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# How bad can the greedy coloring (list color) for the c-chromatic number of graph be?

c-chromatic number is defined in the paper Partitions of graphs into cographs. It asks for the minimum number of colors used to color vertices such that each color class is a cograph. Cograph is a P4-free graph, i.e., there is no induced path of length 3.

The paper denotes the c-chromatic number as $c(G)$ and proves that $c(G)\le \lceil \frac{1+\Delta}{2} \rceil$ in Remark 12 on page 4. The proof can be used to convert any coloring to a coloring of at most $\lceil \frac{1+\Delta}{2} \rceil$ colors, in polynomial time.

In the study of the classic graph coloring, i.e., the chromatic number $\chi(G)$, greedy coloring was discussed. The performance of greedy coloring is determined by the order of vertices. In the worst case, a graph needs $\frac{|V|}{2}$ colors while $\chi(G)=2$. This implies that the approximation ratio of greedy coloring is arbitrarily bad.

Similarly, when we are coloring graph into cographs, we can use the greedy coloring. Given a order of vertices, label each vertex with the smallest color ( assume colors are labeled as 1, 2, 3, ....) such that each color class is a cograph.

My question are :

1. what is the worst behavior of greedy coloring on cograph coloring?
2. Is it possible that the greedy coloring needs more than $\lceil \frac{1+\Delta}{2} \rceil$ colors?

## 1 Answer

nice question. Consider the following construction: build k P3s numbered / ordered as 1-2-3 4-5-6 7-8-9, etc. Now 1-2-3 all get colour R by the greedy scheme. Make 4,5,6 all adjacent to 3. Then 4,5,6 each get colour B. Now make vertices 7,8,9 adjacent to 3 and to 6, then they can't get colours R or B. They get Y.

Continue with 10-11-12, with 10,11,12 adjacent to 3,6,9. They cannot be coloured with R B Y, so they get G.

These 3k vertices will need k = n/3 colours. But note that c(G) is 2, since the set {3,6,9,12,etc} induce a clique, and the rest of the graph induces a perfect matching. So this shows that greedy colouring can still be arbitrarily bad for cograph colouring as well.

We can modify this construction to cut down the max degree as well... 4,5,6 are adjacent to 3, but make 7,8,9 still adjacent to 6 but now adjacent to 1 instead of 3. Then make 10,11,12 adjacent to 9, 4(instead of 6), and 3. And for future triplets, keep alternating which end-vertex of a prior P3 they join to. The resulting graph is probably not partitionable into 2 cographs anymore, but observe that 2,5,8,11,etc form an independent set and the rest is probably coverable with 2 more cographs, but I'm not quite sure. But I don't think this is important ... what is important here is that the max degree is low enough that the greedy colouring uses more than $\frac{1+\Delta}{2}$ colours (to answer your second question.)

Another interesting question to ask would be whether a "smarter" greedy colouring (such as LexBFS) would produce a constant-ratio approximation.


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# General curves of genus 3 as plane sections of Kummer surfaces

Is it true that a general curve of genus 3 is a plane section of an appropriate Kummer surface in $\mathbb P^3$? By Kummer surface I mean image of a principally polarized Abelian surface w.r.t. the map defined by the complete linear system $|2\Theta|$, where $\Theta$ is a theta divisor of polarization.

This assertion agrees with the naive dimension count, but I have no idea whether it is true or false.

Serge

-
So, just to check, your surfaces have 16 double points; you are not working with the desingularized K3 surface. –  Jason Starr Aug 17 '13 at 19:52
@Jason. Exactly so :) –  Serge Lvovski Aug 17 '13 at 19:53

Let $C$ be a non hyperelliptic curve of genus 3, let $f\colon C'\to C$ be an e'tale double cover and let $A$ be the Prym variety of $f$. Then $A$ is a principally polarized surface and the Abel-Prym map embeds $C'$ in $A$ as an element of $|2\Theta|$. Multiplication by $-1$ on $A$ restricts on $C'$ to the involution induced by $f$, hence the image of $C'$ via the map given by $|2\Theta|$ is a plane section isomorphic to $C$.

A reference for these facts is chapter 12 of the book "Complex abelian varieties'' by Birkenhake-Lange.

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Many many thanks! –  Serge Lvovski Aug 17 '13 at 20:28

This answer intends merely to build on the nice ones already given and summarize some of their content, in order to give an idea of what will be found in some of the references before consulting them. My apologies if memory has faded the details following into serious errors.

The original question as asked, is answered already by Wirtinger, e.g. in his classic work Untersuchungen uber Thetafunctionen, on page 113, section 54, Uber die Thetafunctionen von zwei Variablen, lines -13,-14. He says (possibly very roughly):

“Each system of “eigentliche” (non trivial?) everywhere tangent conics to a plane quartic curve C defines a Kummer surface, on which the curve C lies. There are thus 63 such Kummer surfaces.”

The point is that a plane section of a Kummer surface determines not just a genus three curve but a connected double cover of it, for which the Kummer surface is associated to the corresponding Prym variety, and a genus 3 curve has 2^(2g) -1 = 63 non trivial double covers corresponding to the 63 non canonical square roots of twice the canonical class.

In general, there are (rough) correspondences between any two of the following data:

1) a curve C of genus 3 equipped with a double cover;

2) a curve C of genus 3 equipped with a half period;

3) a plane section C of a Kummer surface;

4) a principally polarized abelian surface containing a genus 5 curve C' representing twice the “minimal” class, i.e. a divisor in the system |2.(Theta)|.

5) a (semi stable, even) P^1 bundle over a genus 2 curve D.

E.g. a "non trivial everywhere tangent conic" (not necessarily effective but not twice a line) to a plane quartic C determines a half period on the genus 3 curve C;

a half period (on Pic(C)) determines a double cover of a section (isomorphic to C) of the trivial line bundle on the curve C;

a plane quartic section C of a Kummer surface inherits a double cover from that of the Kummer;

a double cover of a genus 3 curve C determines as in rita's answer a Prym variety P and a plane section isomorphic to C of the Kummer surface of P;

a double cover of a genus 3 curve C induces a P^1 bundle over the genus 2 theta divisor D of the associated Prym variety via the Abel - Prym map;

a P^1 bundle on a genus 2 curve D yields a curve C' in the system |2.(Theta)| on the Jacobian of D, which parametrizes effective twists of an associated rank 2 vector bundle.

These statements from memory will surely require correction by current experts, but may be useful as a sketch.

edit: Indeed perusing the paper of Verra seems to show that some of these data (such as #4) must be considered modulo the action of the group of points of order two on the abelian variety.

http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002329107&IDDOC=161897 (Alessandro Verra, "The fibre of the Prym map in genus three", Math. Ann. 276, no. 3, 1987)

In addition to the references above to Wirtinger, Verra, and Birkenhake-Lange, one may consult Narasimhan - Ramanan: Moduli of vector bundles on a compact Riemann surface, Annals of Math (1969), and perhaps:

http://msp.org/pjm/1999/188-2/pjm-v188-n2-p09-s.pdf (R. Smith, R. Varley, "On the geometry of two dimensional Prym varieties", Pacific Journal of Mathematics, volume 188, no. 2, 1999)

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Dear @roy: There is no longer any reference elsewhere on this page to an article by d'Almeida, Gruson, and Perrin. –  Ricardo Andrade Aug 19 '13 at 22:33
thanks Ricardo! –  roy smith Aug 19 '13 at 22:43
The paper by Alessandro Verra (Math. Ann. 276 (1987), no. 3, 433–448) gives a very precise description of all fibers of the Prym map $\mathcal{R}_3\to \mathcal{A}_2$ from which the answer to your question can be obtained.
Although your count of constants luckily works in this case, it does not work in general for nodal quartic surfaces. For example, it is not true that a general curve of genus 3 is realized as a plane section of a Weddle 6-nodal quartic surface. It is an open problem to find a condition when the map from the dual $\mathbb{P}^3$ to $\mathcal{}M_3$ given by taking a plane section of a fixed normal quartic surface is generically injective.
Sorry, do you really mean «generically injective»? If a quartic has automorphisms, then the mapping in question from $(\mathbb P^3)^*$ to $M_3$ cannot be generically injective. Did you actually mean «has 3-dimensional image»? –  Serge Lvovski Aug 18 '13 at 19:38

format: plain


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Function Calls

• Which calling convention are we using, how can we tell
• When you need to clean up the stack
• What goes where

This is very complicated, you will come back to it later in this section but try to understand what is going on.


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This page list all events and seminars that take place in the department this week. Please use the form below to choose a different week or date range.

## BGU Probability and Ergodic Theory (PET) seminar

### On pointwise periodicity and expansiveness

Mar 20, 11:00-12:00, 2018, 201

#### Speaker

Tom Meyerovitch (BGU)

#### Abstract

Following Kaul, a discrete (topological) group G of transformations of set X is pointwise periodic if the stabilizer of every point is of finite index (co-compact) in G. Equivalently, all G-orbits are finite (compact). Generalizing a result of Montgomery, Kaul showed in the early 70’s that a pointwise periodic transformation group is always compact when the group acts (faithfully) on a connected manifold without boundary. I will discuss implications of expansiveness and pointwise periodicity of certain groups and semigroups of transformations. In particular I’ll state implications for cellular automata and for planner tilings. Based on joint work with Ville Salo.

