar2, 20, String.Empty, ParameterDirection.Output))
cmd.Parameters.Add(New OracleParameter("param6", OracleDbType.Varchar2, 20, String.Empty, ParameterDirection.Output))
cmd.Parameters.Add(New OracleParameter("param7", OracleDbType.Varchar2, 10000, String.Empty, ParameterDirection.Output))
cmd.Parameters.Add(New OracleParameter("param8", OracleDbType.Varchar2, 10, String.Empty, ParameterDirection.Output))
cmd.Parameters.Add(New OracleParameter("param9", OracleDbType.Varchar2, 20, String.Empty, ParameterDirection.Output))
cmd.Parameters.Add(New OracleParameter("param10", OracleDbType.Varchar2, 10, String.Empty, ParameterDirection.Output))
cmd.Parameters.Add(New OracleParameter("param11", OracleDbType.Varchar2, 2, String.Empty, ParameterDirection.Output))
cmd.Parameters.Add(New OracleParameter("param12", OracleDbType.Varchar2, 50, String.Empty, ParameterDirection.Output))
cmd.Parameters.Add(New OracleParameter("param13", OracleDbType.Varchar2, 1500, String.Empty, ParameterDirection.Output))

cmd.ExecuteNonQuery()

SaveJobInfo(djnInfo, cmd)

For Each p As OracleParameter In cmd.Parameters
p.Dispose()
Next

Catch ex As Exception
Return False
End Try
Return True

End Function

Private Shared Sub SaveJobInfo(ByVal djnInfo As JobNumInfo, ByVal cmd As OracleCommand)

If cmd.Parameters(4).Size > 0 Then
djnInfo.UserAccessLevel = cmd.Parameters(4).Value.ToString
Else
djnInfo.UserAccessLevel = String.Empty
End If

If cmd.Parameters(5).Size > 0 Then
djnInfo.HasChildren = cmd.Parameters(5).Value.ToString.ToUpper = "YES"
Else
djnInfo.HasChildren = False
End If

If cmd.Parameters(6).Size > 0 Then
djnInfo.IsReleased = cmd.Parameters(6).Value.ToString.ToUpper = "YES"
Else
djnInfo.IsReleased = False
End If

If cmd.Parameters(7).Size > 0 Then
djnInfo.SpecInfo = cmd.Parameters(7).Value.ToString
Else
djnInfo.SpecInfo = String.Empty
End If

If cmd.Parameters(8).Size > 0 Then
djnInfo.OrgCode = cmd.Parameters(8).Value.ToString
Else
djnInfo.OrgCode = String.Empty
End If

If cmd.Parameters(9).Size > 0 Then
djnInfo.UOMFromOra = cmd.Parameters(9).Value.ToString
Else
djnInfo.UOMFromOra = String.Empty
End If

If cmd.Parameters(10).Size > 0 Then
djnInfo.DrawingNumber = cmd.Parameters(10).Value.ToString
Else
djnInfo.DrawingNumber = String.Empty
End If

If cmd.Parameters(11).Size > 0 Then
djnInfo.ProductRevisionCode = cmd.Parameters(11).Value.ToString
Else
djnInfo.ProductRevisionCode = String.Empty
End If

If cmd.Parameters(12).Size > 0 Then
djnInfo.FirstArticleFamCode = cmd.Parameters(12).Value.ToString
Else
djnInfo.FirstArticleFamCode = String.Empty
End If
' param 13 will be handled differently than the others
End Sub

End Class

• I had about 4 more 'fluffs' i was going to add to this original question so I could could use the formatting. Would it make more sense to make new questions for my other 'fluffs'? – Rose Oct 8 '15 at 15:49
• As long as your new question contains working, non-hypothetical code that you want reviewed, then ask away! :-) – Ethan Bierlein Oct 8 '15 at 15:50
• It would be nice if I could keep it all in one question. I'm not sure how to add to this question and have the option to 'format' my fluff additions? – Rose Oct 8 '15 at 15:59
• Please just ask a new question. It's easier for everyone on the site, and it doesn't invalidate answers. – Ethan Bierlein Oct 8 '15 at 15:59

• My first impression is that this class is doing too much. It is calling the database and manipulating the user interface. This needs to be separated.
• Exceptions are being swallowed and shown to the user, and a boolean value is returned.

