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TowardsMachineLearning

# RetinaNet Model for object detection explanation

## Introduction:-

Lately RetinaNet model for object detection has been buzz word in Deep learning community.

And why should it not ? Object detection is a tremendously important field in computer vision . People are using different object detection methods for autonomous driving, video surveillance, medical applications, and to solve many other business problems.

1. What is RetinaNet
2. Why RetinaNet was needed
3. Architecture of RetinaNet
• Backbone Network
• Subnetwork for object Classification
• Subnetwork for object Regression
4. Focal Loss
5. Final Notes

In this article, I’ll introduce you to the architecture of RetinaNet model & working of it. Cherry on top? In next article , we’ll build a “Face mask detector” using RetinaNet to help us in this ongoing pandemic.

## What is RetinaNet Model: –

Facebook AI research (FAIR ) team has introduced RetinaNet model with aim to tackle dense and small objects detection problem.

For this reason, it has become a popular object detection model that one can use with aerial and satellite imagery also.

Researchers have introduced RetinaNet by making two improvements over existing single stage object detection models –

• Feature Pyramid Networks (FPN)
• Focal Loss

## Need of RetinaNet Model: –

Both classic one stage detection methods, like boosted detectors, DPM & more recent methods like SSD evaluate almost ${ 10 }^{ 4 }$ to ${ 10 }^{ 5 }$ candidate locations per image but only a few locations contain objects (i.e. Foreground) and rest are just background objects. This leads to class imbalance problem.

And this turn out to be the central cause of making performance of one stage detectors inferior to two stage detectors.

Hence , researchers have introduced RetinaNet Model with concept called Focal Loss to fill in for the class imbalances and inconsistencies of the single shot object detectors like YOLO and SSD ,while dealing with extreme foreground-background classes.

## Architecture of RetinaNet Model: –

In essence, we can break down RetinaNet architecture in to 3 following components:

1. Backbone Network (i.e. Bottom up pathway + Top down pathway with lateral connections eg. ResNet + FPN)
2. Sub-network for object Classification
3. Sub-network for object Regression

For better understanding, Let’s understand each component of architecture separately –

#### The backbone Network: –

Bottom up pathway – Bottom up pathway (eg. ResNet) is used for feature extraction. So, It calculates the feature maps at different scales, irrespective of the input image size.

Top down pathway with lateral connections– The top down pathway up samples the spatially coarser feature maps from higher pyramid levels, and the lateral connections merge the top-down layers and the bottom-up layers with the same spatial size.

Higher level feature maps tend to have small resolution though semantically stronger and is therefore more suitable for detecting larger objects; on the contrary, grid cells from lower level feature maps have high resolution and hence are better at detecting smaller objects

So, with combination of the top-down pathway and its lateral connections with bottom up pathway, which do not require much extra computation, every level of the resulting feature maps can be both semantically and spatially strong

Hence this architecture is scale-invariant and can provide better performance both in terms of speed and accuracy.

#### Sub-network for object Classification: –

Fully convolutional network (FCN) is attached to each FPN level for object classification.  As it’s shown in diagram above , This subnetwork incorporates 3*3 convolutional layers with 256 filter followed by another 3*3 convolutional layer with K*A filters. Hence output feature map would be of size W*H*KA , where W & H are proportional to the width and height of input feature map and K & A are number of object class and anchor boxes respectively.

At last , researchers have used Sigmoid layer (not softmax) for object classification.

And reason for last convolution layer to have KA filters is because  , if there’re “A ” number of anchor box proposals for each position in feature map obtained from last convolution layer then each anchor box has possibility to be classified in K number of classes . So the output feature map would be of size KA channels or filters.

#### Sub-network for object Regression: –

The regression subnetwork is attached to each feature map of the FPN in parallel to the classification subnetwork. The design of the regression subnetwork is identical to that of the classification subnet, except that the last convolutional layer is of size 3*3 with 4 filters resulting in output feature map with size of W*H*4A .

Reason for last convolution layer to have 4 filters is because in order to localize the class objects, regression sub-network produces 4 numbers for each anchor box that predict the relative offset (in terms of center coordinates, width and height) between the anchor box and the ground truth box. Therefore, the output feature map of the regression sub-net has 4A filters or channels.

So by now we’ve little clarity on RetinaNet model for object detection architecture. Now let’s understand most discussed topic topic of RetinaNet model for object detection and that is Focal loss.

## Focal Loss : –

Focal Loss (FL) is an improved version of Cross-Entropy Loss (CE)  that tries to handle the class imbalance problem by assigning more weights to hard or easily misclassified examples  (i.e. background with noisy texture or partial object or the object of our interest ) and to down-weight easy examples (i.e. Background objects).

So Focal Loss reduces the loss contribution from easy examples and increases the importance of correcting misclassified examples.)

Focal loss is just an extension of cross entropy loss function that would down-weight easy examples and focus training on hard negatives. So to achieve this researchers have proposed ${ (1-{ p }_{ t }) }^{ \gamma }$ to the cross entropy loss ,with a tunable focusing parameter $\gamma \ge =0$

RetinaNet object detection method uses an α-balanced variant of the focal loss, where α=0.25, γ=2 works the best.

So one can define focal loss as –

$FL({ p }{ t })=\quad -{ \alpha }{ t }{ (1-{ p }{ t }) }^{ \gamma }\ln { ({ p }{ t }) }$

The focal loss is visualized for several values of $\gamma \epsilon \left[ 0,5 \right]$ ,refer Figure 1.

#### Focal Loss characteristics:-

We shall note following properties of the focal loss-

1. When an example is misclassified and ${ p }_{ t }$ is small, the modulating factor is near 1 and the loss is unaffected.
2. As ${ p }_{ t }\rightarrow 1$ ,the factor goes to 0 and the loss for well classified examples is down weighed.
3. The focusing parameter $\gamma$ smoothly adjusts the rate at which easy examples are down-weighted.
4. As $\gamma$ is increased , the effect of modulating factor is likewise increased. ( After a lot of experiments and trails , researchers have found $\gamma =2$ to work best.

Note:- When $\gamma =0$ , FL is equivalent to CE. (Shown blue curve in Figure 1)

You can read about Focal loss in detail in this article , Where I’ve talked about evolution of cross entropy into Focal loss, need of focal loss, comparison of focal loss with Cross entropy.

And cherry on top, I’ve used couple of examples to explain why Focal loss is better than cross entropy.

## End Points: –

Retina Net is a powerful model that uses Feature Pyramid Network & ResNet as its backbone.

In general RetinaNet is a good choice to start an object detection project, in particular if you need to quickly get good results. In next article we’ll build a solution using RetinaNet model.

If you’ve enjoyed this article, leave a few claps, it will encourage me to explore further machine learning opportunities 🙂

## References: –

• http://arxiv.org/abs/1605.06409
• https://arxiv.org/pdf/1708.02002.pdf
• https://developers.arcgis.com/python/guide/how-retinanet-works/
• https://analyticsindiamag.com/what-is-retinanet-ssd-focal-loss/
• https://github.com/fizyr/keras-retinanet
• https://www.freecodecamp.org/news/object-detection-in-colab-with-fizyr-retinanet-efed36ac4af3/
• https://deeplearningcourses.com/
• https://blog.zenggyu.com/en/post/2018-12-05/retinanet-explained-and-demystified/

## Article Credit:-

Name:  Ankush Deshmukh
Designation: MTech(Machine Learning & Signal Processing),NIT Surathkal
Research area: AI and Computer Vision


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## Chinese Remainder Theorem

Published on Sunday, January 22, 2012 in , , , ,

I've talked about the modular arithmetic before, especially as it related to the day for any date feat.

In this post, we're going to take it out of the calendar feat's shadow, and give it a starring role in its own feat!

If you remember doing division before you learned about fractions, you remember doing problems such as 21 ÷ 4 = 5 remainder 1. Modular arithmetic is simply focusing on the remainder exclusively. 21 modulo 4, for example, just equals 1, because when you divide 21 by 4, 1 is the remainder.

If we're talking at 10 AM, and I agree to call you in 5 hours, then you know to expect a call from at 3 PM. You did 10 + 5 = 15, but you know that hours aren't numbered any higher than 12, so you just subtracted 15 - 12 to get 3. This is modular arithmetic, and is also why it has the nickname “clock arithmetic.”

Let's try comparing the modular arithmetic patterns of two numbers, say, 2 and 4. Since 2 times 4 = 8, we'll compare the remainders as they run from 0 to 8:

$\begin{matrix} num&mod\2&mod\4 \\ 0 & 0 & 0\\ 1 & 1 & 1\\ 2 & 0 & 2\\ 3 & 1 & 3\\ 4 & 0 & 0\\ 5 & 1 & 1\\ 6 & 0 & 2\\ 7 & 1 & 3\\ 8 & 0 & 0 \end{matrix}$

What if I were to tell you that I was thinking of a number from 0 to 8. I then gave you a further clue that, when divided by 2, it has a remainder of 1, and when divided by 4, it has a remainder of 3, we run into a problem. Look at the chart. That description fits both 3 and 7, and there's no way to work out which of the two. The problem is that the pattern of remainders, when divided by 2 and 4, repeats 2 times from 0 to 8.

If we try this with, say, 3 and 6, and ran up to 18 (3 × 6) you can see in this chart that there are 3 times where a pattern of 3 remainders repeats.

If we want to identify a number by its remainders alone, is there some way to make sure that no repeating pattern emerges? Notice that, when we used 2 and 4 (and went up to 2 × 4, the remainder patterns repeated 2 times, and that 2 is the largest common factor of 2 and 4. Similarly, when we used 3 and 6 (and went up to 3 × 6), the remainder patterns repeated 3 times, and that 3 is the largest common factor of both 3 and 6.

If we want no repeating patterns, then what we're really saying is that, when performing modulo a and b, running from 0 up to a × b, we would like each number combination to only show up 1 time. For this to be true, we simply have to make sure that the greatest common factor of the numbers involved is 1!