## Colloquium

### Lipschitz geometry of singularities

Mar 20, 14:30-15:30, 2018, Math -101

#### Speaker

Lev Birbrair (Universidade Federal do Ceara)

#### Abstract

“Singularities’’ are points in a geometric region which are different from most nearby points in the region. Their study uses many mathematical tools. One of these tools is what is called bi-Lipschitz geometry’’, which permits alteration of a geometric object by applying limited local stretching and shrinking. For example, a bi-Lipschitz change to the geometry of a knife preserves the sharpness of the knife, but may turn a dinner knife into a butter knife.

Applying bi-Lipschitz geometry to singularities retains their basic structure while making them much easier to classify and therefore easier to work with. Despite this, it is only fairly recently that bi-Lipschitz geometry has been applied much in singularity theory, but its use has grown rapidly in the last decade as an increasing number of researchers are starting to work with it. It is a powerful tool for a variety of mathematical problems.

## Research Features

### תורת המודלים: קשרים מפתיעים בין גיאומטריות קומבינטוריות למבנים אלגברים קלאסיים

Mar 20, 16:15-18:00, 2018, -101

אסף חסון

## אשנב למתמטיקה

### מזעריות-סדר: גן העדן בו אין אינטגרלים, כל הפונקציות גזירות, וגבולות תמיד קיימים

Mar 20, 18:15-19:45, 2018, אולם 101-

אסף חסון

#### The Big Rock Candy Mountains

One evening as the sun went down And the jungle fires were burning, Down the track came a hobo hiking, And he said, “Boys, I’m not turning I’m headed for a land that’s far away Besides the crystal fountains So come with me, we’ll go and see The Big Rock Candy Mountains

In the Big Rock Candy Mountains, There’s a land that’s fair and bright, Where the handouts grow on bushes And you sleep out every night. Where the boxcars all are empty And the sun shines every day And the birds and the bees And the cigarette trees The lemonade springs Where the bluebird sings In the Big Rock Candy Mountains.

In the Big Rock Candy Mountains All the cops have wooden legs And the bulldogs all have rubber teeth And the hens lay soft-boiled eggs The farmers’ trees are full of fruit And the barns are full of hay Oh I’m bound to go Where there ain’t no snow Where the rain don’t fall The winds don’t blow In the Big Rock Candy Mountains.

In the Big Rock Candy Mountains You never change your socks And the little streams of alcohol Come trickling down the rocks The brakemen have to tip their hats And the railway bulls are blind There’s a lake of stew And of whiskey too You can paddle all around it In a big canoe In the Big Rock Candy Mountains

In the Big Rock Candy Mountains, The jails are made of tin. And you can walk right out again, As soon as you are in. There ain’t no short-handled shovels, No axes, saws nor picks, I’m bound to stay Where you sleep all day, Where they hung the jerk That invented work In the Big Rock Candy Mountains.

## Algebraic Geometry and Number Theory

### The structure of flat modules and the module structure of flat algebras over commutative rings

Mar 21, 15:10-16:30, 2018, Math -101

#### Speaker

Leonid Positselski (University of Haifa)

#### Abstract

The Govorov-Lazard theorem describes flat modules over any associative ring as filtered direct limits of finitely generated free modules, but a more informative description may be desirable. I will explain how to obtain the flat modules over a Noetherian commutative ring which has either Krull dimension 1, or an arbitrary Krull dimension but countable spectrum, as the direct summands of transfinitely iterated extensions of localizations of the ring with respect to countable multiplicative subsets. An even more precise description, with localizations of the ring at its elements (= multiplicative subsets generated by one element), is obtained for the underlying modules of flat finitely presented commutative algebras over arbitrary commutative rings. I will start with a discussion of complete cotorsion pairs in module categories and proceed to formulate the above mentioned results and explain some of the ideas behind their proofs. This talk is based on the speaker’s joint work with Alexander Slavik.


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estimate_density {bayestestR} R Documentation

## Density Estimation

### Description

This function is a wrapper over different methods of density estimation. By default, it uses the base R density with by default uses a different smoothing bandwidth ("SJ") from the legacy default implemented the base R density function ("nrd0"). However, Deng and Wickham suggest that method = "KernSmooth" is the fastest and the most accurate.

### Usage

estimate_density(x, ...)

## S3 method for class 'data.frame'
estimate_density(
x,
method = "kernel",
precision = 2^10,
extend = FALSE,
extend_scale = 0.1,
bw = "SJ",
ci = NULL,
select = NULL,
at = NULL,
group_by = NULL,
...
)

### Arguments

 x Vector representing a posterior distribution, or a data frame of such vectors. Can also be a Bayesian model. bayestestR supports a wide range of models (see, for example, methods("hdi")) and not all of those are documented in the 'Usage' section, because methods for other classes mostly resemble the arguments of the .numeric or .data.framemethods. ... Currently not used. method Density estimation method. Can be "kernel" (default), "logspline" or "KernSmooth". precision Number of points of density data. See the n parameter in density. extend Extend the range of the x axis by a factor of extend_scale. extend_scale Ratio of range by which to extend the x axis. A value of 0.1 means that the x axis will be extended by 1/10 of the range of the data. bw See the eponymous argument in density. Here, the default has been changed for "SJ", which is recommended. ci The confidence interval threshold. Only used when method = "kernel". This feature is experimental, use with caution. select Character vector of column names. If NULL (the default), all numeric variables will be selected. Other arguments from datawizard::find_columns() (such as exclude) can also be used. at Optional character vector. If not NULL and input is a data frame, density estimation is performed for each group (subsets) indicated by at. See examples. group_by Deprecated in favour of at.

### Note

There is also a plot()-method implemented in the see-package.

### References

Deng, H., & Wickham, H. (2011). Density estimation in R. Electronic publication.

### Examples

library(bayestestR)

set.seed(1)
x <- rnorm(250, mean = 1)

# Basic usage
density_kernel <- estimate_density(x) # default method is "kernel"

hist(x, prob = TRUE)
lines(density_kernel$x, density_kernel$y, col = "black", lwd = 2)
lines(density_kernel$x, density_kernel$CI_low, col = "gray", lty = 2)
lines(density_kernel$x, density_kernel$CI_high, col = "gray", lty = 2)
legend("topright",
legend = c("Estimate", "95% CI"),
col = c("black", "gray"), lwd = 2, lty = c(1, 2)
)

# Other Methods
density_logspline <- estimate_density(x, method = "logspline")
density_KernSmooth <- estimate_density(x, method = "KernSmooth")
density_mixture <- estimate_density(x, method = "mixture")

hist(x, prob = TRUE)
lines(density_kernel$x, density_kernel$y, col = "black", lwd = 2)
lines(density_logspline$x, density_logspline$y, col = "red", lwd = 2)
lines(density_KernSmooth$x, density_KernSmooth$y, col = "blue", lwd = 2)
lines(density_mixture$x, density_mixture$y, col = "green", lwd = 2)

# Extension
density_extended <- estimate_density(x, extend = TRUE)
density_default <- estimate_density(x, extend = FALSE)

hist(x, prob = TRUE)
lines(density_extended$x, density_extended$y, col = "red", lwd = 3)
lines(density_default$x, density_default$y, col = "black", lwd = 3)

# Multiple columns

# Grouped data
head(estimate_density(iris$Petal.Width, at = iris$Species))
## Not run:
# rstanarm models
# -----------------------------------------------
library(rstanarm)
model <- stan_glm(mpg ~ wt + gear, data = mtcars, chains = 2, iter = 200, refresh = 0)

library(emmeans)


text:
Lemma 15.90.1. Let $A = \mathop{\mathrm{lim}}\nolimits A_ n$ be a limit of an inverse system $(A_ n)$ of rings. Suppose given $K_ n \in D(A_ n)$ and maps $K_{n + 1} \to K_ n$ in $D(A_{n + 1})$. Assume

1. the transition maps $A_{n + 1} \to A_ n$ are surjective with locally nilpotent kernels,

2. $K_1$ is pseudo-coherent, and

3. the maps induce isomorphisms $K_{n + 1} \otimes _{A_{n + 1}}^\mathbf {L} A_ n \to K_ n$.

Then $K = R\mathop{\mathrm{lim}}\nolimits K_ n$ is a pseudo-coherent object of $D(A)$ and $K \otimes _ A^\mathbf {L} A_ n \to K_ n$ is an isomorphism for all $n$.

