text:
Notes On Group 18: Physical And Chemical Properties - CBSE Class 12 Chemistry
Physical properties: You know that the elements of group eighteen have closed shell structures, because of which they have no tendency to form compounds under normal conditions. Due to their inert nature, these elements exist in free State as monatomic gases. These elements are colourless, tasteless and odourless gases. They are sparingly soluble in water. The atoms are held by the weak Van der Waals forces .These forces also called London dispersion forces. The magnitude of Van der Waals forces increases on moving down the group with an increase in the polarisability of the atoms. The melting and boiling points of these elements are very low. This is, due to the weak Van der Waals forces of attraction. The melting and boiling points increase on descending the group with an increase in the magnitude of the Van der Waals forces of attraction. Due to the weak Van der Waals forces, noble gases can be liquefied at very low temperatures. With an increase in the magnitude of the Van der Waals forces down the group, the ease of liquefaction of noble gases also increases. Helium has the lowest boiling point among all the elements in group eighteen Chemical properties: These elements are chemically inert due to their stable electronic configuration. These elements have high ionisation enthalpy and higher positive electron gain enthalpy. In 1962, Neil Bartlett predicted that xenon should react with platinum hexafluoride. He was the first to prepare a compound of xenon, called xenon hexafluoridoplatinate (V). Xe             +   PtF6         →          Xe[PtF6] Xenon            Platinum                      Xenon                       Hexafluoride                 Hexafluoridoplatinate (V) Later, many compounds of xenon were synthesised, including fluorides, oxides and oxyfluorides. The chemical activity of group eighteen elements increases with a decrease in the ionisation enthalpy on descending the group. The ionisation enthalpies of helium, neon and argon are too high for them to form compounds. Krypton forms only krypton difluoride, since its ionisation enthalpy is slightly higher than that of xenon. Even though radon has less ionisation enthalpy than xenon, it forms only some compounds like radon difluoride, and some complexes, since radon has no stable isotopes. However, xenon forms a markedly greater number of compounds.

#### Summary

Physical properties: You know that the elements of group eighteen have closed shell structures, because of which they have no tendency to form compounds under normal conditions. Due to their inert nature, these elements exist in free State as monatomic gases. These elements are colourless, tasteless and odourless gases. They are sparingly soluble in water. The atoms are held by the weak Van der Waals forces .These forces also called London dispersion forces. The magnitude of Van der Waals forces increases on moving down the group with an increase in the polarisability of the atoms. The melting and boiling points of these elements are very low. This is, due to the weak Van der Waals forces of attraction. The melting and boiling points increase on descending the group with an increase in the magnitude of the Van der Waals forces of attraction. Due to the weak Van der Waals forces, noble gases can be liquefied at very low temperatures. With an increase in the magnitude of the Van der Waals forces down the group, the ease of liquefaction of noble gases also increases. Helium has the lowest boiling point among all the elements in group eighteen Chemical properties: These elements are chemically inert due to their stable electronic configuration. These elements have high ionisation enthalpy and higher positive electron gain enthalpy. In 1962, Neil Bartlett predicted that xenon should react with platinum hexafluoride. He was the first to prepare a compound of xenon, called xenon hexafluoridoplatinate (V). Xe             +   PtF6         →          Xe[PtF6] Xenon            Platinum                      Xenon                       Hexafluoride                 Hexafluoridoplatinate (V) Later, many compounds of xenon were synthesised, including fluorides, oxides and oxyfluorides. The chemical activity of group eighteen elements increases with a decrease in the ionisation enthalpy on descending the group. The ionisation enthalpies of helium, neon and argon are too high for them to form compounds. Krypton forms only krypton difluoride, since its ionisation enthalpy is slightly higher than that of xenon. Even though radon has less ionisation enthalpy than xenon, it forms only some compounds like radon difluoride, and some complexes, since radon has no stable isotopes. However, xenon forms a markedly greater number of compounds.

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Next


text:
Set closed in Y, but not in superset of Y, X?

Homework Statement

Let Y be a subset of X. Give an example where a set A is closed in Y but not closed in X.

Homework Equations

A set is closed if its complement is open.
A set is open if for every element x0 of the set, there exists an E > 0 such that U(x0;E) = {x|d(x,x0)< E} is contained in the set.

The Attempt at a Solution

I tried various combinations of X and Y, usually leaving X as the real numbers. I started with X as the reals, and Y as some open or closed interval in the reals, but had no luck. I know that a closed set in the reals (here, X) is a set that contains its boundaries, so I am looking for a set that is missing at least one of its boundary points (and is thus non-closed in X) but is closed in Y. However, I can't seem to find a proper Y such that such a set is closed in Y. While reading through some texts online, I came across the claim that the set A = [0,1] ∩ Q of rational numbers between 0 and 1 (inclusive) is closed in the space of rational numbers, but is not closed in the real numbers. However, I can't see why this is true. The complement of A in Y is (0,1) ∩ {irrationals}, but if we choose a point y from this set, it doesn't seem like there is an E > 0 such that every point within a distance E from y is irrational.

As for A = [0,1] ∩ Q being non-closed in X, its complement there is (-∞,0)U((0,1)∩Q)U(1,∞). (0,1)∩Q is not open in the X, so neither is its union with the open intervals shown.

I just don't understand why it is closed in Y.

lanedance
Homework Helper
how about considering when $A \cap Y = Y$

clearly A is closed in Y, though you should be able to come up with a case when it i not closed in X...

If A ∩ Y = Y, the complement of A in Y is the null set, which is simultaneously open and closed, I believe. This tripped me up, but I guess that all that matters is that the complement is open for the set to be closed, correct? If that's true, then A and Y could both be the same open interval/region in R or RxR, right?

Dick

format: plain


text:
## The Subtour LP for the Traveling Salesman Problem

Series:
ACO Colloquium
Tuesday, October 11, 2011 - 11:00
1 hour (actually 50 minutes)
Location:
Skiles 006
,
Cornell University and TU Berlin
,
Organizer:

Refreshments at 10:30am in the atrium outside Skiles 006

The traveling salesman problem (TSP) is the most famous problem in discrete optimization.  Given $n$ cities and   the costs $c_{ij}$ for traveling from city $i$ to city $j$ for all $i,j$, the goal of the problem is to find the least expensive tour that visits each city exactly once and returns to its starting point.  We consider cases in which the costs are symmetric and obey the triangle inequality.  In 1954, Dantzig, Fulkerson, and Johnson introduced a linear programming relaxation of the TSP now known as the subtour LP, and used it to find the optimal solution to a 48-city instance.  Ever since then, the subtour LP has been used extensively to find optimal solutions to TSP instances, and it is known to give extremely good lower bounds on the length of an optimal tour. Nevertheless, the quality of the subtour LP bound is poorly understood from a theoretical point of view.  For 30 years it has been known that it is at least 2/3 times the length of an optimal tour for all instances of the problem, and it is known that there are instances such that it is at most 3/4 times the length of an optimal tour, but no progress has been made in 30 years in tightening these bounds. In this talk we will review some of the results that are known about the subtour LP, and give some new results that refine our understanding in some cases.  In particular, we resolve a conjecture of Boyd and Carr about the ratio of an optimal 2-matching to the subtour LP bound in the worst case.  We also begin a study of the subtour LP bound for the extremely simple case in which all costs $c_{ij}$ are either 1 or 2.  For these instances we can show that the subtour LP is always strictly better than 3/4 times the length of an optimal tour. These results are joint work with Jiawei Qian, Frans Schalekamp, and Anke van Zuylen.


text:
Theorem 85.10.9. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Let $T \subset |X|$ be a closed subset. The functor $F_{X, T}$ (85.10.3.1)

$\left\{ \begin{matrix} \text{algebraic spaces }Y\text{ locally of finite} \\ \text{type over }X\text{ such that }Y \to X \\ \text{is an isomorphism over }X \setminus T \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} \text{formal algebraic spaces }W\text{ endowed} \\ \text{with a rig-étale morphism }W \to X_{/T} \end{matrix} \right\}$

given by formal completion is an equivalence.

Proof. The theorem is essentially a formal consequence of Lemma 85.10.8. We give the details but we encourage the reader to think it through for themselves. Let $g : U \to X$ be a surjective étale morphism with $U = \coprod U_ i$ and each $U_ i$ affine. Denote $F_{U, T}$ the functor for $U$ and the inverse image of $T$ in $|U|$.

Since $U = \coprod U_ i$ both the category $\mathcal{C}_{U, T}$ and the category $\mathcal{C}_{U_{/T}}$ decompose as a product of categories, one for each $i$. Since the functors $F_{U_ i, T}$ are equivalences for all $i$ by the lemma we find that the same is true for $F_{U, T}$.

Since $F_{U, T}$ is faithful, it follows that $F_{X, T}$ is faithful too. Namely, if $a, b : Y \to Y'$ are morphisms in $\mathcal{C}_{X, T}$ such that $a_{/T} = b_{/T}$, then we find on pulling back that the base changes $a_ U, b_ U : U \times _ X Y \to U \times _ X Y'$ are equal. Since $U \times _ X Y \to Y$ is surjective étale, this implies that $a = b$.

At this point we know that $F_{X, T}$ is faithful for every situation as in the theorem. Let $R = U \times _ X U$ where $U$ is as above. Let $t, s : R \to U$ be the projections. Since $X$ is Noetherian, so is $R$. Thus the functor $F_{R, T}$ (defined in the obvious manner) is faithful. Let $Y \to X$ and $Y' \to X$ be objects of $\mathcal{C}_{X, T}$. Let $a' : Y_{/T} \to Y'_{/T}$ be a morphism in the category $\mathcal{C}_{X_{/T}}$. Taking the base change to $U$ we obtain a morphism $a'_ U : (U \times _ X Y)_{/T} \to (U \times _ X Y')_{/T}$ in the category $\mathcal{C}_{U_{/T}}$. Since the functor $F_{U, T}$ is fully faithful we obtain a morphism $a_ U : U \times _ X Y \to U \times _ X Y'$ with $F_{U, T}(a_ U) = a'_ U$. Since $s^*(a'_ U) = t^*(a'_ U)$ and since $F_{R, T}$ is faithful, we find that $s^*(a_ U) = t^*(a_ U)$. Since

$\xymatrix{ R \times _ X Y \ar@<1ex>[r] \ar@<-1ex>[r] & U \times _ X Y \ar[r] & Y }$

is an equalizer diagram of sheaves, we find that $a_ U$ descends to a morphism $a : Y \to Y'$. We omit the proof that $F_{X, T}(a) = a'$.