If an exception gets thrown in code calling to the database, let the exception bubble up. This is a catastrophic problem. If validating the login info is throwing an exception, write the database code so that it returns a flag if someone's login info isn't correct. Don't use exceptions for flow control.

• The ValidateLoginInfo method calls SaveJobInfo. After validating login info, you shouldn't be saving data to the database. That should be a separate operation.

• I'm also questioning why you need to explicitly dispose of the OracleParameter objects. The CLR should clean those up just fine on its own.

• The SaveJobInfo method is confusing. It's taking stored procedure parameter values and mapping them to a JobInfo object? This should be done after validating the login info. The stored procedure you are calling has a lot of output parameters. The validation of login info and loading data into the JobInfo object should be two separate calls. One does not have anything to do with the other.

@Rose commented:

If I changed the Exception handling and Caught any exceptions and simply Threw them back to whatever was calling ValidateLoginInfo, would this meet the 'bubbling up' definition?

If you are doing this:

Try
// ...
Catch ex As Exception
throw ex
End Try

Then there is no point in using a Try-Catch. Just execute the code and let the exceptions fly.

By "let the exception bubble up" I mean "don't use a Try-Catch if you can't recover from the exception." Unless your ValidateLoginInfo catches an exception and can recover from it, don't catch exceptions.

• Thank you for the feedback I really appreciate it. In regards to an exception being thrown in the code calling to the database and letting the exception 'bubble up'. . . Please see edit to original question. If I changed the Exception handling and Caught any exceptions and simply Threw them back to whatever was calling ValidateLoginInfo, would this meet the 'bubbling up' definition? – Rose Oct 8 '15 at 14:54


text:
# STAT and ATP¶

## STAT¶

STAT (the Stack Trace Analysis Tool) is a highly scalable, lightweight tool that gathers and merges stack traces from all of the processes of a parallel application. The results are then presented graphically as a call tree showing the location that each process is executing.

This is a useful tool for debugging an application that hangs because collected call backtraces can quickly tell you where each process is executing at the moment in the code, providing a hint on where to look further for more detailed analysis.

It supports distributed-memory parallel programming only such as MPI, Coarray Fortran and UPC (Unified Parallel C).

One way to collect backtraces under Slurm is explained below.

1. Start an interactive batch job and launch an application in background. Keep the process ID (PID).

nersc$salloc -N 1 -t 30:00 -q debug [...other flags...] ... nersc$ srun -n 4 [...other flags...] ./jacobi_mpi &
[1] 95298

You can also see the PID by running the 'ps' command:

nersc$ps PID TTY TIME CMD 95018 pts/0 00:00:00 bash 95298 pts/0 00:00:00 srun 95302 pts/0 00:00:00 srun 95325 pts/0 00:00:00 ps  2. Load the module stat nersc$ module load stat

3. Run the stat-cl command on this process. You may want to use the -i flag to gather source line numbers, too:

nersc$stat-cl -i 95298 STAT started at 2016-11-30-07:33:37 Attaching to job launcher (null):95298 and launching tool daemons... Tool daemons launched and connected! Attaching to application... Attached! Application already paused... ignoring request to pause Sampling traces... Traces sampled! ... Resuming the application... Resumed! Merging traces... Traces merged! Detaching from application... Detached! Results written to /global/cscratch1/sd/wyang/parallel_jacobi/stat_results/jacobi_mpi.0004  stat-cl takes several backtrace samples after attaching to the running processes. The result file is created in the stat_results subdirectory under the current working directory. This subdirectory contains another subdirectory whose name is based on your parallel application's executable name that contains the merged stack trace file in DOT format. 4. Then, run the GUI command, STATview, with the file above to visualize the generated *.dot files for stack backtrace information. nersc$ STATview stat_results/jacobi_mpi.0004/00_jacobi_mpi.0004.3D.dot

Please note that, if you're running on Cori KNL nodes, you have to go to a login node for this step (after loading the stat module there). Otherwise, fonts will not be shown correctly.