This is the basic idea of the Chinese Remainder Theorem. Martin Gardner discusses this idea in more detail in his book Aha!: Aha! Insight and Aha! Gotcha (Spectrum). You can find the relevant pages online here and here, thanks to Google Books.

When using two numbers, it's pretty easy to make sure their only common factor is 1. If we use, say, 4 and 5, and go up to 20, we can already know that there won't be any repetitions, because the largest factor common to 4 (factors: 1, 2, 4) and 5 (factors: 1, 5) is 1.

The Chinese Remainder Theorem also tells us we can go further, and even use 3 or more numbers, and they won't repeat (up to a × b × c ×...) as long as their largest common factor is 1! The easiest way to do this, of course, is to turn to our old friend, prime numbers.

In the Martin Gardner book linked about, he talks about a version of a trick where someone thinks of a number from 1 to 1,000, and gives you the remainders after dividing by 7, 11, and 13. Since 7 × 11 × 13 = 1,001, you'll get a unique combination of remainders for any number given. But what about the version he mentions from 1 to 100 with 3, 5, and 7? What's the formula for that?

Let's take the approach in his article and apply it. For the remainder after dividing by 3, we need a multiple of 5 × 7 that's 1 greater than a multiple of 3. 35 doesn't work, because 34 isn't a multiple of 3. 70, being 69 + 1, works perfectly, though. OK, we start with 70 × a (or 70a for short).

What about 5? Let's look at the multiples of 3 × 7. There's 21...perfect! It's already 1 more than a multiple of 5. OK, now we've got 70a + 21b. What about 7? 3 × 5 = 15, and 15 is already 1 more than a multiple of 7. For all three numbers, we now have 70a + 21b + 15c. Divide that total by 105 (3 × 5 × 7), and the remainder will be the number you're looking for!

You could do that on a calculator, but if you're familiar at all with Grey Matters, you'll know that I encourage you to do things like that in your head. However, I understand that it can be tricky.

A magician named Tom Harris, back in 1958, proposed a different approach that required no calculation. You memorize the number combinations with help from the Peg/Major system, linking the combined numbers you get to the unique answer for that combination. For example, if someone gives you the numbers 1, 0, and 3, you would recall the phonetic equivalent “twosome”, and remember that you linked that to the word “toes,” which translates to 10.

This is a bit of work, but we have an advantage over someone trying to do this in 1958. Using a spreadsheet program makes an easy grid, and will handle listing the numbers from 1 to 100 for you, and will even handle working out the remainders for you. To find words for each combination, you can use some of the mnemonic generators listed here (My favorite for this would be pinfruit). Those familiar with the Peg System will already have 100 words ready for the answer numbers.

Naturally, I've developed a Wolfram|Alpha widget that can make things easier on your audience members:

I'll leave you with one last related challenge. James Grime needs help counting his juggling balls. Can you use what you've learned to help? When you're ready, here's the answer video.

format: plain


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# Physics help on pulleys and tension

1. May 6, 2007

### princess_7777

2000M3. A pulley of radius R1 and rotational inertia I1 is mounted on an axle with negligible friction. A light cord passing over the pulley has two blocks of mass m attached to either end, as shown above. Assume that the cord does not slip on the pulley. Determine the answers to parts (a) and (b) in terms of m, R1, I1, and fundamental constants.
a. Determine the tension T in the cord.
b. One block is now removed from the right and hung on the left. When the system is released from rest, the three blocks on the left accelerate downward with an acceleration g/3 . Determine the following.
i. The tension T3 in the section of cord supporting the three blocks on the left
ii. The tension Tl in the section of cord supporting the single block on the right
iii. The rotational inertia I1 of the pulley

c. The blocks are now removed and the cord is tied into a loop, which is passed around the original pulley and a second pulley of radius 2R1 and rotational inertia 16I1. The axis of the original pulley is attached to a motor that rotates it at angular speed w1, which in turn causes the larger pulley to rotate. The loop does not slip on the pulleys. Determine the following in terms of I1, RI, and w1.
i. The angular speed w2 of the larger pulley
ii. The angular momentum L2 of the larger pulley
iii. The total kinetic energy of the system

PHP:

Code (Text):

Code (Text):

Last edited: May 6, 2007
2. May 6, 2007

### siddharth

You need to show your work before you get help. What have you done with this question?

3. May 6, 2007

### princess_7777

well for (a) i tried it by using the equation and got 42,129.91
ii.: 6,006.23

4. May 6, 2007

### AKG

The answer to all the questions is $\pi$.

5. May 6, 2007

### hage567

By using what equation? What values did you use? I don't see any numerical values given in the question.


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Selina Concise Mathematics Class 7 ICSE Solutions Chapter 10 Simple Interest

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 10 Simple Interest

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 10 Simple Interest

Simple Interest Exercise 10 – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Find the S.I. and amount on :
(i) Rs. 150 for 4 years at 5% per year.
(ii) Rs. 350 for 3$$\frac {1 }{ 2 }$$ years at 8% p.a.
(iii) Rs. 620 for 4 months at 8 p. per rupee per month.
(iv) Rs. 3,380 for 30 months at 4 $$\frac { 1 }{ 2 }$$ % p.a.
(v) 600 from July 12 to Dec. 5 at 10% p.a.
(vi) Rs. 850 from 10th March to 3rd August at 2 $$\frac { 1 }{ 2 }$$ % p.a.
(vii) Rs. 225 for 3 years 9 months at 16% p.a.
Solution:

Question 2.
On what sum of money does the S.I. for 10 years at 5% become Rs. 1,600 ?
Solution:

Question 3.
Find the time in which Rs. 2,000 will amount to Rs. 2,330 at 11% p.a. ?
Solution:

Question 4.
In what time will a sum of money double it self at 8% p.a ?
Solution:

Question 5.
In how many years will be ₹870 amount to ₹1,044, the rate of interest being 2$$\frac { 1 }{ 2 }$$ % p.a ?
Solution:

Question 6.
Find the rate percent if the S.I. on ₹275 is 2 years is ₹22.
Solution:

Question 7.
Find the sum which will amount to ₹700 in 5 years at 8% rate p.a.
Solution:

Question 8.
What is the rate of interest, if ₹3,750 amounts to ₹4,650 in 4 years ?
Solution:

Question 9.
In 4 years, ₹6,000 amount to ₹8,000. In what time will ₹525 amount to ₹700 at the same rate ?
Solution:

Question 10.
The interest on a sum of money at the end of 2$$\frac { 1 }{ 2 }$$ years is $$\frac { 4 }{ 5 }$$ of the sum. What is the rate percent ?
Solution:

Question 11.
What sum of money lent out at 5% for 3 years will produce the same interest as Rs. 900 lent out at 4% for 5 years ?
Solution:

Question 12.
A sum of Rs. 1,780 become Rs. 2,136 in 4 years,
Find :
(i) the rate of interest.
(ii) the sum that will become Rs. 810 in 7 years at the same rate of interest ?
Solution:

Question 13.
A sum amounts to Rs. 2,652 in 6 years at 5% p.a. simple interest.
Find :
(i) the sum
(ii) the time in which the same sum will double itself at the same rate of interest.
Solution:

Question 14.
P and Q invest Rs. 36,000 and Rs. 25,000 respectively at the same rate of interest per year. If at the end of 4 years, P gets Rs. 3,080 more interest than Q; find the rate of interest.
Solution:

Question 15.
A sum of money is lent for 5 years at R% simple interest per annum. If the interest earned be one-fourth of the money lent, find the value of R.
Solution:

Question 16.
The simple interest earned on a certain sum in 5 years is 30% of the sum. Find the rate of interest.
Solution:


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Are there cases in which we should consider tensors as equivalence classes?

Usually in texts about Physics that uses tensors defines them as multilinear maps. So if $V$ is a vector space over the field $F$, a tensor is a multilinear mapping:

$$T:V\times\cdots\times V\times V^\ast\times\cdots\times V^\ast\to F.$$

In texts about multilinear algebra, however, a tensor is defined differently. They consider a collection $V_1,\dots,V_k$ of vector spaces over the same field, consider the free vector space $\mathcal{M}=F(V_1\times\cdots\times V_k)$, consider the subspace $\mathcal{M}_0$ genereated by vectors of the form

$$(v_1,\dots,v_i+v_i',\dots,v_k)-(v_1,\dots,v_i,\dots,v_k)-(v_1,\dots,v_i',\dots,v_k)$$

$$(v_1,\dots,kv_i,\dots,v_k)-k(v_1,\dots,v_i,\dots,v_k)$$

And then define the tensor product $V_1\otimes\cdots\otimes V_k = \mathcal{M}/\mathcal{M}_0$ and define tensors as elements of such space, which are equivalence classes of functions with finite support in $V_1\times\cdots\times V_k$.

Now, is there some cases in Physics where it's better to think as tensors as such equivalence classes rather than multilinear mappings? If so, how then we get some physical intuition behind those objects?

• I think you just need to understand that the two constructions give isomorphic spaces. – MBN Oct 30 '13 at 17:50


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# 2.1.2: Forces in Solutions

Learning Objectives

• Learn some terminology involving solutions.
• Explain the significance of the statement "like dissolves like".
• Explain why certain substances dissolve in other substances.

A solution is another name for a homogeneous mixture. A mixture as a material composed of two or more substances. In a solution, the combination is so intimate that the different substances cannot be differentiated by sight, even with a microscope. Compare, for example, a mixture of salt and pepper and another mixture consisting of salt and water. In the first mixture, we can readily see individual grains of salt and the flecks of pepper. A mixture of salt and pepper is not a solution. However, in the second mixture, no matter how carefully we look, we cannot see two different substances. Salt dissolved in water is a solution.

The major component of a solution, called the solvent, is typically the same phase as the solution itself. Each minor component of a solution (and there may be more than one) is called the solute. In most of the solutions we will describe in this textbook, there will be no ambiguity about whether a component is the solvent or the solute.) For example, in a solution of salt in water, the solute is salt, and solvent is water Figure $$\PageIndex{1}$$.

