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Suppose the random variables X and Y have a joint pdf f_{XY}(x,y)=\begin{cases}
Suppose the random variables X and Y have a joint pdff_{XY}(x,y)=\begin{cases}x+y & 0\leq x,y \leq 1\\0 & otherwise\end{cases}Find Pr(X>sqrt{Y})
{f}_{XY}\left(x,y\right)=\left\{\begin{array}{ll}x+y& 0\le x,y\le 1\\ 0& otherwise\end{array}
Pr\left(X>\sqrt{Y}\right)
The joint probability density function of the random variables X and Y is as given below:
{f}_{XY}\left(x,y\right)=\left\{\begin{array}{l}x+y.\text{ }0\le x,y\le 1\\ 0.otherwise\end{array}
P\left(X>\sqrt{Y}\right)
The marginal probability density function of X is obtained as given below:
{f}_{X}\left(x\right)={\int }_{y}{f}_{XY}\left(x,y\right)dy
={\int }_{0}^{1}\left(x+y\right)dy
=x\left[y{\right]}_{0}^{1}+\left[\frac{{y}^{2}}{2}{\right]}_{0}^{1}
{f}_{X}\left(x\right)=x+\frac{1}{2}
P\left(X>\sqrt{Y}\right)
is obtained as given below:
P\left(X>\sqrt{Y}\right)={\int }_{\sqrt{y}}^{1}\left(x+\frac{1}{2}\right)dx
={\int }_{\sqrt{y}}^{1}xdx+{\int }_{\sqrt{y}}^{1}\frac{1}{2}dx
=\left[\frac{{x}^{2}}{2}{\right]}_{\sqrt{y}}^{1}+\left[\frac{x}{2}{\right]}_{\sqrt{y}}^{1}
=\frac{1}{2}-\frac{y}{2}+\frac{1}{2}-\frac{\sqrt{y}}{2}
=1-\frac{y}{2}-\frac{\sqrt{y}}{2}
A diving borad 3.00 m long is supported at a point 1.00 m fromthe end, and a diver weighing 500 N stands at the free end. The diving board is of uniform cross section and weighs 280N.
A. Find the force at the support point.
B. Find the force at the end that is beng held down.
Find the point on the plane
x+2y+3z=13
that is closest to the point (1,1,1). How would you minimize the function?
g\left(x\right)=\surd 3x+1
A thin piece if wire 4.00mm long is located in a plpaneperpendicular to the optical axis and 60.0cm in front of a thinlens. The sharp image of the wire formed on a screen is 2.0mm long.What is the focal length of the lens?
A 3.53-gg sample of aluminum completely reacts with oxygen to form 6.67 gg of aluminum oxide.
Use these data to calculate the mass percent composition of aluminum in aluminum oxide.
Express the percent composition to three significant figures
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Recent Posts – Avital Oliver
Correcting a proof in the InfoGAN paper
The InfoGAN paper has the following lemma: Lemma 5.1. For random variables
X, Y
f(x, y)
under suitable regularity conditions:
\mathbb{E}_{x \sim X, y \sim Y|x}[f(x, y)] = \mathbb{E}_{x \sim X, y \sim Y|x, x' \sim X|y}[f(x', y)]
. The statement is correct, but the proof in the paper is confused – here’s a step where
x
mysteriously becomes
x'
Read more ⟶
Why Mean Squared Error and L2 regularization? A probabilistic justification.
When you solve a regression problem with gradient descent, you’re minimizing some differentiable loss function. The most commonly used loss function is mean squared error (aka MSE,
\ell_2
loss). Why? Here is a simple probabilistic justification, which can also be used to explain
\ell_1
loss, as well as
\ell_1
\ell_2
Correcting a proof in the I... May 29, 2018
Why Mean Squared Error and ... Mar 20, 2017
@avitaloliver
© 2018 Avital Oliver. Powered by Jekyll using a modified version of the Balzac theme.
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Alternatives to Linear Regression Practice Problems Online | Brilliant
Least squares linear regression is probably the most well-known type of regression, but there are many other variants which can minimize the problems associated with it.
A common one is known as ridge regression. This method is very similar to least squares regression but modifies the error function slightly.
Previously, we used the sum of square errors of the regression line as a measure of error, but in ridge regression we seek to minimize the squared values of coefficients as well. This gives the error function
\small \text{Error} = \sum_{i=1}^{n} (y_i - m_1x_{1i} - m_2x_{2i} - … - m_px_{pi} - b)^2 + \lambda \sum_{i=1}^{p} (m_i^2).
Here, the value of lambda changes how aggressively coefficients are dampened. Notice that this error function does not penalize the size of the
y
Alternatives to Linear Regression
A close relative to ridge regression is simply known as “lasso.” This also penalizes the size of coefficients in the error function, but does so based on their linear size instead of their squared size. Therefore, error is given by
\small \text{Error} = \sum_{i=1}^{n} (y_i - m_1x_{1i} - m_2x_{2i} - … - m_px_{pi} - b)^2 + \lambda \sum_{i=1}^{p} \vert m_i \vert.
Below, we use a coding environment to import a LASSO class from the sklearn library and compared its results with those of normal linear regression. Press run to see the the results of each model plotted.
#Generate artificial data for regression lines.
x = np.random.normal(0, 10, (20, 1))
y = 4*x+10+np.random.normal(0, 5, (20, 1))
LASSO = linear_model.Lasso(alpha=50) #50 is a large value of alpha, and is chosen for demonstration purposes.
plot_x = np.transpose(np.array([list(range(-20, 20))]))
plt.plot(plot_x, regr.predict(plot_x))
plt.plot(plot_x, LASSO.predict(plot_x))
plt.legend(["Linear Regression", "LASSO"])
plt.plot(x, y, "ro")
plt.axis([-15, 15, -70, 70])
plt.savefig("Plots.png", format="png")
from sklearn import linear_model import matplotlib.pyplot as plt import numpy as np #Generate artificial data for regression lines. x = np.random.normal(0, 10, (20, 1)) y = 4*x+10+np.random.normal(0, 5, (20, 1)) regr = linear_model.LinearRegression() LASSO = linear_model.Lasso(alpha=50) #50 is a large value of alpha, and is chosen for demonstration purposes. plot_x = np.transpose(np.array([list(range(-20, 20))])) regr.fit(x, y) LASSO.fit(x, y) plt.plot(plot_x, regr.predict(plot_x)) plt.plot(plot_x, LASSO.predict(plot_x)) plt.legend(["Linear Regression", "LASSO"]) plt.plot(x, y, "ro") plt.axis([-15, 15, -70, 70]) plt.savefig("Plots.png", format="png")
It is not at all obvious why lasso would have behavior significantly differing from ridge regression, but there is an interesting geometric reason for the differences. However, to demonstrate this, we must first change the way we view both techniques.
In ridge regression, it turns out that for any values of
\lambda
we pick, it’s possible to find a value for
\lambda_2
such that minimizing
\sum_{i=1}^{n} (y_i - m_1x_{1i} - m_2x_{2i} - … - m_px_{pi} - b)^2 + \lambda \sum_{i=1}^{p} \big(m_i^2\big)
is equivalent to minimizing the SSE when
\sum_{i=1}^{p} \big(m_i^2\big) \leq \lambda_2.
(This can be shown using the method of Lagrange multipliers)
Similarly, for any value of
\lambda
there is some value of
\lambda_2
such that using lasso is equivalent to minimizing the SSE when
\sum_{i=1}^{p} \vert m_i \vert \leq \lambda_2.
A useful way to view the SSE when there are two predictor variables is shown in the pictures below. Here, the
x
y
-axes represent the values of coefficients in a best-fit plane, and the ellipses show all pairs of coefficients which produce a certain value of the SSE for a data set. As the SSE increases, the ellipses get larger.
Also in the pictures are two areas. The diamond represents the coefficient values allowed by lasso. The disk represents the possible coefficient values in ridge regression. Viewing these pictures, which form of linear regression will most likely lead to coefficients of zero?
Lasso Ridge Regression
Lasso generally behaves very similarly to ridge regression but, as we saw in the previous question, there is one crucial difference. It is capable of reducing the weights of predictor variables to zero. This is useful if one wants to cull predictor variables, among other things. Usually, this is done when there are many predictor variables and using too many in the model will cause overfitting or make it overcomplicated.
A group of scientists wants to analyze bacterial growth in Petri dishes. They have done a dozen tests, and each time they have recorded every single detail of the environment. The pH levels of the dishes, sugar content of the food, and even the light levels in the room have been recorded. A total of fourteen variables have been taken into account.
In a classic example of overzealous testing, a rogue scientist has added another variable to the mix, his average mood on a scale from zero to ten. When a best-fit equation is generated with this variable included, how will the SSE most likely change? How will the average error on new data change?
The SSE will decrease while the error on new data increases Everything will remain the same The SSE will increase while the error on new data decreases The SSE and the error on new data will increase
The previous question is a good example of a case where ridge regression or lasso would be very useful. Each of these techniques will penalize a best-fit line for having large coefficients, so they are likely to produce equations that make minimal use of predictor variables that have little sway over the result. Because the equation has a limited “budget,” it can only afford to give large weights to variables which are important.
In the case shown previously, this means that something with as little predictive power as the scientist’s mood the day of the test will be largely ignored. In fact, if lasso is used, the variable will probably be ignored entirely.
Of course, linear regression is just one of many techniques. A non-linear method with comparable simplicity is known as K-nearest neighbors regression.
To use K-nearest neighbors regression, or KNN regression for short, we must start with a data set. As with linear regression, the dataset must take of form of pairs of predictor variables
\vec{x}_i
with resultant variables
y_i
. The goal is to use this dataset to predict the value of a resultant variable
y
from a vector of predictor variables
\vec{x}
To make a prediction for
\vec{x}
, we plot each
\vec{x}_i
in our dataset, ignoring the resultants, and pick out the
k
points geometrically closest to
\vec{x}
. The estimate KNN regression provides is simply the average of the resultant values for these points.
One useful property of KNN regression is that it makes very few assumptions about the data sets it builds on. Unlike linear regression, which assumes linear relationships, KNN regression can accommodate nearly anything.
Additionally, by adjusting the value of
k,
we can change the flexibility of KNN regression. If we want to account for even the smallest trends in our data set, we can pick a very small
k
-value. On the other hand, larger values of
k
will eliminate smaller deviations in favor of larger trends.
Let's try applying KNN regression to a simple example. In the image below, we've plotted a dataset of ten points, where the predictor variable is given by the x-axis and the resultant variable is given by the y-axis. In this case,
y = x^2
for all points in the dataset.
Now, suppose that we have a new point for which
x = 3.5
and we want to predict its value with KNN Regression. If we use
2
as our value for K, what will our estimate be?
Hint: In KNN regression we pick out the K points geometrically closest to
\vec{x}
and average their resultant values.
Below, we have three data sets—A, B, C—represented by either tables or scatter plots. We can analyze one with K-nearest neighbor regression, one with lasso, and one with normal linear regression. Which pairings will give the best results?
Data Set A:
\begin{array}{c|c|c|c|c} x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & y \\ \hline 5 & 8 & 97 & 2 & 0 & 8 & 3 \\ \hline 2 & 7 & 0 & 2 & 1 & 7 & 4\\ \hline 2 & 6 & 4 & 12 & 6 & 3 & 14 \\ \hline 15 & 6 & -20 & 5 & 2 & 2 & 6 \\ \hline 4 & 8 & 2 & 6 & 0 & 3 & 5\\ \end{array}
A - KNN, C - Lasso, B - Linear Regression B - KNN, A - Lasso, C - Linear Regression C - KNN, A - Lasso, B - Linear Regression C - KNN, B - Lasso, A - Linear Regression
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Computer Programming Resources for Beginners | Brilliant Math & Science Wiki
Computer Programming Resources for Beginners
Bruna Torman, Arulx Z, and Jimin Khim contributed
When someone wants to start in programming, discover that there are many resources and things you should learn before you say "I'm a good programmer." So, this wiki is to help you know how to start in this field.
1. Start with the logic
Yes, you should study logic and algorithms before learning how to program. This is important because those will teach you how to think in computer science. Algorithms, in an easy way, will show you what you can do with any programming language.
There are many books and websites for learning algorithms, but I strongly recommend the books "Algorithms Unlocked" (prerequisite for the next book) and "Introduction to Algorithm 3rd Edition," by Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. This book is shared by MIT Press. It is easy to find this book and it's the best book I ever used.
2. Now you know algorithms? Well done, and let the fun begin!
First of all, there are some rules to follow:
You don't need to know all the programming languages because it's impossible. Yes, there are a lot of programming languages and a lot of rules in those languages, so relax and study what you really want to know.
To really learn how to program, you need to do one simple thing: do it everyday. You will never learn programming or even become good at it studying and programming just a few days or just when you want to. It's important to program everyday to learn well.
Never stop studying. I think this is the most important rule you should follow. Technology changes everyday, every month, and every year. Stay in touch with the news.
Now we can start with programming finally!
There are some recommended languages you should start with because they are easy: C, C++, Java, Python. They are an easy introduction to this world of bits.
I start with Java, but you can choose anyone of them. The resources I recommend are as follows:
Codecademy: In this website, they teach you many programming languages and .NET development. They have now the introduction to Java and Python. Have fun at codecademy.com!
Free Code Camp: A free website that teaches you how to become a full stack web developer. You will learn HTML, CSS, JavaScript, Databases, etc. You will get a certificate at the end of our studies and you will be allowed to help a non-profit organization recommended by the site to test your programming skills. Go ahead, grasshopper: freecodecamp.com!
Udacity: Free courses (and some courses you have to pay for) in computer science and its theory, which include some programming languages, mobile development, artificial intelligence, startups, etc. You will get a certificate when you finish every course. Be happy at udacity.com!
CS50 - Introduction to Computer Science by Harvard University on EdX: "An introduction to the intellectual enterprises of computer science and the art of programming." One of the most popular courses on Edx that covers many topics and good challenges. No prior experience is necessary. You can start the course and end anytime till February 2, 2016. You'll get a certificate by the end of your studies. Enjoy yourself here!
MIT OpenCourseWare (aka. Paradise): You can learn anything there. A website covered by MIT that provides you with videos, exams, and happiness for all the undergraduate courses from the university. MIT also has Scratch (for beginners) and the MIT App Inventor (for mobile studies). Go ahead at MIT OpenCourseWare - ocw.mit.edu/index.htm Scratch - scratch.mit.edu MIT App Inventor - appinventor.mit.edu/!
3. Let's challenge yourself.
Competitions are good, and I know you agree. So let's test your programming skills with the following:
HackerRank: Website to compete and solve some problems in CS and programming. You also can get money and a job in companies like Facebook, Google, Microsoft, Asana, etc. It's free and fun.
TopCoder: Good resource for a competitor. I recommend for people with some competition experience. You will get money and jobs here, too.
\text{High School: International Olympiad in Informatics (IOI)}
High school competition recognized worldwide: Participation is individual. The evaluation of the solutions is made after the end of the competition and the score is given to each part of the approved test. Similarly, national and preliminary informatics olympiads exist such as the British Informatics Olympiad in Britain and the OBI (Olympiad of Informatics for Brazilians students) in Brazil.
\text{TopCoder High School Tournament (TCHS)}
High school competition organized by the website of TopCoder competitions.
College: Generally students participate in higher education and the first year of graduate school.
\text{ACM International Collegiate Programming Contest (ACM-ICPC)}
Higher education competition recognized worldwide: Participation is in teams of three students to share a single computer. The evaluation of the solutions is done during the competition, but the score is given only to those who pass all the tests. To participate, it is necessary to be among the first in Brazil in Programming Marathon.
Higher education competition recognized in the country: To participate, it needs to be part of one of the teams of the institute or college. It is held in the molds of ACM-ICPC, but in two stages. In the first stage, the various teams compete locally at sites around the country. The top finishers of each site qualify for the national final, where all teams compete in the same place. Each institute or college can take up to two teams for the final.
\text{TopCoder Collegiate Challenge (TCCC)}
Higher education competition organized by the website of TopCoder competitions.
\text{FREE:}
\text{TopCoder Open (TCO):}
Most respected free competition in the world. Organized by the site of TopCoder competitions.