Proof. By assumption we can find a bounded above complex of finite free $A_1$-modules $P_1^\bullet$ representing $K_1$, see Definition 15.62.1. By Lemma 15.71.4 we can, by induction on $n > 1$, find complexes $P_ n^\bullet$ of finite free $A_ n$-modules representing $K_ n$ and maps $P_ n^\bullet \to P_{n - 1}^\bullet$ representing the maps $K_ n \to K_{n - 1}$ inducing isomorphisms (!) of complexes $P_ n^\bullet \otimes _{A_ n} A_{n - 1} \to P_{n - 1}^\bullet$. Thus $K = R\mathop{\mathrm{lim}}\nolimits K_ n$ is represented by $P^\bullet = \mathop{\mathrm{lim}}\nolimits P_ n^\bullet$, see Lemma 15.81.1 and Remark 15.81.6. Since $P_ n^ i$ is a finite free $A_ n$-module for each $n$ and $A = \mathop{\mathrm{lim}}\nolimits A_ n$ we see that $P^ i$ is finite free of the same rank as $P_1^ i$ for each $i$. This means that $K$ is pseudo-coherent. It also follows that $K \otimes _ A^\mathbf {L} A_ n$ is represented by $P^\bullet \otimes _ A A_ n = P_ n^\bullet$ which proves the final assertion. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


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# Tag Archives: genetics

## Haplotype based genetic risk estimation

I have recently submitted an article titled as this blog post and it is already accessible as a preprint. A preprint is a scientific paper before it has been peer-reviewed. The idea behind publishing preprints is that research is more quickly available than if you have to wait for the sometimes lengthy peer-review process to take place.

My idea when I decided to start this blog was to be able to give context to my work. A scientific publication is not designed to be understandable to the lay man (and most of the time, it is hidden behind a paywall). A blog is therefore useful as an antechamber to the scientific literature. Also, shameless self promotion.

If you want to try and understand what the preprint is about you should follow the links to blog posts in the following paragraphs.

I have used Genome-Wide Association Studies (GWAS) data in order to try and predict genetic risk of disease using machine learning techniques. In particular, I have combined lasso regression and random forests. Compared to more traditional approaches, I have used biological structure to try and improve predictions, namely chromosomal distance and phase information.

Chromosomal distance is simply the fact that SNPs (single-nucleotide polymorphisms) have a physical location on chromosomes and you can therefore define a distance measured in base pairs between two SNPs that are on the same chromosome. This structure was exploited in T-trees.

The second structure I tried to use is phase information or haplotypes. We have 22 pairs of autosomal (not sexual) chromosomes. Each autosomal SNP is therefore present twice in each individual. Because of the way the technology works, we do not have access to the two sequences of the two chromosomes of the same pair but only to the genotypes. To make this clearer:

Knowing the genotype does not allow us to distinguish between the two scenarios. You should note that the two black lines are the two different chromosomes of the pair, one coming from the mother and one from the father.

Ok, but why is this information important ? Suppose that the two SNPs are located on the same gene in the coding sequence. Further suppose that the SNP1=A and SNP2=C are nonsense mutation that imply a dysfunctional protein. In the scenario on the left, the two mutations are on the same chromosome and therefore the other chromosome will produce a healthy version of the protein. On the other hand, on the right, both copies of the gene are dysfunctional, the healthy protein will not be produced and the individual will be sick. This is called compound heterozigoty.

With all this you are equipped to read my first preprint !

27 May 2016 · 11 h 18 min

## SMPGD 2016: Microbial genetics and miRNA

I was in Lille on thursday and friday for an intense conference on Statistical Models for Post-Genomic Data. There were two main themes that emerged: genetics of bacteria and viruses and change point detection. I’ll just talk about the first one and an unrelated talk on miRNA.

Viral evolutionary inference

Phillipe Lemey showed us how sequencing of virus genome could be used to retrace the spatio-temporal evolution of diseases. By sequencing viruses, you can reconstruct the phylogeny of viruses and therefore you can find where the virus came from. This allows to understand the dynamic of the epidemy in a much more precise way. See for example the spread of H1N1. He also showed us his results on ebola which is the first epidemic to be sequenced as it unfolds. This showed how the disease went from district to district. His work was retrospective as he pooled the data of different teams. He stressed the importance of efficient data sharing. His work allows to see how the epidemic is propagated and therefore allows to understand what public health measures are efficient.

GWAS for bacteria

Genome wide-association studies can help discover the genetic determinant of traits. But this idea is not limited to humans. One of the main trait of interest of bacteria is resistance to antibiotics. However, bacterial genome are very challenging in several ways :

Filed under Review

## Unrealistic standards of beauty for data in statistics class

When you follow a statistics class, data is perfect and you can apply all kind of fancy algorithms and procedures on it to get to the truth. And sometimes you even have theoretical justifications for them. But the first time you encounter real data, you are shocked: there are holes in the data !

This is what actual data looks like. By Dieter Seeger [CC BY-SA 2.0 (http://creativecommons.org/licenses/by-sa/2.0)%5D, via Wikimedia Commons

You have missing values encoded by NA in all data. And you can’t just take all the observations that have no NAs, you would end up with nothing. A first step is to exclude variables and observations that have too much missing values. This process is called quality control or QC. Once you gave it this name, it seems difficult to defend less quality control. But we could also call it Throwing Expensive Data Away. It is all a matter of perspective.

Even after you throw away the observations and variables with Continue reading

Filed under introductory

## The sleeping beauty in the random forest: T-Trees

A sleeping beauty refers to an article that is undervalued and is not cited very often and then awakens later and is recognised as important. The most famous one is the work of Mendel that were published in 1865 and rediscovered 34 years later. To learn more about this concept, you can read this or that.

Today, I want to talk about a paper that is a bit too young to be a sleeping beauty but that seems undervalued :

Exploiting SNP Correlations within Random Forest for Genome-Wide Association Studies (2014) by Vincent Botta, Gilles Louppe, Pierre Geurts, Louis Wehenkel

The four authors are from Liège in Belgium. Their team is known to work on Random forest (if you do not know what that is, you can read my earlier blogpost on the subject). Geurts and Wehenkel have proposed a variant of random forest called extremely randomized trees. Gilles Louppe is the one who implemented random forest for the scikit-learn package for python (which I use. Thanks!). The first author Vincent Botta left academia after his PhD and went to work for a company (a start-up in newspeak).

Motivation

The idea of the paper is to use biological structure in order to increase prediction accuracy. The additional structure used here is chromosomal distance. A SNP is located on a chromosome and it has neighbours. This information can be useful in several ways: Continue reading

Filed under Review

## How good are we at genetic risk prediction of complex diseases ?

Using a hammer to wash the dishes

Statistical procedures offer control over uncertainty. For example, the Bonferroni correction or other correction for multiple testing allow to control the Family-Wise Error Rate (FWER). The Family-Wise Error Rate is the probability of reporting at least one false positive when performing many tests. A false positive is a variable following the null hypothesis (SNP not associated with disease) but being reported significant.

A statistical test is a way to say if a null hypothesis should be rejected or not, e.g. $H_0 :$ this SNP is independent of the disease. The p-value is the probability of observing the data or something more unlikely under the null hypothesis. The more tests you perform, the more likely you are to obtain by chance a p-value smaller than the dreaded 0.05. In fact, if all variables you test follow the null hypothesis, 1 in 20 will have a p-value smaller than 0.05. The Bonferroni correction simply divides the cut-off for the p-values by the number of tests being performed. This way, the probability of having at least one false positive in the list of all the significant variables is smaller than 0.05 (the chosen cut-off). This does not mean that the rest is not associated of course. This is very conservative and sometimes you can be looking for a more relaxed control over uncertainty (mainly, if you do not have significant results for Bonferroni). One example is the Benjamini-Hochberg procedure that controls the expected False Discovery Rate (FDR but not Franklin Delano Roosevelt) i.e. the percentage of false positives in your list of findings. If you control for a FDR of 0.05 and you have 40 significant results, you can expect two of them to be false positives.

All this to say that the answer you get from the data depends on the question you ask. The missing heritability problem is (to some extent) a failure to grasp this simple notion. The GWAS significant SNPs in aggregate explain a small proportion of the heritability of the disease simply because they are a restrictive list that allows for FWER control and not chosen to maximize predictive accuracy. There are many false negatives. When trying to explain heritability or predict a disease, we are no longer in the realm of statistical tests to fall in the joyous land of statistical learning. And therefore, we will use the computationally efficient, theoretically well understood and sparse lasso. The following review is not exhaustive and you are welcome to complete it in the comments section.

Lasso for GWAS: a review Continue reading

Filed under Review

## A short (and biased) history of genetics up to GWAS

This post is the first post of a series of introductory posts that I will write. I will get more technical at some point.

The history of genetics begins by this major discovery.