Solutions come in all phases, and the solvent and the solute do not have to be in the same phase to form a solution (such as salt and water). For example, air is a gaseous solution of about 80% nitrogen and about 20% oxygen, with some other gases present in much smaller amounts. An alloy is a solid solution consisting of a metal (like iron) with some other metals or nonmetals dissolved in it. Steel, an alloy of iron and carbon and small amounts of other metals, is an example of a solid solution. Table $$\PageIndex{1}$$ lists some common types of solutions, with examples of each.

Figure $$\PageIndex{1}$$ Making a saline water solution by dissolving table salt (NaCl) in water.

The salt is the solute and the water the solvent. (CC-BY-SA 3.0; Chris 73).

Table $$\PageIndex{1}$$ Types of Solutions

Solvent Phase Solute Phase Example
gas gas air
liquid gas carbonated beverages
liquid liquid ethanol (C2H5OH) in H2O (alcoholic beverages)
liquid solid salt water
solid gas H2 gas absorbed by Pd metal
solid liquid Hg(ℓ) in dental fillings
solid solid steel alloys

Example $$\PageIndex{1}$$: Sugar and Water

A solution is made by dissolving 1.00 g of sucrose ($$\ce{C12H22O11}$$) in 100.0 g of liquid water. Identify the solvent and solute in the resulting solution.

Solution

Either by mass or by moles, the obvious minor component is sucrose, so it is the solute. Water—the majority component—is the solvent. The fact that the resulting solution is the same phase as water also suggests that water is the solvent.

Exercise $$\PageIndex{1}$$

A solution is made by dissolving 3.33 g of $$\ce{HCl(g)}$$ in 40.0 g of liquid methyl alcohol ($$\ce{CH3OH}$$). Identify the solvent and solute in the resulting solution.

solute: HCl(g); solvent: CH3OH

## Like Dissolves Like

A simple way to predict which compounds will dissolve in other compounds is the phrase "like dissolves like". What this means is that polar compounds dissolve polar compounds, nonpolar compounds dissolve nonpolar compounds, but polar and nonpolar do not dissolve in each other.

Even some nonpolar substances dissolve in water but only to a limited degree. Have you ever wondered why fish are able to breathe? Oxygen gas, an nonpolar molecules, does dissolve in water and it is this oxygen that the fish take in through their gills. Or, one more example of a nonpolar compound that dissolves in water is the reason we can enjoy carbonated sodas. Pepsi-cola and all the other sodas have carbon dioxide gas, $$\ce{CO_2}$$, a nonpolar compound, dissolved in a sugar-water solution. In this case, to keep as much gas in solution as possible, the sodas are kept under pressure.

This general trend of "like dissolves like" is summarized in the following table:

Table $$\PageIndex{2}$$ Summary of Solubilities
Solute Solvent Is Solution Formed?
Polar Covalent Polar yes
Non-polar Covalent Non-polar yes
Polar Covalent Non-polar no
Non-polar Covalent Polar no
Ionic Polar yes
Ionic Non-polar no

Note that every time charged particles (ionic compounds or polar substances) are mixed, a solution is formed. When particles with no charges (nonpolar compounds) are mixed, they will form a solution. However, if substances with charges are mixed with other substances without charges a solution does not form.

When an ionic compound is considered "insoluble", it doesn't necessarily mean the compound is completely untouched by water. All ionic compounds dissolve to some extent. An insoluble compound just doesn't dissolve in any noticeable or appreciable amount.

What is it that makes a solute soluble in some solvents but not others?

The answer is intermolecular interactions. The intermolecular interactions include London dispersion forces, dipole-dipole interactions, and hydrogen bonding (as described in the previous section). From experimental studies, it has been determined that if molecules of a solute experience the same intermolecular forces that the solvent does, the solute will likely dissolve in that solvent. So, NaCl—a very polar substance because it is composed of ions—dissolves in water, which is very polar, but not in oil, which is generally nonpolar. Nonpolar wax dissolves in nonpolar hexane, but not in polar water. Liquids that dissolve in one another in all proportions are said to be miscible. Liquids that do not dissolve in one another are called immiscible.

Figure $$\PageIndex{2}$$ Water (clear liquid) and oil (yellow) do not form liquid solutions. (CC BY-SA 1.0 Generic; Victor Blacus)

Example $$\PageIndex{2}$$: Polar and Nonpolar Solvents

Would I2 be more soluble in CCl4 or H2O? Explain your answer.

Solution

I2 is nonpolar. Of the two solvents, CCl4 is nonpolar and H2O is polar, so I2 would be expected to be more soluble in CCl4.

Exercise $$\PageIndex{2}$$

Would C3H7OH be more soluble in CCl4 or H2O? Explain your answer.

H2O because both experience hydrogen bonding

Example $$\PageIndex{3}$$

Water is considered a polar solvent. Which substances should dissolve in water?

1. methanol (CH3OH)
2. sodium sulfate (Na2SO4)
3. octane (C8H18)

Solution

Because water is polar, substances that are polar or ionic will dissolve in it.

1. Because of the OH group in methanol, we expect its molecules to be polar. Thus, we expect it to be soluble in water. As both water and methanol are liquids, the word miscible can be used in place of soluble.
2. Sodium sulfate is an ionic compound, so we expect it to be soluble in water.
3. Like other hydrocarbons, octane is nonpolar, so we expect that it would not be soluble in water.

Exercise $$\PageIndex{3}$$

Toluene (C6H5CH3) is widely used in industry as a nonpolar solvent. Which substances should dissolve in toluene?

1. water (H2O)
2. sodium sulfate (Na2SO4)
3. octane (C8H18)

octane (C8H18) will dissolve. It is also non-polar.

## The Dissolving Process

Water typically dissolves most ionic compounds and polar molecules. Nonpolar molecules, such as those found in grease or oil, do not dissolve in water. We will first examine the process that occurs when an ionic compound, such as table salt (sodium chloride), dissolves in water.

Water molecules move about continuously due to their kinetic energy. When a crystal of sodium chloride is placed into water, the water's molecules collide with the crystal lattice. Recall that the crystal lattice is composed of alternating positive and negative ions. Water is attracted to the sodium chloride crystal because water is polar; it has both a positive and a negative end. The positively charged sodium ions in the crystal attract the oxygen end of the water molecules because they are partially negative. The negatively charged chloride ions in the crystal attract the hydrogen end of the water molecules because they are partially positive. The action of the polar water molecules takes the crystal lattice apart (see figure below).

After coming apart from the crystal, the individual ions are then surrounded by solvent particles in a process called solvation. Note in the figure above that the individual $$\ce{Na^+}$$ ions are surrounded by water molecules with the oxygen atom oriented near the positive ion. Likewise, the chloride ions are surrounded by water molecules with the opposite orientation. Thus, numerous ion-dipole interactions are formed. Hydration is the process of solute particles being surrounded by water molecules arranged in a specific manner. Hydration helps to stabilize aqueous solutions by preventing the positive and negative ions from coming back together and forming a precipitate.

Table sugar is made of the molecular compound sucrose $$\left( \ce{C_{12}H_{22}O_{11}} \right)$$. Solid sugar consists of individual sugar molecules held together by intermolecular attractive forces. When water dissolves sugar, it separates the individual sugar molecules by disrupting the attractive forces, but it does not break the covalent bonds between the carbon, hydrogen, and oxygen atoms. Dissolved sugar molecules are also hydrated. The hydration shell around a molecule of sucrose is arranged so that its partially negative oxygen atoms are near the partially positive hydrogen atoms in the solvent, and vice versa.

## Summary

• Solutions are composed of a solvent (major component) and a solute (minor component).
• “Like dissolves like” is a useful rule for deciding if a solute will be soluble in a solvent.
• Liquids that dissolve in one another in all proportions are said to be miscible.
• Liquids that do not dissolve in one another are called immiscible.
• Ion-dipole interactions are formed when ionic compounds dissolve in water.

• Henry Agnew (UC Davis)

• Libretext: Chemistry for Allied Health (Soult)
• Libretext: The Basics of GOB Chemistry (Ball et al.)


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1. Peroxides and Dioxides. The reactions of the Group 2 metals with air rather than oxygen is complicated by the fact that they all react with nitrogen to produce nitrides. Take the test now! M(OH)2, eg 2Ca(s) + 2H2O(l) ----> Ca(OH)2(aq) + H2(g). Reactions of Period 3 Elements. pH produit concentré : env. The oxides. If, in a 2d, your soreness and stiffness melted away. 1. These reactions are also often called neutralization reactions. Top. Acids and bases. AMPHOTERIC OXIDES. MgO (s) + H2O (l) Mg(OH)2 (s) pH 9 Mg(OH)2 is only slightly soluble in water so fewer free OH-ions are produced and so lower pH Group 2 ionic oxides react with water to form hydroxides MgO (s) + 2 HCl (aq) MgCl2 (aq) + H2O (l) 2HNO3 (a q) + Mg(O H)2 (a q) Mg(N O3)2 (a q)+ 2H2O (l) Reactions of the Oxides of Group 2 elements with water The oxides form more strongly alkaline solutions as you go down the group, because the hydroxides get more soluble. SQA Chemistry. Calcium Hydroxide (Ca(OH)2) is used as lime by farmers and gardeners Reaction of the oxides with water The role of hydrogen bonds, ionic bonds and disulfide bridges in the structure of proteins. of Group 2 elements and compounds; (f) explain the use of Ca(OH)2 in agriculture to neutralise acid soils; 2. Table 3 Solubility products (given as pK sol = −log K sol) of metal hydroxides and oxides at 25 °C. Full size table. more alkaline down the group. 5 Comments. $4 Li + O_2 \rightarrow 2Li_2O \label{19}$ Peroxides: Often Lithium and Sodium reacts with excess oxygen to produce the peroxide, $$M_2O_2$$. ; Together with group 1 (the alkali metals), they form the s block of the periodic table because their highest energy electrons are all in s … relieve indigestion. SQA Chemistry. The outer electrons are easier to remove as they are further from the nucleus and there is more shielding resulting in a lower nuclear attraction. Beryllium oxide isn't fully ionic. L'exemple suivant montre la dissolution d'un oxyde basique dans l'eau. Algebraic Techniques Template Notes Blank, Correlation & Regression Exam Question Pack, Binomial Distribution Template Notes Blank, Constant Acceleration Template Notes Blank, Variable Acceleration Template Notes Blank, Proof by Contradiction Template Notes Blank, Correlation & Regression Template Note Blank, Variable Acceleration Template Note Blank. With the oyxgen exhibiting an oxidation number of -2. Group 2. Dissolving basic oxide in water can increase pH (basicity) of water because basic oxides release hydroxide ions (OH−) to the water. A condensation reaction between two amino acids forms a peptide bond. MO + H 2 O → M(OH) 2 (where M = group 2 metal) Thus, these compounds are often called basic oxides. In Table 3 a compilation of solubility products of some metal hydroxides, oxides, and oxide-hydroxides are given. 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AHSME 1981 #22 - Number of lines that pass through four distinct points

How many lines in a three dimensional rectangular coordinate system pass through four distinct points of the form $(i, j, k)$ where $i$, $j$, and $k$ are positive integers not exceeding four?