\text{Google Code Jam (GCJ):}
Free competition organized by Google. It is a worldwide competition that occurs almost every year. Besides the global version, since there were also regional versions in Europe, Latin America, China and India, it was held along the lines of TopCoder. But the official site says that in 2008 they will use a new platform.
\text{TopCoder Single Round Matches (SRM):}
Site American various programming competitions. Are always promoting Single Round Matches (SRM) in which participants can practice and move up in the standings. Companies seek people on the site and classified to provide jobs. Eventually the sponsors award the best placed with t-shirts and even money. Each year there are two major competitions (TCC and TCCO) divided into several phases, where the finalists travel for free to participate in a centralized end. In general, a competition stage or in an SRM TopCoder is made up of three individual and phase. In the first 1 hour and 15 minutes contestants attempt to solve the problems. The points at issue vary the level of difficulty and the time taken to resolve it. The evaluation of the solutions is not made on time. In the second stage, the contestants have 15 minutes to check opponents programs and bolarem test cases to take them down. In the third step the system performs a battery of automated tests to check programs. It's a much more dynamic and less tiring than traditional competition.
\text{Internet Problem Solving Contest (IPSC):}
This is, strictly speaking, not really a programming competition. It is a competition problem solving, but it is almost always necessary for computer to help solve them. In this competition are given descriptions of problems and the input files and participants must send the answers. Quite different from the traditional and very entertaining. The top 10 receive certificates.
\text{Online Judges:}
The main websites that provide online problems with automatic brokers and also tend to hold competitions periodically. In general there is no prize, but follow the lines of the ACM-ICPC, serving as an excellent workout for the Programming Marathon. The following are the most well-known sites that carry such competitions:
Valladolid Online Judge Contest Hosting Service
Ural State University Problem Set Archive
Zhejiang University Online Judge Contests
Saratov State University Online Contester
ACM NEERC Online Contest Site
USACO Contest Gateway (OBI style)
You can do all of these in any field you want to persue. Programming is a skill for everyone in every field. You can do it. Be your own boss.
Cite as: Computer Programming Resources for Beginners. Brilliant.org. Retrieved from https://brilliant.org/wiki/computer-programming-resources-for-beginners/
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Specific Heat | Brilliant Math & Science Wiki
Specific heat is a very important aspect of physics. To understand this, we must get to know what actually is heat. There are many day-to-day activities which involve the use of heat, but we don't realize its importance. Its uses range from simply rubbing our hands to keep ourselves warm to its use in complex machinery. The concept of specific heat, though it looks very difficult, can be dealt with very easily. So, let's explore the unique world of thermodynamics!
Heat is a form of energy. As discussed earlier, while we rub our hands, the energy spent in overcoming friction is converted into heat energy, or simply the mechanical energy is converted to heat. Matter contains heat in the form of kinetic energy and potential energy. The total mechanical energy, i.e. potential energy + kinetic energy, is hence called thermal energy.
The thermal energy in a particular body is proportional to the amount of inter-molecular vibration. This means that as the vibration increases the thermal energy also increases. So, hot bodies have more thermal energy than cold bodies. We know that when a hot body and a cold body are kept in contact with each other, the vibrations travel from the hot body to the cold.
Let's now define heat considering all the conditions:
Heat is a type of mechanical energy, which is stored in matter in the form of vibrational energy. It is the amount of thermal energy that flows from a hot body to a cold body.
_\square
SI unit: Joule
We know that heat moves from a hot body to a cold body. When do you think will the transfer of heat stop? Or is it a never-ending process?
The transfer of heat energy stops when both objects get equally heated, i.e. the flow of heat continues till the temperatures are equalized.
When two bodies are undergoing the process of transfer of heat, the flow will continue if at least one of the body is hotter than the other. So, when the temperatures of both objects are equalized, the flow stops. In this situation the two bodies are said to be in thermal equilibrium.
_\square
It obeys the zeroth law of thermodynamics.
We have been talking about heat until now. But how do we measure this heat?
Temperature can be termed as a quantitative measure of the magnitude of hotness of a body. Or by using the concept of heat we define it as the average kinetic energy per molecule of a particular substance.
_\square
Temperature is measured using various scales, and the most commonly used ones are the Celsius and Fahrenheit scales. Most of us might be familiar with the formulas of conversion between the two scales, which are as follows:
\begin{aligned} \text C&=\text F-32\times \dfrac59 \\ \text F&=\dfrac95 \times \text C+32, \end{aligned}
C
represents Celsius and
F
represents Fahrenheit. Another scale is the Kelvin scale whose conversion formula is as follows:
\text K=\text C+273.16,
K
represents Kelvin. But to make calculations simple, the value added is only
273.
Kelvin scale is taken as the
\text{SI}
unit for temperature. Another scale which only few of us are familiar with is the Newton scale, in which the boiling point is
33\text{ N},
so the conversion formula is
\dfrac{\text{N}}{0.33}=\text C.
Now, as we have understood the concepts of heat and temperature, we can proceed to the main topic specific heat. We know that there are many factors affecting the increase in temperature per unit time for various bodies. Let's consider the following cases:
Case 1: There are two containers
A
B,
each containing an equal amount of water. Let us suppose that
A
is given more heat than
B,
then we see that the temperature rises faster in
A.
So we conclude that heat energy
\text Q
is directly proportional to the rise in temperature
\Delta \text T
\text Q\propto \Delta \text T.
A
B,
B
contains more water than
A
and they are given an equal amount of heat. Then we see that the temperature rises faster in
A.
So we conclude that as the mass
m
of a body increases, the heat energy required also increases:
\text Q\propto m.
Case 3: For the last time, let
A
contain kerosene, and
B
an equal amount of water. Let them each be provided with an equal amount of heat, then we see that kerosene gets heated up earlier. So we conclude that the heat energy required is dependent on the nature of the substance, too.
Now, from cases (1) and (2), we have
Q\propto \Delta T
Q\propto m
\begin{aligned} \text Q&\propto m \Delta \text T \\ \text Q&=\text Cm\Delta \text T, \end{aligned}
\text C
is a constant of proportionality. This is called the specific heat capacity, whose
\text{SI}
unit is
\text{J kg}^{-1} \text{K}^{-1}.
It doesn't matter. All types of chocolate melt at the same temperature. Dark chocolate Milk chocolate White chocolate
If the chocolate pope is not to be melt on a hot sunny day, what should he be made of?
A penny weighs
2.5 \text{ g}
and is made of
100
\text{Cu}.
The penny starts out at
25^\circ\text{C}
and melts at
1085^\circ\text{C}.
\text{Cu}
's heat of melting is
176 \text{ kJ}/\text{kg}
and its specific heat is
0.386\text{ kJ}/\text{kg}\cdot\text{K}.
The area of the lens is
0.2 \text{ m}^2,
the power of the sun is
1370\text{ W/m}^2,
100
% of the energy goes into heating the coin.
You have a piece of aluminium whose mass is
800\text{ g}
, and suppose it is heated up to
1000^\circ\text{C}.
Now, given that you are cooling it down to
150^\circ\text{C}
, calculate the amount of heat (in Joules) given out during the process of cooling.
The specific heat capacity of aluminium is
\text{C}_\text{Aluminum}=900 \text{ J}/\text{kg K}.
Cite as: Specific Heat. Brilliant.org. Retrieved from https://brilliant.org/wiki/specific-heat/
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Classify each of the studies as either descriptive or inferential.
Classify each of the studies as either descriptive or inferential. Explain your answers.Passed at the advanced level: 41−48 items correct
This data is descriptive because it merely DESCRIBES and collects the pass/fail percentage over the course of 2 years.
{\left({x}^{4}{y}^{5}\right)}^{\frac{1}{4}}{\left({x}^{8}{y}^{5}\right)}^{\frac{1}{5}}={x}^{\frac{j}{5}}{y}^{\frac{k}{4}}
j-k
At a High school, 70% of the seniors have taken an advance calculus course. Of those who have taken Advanced Calculus, 60% will apply for a Pre-Health science major when they apply for college admission. Of those who have not taken advanced calculus, 40% will apply for a Pre-Health science major when they apply for college admission. A senior is selected at random. What is the probability that the senior have taken advanced calculus, given that the senior will apply for a pre-health science major?
In fall 2014, 38% of applicants with a Math SAT of 700 or more were admitted by a certain university, while 12% with a Math SAT of less than 700 were admitted. Further, 32% of all applicants had a Math SAT score of 700 or more. What percentage of admitted applicants had a Math SAT of 700 or more? (Round your answer to the nearest percentage point.)
Which one is a simplified expression for
A\left({A}^{\prime }+{B}^{\prime }\right)\left(B+C\right)\left(B+{C}^{\prime }+D\right)
Find the final form of the median and range for uniform distribution (0,1) for order statistics.
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Least Cost Method | Method to Solve Transportation Problem | Transportation Model | Education Lessons
Least cost Method is one of the method to obtain feasible solution of Transportation Problem. We already understand the North West Corner Method | Method to Solve Transportation Problem | Transportation Model for transportation problem, to obtain feasible solution.
To understand Least Cost Method we will go through the numerical provided as follows (the same numerical we used in North West Corner Method | Method to Solve Transportation Problem | Transportation Model) :
A mobile phone manufacturing company has three branches located in three different regions, say Jaipur, Udaipur and Mumbai. The company has to transport mobile phones to three destinations, say Kanpur, Pune and Delhi. The availability from Jaipur, Udaipur and Mumbai is 40, 60 and 70 units respectively. The demand at Kanpur, Pune and Delhi are 70, 40 and 60 respectively. The transportation cost is shown in the matrix below (in Rs). Use the Least Cost method to find a basic feasible solution (BFS).
Note that all the explanation is provided in “CYAN” colour. You have to write in examination the only thing which are given in this regular colour under each steps(if any), else you can directly solve matrix of the problem as explained here.
Balance the problem meaning we need to check that if;
\color{#32c5d4} \Sigma \text { Supply} = \Sigma \text { Demand}
If this holds true, then we will consider the given problem as a balanced problem.
Now, what if it’s not balanced?
\color{#32c5d4} \text {i.e., } \Sigma \text { Supply} \not = \Sigma \text { Demand}
If such a condition occurs, then we have to add a dummy source or market; whichever makes the problem balanced.
You can watch a video on this type of numerical, which is known as Unbalanced Transportation Problems.
\to
The given transportation problem is balanced.
Step 2: Select the lowest cost from the entire matrix and allocate the minimum of supply or demand.
We are using Least Cost Method here, so we will be indentifying the lowest cell value in this entire matrix.
Here, in this matrix we have 1 (For cell: Jaipur-Delhi) as the lowest value.
So, moving with that cell, and allocating the minimum of demand or supply, i.e. allocating 40 here (as supply value is 40, whereas demand is of 60).
Checking out the first row and not the last column, because we are allocating 40 in the cell for supply, as it is minimum.
Subtracting allocated value (i.e. 40) from corresponding supply and demand.
Step 3: Remove the row or column whose supply or demand is fulfilled and prepare a new matrix
As we have fulfilled the demand or supply for that row or column respectively, remove that row or column and prepare a new matrix, as shown below:
Step 4: Repeat the procedure until all the allocations are over
Repeat the same procedure of allocation of the smallest value in the new generated matrix and check out demand or supply based on the smallest value (of demand or supply) as shown below, until all allocations are over.
Tie in the following step!
You may find tie in selecting cell here in the above matrix, as we have minimun cell value 3 for two cells in the matrix above.
So whenever you find this kind of situation while evaluting matrix by Least Cost Method, you will have multiple solution to the problem, called alternate solutions.
The alternate solution of the this problem is provided at the end of this note-blog. In that we will be selecting the intersecting cell of Udaipur - Kanpur.
Step 5: After all the allocations are over, write the allocations and calculate the transportation cost
Once all allocations are over, prepare the table with all allocations marked and calculate the transportation cost as follows:
\begin{aligned} \to \ \text {Transportation cost} &= (1 \times 40) + (3 \times 40) + (3 \times 20) + (6 \times 30) + (2 \times 40) \\ &= \text {Rs } 480 \end{aligned}
As we have tie in selecting minimum value at this step.
We can select any of this cell as mentioned earlier.
We have already seen above the solution by selecting minimum value as of Udaipur-Delhi intersecting cell, we will select now the another one and continuing the allocation using Least Cost Method as follows:
So, we will be having alternate solution as mentioned above, and alternate transportation cost as follows:
\begin{aligned} \to \ \text {Transportation cost} &= (1 \times 40) + (3 \times 60) + (6 \times 10) + (2 \times 40) + (8 \times 20) \\ &= \text {Rs } 520 \end{aligned}
\to
Thus, we have optimum Transportation cost as Rs. 480.
Find solution of same numerical by:
Q-1) In least cost method, we start by identifying
maximum value in the given matrix
maximum value of supply & demand
minimum value in the given matrix
minimum value of supply & demand
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Error correcting codes | Brilliant Math & Science Wiki
Alexander Katz and Satyabrata Dash contributed
An error correcting code (ECC) is an encoding scheme that transmits messages as binary numbers, in such a way that the message can be recovered even if some bits are erroneously flipped. They are used in practically all cases of message transmission, especially in data storage where ECCs defend against data corruption.
Intuitive explanation through an example
Constructing Hamming codes
Suppose Alice was trying to transmit a message over a noisy telephone to Bob, and wanted to ensure that Bob correctly received the message. To ensure this, Alice has several options, the most obvious being for Bob to repeat the message for Alice to confirm. Unfortunately, this is not always possible (as in one-way communication links) or may be costly if, for example, Alice is distributing the same message to several people simultaneously. Another possible option is for Alice to repeat the message multiple times, but this is fairly inefficient.
Alternatively, Alice could choose to use words that are not easily confused with others, so that even if Bob slightly mishears a word, Bob can figure out the intention from context. This is the concept behind, for instance, the NATO phonetic alphabet: where the alphabet is represented by "Alpha", "Bravo", "Charlie", and so on. These words are specifically chosen to be dissimilar from another, so that, for instance, hearing "vo" is enough to guess the intended letter was B. Note that this is true of practically any pair of words below: almost no pair shares a similar sound, so a single syllable is often good enough to figure out the intended letter.
Error Correcting Codes operate in the same way as Alice does, but with binary numbers rather than words. Each letter (or, more generally, every one of a possible set of message) is encoded as a binary number, and bits are transmitted in succession. In this sense, the "noisy telephone" is equivalent to there being some probability that a bit is incorrectly transmitted (i.e. it is transmitted as a 1 despite actually being a 0, or vice versa). The goal of an ECC is to transmit a message that can be accurately interpreted even if some bits are accidentally flipped.
The simplest example of an ECC is analogous to Alice repeating the message: when a bit is to be transmitted, it is instead transmitted multiple times (usually 3), with the bit transmitted more times being interpreted as the intended message. More specifically:
Received Message Interpreted bit
For example, a message of "10101" might be transmitted as 111010111000110, where the bolded bits have been accidentally flipped. This would be interpreted correctly, as
will be interpreted as "10101", the intended message. Thus this method of transmission results in the correct message despite two bits being incorrectly flipped. More specifically, if the probability of a flipped bit is
p
, the probability of an incorrectly transmitted bit is
p^3+3p^2(1-p)
, which is significantly less than the probability of
p
that would result from just transmitting the bit once as usual (since
p
is very small).
A (very) noisy channel is known to flip bits 20% of the time. Alice decides to send the same bit multiple times, and Bob will interpret the message as the bit he receives more times (e.g. if Bob received 0, 1, 1, he would conclude the message was 1).
If Alice wants to ensure the message is correctly received with probability at least 95%, what is the minimum number of times Alice needs to send each bit?
This code is called the (3,1) repetition code, meaning that three bits are sent, one of which is the desired data bit. It can correct a message when one of the transmitted bits is flipped - a single bit error - meaning that the correcting ability of this code is 1 bit. It can also detect an error occurred if at most two bits are flipped, meaning that the detecting ability of this code is 2 bits.
However, this encoding scheme is fairly inefficient, as it requires the sender to transmit three times the message's length in order to achieve single bit correction. More formally, the code rate of this code is
\frac{1}{3}
- the number of data bits divided by the number of transmitted bits.