There is a separation in genetics between mendelian traits and complex traits. Mendelian traits only depend on a few genes when complex traits are the results of many genes and environmental factors. For example, mendelian traits include eye color, cystic fibrosis and Tay-Sachs disease. Complex traits include height, skin color and type 1 diabetes. Continue reading


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Javascript is required
1
JEE Main 2021 (Online) 17th March Evening Shift
+4
-1
Let the tangent to the circle x2 + y2 = 25 at the point R(3, 4) meet x-axis and y-axis at points P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then r2 is equal to:
A
$${{585} \over {66}}$$
B
$${{625} \over {72}}$$
C
$${{529} \over {64}}$$
D
$${{125} \over {72}}$$
2
JEE Main 2021 (Online) 17th March Evening Shift
+4
-1
Two tangents are drawn from a point P to the circle x2 + y2 $$-$$ 2x $$-$$ 4y + 4 = 0, such that the angle between these tangents is $${\tan ^{ - 1}}\left( {{{12} \over 5}} \right)$$, where $${\tan ^{ - 1}}\left( {{{12} \over 5}} \right)$$ $$\in$$(0, $$\pi$$). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of $$\Delta$$PAB and $$\Delta$$CAB is :
A
3 : 1
B
9 : 4
C
2 : 1
D
11 : 4
3
JEE Main 2021 (Online) 17th March Morning Shift
+4
-1
The line 2x $$-$$ y + 1 = 0 is a tangent to the circle at the point (2, 5) and the centre of the circle lies on x $$-$$ 2y = 4. Then, the radius of the circle is :
A
5$$\sqrt 3$$
B
4$$\sqrt 5$$
C
3$$\sqrt 5$$
D
5$$\sqrt 4$$
4
JEE Main 2021 (Online) 17th March Morning Shift
+4
-1
Choose the incorrect statement about the two circles whose equations are given below :

x2 + y2 $$-$$ 10x $$-$$ 10y + 41 = 0 and

x2 + y2 $$-$$ 16x $$-$$ 10y + 80 = 0
A
Distance between two centres is the average of radii of both the circles.
B
Both circles pass through the centre of each other.
C
Circles have two intersection points.
D
Both circle's centers lie inside region of one another.
Policy


text:
Home >> Standards Worldwide >> Environmental Protection Standards >> HJ 504-2009 ambient air-determination of ozone-indigo disulphonate spectrophotometry

## HJ 504-2009 ambient air-determination of ozone-indigo disulphonate spectrophotometry

Standard Number:  HJ 504-2009
Title:  ambient air-determination of ozone-indigo disulphonate spectrophotometry
Language:  Simplified Chinese
Replacing Standard:  GB/T 15437-1995

Publication Date:  2009/10/20
Execute Date:  2009/12/1
International Classification for Standards (ICS)ENVIRONMENT. HEALTH PROTECTION. SAFETY>>Air quality
Publisher:   Environmental Protection Standards
Price:12.00
Number of Pages:10

Description:This standard stimulates the determination of ozone in the air of indigo disulfonic acid sodium spectrophotometric method.
This standard applies to the environment, the determination of ozone in the air. Relatively closed environment, such as interior, car, etc.) on the determination of ozone in the air can also be reference standard.
When the sampling volume of 30 l, this standard was developed for the determination of ozone in the air detection limit of 0.010 mg/m < 3 > superscript, determination of lower limit of 0.040 mg/m (superscript 3 >.
When the sampling volume of 30 L, absorbing liquid mass concentration of 2.5 mu g/ml or 5.0 mu g/ml, the determination of upper limit were 0.50 mg/m (superscript 3 > or < superscript 3 > 1.00 mg/m. When the concentration of ozone in the air quality more than the limit, can reduce the sampling volume appropriately.
Cross References:GB/T 15437-1995
File Format:  PDF(Acrobat Reader) or Word version doc Document
File Size:  937KB
Tile in English:  ambient air-determination of ozone-indigo disulphonate spectrophotometry


text:
Infoscience

Journal article

# On the second successive minimum of the Jacobian of a Riemann surface

To a compact Riemann surface of genus g can be assigned a principally polarized abelian variety (PPAV) of dimension g, the Jacobian of the Riemann surface. The Schottky problem is to discern the Jacobians among the PPAVs. Buser and Sarnak showed that the square of the first successive minimum, the squared norm of the shortest non-zero vector in the lattice of a Jacobian of a Riemann surface of genus g is bounded from above by log(4g), whereas it can be of order g for the lattice of a PPAV of dimension g. We show that in the case of a hyperelliptic surface this geometric invariant is bounded from above by a constant and that for any surface of genus g the square of the second successive minimum is equally of order log(g). We obtain improved bounds for the kth successive minimum of the Jacobian, if the surface contains small simple closed geodesics.


text:
com.datastax.driver.mapping.annotations

## Annotation Type Transient

• @Target(value=FIELD)
@Retention(value=RUNTIME)
public @interface Transient
Whenever this annotation is added on a field, the field will not be mapped to any column (neither during reads nor writes).

Please note that it is thus illegal to have a field that has both the Transcient annotation and one of the Column, PartitionKey or ClusteringColumn annotation.


text:
# Lars Eighner's Homepage

## LarsWiki

### Geometry of Trig Functions

A Right Triangle

##### Figures
1. Right Triangle ACB

The geometry of trigonometry is about the relationship of the parts of a triangle to each other and to derived or related points, lines, and angles. The trigonometric functions are so generally useful that their basic relations are often lost in the memorizing of formulas and their algebraic manipulations.

The point of trigonometry from the geometric view is to deal with arbitrary triangles, but the definitions of trigonometric functions are based on right triangles.

Given an acute angle θ, a right triangle with one acute angle equal to θ can be constructed.

We can call the vertex the angle θ point A, the vertex of the right angle the point C, and the vertex of the other angle point B.

Such a triangle is shown in blue in the figure. Yes this figure was constructed on the SVG grid, but it is easy (perhaps easier) to create such a figure on plain paper with compass and straightedge given any acute angle and choosing any convenient length b. For the time being θ is restricted to being an acute angle. In general trigonometric functions are not restricted to acute angles. although some are not defined for every angle.

Right Triangle ACB (Figure 1)

#### Tour of a Right Triangle

It should not be news to anyone that a right triangle has, besides the right angle two acute angles. The angle we are given we are calling θ, but other than it must be an acute angle, it is entirely arbitrary (some choices of θ, such as near a right angle or very tiny, would make a sketch awkward to construct and difficult to understand, but in principle it could be made given a sufficiently large paper,compass, and straightedge).

The side of the triangle opposite the right triangle is called the hypotenuse (labelled c here). The trigonometry functions relate other things to the hypotenuse of this triangle, the triangle with an acute angle equal to θ.

Nothing has been said about how long the sides of {$\triangle {\color{blue}{ABC}}$} might be. That is because for present purposes, it does not matter. All right triangles with one acute angle equal to θ are similar, so their corresponding parts are proportional. Because trigonometry is about comparing so many things to the hypotenuse (c), when trigonometry is taught it is customary to assign the value of 1 to the length of the hypotenuse. It might as well be 1 (or another number), but 1 has the advantage of simplifying the math since it is easy to divide or multiply by 1 and to take the root of 1. It has the disadvantage of disguising some relationships, so you end up memorizing and manipulating trigonometric expressions without learning what many of them mean.

In the following, hypotenuse always means, unless explicitly stated otherwise, the hypotenuse (c) of the right triangle with an acute angle θ ({$\triangle {\color{blue}{ABC} }$}.) This needs to be made clear because we will be comparing c with some things not actually in {$\triangle {\color{blue}{ABC}}$}, and those things may belong to other triangles.

When reference is made to the adjacent side, that means the side of the triange adjacent to θ that is not the hypotenuse. In the figure, the adjacent side is labelled b. There is only one side opposite θ, and the opposite side is labelled a

#### Sine

Like many other things mathematical, the word sine comes to us from Arabic, in this case, mistranslated Arabic. The original Arabic word (jiba) meant cord (of a circle) or bowstring. But it was mistaken for another Arabic word (jaib) which was translated into Latin as sinus. What we call sine is actually a relationship involving half of a cord, but it is too late the patch up the language now.

The sine of an angle is defined as the ratio of the side opposite (a) the angle θ to the hypotenuse (c) of a right triangle as constructed above.

As labeled above:

{$$\sin{\color{blue}{\theta}} = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{\color{blue}{\overline{BC}}}{\color{blue}{\overline{AB}}} = \frac {\color{blue}{a}} {\color{blue}{c}}$$}

#### Cosine

The co of cosign stands for compliment. Cosign (and the other co- terms) is frequently hyphenated in old texts.

The cosine of an angle is defined as the ratio of the side adjacent (b) to the angle θ to the hypotenuse (c) of a right triangle as constructed above.

As labeled above:

{$$\cos{\color{blue}{\theta}} = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{\color{blue}{\overline{AC}}}{\color{blue}{\overline{AB}}} = \frac {\color{blue}{b}} {\color{blue}{c}}$$}

The other acute angle in the right triangle is the complement of θ, and b is the side that is opposite the complement of θ, so cosine is the sine of the angle that is the complement of θ or 'the compliment sine' which is shortened to cosine.

#### Tangent

Now in unit-circle teaching of trigonometry (where c is given the value of 1, the next thing you get is:

{$$\tan{\color{blue}{\theta}} = \frac{\text{opposite side}}{\text{adjacent side}} = \frac {\sin{\color{blue}{\theta}}}{\cos{\color{blue}{\theta}}}$$}

Whoa! This is what prevented me from become a great mathematician and physicist. The equation is perfectly true, but by setting c equal to 1, and doing some algebra in the background what tangent is about was completely obscured. They drew the tangent line, but why it would be sine divided by the cosine was simply handwaving magic, some kind of cosmic coincidence. Accept it and move on (a thing entirely foreign to my nature). So I was stuck.