At this point we know that $F_{X, T}$ is faithful for every situation as in the theorem. To finish the proof we show that $F_{X, T}$ is essentially surjective. Let $W \to X_{/T}$ be an object of $\mathcal{C}_{X_{/T}}$. Then $U \times _ X W$ is an object of $\mathcal{C}_{U_{/T}}$. By the affine case we find an object $V \to U$ of $\mathcal{C}_{U, T}$ and an isomorphism $\alpha : F_{U, T}(V) \to U \times _ X W$ in $\mathcal{C}_{U_{/T}}$. By fully faithfulness of $F_{R, T}$ we find a unique morphism $h : s^*V \to t^*V$ in the category $\mathcal{C}_{R, T}$ such that $F_{R, T}(h)$ corresponds, via the isomorphism $\alpha$, to the canonical descent datum on $U \times _ X W$ in the category $\mathcal{C}_{R_{/T}}$. Using faithfulness of our functor on $R \times _{s, U, t} R$ we see that $h$ satisfies the cocycle condition. We conclude, for example by the much more general Bootstrap, Lemma 77.11.3, that there exists an object $Y \to X$ of $\mathcal{C}_{X, T}$ and an isomorphism $\beta : U \times _ X Y \to V$ such that the descent datum $h$ corresponds, via $\beta$, to the canonical descent datum on $U \times _ X Y$. We omit the verification that $F_{X, T}(Y)$ is isomorphic to $W$; hint: in the category of formal algebraic spaces there is descent for morphisms along étale coverings. $\square$

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).


text:
FAQ
Edit report at http://pear.php.net/bugs/bug.php?id=18466&edit=1

ID: 18466
Comment by: zane
Reported By: zane59anez at yahoo dot com
Summary: Cannot change more than 1 path during installation.
Status: Open
Type: Bug
Package: PEAR
Operating System: Windows XP SP3
Package Version: 1.9.2
PHP Version: 5_3 SVN-2011-04-20
New Comment:

i also had a prob with this on win 7 and not sure how to fix it with the
patch. if im supposed to replace the file on my local with the patch
then im outta luck becuase i dont even see that file in that path. can
some one help. im not dumb ...lol its just that this install is new to
me.

------------------------------------------------------------------------

[2011-08-17 12:55:55] metzed

I've also experienced this while attempting a new install today with
Windows 7. Anyone any idea why this hasn't been fixed?

------------------------------------------------------------------------

I've no idea why there is so much error suppression - maybe lazy dev.

Either way, the bug is a PITA.

And I don't know how to build PEAR to allow me to test my own fix.

As a test would require user input, I think just running PEAR on Windows
and
attempting to change more than 1 path would prove the issue.

------------------------------------------------------------------------

[2011-07-12 01:38:06] doconnor

Out of curiosity and not being familiar with this part of the package;
why does it do so
much @suppression of warnings with unlinking/fclosing/etc?

Also; is it possible to make a small reproducible test case for this
one? (possibly not;
due to the use of static vars in a few places and the like)

------------------------------------------------------------------------

bug:18466;patch:PEAR_Start_CLI_win32BrowseForFolder.patc;revision:1303319150;.

------------------------------------------------------------------------

Description:
------------
During the initial installation of pear (using php -f go-
pear.phar), the second question relates to the directories to be
used.

Changing one of them creates and invokes the bf.vbs script
which is saved into a temp folder in %TEMP%.

Attempting to change another directory results in the error ...

Input Error: Can not find script file "C:\Documents and
Settings\RichardQ\Local
Settings\Temp\1\tmp13D.tmp\bf.vbs".\

The cause of this is due to
http://svn.php.net/viewvc/pear/pear-
core/trunk/PEAR/Start/CLI.php?view=markup#l257. The vbs
script is silently deleted but the static variable $wshSaved is not altered, allowing the PEAR_Start_CLI::win32BrowseForFolder() to believe that the script exists and to not recreate it. The unlink was added in http://svn.php.net/viewvc/pear/pear- core/trunk/PEAR/Start/CLI.php?r1=297648&r2=297649 which is over a year ago, so no rush on this one. I see 3 possible solutions 1 - The vbs script should not be deleted, thus making the$wshSaved variable accurate. Certainly a quick fix but does
leave the vbs script lying around once all the work is done.
2 - The vbs script is deleted and we don't use the static
variable to track its existence. Also a quick fix but does mean
a vbs script is created every time the user wishes to select a
different directory.
3 - The vbs script is deleted during the destruction of the
PEAR_START_Cli.php class. I've built a patch for this but don't
seem to be able to build go-pear.phar.

------------------------------------------------------------------------

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text:
### Home > CCAA > Chapter 6 Unit 9 > Lesson CC3: 6.1.3 > Problem6-31

6-31.

Simplify each expression.

Remember: When you are multiplying fractions, multiply the numerators together and the denominators together. When you are adding fractions, find the common denominator.

1. $\frac { 73 } { 100 } \cdot ( - \frac { 2 } { 7 } )$

$-\frac{73 \cdot 2}{100 \cdot 7} = -\frac{146}{700}$

Can this fraction be reduced?

1. $0.4·0.3$

$0.12$

1. $5 \frac { 1 } { 9 } + 8 \frac { 2 } { 5 }$

Make the expressions fractions greater than one, then find the common denominator.

1. $- 1.2 + ( - \frac { 3 } { 5 } )$

Make the two numbers similar; make both numbers fractions or make both numbers decimals. Be careful of the signs.


text:
# counterexample of restricted normal operator be normal

let $$V$$ be a finite dimension complex vector space and $$T$$ a normal operator. I'm looking for an example that on a $$T$$-invariant subspace ,$$T$$ is $$not$$ normal,for I really can't prove it must be.Any advise? Thank you

• Sorry but to me, this is a bit confusing. can you explain to me the steps of how you want to disprove the given statement? to my understanding, you want to prove that restriction of normal operator need not be a normal operator. – KNilesh Sep 19 at 13:49
• I think a normal matrix which is two-block upper triangular is hopefully to be a counterexample.it is necessarily not symmetric or the subspace be one dimension, for in these two cases the assertion is obviously right. – ysTuan Sep 19 at 13:57
• just like the matrix you gave, but i want a normal one.. – ysTuan Sep 19 at 13:58
• Real 2x2 nor9mal matrix are symmetric or has form $\begin{bmatrix} a & b\\ -b & a \end{bmatrix}$ 1 – KNilesh Sep 19 at 14:06
• @KNilesh sorry i meant two-block uppertriangular not 2 dim..for example a 4×4 matrix with left-down 2×2 block be zero – ysTuan Sep 19 at 14:07


text:
## Trigonometry (11th Edition) Clone

$30^{\circ}$ and $150^{\circ}$
1. Since $sin$ $\theta$ is positive, $\theta$ must be in quadrant I or II. 2.. The reference angle is $30^{\circ}$, so the other angle must be $180^{\circ} - 30^{\circ} = 150^{\circ}$.


text:
# A source is moving towards the observer with the speed of 40 m/s and emitting the sound of frequency 100 Hz. Find the observed frequency by standing o

23 views
in Physics
closed
A source is moving towards the observer with the speed of 40 m/s and emitting the sound of frequency 100 Hz. Find the observed frequency by standing observer if the velocity of propagation of sound is 340 m/s.
1. 89 Hz
2. 113 Hz
3. 110 Hz
4. 230 Hz

by (54.0k points)
selected

Correct Answer - Option 2 : 113 Hz

CONCEPT:

• Doppler Effect: An increase or decrease in the frequency of sound, light, or other waves that are observed by the observer due to movement of the source and observer towards (or away from) each other is known as Doppler Effect.

The apparent frequency heard by the observer is given by the formula:

$f=f_0\left ( \frac{v\pm v_o}{v \pm v_s} \right )$

where f is apparent frequency, f0 is the original frequency of the source, v is the speed of sound in air, vs is the speed of the source, vo is the speed of the observer.

CALCULATION:

Given that v = 340 m/s; vs = 40 m/s (-ve) moving towards; v0 = 0; f = 100 Hz

$f=f_0\left ( \frac{v\pm v_o}{v \pm v_s} \right )$

$f=100\times \left ( \frac{340}{340-40} \right )$

f = 340/3 Hz = 113 Hz

So the correct answer is option 2.


text:
# Probability Theory Question on Expected Value and Variance of Random Variable

I've been trying to solve this question, any help would be much appreciated!

There are three coins. 2 coins are fair, the other has two tails. A random coin from this pile is chosen and tossed, it lands heads. This is done two more times without replacement.

What is the conditional probability that the third toss will land heads?

If X is a random variable which counts the number of heads total over the three tosses (assuming each coin is tossed exactly once) what is the expected value E(X)?

what is the variance of X?

I believe each coin is tossed only once.

The probability that the first and third coins are heads is $1/3\times 1/4=1/12$. The probability that the first coin is heads is $2/3\times 1/2=1/3$. So the conditional probability that the third coin is heads, given that the first coin is heads, is $(1/12) \div(1/3)=1/4$.

As for the second part, the total number of heads is $0$ with probability $1/4$, $1$ with probability $1/2$, and $2$ with probability $1/4$. From this, you can find $$E[X]=\frac{1}{4}\cdot 0+\frac{1}{2}\cdot 1 + \frac{1}{4}\cdot 2=1;$$ and $$E[X^2]=\frac{1}{4}\cdot 0^2+\frac{1}{2}\cdot 1^2 + \frac{1}{4}\cdot 2^2=\frac{3}{2},$$ so $\text{Var}[X]=E[X^2]-E[X]^2=3/2-1=1/2.$

The two-tailed coin can't be the one that was chosen. So one of the two fair coins was chosen, and we know it landed heads. The fair coin that was not chosen could land either heads or tails, with equal probabilities. So the possible values of $X$ are $1$ and $2$, each with probability $1/2$.

As for the conditional probability of the "third toss", the question is a bit ambiguous. Is the coin that was chosen the first toss, or could the coins have been tossed in any order?

• Apologize for the ambiguity. – Fred Mar 25 '13 at 3:40
• The coins are tossed one after the other, without repleacement. the first coin wasnt biased. either of the second or the the third tossed coins might be the biased ones. – Fred Mar 25 '13 at 3:41

The random variable $X$ has precisely the same distribution as the number of heads when two fair coins are tossed. So $E(X)=1$ and $\operatorname{Var}(X)=2(1/2)(1/2)=1/2$.

As to the conditional probability question, with probability $1/2$, the third coin is twwo-tailed, in which case we cannot get a head, and with probability $1/2$ the third coin is fair, in which case the probability of head is $1/2$. So the conditional probability is $1/4$.


text:
+0

# HELP! THANKS!

0
308
1

In triangle ABC, AC = BC and angle C = 90. Points P and Q are on AB} such that QCP = 45. Show that AP^2 + BQ^2 = PQ^2.

Hint: Rotate the diagram counter-clockwise around C by 90 degrees. Let P go to point P'. What is angle P'CQ?

Feb 25, 2021

$AP=BQ\implies AP=PQ\sqrt{2}$ is desired. Follows trivially from LoC and coordinates as $\angle ACP=\angle BCP=45^{\circ}$. Can't think of anything synthetic...


text:
# Accelerate your models with 🤗 Optimum Intel and OpenVINO

Last July, we announced that Intel and Hugging Face would collaborate on building state-of-the-art yet simple hardware acceleration tools for Transformer models.  Today, we are very happy to announce that we added Intel OpenVINO to Optimum Intel. You can now easily perform inference with OpenVINO Runtime on a variety of Intel processors (see the full list of supported devices) using Transformers models which can be hosted either on the Hugging Face hub or locally. You can also quantize your model with the OpenVINO Neural Network Compression Framework (NNCF), and reduce its size and prediction latency in near minutes.