The above call tree diagram reveals that rank 0 is in the 'init_fields' routine (line 172 of jacobi_mpi.f90), rank 3 in the 'set_bc' routine (line 214 of the same source file), and the other ranks (1 and 2) are in the MPI_Sendrecv function. If this pattern persists, it means that the code hangs in these locations. With this information, you may want to use a full-fledged parallel debugger such as DDT or TotalView to find out why your code is stuck in these places.

## ATP¶

Another useful tool in the same vein is ATP (Abnormal Termination Processing) that Cray has developed. ATP gathers stack backtraces when the code crashes, by running STAT before it exits.

The 'atp' module is load by default on Cray systems, but it is not enabled. To enable it so that it generates stack backtrace info upon a failure, set the following environment variable before your aprun command in your batch script:

setenv ATP_ENABLED 1          # for csh/tcsh

export ATP_ENABLED=1          # for bash/sh/ksh

In addition, you need to set the FOR_IGNORE_EXCEPTIONS environment variable if you're using Fortran and you have built with the Intel compiler:

setenv FOR_IGNORE_EXCEPTIONS true   # for csh/tcsh

export FOR_IGNORE_EXCEPTIONS=true   # for bash/sh/ksh

If your Fortran code is built with the GNU compiler, you will need to link with the -fno-backtrace option.

When atp is loaded no core file will be generated. However, you can get core dumps (core.atp.<apid>.<rank>) if you set coredumpsize to unlimited:

unlimit coredumpsize   # for csh/tcsh

ulimit -c unlimited    # for bash/sh/ksh

More information can be found in the man page: type man intro_atp or, simply, man atp.

The following is to test ATP using an example code available in the ATP distribution package:

nersc$cp$ATP_HOME/demos/testMPIApp.c .
nersc$cc -o testMPIApp testMPIApp.c nersc$ cat runit
#!/bin/bash
#SBATCH -N 1
#SBATCH -t 5:00
#SBATCH -q debug

export ATP_ENABLED=1
srun -n 8 ./testMPIApp 1 4

nersc$sbatch runit Submitted batch job 3044170 nersc$ cat slurm-3044710.out
[snip]
testApp: (31929) starting up...
testApp: (31932) starting up...
testApp: (31930) starting up...
testApp: (31933) starting up...
Application 3044170 is crashing. ATP analysis proceeding...

ATP Stack walkback for Rank 3 starting:
_start@start.S:122
__libc_start_main@libc-start.c:285
main@testMPIApp.c:76
raise@pt-raise.c:37
ATP Stack walkback for Rank 3 done
Process died with signal 4: 'Illegal instruction'
View application merged backtrace tree with: STATview atpMergedBT.dot
You may need to: module load stat

srun: error: nid00009: tasks 0-3: Killed
srun: Terminating job step 3044170.0
srun: Force Terminated job step 3044170.0
[snip]

ATP creates a merged backtrace files in DOT fomat in atpMergedBT.dot and atpMergedBT_line.dot. The latter shows source line numbers, too. To view the collected backtrace result, you need to load the stat module and run STATview:

nersc$module load stat nersc$ STATview atpMergedBT.dot

Note

The following feature doesn't work at the moment (as of March 22, 2018). Cray is working on the issue.

ATP can be a useful tool in debugging a hung application, too. You can force ATP to generate backtraces for a hung application by killing the application. To do that, you should have done necessary preparational work such as setting the ATP_ENABLED environment variable, etc. in the batch script for the job in question.