An even simpler example of a repetition code is the parity code, in which each bit is transmitted twice:
It can detect, but not correct, single-bit errors. The importance of this code lies in the concept of a parity bit, which is a bit added to make the number of 1s in each encoding even. By doing so, any message with an odd number of 1s can immediately be recognized as erroneous. More generally, parity bits can be added to make the number of 1s in specific positions even, also making any message that has an odd number of ones in those positions immediately recognizable as erroneous.
A more efficient encoding scheme is a Hamming code, which is analogous to the phonetic alphabet from the opening section. In a Hamming code, every possible message string is encoded as a certain binary number, with the set of numbers specifically chosen so that they are all significantly different in some sense; in other words, every pair of encoded messages are substantially different by some measure.
That measure is Hamming distance. The Hamming distance between 2 numbers is the number of bits they differ at. For example, 1101010 and 1111000 are a hamming distance of 2 apart:
The key here is that if any pair of encodings are sufficiently far apart in terms of Hamming distance, errors can be detected and corrected by seeing which of the codewords is closest to the transmitted message. For example, consider the encoding
In this encoding, the minimum Hamming distance between encodings is 2, which means that single-bit errors can be detected -- i.e. if a single bit is flipped during transmission of a letter, it can be determined that an error was made. It cannot, however, be determined what the original message was; for example, a transmitted message of "010" could have been a single-bit error resulting from sending an "A", "B", or "D".
In this encoding, however:
the minimum Hamming distance between encodings is 3, which means that single-bit errors can be corrected, and double-bit errors detected. This is the (3,1) code from the previous section.
Generally speaking, an encoding can detect
k
-bit errors if the minimum Hamming distance is at least
k+1
, and correct
k
2k+1
A noisy channel is known to flip bits with low frequency (so it can be safely assumed that double-bit errors will not occur). Alice has built the following partial encoding:
what value should Alice encode D as in order to achieve single-bit correction?
Hamming codes take this idea, combined with the idea of parity bits, and allow the parity bits to overlap. More specifically, they follow this algorithm:
When transmitting a binary number, use the 1st, 2nd, 4th, 8th, etc. (all powers of two) bits as parity bits, and all other positions as data bits.
The parity bit at
2^n
is the sum of all bits (taken modulo 2) whose positions have the
n
th least significant bit set to 1. For example, the parity bit at 2 is the sum of the bits at positions
3=11_2, 6=110_2, 7=111_2, 10=1010_2, 11=1011_2, \ldots
, since these positions all have the 2nd rightmost bit set to 1.
Transmit this entire message. If an error in a data bit occurred, some of the parity bits will show an error (since there would be an odd number of 1s in some of these sets). Their sum is the location of the error. If only one parity bit shows an error, that parity bit was in fact in error.
The key point is that every bit is encoded in a unique set of parity bits: specifically, the ones for which the binary representation of the bit's position is a 1. For example, the 14th bit is included in the parity bits at 8, 4, and 2, since
14=1110_2
. If parity bits 8, 4, and 2 show an error, then the 14th bit is in error.
Generally speaking, with
n
parity bits,
(2^n-1)-n=2^n-n-1
bits are usable for data, meaning that up to
2^{2^n-n-1}
different members of an alphabet can be encoding using just
n
parity bits. This is a vast improvement on repetition codes when
n>2
; for example, an encoding in which every word needs 4 bits of information (thus up to 16 codewords can be encoded) can be transmitted with 3 parity bits for a total of 7 bits, rather than
4 \cdot 3=12
bits from the repetition scheme in the previous section. Since it can correct single-bit errors and detect double-bit errors, this makes Hamming codes far more efficient than repetition codes while achieving the same purpose.
Hamming codes can be constructed by hand, but they are much easier to construct using linear algebra. Consider a matrix
M
whose columns are all the possible nonzero vectors of parity bits. For example, with 3 parity bits, the matrix is
M=\begin{pmatrix}1&1&0&1&1&0&0\\1&0&1&1&0&1&0\\0&1&1&1&0&0&1\end{pmatrix}
The order of the columns doesn't matter, but it is convenient to write
M
in this form (with the rightmost 3 columns the 3-identity matrix) as the true goal is to find generator matrix
G
MG^T=0
. This can be formed by taking the transpose of the left hand side of
M
(the part distinct from the identity matrix), and adding the appropriate identity matrix to the left:
G=\begin{pmatrix}1&0&0&0&1&1&0\\0&1&0&0&1&0&1\\0&0&1&0&0&1&1\\0&0&0&1&1&1&1\end{pmatrix}
It can easily be checked that this algorithm results in
MG^T=0
as desired. A Hamming code then takes a bit-vector
\overrightarrow{x}
\overrightarrow{x}G
; e.g.
\begin{pmatrix}1&0&1&1\end{pmatrix} \rightarrow \begin{pmatrix}1&0&1&1\end{pmatrix}\begin{pmatrix}1&0&0&0&1&1&0\\0&1&0&0&1&0&1\\0&0&1&0&0&1&1\\0&0&0&1&1&1&1\end{pmatrix}=\begin{pmatrix}1&0&1&1&0&1&0\end{pmatrix}
so "1011" is encoded as "1011010".
This results in the Hamming[7,4] code, the original and most famous of the Hamming codes.
Cite as: Error correcting codes. Brilliant.org. Retrieved from https://brilliant.org/wiki/error-correcting-codes/
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Unit Name: Polynomial and Rational Functions Solve the following inequality algebraically.
Unit Name: Polynomial and Rational Functions
Solve the following inequality algebraically. Explain your process.
8x-3\le 2x+1\le 17x-8
Split the inequality into two:
8x-3\le 2x+1\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }2x+1\le 17x-8
Isolate x from each inequality using inverse operations.
8x-3\le 2x+1
8x-2x\le 3+1
6x\le 4
x\le \frac{2}{3}
2x+1\le 17x-8
2x-17x\le -8-1
-15x+15x\le -9+15x
0\le -9+15x
\frac{3}{5}\le x
Therefore, the solution is:
\frac{3}{5}\le x\le \frac{2}{3}
Solve the following inequality algebraically.
8x-3\le 2x+1\le 17x-8
We've
8x-3\le 2x+1\le 17x-8
⇒8x-3-1\le 2x+1-1\le 17x-8-1
⇒8x-4\le 2x\le 17x-9
Now dividing above equation by 2 we get
⇒\frac{8x-4}{2}\le \frac{2x}{2}\le \frac{17}{2}x-\frac{9}{2}
⇒4x-2\le x\le \frac{17}{2}x-\frac{9}{2}
Now from equation (2) we've two cases.
x\ge 4x-2
⇒
adding 2 to both sides we get
⇒x+2\ge 4x-2+2
⇒x+2\ge 4x
Subtracting x from both sides
⇒x+2-x\ge 4x-x
⇒2\ge 3x
⇒x\le \frac{2}{3}
x\le \frac{17}{2}x-\frac{9}{2}
\frac{9}{2}
to both sides we get
⇒x+\frac{9}{2}\le \frac{17}{2}x-\frac{9}{2}+\frac{9}{2}
⇒x+\frac{9}{2}\le \frac{17}{2}x
⇒x+\frac{9}{2}-x\le \frac{17}{2}x-x
⇒\frac{q}{2}\le \frac{17x-2x}{2}
⇒\frac{9}{2}\le \frac{15x}{2}
Multiplying 2 to both sides we get
\text{Undefined control sequence \cancel}
⇒9\le 15x
⇒x\ge \frac{3}{5}
From equation (3) and equation (4) we get
\frac{3}{5}\le x\le \frac{2}{3}
⇒0.6\le x\le 0.6666
What is negative one hundred forty-one times one hundred seventeen?
What is the least perfect square, which is divisible by each of 21,36 and 66?
A computer repair technician charges $50 per visit plus $30/h for house calls.
a) Write an algebraic expression that describes the service charge for one household visit.
b) Use your expression to find the total service charge for a 2.5-h repair job.
5{x}^{2}\left(3{x}^{2}-x+2\right)
The written algebraic expression that correctly translates 10y
Airlines schedule about 5.5 hours of flying time for an A320 Airbus to fly from Dulles International Airport near Washington, D.C., to Los Angeles International Airport. Airlines schedule about 4.5 hours of flying time for the reverse direction. The distance between these airports is about 2,300 miles. They allow about 0.4 hour for takeoff and landing.
a. From this information, estimate (to the nearest 5 mph) the average wind speed the airlines assume in making their schedule.
b. What average airplane speed (to the nearest 5 mph) do the airlines assume in making their schedule?
To perform: the given operation
0.946\text{ }L-210\text{ }ml
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Options for training deep learning neural network - MATLAB trainingOptions - MathWorks France
\begin{array}{l}{\mu }^{*}={\lambda }_{\mu }\stackrel{^}{\mu }+\left(1-{\lambda }_{\mu }\right)\mu \\ {\sigma }^{2}{}^{*}={\lambda }_{{\sigma }^{2}}\stackrel{^}{{\sigma }^{2}}\text{}\text{+}\text{}\text{(1-}{\lambda }_{{\sigma }^{2}}\right)\text{}{\sigma }^{2}\end{array}
{\mu }^{*}
{\sigma }^{2}{}^{*}
{\lambda }_{\mu }
{\lambda }_{{\sigma }^{2}}
\stackrel{^}{\mu }
\stackrel{^}{{\sigma }^{2}}
\mu
{\sigma }^{2}
{\theta }_{\ell +1}={\theta }_{\ell }-\alpha \nabla E\left({\theta }_{\ell }\right),
\ell
\alpha >0
\theta
E\left(\theta \right)
\nabla E\left(\theta \right)
{\theta }_{\ell +1}={\theta }_{\ell }-\alpha \nabla E\left({\theta }_{\ell }\right)+\gamma \left({\theta }_{\ell }-{\theta }_{\ell -1}\right),
\gamma
determines the contribution of the previous gradient step to the current iteration. You can specify this value using the Momentum training option. To train a neural network using the stochastic gradient descent with momentum algorithm, specify 'sgdm' as the first input argument to trainingOptions. To specify the initial value of the learning rate α, use the InitialLearnRate training option. You can also specify different learning rates for different layers and parameters. For more information, see Set Up Parameters in Convolutional and Fully Connected Layers.
{v}_{\ell }={\beta }_{2}{v}_{\ell -1}+\left(1-{\beta }_{2}\right){\left[\nabla E\left({\theta }_{\ell }\right)\right]}^{2}
{\theta }_{\ell +1}={\theta }_{\ell }-\frac{\alpha \nabla E\left({\theta }_{\ell }\right)}{\sqrt{{v}_{\ell }}+ϵ}
{m}_{\ell }={\beta }_{1}{m}_{\ell -1}+\left(1-{\beta }_{1}\right)\nabla E\left({\theta }_{\ell }\right)
{v}_{\ell }={\beta }_{2}{v}_{\ell -1}+\left(1-{\beta }_{2}\right){\left[\nabla E\left({\theta }_{\ell }\right)\right]}^{2}
{\theta }_{\ell +1}={\theta }_{\ell }-\frac{\alpha {m}_{l}}{\sqrt{{v}_{l}}+ϵ}
E\left(\theta \right)
{E}_{R}\left(\theta \right)=E\left(\theta \right)+\lambda \Omega \left(w\right),
w
\lambda
\Omega \left(w\right)
\Omega \left(w\right)=\frac{1}{2}{w}^{T}w.
\lambda
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I took an informal poll of my friends and it turns out we all intially learned a method of matrix multiplication in school which I call row oriented multiplication. However, the friends who had been working with linear algebra for a long time frequently used a more intuitive method which I call column oriented multiplication. Here I’ll explain the difference and try to show why the column oriented point of view should be a useful tool in your toolbox, especially if you are working in a field that is heavy in linear algebra (e.g. robotics).
Row Oriented Multiplication
When I learned how to multiply vectors and matrices, it went something like this.
\begin{bmatrix} & & \\ \rightarrow & \rightarrow & \rightarrow \\ & & \end{bmatrix} \begin{bmatrix} \downarrow \\ \downarrow \\ \downarrow \end{bmatrix} =\begin{bmatrix} \\ \cdot \\ \\ \end{bmatrix}
Or if you prefer formulas, it went like this.
A x = \begin{bmatrix} ... \\ A_{i1} x_1 + A_{i2} x_2 + ... + A_{i n} x_n \\ ... \end{bmatrix}
I call this row oriented multiplication. It definitely works but it often hides the geometrical content of a matrix behind nasty algebra. Consider this
3D
rotation matrix.
R = \begin{bmatrix} \frac{\sqrt 3}2 & -\frac 1 2 & 0 \\ \frac 1 2 & \frac{\sqrt 3}2 & 0 \\ 0 & 0 & 1\end{bmatrix}
We can verify that the axis of rotation is the z-axis with row oriented multiplication by seeing that a point on the z axis
\begin{bmatrix}0 & 0 & z\end{bmatrix}^T
is unaffected by multiplication with
R
R \begin{bmatrix}0 \\ 0 \\ z\end{bmatrix} = \begin{bmatrix} \frac{\sqrt 3}2 & -\frac 1 2 & 0 \\ \frac 1 2 & \frac{\sqrt 3}2 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}0 \\ 0 \\ z\end{bmatrix} = \begin{bmatrix} \frac{\sqrt 3}2\cdot 0 + -\frac 1 2\cdot 0 + 0 \cdot z \\ \frac 1 2 \cdot 0 + \frac{\sqrt 3}2\cdot 0 + 0 \cdot z \\ 0 \cdot 0 + 0\cdot 0 + 1\cdot z\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ z\end{bmatrix}
Now I’ll introduce column oriented multiplication and how it not only simplifies the previous calculation but also reveals the geometrical content of the matrix.
Column Oriented Multiplication
First let’s establish the notation that for any matrix
A
A_i
means the
i^{th}
A
. We obviously have
A = \begin{bmatrix} A_1 | A_2 | ... | A_n\end{bmatrix}
Column oriented multiplication of
A x
works like this. We use the columns of
A
as vectors.
A x = \begin{bmatrix} A_1 | A_2 | ... | A_n\end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\ ... \\ x_n \end{bmatrix} = x_1 A_1 + x_2 A_2 + ... + x_n A_n
There are multiple ways to verify that this leads to the same result as row oriented multiplication. I’ll show a really easy way in the appendix. But for now let’s see how column oriented multiplication can be applied.
Using Column Oriented Multiplication
First let’s redo our previous calculation involving our rotation matrix
R
using column oriented multiplication.
R \begin{bmatrix}0 \\ 0 \\ z\end{bmatrix} = 0 \cdot R_1 + 0 \cdot R_2 + z R_3 = z R_3
We can completely ignore the first two columns of
R
, and compute the answer simply by reading off the third column of
R
z
. We arrive at the same answer we got with row oriented multiplication.
z R_3 = \begin{bmatrix}0 \\ 0 \\ z\end{bmatrix}
Now recall the notation for the standard basis vectors.
e_1 = \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}, \,\,\,\, e_2 = \begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}, \,\,\,\, e_3 = \begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}
e_1, e_2, e_3
as points on the x, y, and z axis respectively. Using column oriented multiplication we can quickly see these relationships.
R e_1 = R_1, \,\,\,\, \ R e_2 = R_2, \,\,\,\, R e_3= R_3
This means that the geometrical content of
R
— the way it acts on the coordinate axes — can be entirely read off from its columns!
e_1 \rightarrow \begin{bmatrix} \frac {\sqrt 3} 2 \\ \frac 1 2 \\ 0\end{bmatrix}, \,\,\,\, \ e_2 \rightarrow \begin{bmatrix} -\frac 1 2 \\ \frac{\sqrt 3} 2 \\ 0\end{bmatrix}, e_3 \rightarrow e_3
(read the arrows as “rotates to”)
Try carrying out this rotation on your thumb as
e_1
, your index finger as
e_2
, and your middle finger as
e_3
e_2
is redundant since it must be perpendicular to
e_1
e_3
The story doesn’t end here. Once I got into the habit of seeing matrix multiplication this way, I started to have insights into the nature of other types of matrices, including Jacobian matrices which are ubiquitous in robotics algorithms. But I’ll save that for another post.
