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# Tag Info

5

Strong, Weak, and Higgs "forces" are a metaphor to mollify the public; these are short-range interactions, and lack long-range effects of the electromagnetism or gravity type. An implicit conceptual "bridge" through electromagnetism is implied, to help non-HEP physicists find their bearings. It routinely confuses laymen, and might as well ...

3

This is indeed a very confusing question and I have spent a lot of time parsing the literature looking for an answer. The references I found most useful are section 3 of the paper by Harlow and Ooguri, a set of lectures notes, and a study of SSB in superconductors(1) by van Wezel and van den Brink. Below, I summarize my current understanding; any comments ...

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Depending on what quanta dark matter truly is made of, the Higgs boson may or may not be able to decay into those quanta. Being further nit-picky, at colliders, where the Higgs boson can be studied, dark matter may or may not be produced, the collider experiment will never be able to tell whether a new particle has a long enough lifetime to be the ...

1

Just to augment @ChiralAnomaly's answer with some algebra, we can compute a squared mass$$\frac{\partial^2V}{\partial\phi\partial\phi^\ast}=4\lambda\phi^\ast\phi-\mu^2$$for each solution of $\frac{\partial V}{\partial\phi}=0$. If $\mu^2\ge0$, $|\phi|^2=\frac{\mu^2}{2\lambda}$; if $\mu^2<0$, $\phi=0$. In both cases\frac{\partial^2V}{\partial\phi\partial\...

1

People often use the name "mass term" for a quadratic term in the lagrangian. That careless habit comes from free field theories, where it really is a mass term. But in general, the theory's physical predictions are determined by the whole theory, not just by one term in the lagrangian. The mass of the Higgs particle is real and positive. It's not ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# Is it true that frequency deviation over 2*pi is the phase deviation of an FM/PM signal?

for ∆f frequency deviation if for FM only and PM has Phase deviation. and Kf= ∆f (Hz) of FM? Kf = ∆f (rad/s) for PM? and for th rules:

∆f=(kfmp)/(2p) for FM where mp = mp max-mp min ∆f=(kfmp)/(2p) for PM where mp' =mp'max-mp'min

does this rule apply for frequency deviation of the carrier and the former of Kf and Kp is for the frequency deviation of the message signal?

and do if it is correct that PM has frequency deviation, does FM has phase deviation?

I am so confused and some one clarify me this concept.

The modulation index at a single modulation frequency is $\frac{FrequencyDeviation}{ModulationFrequency}$

format: plain


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# Latex error file not found

Sep 21, - When you run LaTeX (pdflatex) you need write permission to write the PDF. So your LaTeX compiler is not able to generate a pdf smoking-quit.info source. smoking-quit.info file was not found in you case, hence the error from TeXmaker. Try working on a local drive where you have full write permissions for the smoking-quit.info - LaTeX Error: File `smoking-quit.info' not found - TeX.

Hi Christopher, If you omit the final slash, it will probably work. I just tried and the / gives indeed error. Besides, if the file is in the same directory Texmaker Error log file not found - smoking-quit.info This error is related to a file which cannot be found by LaTeX. This error usually appears when See the LaTeX manual or LaTeX Companion for explanation.

## Message latex error file not found can look for

Aug 23, - Hi, I observed similar error while working with Latex. There can be several reasons for this error. In my case, I forgot to close the caption bracket. Ok so I'm sure these type of posts get put up all the time, but all the solutions I've followed online haven't helped me so far. I'm trying to LaTeX Error: File 'smoking-quit.info' not found.

Jul 29, - Problem: You're using latex or pdflatex to compile smoking-quit.info file, but you get an error message similar to this one (the solution will work for any. B File not found. This error will be caused by, say,. ✘ Make sure that the file is in the same directory as your document or in the LaTeX path. If the file is in.