$\text{(A)} \ 60 \qquad \text{(B)} \ 64 \qquad \text{(C)} \ 72 \qquad \text{(D)} \ 76 \qquad \text{(E)} \ 100$

I am interested in knowing why my solution below is incorrect:

Any line in 3D space is characterized by an initial point and a vector.

Starting with an initial point of form $(x, y, z)$ and a vector of $(\Delta x, \Delta y, \Delta z)$ you can get other points on the line by adding the vector to the initial point. Since we are interested in a line that passes through four points, the fourth point on this line will be $(x + 3\Delta x, y + 3\Delta y, z + 3\Delta z)$

You need each of these points to be between 1 and 4 (inclusive). If the initial and final points are within the boundary then all points inbetween will also be.

Thus if, $$1 \leq x\leq 4$$

and $$1 \leq x + 3\Delta x \leq 4$$ then you have the following choices:

$x = 1$ and $\Delta x = 0, 1$

$x = 2$ and $\Delta x = 0$

$x = 3$ and $\Delta x = 0$

$x = 4$ and $\Delta x = 0$

Resulting in a total of $1\cdot2 + 1\cdot1 + 1\cdot1 + 1\cdot1 = 5$ valid for $x$ and $\Delta x$. The same holds true for the $y$ and $z$ cases, resulting in a total of $5^3$ total choices.

However, because the points need to be distinct a vector of $(0, 0, 0)$ is not allowed, and thus we must subtract all cases where $\Delta x = 0$, which make up $(1\cdot1 + 1\cdot1 + 1\cdot1 + 1\cdot1)^3$ cases.

The total is thus $5^3 - 4^3 = 61$. However, the correct answer is 76. Why did I not get the right answer?

You’re assuming that $\Delta x,\Delta y,\Delta z\ge 0$ but that doesn’t cover all of the possibilities. The diagonal that runs from $\langle 1,4,4\rangle$ to $\langle 4,1,1\rangle$, for instance, isn’t covered by any of your cases.

• That adds the case of $x = 4$, $\Delta x = -1, 0$. Which still does not lead to the right answer, however. – 1110101001 Dec 12 '14 at 5:43
• @1110101001: That’s not the only case that it adds. You’re missing every diagonal along which at least one coordinate increases while another decreases. – Brian M. Scott Dec 12 '14 at 5:45
• I don't quite get what you mean. When you add in the cases of $x = 4$ and $\Delta x = -1$ to the others to get a total of 6 cases and then cube it to account for the y and z, are you not including all possibilities? Referring to the table posted by @turkeyhundt I don't see any cases which are not covered by this – 1110101001 Dec 12 '14 at 5:50
• @1110101001: If you're making the calculation that I think you're making, you're now counting some diagonals twice, once from each end. – Brian M. Scott Dec 12 '14 at 5:58
• @1110101001: You're welcome! – Brian M. Scott Dec 12 '14 at 6:08

You can think of it in a physical sense instead of math.

For each pair of opposite faces, there are 16 lines that can go through the $4\times4$ cube perpendicularly. So that's $3\times 16=48$.

And for each pair of diagonally opposite edges, there are 4 lines that will go through 4 lattice points. So that's $6\times 4 = 24$

Then for each opposite pair of vertices there is one line. $4\times 1$

Add them up and you get 76 :)

ETA: Doesn't really address your question, but I think these are all the point vector solutions.

\begin{array}{|c|c|c|} \hline (x,y,z)& (\Delta x, \Delta y, \Delta z) & \text{count}\\ \hline (1,y,z) & (1, 0, 0) & 16\\ \hline (x,1,z) & (0, 1, 0) & 16\\ \hline (x,y,1) & (0, 0, 1) & 16\\ \hline (1,1,z) & (1, 1, 0) & 4\\ \hline (4,1,z) & (-1, 1, 0) & 4\\ \hline (1,y,1) & (1, 0, 1) & 4\\ \hline (4,y,1) & (-1, 0, 1) & 4\\ \hline (x,1,1) & (0, 1, 1) & 4\\ \hline (x,4,1) & (0, -1, 1) & 4\\ \hline (1,1,1) & (1, 1, 1) & 1\\ \hline (1,4,4) & (1, -1, -1) & 1\\ \hline (4,1,4) & (-1, 1, -1) & 1\\ \hline (4,4,1) & (-1, -1, 1) & 1\\ \hline \end{array}

• I am interested as to why my solution was incorrect. I already know the answer and how to get it another way. – 1110101001 Dec 12 '14 at 5:02
• Ah, my bad. Misread the question. – turkeyhundt Dec 12 '14 at 5:02
• It's fine. Still keep your answer up though. – 1110101001 Dec 12 '14 at 5:03

Just reading your solution, when I got to the part about $x = 4$, then $\Delta x = -1$ is a permissible condition and I stopped right there.

Reading a bit further, we can also immediately see that your enumeration is flawed in a second way, which is that the choices of $(x, \Delta x)$, $(y, \Delta y)$, and $(z, \Delta z)$ are not independent of each other, which is what you are assuming when writing $5^3$.

• wouldn't that case already be accounted for with $x=1$ and $\Delta x = 1$ – turkeyhundt Dec 12 '14 at 5:16
• No. For what choices of $(x, y, z)$ and $(\Delta x, \Delta y, \Delta z)$ where none of the $\Delta$ are negative, can you obtain the line passing through $(4,1,1), (3,2,2), (2,3,3), (1,4,4)$? – heropup Dec 12 '14 at 5:20
• Ah yes. I was definitely missing that. – turkeyhundt Dec 12 '14 at 5:21
• Why aren't the choices of (x, dx)... independent of each other? Doesn't the value you choose for (x, dx) have no impact on the others? – 1110101001 Dec 12 '14 at 5:39


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# ; (Command Separator)

The semicolon ( ; ) character is used to separate multiple commands on a single line.

``````    Command1 ; Command2 [; Command3 ...]
``````

## Parameters

Command1, Command2, ...
The commands to be executed.

## Remarks

Commands are executed sequentially from left to right. All commands on a single line refer to the current thread, unless otherwise specified. If a command causes the thread to execute, the remaining commands on the line will be deferred until that thread stops on a debug event.

A small number of commands cannot be followed by a semicolon, because they automatically take the entire remainder of the line as their argument. These include as (Set Alias), \$< (Run Script File), \$>< (Run Script File), and any command beginning with the * (Comment Line Specifier) token.

Here is an example. This executes the current program to source line 123, prints the value of counter, then resumes execution:

``````0:000> g `:123`; ? poi(counter); g
``````


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# Proof of finite abelian groups structure theorem using representation theory

While looking at the french wikipedia page for the finite abelian groups structure theorem (that is, every finite abelian group $G$ is isomorphic to a group of the form $\mathbb{Z}/a_1\mathbb{Z} \times \dotsb \times \mathbb{Z}/a_n\mathbb{Z}$, with $a_1 \mid \dotsb \mid a_n$), which can be found here for french-speaking people, I found written that there are short proofs of the theorem relying on representation theory.

But I have not found a proof of this kind anywhere; all the classical books prove it with the usual generalization to finitely generated $A$-modules with $A$ a principal ring.

Would someone know anything about such a proof ? If so, could someone give me a reference where it would be done, or write the proof here ?

• I was under the impression that the structure theorem for modules over a principal ideal domain is the way of proving this with representation theory (representation theory of the principal ideal domain $\mathbb{Z}$). – Joppy Nov 2 '17 at 20:12

Fact. Let $G$ be a finite abelian group and let $g\in G$ have order $n$. Then there is a linear character $\lambda\colon G\to \mathbb{C}^*$ which sends $g$ to a primitive $n$th root of unity.
Proof. Sending $g$ to $e^{2\pi i/n}$ defines a character of $\langle g \rangle$, and there must be some irreducible (i. e., linear) character of $G$ lying over this character. (When $g$ has prime power order, one can also argue that there must be some character which is nontrivial on an element of prime order contained in $\langle g \rangle$.)
Now among all linear characters $\lambda\colon G\to \mathbb{C}^*$, choose one such that the image $\lambda(G)$ has maximal order. Since $\lambda(G)$ is a finite subgroup of $\mathbb{C}^*$, it is cyclic, say $\lambda(G)= \langle \lambda(g_1)\rangle$ for some $g_1\in G$. I claim that $G= \langle g_1 \rangle \times \ker \lambda$ and that the order of $g_1$ is the exponent of $G$. Then one can continue by induction.
For any $g\in G$ we have $\lambda(g) = \lambda(g_1)^k = \lambda(g_1^k)$ for some $k$ and thus $gg_1^{-k}\in \ker\lambda$. As $g\in G$ was arbitrary, this shows $G=\langle g_1 \rangle \ker \lambda$. By the above Fact, we see that $\lambda(g_1)$ and $g_1$ must have the same order. Thus $\langle g_1 \rangle \cap \ker \lambda = \{1\}$ and we have shown $G=\langle g_1 \rangle \times \ker \lambda$.
Finally, let $\mu$ be a linear character of $\ker \lambda$. Then $(\lambda\times \mu)$ is a character of $G$ and $(\lambda \times \mu)(G)$ contains $\lambda(g_1)$, and thus by maximality we have $(\lambda\times \mu)(G) = \langle \lambda(g_1) \rangle$. This yields $\mu(\ker\lambda) \subseteq \langle \lambda(g_1)\rangle$. So again from the Fact, the exponent of $\ker \lambda$ divides the order of $g_1$.