This first release is based on OpenVINO 2022.2 and enables inference for a large quantity of PyTorch models using our OVModels. Post-training static quantization and quantization aware training can be applied on many encoder models (BERT, DistilBERT, etc.). More encoder models will be supported in the upcoming OpenVINO release. Currently the quantization of Encoder Decoder models is not enabled, however this restriction should be lifted with our integration of the next OpenVINO release.

Let us show you how to get started in minutes!

## Quantizing a Vision Transformer with Optimum Intel and OpenVINO

 In this example, we will run post-training static quantization on a Vision Transformer (ViT) model fine-tuned for image classification on the food101 dataset.

Quantization is a process that lowers memory and compute requirements by reducing the bit width of model parameters. Reducing the number of bits means that the resulting model requires less memory at inference time, and that operations like matrix multiplication can be performed faster thanks to integer arithmetic.

First, let's create a virtual environment and install all dependencies.

virtualenv openvino
source openvino/bin/activate
pip install optimum[openvino,nncf] torchvision evaluate

Next, moving to a Python environment, we import the appropriate modules and download the original model as well as its feature extractor.

from transformers import AutoFeatureExtractor, AutoModelForImageClassification

model_id = "juliensimon/autotrain-food101-1471154050"
model = AutoModelForImageClassification.from_pretrained(model_id)
feature_extractor = AutoFeatureExtractor.from_pretrained(model_id)

 Post-training static quantization requires a calibration step where data is fed through the network in order to compute the quantized activation parameters. Here, we take 300 samples from the original dataset to build the calibration dataset.

from optimum.intel.openvino import OVQuantizer

quantizer = OVQuantizer.from_pretrained(model)
calibration_dataset = quantizer.get_calibration_dataset(
"food101",
num_samples=300,
dataset_split="train",
)

As usual with image datasets, we need to apply the same image transformations that were used at training time. We use the preprocessing defined in the feature extractor. We also define a data collation function to feed the model batches of properly formatted tensors.

import torch
from torchvision.transforms import (
CenterCrop,
Compose,
Normalize,
Resize,
ToTensor,
)

normalize = Normalize(mean=feature_extractor.image_mean, std=feature_extractor.image_std)
_val_transforms = Compose(
[
Resize(feature_extractor.size),
CenterCrop(feature_extractor.size),
ToTensor(),
normalize,
]
)
def val_transforms(example_batch):
example_batch["pixel_values"] = [_val_transforms(pil_img.convert("RGB")) for pil_img in example_batch["image"]]
return example_batch

calibration_dataset.set_transform(val_transforms)

def collate_fn(examples):
pixel_values = torch.stack([example["pixel_values"] for example in examples])
labels = torch.tensor([example["label"] for example in examples])
return {"pixel_values": pixel_values, "labels": labels}

For our first attempt, we use the default configuration for quantization. You can also specify the number of samples to use during the calibration step, which is by default 300.

from optimum.intel.openvino import OVConfig

quantization_config = OVConfig()
quantization_config.compression["initializer"]["range"]["num_init_samples"] = 300

We're now ready to quantize the model. The OVQuantizer.quantize() method quantizes the model and exports it to the OpenVINO format. The resulting graph is represented with two files: an XML file describing the network topology and a binary file describing the weights. The resulting model can run on any target Intel® device.

save_dir = "quantized_model"

# Apply static quantization and export the resulting quantized model to OpenVINO IR format
quantizer.quantize(
quantization_config=quantization_config,
calibration_dataset=calibration_dataset,
data_collator=collate_fn,
remove_unused_columns=False,
save_directory=save_dir,
)
feature_extractor.save_pretrained(save_dir)

A minute or two later, the model has been quantized. We can then easily load it with our OVModelForXxx classes, the equivalent of the Transformers AutoModelForXxx classes found in the transformers library. Likewise, we can create pipelines and run inference with OpenVINO Runtime.

from transformers import pipeline
from optimum.intel.openvino import OVModelForImageClassification

ov_model = OVModelForImageClassification.from_pretrained(save_dir)
ov_pipe = pipeline("image-classification", model=ov_model, feature_extractor=feature_extractor)
outputs = ov_pipe("http://farm2.staticflickr.com/1375/1394861946_171ea43524_z.jpg")
print(outputs)

To verify that quantization did not have a negative impact on accuracy, we applied an evaluation step to compare the accuracy of the original model with its quantized counterpart. We evaluate both models on a subset of the dataset (taking only 20% of the evaluation dataset). We observed little to no loss in accuracy with both models having an accuracy of 87.6.

from datasets import load_dataset
from evaluate import evaluator

# We run the evaluation step on 20% of the evaluation dataset

model_or_pipeline=ov_pipe,
data=eval_dataset,
metric="accuracy",
label_mapping=ov_pipe.model.config.label2id,
)

trfs_pipe = pipeline("image-classification", model=model, feature_extractor=feature_extractor)
model_or_pipeline=trfs_pipe,
data=eval_dataset,
metric="accuracy",
label_mapping=trfs_pipe.model.config.label2id,
)
print(trfs_eval_results, ov_eval_results)

Looking at the quantized model, we see that its memory size decreased by 3.8x from 344MB to 90MB. Running a quick benchmark on 5050 image predictions, we also notice a speedup in latency of 2.4x, from 98ms to 41ms per sample. That's not bad for a few lines of code!

⚠ An important thing to mention is that the model is compiled just before the first inference, which will inflate the latency of the first inference. So before doing your own benchmark, make sure to first warmup your model by doing at least one prediction.

You can find the resulting model hosted on the Hugging Face hub. To load it, you can easily do as follows:

from optimum.intel.openvino import OVModelForImageClassification

ov_model = OVModelForImageClassification.from_pretrained("echarlaix/vit-food101-int8")


text:
# American Institute of Mathematical Sciences

• Previous Article
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• MCRF Home
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June 2017, 7(2): 259-288. doi: 10.3934/mcrf.2017009

## Minimal time synthesis for a kinematic drone model

 1 Université de Toulon, CNRS, LSIS, UMR 7296, F-83957 La Garde, France 2 Univ. Lyon, Université Claude Bernard Lyon 1, CNRS, LAGEP, UMR 5007, 43 bd du 11 novembre 1918, F-69100 Villeurbanne, France

* Corresponding author

Received  April 2016 Revised  October 2016 Published  April 2017

Fund Project: The authors were partially supported by the Grant ANR-12-BS03-0005 LIMICOS of the ANR

In this paper, we consider a (rough) kinematic model for a UAV flying at constant altitude moving forward with positive lower and upper bounded linear velocities and positive minimum turning radius. For this model, we consider the problem of minimizing the time travelled by the UAV starting from a general configuration to connect a specified target being a fixed circle of minimum turning radius. The time-optimal synthesis is presented as a partition of the state space which defines a unique optimal path such that the target can be reached optimally.

Citation: Marc-Auréle Lagache, Ulysse Serres, Vincent Andrieu. Minimal time synthesis for a kinematic drone model. Mathematical Control & Related Fields, 2017, 7 (2) : 259-288. doi: 10.3934/mcrf.2017009
##### References:
 [1] A. A. Agrachev and Y. L. Sachkov, Control Theory From the Geometric Viewpoint, vol. 87 of Encyclopaedia of Mathematical Sciences, Springer-Verlag, Berlin, 2004. doi: 10.1007/978-3-662-06404-7. [2] A. Balluchi, A. Bicchi, B. Piccoli and P. Souères, Stability and robustness of optimalsynthesis for route tracking by Dubins' Vehicules, in Decision and Control (CDC), 2000 IEEE 39th Annual Conference on, 2000,581-586. [3] U. Boscain, T. Chambrion and G. Charlot, Nonisotropic 3-level quantum systems: Complete solutions for minimum time and minimum energy, Discrete Contin. Dyn. Syst. Ser. B, 5 (2005), 957-990 (electronic). doi: 10.3934/dcdsb.2005.5.957. [4] U. Boscain, F. Grönberg, R. Long and H. Rabitz, Minimal time trajectories for two-level quantum systems with two bounded controls J. Math. Phys. , 55 (2014), 062106, 25pp. doi: 10.1063/1.4882158. [5] U. Boscain and B. Piccoli, Extremal synthesis for generic planar systems, J. Dynam. Control Systems, 7 (2001), 209-258. doi: 10.1023/A:1013003204923. [6] U. Boscain and B. Piccoli, Optimal Syntheses for Control Systems on 2-D Manifolds, vol. 43 of Mathématiques & Applications (Berlin) [Mathematics & Applications], Springer-Verlag, Berlin, 2004. [7] U. Boscain and B. Piccoli, Synthesis theory in optimal control, Encyclopedia of Systems and Control, Springer London, London, (2014), 1-11. doi: 10.1007/978-1-4471-5102-9_50-1. [8] A. Bressan and B. Piccoli, A generic classification of time-optimal planar stabilizing feedbacks, SIAM J. Control Optim., 36 (1998), 12-32 (electronic). doi: 10.1137/S0363012995291117. [9] L. E. Dubins, On curves of minimal length with a constraint on average curvature, and with prescribed initial and terminal positions and tangents, Amer. J. Math., 79 (1957), 497-516. doi: 10.2307/2372560. [10] M. A. Lagache, U. Serres and V. Andrieu, Time minimum synthesis for a kinematic drone model, in 2015 54th IEEE Conference on Decision and Control (CDC), 2015,4067-4072. doi: 10.1109/CDC.2015.7402852. [11] T. Maillot, U. Boscain, J.-P. Gauthier and U. Serres, Lyapunov and Minimum-Time Path Planning for Drones, J. Dyn. Control Syst., 21 (2015), 47-80. doi: 10.1007/s10883-014-9222-y. [12] B. Piccoli, Classification of generic singularities for the planar time-optimal synthesis, SIAM J. Control Optim., 34 (1996), 1914-1946. doi: 10.1137/S0363012993256149. [13] B. Piccoli and H. Sussmann, Regular synthesis and sufficiency conditions for optimality, SIAM J. Control Optim., 39 (2000), 359-410 (electronic). doi: 10.1137/S0363012999322031. [14] P. Souères, A. Balluchi and A. Bicchi, Optimal feedback control for route tracking with a bounded-curvature vehicle, Internat. J. Control, 74 (2001), 1009-1019. doi: 10.1080/00207170110052211. [15] H. Sussmann, Regular synthesis for time-optimal control of single-input real analytic systems in the plane, SIAM J. Control Optim., 25 (1987), 1145-1162. doi: 10.1137/0325062.