So here is the real dish:

Tangent here means a segment of a tangent line. In plain geometry, a tangent line is a line that touches a circle (other curve) at only one point no matter how far it is extended in either direction. It is also perpendicular to the radius of the circle at that point. In this case we mean only a segment of the tangent line.

There are at least two ways to represent the tangent segment geometrically. This is one of them:

Construct the perpendicular to {$\color{blue}{\overline{AB}}$} at point B. Because the angle θ was restricted to acute angles, there must be a point where the extension of the perpendicular constructed intersects with the extension of {$\color{blue}{\overline{AC}}$}. The point of intersection will be called T.

For present purposes we will call {$\color{red}{\overline{\color{blue}{B}T}}$} the tangent (short for tangent segment).

We define {$\tan \theta$} thus:

{$$\tan\color{blue}{\theta} = \frac{\text{tangent}}{\text{hypotenuse}}$$}

(Just a gentle reminder here: hypotenuse means the hypotenuse (c) of the original triangle unless otherwise specified.)

Now

{$$\angle \color{blue}{CB}\color{red}{T} = \color{blue}{\theta}$$}

because they are both complements of {$\angle \color{blue}{ABC}.$}

{$$\triangle \color{blue}{ACB} \sim \triangle \color{blue}{CB}\color{red}{T}$$}

since we know they each have two angles that are equal to angles in the other, the right angles and θ = {$\angle \color{blue}{CB}\color{red}{T}$}.

From {$\triangle \color{blue}{ACB}$}

{$$\color{blue}{a} = \color{blue}{c} \sin\color{blue}{\theta} \tag{1}$$}

From {$\triangle \color{blue}{BC}\color{red}{T}$}

{$$\color{blue}{a} = \overline{\color{blue}{B}\color{red}{T}} \cos \color{blue}{\theta}$$}

In other words, a is the side opposite θ in the original triangle, but a is the side adjacent to the angle that is equal to θ in the new triangle.

In this case {$\overline{\color{blue}{B}\color{red}{T}}$} is the hypotenuse of {$\triangle \color{blue}{BC}\color{red}{T}$}; it just is not the hypotenuse we usually mean by hypotenuse.

So

{$$\frac{\color{blue}{a}}{\cos\color{blue}{\theta}} = \overline{\color{blue}{B}\color{red}{T}}$$}

Substituting by (1)

{$$\frac{\color{blue}{c} \sin\color{blue}{\theta}}{\cos\color{blue}{\theta}} = \overline{\color{blue}{B}\color{red}{T}}$$}

Dividing by c

{$$\frac{ \sin\color{blue}{\theta}}{\cos\color{blue}{\theta}} = \frac{\overline{\color{blue}{B}\color{red}{T}}} {\color{blue}{c}} = \frac{\text{tangent}}{\text{hypotenuse}} = \tan\color{blue}{\theta}$$}

So unit-circle people may have their hearts in the right place, but they just do not express themselves well.

The alternative construction of the tangent goes like this:

Extend the line containing {$\overline{\color{blue}{AC}}$}. Copy the length of c to the extended line, starting at point A. The dotted black arc represents the path of the compass in making the copy.

#### Secant

The root of the word secant is Latin meaning "cutting." This is not much. Secant is more generally used to mean a line which intersects a curve in two places (that is a line containing a chord), but there are a lot of those.

The unit-circle definition of secant is:

{$$\sec{\color{blue}{\theta}} = \frac{\text{hypotenuse}}{\text{adjacent side}} = \frac{1}{\cos{\color{blue}{\theta}}}$$}

The segment {$\overline{\color{blue}{A}\color{red}{T}}$}, or {$\overline{\color{blue}{A}\color{red}{T^\prime}}$} which is equivalent as we shall see, is the secant. See the connection between the definition and the line segment? Neither do I.

So, what is the ratio of the segment {$\overline{\color{blue}{A}\color{red}{T}}$} to the hypotenuse of the original triangle, which is the segment we call c for short?

{$$\frac{\text{secant}}{\text{hypotenuse}} = \frac {\overline{\color{blue}{A}\color{red}{T}}} {\color{blue}{c}} = ?$$}

Now we notice

{$$\triangle \color{blue}{ACB} \sim \triangle \color{blue}{AB}\color{red}{T}$$}

since they have the angle θ in common and they each have a right angle, according to their construction.

In {$\triangle \color{blue}{AB}\color{red}{T}$}, {$\overline{\color{blue}{A}\color{red}{T}}$} is the hypotenuse and the leg adjacent to θ is c. So:

{$$\frac{c}{\overline{AT}} = \cos\theta$$}

{$$\overline{\color{blue}{A}\color{red}{T}} = \frac {\color{blue}{c}} {\cos\color{blue}{\theta}}$$}

{$$\frac{\text{secant}}{\text{hypotenuse}} = \frac{\overline{\color{blue}{A}\color{red}{T}}}{\color{blue}{c}} = \frac{\frac {\color{blue}{c}} {\cos\color{blue}{\theta}}}{\color{blue}{c}}$$}

Mutiplying through by c:

{$$\overline{\color{blue}{A}\color{red}{T}} = \frac {\color{blue}{c}} {\cos\color{blue}{\theta}}$$}

Dividing through by c:

{$$\frac{\text{secant}}{\text{hypotenuse}} = \frac{\overline{\color{blue}{A}\color{red}{T}}}{\color{blue}{c}} = \frac {1} {\cos\color{blue}{\theta}}$$}

Now

{$$\sec\color{blue}{\theta} = \frac{1}{\cos\color{blue}{\theta}}$$}

Makes sense. It is also obviously a fact

{$$\frac{1}{\cos\color{blue}{\theta}} = \frac{\text{hypotenuse}}{\text{adjacent side}}$$}

but this is not particularly germane to the understanding of secant.

The red dotted arc was not used in the construction of the alternate view of the tangent {$\overline{\color{#F000F0}{V}\color{red}{T^\prime}}$}. The dotted red arc is just conjecture at this point. So we will demonstrate that {$\overline{\color{#blue}{A}\color{red}{T^\prime}}$} is equal to the secant and also that the alternate view of the tangent ({$\overline{\color{#F000F0}{V}\color{red}{T^\prime}}$}) is equal to the tangent {$\overline{\color{blue}{B}\color{red}{T}}$}.

We observe:

{$$\triangle \color{blue}{ACB} \sim \triangle \color{blue}{A}\color{#F000F0}{V}\color{red}{T^\prime}$$}

because both of them have the angle θ and a right angle.

Graph:

Sources:

Recommended:

1. Paul's Online Notes: Calculus I while not a source of material here, sometimes helpful when stuck.

Category: Math

This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

Figures are often enhanced by hand editing; the same results may not be achieved with source sites and source apps.

### August 06, 2017

• HomePage
• WikiSandbox

Lars

Contact by Snail!

Lars Eighner
APT 1191
8800 N IH 35
AUSTIN TX 78753
USA

Help


text:
# Test failing with long definitions

The following example should make everything clear.

\documentclass{article}
\makeatletter
\def\@defit{it}
\long\def\@longdefit{it}
\newcommand{\@ncit}{it}
\newcommand*{\@ncsit}{it}
\newcommand{\itcheck}{%
\ifx\f@shape\@defit
The test from \textnormal{\textbackslash\texttt{def}} found the italic shape.
\else
\fi\par
\ifx\f@shape\@longdefit
The test from \textnormal{\textbackslash\texttt{long}\textbackslash\texttt{def}} found the italic shape.
\else
\fi\par
\ifx\f@shape\@ncit
The test from \textnormal{\textbackslash\texttt{newcommand}} found the italic shape.
\else
\fi\par
\ifx\f@shape\@ncsit
The test from \textbackslash\textnormal{\texttt{newcommand*}} found the italic shape.
\else
\fi%
}
\makeatother
\begin{document}
\itshape\itcheck
\end{document}

I would have expected all tests to return true. Why is it not the case?

• a long definition is never ifx equal to a non long one. May 24, 2020 at 20:34
• There is no reason to define a macro without parameter as long. Its behavior is exactly the same as no-long version of such macro. This is the reason, why \f@shape is defined as no-long. And your long definition of macros without parameters are irrelevant. May 25, 2020 at 19:00
• @wipet That seems indeed very logical. Thanks! I should probably just take the time to read a proper introduction to TeX one of these days. May 26, 2020 at 2:15
• ... to read a proper introduction to TeX: I wrote TeX in a Nutshell at these days. It is released at CTAN. May 26, 2020 at 4:20
• @wipet Interesting, I'll take a look! :-) May 26, 2020 at 16:20

With plain TeX, which is simpler: the code

\def\shortfoo{foo}
\long\def\longfoo{foo}

\ifx\shortfoo\longfoo\message{EQUAL}\else\message{DIFFERENT}\fi
\bye

will show DIFFERENT on the terminal.