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# Category:Thermodynamics

Jump to: navigation, search

This category is mainly for the Thermodynamics project. Also other entries related to the subject.

## Subcategories

This category has the following 2 subcategories, out of 2 total.


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# [tex-live] bug: misleading message "TeX capacity exceeded" in beamer

Zdenek Wagner zdenek.wagner at gmail.com
Sat Jul 21 17:55:27 CEST 2012

2012/7/18 Camilo La Rota <camilo.larota2 at gmail.com>:
> The following error message occurs:
>
> TeX capacity exceeded in beamer, sorry [hash size=5000]
>
> in beamer, when a \block statement lacks the name of the block,
> and that the block contains a list of items:
>
> \begin{block}
> \begin{itemize}
> \item ...
> \end{itemize}
> \end{block}
>
> LaTeX didn't figure out that the name of the block was not there and print
> this error message without giving any hint of what has really happened.

1. This is not a TeX Live issue

2. I do not know beamer but I assume that the code should be
\begin{block}{somename} and you complain that in your case LaTeX does
not complain that the name is missing. LaTeX cannot do that because
the name is present, the problem is that the name makes no sense. The
block name is \begin (as if you wrote \begin{block}{\begin}{itemize})
- read the TeXbook in order to understand ho TeX is interpreting the
input. Since \begin is a macro, it gets expanded. It looks for an
argument (the name of the environment) but it cannot find {itemize}
because it is expanded in the middle of interpretation of
\begin{block}. It therefore takes some other nonsense (possibly an
internal macro) as an environment name. There is no surprise that
memory is exhausted before LaTeX gets any chance to run other tests.

--
Zdeněk Wagner
http://hroch486.icpf.cas.cz/wagner/
http://icebearsoft.euweb.cz


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# Standard normal distribution inequality

I want to know how to prove the following inequality that seems to be true numerically.

Let $n(x)$ be the density of the standard normal, and $N(x)$ be the cdf of standard normal. Then, for $x\geq 0$,

$\left(\frac{n(x)}{1-N(x)}-x \right)\left(\frac{2n(x)}{1-N(x)}-x\right)\geq 1$

Thanks, and sorry that I don't know how to write the math symbols in this environment.

• Typeset in LaTeX please, it's unreadable – james42 Jul 8 '15 at 22:26
• Thanks Ale42, I just edited the inequality. – Mehmet Jul 8 '15 at 23:39
• Excuse me but why am I voted down? I am a new user and please advise. – Mehmet Jul 9 '15 at 0:05
• Well I don't know... Maybe somebody doesn't like your question. Anyway I'll give you one upvote :) – james42 Jul 9 '15 at 0:09

I get that $\left(\frac{n(x)}{M(x)}-x \right)\left(\frac{2n(x)}{M(x)}-x\right) - 1 \approx \frac{3}{x^2}$ for large $x$.

Since $N(x) =\int_{-\infty}^x n(t)dt$, $1-N(x) =\int_x^{\infty} n(t)dt$. Call this $M(x)$.

The inequality becomes $\left(\frac{n(x)}{M(x)}-x \right)\left(\frac{2n(x)}{M(x)}-x\right)\geq 1$.

Multiplying by $M^2(x)$, I get $(n(x) - xM(x))(2n(x) - xM(x)) \ge M^2(x)$ or $2n^2(x) -3xn(x)M(x)+x^2M^2(x) \ge M^2(x)$ or $2n^2(x) -3xn(x)M(x)+(x^2-1)M^2(x) \ge 0$.

The purported inequality above thus becomes $P(x) =2n^2(x) -3xn(x)M(x)+(x^2-1)M^2(x) \ge 0$.

Since $M(0) = 1/2$ and $n(0) =1/\sqrt{2\pi} \approx 0.4$, $P(0) \approx 0.32-1/4 = 0.07 > 0$.

Let's see what happens for large $x$.

Asymptotically, $M(x) \approx n(x)/x$.

Therefore, for large $x$,

$\begin{array}\\ P(x) &= 2n^2(x) -3xn(x)M(x)+(x^2-1)M^2(x)\\ &\approx 2n^2(x) -3xn(x)(n(x)/x)+(x^2-1)(n(x)/x)^2\\ &= 2n^2(x) -3n^2(x)+(1-1/x^2)n^2(x)\\ &= -n^2(x)/x^2\\ \end{array}$

Therefore, modulo errors on my part, $P(x) \approx -n^2(x)/x^2$, so it is negative and small.

Note: A comment said that this was not correct, so I'll take an additional term in the expansion and see what happens.

Asymptotically, $M(x) \approx \frac{n(x)}{x}(1-\frac1{x^2})$.

Therefore, for large $x$,

$\begin{array}\\ P(x) &= 2n^2(x) -3xn(x)M(x)+(x^2-1)M^2(x)\\ &\approx 2n^2(x) -3xn(x)(n(x)/x)(1-1/x^2)+(x^2-1)(n(x)/x)^2(1-1/x^2)^2\\ &= 2n^2(x) -3n^2(x)(1-1/x^2)+n^2(x)(1-1/x^2)^3\\ &= 2n^2(x) -3n^2(x)+3n^2(x)/x^2+n^2(x)(1-3/x^2+3/x^4-1/x^6)\\ &= -n^2(x)+3n^2(x)/x^2+n^2(x)-n^2(x)(3/x^2-3/x^4+1/x^6)\\ &= n^2(x)(3/x^4-1/x^6)\\ &\approx 3n^2(x)/x^4\\ \end{array}$

Therefore, modulo errors on my part, $P(x) \approx 3n^2(x)/x^4$, so it is positive and small.

Looks like the comment was correct.

Since I multiplied by $M^2(x)$ to get this inequality, the difference has to be divided by $M^2(x)$, so it is $\frac{3n^2(x)/x^4}{M^2(x)} \approx \frac{3n^2(x)/x^4}{(n(x)/x)^2} =\frac{3}{x^2}$.

• Thanks Marty but I think your answer is not correct. Your approximation has small errors that are at the same order as $\frac{n(x)^2}{x^2}$. – Mehmet Jul 9 '15 at 2:52
• Not surprised . I might try more terms later. – marty cohen Jul 9 '15 at 5:43
• You were right. – marty cohen Jul 9 '15 at 19:03


text:
# B-convergence

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

The theory of B-convergence predicts the convergence properties of discretization methods, implicit Runge–Kutta methods in particular, applied to initial value problems for systems of non-linear ordinary differential equations (cf. Runge–Kutta method; Initial conditions; Differential equation, ordinary).

## Problem class and historical background.

Concerning the class of ordinary differential equations considered, the underlying assumption is that satisfies a one-sided Lipschitz condition

If (the so-called one-sided Lipschitz constant of ) is of moderate size, the initial value problem is easily proved to be well-conditioned throughout. Of particular interest are stiff problems (cf. also Stiff differential system). Stiffness means that that the underlying ordinary differential equation admits smooth solutions, with moderate derivatives, together with non-smooth ( "transient" ) solutions rapidly converging towards smooth ones. In the stiff case, the conventional Lipschitz constant of inevitably becomes large; therefore the classical error bounds for discretization methods (which depend on ) are of no use for an appropriate characterization and analysis of methods which are able to efficiently integrate a stiff problem. Thus there is a need for a special convergence theory applicable in the presence of stiffness.

The idea to use (instead of ) as the problem-characterizing parameter goes back to [a7], where it was used in the analysis of multi-step methods. The point is that stiffness is often compatible with moderate values of , while . In the same spirit, the concept of B-stability was introduced in [a4], [a5], [a6] in the context of implicit Runge–Kutta methods, and an algebraic criterion on the implicit Runge–Kutta coefficients entailing B-stability was derived ( "algebraic stability in numerical analysisalgebraic stability" ). The notion of -stability enables realistic estimates of the propagation of inevitable perturbations like local discretization errors.

## The concept of B-convergence.

In the convergence theory of implicit Runge–Kutta methods applied to stiff problems, stability is essential but also the analysis of local errors is non-trivial: straightforward estimates are affected by and do not reflect reality. For a simple scalar model class, the local error of implicit Runge–Kutta schemes was studied in [a13]. It turned out that the order observed for practically relevant stepsizes is usually reduced compared to smooth, non-stiff situations.

In [a9], [a10], [a11], the convergence properties of implicit Runge–Kutta schemes are studied and the notion of B-convergence is introduced. A B-convergence result is nothing but a realistic global error estimate based on the parameter but unaffected by . Besides relying on B-stability, the essential point are sharp local error estimates which require a special internal stability property called BS-stability. The latter can be concluded from a certain algebraic condition on the Runge–Kutta coefficients ( "diagonal stability in numerical analysisdiagonal stability" ). Explicit error bounds have been derived for Gauss, Radau-IA and Radau-IIA schemes; the corresponding "B-convergence order" is in accordance with the observations from [a13]. B-convergence results for Lobatto-IIIC schemes are given in [a14].

An overview on the "B-theory" of implicit Runge–Kutta methods is presented in [a8]. Another relevant text is [a12].