nersc$sacct -j 3169879 # find job step id for the application - it's 3169879.0 JobID JobName Partition Account AllocCPUS State ExitCode ------------ ---------- ---------- ---------- ---------- ---------- -------- 3169879 runit knl mpccc 544 RUNNING 0:0 3169879.ext+ extern mpccc 544 RUNNING 0:0 3169879.0 jacobi_mp+ mpccc 4 RUNNING 0:0 3169879.1 cti_dlaun+ mpccc 2 RUNNING 0:0 nersc$ ssh cmom02                      # You have to be on a MOM node

nersc$scancel -s ABRT 3169879.0 # Kill the application nersc$ exit                            # Back to the login node

nersc$cat slurm-3169879.out Application 3169879 is crashing. ATP analysis proceeding... ATP Stack walkback for Rank 0 starting: _start@start.S:122 __libc_start_main@libc-start.c:285 main@0x40a59d MAIN__@jacobi_mpi.f90:174 ATP Stack walkback for Rank 0 done Process died with signal 6: 'Aborted' View application merged backtrace tree with: STATview atpMergedBT.dot You may need to: module load stat [snip] nersc$ STATview atpMergedBT_line.dot

Note

Cori's MOM node names are cmom02 and cmom05.

The above example is to use SIGABRT in killing the application. There are other signals accepted by ATP. For info, please read the atp man page.


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# A system has a receiver noise reactance of 50 Ω. It is connected to an antenna with an output resistance of 50 Ω. The noise figure of the system is:

This question was previously asked in
ESE Electrical 2013 Paper 2: Official Paper
View all UPSC IES Papers >
1. 1
2. 2
3. 50
4. 101

Option 2 : 2
Free
CT 3: Building Materials
2480
10 Questions 20 Marks 12 Mins

## Detailed Solution

Concept:

Noise figure (NF) or noise factor (F) are measures of degradation of the signal-to-noise ratio (SNR), caused by components in a signal chain.

The noise figure is given as:

$$NF= 1 + \frac{{{R_{eq}}}}{{{R_s}}}$$

Req = Equivalent resistance of antenna

Rs = Noise resistance of the system

Calculation:

With Rs = 50 Ω, and Req = 50 Ω, the noise figure will be:

$$NF= 1 + \frac{{{50}}}{{{50}}}$$

NF = 2


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## Irrationality proofs using modular forms.(English)Zbl 0613.10031

Journées arithmétiques, Besançon/France 1985, Astérisque 147/148, 271-283 (1987).
[For the entire collection see Zbl 0605.00004.]
Partly from the author’s introduction: In the years following Apéry’s discovery of his irrationality proofs for $$\zeta$$ (2), $$\zeta$$ (3) it has become clear that these proofs do not only have significance as irrationality proofs, but the numbers that occur in them serve as interesting examples for several phenomena in algebraic geometry and modular form theory (see the citations in the paper). Furthermore it turns out that Apéry’s proofs themselves are simple consequences of elementary complex analysis on spaces of certain modular forms. Although the use of modular forms looks promising at first sight, the yield of new irrationality proofs is disappointingly low. However it is easy to overlook some simple tricks that may give new interesting results.
This paper describes the general framework of the proofs and gives, using the present approach, a very detailed proof of the irrationality of $$\zeta$$ (3) as well as more summary proofs of additional results. This work goes some considerable distance towards giving a sophisticated answer to the reviewer’s question about Apéry’s proofs [Math. Intell. 1, 195-203 (1979; Zbl 0409.10028)]: ”What on earth is going on here?”

### MSC:

 11J81 Transcendence (general theory) 11F11 Holomorphic modular forms of integral weight

### Citations:

Zbl 0605.00004; Zbl 0409.10028


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# Besov and Triebel-Lizorkin spaces