In all matrix multiply expressions that follow, assume that we are using row oriented multiplication.
A x = A \begin{bmatrix}x_1 \\ x_2 \\ ... \\ x_n \end{bmatrix} = A (\sum_{i=1}^n x_i e_i)
= A (\sum_{i=1}^n x_i e_i)
= \sum_{i=1}^n x_i A e_i
= \sum_{i=1}^n x_i A_i
(verify, with row oriented multiplication that
A e_i = A_i
So we end up with the formula for column oriented multiplication
Ax = \sum_{i=1}^n x_i A_i
← Deriving the Kalman Filter
The Multivariate Gaussian For Programmers →
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Use of Colloidal Graphite Coating to Reduce Magnetic Resonance Imaging Artifacts Caused by Metallic Objects | J. Med. Devices | ASME Digital Collection
N. Knutson,
N. Knutson
S. McDonald,
Knutson, N., McDonald, S., and Erdman, A. (July 24, 2009). "Use of Colloidal Graphite Coating to Reduce Magnetic Resonance Imaging Artifacts Caused by Metallic Objects." ASME. J. Med. Devices. June 2009; 3(2): 027545. https://doi.org/10.1115/1.3136432
Magnetic susceptibility mismatch, between human tissue and a foreign metallic object, is one of several factors responsible for image distortions in magnetic resonance imaging (MRI). Combining diamagnetic materials such as bismuth or carbon with paramagnetic materials such as nitinol or titanium can reduce the mismatch in bulk susceptibility of a foreign object and the surrounding tissue. Muller-Bierl et al. have succeeded in reducing MRI field distortion by coating titanium wire with bismuth. Wilson et al. used a pyrolytic graphite mouth shim to improve brain functional MRI performance. Conolly et al. have successfully used pyrolytic graphite in foam to reduce image artifacts at air-tissue interfaces. In this study, it was hypothesized that coating a metallic object with carbon particles suspended in a polymer can reduce the size of image artifacts. Four 6Al-4V titanium discs
(2.3mm×9.5mm∅)
were encapsulated in an epoxy-graphite mixture. Mixtures of graphite and epoxy were poured around the titanium discs in molds and allowed to cure. A specimen of titanium was encapsulated in plain epoxy to serve as the control sample. Polycrystalline graphite was mixed at mass ratios of 1:2 and 1:1 to epoxy for two of the samples. Pyrolytic graphite flakes were mixed at a 1:2 mass ratio to epoxy. The sample discs were placed in an aqueous solution of copper sulfate and gadolinium contrast agent inside a wrist imaging coil at the isocenter of a 3 Tesla MRI machine; disc axes were perpendicular to the
B0
direction. A T2-weighted gradient echo MRI image was taken in the coronal plane. Echo time, relaxation time, flip angle, and phase encode direction set to 71 ms, 3430 ms, 80 degrees, and right to left respectively. The control sample produced an arrowhead artifact sweeping in the same direction as the static magnetic field vector,
B0
. The two samples containing powdered polycrystalline graphite produced arrowhead shaped artifacts. The direction of image distortion, however, was opposite from that of the control sample. The change in direction of the image artifact is attributed to the change in bulk magnetic susceptibility of the sample from paramagnetic behavior of titanium encapsulated in plain epoxy to a diamagnetic behavior from the added carbon powder. The titanium sample encapsulated in the pyrolytic graphite-epoxy mixture produced an artifact with irregular outline and no discernable directional bias relative to
B0
. The hypothesized cause for this difference in artifact shape between the polycrystalline and pyrolytic graphite samples is an increase in air bubble entrapment due to the planar structure of the pyrolytic graphite flakes during the epoxy mixing process. Further study is underway to find a specific carbon-polymer mass ratio and coating thickness that will reduce MR image artifacts that would otherwise appear due to the presence of a metallic object in the MRI region of interest. This work is supported by MIMTeC, a National Science Foundation Industry University Collaborative Research Center and by NIH Grant P30NS057091.
biomedical MRI, colloids, curing, diamagnetic materials, graphite, magnetic susceptibility, mixtures, paramagnetic materials, polymers, titanium
Coating processes, Coatings, Graphite, Magnetic resonance imaging, Magnetic susceptibility, Polymers, Titanium
Study of MRI Susceptibility Artifacts for Nanomedical Applications
Oxygen Separation/Enrichment From Atmospheric Air Using Magnetizing Force
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In this article, we learn about non-deterministic turing machines - a generalization of the standard deterministic turing machine from a uniquely determined sequence of computation steps to several possible sequences of computational steps.
A formal definition.
Properties of non-deterministic turing machines.
The complexity of non-deterministic turing machines.
A turing machine is a theoretical machine that manipulates symbols on a strip of tape according to the rule specified in a table. It is as simple as it sounds but turing machines can simulate any logic of any algorithm.
A non-deterministic turing machine is a generalization of a standard deterministic turing machine, in that, we move from a uniquely determined sequence of computation steps to several possible sequences. Although the set of computable functions f(n) doesn't change, the computational complexity differs between deterministic and non-deterministic turing machines.
A comparison between deterministic and non-deterministic turing machines.
By reducing the amount of computational work from the deterministic paradigm, non-deterministic turing machines pave way for artificially intelligent computing whereby computers learn to solve complex problems and think more like humans.
An example of a non-deterministic turing machine is a probabilistic turing machine whereby an array of actions are determined through a probability distribution. This means that when the machine has more than a single choice, it takes a probabilistic model, analyzes it, and chooses accordingly.
An example of a non-deterministic turing machine model is the following:
the computer follows paths of logic until an accepted or rejected state is reached then goes back and chooses an action accordingly.
A non-deterministic turing machine is 6-tuple, that is M= (Q, X, ∑, δ, q0, B, F) where;
X is the tape alphabet.
δ is a transition function.
B is the blank symbol.
The process of computation of a non-deterministic turing machine is interpreted as a deterministic turing machine with the capability to create and start other turing machines processing alternative branches of the calculation. Therefore, a machine T' can simulate a non-deterministic machine T by iterating all branches of the calculation of T up to a specific depth d, after which d is incremented, and the iteration repeats.
The above is done until an accepting branch in the tree is found or no branch can be continued.
Note that this simulation may be infinite since the paths of the tree might not be finite.
Looking at this algorithmically, T' performs a breadth-first search that ensures paths with infinite lengths are handled appropriately.
\mathrm{L}\subseteq {\mathrm{\Sigma }}^{*}
, we determine the time t(x) needed by a non-deterministic machine to process input x ∈
{\mathrm{\Sigma }}^{*}
in the following way;
First, we specify two cases. The case x ∈ L, is defined as the length of the shortest path in the computation tree accepting x.
In another case, x ∉ L, the value of t(x) is defined as the length of the shortest path in the computation tree.
Time t(n) needed by T for processing input x ∈
{\mathrm{\Sigma }}^{*}
with the length |x| = n ∈
\mathbb{N}
is defined as the maximum of all finite t(x) for x ∈
{\mathrm{\Sigma }}^{*}
with |x| = n.
Now, a language
\mathrm{L}\subseteq {\mathrm{\Sigma }}^{*}
belongs to the complexity class NP since it exists in a non-deterministic turing machine T and a polynom p such that T accepts L wit O(p(|x|)) computation steps for all x ∈ L.
The corresponding complexity class for deterministic machines is P and since these machines are a special case of non-deterministic machines, it holds that P ⊆ NP.
Whether the above statement can be strengthened to P = NP is an open problem in computer science.
Reducing the amount of computational work from the deterministic paradigm allows non-deterministic turing machines to pave way for artificially intelligent computing whereby computers learn to solve complex problems and think more like humans.
A non-deterministic turing machine is a generalization of a standard deterministic turing machine from a uniquely determined sequence of computation steps to several possible sequences.
With this article at OpenGenus, you must have the complete idea of Non Deterministic Turing Machines.
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How can you determine all of the zeroes (real and
How can you determine all of the zeroes (real and imaginary) of the polynomial function:
P\left(x\right)={x}^{4}+2{x}^{3}-9{x}^{2}-10x-24?
You can expect any factors to divide -24, the constant term over the lead coefficient. So check 1, 2, 3, 4, 6,12, 24 and their negatives. Turns out that 3 and -4 work, so you are down to a quadratic
P\left(x\right)=\left(x-3\right)×\left(x+4\right)×\left({x}^{2}+x+2\right)
so the roots are
x=\left\{3,\text{ }-4,\frac{-1-\sqrt{7}×i}{2},\text{ }\frac{-1+\sqrt{7}×i}{2}\right\}
The basic trick is to try finding 3 numbers k,m,n which would enable us to write:
{x}^{4}+2{x}^{3}-9{x}^{2}-10x-24=\left({x}^{2}+x+k{\right)}^{2}-\left(mx+n{\right)}^{2}
{x}^{4}+2{x}^{3}+\left(1+2k-{m}^{2}\right){x}^{2}+2\left(k-mn\right)x+\left({k}^{2}-{n}^{2}\right)={x}^{4}+2{x}^{3}-9{x}^{2}-10x-24
1+2k-{m}^{2}=-9\phantom{\rule{0ex}{0ex}}2k-2mn=-10\phantom{\rule{0ex}{0ex}}{k}^{2}-{n}^{2}=-24
Rearrange these equations and get
{m}^{2}=2\left(k+5\right)\phantom{\rule{0ex}{0ex}}mn=k+5\phantom{\rule{0ex}{0ex}}{n}^{2}={k}^{2}+24
From the 2 equations on the left we get
{m}^{2}=2mn
and therefore one possible solution would be:
m=0\phantom{\rule{0ex}{0ex}}k=-5\phantom{\rule{0ex}{0ex}}{n}^{2}=49={7}^{2}
Hence we get the following factorization of the given polynomial:
{x}^{4}+2{x}^{3}-9{x}^{2}-10x-24=\left({x}^{2}+x-5{\right)}^{2}-{7}^{2}=\phantom{\rule{0ex}{0ex}}=\left({x}^{2}+x+2\right)\left({x}^{2}+x-12\right)
So, all the problem has now been reduced into finding the 4 roots of the 2 quadratic polynomials, which are
{x}_{1}=-\frac{1}{2}+\frac{\sqrt{7}}{2}i\phantom{\rule{0ex}{0ex}}{x}_{2}=-\frac{1}{2}-\frac{\sqrt{7}}{2}i
{x}_{3}=-\frac{1}{2}+\frac{7}{2}=3\phantom{\rule{0ex}{0ex}}{x}_{4}=-\frac{1}{2}-\frac{7}{2}=-4
a) What is the sum of:
\left(3-4i\right)
\left(-21-6i\right)
-24-10i
18-2i
-18-10i
24+2i
b) Simplify
\frac{2-3i}{2+2i}
-\frac{1}{24};\text{ }-\left(\frac{5}{4}\right)i
-\frac{1}{24};\text{ }+\left(\frac{5}{4}\right)i
\frac{1}{22};\text{ }-\frac{1}{22};\text{ }i
\frac{1}{22};\text{ }+\frac{1}{22};i
\left(4+15i\right)-\left(\frac{4}{3}+\frac{9i}{8}\right)
X=4\mathrm{\angle }{40}^{\circ }\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }Y=20\mathrm{\angle }-{30}^{\circ }
. Evaluate the following quantities and express your results in polar form: (X - Y)
What are complex numbers?
\mathrm{ln}\left(1-i\right)
{z}_{1}{z}_{2}
\frac{{z}_{1}}{{z}_{2}}
{z}_{1}=\frac{7}{8}\left(\mathrm{cos}\left({35}^{\circ }\right)+i\mathrm{sin}\left({35}^{\circ }\right)\right),{z}_{2}=\frac{1}{8}\left(\mathrm{cos}\left({135}^{\circ }\right)+i\mathrm{sin}\left({135}^{\circ }\right)\right)
{z}_{1}{z}_{2}=?
\frac{{z}_{1}}{{z}_{2}}=?
\left(\frac{5}{3}+2i\right)+\left(13+\frac{7i}{9}\right)
z×zm=1
z=1+i
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Is this the recursive formula for geometric sequences? \begin{cases}a_1=first\ term\\a_n=r\times a_{n-1}\end{cases}
Is this the recursive formula for geometric sequences?\begin{cases}a_1=first\ term\\a_n=r\times a_{n-1}\end{cases}
Is this the recursive formula for geometric sequences?
\left\{\begin{array}{l}{a}_{1}=first\text{ }term\\ {a}_{n}=r×{a}_{n-1}\end{array}
\left\{\begin{array}{l}{a}_{1}=first\text{ }term\\ {a}_{n}=r×{a}_{n-1}\end{array}\phantom{\rule{0ex}{0ex}}\text{The second term of this sequence is}\phantom{\rule{0ex}{0ex}}{a}_{2}=r×{a}_{2-1}\phantom{\rule{0ex}{0ex}}{a}_{2}=r×{a}_{1}\text{ }\left(2\right)\phantom{\rule{0ex}{0ex}}\text{The third term:}\phantom{\rule{0ex}{0ex}}{a}_{3}=r×{a}_{3-1}\phantom{\rule{0ex}{0ex}}{a}_{3}=r×{a}_{2}\phantom{\rule{0ex}{0ex}}\text{Put value of}\text{ }{a}_{2}\text{ }\text{from equation 2}\phantom{\rule{0ex}{0ex}}⇒{a}_{3}=r×\left(r×{a}_{1}\right)\phantom{\rule{0ex}{0ex}}⇒{a}_{3}={r}^{2}×{a}_{1}\left(3\right)\phantom{\rule{0ex}{0ex}}\text{Similarly, you can calculate}\text{ }{a}_{4}\phantom{\rule{0ex}{0ex}}{a}_{4}=r×{a}_{4-1}\phantom{\rule{0ex}{0ex}}{a}_{4}=r×{a}_{3}\phantom{\rule{0ex}{0ex}}\text{Put a value of}\text{ }{a}_{3}\text{ }\text{from the equation}\text{ }\left(3\right)\phantom{\rule{0ex}{0ex}}⇒{a}_{4}=r×\left({r}^{2}×{a}_{1}\right)\phantom{\rule{0ex}{0ex}}⇒{a}_{4}={r}^{3}×{a}_{1}\text{ }\left(4\right)\phantom{\rule{0ex}{0ex}}\text{And so on. Thus, we have the following}\phantom{\rule{0ex}{0ex}}{a}_{1},r×{a}_{1},{r}^{2}×{a}_{1},....\phantom{\rule{0ex}{0ex}}\text{So we have observed that}\phantom{\rule{0ex}{0ex}}\frac{{a}_{2}}{{a}_{1}}=\frac{{a}_{3}}{{a}_{2}}=\frac{{a}_{4}}{{a}_{3}}=...=\frac{{a}_{n}}{{a}_{n-1}}=r
Thus, it's the formula of geometric sequences.
What is the sum of the first five terms of the following geometric sequences.
1) h 1h 2t…
2) 3 12 ht…
4) 1 t 3t…
5) 5h 1t t…
Determine the convergence or divergence of the following sequences. If it converges give it its limit
{a}_{n}=\frac{\mathrm{ln}\left({n}^{3}\right)}{2n}
\left\{k{\left(\frac{1}{2}\right)}^{k}\right\}
Write a recursive rule for the sequence. 2, 2, 4, 12, 48, ....
We know that convergent sequences approximate certain numbers. For example, the sequence
\frac{1}{2},\frac{1}{2}+\frac{1}{4},\frac{1}{2}+\frac{1}{4}+\frac{1}{8},\dots
etc approximates the number 1.
Are there sequences that approximate sequences?
|
Translate the following verbal statement into an algebraic
Translate the following verbal statement into an algebraic equation and then solve: Use xx for your variable. The difference of four times a number and seven is ten more than the number.
difference of four times a number and seven is ten mote than the numbers.
4x-7=10+x
3x-17=0
--- equation
3x=17
x=\frac{17}{3}
What is the value of the digit in the hundredths place for 0.44?
Write and solve an equation to find the number. Four times the difference of a number and 7 is 12.