## Further developments.

Concerning the relevance of the B-theory for stiff problems, there remains a gap. The point is that for most stiff problems there is a strong discrepancy between the local and the global condition: Neighbouring solutions may locally strongly diverge, such that the problem is locally ill-conditioned. This is a transient effect, leaving the good global condition unaffected. But it inevitably implies that the one-sided Lipschitz constant is strongly positive (like ). For details, cf. [a1], where it is shown that remains moderate only for a restricted class of stiff problems, namely with Jacobians that are "almost normal" . In general, however, is large and positive, and the B-convergence bounds based on become unrealistically large.

As a consequence, not even linear stiff problems are satisfactorily covered. In [a2] the B-theory is extended to semi-linear stiff problems of the form , where has a smoothly varying eigensystem and is smooth. However, this does not cover a sufficiently large class of non-linear problems. Current work concentrates on a more natural, geometric characterization of stiffness; cf., e.g., [a3].

#### References

 [a1] W. Auzinger, R. Frank, G. Kirlinger, "A note on convergence concepts for stiff problems" Computing , 44 (1990) pp. 197–208 [a2] W. Auzinger, R. Frank, G. Kirlinger, "An extension of B-convergence for Runge–Kutta methods" Appl. Numer. Math. , 9 (1992) pp. 91–109 [a3] W. Auzinger, R. Frank, G. Kirlinger, "Extending convergence theory for nonlinear stiff problems, Part I" BIT , 36 (1996) pp. 635–652 [a4] K. Burrage, J.C. Butcher, "Stability criteria for implicit Runge–Kutta methods" SIAM J. Numer. Anal. , 16 (1979) pp. 46–57 [a5] J.C. Butcher, "A stability property of implicit Runge–Kutta methods" BIT , 15 (1975) pp. 358–361 [a6] M. Crouzeix, "Sur la B-stabilité des méthodes de Runge–Kutta" Numer. Math. , 32 (1979) pp. 75–82 [a7] G. Dahlquist, "Error analysis for a class of methods for stiff nonlinear initial value problems" , Numerical Analysis , Lecture Notes in Mathematics , 506 (1976) pp. 60–72 [a8] K. Dekker, J.G. Verwer, "Stability of Runge–Kutta methods for stiff nonlinear differential equations" , North-Holland (1984) [a9] R. Frank, J. Schneid, C.W. Ueberhuber, "The concept of B-convergence" SIAM J. Numer. Anal. , 18 (1981) pp. 753–780 [a10] R. Frank, J. Schneid, C.W. Ueberhuber, "Stability properties of implicit Runge–Kutta methods" SIAM J. Numer. Anal. , 22 (1985) pp. 497–515 [a11] R. Frank, J. Schneid, C.W. Ueberhuber, "Order results for implicit Runge–Kutta methods applied to stiff systems" SIAM J. Numer. Anal. , 22 (1985) pp. 515–534 [a12] E. Hairer, G. Wanner, "Solving ordinary differential equations" , II: stiff and differential-algebraic problems , Springer (1991) [a13] A. Prothero, A. Robinson, "On the stability and accuracy of one-step methods for solving stiff systems of ordinary differential equations" Math. Comp. , 28 (1974) pp. 145–162 [a14] J. Schneid, "B-convergence of Lobatto IIIC formulas" Numer. Math. , 51 (1987) pp. 229–235
How to Cite This Entry:
B-convergence. W. AuzingerR. Frank (originator), Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=B-convergence&oldid=17814
This text originally appeared in Encyclopedia of Mathematics - ISBN 1402006098


text:
# 4.1 Ionisation energy and trends

 Page 1 / 3

## Ions

In the previous section, we focused our attention on the electron configuration of neutral atoms. In a neutral atom, the number of protons is the same as the number of electrons. But what happens if an atom gains or loses electrons? Does it mean that the atom will still be part of the same element?

A change in the number of electrons of an atom does not change the type of atom that it is. However, the charge of the atom will change. If electrons are added, then the atom will become more negative . If electrons are taken away, then the atom will become more positive . The atom that is formed in either of these cases is called an ion . Put simply, an ion is a charged atom.

Ion

An ion is a charged atom. A positively charged ion is called a cation e.g. ${\mathrm{Na}}^{+}$ , and a negatively charged ion is called an anion e.g. ${\mathrm{F}}^{-}$ . The charge on an ion depends on the number of electrons that have been lost or gained.

But how do we know how many electrons an atom will gain or lose? Remember what we said about stability? We said that all atoms are trying to get a full outer shell. For the elements on the left hand side of the periodic table the easiest way to do this is to lose electrons and for the elements on the right of the periodic table the easiest way to do this is to gain electrons. So the elements on the left of the periodic table will form cations and the elements on the right hand side of the periodic table will form anions. By doing this the elements can be in the most stable electronic configuration and so be as stable as the noble gases.

Look at the following examples. Notice the number of valence electrons in the neutral atom, the number of electrons that are lost or gained and the final charge of the ion that is formed.

## Lithium

A lithium atom loses one electron to form a positive ion: In this example, the lithium atom loses an electron to form the cation ${\mathrm{Li}}^{+}$ .

## Fluorine

A fluorine atom gains one electron to form a negative ion:

You should have noticed in both these examples that each element lost or gained electrons to make a full outer shell.

## Investigation : the formation of ions

1. Use the diagram for lithium as a guide and draw similar diagrams to show how each of the following ions is formed:
1. ${\mathrm{Mg}}^{2+}$
2. ${\mathrm{Na}}^{+}$
3. ${\mathrm{Cl}}^{-}$
4. ${\mathrm{O}}^{2+}$
2. Do you notice anything interesting about the charge on each of these ions? Hint: Look at the number of valence electrons in the neutral atom and the charge on the final ion.

## Observations:

Once you have completed the activity, you should notice that:
• In each case the number of electrons that is either gained or lost, is the same as the number of electrons that are needed for the atoms to achieve a full outer energy level.
• If you look at an energy level diagram for sodium ( $\mathrm{Na}$ ), you will see that in a neutral atom, there is only one valence electron. In order to achieve a full outer energy level, and therefore a more stable state for the atom, this electron will be lost .
• In the case of oxygen ( $\mathrm{O}$ ), there are six valence electrons. To achieve a full energy level, it makes more sense for this atom to gain two electrons. A negative ion is formed.

How we are making nano material?
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What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
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Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
The fundamental frequency of a sonometer wire streached by a load of relative density 's'are n¹ and n² when the load is in air and completly immersed in water respectively then the lation n²/na is
Properties of longitudinal waves


text:
# A 50 μA basic d’Arsonval movement with an internal resistance of 500 Ω is to be used as a voltmeter. The value of the multiplier resistance required to measure a full-scale voltage range of 0-5 volts is

This question was previously asked in
ESE Electrical 2015 Paper 1: Official Paper
View all UPSC IES Papers >
1. 999.5 kΩ
2. 99.5 kΩ
3. 9.99 kΩ
4. 0.99 kΩ

Option 2 : 99.5 kΩ
Free
CT 3: Building Materials
2565
10 Questions 20 Marks 12 Mins

## Detailed Solution

Concept:

To increase the range of a voltmeter, we need to the series resistance and it is given by

$${R_{se}} = {R_m}\left( {\frac{V}{{{V_m}}} - 1} \right)$$

Where V is the required voltmeter range

Vm is the voltmeter range

Rm is the meter internal resistance

Calculation:

Given that,

Meter full scale current reading (Im) = 50 μA

Internal resistance (Rm) = 500 Ω

Voltmeter range (V) = I Rm = 50 × 10-6 × 500 = 0.025 V

Required voltmeter range (V) = 5 V

$${R_{se}} = 500\left( {\frac{{5}}{0.025} - 1} \right) = 99.5\;{\rm{kΩ }}$$


text:
# SuanShu, a Java numerical and statistical library

com.numericalmethod.suanshu.stats.test.distribution.normality

## Class Lilliefors

• public class Lilliefors
extends HypothesisTest
Lilliefors test tests the null hypothesis that data come from a normally distributed population with an estimated sample mean and variance. The test statistic is the maximal absolute difference between empirical and the hypothetical Normal distribution function. Since the hypothesized cdf has been moved closer to the data by estimation based on the data, the "null distribution" of the test statistic, i.e. its probability distribution assuming the null hypothesis is true, is stochastically smaller than the Kolmogorov-Smirnov distribution, hence the Lilliefors distribution.

In this implementation, we first compute the p-value from the Dallal-Wilkinson (1986) formula, which is claimed to be only reliable when the p-value is smaller than 0.1. When this p-value is > 0.1, we compute again the p-value from the distribution of the modified statistic (Stephen 1974).

The R equivalent function is lillie.test in package nortest.

• G. E. Dallal, L. Wilkinson, "An analytic approximation to the distribution of Lilliefors' test for normality," The American Statistician, 40, 294-296, 1986.
• M. A. Stephens, "EDF statistics for goodness of fit and some comparisons," Journal of the American Statistical Association, 69, 730-737, 1974.
• Wikipedia: Lilliefors test
• ### Constructor Summary

Constructors
Constructor and Description
Lilliefors(double[] sample)
Perform the Lilliefors test to test for the null hypothesis that data come from a normally distributed population with an estimated sample mean and variance.
• ### Method Summary

All Methods
Modifier and Type Method and Description
String getAlternativeHypothesis()
Get the description of the alternative hypothesis.
String getNullHypothesis()
Get a description of the null hypothesis.
double pValue()
Get the p-value for the test statistics.
double statistics()
Get the test statistics.
• ### Methods inherited from class com.numericalmethod.suanshu.stats.test.HypothesisTest

isNullRejected, nGroups, nObs, oneSidedPvalue
• ### Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
• ### Constructor Detail

• #### Lilliefors

public Lilliefors(double[] sample)
Perform the Lilliefors test to test for the null hypothesis that data come from a normally distributed population with an estimated sample mean and variance.
Parameters:
sample - a sample
• ### Method Detail

• #### getNullHypothesis

public String getNullHypothesis()
Description copied from class: HypothesisTest
Get a description of the null hypothesis.
Specified by:
getNullHypothesis in class HypothesisTest
Returns:
the null hypothesis description
Wikipedia: Null hypothesis
• #### getAlternativeHypothesis

public String getAlternativeHypothesis()
Description copied from class: HypothesisTest
Get the description of the alternative hypothesis.
Specified by:
getAlternativeHypothesis in class HypothesisTest
Returns:
the alternative hypothesis description
Wikipedia: Alternative hypothesis
• #### statistics

public double statistics()
Description copied from class: HypothesisTest
Get the test statistics.
Specified by:
statistics in class HypothesisTest
Returns:
the test statistics
Wikipedia: Test statistic
• #### pValue

public double pValue()
Description copied from class: HypothesisTest
Get the p-value for the test statistics.
Specified by:
pValue in class HypothesisTest
Returns:
the p-value


text:
# Problem 10-12 Variance Analysis in a Hospital [LO10-1, LO10-2, LO10-3] John Fleming, chief ad...