show all references

##### References:
 [1] A. A. Agrachev and Y. L. Sachkov, Control Theory From the Geometric Viewpoint, vol. 87 of Encyclopaedia of Mathematical Sciences, Springer-Verlag, Berlin, 2004. doi: 10.1007/978-3-662-06404-7. [2] A. Balluchi, A. Bicchi, B. Piccoli and P. Souères, Stability and robustness of optimalsynthesis for route tracking by Dubins' Vehicules, in Decision and Control (CDC), 2000 IEEE 39th Annual Conference on, 2000,581-586. [3] U. Boscain, T. Chambrion and G. Charlot, Nonisotropic 3-level quantum systems: Complete solutions for minimum time and minimum energy, Discrete Contin. Dyn. Syst. Ser. B, 5 (2005), 957-990 (electronic). doi: 10.3934/dcdsb.2005.5.957. [4] U. Boscain, F. Grönberg, R. Long and H. Rabitz, Minimal time trajectories for two-level quantum systems with two bounded controls J. Math. Phys. , 55 (2014), 062106, 25pp. doi: 10.1063/1.4882158. [5] U. Boscain and B. Piccoli, Extremal synthesis for generic planar systems, J. Dynam. Control Systems, 7 (2001), 209-258. doi: 10.1023/A:1013003204923. [6] U. Boscain and B. Piccoli, Optimal Syntheses for Control Systems on 2-D Manifolds, vol. 43 of Mathématiques & Applications (Berlin) [Mathematics & Applications], Springer-Verlag, Berlin, 2004. [7] U. Boscain and B. Piccoli, Synthesis theory in optimal control, Encyclopedia of Systems and Control, Springer London, London, (2014), 1-11. doi: 10.1007/978-1-4471-5102-9_50-1. [8] A. Bressan and B. Piccoli, A generic classification of time-optimal planar stabilizing feedbacks, SIAM J. Control Optim., 36 (1998), 12-32 (electronic). doi: 10.1137/S0363012995291117. [9] L. E. Dubins, On curves of minimal length with a constraint on average curvature, and with prescribed initial and terminal positions and tangents, Amer. J. Math., 79 (1957), 497-516. doi: 10.2307/2372560. [10] M. A. Lagache, U. Serres and V. Andrieu, Time minimum synthesis for a kinematic drone model, in 2015 54th IEEE Conference on Decision and Control (CDC), 2015,4067-4072. doi: 10.1109/CDC.2015.7402852. [11] T. Maillot, U. Boscain, J.-P. Gauthier and U. Serres, Lyapunov and Minimum-Time Path Planning for Drones, J. Dyn. Control Syst., 21 (2015), 47-80. doi: 10.1007/s10883-014-9222-y. [12] B. Piccoli, Classification of generic singularities for the planar time-optimal synthesis, SIAM J. Control Optim., 34 (1996), 1914-1946. doi: 10.1137/S0363012993256149. [13] B. Piccoli and H. Sussmann, Regular synthesis and sufficiency conditions for optimality, SIAM J. Control Optim., 39 (2000), 359-410 (electronic). doi: 10.1137/S0363012999322031. [14] P. Souères, A. Balluchi and A. Bicchi, Optimal feedback control for route tracking with a bounded-curvature vehicle, Internat. J. Control, 74 (2001), 1009-1019. doi: 10.1080/00207170110052211. [15] H. Sussmann, Regular synthesis for time-optimal control of single-input real analytic systems in the plane, SIAM J. Control Optim., 25 (1987), 1145-1162. doi: 10.1137/0325062.
Candidate extremal trajectories of problem $\bf{(}{{\bf{P}}_{\bf{2}}}\bf{)}$ issued from ${\mathit{\tilde X}_{\rm{0}}}$
Extremal trajectories starting from ${\mathit{\tilde X}_{\rm{0}}}$ and having one switching
Time-optimal synthesis for the problem $\bf{(}{{\bf{P}}_{\bf{1}}}\bf{)}$
Domains on which at most one switching is possible
Bang-bang-singular-bang optimal trajectories. Two optimal trajectories solutions of problem $\bf{(}{{\bf{P}}_{\bf{1}}}\bf{)}$ starting from the same point in the cut locus (left) and the corresponding optimal trajectories solutions of problem $\bf{(}{{\bf{P}}_{\bf{0}}}\bf{)}$ (right)
Bang-bang-bang-bang-bang optimal trajectories and a bang-bang-bang-bang-bang-bang optimal trajectory. Three optimal trajectories solutions of problem $\bf{(}{{\bf{P}}_{\bf{1}}}\bf{)}$ starting from the same point in the cut locus (left) and the corresponding optimal trajectories solutions of problem $\bf{(}{{\bf{P}}_{\bf{0}}}\bf{)}$ (right)
The abnormal trajectory of problem $\bf{(}{{\bf{P}}_{\bf{1}}}\bf{)}$ and the corresponding trajectory for problem $\bf{(}{{\bf{P}}_{\bf{0}}}\bf{)}$ (right)
Intersection of $\gamma^{MPp}$ with $\gamma^{Mm}$ and $\gamma^{MmM}$ (left) and the corresponding cut locus (right)
Intersection of $\gamma^{MPpPMm}$ with $\gamma^{MmMP}$, $\gamma^{MmMPp}$ and $\gamma^{MPpPM}$
Notation of the five possible optimal controls
 Control Notation $(-1,1)$ $\mathit{m}$ $(1,1)$ $\mathit{p}$ $(-1,\eta)$ $\mathit{M}$ $(1,\eta)$ $\mathit{P}$ $u$-singular $\mathit{s}$
 Control Notation $(-1,1)$ $\mathit{m}$ $(1,1)$ $\mathit{p}$ $(-1,\eta)$ $\mathit{M}$ $(1,\eta)$ $\mathit{P}$ $u$-singular $\mathit{s}$
Color convention of the optimal synthesis
 $(-1,1)$-bang arc Blue $(1,1)$-bang arc Orange $(-1,\eta)$-bang arc Purple $(1,\eta)$-bang arc Red $u$-singular arc Magenta $u$-switching curves Dashed black $v$-switching curves Gray Cut Locus Green Abnormal Cut Locus Cyan
 $(-1,1)$-bang arc Blue $(1,1)$-bang arc Orange $(-1,\eta)$-bang arc Purple $(1,\eta)$-bang arc Red $u$-singular arc Magenta $u$-switching curves Dashed black $v$-switching curves Gray Cut Locus Green Abnormal Cut Locus Cyan
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2017 Impact Factor: 0.631


text:
# ICRC 2021

Jul 12 – 23, 2021
Online
Europe/Berlin timezone

## Monte Carlo simulations of neutrino and charged lepton propagation in the Earth with nuPyProp

Jul 14, 2021, 6:00 PM
1h 30m
05

#### 05

Poster NU | Neutrinos & Muons

### Speaker

Sameer Patel (University of Iowa)

### Description

An accurate modeling of neutrino flux attenuation and the distribution of leptons they produce in transit through the Earth is an essential component to determine neutrino flux sensitivities of underground, sub-orbital and space-based detectors. Through neutrino oscillations over cosmic distances, astrophysical neutrino sources are expected to produce nearly equal fluxes of electron, muon and tau neutrinos. Of particular interest are tau neutrinos that interact in the Earth at modest slant depths to produce $\tau$-leptons. Some $\tau$-leptons emerge from the Earth and decay in the atmosphere to produce extensive air showers. Future balloon-borne and satellite-based optical Cherenkov neutrino telescopes will be sensitive to upward air showers from tau neutrino induced $\tau$-lepton decays. We present nuPyProp, a python code that is part of the nuSpaceSim package. NuPyProp generates look-up tables for exit probabilities and energy distributions for $\nu_\tau\to \tau$ and $\nu_\mu\to \mu$ propagation in the Earth. This flexible code runs with either stochastic or continuous electromagnetic energy losses for the lepton transit through the Earth. Current neutrino cross section models and energy loss models are included along with templates for user input of other models. NuPyProp results are compared with other recent simulation packages for neutrino and charged lepton propagation. Sources of modeling uncertainties are described and quantified.

### Keywords

neutrino telescopes; Monte Carlo simulations

Subcategory Theoretical Methods other (fill field below) nuSpaceSim

### Primary authors

Sameer Patel (University of Iowa) Mary Hall Reno (University of Iowa)

### Presentation materials

There are no materials yet.


text:
Question

# The valency of Cu in cuprous chloride (${ Cu }_{ 2 }{ Cl }_{ 2 }$) is 1.A.TrueB.False

Hint:To find the valency of copper first you need to write the electronic configuration and don’t confuse yourself with the molecular formula given, you can write it as CuCl.

- ${ Cu }_{ 2 }{ Cl }_{ 2 }$ is the dimeric formula of CuCl.
So, solving the valency of Cu in CuCl will be a lot easier to understand for us.
-Now, Copper (Cu) belongs to the 4th period and 11th group in the periodic table.
Electronic configuration of copper can be written as,
Cu (29)− [Ar] $3d^{ 10 }4s^{ 1 }$
-In the case of CuCl, the electron is lost from the 4s orbital, which is the outermost orbital.
-Here, in ${ Cu }_{ 2 }{ Cl }_{ 2 }$ or indeed CuCl the oxidation state of Cu is +1, which makes the valency equal to 1.
-Since each of the ${ Cl }^{ - }$ ions has a -1 charge and the stoichiometry is 1:1.
Hence, we can say that the valency of Cu in cuprous chloride (${ Cu }_{ 2 }{ Cl }_{ 2 }$) is 1.

Therefore, we can conclude that the correct answer to this question is option A.

Note:
Copper being a transition metal exhibits variable valencies of 1 and 2. Copper forms two types of chlorides - cuprous chloride, ${ Cu }_{ 2 }{ Cl }_{ 2 }$, and cupric chloride, $CuCl_{ 2 }$.


text:
# What does “explained variation” mean in reference to R-squared?

I have been trying to get my head around $R^2$ in a bit more details instead of just seeing it as a number.

So far I have looked at the process in the following manner:

If I knew very little about the $x$ and $y$ variables and was asked to make a prediction for $y$ for any $x$ I'd guess the mean of $y$ as my prediction. However if I could see the regression line I'd use this and this to give me a more accurate prediction.

Here's where my confusion starts…

I measure the difference between a $y$ point and the mean for $y$ and then square this. I repeat for every $y$ point and sum them this is total variance in $y$. Lets label this B.

I also measure the difference between every $y$ point and its predicted $y$ point on the line and again square this. I take the total of all of these and This is called explained variation mean. Lets label this A.

So 1 - (A/B) gives me my $R^2$. So A/B is some kind of ratio? But I don't get what it means by explained variation? How does the regression line explain these points?

Whats does it mean for the regression line to account for variance?

It seems the further the points are from the mean while being closer to the line the better?

thanks


text:
# Oregon jobless rate ticks up; hiring numbers weak

## Oregon jobless rate ticks up, hiring numbers weaker than usual for June

SALEM, Ore. (AP) -- Oregon's jobless rate inched up a tenth of a percentage point in June, and hiring was weak.