Let me start with a couple of notational reminders. For $\xi\in \mathbb R^n$, $$1=\varphi_{0}(\xi)+\sum_{\nu \ge 1}\varphi_{\nu}(\xi),\quad \varphi_{0}\in C^\infty_c(\mathbb R^{n}),\quad \varphi_{\nu}=\varphi(\frac{\xi}{2^{\nu}}),$$ where $\varphi\in C^\infty_c(\mathbb R^{n}), \text{supp}\varphi=\{1/4\le \vert \xi\vert \le 4\}\supset\varphi^{-1}(\{1\})= \{1/2\le \vert \xi\vert \le 2\}$. We have \begin{align} \Vert{u}\Vert_{B^{s}_{p,q}}=\bigl\Vert{(2^{\nu s}\Vert{\varphi_{\nu}(D) u}\Vert_{L^{p}})_{\nu\ge 0}}\bigr\Vert_{\ell^{q}(\mathbb N)} \quad\text{(Besov spaces)},\\ \Vert{u}\Vert_{F^{s}_{p,q}}=\bigl\Vert{ \Vert(2^{\nu s}\vert{\varphi_{\nu}(D) u}\vert )_{\nu\ge 0}\Vert_{ \ell^{q}(\mathbb N) }}\bigr\Vert_{L^{p}(\mathbb R^n)} \quad\text{(Triebel-Lizorkin spaces).} \end{align} It is easy to see that for $1<q\le p<+\infty$, we have $B^s_{p,q}\subset F^s_{p,q}\subset B^s_{p,p}$ and thus $F^s_{p,p}=B^s_{p,p}$.

My question: is it true that $$B^s_{p,p}=W^{s,p}=\{u\in \mathscr S'(\mathbb R^n), (1+\vert D\vert^2)^{s/2} u\in L^p\} ?$$

Your space $$W^{s,p}$$ is the same as $$F^s_{p2}$$, not $$F^s_{pp}$$.
See these lecture notes. The definition of $$H^{s,p}$$ (which you call $$W^{s,p}$$) is given on page 34 and the result is on page 52.


text:
Comment
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Q)

# What is molarity of a solution contain $5.84\;gms$ of Nacl in 200 ml of solution ?

$(a)\;0.5\;m \\ (b)\;25\;m \\(c)\;90\;m \\(d)\;0.1\;m$


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# The distance between two stations is $240\ km$. A train takes 4 hours to cover this distance. Calculate the speed of the train.

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Given: The distance between two stations is 240 km and a train takes 4 hours to cover the distance.

To find: To find the speed of the train.

Solution:

Speed is the distance covered by a moving object per unit of time.

The formula for calculating speed:

$\boxed{Speed=\frac{distance}{time}}$

Here, the distance traveled by train $=240\ km$

Time taken to cover the distance $=4\ hours$

Therefore, the speed of the train $=\frac{240\ km}{4\ hr}$

$=60\ km/h$

Hence, the speed of the train is $60\ km/h$.

Updated on 10-Oct-2022 13:25:51


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# What is the difference between melting and dissolving?

What is the difference between melting and dissolving? I am looking some general features.

The answer should be adaptable to the melting/dissolving of ice cube (water) in a class of pure alcohol (ethanol) just below (or at the) melting point of ice, or similar phenomena.

I am now assuming that the ice is dissolving and melting at the same time.

In other words which reaction energy is higher in the following reactions:

\begin{align} \ce{H_2O(s) &\to H_2O(l)\\ H_2O(s) + n EtOH(l) &\to H_2O \centerdot (EtOH)_{n}} \end{align}

Or, are there substances that release more energy in dissolvation than consume in melting?

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Hi Juha! Melting and dissolution are two very different physicochemical processes: for one thing, melting is a pure-component equilibrium between the solid phase and liquid phase of the same component, while dissolution involves a solid and a solvent, which is a different chemical compounds. As it currently is, I'm afraid your question is too broad… Could you read up a bit on both melting and dissolution, and edit it to something more specific? –  F'x May 9 '12 at 9:59
Ice melts, and salt dissolves in water. Ice is made of $H_2 O$ which is the same as water. –  picakhu May 9 '12 at 12:20
What I was actually wondering that is there a substance that exists in two different phases and the first dissolves to the second? or is this always a phase transition (e.g. melting)? In other words, the only difference between melting and dissolving is that the substances are different. –  Juha May 9 '12 at 20:13
One more question: Can alcohol dissolve ice? –  Juha May 9 '12 at 20:27
@Juha regarding melting being endothermic, Wikipedia says “ The enthalpy of fusion is almost always a positive quantity; helium is the only known exception.[1] Helium-3 has a negative enthalpy of fusion at temperatures below 0.3 K. Helium-4 also has a very slightly negative enthalpy of fusion below 0.8 K. This means that, at appropriate constant pressures, these substances freeze with the addition of heat.” I didn't know of this exception, and doubt most chemists would. In short, for a chemist: melting is endothermic. –  F'x May 9 '12 at 21:25

## 4 Answers

Juha invited me to write a summary (see comment on my previous answer ) of the differences between melting and dissolving. I’ll try to outline this roughly in the same order as his. I'm giving this as a new answer since my last answer was quite long as it was.