Sun-Yi estimated 270 + 146 and got 320. Is her estimate reasonable? Explain.
A student bought a calculator and a textbook for a course in algebra. He told his friend that the total cost was $145(without tax) and that the calculator cost $20 more than four times the cost of the textbook. What was the cost of each item? Let
x=
the cost of a calculator and
y=
the cost of the textbook. The corresponding modeling system is
x+y=145
x=4y+20
Solve the system by using the method of substitution and enter the ordered pair
4\left(\left(\frac{t}{8}\right)+5\right)
Which digit is in the ten thousands place in 782,341,693?
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Triangles - Orthocenter Practice Problems Online | Brilliant
The orthocenter is the intersection of which 3 lines in a triangle?
Angle Bisectors of triangle Perpendicular Bisector of sides of triangle Altitudes of triangle Medians of triangle
Where is the orthocenter of an obtuse triangle?
Outside of the triangle Inside of the triangle On a side of the triangle On a vertex of the triangle
Where is the orthocenter of a right triangle?
Inside of the triangle On a vertex of the triangle On a side of the triangle Outside of the triangle
Does every triangle have its orthocenter?
No, obtuse triangles do not have their orthocenter No, right triangles do not have their orthocenter Yes, every triangle has its orthocenter No, some scalene triangles do not have their orthocenter
H
\triangle{ABC}
\angle{A}=59^\circ
\angle{B}=60^\circ
\angle{C}=61^\circ
\angle{BHD}
58^\circ
61^\circ
59^\circ
60^\circ
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LMIs in Control/Matrix and LMI Properties and Tools/Change of Subject - Wikibooks, open books for an open world
LMIs in Control/Matrix and LMI Properties and Tools/Change of Subject
A Bilinear Matrix Inequality (BMI) can sometimes be converted into a Linear Matrix Inequality (LMI) using a change of variables. This is a basic mathematical technique of changing the position of variables with respect to equal signs and the inequality operators. The change of subject will be demonstrated by the example below.
{\displaystyle A\in \mathbb {R} ^{n\times n},B\in \mathbb {R} ^{n\times m},K\in \mathbb {R} ^{m\times n}}
{\displaystyle Q\in \mathbb {S} ^{n}}
{\displaystyle Q>0}
The matrix inequality given by:
{\displaystyle QA^{T}+AQ-QK^{T}B{T}-BKQ<0}
is bilinear in the variables
{\displaystyle Q}
{\displaystyle K}
Defining a change of variable as
{\displaystyle F=KQ}
{\displaystyle QA^{T}+AQ+-F^{T}B^{T}-BF<0}
which is an LMI in the variables
{\displaystyle Q}
{\displaystyle F}
Once this LMI is solved, the original variable can be recovered by
{\displaystyle K=FQ^{-1}}
It is important that a change of variables is chosen to be a one-to-one mapping in order for the new matrix inequality to be equivalent to the original matrix inequality. The change of variable
{\displaystyle F=KQ}
from the above example is a one-to-one mapping since
{\displaystyle Q^{-1}}
is invertible, which gives a unique solution for the reverse change of variable
{\displaystyle K=FQ^{-1}}
Retrieved from "https://en.wikibooks.org/w/index.php?title=LMIs_in_Control/Matrix_and_LMI_Properties_and_Tools/Change_of_Subject&oldid=3969279"
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Working with MAD Portfolio Constraints Using Defaults - MATLAB & Simulink - MathWorks í•œêµ
Xâ{R}^{n}
When setting up your portfolio set, ensure that the portfolio set satisfies these conditions. The most basic or “default†portfolio set requires portfolio weights to be nonnegative (using the lower-bound constraint) and to sum to 1 (using the budget constraint). For information on the workflow when using PortfolioMAD objects, see PortfolioMAD Object Workflow.
The “default†MAD portfolio problem has two constraints on portfolio weights:
LowerBound: [20×1 double]
BoundType: [20×1 categorical]
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Joint torques that cancel velocity-induced forces - MATLAB velocityProduct - MathWorks Nordic
\frac{d}{dt}\left[\begin{array}{c}q\\ \stackrel{˙}{q}\end{array}\right]=\left[\begin{array}{c}\stackrel{˙}{q}\\ M{\left(q\right)}^{-1}\left(-C\left(q,\stackrel{˙}{q}\right)\stackrel{˙}{q}-G\left(q\right)-J{\left(q\right)}^{T}{F}_{Ext}+\tau \right)\end{array}\right]
M\left(q\right)\stackrel{¨}{q}=-C\left(q,\stackrel{˙}{q}\right)\stackrel{˙}{q}-G\left(q\right)-J{\left(q\right)}^{T}{F}_{Ext}+\tau
M\left(q\right)
C\left(q,\stackrel{˙}{q}\right)
\stackrel{˙}{q}
G\left(q\right)
J\left(q\right)
{F}_{Ext}
\tau
q,\stackrel{˙}{q},\stackrel{¨}{q}
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Maurice and Lester are twins who have just graduated from
Maurice and Lester are twins who have just graduated from college.They have both been offered jobs where their take-home pay would be $2500 per month
Maurice and Lester are twins who have just graduated from college. They have both been offered jobs where their take-home pay would be $2500 per month. Their parents have given Maurice and Lester two options for a graduation gift. Option 1: If they choose to pursue a graduate degree, their parents will give each of them a gift of $35,000. However, they must pay for their tuition and living expenses out of the gift. Option 2: If they choose to go directly into the workforce, their parents will give each of them a gift of $5000. Maurice decides to go to graduate school for 2 years. He locks in a tuition rate by paying $11,500 for the 2 years in advance, and he figures that his monthly expenses will be $1000. Lester decides to go straight into the workforce. Lester finds that after paying his rent, utilities, and other living expenses, he will be able to save $200 per month. Their parents deposit the appropriate amount of money in a money market account for each twin. The money market accounts are currently paying a nominal interest rate of 3 percent, compounded monthly. Lester works during the time that Maurice attends graduate school. Each month, Lester saves $200 and deposits this amount into the $5000 money market account that his parents set up for him when he graduated. If Lester's initial balance is u0=5000,un is the current month's balance, and un−1 is last month's balance, write an expression for un in terms of un−1.
We know from item 10 that:
\left(\text{Account balance any month}={u}_{n}\right)=\left(\text{account balance the month before}={u}_{n-1}\right)×1.0025+200
So, the equation becomes:
{u}_{n}={u}_{n-1}\cdot 1.0025+200
W\left(\alpha \right)=1.9\alpha +2.3
0\le \alpha \le 6
The yield strength of CP titanium welds was measured for welds cooled at rates of \(10^{\circ}C/s,15^{\circ}C/s\ \text{and}\ 28^{\circ}C/s\). The results are presented in the following table. (Based on the article “Advances in Oxygen Equivalence Equations for Predicting the Properties of Titanium Welds,” D. Harwig, W. Ittiwattana, and H. Castner, The Welding Journal, 2001:126s-136s.)
\(\begin{array}{} \hline \text{Cooling Rate}&\text{Yield Strenghts}\\ \hline 10&71.00&75.00&79.67&81.00&75.50&72.50&73.50&78.50\\ 15&63.00&68.00&73.00&76.00&79.67&81.00\\ 28&68.65&73.70&78.40&84.40&91.20&87.15&77.20&80.70&84.85&88\\ \hline \end{array}\)
a. Construct an ANOVA table. You may give a range for the P-value. b. Can you conclude that the yield strength of CP titanium welds varies with the cooling rate?
Find the curvature K of the curve at the point
r\left(t\right)={e}^{t}\mathrm{cos}t\cdot i+{e}^{t}\mathrm{sin}t\cdot j+{e}^{t}k,P\left(1,0,1\right)
Paulie Gone has a shiny red wagon. When he puts his puppy, Hypotenuse, in the wagon, the wagon weighs in at 40 pounds. When Paulie rides the wagon by himself, the wagon weighs in at 64 pounds. Paulie weighs 3 times as much as Hypotenuse, how much does the empty wagon weigh?
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\text{Misplaced \hline}
The ratio of consecutive values is not constant. therefore, it is not an exponential function. (Note that the difference of consecutive values is constant. therefore, this is a linear function.)
A=33.1e0.009t
A=28.2e0.034t
Graph each function and tell whether it represents exponential growth, exponential decay, or neither.
y={\left(2.5\right)}^{x}
x{}^{″}+2{x}^{\prime }+26x=82\mathrm{cos}4t
A=33.1{e}^{0.009}t
A=28.2{e}^{0.034}t
A researcher is trying to determine the doubling time fora population of the bacterium Giardia lamblia. He starts a culture in a nutrient solution and estimates the bacteria count every four hours. His data are shown in the table.
\begin{array}{|cc|}\hline \text{Time (hours)}& \text{Bacteria count (CFU/mL)}\\ 0& 37\\ 4& 47\\ 8& 63\\ 12& 78\\ 16& 105\\ 20& 130\\ 24& 173\\ \hline\end{array}
Use a graphing calculator to find an exponential curve
f\left(t\right)=a×{b}^{t}
that models the bacteria population t hours later.
Find The Exponential Function
f\left(x\right)={a}^{x}
Whose Graph Is Given.
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North West Corner Method | Method to Solve Transportation Problem | Transportation Model | Education Lessons
The North West corner method is one of the methods to obtain a basic feasible solution of the transportation problems (special case of LPP).
We will now see how to apply this very simple method to a transportation problem. We will study steps of this method while applying it in the problem itself.
A mobile phone manufacturing company has three branches located in three different regions, say Jaipur, Udaipur and Mumbai. The company has to transport mobile phones to three destinations, say Kanpur, Pune and Delhi. The availability from Jaipur, Udaipur and Mumbai is 40, 60 and 70 units respectively. The demand at Kanpur, Pune and Delhi are 70, 40 and 60 respectively. The transportation cost is shown in the matrix below (in Rs). Use the North-West corner method to find a basic feasible solution (BFS).
Note that all the explanation is provided in “CYAN” colour. You have to write in examination the only thing which are given in this regular colour under each steps(if any), else you can directly solve matrix of the problem as explained here
\color{#32c5d4} \Sigma \text { Supply} = \Sigma \text { Demand}
If this holds true, then we will consider the given problem as a balanced problem.
\color{#32c5d4} \text {i.e., } \Sigma \text { Supply} \not = \Sigma \text { Demand}
If such a condition occurs, then we have to add a dummy source or market; whichever makes the problem balanced. You can watch a video on this type of numerical, which is known as Unbalanced Transportation Problems.
\to
Q-1) In North-West corner method, we start by allocating cell which is at
A. left top most corner
B. right top most corner
C. left bottom most corner
D. right bottom most corner
Step 2: Start allocating from North-West corner cell
We will start the allocation from the left hand top most corner (north-west) cell in the matrix and make allocation based on availability and demand.
Now, verify the smallest among the availability (Supply) and requirement (Demand), corresponding to this cell. The smallest value will be allocated to this cell and check out the difference in supply and demand, representing that supply and demand are fulfilled, as shown below.
As we have fulfilled the availability or requirement for that row or column respectively, remove that row or column and prepare a new matrix, as shown below.
Repeat the same procedure of allocation of the new North-west corner so generated and check based on the smallest value as shown below, until all allocations are over.
Once all allocations are over, prepare the table with all allocations marked and calculate the transportation cost as follows.
\begin{aligned} \to \ \text {Transportation cost} &= (40 \times 40) + (3 \times 30) + (4 \times 30) + (2 \times 10) + (8 \times 60) \\ &= \text {Rs } 870 \end{aligned}
left top most corner
right top most corner
left bottom most corner
right bottom most corner
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Key Strategies Practice Problems Online | Brilliant
In addition to learning about various methods and formulas, this course will also expose you to meta-strategies for problem-solving. Some of these apply generally; others are specifically for math contests at the level of the AMC.
Often, it's useful to think about an extreme case of a problem. Then, you can check if that extreme case is possible, and if not, how close you can get.
What is the maximum number of coins that can be placed on squares in an
8\times 8
chessboard such that each row, each column, and each long diagonal contains at most 4 coins? (Note: Only 1 coin is allowed per square.)
28
29
30
31
32
When solving questions about a general variable, it can be useful to test small cases and then make a conjecture about the overall problem.
p
is a prime number such that
p^2+11
has exactly six different positive divisors
(
1
and the number itself
).
What is the sum of all possible values of
p?
(Note: There may be more than one possible value.)
2
3
5
A finite number greater than
5
\infty
When a problem involves some sort of symmetry, make use of it!
6
fair coins are flipped, what is the probability that there are more heads than tails?
Hint: What is the probability that the same number of heads and tails are flipped, and what are the other possibilities?
\frac{11}{32}
\frac{3}{8}
\frac{7}{16}
\frac{1}{2}
\frac{11}{16}
In math contests, eliminating impossible or unlikely choices makes guessing much more beneficial.
You encounter the question below, but your paper has been smudged. Nonetheless, you can still answer the question correctly!
n,
the expression above is always divisible by all of the following positive integers except one. Which one?
2
3
6
10
20
Casework is a wide-reaching strategy of breaking a complex problem down into multiple cases - each of which is relatively simple - and then combining them to obtain the overall result.
\large (x^2+5x+5)^{x^2-10x+21}=1 .
Hint: Don't forget about the case where the base is
-1
and the exponent is even!
These introductory quizzes have given you just a glimpse of what's to come in Contest Math II.
Whether you're looking for a comprehensive guide
(
with over
50
quizzes in a wide range of topics
)
or just looking to brush up on some specific techniques, there's something here for you.
The last two quizzes of this introduction involve deep problem solving of the sort you'll find on the AMC and other contest math situations.
Dive in, and good luck!
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4.7 Markers | IOTA Wiki
This section specifies the requirements for the marker tool. A tool as defined here is a feature that adds functionality to the node software but is not an essential component. The marker tool improves the efficiency with which certain properties can be checked or certain metrics calculated.
The potential issues addressed by the use of the marker tool are in handling potentially numerically expensive operations. Specifically, the following operations can become numerically expensive:
Future- or past cone inclusion. For certain applications it is necessary to know whether a certain message is in the past or future cone of another message. In the default approach the Tangle has to be walked until a given message is found.
Approval weight. In order to compute the approval weight of a given message the node software needs to traverse the Tangle from that message to the tips and sum up the active consensus Mana of all the messages in its future cone, see also the section on approval weight.
The marker tool allows a node to efficiently determine whether certain markers are in the past or future cone of a given message, by reducing the proportion of the Tangle that needs to be traversed.
The marker tool achieves this by defining a parallel internal data structure, consisting of additional metadata applied to specific messages in the Tangle. Specifically, the marker tool "marks" certain messages, which form a subDAG which approximates the topological structure of the Tangle. Furthermore, the markers are grouped into sequences (which themselves form yet another DAG), which allow the node to quickly determine which markers reference each other.
Note, that we shall require that markers are assigned when booking a message. Thus, for that part of the message DAG that is already booked the corresponding marker DAG does not change anymore.
The Markers tool specification depends on:
4.1 - The Tangle.
5.2 - Ledger State.
The following terms are defined in relation to markers:
UTXO branch: This is a set of outputs that spawn off from a conflict transaction. Each UTXO branch by itself is conflict free. See also Section 5.1 - UTXO and Section 5.2 - Ledger State for a more complete discussion on UTXO and its branches.
Aggregated branch: The aggregation of a combination of several branches.
Branch identifier (BID): The unique identifier of a branch or aggregated branch.
Main branch: The part of the UTXO DAG, in which all outputs are considered to be good in the sense that all conflicts in their past have been resolved, either by a given conflict being accepted or rejected.
Rank: The length of the longest directed path in DAG terminating in a given vertex/object. Specifically, if a vertex
A
directly references only
B
C
rank(A)=max(rank(B),rank(C))+1
Marker: A message that is assigned additional properties locally on the node, and that tracks a particular UTXO branch.
Marker identifier (MID): The unique identifier of the marker.
Marker DAG: The collection of all markers.
Marker rank (MR): The rank of a marker in the marker DAG.
Marker-sequence: A marker-sequence is a group of markers. Each marker-sequence maps to a UTXO branch; see Section 5.2 - Ledger State.