Aug 3, - R Studio R v Windows MikTex Knitr Pandoc I only used a template file and changed the bib. Apr 1, - The Rmd is not rendered although rendering is fine with html! There is a similar LaTeX Error: File `smoking-quit.info' not found.! Emergency stop. https://smoking-quit.info/orgy/

Dec 28, - File not found error. ∗. Frank Mittelbach. May 9, This file is maintained by the LATEX Project team. Bug reports can be opened (category. Jump to LaTeX Error: File ` ' not found! - This error is serious because latex will terminate without finishing the document.

### Seems excellent latex error file not found

I am using TeXworks r with MikTeX on Windows 7 bit. When I add \usepackage{datetime}. and try to typeset the document, I get a dialog. Jul 30, - According to Ubuntu Packges this file is in a package called texlive-fonts-extra. To install this package use the command: sudo apt-get install.

This has been resolved at smoking-quit.info and. [KILE] File not found "" \includegraphics (I'm assuming you are using LaTeX.) Note the double curley braces. It should also work with I'm afraid that doing that has now given me 66 errors. Beneath my use.


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In this lesson we will cover classification models, starting with an introduction to logistic regression.

We will learn how to assess the accuracy and performance of a classification model using confusion matrices, ROC curves, and other metrics.

We will also learn how to fit logistic regression models with the tidymodels package.

Next Steps

Please head over to the R tutorial where you will learn how to fit logistic regression models with tidymodels.


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# Finding dimension of a kernel from a linear transformation.

I'm trying to solve an exercise where I'm asked to find this linear transformation's kernel and its dimension:

$$f(x_1,x_2,x_3,x_4)=(x_1-x_2,2x_3+x_4)$$

So, from the definition of kernel and converting to parametric equations, I found $x_1=x_2=\lambda$, $x_3=-\frac{1}{2}\mu$ and $x_4=\mu$. Therefore, I assumed: $$\ker(f)=\{(\lambda,\lambda,-\frac{1}{2}\mu,\mu): \lambda,\mu \in \mathbb{R}\}$$

And since the kernel can be described with only one vector (with two parameters), I thought $\dim(\ker(f))=1$, but my textbook says it's actually equal to $2$. Where is my reasoning flawed?

• A basis for the kernel is $\{(1,1,0,0), (0,0,-\frac{1}{2}, 1) \}$ – leibnewtz Jan 9 '18 at 23:20
• @leibnewtz Hm, so that means I can only 'fix' one parameter at once? I thought it'd be possible to establish something like $\lambda=1$, $\mu=2$ so that a basis for the kernel is $\{(1,1,-1,2)\}$. Why isn't that possible? – Manuel Jan 9 '18 at 23:22
• Of course you can do what you say...but that way you'll only get one vector, and you already know you need two lin. ind. vectors to have a basis... – DonAntonio Jan 9 '18 at 23:30
• Since $f:\mathbb R^4\to\mathbb R^2$, its image is at most two-dimensional, so you know that its kernel must be at least two-dimensional. – amd Jan 9 '18 at 23:35
• @Manuel Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Feb 4 '18 at 0:09

It is all correct, the dimension of the ker(f) is 2 because you have 2 free parameters that define it $(\lambda,\mu)$.
• @Manuel They are equivalent concepts. As noted in the comments you can set $\lambda=1$ and $\mu=0$ and define a basis vector, then $\lambda=0$ and $\mu=1$ and define a second vector linearly independent from the first, thus the dimension is 2. This is true for any number of free parameter (EG a line or a plane in $\mathbb{R^3}$). – user Jan 9 '18 at 23:25
• @Manuel There are two vectors in your description of the kernel: $\left(\lambda,\lambda,-\frac12\mu,\mu\right)=\lambda(1,1,0,0)+\mu\left(0,0,-\frac12,1\right)$; every vector in the kernel is a linear combination of these two linearly-independent vectors. – amd Jan 9 '18 at 23:36