Problem 10-12 Variance Analysis in a Hospital [LO10-1, LO10-2, LO10-3]

 John Fleming, chief administrator for Valley View Hospital, is concerned about the costs for tests in the hospital’s lab. Charges for lab tests are consistently higher at Valley View than at other hospitals and have resulted in many complaints. Also, because of strict regulations on amounts reimbursed for lab tests, payments received from insurance companies and governmental units have not been high enough to cover lab costs.
 Mr. Fleming has asked you to evaluate costs in the hospital’s lab for the past month. The following information is available:
 a. Two types of tests are performed in the lab—blood tests and smears. During the past month, 1,450 blood tests and 4,300 smears were performed in the lab. b. Small glass plates are used in both types of tests. During the past month, the hospital purchased 21,500 plates at a cost of $53,105. 3,300 of these plates were unused at the end of the month; no plates were on hand at the beginning of the month. c. During the past month, 3,300 hours of labor time were recorded in the lab at a cost of$37,785. d. The lab’s variable overhead cost last month totaled $29,370. Valley View Hospital has never used standard costs. By searching industry literature, however, you have determined the following nationwide averages for hospital labs:  Plates: Three plates are required per lab test. These plates cost$2.60 each and are disposed of after the test is completed. Labor: Each blood test should require 0.8 hours to complete, and each smear should require 0.40 hours to complete. The average cost of this lab time is $12.50 per hour. Overhead: Overhead cost is based on direct labor-hours. The average rate for variable overhead is$8.40 per hour.
 Required:
 1. Compute a materials price variance for the plates purchased last month and a materials quantity variance for the plates used last month. (Input all amounts as positive values. Indicate the effect of each variance by selecting "F" for favorable, "U" for unfavorable, and "None" for no effect (i.e, zero variance).)

 2 For labor cost in the lab:
 a. Compute a labor rate variance and a labor efficiency variance. (Input all amounts as positive values. Indicate the effect of each variance by selecting "F" for favorable, "U" for unfavorable, and "None" for no effect (i.e, zero variance).)

b.

In most hospitals, one-half of the workers in the lab are senior technicians and one-half are assistants. In an effort to reduce costs, Valley View Hospital employs only one-fourth senior technicians and three-fourths assistants. Would you recommend that this policy be continued?

 Yes No

 3-a. Compute the variable overhead rate and efficiency variances. (Input all amounts as positive values. Indicate the effect of each variance by selecting "F" for favorable, "U" for unfavorable, and "None" for no effect (i.e, zero variance).)

3-b.

Is there any relation between the variable overhead efficiency variance and the labor efficiency variance?

 Yes No

rev: 06_13_2016_QC_CS-53459


text: 2. Both of your ideas are good, just go ahead: $\vec{u}\times\vec{v}$ is a vector orthogonal to both $\vec{u}$ and $\vec{v}$. Since the orthogonal complement of a plane in $\mathbb{R}^3$ is a line, $\vec{u}\times\vec{v}$ is a directing vector for the line you want. So a parametric equation for this line is: $\lambda\mapsto (2,3,0)+\lambda \vec{u}\times\vec{v}$, defined for $\lambda\in\mathbb{R}$.


text:
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## Description

 Title: Quantum saturation and condensation of excitons in copper (I) oxide Author(s): Kavoulakis, Georgios Doctoral Committee Chair(s): Baym, Gordon A. Department / Program: Physics, Condensed Matter Discipline: Physics, Condensed Matter Degree Granting Institution: University of Illinois at Urbana-Champaign Degree: Ph.D. Genre: Dissertation Subject(s): Physics, Condensed Matter Abstract: Recent experiments on high-density excitons in Cu$\sb2$O provide evidence for degenerate quantum statistics and Bose-Einstein condensation of this nearly ideal gas. We model the time dependence of this bosonic system including exciton decay mechanisms, energy exchange with phonons, and interconversion between ortho (triplet-state) and para (singlet-state) excitons, using parameters for the excitonic decay, the coupling to acoustic and low-lying optical phonons, Auger recombination, and ortho-para interconversion derived from experiment. The single adjustable parameter in our model is the optical-phonon cooling rate for Auger and laser-produced hot excitons. We show that the orthoexcitons move along the phase boundary without crossing it (i.e., exhibit a "quantum saturation"), as a consequence of the balance of entropy changes due to cooling of excitons by phonons and heating by the non-radiative Auger two-exciton recombination process. Our study implies that the Auger annihilation rate for para-para collisions is much smaller than that for ortho-para and ortho-ortho collisions due to the band structure of the material, explaining why, under the given experimental conditions, the paraexcitons condense while the orthoexcitons fail to do so. Furthermore, we review the band structure underlying the properties of excitons in Cu$\sb2$O and connect these properties to the Auger recombination process of excitons. Issue Date: 1996 Type: Text Language: English URI: http://hdl.handle.net/2142/23470 ISBN: 9780591088007 Rights Information: Copyright 1996 Kavoulakis, Georgios Date Available in IDEALS: 2011-05-07 Identifier in Online Catalog: AAI9702557 OCLC Identifier: (UMI)AAI9702557


text:
# Approximation by polynomials

Let $f:[a,b] \rightarrow \mathbb{R}$ be of class $C^n$. Let $x_0, ..., x_m$ be different numbers from $[a,b]$.

Does for each $\varepsilon >0$ there exist a polynom $P$ such that $P^{(k)}(x_i)=f^{(k)}(x_i)$ for $i=0,...,m$, $k=0,...,n$ and $sup_{x \in [a,b]} |f(x)-P(x)|< \varepsilon$?

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I have never found an adequate reference for what one would call a "differentiable Stone--Weierstrass theorem", but I would assume that such a thing exists. However, I do happen to know a theorem found in Real Algebraic Manifolds by J. Nash that say: Let $f: U \rightarrow \R^n$ be analytic, let $A\subseteq U$ be compact and bounded by an analytic manifold. Then there exists a sequence of polynomials $\{f_n\}$ so that for any $\alpha$: [ D^\alpha f_n \rightarrow D^\alpha f ] I don't know much about Fourier series, which appear in the proof. Maybe it works with $C^n$ in place of $C^\omega$. –  Malte Mar 13 '12 at 21:27
@Malte: this is true. See mathoverflow.net/questions/85153/…. But I think that the OT wants something different: a uniform approximation of the function and exact value of the function and derivatives at the given points. It is not required that the derivative converges uniformly. –  Valerio Capraro Mar 13 '12 at 21:38

The problem may be split into two independent and classical ones: the Hermite interpolation, and the Weierstrass approximation.

First, we want a polynomial $p\in \mathbb{R}[x]$ with given derivatives at some given nodes $x _ 0,\dots, x _ m$. This is an instance of the Hermite interpolation problem; yours has exactly one solution $p$ with $\operatorname{deg}(p) < (m+1)(n+1)$ (the degree one would expect in terms of number of linear conditions). So, given your $f\in C^n$, you can find a polynomial with $p^{(j)}(x _ i)=f^{(j)}(x _ i)$ for all $0 \le i \le m$ and $0 \le j \le n$.

Second, as a consequence,
$$\frac{f(x)-p(x)}{\prod _ {i=0}^ m(x- x _ i)^n}$$ is (extends to) a continuous function on $[a,b]$ that vanishes on the points $x _ i$. By the Stone-Weierstrass approximation theorem there is a polynomial vanishing on the points $x _ i$ as well, whose uniform distance from that function on the interval $[a,b]$ is less than, say, $\epsilon (b-a)^{-n(m+1)}$. In other words, there is a polynomial $q\in \mathbb{R}[x]$ such that

$$\bigg\| \frac{f(x)-p(x)}{\prod _ {i=0}^ m(x- x _ i)^n} - q(x) \prod _ {i=0} ^ m(x- x _ i) \bigg\|_{\infty, [a,b]} < \epsilon (b-a)^{-n(m+1)}\ ,$$

therefore the polynomial $P(x):= p(x)+ q(x) \prod _ {i=0} ^ m(x- x _ i) ^{n+1}$ fullfills the requirements, for $P^{(j)}(x _ i)=p^{(j)}(x _ i)=f^{(j)}(x _ i)$ for all $0 \le i \le m$ and $0 \le j \le n$, and $$\|f- P\|_{\infty,[a,b]} < \epsilon\ .$$

btw. Incidentally, some time ago I happen to notice that one can find the solution of the Hermite interpolation problem as an application of the Chinese Remainder Theorem in the ring of polynomials, and wrote here the details.

edit. As to why The set $A$ of all polynomial functions on $[a,b]$ that vanish at given points $x_0,\dots, x_m$ is dense in all continuous functions on $[a,b]$ that vanish in $x_0,\dots, x_r$. One way, a bit abstract but quite immediate is, to see it as a corollary of the Stone-Weierstrass theorem (A separating closed algebra of real valued functions on a compact space $X$ is either $C(X)$ or a maximal ideal $M_x\subset C(X)$, the set of all functions vanishing at $x$). Consider $X=$ the topological quotient of $[a,b]$ obtained identifying all points $x_i$ to a point $\xi$. All functions in $A$ factor through to the quotient map, and define a closed separating algebra of continuous functions on $X$ that vanish on the identified point $\xi$. Thus, this algebra contains all continuous functions on $X$ that vanish on $\xi$, which is the thesis read on the quotient.