Marker-sequence identifier (SID): A marker-sequence identifier is a number that uniquely identifies a marker-sequence.
Marker-sequence rank (SR): The rank of a marker-sequence in the marker-sequence DAG.
Future marker (FM): This field in the message metadata is (potentially) updated when a new marker is generated in the future cone of the message, following the rules defined in Section "Message Metadata". Essentially it contains the list of markers for which there is no other marker between the marker in question and the message, or in more mathematical terms, the minimal markers in the future cone.
Past marker (PM): A past marker of a message is a most recent past marker of the parents (with respect to MR). The past marker of a marker is set to itself.
4.7.3 The Markers
A marker consists of the following data:
MID uint64 Unique identifier of the marker
SID uint64 Unique identifier of the marker-sequence
MR uint64 Marker rank
Table 4.7.1: Marker data.
A new marker shall be created by the marker tool when any of the following conditions are met:
a new UTXO branch is created and the message that would get a marker assigned is not yet booked. This also creates a new marker-sequence.
more than a certain number of messages (maxMsgPerMarker) have been received since the last marker. This rule must be applied per marker-sequence. I.e. for each marker-sequence with more than maxMsgPerMarker since the last marker in that marker-sequence, the rule shall be applied independently.
a certain time window (maxTimePerMarker) has passed since the last marker.
A marker is created with a MID, an this MID must be unique.
To set a new marker within a marker-sequence, the marker tool randomly selects from strong tips set a message whose past marker is the last marker in the sequence. The next marker will then reference that transaction. If there is no strong tip with the appropriate past marker, the selection shall be from message in the weak tips set. The rank of the new marker should be one greater than the rank of all the past markers of the message.
\texttt{MR}(x)=1+\max \limits_{y: x\text{ references }y}\texttt{MR}(y)
, marker ranks are monotonically non-decreasing such that
\forall x \in fc(y) \Rightarrow \texttt{MR}_x > \texttt{MR}_y
fc(y)
is the future cone of
y
4.7.4 The Marker-sequence
Marker-sequences are used to track the UTXO DAG branches. Each branch corresponds to a marker-sequence with a unique SID, and the marker-sequences form a DAG.
4.7.4.1 Marker-sequence data
Each marker-sequence is associated with the following data:
SID unit64 The marker-sequence identifier
SR unit64 The rank of a marker-sequence in the marker-sequence DAG
MRMax unit64 The highest MR in the marker-sequence
MRMin unit64 The lowest MR in the marker-sequence
ParentReferences map[Marker] Marker Relationship map from parent marker-sequences to markers (*)
Table 4.7.2: Data associated to a marker-sequence.
*The field ParentReferences models the relationship between marker-sequences. This maps which marker in this marker-sequence references which other markers from other marker-sequences.
Whenever a new marker is added that is a member of a given marker-sequence, MR_max and ParentReferences for that marker-sequence shall be updated.
4.7.4.2 Creation of Marker-sequences
A new marker-sequence shall be created when:
there's a transaction that creates a new conflict, i.e. creates a new UTXO branch.
Each new marker-sequence shall start with a new marker. Hence, with the creation of a new marker-sequence also a new marker must be assigned to the message that caused one of the three above events.
Whenever a new marker-sequence is created, the marker tool shall assign:
a new SID, created by the rule
\text{new }\texttt{SID}=1+\text{last } \texttt{SID}
. A new created SID must be unique.
a new
\texttt{SR}=1+max(\text{referenced }\texttt{SR})
. To prevent assigning a new SID when combining the same marker-sequences at different times, the marker tool shall build parent-child relationships in a map whenever a new marker-sequence is created.
For further details about the UTXO model, please refer to the section on UTXO.
4.7.5 Message Metadata
For each message in the Tangle, the marker tool shall maintain metadata that provides information about the markers that are closest in the past or future cone of that message, as well as whether the message itself is a marker and what rank the message has. The following message metadata shall be defined in the marker tool to support that requirement:
IsMarker bool A flag to indicate whether a message is a marker.
PastMarkers map[SID]MID A list of the closest markers from different marker-sequences in the past cone of the message.
FutureMarkers map[SID]MID A list of the closest markers from different marker-sequences in the future cone of the message.
MarkerBranchID BID The branch ID to which the marker is mapped, or nil if the message is no marker.
PayloadBranchID BID The branch ID to which the Payload is mapped in case it is a conflict, or nil otherwise.
IndividualBranchID BID The branch ID if there is need for mapping the message individually to a branch ID, or nil otherwise.
Table 4.7.3: Markers metadata.
The PastMarkers field contains
only the marker identifier of itself, if the message is marked as a marker.
the marker identifier of its closest past markers (PMs), i.e. from each referenced marker-sequence only the markers with the highest marker rank (MR). Markers which are referenced by other markers in this list shall be removed.
The FutureMarkers list shall be empty at the start and shall be updated when a new marker directly or indirectly references that list.
The propagation of a FM to its past cone (i.e. the update of the FutureMarkers list in the encountered messages) shall not continue beyond a message if:
FutureMarkers of a message includes a previous marker of the same marker-sequence; the message that includes such a marker shall not get updated.
the message is the marker in a different marker-sequence. Then the FutureMarkers shall be updated for that marker only.
Through this approach past and future markers do not cross weak parents. It also prevents the lists from growing unboundedly.
The fields MarkerBranchID, PayloadBranchID and IndividualBranchID allow for making connections between the marker DAG, the message DAG and the UTXO branch DAG. When a new Sequence is created the MarkerBranchID is set to the branch that creates the sequence.
4.7.5.1 Update of Already Booked Messages on Double Spends
If a transaction arrives that double spends an already booked transaction, a new marker-sequence shall be created for the newly arrived message (containing the transaction), see Section Creation of marker-sequences.
For the already booked conflicting transaction no new marker or marker Sequence shall be created. This is because the marker DAG and Sequence DAG shall not be changed post-booking a message. However a new UTXO branch is created.
First, assume the existing booked transaction is a Marker itself. Then the marker gets mapped onto the new branch by updating the field MarkerBranchID in the message metadata. Furthermore, the PayloadBranchID is updated to the new branch. For all FM in the same sequence the MarkerBranchID gets updated to the new branch. Furthermore, for every sequence that directly or indirectly references the sequence in which the double-spend occurs, the first marker is remapped to the new branch as well.
Second, assume the existing transaction is not a marker. Then all messages between the transaction and the following future markers (including the transaction itself) get mapped individually to the new branch mapping using the field IndividualBranchID. From the future markers onwards, the same applies as in the first scenario.
For an example implementation of these scenarios also visit the example here.
4.7.6 Marker Application Description
Figure 1 shows an example of how the markers and marker-sequences (here also called Sequence) would look in the Tangle from the perspective of the Message DAG, the marker DAG and the marker-sequence DAG. The purple colored messages are markers:
Image 4.7.4: Markers and marker-sequences in the Tangle.
4.7.6.1 Example Implementation
An illustrative example of the markers tool in action is provided here for the prototype implementation.
4.7.6.2 Approval Weight Approximation
To approximate the approval weight of a message, the markers tool retrieves the approval weight of FutureMarkers. Since a given message is in the past cone of its FMs, the approval weight and thus the finality of the message will be at least the same as the maximum weight of its FMs. This gives a lower bound (which is the “safe” bound), and if the markers are set frequently enough, this provides a good approximation of that bound.
4.7.6.3 Past Cone Check
By comparing the PastMarkers of a message with the FutureMarkers of another message, the markers tool can determine if that message is in the past cone of the other. For example, consider two messages X and Y that are members in the same marker-sequence. Then if PM(X)>FM(Y), then X is in the future of Y.
One way in which this check can be carried out is by traversing the marker DAG while remaining in the bounds of the marker ranks.
A potential optimization is that the marker-sequence DAG can be traversed while considering the marker-sequence ranks, prior to any traversal of the marker DAG.
It is possible that the marker DAG does not cover certain areas of the message DAG at a given point in time. In this case, a check on this question can return one of the following three values:
If the check returns a N/A, then the Message DAG must be searched via a search algorithm.
For an example implementation of the algorithm for the past cone check visit GoShimmer markers.
4.7.3 The Markers
4.7.4 The Marker-sequence
4.7.4.1 Marker-sequence data
4.7.4.2 Creation of Marker-sequences
4.7.5 Message Metadata
4.7.5.1 Update of Already Booked Messages on Double Spends
4.7.6 Marker Application Description
4.7.6.1 Example Implementation
4.7.6.2 Approval Weight Approximation
4.7.6.3 Past Cone Check
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Radiation astronomy/Mesons - Wikiversity
{\displaystyle p^{+}+{\bar {p}}^{-}\rightarrow \pi ^{+}+\pi ^{-},}
{\displaystyle {\pi }^{+}\rightarrow {\mu }^{+}+{\nu }_{\mu }\rightarrow e^{+}+{\nu }_{e}+{\bar {\nu }}_{\mu }+{\nu }_{\mu },}
{\displaystyle D_{S}\rightarrow \tau +{\bar {\nu }}_{\tau }\rightarrow \nu _{\tau }+{\bar {\nu }}_{\tau }.}
9.1 JP = 1/2+ Sigma baryons
A meson is a subatomic particle that is intermediate in mass between an electron and a proton and transmits the strong interaction that binds nucleons together in the atomic nucleus.
Mesons travel at speeds slower than the speed of light.
Each type of meson has a corresponding antiparticle (antimeson) in which by theory quarks are replaced by their corresponding antiquarks and vice-versa.
At right a person works lower center left in front of the huge ATLAS detector, one of six detectors attached to the Large Hadron Collider at CERN. A hadron, like an atomic nucleus, is a composite particle held together by the strong force. Hadrons are categorized into two families: baryons (such as protons and neutrons) and mesons.
Def. a composite subatomic particle bound together by the strong interaction "intermediate between electrons and protons"[7] is called a meson.
"The K0-K0 bar, D0-D0 bar, and B0-B0 bar oscillations are extremely sensitive to the K0 and K0 bar energy at rest. The energy is determined by the values mc2 with the related mass as well as the energy of the gravitational interaction. Assuming the CPT theorem for the inertial masses and estimating the gravitational potential through the dominant contribution of the gravitational potential of our Galaxy center, we obtain from the experimental data on the K0-K0 bar oscillations the following constraint: |(mg/mi)K0 - (mg/mi)K0 bar| ≤ 8·10-13, CL=90%. This estimation is model dependent and in particular it depends on a way we estimate the gravitational potential. Examining the K0-K0 bar, B0-B0 bar, and D0-D0 bar oscillations provides us also with weaker, but model independent constraints, which in particular rule out the very possibility of antigravity for antimatter."[8]
"In spite of the apparent parity non-invariance of the ordinary particles, the universe could still be left-right symmetric if [charge conjugation parity] CP were an exact symmetry[11]. But this option is [...] ruled out by experiments on kaons and B-mesons!)."[9]
{\displaystyle D_{S}\rightarrow \tau +{\bar {\nu }}_{\tau }\rightarrow \nu _{\tau }+{\bar {\nu }}_{\tau }.}
{\displaystyle 0.256_{-0.022}^{+0.024}}
Sigma[36] Σ+
{\displaystyle \Phi ^{0}}
{\displaystyle \Phi ^{0}\rightarrow \mathrm {K} ^{+}+\mathrm {K} ^{-}or}
{\displaystyle \Phi ^{0}\rightarrow \mathrm {K} _{S}^{0}+\mathrm {K} _{L}^{0}.}
{\displaystyle \rho }
{\displaystyle \phi }
{\displaystyle \phi }
{\displaystyle p+d\rightarrow He^{3}+\omega ,}
{\displaystyle {\bar {p}}+p\rightarrow \omega +\eta +\pi _{0},}
{\displaystyle \pi ^{-}+p\rightarrow \omega +n,}
{\displaystyle p+{\bar {p}}\rightarrow \mathrm {K} ^{+}+\mathrm {K} ^{-}+\omega ,}
{\displaystyle p+{\bar {p}}\rightarrow \mathrm {K} 1+\mathrm {K} 1+\omega ,}
{\displaystyle \omega \rightarrow \pi ^{+}+\pi ^{-}+\pi ^{0},}
{\displaystyle \omega \rightarrow \pi ^{0}+\gamma ,}
{\displaystyle \omega \rightarrow \pi ^{+}+\pi ^{-},}
{\displaystyle \omega \rightarrow neutrals(excluding:\pi ^{0}+\gamma ),}
{\displaystyle \omega \rightarrow \eta +\gamma ,}
{\displaystyle \omega \rightarrow \pi ^{0}+e^{+}+e^{-},}
{\displaystyle \omega \rightarrow \pi ^{0}+\mu ^{+}+\mu ^{-},}
{\displaystyle \omega \rightarrow \eta +e^{+}+e^{-},}
{\displaystyle \omega \rightarrow e^{+}+e^{-},}
{\displaystyle \omega \rightarrow \pi ^{+}+\pi ^{-}+\pi ^{0}+\pi ^{0},}
{\displaystyle \omega \rightarrow \pi ^{+}+\pi ^{-}+\gamma ,}
{\displaystyle \omega \rightarrow \pi ^{+}+\pi ^{-}+\pi ^{+}+\pi ^{-},}
{\displaystyle \omega \rightarrow \pi ^{0}+\pi ^{0}+\gamma ,}
{\displaystyle \omega \rightarrow \eta +\pi ^{0}+\gamma ,}
{\displaystyle \omega \rightarrow \mu ^{+}+\mu ^{-},}
{\displaystyle \omega \rightarrow 3\gamma ,}
{\displaystyle \omega \rightarrow \eta +\pi ^{0},}
{\displaystyle \omega \rightarrow 2\pi ^{0},and}
{\displaystyle \omega \rightarrow 3\pi ^{0}.}
{\displaystyle \gamma _{\rm {CMB}}}
{\displaystyle \Delta }
{\displaystyle \gamma _{\rm {CMB}}+p\rightarrow \Delta ^{+}\rightarrow p+\pi ^{0},}
{\displaystyle \gamma _{\rm {CMB}}+p\rightarrow \Delta ^{+}\rightarrow n+\pi ^{+}.}
An "analysis of the energy-loss distributions in the GRS HEM during the impulsive phase of this event indicates that γ-rays from the decay of π0 mesons were detected [...] The production of pions, which is accompanied (on average) by neutrons, has an energy threshold of ~290 MeV for p-p and ~180 MeV for p-α interactions, giving, therefore, a lower limit to the maximum energy of the particles accelerated at the Sun."[60]
{\displaystyle p^{+}+{\bar {p}}^{-}\rightarrow \pi ^{+}+\pi ^{-}}
{\displaystyle {\pi }^{+}\rightarrow {\mu }^{+}+{\nu }_{\mu }\rightarrow e^{+}+{\nu }_{e}+{\bar {\nu }}_{\mu }+{\nu }_{\mu }.}
When analyzing the outcome of some experiments with muons incident on a hydrogen bubble chamber in 1956, muon-catalysis of exothermic p-d, proton and deuteron, nuclear fusion was observed, which results in a helion, a gamma ray, and a release of about 5.5 MeV of energy.[67]
{\displaystyle \eta \rightarrow \gamma +\gamma ,}
{\displaystyle \eta \rightarrow \pi ^{0}+\pi ^{0}+\pi ^{0},or}
{\displaystyle \eta \rightarrow \pi ^{+}+\pi ^{0}+\pi ^{-},}
{\displaystyle \eta ^{'}\rightarrow \pi ^{+}+\pi ^{-}+\eta or}
{\displaystyle \eta ^{'}\rightarrow \pi ^{0}+\pi ^{0}+\gamma ,}
Although subluminal mesons are short-lived, they can be accelerated to near-luminal speeds so that they are detectable either directly or by their decay products.
↑ 1.0 1.1 Eberhard Klempt; Chris Batty; Jean-Marc Richard (July 2005). "The antinucleon-nucleon interaction at low energy: annihilation dynamics". Physics Reports 413 (4-5): 197-317. doi:10.1016/j.physrep.2005.03.002. http://adsabs.harvard.edu/abs/2005PhR...413..197K. Retrieved 2014-03-09.