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MathSciNet bibliographic data MR1974337 43A85 (22E30) Andersen, Nils Byrial \$L\sp p\$$L\sp p$ versions of Hardy's uncertainty principle on hyperbolic spaces. Proc. Amer. Math. Soc. 131 (2003), no. 9, 2797–2807 (electronic). Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
Question: SC3 gene clusters
1
20 months ago by
ahmed.elewa20
Karolinska Institute
ahmed.elewa20 wrote:

Greetings,

SC3 provides an expression matrix after clustering genes by kmeans with k = 100. How does one extract the cluster to which each gene was clustered? I checked the features data (fData) but this data isn't included. Oddly, the column sc3_N_markers_clusts has numbers ranging from 1 to N, where N is the k used to cluster cells, not genes. I'm not sure how a gene can be discretely associated with a cell cluster and I think perhaps this column was intended for the gene clusters instead.

Ahmed

modified 20 months ago by Martin Morgan ♦♦ 23k • written 20 months ago by ahmed.elewa20
Answer: SC3 gene clusters
0
20 months ago by
Sanger Institute, Cambridge, UK

Hi Ahmed,

SC3 does not perform gene clustering. It only clusters cells. Therefore there is no column in fData for that. sc3_N_markers_clusts column defines which cell cluster each gene is a marker for.

Hope this helps,

Cheers,

Thanks for your reply. I'm interested in interrogating the dendogram on the left when I use the command sc3_plot_expression(scesetk = N). The top dendogram is for the cells, but what is the dendogram on the left?

## 5.3 Expression Matrix

The expression panel represents the original input expression matrix (cells in columns and genes in rows) after cell and gene filters. Genes are clustered by kmeans with k = 100 (dendrogram on the left) and the heatmap represents the expression levels of the gene cluster centers after log2-scaling.

Ahmed

Hi Ahmed,

Oh sorry, now I see what confused you. Clustering of genes on all heatmaps is done by default functionality of the pheatmap package (it has nothing to do with SC3):

?pheatmap::pheatmap()

kmeans_k - the number of kmeans clusters to make, if we want to agggregate the rows before drawing heatmap. If NA then the rows are not aggregated.

I just used it to reduce the complexity of the heatmaps, visualising just 100 cluster centers instead of plotting all genes. Unfortunately, this clusterings are not saved anywhere. But you can always run a simple k-means on your genes and should get the same result.

Cheers,

1

AE

2

And if you do run 'tmp <- pheatmap(...)', then tmp$kmeans$cluster will give you the clusters (though you might want to set.seed(1) so that you always get the same results!)


text:
José A. CañizoResearch · Publications · Teaching · Other

General Topology (MSM3P22 / 4P22) - 2012 - University of Birmingham

This page contains the information for a topology course I taught at the University of Birmingham during the 2012/2013 academic year. I taught the same course again in 2013 and 2014, but the material for those courses was published only in the internal online system of the university. This page is archived as it was at the end of the course.

New: some solutions are available below.

Here is the material and information available for this course. Notice that I am not using WebCT for the moment; all of the material is here.

Course information - Program and syllabus

Course notes - Handouts

Errata: 1/4/2013: In Handout 9 there was a mistake in Definition 9.3: it should be “contains” rather than “equals”. (Corrected now). 4/4/2013: In Problem Sheet 4, Exercise 4.1, $f$ should be also surjective (corrected).

Solutions to problem sheets

The fourth year students taking MSM4P22 need to complete an additional task for the subject. Here are some details and suggestions for it.

Lecture rooms

Since there were more people registered for the class than expected, the original lecture room was not big enough to seat everyone. A new arrangement had to be made, and due to lack of space is was not possible to reserve the same room for every class. The conclusion is that lectures now take place in the following rooms:

• Tuesdays 1pm - 2pm at Watson Lecture Room B (there is class only in alternate weeks)
• Tuesdays 2pm - 3pm at Gisbert Kapp 225
• Fridays 4pm - 5pm at Nuffield G13

Continuous assessment

Since some people have told me they don’t find the way continuous assessment is evaluated very clear, here it is. Continuous assessment is worth 10% of the total module mark; since General Topology is half the module (the other half being Further Complex Variable), continuous assessment in General Topology is worth 5% of the total module mark.