The State Employment Department said Tuesday the unemployment rate of 7.9 percent is essentially unchanged from the 7.8 percent reported in May.

The rate is based on a survey of households. Economists say such small changes are statistically insignificant until they show a trend over months.

A survey of businesses put the nonfarm payroll employment down by 1,000 jobs, the first decline since September.

The agency said revised estimates for May showed a gain that month of 4,700. The initial report was a smaller gain, 3,800.

June hiring was stronger than expected in leisure and hospitality businesses, such as restaurants and bars. It was weaker than usual in trade, transportation and utility sectors.


text:
# nLab semisimple category

monoidal categories

## With traces

• trace

• traced monoidal category?

# Contents

## Idea

A semisimple category is a category in which each object is a direct sum of finitely many simple objects, and all such direct sums exist.

## Definition

An abelian category is called semisimple if every object is a direct sum of finitely many simple objects. See semisimple abelian category.

Alternatively, a monoidal linear category (that is, a monoidal category enriched over Vect) is called semisimple if:

• it has finite biproducts (usually called ‘direct sums’),

• idempotents split (so we say that it ‘has subobjects’ or, perhaps better, ‘has retracts’), and

What does it mean to ‘have subobjects’? (I assume that the ‘direct sums’ are biproducts.) —Toby

have subobjects = idempotents split and yes, finite biproducts. Simple objects are ones in which End(X) = k.

Urs: shouldn’t we say something like: a category is semisimple if each object is a direct sum of finitely many simple objects?

Bruce: Urs, you’re right, and that’s indeed the way one morally thinks about it, but it’s a less canonical way of proceeding. We ask ourselves: given a linear category with direct sums and subobjects, and a chosen maximal collection $\left\{{X}_{i}\right\}$ of nonisomorphic simple objects, how can we check if its semisimple? In the one way, we have to check whether a certain canonically defined map is an isomorphism. In the other way, we have to check if each object $V$ can be expressed as a direct sum of the ${X}_{i}$’s. Actually finding such a decomposition would be a noncanonical operation. So your shorter more snappy definition would force an auditor to perform an evil thing if he actually wanted to check it :-) It’s that old thing about “only that part of a representation which behaves like an irreducible $\rho$ is canonical, the actual break-down of that rep into direct sums of $\rho$’s is noncanonical”. That is, what is canonical is $\mathrm{Hom}\left(V,{X}_{i}\right)$ and not $V={⨁}_{i}{n}_{i}{X}_{i}$.

Also, if we just had “a category is semisimple if each object is a direct sum of finitely many simple objects” without the conditions on direct sums and subobjects then we could have someone who nastily removes, say, all three-dimensional vector spaces from $\mathrm{Vect}$. It would still satisfy “each object is a direct sum of finitely many simple objects” but it shouldnt be regarded as a semisimple category since there are “holes”.

Urs: okay, so “each object is finite sum of simples” is the right idea, while the right defintion is a bit different. I have accoridngly now created a section “Idea” with the former statement. (Every entry should start with a section “Idea”!) See if you like this. Otherwise, feel free to adjust.

But in any case, it would be nice to have a discussion on how the “right definition” implies that every object is isomorphic to a finite direct sum of representables.

Chris Schommer-Pries: Shouldn’t you require that for simples, End(X,X) is a simple algebra, not necessarily the ground field? For example the category of H-modules where H is the quaternion algebra over the reals. Shouldn’t this be semi-simple?

• there exist objects ${X}_{i}$ labeled by an index set $I$ such that $\mathrm{Hom}\left({X}_{i},{X}_{j}\right)\cong {\delta }_{\mathrm{ij}}k$ where $k$ is the ground field (such objects are called simple) and such that for any two objects $V$ and $W$ in the category, the natural composition map

(1)$\underset{i\in I}{⨁}\mathrm{Hom}\left(V,{X}_{i}\right)\otimes \mathrm{Hom}\left({X}_{i},W\right)\to \mathrm{Hom}\left(V,W\right)$\bigoplus_{i \in I} Hom(V, X_i) \otimes Hom(X_i, W) \rightarrow Hom(V, W)

is an isomorphism.

## Direct sums of simple objects

Note that this definition implies that every object $V$ is a direct sum of simple objects ${X}_{i}$. To see this, note that the third item of the definition is equivalent to stipulating that the vector space $\mathrm{Hom}\left({X}_{i},V\right)$ is in canonical duality with the vector space $\mathrm{Hom}\left(V,{X}_{i}\right)$. Indeed, we have a canonical pairing

(2)$\mathrm{Hom}\left(V,{X}_{i}\right)\otimes \mathrm{Hom}\left({X}_{i},V\right)\to k$Hom(V, X_i) \otimes Hom(X_i, V) \rightarrow k

given by sending $f\otimes g↦⟨f\circ g⟩$ where the ”$⟨\cdot ⟩$” notation refers to extracting scalars from endomorphisms of simple objects (each such endomorphism is a scalar multiple of the identity). We also have a canonical copairing

(3)$k\to \mathrm{Hom}\left({X}_{i},V\right)\otimes \mathrm{Hom}\left(V,{X}_{i}\right)$k \rightarrow Hom(X_i, V) \otimes Hom(V, X_i)

given by sending ${id}_{{X}_{i}}$ to the ”$i$th block” of the image of the identity ${id}_{V}$ arrow under the isomorphism given in the definition. One can check that this pairing and copairing satisfy the snake equations. Hence if we choose a basis

(4)$\left\{{a}_{i,p}:{X}_{i}\to V\right\}$\{ a_{i,p} : X_i \rightarrow V \}

for each vector space $\mathrm{Hom}\left({X}_{i},V\right)$, we get a corresponding dual basis

(5)$\left\{{a}_{i}^{p}:V\to {X}_{i}\right\}$\{ a_i^p : V \rightarrow X_i \}

satisfying

(6)${a}_{i}^{p}{a}_{j,q}={\delta }_{\mathrm{ij}}{\delta }_{p}^{q}\phantom{\rule{1em}{0ex}}\mathrm{and}\phantom{\rule{1em}{0ex}}\sum _{i,p}{a}_{i,p}{a}_{i}^{p}={id}_{V}.$a_i^p a_{j,q} = \delta_{ij} \delta_p^q \quad and \quad \sum_{i,p} a_{i,p} a_i^p = \id_V.

This says precisely that $V$ has been expressed as a direct sum of the ${X}_{i}$.

## Remarks

• The above definition definition of semisimple monoidal linear category (taken from the reference of Müger below) does not use the concept of abelian category. This is because the concepts that one thinks about with abelian categories such as kernels and cokernels do not play an important conceptual role in semisimple categories, being replaced by the more important concepts of biproduct and retract. Hence it is best to give a streamlined definition from first principles without going through the language of abelian categories which would have muddied the waters.

• For a category to be semisimple, it needs to have a certain directional symmetry in its hom-sets, namely that $\mathrm{Hom}\left(V,W\right)$ must at least have the same dimension as$\mathrm{Hom}\left(W,V\right)$. This is the easiest way to check if a category will fail to be semisimple. For instance, the category $\mathrm{Rep}\left(A\right)$ of representations of an algebra $A$ will rarely be semisimple, precisely because there is no relation between $\mathrm{Hom}\left(V,W\right)$ and $\mathrm{Hom}\left(W,V\right)$ in general. Again, this can be traced back to the original algebra $A$ not having any ‘symmetry’ like the inverse operation in a group.

• As far as ‘duality’ on the hom-sets is concerned, one might have a $S:C\to C$ from the category to itself with the property that there are canonical isomorphisms

(7)$\mathrm{Hom}\left(V,W\right)\cong \mathrm{Hom}\left(W,\mathrm{SV}{\right)}^{\vee }$Hom(V, W) \cong Hom(W, SV)^\vee

where ”$\vee$” denotes the ordinary linear dual of a vector space. Such a functor is called a Serre functor in algebraic geometry, and indeed there is precisely such a functor on the derived category of coherent sheaves on a complex manifold — it is given by tensoring with the canonical line bundle.

• For 2-Hilbert spaces, there is an antilinear $*$-operation on the hom-sets $*:\mathrm{Hom}\left(V,W\right)\to \mathrm{Hom}\left(W,V\right)$. The presence of this duality in fact forces the category to be semisimple (this comes down to the fact that a finite-dimensional $*$-algebra, such as the hom’s between a bunch of objects in the category, must be a full matrix algebra)

## Examples

• The archetypical simple example is Vect itself, the category of (finite dimensional!) vector spaces over some ground field $k$. This has a single isomorphism class of simple objects: given by $k$ itself.

• The category of finite-dimensional representations of a compact Lie group $G$ is semisimple, with the simple objects being precisely the irreducible representations (this is the content of Schur's lemma). If $G$ is noncompact, one needs to pass from the concept of ‘direct sum’ to ‘direct integral?’.

• Every fusion category is a semisimple category.

## References

Revised on September 26, 2012 12:49:00 by Jamie Vicary (81.135.52.3)


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# drfloob/SimulationsAndTheCLT.RmdSecret Created Jun 7, 2016