Differences:

• Melting and dissolving are completely different processes on the molecular/atomic level that could not be mistaken for each other if you could observe what was happening at that scale.
• If you cool the liquid that arose from melting the solid, to a temperature below it’s melting point, you would see the entire sample solidify. If you cool the solution (dissolved solid + solute) to below the melting point of the solid (solute) in it, you would see no change. (Unless you had a saturated or nearly saturated solution.)
• Melting requires only a single substance and energy input while dissolving requires a solvent and a solute that are compatible (“like forces”). (This is actually a pretty huge difference and would potentially affect all of the physical and chemical properties.) Dissolving a solid can be either endo- or exo-thermic.
• "Phase change" (wording from the previous summary) In each case you end up with a liquid. Melting caused a phase change (composition of the substance didn’t change) while dissolving a solid in a liquid is not considered a phase change since a change in composition occurred.

Similarities:

• In each case, forces between the particles that comprise the solid are disrupted and that takes energy. (Whether it’s chemical bonds or intermolecular forces depends on the process and on the solid and on your definitions. (See this question.) But melting (rare exception noted in previous comments) is endothermic and dissolving can be either endo- or exo-thermic.
• In each case you end up with a liquid. Macroscopically, if you walked into a room and saw the liquid on the table, it would be difficult to say whether this liquid came from a solid that had melted or a solid that had dissolved in a solute and made a solution. But it would be very easy to determine which you had experimentally in a “dozen” different ways.
• Both melting and dissolving require interaction among groups of atoms, molecules, or ions.

There are probably more differences that could be given (how to handle thermodynamic calculations, complexity of the system etc.), and possibly more similarities, but that's enough for me on this topic.

-

Some additional information relevant to this question.

I'm going to use "forces" and "particles" in this explanation to generalize it. Particles can be ions (ionic compounds held together by strong electrostatic forces), metal atoms (held together by electrostatic forces from attraction of the metal atom nuclei for the "sea of electrons") or molecules (for covalent compounds held together in the solid or liquid by intermolecular forces: London dispersion forces, dipole-dipole attractions, and hydrogen bonding. Note: we are not talking about the intramolecular covalent bonds in this explanation.) So, solids and liquids are held together in that form by forces between the particles.

The reason that melting is always endothermic (except for the exceptions given by F'x...which I also did not know about) is that when you melt a substance, you have to break up some of the forces holding the particles together in the solid so that they can move past each other and form a liquid. This takes energy, the "heat of fusion" (usually given on a per mole basis) mentioned above. The stronger the forces, the more energy it would take to melt the substance. As long as you are melting a substance and holding the temperature at the melting point (i.e. not adding enough energy to raise the temperature) then you will have an equilibrium between the solid and liquid.

I think there is some confusion in the terms you are using. Desolvation means removing the solvent from a solute-solvent mixture. (This can happen in crystallization, for example.) Solvation (which I think you're asking about) is the interaction of solvent particles with solute particles. Dissolving or dissolution is the process in which a solid, liquid, or gas forms a solution with a solvent.

So, as answered by Ashu, dissolution and be either exothermic or endothermic as he's noted above. This is true because in order to dissolve a substance in a solute, it takes energy to break up the solvent-solvent interactions, it takes energy to break up the solute-solute interactions, and you get energy back from new solvent-solute interactions. If you have stronger attractions between the solvent and solute than you had originally, then the process releases energy and is exothermic. The same concepts can be applied to mixing two liquids...which is usually referred to by saying they are "miscible" instead of speaking of one dissolving in the other. Ethanol and water are miscible in all proportions.