↑ 2.0 2.1 Eli Waxman; John Bahcall (December 14, 1998). "High energy neutrinos from astrophysical sources: An upper bound". Physical Review D 59 (2): e023002. doi:10.1103/PhysRevD.59.023002. http://prd.aps.org/abstract/PRD/v59/i2/e023002. Retrieved 2014-03-09.
↑ 3.0 3.1 K. Kodama; N. Ushida1; C. Andreopoulos; N. Saoulidou; G. Tzanakos; P. Yager; B. Baller; D. Boehnlein et al. (April 12, 2001). "Observation of tau neutrino interactions". Physics Letters B 504 (3): 218-24. http://www.sciencedirect.com/science/article/pii/S0370269301003070. Retrieved 2014-03-10.
↑ "Particle Data Groups: 2006 Review of Particle Physics – Xi0" (PDF). Retrieved 2008-04-20.
↑ "Particle Data Groups: 2006 Review of Particle Physics – Xi-" (PDF). Retrieved 2008-04-20.
↑ 25.0 25.1 "Particle Data Groups: 2006 Review of Particle Physics – Xi(1530)" (PDF). Retrieved 2008-04-20.
↑ "Particle Data Groups: 2006 Review of Particle Physics – Sigma+" (PDF). Retrieved 2008-04-20.
↑ "Particle Data Groups: 2006 Review of Particle Physics – Sigma0" (PDF). Retrieved 2008-04-20.
↑ "Particle Data Groups: 2006 Review of Particle Physics – Sigma-" (PDF). Retrieved 2008-04-20.
↑ 39.0 39.1 39.2 Amsler, C. (2008). "Σ
↑ 41.0 41.1 41.2 Amsler, C. (2008). "Σ(1385)" (PDF). Particle Data Group. Particle listings. Lawrence Berkeley Laboratory.
↑ 47.0 47.1 G. F. Burgio; A. Drago; G. Pagliara; H.-J. Schulze; J.-B. Wei (21 June 2018). "Are Small Radii of Compact Stars Ruled out by GW170817/AT2017gfo?". The Astrophysical Journal 860 (2): 139. doi:10.3847/1538-4357/aac6ee. https://iopscience.iop.org/article/10.3847/1538-4357/aac6ee/meta. Retrieved 12 July 2019.
↑ 53.0 53.1 53.2 53.3 S. Nakayama; C. Mauger; M.H. Ahn; S. Aoki; Y. Ashie; H. Bhang; S. Boyd; D. Casper et al. (July 2005). "Measurement of single π0 production in neutral current neutrino interactions with water by a 1.3 GeV wide band muon neutrino beam". Physics Letters B 619 (3-4): 255-62. http://arxiv.org/pdf/hep-ex/0408134.pdf?origin=publication_detail. Retrieved 2014-03-22.
↑ 54.0 54.1 Forrest D. J.; Vestrand W. T.; Chupp E. L.; Rieger E.; Cooper J. F.; Share G. H. (August 1985). Neutral Pion Production in Solar Flares, In: 19th International Cosmic Ray Conference. 4. NASA. pp. 146-9. Bibcode: 1985ICRC....4..146F. http://adsabs.harvard.edu/full/1985ICRC....4..146F. Retrieved 2014-10-01.
↑ Y. Z. Fan; Bing Zhang; D. M. Wei (August 10, 2005). "Early photon-shock interaction in a stellar wind: A sub-GeV photon flash and high-energy neutrino emission from long gamma-ray bursts". The Astrophysical Journal 629 (1): 334-40. doi:10.1086/431473. http://iopscience.iop.org/0004-637X/629/1/334. Retrieved 2013-11-07.
↑ W.B. Atwood; P.F. Michelson; S.Ritz (2008). "Una Ventana Abierta a los Confines del Universo". Investigación y Ciencia 377: 24–31. https://opacbiblioteca.intec.edu.do/cgi-bin/koha/opac-detail.pl?biblionumber=51007.
↑ 65.00 65.01 65.02 65.03 65.04 65.05 65.06 65.07 65.08 65.09 65.10 65.11 65.12 65.13 65.14 V Khachatryan; AM Sirunyan; A Tumasyan; W Adam (August, 2010). "Measurement of the charge ratio of atmospheric muons with the CMS detector". Physics Letters B 692 (2): 83-104. http://www.sciencedirect.com/science/article/pii/S0370269310008725. Retrieved 2014-03-22.
↑ B. Pontecorvo (1957). [inspirehep.net/record/2884 "Mesonium and anti-mesonium"]. Zh. Eksp. Teor. Fiz. 33: 549–551. inspirehep.net/record/2884. reproduced and translated in Sov. Phys. JETP 6: 429. 1957. inspirehep.net/record/2884. and B. Pontecorvo (1967). [inspirehep.net/record/51319?ln=en "Neutrino Experiments and the Problem of Conservation of Leptonic Charge"]. Zh. Eksp. Teor. Fiz. 53: 1717. inspirehep.net/record/51319?ln=en. reproduced and translated in Sov. Phys. JETP 26: 984. 1968.
↑ "Particle Data Groups: 2006 Review of Particle Physics – Omega-" (PDF). Retrieved 2008-04-20.
D.C. Hom (1977). "Observation of a Dimuon Resonance at 9.5 Gev in 400-GeV Proton-Nucleus Collisions". Physical Review Letters 39: 252–255. doi:10.1103/PhysRevLett.39.252. http://lss.fnal.gov/archive/1977/pub/Pub-77-058-E.pdf.
J. Yoh (1998). "The Discovery of the b Quark at Fermilab in 1977: The Experiment Coordinator's Story". AIP Conference Proceedings 424: 29–42. http://lss.fnal.gov/archive/1997/conf/Conf-97-432-E.pdf.
S. Eidelman (Particle Data Group) (2004). "Review of Particle Physics – ϒ meson". Physics Letters B 592: 1. doi:10.1016/j.physletb.2004.06.001. http://pdg.lbl.gov/2005/listings/m049.pdf.
Joseph I. Kapusta; Berndt Müller; Johann Rafelski (2003). Joseph I. Kapusta. ed. Electromagnetic Signals, In: Quark-gluon Plasma: Theoretical Foundations : an Annotated Reprint Collection. Gulf Professional Publishing. pp. 432-70. ISBN 0444511105. http://books.google.com.au/books?id=8AD3GDoVaMkC&hl=en&sa=X&ei=y40kVKjHOI-A8QXpioC4Bw&ved=0CC8Q6AEwAA#v=onepage&f=false. Retrieved 25 September 2014.
Shared Physics prize for elementary particle. The Royal Swedish Academy of Sciences. 18 October 1976. http://www.nobelprize.org/nobel_prizes/physics/laureates/1976/press.html. Retrieved 2012-04-23.
Zielinski, L (8 August 2006). Physics Folklore. QuarkNet. http://ed.fnal.gov/samplers/hsphys/folklore.html. Retrieved 2009-04-13.
Matsui, T; Satz, H (1986). "J/ψ suppression by quark-gluon plasma formation". Physics Letters B 178 (4): 416. doi:10.1016/0370-2693(86)91404-8.
R. L. Thews; M. Schroedter; J. Rafelski (2001). "Enhanced J/ψ production in deconfined quark matter". Physical Review C 63 (5): 054905. doi:10.1103/PhysRevC.63.054905.
M. Schroedter; R. L. Thews; J. Rafelski (2000). "Bc-meson production in ultrarelativistic nuclear collisions". Physical Review C 62 (2): 024905. doi:10.1103/PhysRevC.62.024905.
L. P. Fulcher; J. Rafelski; R. L. Thews (1999). Bc mesons as a signal of deconfinement. https://arxiv.org/pdf/hep-ph/9905201.
Learn more about Mesons
Retrieved from "https://en.wikiversity.org/w/index.php?title=Radiation_astronomy/Mesons&oldid=2368919"
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Drift-rate model component - MATLAB - MathWorks India
Create a drift Object
Drift-rate model component
The drift object specifies the drift-rate component of continuous-time stochastic differential equations (SDEs).
The drift-rate specification supports the simulation of sample paths of NVars state variables driven by NBrowns Brownian motion sources of risk over NPeriods consecutive observation periods, approximating continuous-time stochastic processes.
The drift-rate specification can be any NVars-by-1 vector-valued function F of the general form:
F\left(t,{X}_{t}\right)=A\left(t\right)+B\left(t\right){X}_{t}
And a drift-rate specification is associated with a vector-valued SDE of the form
d{X}_{t}=F\left(t,{X}_{t}\right)dt+G\left(t,{X}_{t}\right)d{W}_{t}
A and B are model parameters.
The drift-rate specification is flexible, and provides direct parametric support for static/linear drift models. It is also extensible, and provides indirect support for dynamic/nonlinear models via an interface. This enables you to specify virtually any drift-rate specification.
DriftRate = drift(A,B)
DriftRate = drift(A,B) creates a default DriftRate model component.
Specify required input parameters A and B as one of the following types:
The drift object that you create encapsulates the composite drift-rate specification and returns the following displayed parameters:
Rate — The drift-rate function, F. Rate is the drift-rate calculation engine. It accepts the current time t and an NVars-by-1 state vector Xt as inputs, and returns an NVars-by-1 drift-rate vector.
A — Access function for the input argument A.
B — Access function for the input argument B.
A — A represents the parameter A
array or deterministic function of time
A represents the parameter A, specified as an array or deterministic function of time.
If you specify A as an array, it must be an NVars-by-1 column vector of intercepts.
As a deterministic function of time, when A is called with a real-valued scalar time t as its only input, A must produce an NVars-by-1 column vector. If you specify A as a function of time and state, it must generate an NVars-by-1 column vector of intercepts when invoked with two inputs:
B — B represents the parameter B
B represents the parameter B, specified as an array or deterministic function of time.
If you specify B as an array, it must be an NVars-by-NVars two-dimensional matrix of state vector coefficients.
As a deterministic function of time, when B is called with a real-valued scalar time t as its only input, B must produce an NVars-by-NVars matrix. If you specify B as a function of time and state, it must generate an NVars-by-NVars matrix of state vector coefficients when invoked with two inputs:
Rate — Composite drift-rate function
value stored from drift-rate function (default) | function accessible by F(t,Xt)
Composite drift-rate function, specified as F(t,Xt). The function stored in Rate fully encapsulates the combined effect of A and B, where A and B are:
Create a drift-rate function F:
F = drift(0, 0.1) % Drift rate function F(t,X)
The drift object displays like a MATLAB® structure and contains supplemental information, namely, the object's class and a brief description. However, in contrast to the SDE representation, a summary of the dimensionality of the model does not appear, because the drift class creates a model component rather than a model. F does not contain enough information to characterize the dimensionality of a problem.
When you specify the input arguments A and B as MATLAB arrays, they are associated with a linear drift parametric form. By contrast, when you specify either A or B as a function, you can customize virtually any drift-rate specification.
Accessing the output drift-rate parameters A and B with no inputs simply returns the original input specification. Thus, when you invoke drift-rate parameters with no inputs, they behave like simple properties and allow you to test the data type (double vs. function, or equivalently, static vs. dynamic) of the original input specification. This is useful for validating and designing methods.
When you invoke drift-rate parameters with inputs, they behave like functions, giving the impression of dynamic behavior. The parameters A and B accept the observation time t and a state vector Xt, and return an array of appropriate dimension. Specifically, parameters A and B evaluate the corresponding drift-rate component. Even if you originally specified an input as an array, drift treats it as a static function of time and state, by that means guaranteeing that all parameters are accessible by the same interface.
diffusion | sdeddo
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Centrum Wiskunde & Informatica: The itinerant list update problem
N.K. Olver (Neil), K. Pruhs (Kirk), K. Schewior (Kevin), R.A. Sitters (René) and L. Stougie (Leen)
Presented at the Workshop on Approximation and Online Algorithms (August 2018), Helsinki, Finland
We introduce the itinerant list update problem (ILU), which is a relaxation of the classic list update problem in which the pointer no longer has to return to a home location after each request. The motivation to introduce ILU arises from the fact that it naturally models the problem of track memory management in Domain Wall Memory. Both online and offline versions of ILU arise, depending on specifics of this application. First, we show that ILU is essentially equivalent to a dynamic variation of the classical minimum linear arrangement problem (MLA), which we call DMLA. Both ILU and DMLA are very natural, but do not appear to have been studied before. In this work, we focus on the offline ILU and DMLA problems. We then give an
O\left({\mathrm{log}}^{2}n\right)
-approximation algorithm for these problems. While the approach is based on well-known divide-and-conquer approaches for the standard MLA problem, the dynamic nature of these problems introduces substantial new difficulties. We also show an
\Omega \left(\mathrm{log}n\right)
lower bound on the competitive ratio for any randomized online algorithm for ILU. This shows that online ILU is harder than online LU, for which O(1)-competitive algorithms, like Move-To-Front, are known.
Conference Workshop on Approximation and Online Algorithms
Olver, N.K, Pruhs, K, Schewior, K, Sitters, R.A, & Stougie, L. (2018). The itinerant list update problem. In Proceedings of the 16th Workshop on Approximation and Online Algorithms (WAOA 2018) (pp. 310–326). doi:10.1007/978-3-030-04693-4_19
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Abstract Algebra/Group Theory/Subgroup/Coset/a Group is Partitioned by Cosets of Its Subgroup - Wikibooks, open books for an open world
Abstract Algebra/Group Theory/Subgroup/Coset/a Group is Partitioned by Cosets of Its Subgroup
2.1 Cosets of H are Subsets of G
2.2 Each Element of G is in a Coset of H
2.3 The Cosets of H are Disjoint
Let G be a Group. Let H be a Subgroup of G.
Then, Cosets of Subgroup H partition Group G.
Overview: G is partition by the cosets if
The cosets are subsets of G
Each element of G is in one of the cosets.
The cosets are disjoint
Cosets of H are Subsets of G Edit
{\displaystyle g\in G}
{\displaystyle k\in gH}
By definition of gH
{\displaystyle \exists \;h\in H:k=g\ast h}
As Subgroup H is Subset of G
{\displaystyle h\in G}
By 2., and Closure on G justified by 0. and 3.,
{\displaystyle k=g\ast h\in G}
Each Element of G is in a Coset of H Edit
{\displaystyle e_{G}\in H}
subgroup inherits identity (usage 2)
{\displaystyle g\in G}
{\displaystyle g\ast e_{G}\in gH}
definition of gH
{\displaystyle g=g\ast e_{G}\in gH}
eG is identity of G (usage 3)
The Cosets of H are DisjointEdit
0. Suppose 2 different cosets of H are not disjoint
1. Let the 2 cosets be g1H and g2H where
{\displaystyle {\color {Blue}g_{1}},{\color {OliveGreen}g_{2}}\in G}
Since they are not disjoint
{\displaystyle \exists u\in G:u\in {\color {Blue}g_{1}}H{\text{ and }}u\in {\color {OliveGreen}g_{2}}H}
By Definition of the Cosets,
{\displaystyle \exists {\color {Blue}h_{1}},{\color {OliveGreen}h_{2}}\in H:{\color {Blue}g_{1}}\ast {\color {Blue}h_{1}}=u={\color {OliveGreen}g_{2}}\ast {\color {OliveGreen}h_{2}}}
{\displaystyle z={\color {OliveGreen}g_{2}^{-1}}\ast {\color {Blue}g_{1}}={\color {OliveGreen}h_{2}}\ast {\color {Blue}h_{1}^{-1}}\in H}
{\displaystyle k\in g_{1}H}
By Definition of g1H
{\displaystyle \exists \;h_{k}\in H:k=g_{1}\ast h_{k}}
{\displaystyle z\ast h_{k}\in H}
{\displaystyle g_{2}\ast (z\ast h_{k})\in g_{2}H}
{\displaystyle g_{2}\ast g_{2}^{-1}\ast g_{1}\ast h_{k}\in g_{2}H}
{\displaystyle g_{1}\ast h_{k}\in g_{2}H}
{\displaystyle k\in g_{2}H}
{\displaystyle g_{1}H\subseteq g_{2}H}
As we can exchange g_1 and g_2 and apply the same procedure
{\displaystyle g_{2}H\subseteq g_{1}H}
{\displaystyle g_{1}H=g_{2}H}
contradicting that the two coset are different (0.)