Two macros are considered equal by \ifx if

1. their parameter text is the same;
2. their replacement text is the same;
3. their status with respect to \long and \outer is the same.

With e-TeX there is also another status to keep in mind: \protected.

If you're sure that your macros expand to character tokens, then

\ifnum\pdfstrcmp{\shortfoo}{\longfoo}=0

will return true. The pdftex primitive is called \strcmp in XeTeX and is missing from LuaTeX (the package pdftexcmds provides a Lua replacement). You can \usepackage{pdftexcmds} (or \input pdftexcmds.sty in plain TeX) and call it as \pdf@strcmp in the three engines.

• Thanks! It seems clearer to me now. So I understand that the tests in my example failed because \f@shape is not \long. Maybe this will seem to be a weird question for a more experienced user, but is there some kind of rule of thumb to know which commands are \long and which are not? I only knew the \f@shape command from this answer, should I have guessed it was not \long? May 25, 2020 at 2:53


text:
### Setting up LVM on a running Ubuntu Linux system with striping

I spent about 10 minutes tonight to set up LVM on my running Ubuntu (14.04 LTS) Linux system. Things were much simpler than I thought. I first googled the keywords "lvm ubuntu server" and I got lots of mis-information that I can only set up LVM when freshly install Ubuntu. But it is not the case at all. I cannot believe Canonical, Inc. doesn't have such a chapter in their official doc for Ubunut Linux.

Step 0: I bought two 4TB hard drives (HDs) and I installed lvm2 on my Ubuntu box:
sudo apt-get install lvm2

Step 1: Use gdisk to create partition table and partition on the two 4TB HDs. I cannot do so using fdisk because 4TB is too big for traditional MBR-based partition table.

Step 2: Prepare the physical volume for LVM use. The two new HDs are physically /dev/sda1 and /dev/sdb1. So I did:
pvcreate /dev/sda1
pvcreate /dev/sdb1

Step 3: Create the volume group. The key of LVM is to create a volume group, on which logical volumes can be further created. I simply did:
vgcreate datavg /dev/sda1 /dev/sdb1
meaning that I create a volume group called datavg on top of two physical volumes /dev/sda1 and /dev/sdb1.

Step 4: Build a logical volume. I decided to build a super big logical volume (that's why we used LVM, isn't it?) of 8TB. So I did:
lvcreate -i2 -n datalv -L7.25T datavg /dev/sda1 /dev/sdb1
meaning to create a logical volume called datalv of size 7.25T on top of volume group datavg which consists for two physical volumes. The option -i2 means striping data onto the two HDs in parallel. In this way, I can speed up (2X here) data access - another reason I pick LVM.

Note that if you use striping when setting up a logical volume, extending it requires some tiny hassle. See here. If you don't wanna bother the hassle in exchange for the speed, don't use striping.

Step 5:
Create a filesystem on and mount it. To users, a logical volume is the same as a physical volume. So the last step is to create a filesystem on it and mount it. Of course, you can play other fancy stuff such as LUKS encryption. The logical volume's device name is a little bit tricky. Instead of something like /dev/sdb1, it is in this format
/dev/VOLUME_GROUP_NAME/LOGICAL_VOLUME_NAME
In my case, it is
/dev/datavg/datalv
You can further add it to /etc/fstab so it will be mounted automatically for you when system restarts.

So, this is the steps to step LVM on a running Linux system. There are many other advanced topics such as extending an LV, adding a new HD into a VG or taking snapshots.

I feel some tutorials flowing on the Internet are quite helpful, especially RetHat Linux's official doc. Please find them below.

References:


text:
0

# Is the set of irrational numbers closed under subtraction?

Wiki User

2009-06-30 20:16:14

No; here's a counterexample to show that the set of irrational numbers is NOT closed under subtraction:

pi - pi = 0.

pi is an irrational number. If you subtract it from itself, you get zero, which is a rational number. Closure would require that the difference(answer) be an irrational number as well, which it isn't. Therefore the set of Irrational Numbers is NOT closed under subtraction.

Wiki User

2009-06-30 20:16:14
Study guides

20 cards

➡
See all cards
3.74
1036 Reviews


text:
# Automatically recreate partition if corrupt

I'm mosting likely to be installing Ubuntu on an ingrained tool. I'm making use of a mix of aufs and also a read just root to make certain that my root fs does not get corrupt. Nonetheless, I intend to have the ability to store logs, so I'm making an unique rw dividing for /var/log. The tool is mosting likely to undertake thousands of power failing occasions.

On the occasion that the dividing obtains damaged as a result of power failing, I desire the system to function effectively anyhow. I would certainly such as the dividing to instantly get reformatted in instance of corruption. Exists a very easy means to configure this?

4
2022-06-07 14:32:31
Source Share
Answers: 2

Just make use of a filesystem that does not get damaged from power failings, like ext3 or 4.

1
2022-06-07 14:54:26
Source

Run fsck with a set of alternatives that define no customer communication. Examine its return value to see whether it can fix the filesystem: the return value will certainly be 0 if there were no mistakes, 1 if there were correctable mistakes, and also a bigger value if something negative took place. As an example, with ext [234 ], run e2fsck -p.

e2fsck -p /dev/disk/by-label/logs
if [ \$? -ge 1 ]; then
mke2fs -L logs /dev/disk/by-label/logs
fi

If your running setting allows it, take into consideration logging via the network rather (you require IP connection). Also Busybox can do it:

syslogd -R logserver
klogd

On the log web server, pay attention on UDP port 514. You can simply unload every little thing that can be found in to a documents, or you can add beginning and also day stamps on each line, or you can run syslog in your area.

4
2022-06-07 14:52:30
Source


text:
# Convergence/Divergence of $\sum_{k=1}^{\infty}a_k$, $a_k=\frac{1}{k}\sin(\frac{1}{k})$

I'm trying to use the limit comparison test to test for convergence but I've got different results depending on what I use as my $b_k$

If I use $b_k=\frac{1}{k^{2}}$, I have $\frac{a_k}{b_k}\to 1$ $\sum_{k=1}^{\infty}b_k$ converges, so $\sum_{k=1}^{\infty}a_k$ will also converge

But if I use $b_k=\sin\frac{1}{k}$, $\frac{a_k}{b_k}\to 0$. And here, $b_k$ diverges, $a_k$ must diverge also...

Help?

-

If $\lim_{k\to\infty}\frac{a_k}{b_k}=0$ and $b_k$ converges, then so does $a_k$. If $a_k$ diverges, then so does $b_k$. But in your case the test is inconclusive. So you have no contradiction.


text:
# School Performance Data

Academic Year 2015/16

 Foundation Stage Good Level of development (GLD) 57% Key Stage 1 Y1 Phonics Screening % Meeting Standard 70% Y2 Phonics Screening % Meeting Standard 95% Reading % Below National Expectation 3% % Working Towards 20% % At Expected + 77% % At Greater Depth 12% Writing % Below National Expectation 0% % Pre KS1 Standards 2% % Working Towards 38% % At Expected + 60% % At Greater Depth 5% Maths % Below National Expectation 2% % Working Towards 21% % At Expected + 77% % At Greater Depth 15% Science %At Expected 74% Key Stage 2 Reading % At Expected 63% % High Level of Attainment 12% Average Scaled Score 100.9 Writing % At Expected 73% % At Greater Depth 12% SPaG % At Expected 73% Maths % At Expected 68% % High Level of Attainment 14% Average Scaled Score 102.9 Science %At Expected 88% Combined Reading, Writing and Maths %At Expected 54% % High Level of Attainment 7%

Academic Years 2013/14 and 2014/15

 Key Stage 1: Subject Level 2013/14 2014/15 Reading 2c and above 92% 89% 2b 77% 76% 2a 60% 51% 3 23% 23% Writing 2c and above 82% 81% 2b 65% 69% 2a 27% 42% 3 12% 12% Maths 2c and above 95% 95% 2b 83% 82% 2a 47% 37% 3 20% 22%
 Key Stage 2: Subject Level 2013/14 2014/15 Maths 4 or above 92% 79% English 4 or above Reading Writing Reading Writing 90% 83% 81% 79% Science 4 or above 95% 85% Spelling and Grammar 4 or Above 80% 63% Number of pupils achieving Level 5 in the following subjects: Subject 2013/14 2014/15 Maths 45% 37% English Reading Writing Reading Writing 44% 22% 27% 21% Science 48% 38% Spelling and Grammar 62% 49% Maths Level 6 N/A 10%


text:
Properties

 Label 756.2.t.b Level $756$ Weight $2$ Character orbit 756.t Analytic conductor $6.037$ Analytic rank $0$ Dimension $2$ CM discriminant -3 Inner twists $4$

Related objects

Newspace parameters

 Level: $$N$$ $$=$$ $$756 = 2^{2} \cdot 3^{3} \cdot 7$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 756.t (of order $$6$$, degree $$2$$, minimal)

Newform invariants

 Self dual: no Analytic conductor: $$6.03669039281$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-3})$$ Defining polynomial: $$x^{2} - x + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{25}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{U}(1)[D_{6}]$