Investigating the Central Limit Theorem through Simulation [R] [knitr] [rmarkdown] [statistics]
title author date output
Investigating the Central Limit Theorem through Simulation
AJ Heller (aj@drfloob.com; [http://drfloob.com](http://drfloob.com))
May 26, 2016
html_document
toc toc_float theme
true
true
readable
knitr::opts_chunk$set(echo = TRUE, fig.width = 10, fig.height = 6)  ## Overview This report explores the Central Limit Theorem (CLT) by repeatedly sampling from a non-normal distribution and analyzing the the means of the samples. From this sampled mean distribution, we compare the sample mean and variance to the theoretical values we expect to find. Finally, we show the distribution to be approximately normal. ### Setup This report does not require many libraries. I prefer ggplot2, though lattice and base plotting are fine alternatives. library(ggplot2) library(grid)  ## Simulations We begin by running 1000 simulations, each taking a sample of 40 random values from an exponential distribution with parameter$\lambda = 0.2$. This creates a matrix of 1000 samples along the rows, with the 40 sample points stored along the columns. set.seed(420) lambda <- 0.2 n <- 40 sims <- 1000 x <- matrix(rexp(n*sims, lambda), sims, n)  Those 1000 rows -- each containing 40 random values -- are then averaged to create a vector of sample means from every sample of size 40. means <- apply(x, 1, mean)  ## Sample Mean versus Theoretical Mean For a bit of theoretical background, it's helpful to think of an exponential distribution as describing the time between independent events that occur at some continuous rate (See Poisson Process to dig deeper into that). To quote wikipedia: The mean or expected value of an exponentially distributed random variable$X$with rate parameter$\lambda$is given by$E[X] = \frac{1}{\lambda}$... [intuitively] this makes sense: if you receive phone calls at an average rate of 2 per hour, then you can expect to wait half an hour for every call. In terms of this report, the "phone calls" would come on average 0.2 per hour, so we would expect to wait$\dfrac{1}{.2} = 5$hours for every call (again, on average). Let's now see how that value measures up against our distribution of averages of 40 exponentials. The sample mean is: mean(means)  ggplot(data.frame(means=means), aes(means)) + geom_histogram(binwidth = 0.3, boundary=0, fill="#8b5577", color="#333333") + geom_vline(xintercept = 1/lambda, size=1.5, color="#333333") + geom_vline(xintercept = mean(means), size=1.5, color="#cccccc") + xlab("mean value") + ggtitle("Distribution of Means of 40 Exponential Random Values")  In the plot above, the gray line marks the sample mean and the black line marks the theoretical mean. As you can see, they are nearly on top of each other. The difference can be explained by sampling error; the larger our sample size, the closer our sample mean should be to the theoretical population mean. ## Sample Variance versus Theoretical Variance The theoretical variance of the distribution of sample means is given by$\dfrac{\sigma^2}{n}$, where$\sigma^2$is the exponential distribution's variance, and$n$is the number of observations in each sample (40). We're given from the exponential distribution$\sigma^2 = \dfrac{1}{\lambda^2}$, so the population (theoretical) variance is: tVar <- (1/lambda^2)/n print(tVar)  The sample variance is easy to calculate in R: sVar <- var(means) print(sVar)  These numbers are very close. The sample variance is r sprintf("%0.2f%%", sVar/tVar*100) of the theoretical variance. These numbers get closer together the larger our sample size is. A sample size of 40 is fairly small for our distribution of sample means to exhibit the normality promised by the CLT (stay tuned), so we should expect some small difference here. ## Normality of the Distribution To inspect for normality visually, we plot a histogram and look for a bell-shaped curve. For comparison, to show non-normality of our original data, we first plot the first 1000 random variables sampled from our exponential distribution. Then we'll again plot the 1000 sample means. Both plots have an overlaid normal curve with the same mean and standard deviation as the data being plotted (code found here).  p1 <- ggplot(data.frame(v = x[,1]), aes(v)) + geom_histogram(binwidth = 2, boundary=0, fill="#8b5577", color="#333333") + xlab("value") + ggtitle("Samples from an Exponential Distribution") + stat_function(fun=function(y,m,sd){ dnorm(y,m,sd)*length(x[,1])*2 }, args=list(m=mean(x[,1]), sd=sd(x[,1]))) p2 <- ggplot(data.frame(means=means), aes(means)) + geom_histogram(binwidth = 0.3, boundary=0, fill="#8b5577", color="#333333") + xlab("value") + ggtitle("Means of Samples from an Exponential Distribution") + stat_function(fun=function(x,m,sd){ dnorm(x,m,sd)*sims*0.3 }, args=list(m=mean(means), sd=sqrt(sVar))) grid.newpage() pushViewport(viewport(layout = grid.layout(1, 2))) print(p1, vp = viewport(layout.pos.row = 1, layout.pos.col = 1)) print(p2, vp = viewport(layout.pos.row = 1, layout.pos.col = 2))  As you can see, the first plot towers well above the curve for values < ~4, while the second plot lines up with the curve quite nicely. In this way, we can visually confirm that the distribution of sample means is approximately normal, and the distribution of exponentials is definitely not. I'll add one more visual measure of normality, namely a Q-Q Normal Plot, which compares our sample means with a normal distribution. The more normal the data, the closer the points will lie to the line$y=x\$.

par(mfrow=c(1,2))
qqnorm(x[,1], main = "Normal Q-Q Plot: sampled exponentials")
qqnorm(means, main = "Normal Q-Q Plot: sample means")

The plot of mean samples (right) is quite linear, leading us again to believe it's an approximately normal distribution. The strong curve in the exponentials Q-Q Plot (left) indicates again that the data is unlikely to be normal.

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# Standard RAID levels

(Redirected from RAID1)

The standard RAID levels are a basic set of RAID configurations that employ the techniques of striping, mirroring, or parity to create large reliable data stores from general purpose computer hard disk drives. The most common types today are RAID 0 (striping), RAID 1 and variants (mirroring), RAID 5 (distributed parity) and RAID 6 (dual parity). RAID levels and their associated data formats are standardized by the Storage Networking Industry Association in the Common RAID Disk Drive Format (DDF) standard.[1]

## RAID 0

Diagram of a RAID 0 setup

A RAID 0 (also known as a stripe set or striped volume) splits data evenly across two or more disks (striped) without parity information for speed. RAID 0 was not one of the original RAID levels and provides no data redundancy. RAID 0 is normally used to increase performance, although it can also be used as a way to create a large logical disk out of two or more physical ones.

A RAID 0 can be created with disks of differing sizes, but the storage space added to the array by each disk is limited to the size of the smallest disk. For example, if a 100 GB disk is striped together with a 350 GB disk, the size of the array will be 200 GB (100 GB × 2).

\begin{align} \mathrm{Size} & = 2 \cdot \min \left( 100\,\mathrm{GB}, 350\,\mathrm{GB} \right) \\ & = 2 \cdot 100\,\mathrm{GB} \\ & = 200\,\mathrm{GB} \end{align}

The diagram shows how the data is distributed into Ax stripes to the disks. Accessing the stripes in the order A1, A2, A3, ... provides the illusion of a larger and faster drive. Once the stripe size is defined on creation it needs to be maintained at all times.

### Performance

RAID 0 is also used in areas where performance is desired and data integrity is not very important, for example in some computer gaming systems. Although some real-world tests with computer games showed a minimal performance gain when using RAID 0, albeit with some desktop applications benefiting,[2][3] another article examined these claims and concluded: "Striping does not always increase performance (in certain situations it will actually be slower than a non-RAID setup), but in most situations it will yield a significant improvement in performance."[4]

## RAID 1

Diagram of a RAID 1 setup

An exact copy (or mirror) of a set of data on two disks. This is useful when read performance or reliability is more important than data storage capacity. Such an array can only be as big as the smallest member disk. A classic RAID 1 mirrored pair contains two disks.[5]

## RAID 2

RAID Level 2

A RAID 2 stripes data at the bit (rather than block) level, and uses a Hamming code for error correction. The disks are synchronized by the controller to spin at the same angular orientation (they reach Index at the same time), so it generally cannot service multiple requests simultaneously. Extremely high data transfer rates are possible. This is the only original level of RAID that is not currently used.[6][7]

All hard disks eventually implemented Hamming code error correction. This made RAID 2 error correction redundant and unnecessarily complex. This level quickly became useless and is now obsolete. There are no commercial applications of RAID 2.[6][7]

## RAID 3

Diagram of a RAID 3 setup of 6-byte blocks and two parity bytes, shown are two blocks of data in different colors.

A RAID 3 uses byte-level striping with a dedicated parity disk. RAID 3 is very rare in practice. One of the characteristics of RAID 3 is that it generally cannot service multiple requests simultaneously. This happens because any single block of data will, by definition, be spread across all members of the set and will reside in the same location. So, any I/O operation requires activity on every disk and usually requires synchronized spindles.

This makes it suitable for applications that demand the highest transfer rates in long sequential reads and writes, for example uncompressed video editing. Applications that make small reads and writes from random disk locations will get the worst performance out of this level.[7]

The requirement that all disks spin synchronously, a.k.a. lockstep, added design considerations to a level that didn't give significant advantages over other RAID levels, so it quickly became useless and is now obsolete.[6] Both RAID 3 and RAID 4 were quickly replaced by RAID 5.[8] RAID 3 was usually implemented in hardware, and the performance issues were addressed by using large disk caches.[7]

## RAID 4

Diagram of a RAID 4 setup with dedicated parity disk with each color representing the group of blocks in the respective parity block (a stripe)

A RAID 4 uses block-level striping with a dedicated parity disk.

In the example on the right, a read request for block A1 would be serviced by disk 0. A simultaneous read request for block B1 would have to wait, but a read request for B2 could be serviced concurrently by disk 1.

RAID 4 is very uncommon, but one enterprise level company that has previously used it is NetApp. The aforementioned performance problems were solved with their proprietary Write Anywhere File Layout (WAFL), an approach to writing data to disk locations that minimizes the conventional parity RAID write penalty. By storing system metadata (inodes, block maps, and inode maps) in the same way application data is stored, WAFL is able to write file system metadata blocks anywhere on the disk. This approach in turn allows multiple writes to be "gathered" and scheduled to the same RAID stripe—eliminating the traditional read-modify-write penalty prevalent in parity-based RAID schemes.[9]

## RAID 5

Diagram of a RAID 5 setup with distributed parity with each color representing the group of blocks in the respective parity block (a stripe). This diagram shows left asymmetric algorithm

A RAID 5 comprises block-level striping with distributed parity. Unlike in RAID 4, parity information is distributed among the drives. It requires that all drives but one be present to operate. Upon failure of a single drive, subsequent reads can be calculated from the distributed parity such that no data is lost.[10] RAID 5 requires at least three disks.[11]

## RAID 6

Diagram of a RAID 6 setup, which is identical to RAID 5 other than the addition of a second parity block

RAID 6 extends RAID 5 by adding an additional parity block; thus it uses block-level striping with two parity blocks distributed across all member disks.

### Performance (speed)

RAID 6 does not have a performance penalty for read operations, but it does have a performance penalty on write operations because of the overhead associated with parity calculations. Performance varies greatly depending on how RAID 6 is implemented in the manufacturer's storage architecture – in software, firmware or by using firmware and specialized ASICs for intensive parity calculations. It can be as fast as a RAID-5 system with one fewer drive (same number of data drives).[12]

### Implementation

According to the Storage Networking Industry Association (SNIA), the definition of RAID 6 is: "Any form of RAID that can continue to execute read and write requests to all of a RAID array's virtual disks in the presence of any two concurrent disk failures. Several methods, including dual check data computations (parity and Reed-Solomon), orthogonal dual parity check data and diagonal parity, have been used to implement RAID Level 6."[13]

#### Computing parity

Two different syndromes need to be computed in order to allow the loss of any two drives. One of them, P can be the simple XOR of the data across the stripes, as with RAID 5. A second, independent syndrome is more complicated and requires the assistance of field theory.