In order to obtain the full mark you need to turn in a total of 12 problems before the last day of term (December 7th 2012.) If you turn in fewer of them you will get the corresponding proportion of the mark.

Which problems can you turn in? They are:

1. Any of the marked problems in the problem sheets. They indicate how much they are worth (usually 1 point, possibly more according to their difficulty.) You can give these to me any day up to, and including, December 7th 2012.
2. Any of the exercises in the handouts given in class. These are worth 1 point. You have to give these to me before I explain them in class. This will usually mean before the next class.
3. You can also get one point by explaining on the blackboard an unmarked problem in any of the problem sheets. Please tell me in advance if you would like to do one of them.


text:
CEA Slams DOE TV Set Testing Mandate - Multichannel

# CEA Slams DOE TV Set Testing Mandate

Says It's Duplicative And Could Threaten Existing Industry Energy Ratings Programs
Author:
Publish date:

The Consumer Electronics Association calls a new DOE test procedure for measuring TV set power consumption "wasteful and unnecessary."

The adoption of the new procedure was not immediately accessible on the DOE site, which is not being updated during the government shutdown, but CEA said the department has come out with a final rule, and it is not happy.

CEA had asked DOE to defer to the existing industry test procedure, saying that a government mandate would tie the industry's hands and threaten existing programs like Energy Star and EnergyGuide. DOE has argued that it used the CEA CEA-2037-A draft  industry standard as its starting point for its own recommended testing process and incorporated parts of that into the DOE standard.

"DOE's test procedure may be very similar to the industry standard, but not exactly the same, and because it is locked into law, it will not keep pace with technology changes relevant to TV energy use and disclosures of energy use to consumers," said CEA in response to the mandate.

DOE signaled its intent to establish the new testing process in January.


text:
If \begin{bmatrix}4\\ 1 \\3 \end{bmatrix} A = \begin{bmatrix}-1 & | Filo

Class 12

Math

Algebra

Matrices

579
150

If  A =  Find A

Solution: We have XA = 33 matrix where X is 31. Now XA is conformable for multiplication if the number of columns in pre - factor is equal to the number of rows in the post factor . Keeping in view that the product is  33 matrix , we must have matrix A of 13 type i.e., A has one row and 3 columns i.e., A is row matrix
Let A = [p.q.r] say of 13 type
Now  [p q r] =
Clearly p = -1, q = 2 and r = 1 satisfy above
Hence matrix A = (-1 2 1)
579
150

Connecting you to a tutor in 60 seconds.


text:
# Probabilty of choosing 2 items

1. Jul 27, 2014

### desmond iking

1. The problem statement, all variables and given/known data
Two items are chosen at random from a batch of items containing 90 good and 2 defective ones.
find the probability of at least one out of two items is defective

2. Relevant equations

3. The attempt at a solution

P(defective , good) +P(good , defective) + P(defective , defective) =
((10/100) x (90/99)) + ((90/100) x (10/99)) + ((10/100) x (9/99))= 0.1909 (correct ans)

why cant i do on this way?
P( 1 defective , i good ) +P (2 defective) = ((2c1 x 90c1)/(92c2)) + (2c2/92c2)= 0.0452

2. Jul 27, 2014

### Mentallic

Because you're answering a different question. You're looking to solve for 90 good and 10 defective items, as opposed to 2. Your solution would work if you account for this.

3. Jul 27, 2014

### HallsofIvy

Staff Emeritus
Did you copy the problem correctly? With denominators 100 and 99, it looks like they are saying there were 90 good and 10 defective items, not 90 good and 2 defective.