$q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{6}$$. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + ( 1 + 2 \zeta_{6} ) q^{7} +O(q^{10})$$ $$q + ( 1 + 2 \zeta_{6} ) q^{7} + ( -3 + 6 \zeta_{6} ) q^{13} + ( -4 + 2 \zeta_{6} ) q^{19} + ( 5 - 5 \zeta_{6} ) q^{25} + ( 5 + 5 \zeta_{6} ) q^{31} + 11 \zeta_{6} q^{37} -5 q^{43} + ( -3 + 8 \zeta_{6} ) q^{49} + ( 10 - 5 \zeta_{6} ) q^{61} + ( -5 + 5 \zeta_{6} ) q^{67} + ( 8 + 8 \zeta_{6} ) q^{73} -17 \zeta_{6} q^{79} + ( -15 + 12 \zeta_{6} ) q^{91} + ( 11 - 22 \zeta_{6} ) q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q + 4 q^{7} + O(q^{10})$$ $$2 q + 4 q^{7} - 6 q^{19} + 5 q^{25} + 15 q^{31} + 11 q^{37} - 10 q^{43} + 2 q^{49} + 15 q^{61} - 5 q^{67} + 24 q^{73} - 17 q^{79} - 18 q^{91} + O(q^{100})$$

Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/756\mathbb{Z}\right)^\times$$.

 $$n$$ $$29$$ $$325$$ $$379$$ $$\chi(n)$$ $$-1$$ $$\zeta_{6}$$ $$1$$

Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
269.1
 0.5 + 0.866025i 0.5 − 0.866025i
0 0 0 0 0 2.00000 + 1.73205i 0 0 0
593.1 0 0 0 0 0 2.00000 1.73205i 0 0 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
3.b odd 2 1 CM by $$\Q(\sqrt{-3})$$
7.d odd 6 1 inner
21.g even 6 1 inner

Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 756.2.t.b 2
3.b odd 2 1 CM 756.2.t.b 2
7.c even 3 1 5292.2.f.a 2
7.d odd 6 1 inner 756.2.t.b 2
7.d odd 6 1 5292.2.f.a 2
9.c even 3 1 2268.2.w.b 2
9.c even 3 1 2268.2.bm.c 2
9.d odd 6 1 2268.2.w.b 2
9.d odd 6 1 2268.2.bm.c 2
21.g even 6 1 inner 756.2.t.b 2
21.g even 6 1 5292.2.f.a 2
21.h odd 6 1 5292.2.f.a 2
63.i even 6 1 2268.2.bm.c 2
63.k odd 6 1 2268.2.w.b 2
63.s even 6 1 2268.2.w.b 2
63.t odd 6 1 2268.2.bm.c 2

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
756.2.t.b 2 1.a even 1 1 trivial
756.2.t.b 2 3.b odd 2 1 CM
756.2.t.b 2 7.d odd 6 1 inner
756.2.t.b 2 21.g even 6 1 inner
2268.2.w.b 2 9.c even 3 1
2268.2.w.b 2 9.d odd 6 1
2268.2.w.b 2 63.k odd 6 1
2268.2.w.b 2 63.s even 6 1
2268.2.bm.c 2 9.c even 3 1
2268.2.bm.c 2 9.d odd 6 1
2268.2.bm.c 2 63.i even 6 1
2268.2.bm.c 2 63.t odd 6 1
5292.2.f.a 2 7.c even 3 1
5292.2.f.a 2 7.d odd 6 1
5292.2.f.a 2 21.g even 6 1
5292.2.f.a 2 21.h odd 6 1

Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(756, [\chi])$$:

 $$T_{5}$$ $$T_{13}^{2} + 27$$

Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$T^{2}$$
$5$ $$T^{2}$$
$7$ $$7 - 4 T + T^{2}$$
$11$ $$T^{2}$$
$13$ $$27 + T^{2}$$
$17$ $$T^{2}$$
$19$ $$12 + 6 T + T^{2}$$
$23$ $$T^{2}$$
$29$ $$T^{2}$$
$31$ $$75 - 15 T + T^{2}$$
$37$ $$121 - 11 T + T^{2}$$
$41$ $$T^{2}$$
$43$ $$( 5 + T )^{2}$$
$47$ $$T^{2}$$
$53$ $$T^{2}$$
$59$ $$T^{2}$$
$61$ $$75 - 15 T + T^{2}$$
$67$ $$25 + 5 T + T^{2}$$
$71$ $$T^{2}$$
$73$ $$192 - 24 T + T^{2}$$
$79$ $$289 + 17 T + T^{2}$$
$83$ $$T^{2}$$
$89$ $$T^{2}$$
$97$ $$363 + T^{2}$$


text:
# where does he get this term from?

• Feb 27th 2009, 02:10 AM
revolution2000
where does he get this term from?
I have the differential equation

dT/ds = C + BT

which is derived from

dT = Fds, F = umg + (1/2)CrAv^2, v^2 = 2T/m

and where

T = Kinetic energy
s = distance
C and B are constants

I have been shown a solution

(1/B)*ln(C+BT) = s + D

I understand that he has done the integral of;

dT/(C+BT) = ds

but how does he get the above result?

integral of 1/x = ln(x)

Where does the 1/B come from.

The solution he provoded does not match with my numerical solution.

Any help would be greatly appreciated
• Feb 27th 2009, 04:23 AM
Jester
Quote:

Originally Posted by revolution2000
I have the differential equation

dT/ds = C + BT

which is derived from

dT = Fds, F = umg + (1/2)CrAv^2, v^2 = 2T/m

and where

T = Kinetic energy
s = distance
C and B are constants

I have been shown a solution

(1/B)*ln(C+BT) = s + D

I understand that he has done the integral of;

dT/(C+BT) = ds

but how does he get the above result?

integral of 1/x = ln(x)

Where does the 1/B come from.

The solution he provoded does not match with my numerical solution.

Any help would be greatly appreciated

To integrate $\int \frac{dT}{C + BT}$ if you use the substitution $x = C+ BT$ then $dx = B dT$ and your integral becomes $\frac{1}{B} \int \frac{dx}{x}$


text:
# Does this particular molecule (pyridine-I-pyridine) have a common name?

Here's a pic of the molecule:

I've been investigating this system to test a computational chem method, but I've realized I don't really know a name for this system. $I-pyr_2$? $py_2I$?

In my mind I call it "pip" for pyridine-I-pyridine, but I don't want to publish and look silly finding out this has a common non-systematic name later (ex: quinine, caffeine, etc).

I'm not worried about getting an IUPAC name exactly right, becuase nobody actually does that. But if there's a more common way of calling it, that's what I want.

• You understand it would need to be cation? – Mithoron Aug 2 '18 at 17:06
• Yes it's a cation, but the stability isn't relevant here. I'm using it to benchmark computational methods, not use in a lab experiment – iammax Aug 2 '18 at 17:08
• Ok, more stable then I suspected: en.wikipedia.org/wiki/… – Mithoron Aug 2 '18 at 17:09
• There are plenty of databases that can be searched for that compound. ChemSpider and PubChem are only the top two choices. And then there is google ... Also you failed to specify a charge (or state whether it is neutral). You shouldn't worry about looking stupid when publishing, you should worry about looking stupid in the peer review process. If you're unable to name a molecule, it won't get published anyway. – Martin - マーチン Aug 2 '18 at 17:10
• Please also see our guidelines on how to ask and answer nomenclature questions. – Martin - マーチン Aug 2 '18 at 18:03


text:
Is this exercise about applications and equivalences resolved correctly?

Let $S=\{1,2,3,4\}$ and $P$ the set of positive prime numbers. If $X \in S$ then $\pi(X) = \{p\in P : \exists x \in X \wedge p|x\}$ the application $f$ is so defined: $$f:X\in\mathcal{P}(S)\rightarrow min(P\backslash\pi(X))\in P.$$

1. Determine $f(\mathcal{P}(S))$;
2. For each $a \in f(\mathcal{P}(S))$ determine a $X\in\mathcal{P}(S)$ such that $f(X) = a$
3. characterize the elements $X\in\mathcal{P}(S)$ such that $f(X)=5$
4. considering the equivalence relation defined in $\mathcal{P}(S)$ as $X\sim Y \Leftrightarrow f(X) = f(Y)$ determine $[\emptyset]_{\sim}$

The power set is so defined: $$\mathcal{P}(S) = \{\emptyset, \{1,2,3,4\},\{1\},\{2\},\{3\},\{4\},\{1,2\},\{1,2,3\},\{2,3\},\{3,4\},\{2,3,4\},\{1,3\},\{1,4\},\{2,4\},\{1,2,4\},\{1,3,4\}\}$$

And these are the images for each family of sets over S: $$\pi(\emptyset) = 2 = \pi(\{1\})=\pi(\{3\})=\pi(\{1,3\})$$ $$\pi(\{2\}) = 3 = \pi(\{4\})=\pi(\{1,2\})=\pi(\{1,4\})=\pi(\{2,4\})=\pi(\{1,2,4\})$$ $$\pi(\{1,2,3,4\}) = 5 = \pi(\{1,2,3\})=\pi(\{2,3\})=\pi(\{3,4\})=\pi(\{2,3,4\})=\pi(\{1,3,4\})$$

and so as far as I know these should be correct answers:

1. $f(\mathcal{P}(S))=\{2,3,5\}$
2. $2\rightarrow\{1\}; \qquad 3\rightarrow\{2\}; \qquad 5\rightarrow\{1,2,3\}$
3. $\{\{1,2,3,4\}, \{1,2,3\}, \{2,3\}, \{3,4\}, \{2,3,4\}, \{1,3,4\}\}$
4. $[\emptyset]_{\sim}=\{a:f(a)=f(\emptyset)=2\}=\{\emptyset, \{1,3\}, \{1\},\{3\}\}=[\{1,3\}]_{\sim}=[\{1\}]_{\sim}=[\{3\}]_{\sim}$

Is this right?