Can alcohol dissolve ice ("at or just below the melting point of ice" as phrased in your original question)? There are a couple of ways to answer this. If you put ice in ethanol at 0 degrees (its melting point) it will melt because the melting point of ice in a mixture of water and ethanol is lower than the melting point of pure water. The melting point depends on the proportions of water and ethanol.(That is to say, as some melted, it could not refreeze as you would expect to happen in an equilibrium mixture, so it would continue to melt.) See colligative properties for more information.

You've given two equations, H2O(s)→H2O(l) H2O(s)+nEtOH(l)→H2O⋅EtOHn In either case, you will need to provide the heat of fusion to melt the ice. So the question you asked is really about the energy of solvation of water in ethanol. When you mix water and ethanol, the flask feels warmer, so this is an exothermic process; you get more energy back than you put into it because of good interactions (hydrogen bonding) between water and ethanol particles. That solvation energy would provide some of the energy needed to melt the ice.

For the broader question "can something happen at a given temperature", you're generally asking if it happens spontaneously. To answer this you'd have to calculate the Gibbs Free Energy. This would take into account not only the energy considerations spoken of above, but also entropy considerations (which would matter here). If the Gibbs Free Energy value is negative, then the process or reaction is spontaneous at the temperature given. So, the answer to which would be the preferred process would also depend on entropy considerations.

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If I make some summary: In both cases there is an Energy barrier that you have to break. In melting, only one exothermic reaction is known. To me the processes look very similar. If you have two exothermic reactions, one where a solid material X melts in a liquid Y into mixture of liquids X and Y and a second where solid material X dissolves into liquid Y forming a solution of X and Y. How can you tell did it melt or dissolve? Is it just that the reaction components are different (mixture of liquids vs. solution)? –  Juha May 11 '12 at 11:56
@Juha: www.thefreedictionary.com/dissolvedis·solve (d -z lv ). v. dis·solved, dis·solv·ing, dis·solves. v.tr. 1. To cause to pass into solution: dissolve salt in water. 2. To reduce (solid matter) to liquid form; melt. –  Janice DelMar May 11 '12 at 15:44
@Juha..So, perhaps the distinction is whether or not there is a solvent present with a bit of "are you above the solids melting point" thrown in. Sugar dissolves in water at room temperature, even if you hold that temperature constant...it doesn't melt. So, if I'm understanding you, your second equation was asking if ice (solid water) would dissolve in ethanol (mp -114) at something like -50 degrees C? –  Janice DelMar May 11 '12 at 15:52
Rereading your comment more carefully: If you were near the melting point of the solid and the two liquids mixed, it would be difficult (impossible?) to tell and could be a mixture of both. (If the liquids were immiscible, then you could watch the solid melt as a separate layer.) If you were at a temperature much lower than the melting point of the solid, and it went into solution, then it dissolved. –  Janice DelMar May 11 '12 at 21:15
So, can I say that melting and dissolving are so similar in some cases that they cannot be separated from each other? And if you would want to demonstrate this, which materials you would use and why? (I can ask this in another thread if this becomes too broad for this) –  Juha May 12 '12 at 16:34

To summarize the differences and similarities, here is a list (I will update the list, if new comments arrive):

differences

• melting is (almost) always exothermic, dissolving can be endo- or exothermic
• melting is within one substance, dissolvation is between solute and solvent that are different substances
• melting is a collective phenomenon that requires $N_a$ particles, dissolvation can happen one atom at the time

similarities

• In both cases, bonds are broken by crossing an energy barrier
• phase of matter changes
• You can create a situation where it is difficult to say whether the substance is melting or dissolving.
-

Is dissolving always exothermic?

No, it is not always exothermic. For example, dissolving $\ce{NaOH}$ is an exothermic reaction, while dissolving $\ce{NH4NO3}$ is an endothermic reaction (This is from my personal experience)

Is melting always endothermic?