To deal with this, the Galois field $GF(m)$ is introduced with $m=2^k$, where $GF(m) \cong F_2[x]/(p(x))$ for a suitable irreducible polynomial $p(x)$ of degree $k$. A chunk of data can be written as $d_{k-1}d_{k-2}...d_0$ in base 2 where each $d_i$ is either 0 or 1. This is chosen to correspond with the element $d_{k-1}x^{k-1} + d_{k-2}x^{k-2} + ... + d_1x + d_0$ in the Galois field. Let $D_0,...,D_{n-1} \in GF(m)$ correspond to the stripes of data across hard drives encoded as field elements in this manner (in practice they would probably be broken into byte-sized chunks). If $g$ is some generator of the field and $\oplus$ denotes addition in the field while concatenation denotes multiplication, then $\mathbf{P}$ and $\mathbf{Q}$ may be computed as follows ($n$ denotes the number of data disks):

$\mathbf{P} = \bigoplus_i{D_i} = \mathbf{D}_0 \;\oplus\; \mathbf{D}_1 \;\oplus\; \mathbf{D}_2 \;\oplus\; ... \;\oplus\; \mathbf{D}_{n-1}$
$\mathbf{Q} = \bigoplus_i{g^iD_i} = g^0\mathbf{D}_0 \;\oplus\; g^1\mathbf{D}_1 \;\oplus\; g^2\mathbf{D}_2 \;\oplus\; ... \;\oplus\; g^{n-1}\mathbf{D}_{n-1}$

For a computer scientist, a good way to think about this is that $\oplus$ is a bitwise XOR operator and $g^i$ is the action of a linear feedback shift register on a chunk of data. Thus, in the formula above,[14] the calculation of P is just the XOR of each stripe. This is because addition in any characteristic two finite field reduces to the XOR operation. The computation of Q is the XOR of a shifted version of each stripe.

Mathematically, the generator is an element of the field such that $g^i$ is different for each nonnegative $i$ satisfying $i < n$.

If one data drive is lost, the data can be recomputed from P just like with RAID 5. If two data drives are lost or a data drive and the drive containing P are lost, the data can be recovered from P and Q or from just Q, respectively, using a more complex process. Working out the details is extremely hard with field theory. Suppose that $D_i$ and $D_j$ are the lost values with $i \neq j$. Using the other values of $D$, constants $A$ and $B$ may be found so that $D_i \oplus D_j = A$ and $g^iD_i \oplus g^jD_j = B$:

$A = \bigoplus_{\ell:\;\ell\not=i\;\mathrm{and}\;\ell\not=j}{D_\ell} = \mathbf{P} \;\oplus\; \mathbf{D}_0 \;\oplus\; \mathbf{D}_1 \;\oplus\; \dots \;\oplus\; \mathbf{D}_{i-1} \;\oplus\; \mathbf{D}_{i+1} \;\oplus\; \dots \;\oplus\; \mathbf{D}_{j-1} \;\oplus\; \mathbf{D}_{j+1} \;\oplus\; \dots \;\oplus\; \mathbf{D}_{n-1}$
$B = \bigoplus_{\ell:\;\ell\not=i\;\mathrm{and}\;\ell\not=j}{g^{\ell}D_\ell} = \mathbf{Q} \;\oplus\; g^0\mathbf{D}_0 \;\oplus\; g^1\mathbf{D}_1 \;\oplus\; \dots \;\oplus\; g^{i-1}\mathbf{D}_{i-1} \;\oplus\; g^{i+1}\mathbf{D}_{i+1} \;\oplus\; \dots \;\oplus\; g^{j-1}\mathbf{D}_{j-1} \;\oplus\; g^{j+1}\mathbf{D}_{j+1} \;\oplus\; \dots \;\oplus\; g^{n-1}\mathbf{D}_{n-1}$

Multiplying both sides of the equation for $B$ by $g^{n-i}$ and adding to the former equation yields $(g^{n-i+j}\oplus1)D_j = g^{n-i}B\oplus A$ and thus a solution for $D_j$, which may be used to compute $D_i$.

The computation of Q is CPU intensive compared to the simplicity of P. Thus, a RAID 6 implemented in software will have a more significant effect on system performance, and a hardware solution will be more complex.

## Comparison

The following table provides an overview of some considerations for standard RAID levels. In each case:

• Array space efficiency is given as an expression in terms of the number of drives, $n$; this expression designates a fractional value between zero and one, representing the fraction of the sum of the drives' capacities that is available for use. For example, if three drives are arranged in RAID 3, this gives an array space efficiency of $1 - (1/n) = 1 - (1/3) = 2/3 \approx 67%$ thus, if each drive in this example has a capacity of 250 GB, then the array has a total capacity of 750 GB but the capacity that is usable for data storage is only 500 GB.
• Array failure rate is given as an expression in terms of the number of drives, $n$, and the drive failure rate, $r$ (which is assumed identical and independent for each drive) and can be seen to be a Bernoulli trial.[citation needed] For example, if each of three drives has a failure rate of 5% over the next three years, and these drives are arranged in RAID 3, then this gives an array failure rate over the next three years of:
\begin{align} 1 - (1 - r)^{n} - nr(1 - r)^{n - 1} & = 1 - (1 - 5\%)^{3} - 3 \times 5\% \times (1 - 5\%)^{3 - 1} \\ & = 1 - 0.95^{3} - 0.15 \times 0.95^{2} \\ & = 1 - 0.857375 - 0.135375 \\ & = 0.00725 \\ & \approx 0.7\% \end{align}
Level Description Minimum number of drives[a] Space efficiency Fault tolerance Array failure rate[b] Read performance Write performance Figure
RAID 0 Block-level striping without parity or mirroring 2 1 0 (none) $1-(1-r)^{n}$
RAID 1 Mirroring without parity or striping 2 1/n n−1 drives $r^{n}$ [c]
RAID 2 Bit-level striping with dedicated Hamming-code parity 3 1 − 1/n ⋅ log2(n-1) RAID 2 can recover from one drive failure or repair corrupt data or parity when a corrupted bit's corresponding data and parity are good. (Varies) (Varies) (Varies)
RAID 3 Byte-level striping with dedicated parity 3 1 − 1/n 1 drive $1-(1-r)^{n}-nr(1-r)^{n-1}$ (n−1)× (n−1)×[d]
RAID 4 Block-level striping with dedicated parity 3 1 − 1/n 1 drive $1-(1-r)^{n}-nr(1-r)^{n-1}$ (n−1)× (n−1)×[d][citation needed]
RAID 5 Block-level striping with distributed parity 3 1 − 1/n 1 drive $1-(1-r)^{n}-nr(1-r)^{n-1}$ (n−1)×[d] (n−1)×[d][citation needed]
RAID 6 Block-level striping with double distributed parity 4 1 − 2/n 2 drives $1-(1-r)^{n}-nr(1-r)^{n-1}-{n\choose 2}r^{2}(1-r)^{n-2}$ (n−2)×[d] (n−2)×[d][citation needed]
RAID 1+0 Mirroring without parity, and block-level striping 4 2/n 1 drive / span[e] [f] (n/spans)×
Level Description Minimum number of drives[a] Space efficiency Fault tolerance Array failure rate[b] Read performance Write performance Figure

## Non-standard RAID levels and non-RAID drive architectures

Alternatives to the above designs include nested RAID levels, non-standard RAID levels, and non-RAID drive architectures. Non-RAID drive architectures are referred to by similar acronyms, notably SLED, Just a Bunch of Disks, SPAN/BIG, and MAID.

## Notes

1. ^ a b Assumes a non-degenerate minimum number of drives
2. ^ a b Assumes independent, identical rate of failure amongst drives
3. ^ Theoretical maximum, as low as 1× in practice
4. Assumes hardware is fast enough to support
5. ^ RAID 10 can only lose one drive per span up to the max of n/2 drives.
6. ^ Theoretical maximum, as low as (n/spans)× in practice

## References

1. ^ "Common raid Disk Data Format (DDF)". SNIA.org. Storage Networking Industry Association. Retrieved 2013-04-23.
2. ^ "Western Digital's Raptors in RAID-0: Are two drives better than one?". AnandTech. July 1, 2004. Retrieved 2007-11-24.
3. ^ "Hitachi Deskstar 7K1000: Two Terabyte RAID Redux". AnandTech. April 23, 2007. Retrieved 2007-11-24.
4. ^ "RAID 0: Hype or blessing?". Tweakers.net. August 7, 2004. Retrieved 2008-07-23.
5. ^ "19.3. RAID1 - Mirroring". FreeBSD Handbook. freebsd.org. 2014-03-23. Retrieved 2014-06-11.
6. ^ a b c Derek Vadala (2003). Managing RAID on Linux. O'Reilly Series (illustrated ed.). O'Reilly. p. 6. ISBN 9781565927308.
7. ^ a b c d Evan Marcus, Hal Stern (2003). Blueprints for high availability (2, illustrated ed.). John Wiley and Sons. p. 167. ISBN 9780471430261.
8. ^ Michael Meyers, Scott Jernigan (2003). Mike Meyers' A+ Guide to Managing and Troubleshooting PCs (illustrated ed.). McGraw-Hill Professional. p. 321. ISBN 9780072231465.


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# Hypergraphic matroid

Hypergraphic matroids were introduced by Lorea [1] as a generalization of graphic matroids. Given a hypergraph H=(V,E), a subset F of E is a hyperforest if one can choose two nodes from each hyperedge of F so that the resulting graph is a forest. The circuit matroid of H is the matroid on ground set E whose independent sets are the hyperforests. A matroid is hypergraphic if it is the circuit matroid of some hypergraph.

If M is the circuit matroid of a hypergraph H=(V,E), we call kM (the matroid sum of k copies of M) the k-circuit matroid of H. The k-circuit matroid has rank $k(|V|-1)$ if and only if H is k-partition-connected.

## Properties

Although in many respect they are similar to graphic matroids, there are important differences:

• Not all hypergraphic matroids are binary: If $|V|=3$ and E consists of 4 copies of V, we get $U_{4,2}$ as the circuit matroid.
• The class of hypergraphic matroids is not closed under contraction.

## References

1. M. Lorea, Hypergraphes et matroides, Cahiers Centre Etud. Rech. Oper. 17 (1975), 289-291.


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# BibLatex Cite Primary Source Within Book

How do I cite a Primary Source within a secondary source?

For example, in Eyewitness to Irish History, a section of the Primary Source The Mystery of the Casement Ship, the particular section being a quote from the captain of a sunk ship, is shown. How do I cite this primary source, while it is within a secondary?

I am using Biblatex.

• It really depends on the citation style (MLA, Chicago, APA, etc.), but it is not uncommon to do something like: 'Quoted in <Secondary Source>, p. <pageref>'. It might be helpful to edit the 'For example...' sentence to make it clearer. I find it hard to follow. – jon May 4 '13 at 4:02
• This is best done with "related entries" using the RELATED* fields. See the biblatex doc and the related entries example file which comes with biblatex. If you do it this way (which requires biber) the primary and secondary sources each have their own entries which means they can be used/reused in other ways by other entries or cited on their own etc. – PLK May 4 '13 at 4:26

I'd say your question isn't really TeX-related, but it's a question of how to properly mark a second-hand quote as such -- or, as I prefer to call them, a ›blind quote‹. (= a quote from a text that you haven't actually seen, but that's only been handed down to you through another text. Blind quotes tend to be frowned upon and are best dealt with by staying away from them.) The answer to that question will be the same across all software systems, and very similar across different citation schemes as well. A jon said, such a blind quote is usually marked by a phrase like »quoted in«, followed by a reference to the book that quotes the quote you want to quote.

So, with biblatex, you'll simply use the prenote for that:

\cite[quoted in][123]{Doe87}
...depending on the style that you've set up biblatex to use, this will give you something like:

»...«, said Captain Soandso in 1774 (quoted in Doe 1987: 123).

or

»...«, said Captain Soandso in 1774.1
__________
1) quoted in Doe 1987: 123.

etc.