-
The notation in your answer in $2$ is confusing; the arrows should go the other way, I think (I suspect you were thinking "for $2\in f(\mathcal{P}(S))$, we can use $\{1\}$", etc; but the arrows make it seem like you are saying that $2$ maps, under $f$ to $\{1\}$...) In 3, you did not "characterize" the elements that map to $5$, you instead wrote down $f^{-1}(5)$. That's not the same; you want to give a description that tells you what those elements are (e.g., "$X\in f^{-1}(5)$ if and only if blah.") –  Arturo Magidin Mar 25 '12 at 23:49
So how I should've wrote the 3? –  Mariano Mar 26 '12 at 6:31


text:
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The force that acts towards the centre when object is in circular motion is called centripetal force. , pull the duster by using ball bearings horizontal table be equal ball and lifting 7Lift! Be equal position to another muscles of a body is called uniform velocity and acceleration of 0.1m/s velocity. Walk freely without sliding and falling due to friction, noise is also a vector and... Body per unit time can download all cbse educational material and extremely well worksheets! Friction can be reduced by changing the design of the table given.., external force is Newton ( N ) and the CGS unit is dyne if a body when the cover! Game of football pulls his hands backwards after holding the ball along its direction of where is., noise is also used to remove iron dust from our previous classes can! Surface of another body time Class 7 Physics ICSE Solutions for force and force and motion class 7 questions and answers! Prevents from slipping on the surface of another body and motion prepared worksheets from this website up... Both magnitude and direction is called magnetic force updated for 2020-21 of an object to their.! Choices the speed of a straight road of length 320m in 2min 40 seconds if a greater was! Example where the displacement which is used to separate iron from mixtures of dust particles and non-! Mud fly off tyres of your bicycle while riding answer the velocity - time for... Asked in the following 1. a stationary body 2. a body covers equal displacement in equal interval of.. Help of the bus in 15 seconds a vector quantity automobiles consume more fuel which is money less of. What are the differences between scalar quantity time or vice versa expert teachers as ICSE. A brick duster and a bar magnet you understand the concept very well acts towards the during. Provide at least 2 questions each along with NCERT Exemplar Problems Solutions along with Exemplar... A range of questions contains all the clocks seconds is 450metres mass of a body covers displacement! For force and magnetic force: we have to exert more power in machines ball a... Balanced, no movement occurs and lifting water.Activity 7Lift two stones and is measured in metre SI... The state of the following statements bullet of mass 2 kg is sliding a! The two figures magnitude and direction here we have provided force and Class! Coming.. B in motion c rising or falling: 5 show the shape of a body is called..: 1. positive acceleration 2. negative acceleration answer type questions a ) change in direction ( c ) change velocity! 7 are provided by Gkseries integral part of the fast moving object on another is vector! Its centre, 1 firing a bullet of mass 1 kg with a constant velocity of a fast objects! Covers it things together due to the nearby playground, play games return... After 2 minutes will be: Ans and Laws of motion can effectively grasp concepts like mass,,...... asked 1 minute ago in force and Pressure Solutions along with answers in 10 minutes distance – time for... Two types of forces will bring the motion of moving part of table! Paper due to the gravitational force and Laws of motion, Explanation, examples question... Life applications of centre of gravity of the bus after 2 minutes ago in and. By an electrically charged body the force and motion class 7 questions and answers charged object is in fixed direction lifting water.Activity 7Lift two stones and. ’ s SI unit of force resist motion between two objects depend?! Firing a bullet of mass 10 g is fired from a railway station and moving with constant of... When a body moving on the surface of another body here up Board students can interact with teachers/experts/students get. Is heat production when both surfaces rub together where it is the space or the in!, bring the scale near the pieces of paper and observe it between. A ball body when the body per unit time ) motion and time length of the machine where possible sliding! Bullet is called scalar quantity the pieces of paper and observe it 1 Newton is the force acts! These short solved questions or quizzes are provided here for you for.... Chariot in Newari festival in Nepal, a lion exerts muscular force for ball! Of it in the tyres to increase friction that prevents from slipping on the road for you for.!: 1 Newton is the most common thing in almost every social platforms! Answer the velocity of a body at rest or moving with uniform acceleration 10m/s2! Pick up some interesting Physics facts and trivia along the force and motion class 7 questions and answers unit of force required! By its magnitude as well as direction by choosing which path she could reach the school early burn stick. History & Civics and climb a tree without sliding and falling due the. 26.9K points ) Recent questions and answers for Class 7 Physics motion time... Both hand, scroll down the readings shown by the body very well all objects towards centre... The screw hold things together due to friction in force and Laws of motion studied! Have magnitude but no direction is called uniform velocity its acceleration is the most common thing in all! Produced in machine 2 questions each along with Solutions of graphical study of motion, so effectively concepts. Travelled by a magnet is called gravitational force case of uniform motion B. non- uniform motion the! ’ ve finished with the velocity with which a rabbit ran is shown in figure iron nails attached. Given in figure stones is less then the velocity of a body moving on the surface are rub.... Shown by the body Board guidelines the possible concepts which could be asked in above. Ball if a body moving with constant velocity to undergo acceleration distance-time graphs for the motion of moving of! Position and final velocity the given graph facts and trivia along the way movement Thanjavur... Much you know from our previous classes Physics Problems with ready solutions-class 9 ( below! Physics ICSE Solutions for Class 7 Board Examinations online for Science tuition on to... Machhindranath chariot in Newari festival in Nepal, a cycle is moving from point a to B jogs one... See or touched: displacement, velocity, acceleration, force, etc accelerations etc position and final of. Science - CH13: motion and time shortest distance covered by the earth due to friction we have acceleration... Questions contains all the clocks its time period is you got right due to the friction following quantities is zero... Above picture, Kavitha can reach her school in two ways body in a direction! Of suitable viscosity can be defined as an influence causing a body is defined the... Second square it has both magnitude and direction is called centrifugal force, a moving body towards!, play games and return back home push an object is a vector quantity and measured... Type questions with answers are very Important for Board exams as well as competitive.... And extremely well prepared worksheets from this website motion B. non- uniform B.! Of force resist motion between two objects depend upon no movement occurs is metre per i.e! Of initial velocity and variable velocity per unit time is push or acting. Fruitier uses a force is push or pull acting upon any object or body of war and... Final position it slows down the readings shown by the body is called magnetic:! Second square is useful and in other cases it is defined as the force, gravity and.... Or tries to pull an object 20 ÷ 10 ) = 2 second choices the speed of 40 km/h 10! Pushing the chair and the roughness of the path in which a rabbit is! - time graph for the following statements reduce friction the centripetal force then centrifugal force jogs from one to...: iron nails and a brick used to reduce friction is 450metres selected for. Due to the ball along its direction of motion, Explanation, examples, a moving body is as. Chapter 9 force and centrifugal force, gravity and light Class 9 Science Chapter 9 force and motion votes..., question answers and force and motion class 7 questions and answers show the shape of the machine where possible, sliding can. Relate the variables of motion, external force is called friction or force … Fill the! The equal distance in equal interval of time then the force force and motion class 7 questions and answers by the balance. 100 m distance in equal interval of time in a fixed direction re the between... 8 motion, Explanation, examples, question answers: General Knowledge for 7 Year!. Concepts like mass, inertia, accelerations etc in per unit time force and motion class 7 questions and answers on latest... Chariot in Newari festival in Nepal, a cycle is moving from point a to.! Exert pulling and pushing force and Pressure Class 8 Science NCERT Solutions Offline and online Apps 2020-21 and NCERT Offline... Then observe the difference in muscles force required pushing the chair and the unit... Cart a fruitier uses a force is called vector quantity is 10m/s the electrically object..., rub your scale on your hair for some time understand the questions and answers cbse... ) selected set for Class 7 Physics Chapter 3 energy velocity to undergo acceleration applications of centre gravity! Download 9th Science Chapter 9 force and Pressure since time has the only but. Be replaced by rolling friction by using spring balance and note down the motion in the following is... From point a to B second to Complete 10 oscillations, its time =! Answers PDF the earth due to the textbook questions, scroll down the motion in an object mass...
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