Note that the same construction holds in general, and provides a characterization of all closed algebras $A$ of continuous functions on a compact space $X$: identifying all points that are not distinguished by the functions of $A$ (that is, under the equivalence relation $x R_A y$ iff $f(x)=f(y)$ for all $f\in A$) one gets a Hausdorff compact quotient space (whether or not $X$ is Hausdorff), and the quotient map $\pi: X\to X/{R _ A}$ induces an isometric isomorphism of algebras $f\mapsto f\circ \pi$ of either $C(X/{R_A})$ or a maximal ideal of it onto $A$; conversely, any Hausdorff quotient of $X$ produces a closed sub-algebra of $C(X)$ this way.

Another way to see it is as a corollary of the classic Weierstrass theorem: Consider $P$ as in your comment below, then add a perturbation $L$ that makes $P+L$ vanish on the points $\{x_i\}$; this has been clearly explained in Ilya Bogdanov's answer. Here you don't have derivatives and $L$ is just a Lagrange interpolation polynomial, which is small in the uniform norm because it is small on the points $\{x_i\}$.

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Very thanks. I have one question. You used the following fact: if a continuous function $g:[a,b]\rightarrow R$ vanishes at points $x_0,...,x_m$ then for every $\delta >0$ there exists a polynomial $Q$ which vanishes at $x_0,...,x_m$ such that $\|g(x)-Q(x)\|_{sup} <\delta$. It is easy for $m=0$, because by Weierstrass threorem we find polynomial $P$ such that $\|f-P\| <\frac{\delta}{2}$ and polynomial $Q:=P-P(x_0)$ will be good. How to prove its for $m\geq 1$? –  arc Mar 14 '12 at 13:59
I've edit and added a few lines on this point. –  Pietro Majer Mar 14 '12 at 17:02

For the sake of simplicity, let us assume that $[a,b]=[0,1]$. $C^k$ always means $C^k[0,1]$. We will even approximate $f$ in $C^n$-norm satisfying your additional condition.

1. As was mentioned in the comments, you can easily approximate $f$ together with all its derivatives up to $n$th uniformly by a polynomial. In fact, it is enough to approximate $f^{(n)}$ with an adequate accuracy: if $||f'-P'||_C<\varepsilon$ and $f(0)=P(0),$ then $||f-P||_C<\varepsilon.$

2. Now take the polynomials $Q_{ik}(x)$ such that $Q_{ik}^{(d)}(x_j)=0$ for all $d=0,\dots,n$ and $j=0,\dots,m$ except that $Q_{ik}^{(k)}(x_i)=1$. Such polynomials are easy to construct: for instance, one may take $$Q_{ik}(x)=c_{ik}(x-x_i)^k\prod_{j\neq i}\left((x-x_i)^{n+1}-(x_j-x_i)^{n+1}\right)^{n+1}\;\;$$ for a suitable constant $c_{ik}.$ Let $M=\max_{i,k}||Q_{ik}||_{C^n}.$ Then, let the approximation in the previous paragraph be $\delta$-accurate with $\delta=\varepsilon/(2M(m+1)(n+1)).$ To correct the values of the polynomial and its derivatives at $x_i,$ it is enough to add the polynomials $Q_{ik}$ multiplied by the coefficients with absolute values $\leq\delta,$ hence the total error will be not more that $\delta+(m+1)(n+1)M\delta<\varepsilon.$

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nice ! –  Pietro Majer Mar 14 '12 at 5:49


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Session 8 - Distance Indicators.
Display session, Monday, January 15
North Banquet Hall, Convention Center

## [8.01] The Distance to the Cluster of Galaxies Abell 2634

M. Scodeggio, R. Giovanelli, M. P. Haynes (Cornell)

The Tully-Fisher and D_n - \sigma techniques are most commonly used to obtain redshift-independent estimates of galaxy distances. In some cases, however, the two methods yield conflicting results. The most notorious of such cases refers to the cluster of galaxies Abell 2634: while the Tully-Fisher relation applied to a sample of 11 spiral galaxies gave a distance estimate corresponding to essentially zero peculiar velocity, the D_n - \sigma relation applied to a sample of 18 elliptical and lenticular galaxies gave a distance estimate corresponding to a peculiar velocity of --3400 km s^-1. This difference has been used as evidence for the possible non-universality of the two relations, which would result in spurious peculiar velocities, and, in turn, in inadequate reconstructions of the local Universe mass density field.

As part of a larger project to compare Tully-Fisher and D_n - \sigma redshift-independent distance estimates for nearby clusters of galaxies, we have obtained I band CCD photometry, and either 21cm line rotational velocities (for spirals) or Mg lines internal velocity dispersions (for E and S0's) for a large sample of galaxies in the region of A2634 (45 spirals and 52 E and S0's). This data-set, combined with a detailed study of cluster membership, is used to obtain an accurate determination of the distance to A2634.


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# 2:06 LTED Question 10 of 20 Submit How many liters of a 0.209 MKI solution is needed to completely react

###### Question:

2:06 LTED Question 10 of 20 Submit How many liters of a 0.209 MKI solution is needed to completely react with 2.43 g of Cu(NO3)2 according to the balanced chemical reaction: 2 2 Cu(NO3)2(aq) + Cul(aq) + 12(s) + 4 4 Kl(aq) → KNO3(aq) STARTING AMOUNT ADD FACTOR ANSWER RESET 2.43 2 0.124 6.022 x 1023 187.57 0.00542 0.0310 0.0620 0.209 166 4 Tap here or pull up for additional resources

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# Why is the residual proton resonance in deuterated benzene a singlet?

In deuterated benzene ($99\%$) the residual protio peak is a singlet at $\pu{7.16 ppm}$. This however doesn't make sense since the protons will be randomly distributed among various benzene molecules. Therefore there won't be a mixture of $99$ $\ce{C6D6}$ molecules and $1$ $\ce{C6H6}$ but $94$ $\ce{C6D6}$ and $6$ $\ce{C6D5H}$. Considering that deuterium has a spin of $1$ the residual solvent peak in D6-benzene should be a septet. Any thoughts?

To a very good approximation, $$J_\ce{HD}$$ in a deuterated molecule is related to $$J_\ce{HH}$$ in the corresponding protonated molecule by the gyromagnetic ratios:

$$\frac{J_\ce{HD}}{J_\ce{HH}} = \frac{\gamma_\ce{D}}{\gamma_\ce{H}} = \frac{\pu{41.066 * 10^6 rad s-1 T-1}}{\pu{267.522 * 10^6 rad s-1 T-1}} = 0.15$$

In benzene where $$^3J_\ce{HH} = \pu{7.6 Hz}$$ and $$^4J_\ce{HH} = \pu{1.4 Hz}$$ (Reich NMR tables), we would thus have $$^3J_\ce{HD} \approx \pu{1.2 Hz}$$ and $$^4J_\ce{HD} \approx \pu{0.2 Hz}$$. The four-bond meta coupling would be all but invisible, and the three-bond ortho coupling if resolved would give rise to a $$1:2:3:2:1$$ quintet.

On a lower-field spectrometer this relatively small coupling is usually not properly resolved. A suitable window multiplication, though, will reveal all. I don't have a d6-benzene spectrum on hand, but I do have a $$\pu{500 MHz}$$ d8-toluene spectrum here. The two spectra on the left are the aromatic residual peaks, but at the bottom-left a Gaussian window function has been used to improve resolution (at the cost of signal:noise ratio). This allows us to resolve the three-bond ortho couplings of $$\pu{1.1 Hz}$$.

At the bottom-right is the signal corresponding to residual $$\ce{C6D5CD2H}$$. Since the two-bond H–D coupling has a larger magnitude ($$^2J_\ce{HD} = \pu{2.2 Hz}$$), it is much easier to resolve, and can be seen even with the normal exponential window function (which sacrifices resolution for signal:noise ratio improvement).

Similarly, in d6-DMSO (shown below at $$\pu{500 MHz}$$), the two-bond coupling is more easily resolved.


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# Is this dynamic optimization?

I would like to know what I should know to understand this IMF paper. What kind of optimization is used to maximize the utility function on page 9 (number 1) subject to constraints (2) and (3)?

The function I must maximize is

$U_0^i=E_0\sum_{t=0}^{\infty} \beta_i^t \left[ \dfrac{(c_t^i - c_{min}^i)^{(1-\frac{1}{\sigma_i})}}{\left(1-\frac{1}{\sigma_i}\right)} + \xi_d log (d_t (1-(1-\gamma_\ell)\pi_t)) + \xi_k log (\overline{k} + k_t(1-(1-\gamma_k)\pi_t))\right]$

The constraints are

$k_t=(1-\delta)\Delta_{k_t} k_{t-1}+I_t$

and

$d_t q_t = \Delta_{\ell_t} d_{t-1}+r_t^k \Delta_{k_t}k_{t-1}-c_t^i-I_t$

The optimality conditions for $c$, $d$ and $k$ are:

$(c_t^i-c_{min}^i)^{-\frac{1}{\sigma_i}}=\lambda_t^i$

$1=\beta_i E_t \left( \dfrac{\lambda_{t+1}^i}{\lambda_t^i} \right) \dfrac{1-(1-\gamma_\ell)\pi_t}{q_t} + \dfrac{\xi_d}{\lambda_t^i d_t q_t}$

and

$1=\beta_i E_t \left( \dfrac{\lambda_{t+1}^i}{\lambda_t^i} \right) (r_{t+1}^k+1-\delta)(1-(1-\gamma_k)\pi_t)+\dfrac{\xi_k(1-(1-\gamma_k)\pi_t)}{\lambda_t^i(\overline{k}+k_t (1-(1-\gamma_k)\pi_t))}$

What should I study to understand this maximization? Is this dynamic optimization?

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