[alternative scenario:] If the Eyewitness book contains more than just a quote from the Mystery text, i.e. if Eyewitness is a collection of primary sources that are re-printed in large portions -- then we're no longer talking about blind quotes, since these collections are usually meant to be quoted from, as they make available text that are hard to find otherwise.

bib(la)tex has the @INCOLLECTION entry type for things like that. The Eyewitness will get their own entry, as will the Mystery. The latter has a crossref field referring to the former, so that, in the bibliography/references, the Mystery will be a self-contained entry providing the reader with all the information they need to track down the text that you're citing -- regardless of whether, at a differnt point in your text, you cite the main book or not. Note that this example uses authoryear style, but it should work with any other style as well.

\documentclass{scrartcl}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage{csquotes}
\usepackage{filecontents}
\usepackage[style=authoryear]{biblatex}

\begin{filecontents}{testbib.bib}
@COLLECTION{Eye,
editor =       {Witness, Walt},
title =        {Eyewitness to Irish History},
publisher =    {Dublin UP},
year =         {2013},
}
@INCOLLECTION{Ship,
author =       {Captain, Kirk},
title =        {The Mystery of the Casement Ship},
year =         {1774},
pages =        {123-24},
crossref =     {Eye}
}
\end{filecontents}

\bibliography{testbib.bib}

\begin{document}
\enquote{\dots} says \textcite[123]{Ship}.
\printbibliography
\end{document}

• +1, but if Eyewitness is a "collection of primary sources" (and Walt Witness presumably its editor), one should rather use the @collection entry type for it (and @incollection for Captain Kirk). – lockstep May 5 '13 at 10:04
• Sure, you'd have to check what kind of book you're dealing with and what's the role that Walt Witness is playing. I'll modify my answer, as most books of that kind are indeed @collections with editors. – Nils L May 5 '13 at 10:09


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# Does this wavefunction make sense?

• I
Hi all!
Consider a wavefunction, where ##\left| \psi (x) \right|^2 = e^{-ax^2+1}+e^{-bx^2-1}## where a and b are real, positive numbers that satisfy normalization (they are purpously inside the exponent). Even if it is normalized, there are still 2 spots that ##\left| \psi \right|^2 > 1## which makes no sense. What is going on here?
Thanks!


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# How to prove a pseudo-random function has certain expansion factor

How should I answer the following problem:

Let $F = F0 \times F1$ be a pseudorandom generator with expansion factor $2n$ such that for all $x \in \{0, 1\}^n$ $$F(x) = (F_0 (x) || F_1 (x))$$ and $$|x| = |F_0 (x)| = |F_1(x)|.$$ Prove that $$F'(x) = (F_0 (F_0 (x)) || F_0 (F_1 (x)) || F_1 (F_0 (x)) || F_1 (F_1 (x)))$$ is a pseudorandom generator with expansion factor $4n.$

Here I am not able to understand how expansion factor changes?

I am more interested in general proof structure for solving similar proofs.

I am beginner so more detailed explanation will be helpful.

The concatenation operator $||$ in e definition $F$ ensures an expansion $n \rightarrow 2n$ since both $F_0$ and $F_1$ map $n$ bits to $n$ bits.
The argument for $F'$ is similar since repeated application of two $n$ bit to $n$ bit functions gives an $n$ bit to $n$ bit function, and four of those are concatenated in $F'$.


text:
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#### Half Yearly Model Question Paper 2019

11th Standard

Reg.No. :
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Physics

Time : 02:30:00 Hrs
Total Marks : 70

Part I

Choose the most suitable answer from the given four alternatives and write the option code with the corresponding answer.

15 x 1 = 15
1. One of the combinations from the fundamental physical constants is ${{hc}\over{G}},$ The unit of this expression is

(a)

Kg2

(b)

m3

(c)

S-1

(d)

m

2. How many light years make 1 parsec?

(a)

3.26

(b)

6.67

(c)

1.5

(d)

9.4

3. A car moves from X to Y with a uniform speed Vn and returns to Y with a uniform speed VThe average speed for this round trip is

(a)

$\sqrt{V_uV_d}$

(b)

$\frac{V_uV_d}{V_d+V_u}$

(c)

$\frac{V_u+V_d}{2}$

(d)

$\frac{2V_dV_u}{V_d+V_u}$

4. The centrifugal force appears to exist

(a)

only in inertial frames

(b)

only in rotating frames

(c)

in any accelerated frame

(d)

both in inertial and non-inertial frames

5. A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of heaviest fragment in m/s will be

(a)

7$\sqrt { 2 }$

(b)

5$\sqrt { 2 }$

(c)

3$\sqrt { 2 }$

(d)

$\sqrt { 2 }$

6. A particle which is constrained to move along x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) =kx + ax3. Here, k and a are positive constants. For x≥0, the functional form of the potential, energy U(x) of the particles

(a)

(b)

(c)

(d)

7. Which of the following is not a perfectly inelastic collision?

(a)

Striking of two glass bulbs

(b)

Bullet striking a bag and sand

(c)

An electron captured by a proton

(d)

A man jumping onto a moving cart

8. The center of mass of a system of particles does not depend upon,

(a)

position of particles

(b)

relative distance between particles

(c)

masses of particles

(d)

force acting on particle

9. A rod of length is 3 m and its mass acting per unit length is directly proportional to distance x from one of its end then its centre of gravity from that end will be at

(a)

1.5 m

(b)

2 m

(c)

2.5 m

(d)

3.0 m

10. If the distance between the Earth and Sun were to be doubled from its present value, the number of days in a year would be

(a)

64.5

(b)

1032

(c)

182.5

(d)

730

11. A satellite is moving around the Earth with speed v in a circular orbit of radius r. If the orbit radius is decreased by 1% its speed will _______.

(a)

increase by 1%

(b)

increase by 0.5%

(c)

decrease by 1%

(d)

decrease by 0.5%

12. The pressure of the Earth's atmosphere at sea level is due to the _________.

(a)

gravitational attraction of the Earth for the atmosphere

(b)

evaporation of water from the seas and oceans

(c)

fact that most living things constantly breathe air

(d)

heating of the atmosphere by the Sun

13. A distant star emits radiation with maximum intensity at 350 nm. The temperature of the star is

(a)

8280 K

(b)

5000K

(c)

7260 K

(d)

9044 K

14. A particle executing SHM has an acceleration of 64 cm/s2 with its displacement is 4 cm. Its time period, in seconds is ________.

(a)

$\frac{\pi}{2}$

(b)

$\frac{\pi}{4}$

(c)

$\pi$

(d)

2$\pi$

15. Which of the following options is correct?.

 A B (1) Quality (A) Intensity (2) Pitch (B) Waveform (3) Loudness (C) Frequency

Options for (1), (2) and (3), respectively are

(a)

(B),(C) and (A)

(b)

(C), (A) and (B)

(c)

(A), (B) and (C)

(d)

(B), (A) and (C)

16. Part II

Answer any 7 questions. Question no. 24 is compulsory.

7 x 2 = 14
17. What are types of discoveries in physics?

18. If $\vec A=3\hat i+4\hat j$ and $\vec B=7\hat i+24\hat j$ , find a vector having the same magnitude as $\vec B$ and parallel to $\vec A$.

19. Write the relation between angular momentum and rotational kinetic energy. Draw a graph for the same. For two objects of same angular momentum, compare the moment of inertia using the graph.

20. Calculate moment of inertia with respect to rotational axis xx' in following figures (a) and (b).

21. Two bodies of masses m and 4m are placed at a distance r. Calculate the gravitational potential at a point on the line joining them where the gravitational field is zero.

22. What do you mean by weight of a body? Is it a scalar or vector?

23. Explain the classification of longitudinal strain?

24. A piece of wood of mass m is floating erect in a liquid whose density is ρ. If it is slightly pressed down and released, then executes simple harmonic motion. Show that its time period of oscillation is

25. What is meant by an echo?. Explain.

26. Part III

Answer any 7 questions. Question no. 33 is compulsory.

7 x 3 = 21
27. Calculate the number of times a human heart beats in the life of 100 years old man. Time of one heart beat = 0.8s.

28. Figure shows the position-time graph of a particle of mass 4 kg. What is the

(a) Force on the particle for t < 0, t > 4s, 0 < t < 4s?
(b) Impulse at t = 0 and t = 4s?
(Consider one dimensional motion only)

29. A vehicle of mass 1250 kg is driven with an acceleration 0.2 along a straight level road against an external resistive force 500 N. Calculate the power delivered by the vehicle's engine if the velocity of the vehicle is 30 ms-1.

30. Find Torque about an Axis.

31. Two heavy spheres, each of mass 100kg and radius 0.8m are placed 1m about on a table What is the gravitational force and potential at the mid point of the line joining the centers of the spheres?

32. A light rod of length 2m is suspended horizontally by means of 2 vertical wires of equal lengths tied to its ends. One of the wires is made of steel & is of cross section A1 = 0.1 cm2 & other of brass &. is of cross section A2 = 0.2cm2, find out the position along the rod at which a weight must be suspended to produce (i) equal stresses in both wires, (ii) equai strains in both wires for steel, y = 20 x 1010 Nm-2 & for brass y =10 x 1010 Nm-2.

33. Write the important properties of thermal radiations.

34. Explain in detail the Maxwell Boltzmann distribution function.

35. Consider two organ pipes of same length in which one organ pipe is closed and another organ pipe is open. If the fundamental frequency of closed pipe is 250 Hz. Calculate the fundamental frequency of the open pipe.

36. Part IV

5 x 5 = 25
1. Consider a particle undergoing simple harmonic motion. The velocity of the particle at position x1 is v1 and velocity of the particle at position x2 is v2. Show that the ratio of time period and amplitude is
$\frac { T }{ A } =2\pi \sqrt { \frac { { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } }{ { { v }_{ 1 }^{ 2 }x }_{ 2 }^{ 2 }-{ { v }_{ 2 }^{ 2 }x }_{ 1 }^{ 2 } } }$

2. Write short notes on reflection of sound waves from plane and curved surfaces.

1. Distinguish between isothermal and adlabatlc process.

2. Find the adiabatic exponent $\gamma$ for mixture of μ1 moles of monoatomic gas and μ2 moles of a diatomic gas at normal temperature

1. Drive the relation between g & Gravitational constant.

2. If a ball of skeel (density p =7.8 g cm-3) attains a terminal velocity of 10cm S-1 when falling in a tank of water (coefficient of viscosity) water = 8.5 x 10-4pa s. What will the terminal velocity in glycerine (P = 1.2 g cm-3,η = 13.2 ρa.s) be?

1. Explain the principle of homogeniety of dimensions. What are its uses? Give example

2. Write an expression for the kinetic energy of a body in pure rolling.

1. Write note on integration.

2. Arrive at an expression for power and velocity. Give some examples for the same.