Yes, it always is (except for the exception F'x gave). This is because during melting the body to be melted takes in "latent heat of fusion" from the surroundings, thus bringing down the temperature. The latent heat of fusion is required to break the intermolecular attraction forces that define a solid. Breaking such bonds requires energy, hence the reaction is endothermic.

Can alcohol dissolve ice?

Lower alcohols like methanol and ethanol may be able to dissolve ice. They can effectively form hydrogen-bonds with water, but higher alcohols may not be able to do that.

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### The Impact Of Network Size And Mobility On Information Delivery In Cognitive Radio Networks

5

4

3

2

1

Basic infomation
Authors: Lei SunWenye WangYujin Li
Published in: IEEE Transactions On Mobile Computing
Year: 2016
Citations: 0
EI: NO
J-Index: 0
Abstract:
there have been extensive works on the design of opportunistic spectrum access and routing schemes to improve spectrum efficiency in cognitive radio networks \( crns \) , which becomes an integral component in the future communication regime nonetheless , the potentials of crns in boosting network performance yet remain to be explored to reach the full benefits of such a phenomenal technique in this paper , we study the end to end latency in crns in order to find the sufficient and necessary conditions for real time applications in finite networks and large scale deployments we first provide a general mobility framework which captures most characteristics of the existing mobility models and takes spatial heterogeneity into account under this general mobility framework , secondary users are mobile with an mobility radius alpha , which indicates how far a mobile node can reach in spatial domain we find that there exists a cutoff point on alpha , below which the latency has a heavy tail and above which the tail of the latency is bounded by some gamma distributions as the network grows large , the latency is asymptotically scalable \( linear \) with respect to the dissemination distance \( e g , the number of hops or euclidean distance \) an interesting observation is that although the density of primary users adversely impacts the expected latency , it makes no influence on the dichotomy of the latency tail in finite networks and the linearity of latency in large networks our results encourage crn deployment for real time and large applications , when the mobility radius of secondary u ers is large enough
Popular Words:


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+0

# graphing lines

0
41
1
+316

Given points $P(-2,7)$ and $Q(4,y)$ in a coordinate plane, for what value of $y$ is the slope of the line through $P$ and $Q$ equal to $\frac{-3}{2}$?

Jul 4, 2020

#1
+640
+1

We plug in what is given into the slope formula and solve for y.

$$\frac{y-7}{4-(-2)}=-\frac{3}{2}\\ \frac{y-7}{6}=-\frac{3}{2}\\ y-7=-9\\ \boxed{y=-2}.$$

Jul 4, 2020


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# nLab Lev Pontrjagin

Lev Semёnovič Pontrjagin (Лев Семёнович Понтрягин) was an influential Soviet mathematician working in Moscow. His main works till the late 1940s were in topology, especially the study of topological groups (including study of Pontrjagin duality) and algebraic topology in which he also explored applications of differentiable manifolds for computation of homotopy groups; including the introduction of framed cobordism in that work. After the success of the French mathematical school in the late 1940s in introducing new methods of sheaf theory, which Pontrjagin did not like as less direct than the more intuitive geometric methods he was master of, and after some pressure from the government, Pontrjagin switched into applied mathematics where he did some fundamental work, especially in optimization theory.

Pontrjagin was a full member of the (Soviet) Academy of Sciences. He was blind since the age of 14.

## Selected writings

Introducing Pontryagin's theorem – the bijection between cobordism classes of normally framed submanifolds of a closed smooth manifold and its Cohomotopy-sets, established via the Pontryagin-Thom collapse/Cohomotopy charge – and its application to the computation of

all three of which are available in English translation in Gamkrelidze 86.

A comprehensive account of this cobordism theory/stable homotopy theory via Pontryagin's theorem (now mostly: “Pontryagin-Thom theorem”) is then given in:

Introducing the Pontrjagin product:

• Lev Pontrjagin, Homologies in compact Lie groups, Rec. Math. [Mat. Sbornik] N.S., 1939 Volume 6(48), Number 3, Pages 389–422 (mathnet:5835)

Selected articles:

category: people

Last revised on February 4, 2021 at 06:20:49. See the history of this page for a list of all contributions to it.