First, the problem should be, as I said, 90 bad and 10 defective items. Either you or the person who put the problem in the book confused the number of defective items with the number sampled. But also, by using the same denominator, $^{92}C_2$, for both probabilities, drawing a bad item first and a good item second, you are assuming sampling with replacement which is clearly not intended here. Because there are two ways of choosing 1 good and 1 defective item: assuming there were 90 good items and 10 defective, so a total of 100 items, the probability of choosing a good item first is 90/100. Then there are 89 good items left and 10 bad for a total of 99 items. The probability of choosing a defective item is 10/99 so the probability of choosing a good item first then a bad item is (90/100)(10/99), the first fraction above. But you could also choose a bad item first. The probability of that is 10/100. The there are still 90 good items left and 9 bad for a total of 99 so the probability of choosing a good item second is 90/99 and the probability of choosing a bad item first, then a good item is (10/100)(90/99).

Of course, those two products, (90/100)(10/99) and (10/100)(90/99) give the same thing. What's normally done is to note that there are n! ways of ordering n items so multiply by n!. 2!= 2 so instead of (90/100)(10/99)+ (10/100)(90/99), I would write that as 2(90/100)(10/99).

4. Jul 27, 2014

### desmond iking

deleted

Last edited: Jul 27, 2014
5. Jul 27, 2014

### desmond iking

deleted

6. Jul 27, 2014

### desmond iking

sorry . it should be 10 defective ones, do you mean 92C2 is for replacement case? which means after I pick 1 item out of 92 items , then when i choose the 2nd item, i still have 92 options , which means the first ball chosen is put back ?

can i do on this way?
(10C1 x 9C1)/100C2) +( 10C2/100C2) = 0.1909 , but by doing so , i am assuming , the selection is with replacement am i right? but suprisingly, i also gt the same ans

7. Jul 27, 2014

### CAF123

If $X \in \left\{0,1,2\right\}$ is the number of defective items chosen, then one way to obtain the answer is to simply compute $P(X \geq 1) = 1 - P(X < 1) = 1 - P(X=0)$. This is quite a standard practice employed when you see the words 'at least'.

8. Jul 27, 2014

### Ray Vickson

This is a problem involving the hypergeometric distribution; it is something you will see over and over again, so is worth learning about once and for all. You already have the germ of the idea in some of your workings.

If you have a population of $N_1$ items of type 1 and $N_2$ items of type 2 (with $N = N_1 + N_2$ items altogether), say you pick $n$ items at random without replacement. Then the probability of having $k$ type-1 items in your sample is
$$P\{ k\: \text{ type 1 }\} = \frac{{N_1 \choose k} {N_2 \choose n-k}}{{N \choose n}}= \frac{_{N_1}C_k \, \times \, _{N_2}C_{n-k}}{_NC_n}$$
This is not hard to get: what is the probability of 11...122...2 in that specific order (with k type 1s and n-k type 2s)? Well, it is
$$p = \frac{N_1}{N} \frac{N_1 - 1}{N-1} \cdots \frac{N_1 - k+1}{N - k+1} \frac{N_2}{N-k} \frac{N_2-1}{N-k-1} \cdots \frac{N_2 - (n-k)+1}{N- n+1}$$
The probability of any other string of k 1s and (n-k)2s is the same; so for example, the probability of 1221...122211...2 is also p, etc. This is because for each such string, when we compute the probability we have numerators $N_1, N_1-1, \ldots, N_1 - k+1$ and $N_2, N_2 -1 , \ldots, N_2 - (n-k)+1$ all multiplied together in some order, and we have denominators $N, N-1, \ldots N-n+1$ all multiplied together. So, the total numerators and denominators are the same, hence the probability is p. So, to get the probability of k type Is we need only multiply p by the number of different strings of k 1s and n-k 2s, which is $_nC_k$. By manipulation, you end up with the formula given above.

For more details about the hypergeometric distribution, see, eg.,
http://en.wikipedia.org/wiki/Hypergeometric_distribution or
http://www.math.uah.edu/stat/urn/Hypergeometric.html

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