Thus, two Cosets of H are either identical or are disjoint. Hence, the Cosets of H are disjoint.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Abstract_Algebra/Group_Theory/Subgroup/Coset/a_Group_is_Partitioned_by_Cosets_of_Its_Subgroup&oldid=3655737"
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DC-to-DC converter that supports bidirectional boost and buck - Simulink - MathWorks América Latina
Bidirectional DC-DC
VoltCmd
SrcVolt
LdVolt
SrcCurr
Converter response time constant
Converter response initial voltage, Vinit
Converter power limit, Plimit
Overall DC to DC converter efficiency, eff
Vector of voltages (v) for tabulated loss, v_loss_bp
Vector of currents (i) for tabulated loss, i_loss_bp
Vector of voltages (v) for tabulated efficiency, v_eff_bp
Vector of currents (i) for tabulated efficiency, i_eff_bp
DC-to-DC converter that supports bidirectional boost and buck
Powertrain Blockset / Energy Storage and Auxiliary Drive / DC-DC
The Bidirectional DC-DC block implements a DC-to-DC converter that supports bidirectional boost and buck (lower) operation. Unless the DC-to-DC conversion limits the power, the output voltage tracks the voltage command. You can specify electrical losses or measured efficiency.
Depending on your battery system configuration, the voltage might not be at a potential that is required by electrical system components such has inverters and motors. You can use the block to boost or buck the voltage. Connect the block to the battery and one of these blocks:
To calculate the electrical loss during the DC-to-DC conversion, use Parameterize losses by.
Electrical loss calculated using a constant value for conversion efficiency.
Electrical loss calculated as a function of load current and voltage. DC-to-DC converter data sheets typically provide loss data in this format. When you use this option, provide data for all the operating quadrants in which the simulation will run. If you provide partial data, the block assumes the same loss pattern for other quadrants. The block does not extrapolate loss that is outside the range voltage and current that you provide. The block allows you to account for fixed losses that are still present for zero voltage or current.
Electrical loss calculated using conversion efficiency that is a function of load current and voltage. When you use this option, provide data for all the operating quadrants in which the simulation will run. If you provide partial data, the block assumes the same efficiency pattern for other quadrants. The block:
Assumes zero loss when either the voltage or current is zero.
Uses linear interpolation to determine the loss. At lower power conditions, for calculation accuracy, provide efficiency at low voltage and low current.
The block does not support inversion. The polarity of the input voltage matches the polarity of the output voltage.
The Bidirectional DC-DC block uses the commanded voltage and the actual voltage to determine whether to boost or buck (lower) the voltage. You can specify a time constant for the voltage response.
Voltcmd > SrcVolt Boost
Voltcmd < SrcVolt Buck
The Bidirectional DC-DC block uses a time constant-based regulator to provide a fixed output voltage that is independent of load current. Using the output voltage and current, the block determines the losses of the DC-to-DC conversion. The block uses the conversion losses to calculate the input current. The block accounts for:
Source to load — Battery discharge
Load to source — Battery charge
Rated power limits
The block provides voltage control that is power limited based on these equations. The voltage is fixed. The block does not implement a voltage drop because the load current approximates DC-to-DC conversion with a bandwidth that is greater than the load current draw.
DC-to-DC converter load voltage
\begin{array}{l}LdVol{t}_{Cmd}=\mathrm{min}\left(Vol{t}_{Cmd},\frac{{P}_{limit}}{L{d}_{Amp}},0\right)\\ LdVolt=LdVol{t}_{Cmd}\cdot \frac{1}{\tau s+1}\end{array}
Pw{r}_{Loss}=\frac{100-Eff}{Eff}\cdot L{d}_{Volt}\cdot L{d}_{Amp}
Pw{r}_{Loss}=\frac{100-Eff}{Eff}\cdot |L{d}_{Volt}\cdot L{d}_{Amp}|
Pr{w}_{Loss}=f\left(L{d}_{Volt},L{d}_{Amp}\right)
Source current draw from DC-to-DC converter
Sr{c}_{Amp}=\frac{L{d}_{Pwr}+Pr{w}_{Loss}}{Sr{c}_{Volt}}
Source power from DC-to-DC converter
Sr{c}_{Pwr}=Sr{c}_{Amp}\cdot Sr{c}_{Volt}
PwrBusSrc
Source power to DC-to-DC converter
{P}_{src}= SrcPwr
PwrBusLd
Load power from DC-to-DC converter
{P}_{bus}= -LdVolt
Converter power loss
{P}_{loss}= PwrLoss
DC-to-DC converter commanded output voltage
Source input voltage to DC-to-DC converter
LdAmp
Load current of DC-to-DC converter
Load voltage of DC-to-DC converter
SrcAmp
Conversion time constant
Initial load voltage of the DC-to-DC converter
Output power limit for DC-to-DC converter
Input to output efficiency
LdVoltCmd
Commanded load voltage of DC-to-DC converter before application of time constant
VoltCmd — Commanded voltage
DC-to-DC converter commanded output voltage, VoltCmd, in V.
SrcVolt — Input voltage
Source input voltage to DC-to-DC converter, SrcVolt, in V.
LdCurr — Load current
Load current of DC-to-DC converter, LdAmp, in A.
LdVoltCmd V
LdVolt — Load voltage
Load voltage of DC-to-DC converter, LdVolt, in V.
SrcCurr — Source current
Source current draw from DC-to-DC converter, SrcAmp, in A.
Converter response time constant — Constant
Converter response time, τ, in s.
Converter response initial voltage, Vinit — Voltage
Initial load voltage of the DC-to-DC converter, Vinit, in V.
Converter power limit, Plimit — Power
Initial load voltage of the DC-to-DC converter, Plimit, in W.
Parameterize losses by — Loss calculation
This table summarizes the loss options used to calculate electrical options.
Overall DC to DC converter efficiency, eff — Constant
Overall conversion efficiency, Eff, in %.
To enable this parameter, for Parameterize losses by, select Single efficiency measurement.
Vector of voltages (v) for tabulated loss, v_loss_bp — Breakpoints
Tabulated loss breakpoints for M load voltages, in V.
Vector of currents (i) for tabulated loss, i_loss_bp — Breakpoints
Tabulated loss breakpoints for N load currents, in A.
Corresponding losses, losses_table — 2-D lookup table
Electrical loss map, as a function of N load currents and M load voltages, in W.
Vector of voltages (v) for tabulated efficiency, v_eff_bp — Breakpoints
Tabulated efficiency breakpoints for M load voltages, in V.
Vector of currents (i) for tabulated efficiency, i_eff_bp — Breakpoints
Tabulated efficiency breakpoints for N load currents, in A.
Corresponding efficiency, efficiency_table — 2-D lookup table
Electrical efficiency map, as a function of N load currents and M load voltages, in %.
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Brake — Wikipedia Republished // WIKI 2
Mechanical device that inhibits motion
This article is about the vehicle component. For other uses, see Brake (disambiguation).
Not to be confused with Break.
Disc brake on a motorcycle
A brake is a mechanical device that inhibits motion by absorbing energy from a moving system.[1] It is used for slowing or stopping a moving vehicle, wheel, axle, or to prevent its motion, most often accomplished by means of friction.[2]
Basic CDL Air Brake Components
How the braking system works? Types of brake disc (Car Part 4) ABS
2.1 Frictional
2.2 Pumping
3.1 Foundation components
3.2 Brake boost
4 Noise
5 Fires
6 Inefficiency
7.1 Early brake system
7.2 Electronic brake system
Most brakes commonly use friction between two surfaces pressed together to convert the kinetic energy of the moving object into heat, though other methods of energy conversion may be employed. For example, regenerative braking converts much of the energy to electrical energy, which may be stored for later use. Other methods convert kinetic energy into potential energy in such stored forms as pressurized air or pressurized oil. Eddy current brakes use magnetic fields to convert kinetic energy into electric current in the brake disc, fin, or rail, which is converted into heat. Still other braking methods even transform kinetic energy into different forms, for example by transferring the energy to a rotating flywheel.
Since kinetic energy increases quadratically with velocity (
{\displaystyle K=mv^{2}/2}
), an object moving at 10 m/s has 100 times as much energy as one of the same mass moving at 1 m/s, and consequently the theoretical braking distance, when braking at the traction limit, is up to 100 times as long. In practice, fast vehicles usually have significant air drag, and energy lost to air drag rises quickly with speed.
Almost all wheeled vehicles have a brake of some sort. Even baggage carts and shopping carts may have them for use on a moving ramp. Most fixed-wing aircraft are fitted with wheel brakes on the undercarriage. Some aircraft also feature air brakes designed to reduce their speed in flight. Notable examples include gliders and some World War II-era aircraft, primarily some fighter aircraft and many dive bombers of the era. These allow the aircraft to maintain a safe speed in a steep descent. The Saab B 17 dive bomber and Vought F4U Corsair fighter used the deployed undercarriage as an air brake.
Friction brakes on automobiles store braking heat in the drum brake or disc brake while braking then conduct it to the air gradually. When traveling downhill some vehicles can use their engines to brake.
When the brake pedal of a modern vehicle with hydraulic brakes is pushed against the master cylinder, ultimately a piston pushes the brake pad against the brake disc which slows the wheel down. On the brake drum it is similar as the cylinder pushes the brake shoes against the drum which also slows the wheel down.
Single pivot side-pull bicycle caliper brake
Frictional brakes are most common and can be divided broadly into "shoe" or "pad" brakes, using an explicit wear surface, and hydrodynamic brakes, such as parachutes, which use friction in a working fluid and do not explicitly wear. Typically the term "friction brake" is used to mean pad/shoe brakes and excludes hydrodynamic brakes, even though hydrodynamic brakes use friction. Friction (pad/shoe) brakes are often rotating devices with a stationary pad and a rotating wear surface. Common configurations include shoes that contract to rub on the outside of a rotating drum, such as a band brake; a rotating drum with shoes that expand to rub the inside of a drum, commonly called a "drum brake", although other drum configurations are possible; and pads that pinch a rotating disc, commonly called a "disc brake". Other brake configurations are used, but less often. For example, PCC trolley brakes include a flat shoe which is clamped to the rail with an electromagnet; the Murphy brake pinches a rotating drum, and the Ausco Lambert disc brake uses a hollow disc (two parallel discs with a structural bridge) with shoes that sit between the disc surfaces and expand laterally.
A drum brake is a vehicle brake in which the friction is caused by a set of brake shoes that press against the inner surface of a rotating drum. The drum is connected to the rotating roadwheel hub.
The disc brake is a device for slowing or stopping the rotation of a road wheel. A brake disc (or rotor in U.S. English), usually made of cast iron or ceramic, is connected to the wheel or the axle. To stop the wheel, friction material in the form of brake pads (mounted in a device called a brake caliper) is forced mechanically, hydraulically, pneumatically or electromagnetically against both sides of the disc. Friction causes the disc and attached wheel to slow or stop.
Pumping brakes are often used where a pump is already part of the machinery. For example, an internal-combustion piston motor can have the fuel supply stopped, and then internal pumping losses of the engine create some braking. Some engines use a valve override called a Jake brake to greatly increase pumping losses. Pumping brakes can dump energy as heat, or can be regenerative brakes that recharge a pressure reservoir called a hydraulic accumulator.
Main article: Electromagnetic brake
Electromagnetic brakes are likewise often used where an electric motor is already part of the machinery. For example, many hybrid gasoline/electric vehicles use the electric motor as a generator to charge electric batteries and also as a regenerative brake. Some diesel/electric railroad locomotives use the electric motors to generate electricity which is then sent to a resistor bank and dumped as heat. Some vehicles, such as some transit buses, do not already have an electric motor but use a secondary "retarder" brake that is effectively a generator with an internal short circuit. Related types of such a brake are eddy current brakes, and electro-mechanical brakes (which actually are magnetically driven friction brakes, but nowadays are often just called "electromagnetic brakes" as well).
Electromagnetic brakes slow an object through electromagnetic induction, which creates resistance and in turn either heat or electricity. Friction brakes apply pressure on two separate objects to slow the vehicle in a controlled manner.
Continuous power dissipation – Brakes typically get hot in use and fail when the temperature gets too high. The greatest amount of power (energy per unit time) that can be dissipated through the brake without failure is the continuous power dissipation. Continuous power dissipation often depends on e.g., the temperature and speed of ambient cooling air.
Fade – As a brake heats, it may become less effective, called brake fade. Some designs are inherently prone to fade, while other designs are relatively immune. Further, use considerations, such as cooling, often have a big effect on fade.
Smoothness – A brake that is grabby, pulses, has chatter, or otherwise exerts varying brake force may lead to skids. For example, railroad wheels have little traction, and friction brakes without an anti-skid mechanism often lead to skids, which increases maintenance costs and leads to a "thump thump" feeling for riders inside.
Power – Brakes are often described as "powerful" when a small human application force leads to a braking force that is higher than typical for other brakes in the same class. This notion of "powerful" does not relate to continuous power dissipation, and may be confusing in that a brake may be "powerful" and brake strongly with a gentle brake application, yet have lower (worse) peak force than a less "powerful" brake.
Durability – Friction brakes have wear surfaces that must be renewed periodically. Wear surfaces include the brake shoes or pads, and also the brake disc or drum. There may be tradeoffs, for example, a wear surface that generates high peak force may also wear quickly.
Weight – Brakes are often "added weight" in that they serve no other function. Further, brakes are often mounted on wheels, and unsprung weight can significantly hurt traction in some circumstances. "Weight" may mean the brake itself, or may include additional support structure.
Brake booster from a Geo Storm.
Most modern passenger vehicles, and light vans, use a vacuum assisted brake system that greatly increases the force applied to the vehicle's brakes by its operator.[4] This additional force is supplied by the manifold vacuum generated by air flow being obstructed by the throttle on a running engine. This force is greatly reduced when the engine is running at fully open throttle, as the difference between ambient air pressure and manifold (absolute) air pressure is reduced, and therefore available vacuum is diminished. However, brakes are rarely applied at full throttle; the driver takes the right foot off the gas pedal and moves it to the brake pedal - unless left-foot braking is used.
Because of low vacuum at high RPM, reports of unintended acceleration are often accompanied by complaints of failed or weakened brakes, as the high-revving engine, having an open throttle, is unable to provide enough vacuum to power the brake booster. This problem is exacerbated in vehicles equipped with automatic transmissions as the vehicle will automatically downshift upon application of the brakes, thereby increasing the torque delivered to the driven-wheels in contact with the road surface.
Heavier road vehicles, as well as trains, usually boost brake power with compressed air, supplied by one or more compressors.
Brake lever on a horse-drawn hearse
Main article: Roadway noise
Although ideally a brake would convert all the kinetic energy into heat, in practice a significant amount may be converted into acoustic energy instead, contributing to noise pollution.
For road vehicles, the noise produced varies significantly with tire construction, road surface, and the magnitude of the deceleration.[5] Noise can be caused by different things. These are signs that there may be issues with brakes wearing out over time.
Railway brake malfunctions can produce sparks and cause forest fires.[6] In some very extreme cases, disc brakes can become red hot and set on fire. This happened in the Tuscan GP, when the Mercedes car, the W11 had its front carbon disc brakes almost bursting into flames, due to low ventilation and high usage.[7] These fires can also occur on some Mercedes Sprinter vans, when the load adjusting sensor seizes up and the rear brakes have to compensate for the fronts.[8]
A significant amount of energy is always lost while braking, even with regenerative braking which is not perfectly efficient. Therefore, a good metric of efficient energy use while driving is to note how much one is braking. If the majority of deceleration is from unavoidable friction instead of braking, one is squeezing out most of the service from the vehicle. Minimizing brake use is one of the fuel economy-maximizing behaviors.
While energy is always lost during a brake event, a secondary factor that influences efficiency is "off-brake drag", or drag that occurs when the brake is not intentionally actuated. After a braking event, hydraulic pressure drops in the system, allowing the brake caliper pistons to retract. However, this retraction must accommodate all compliance in the system (under pressure) as well as thermal distortion of components like the brake disc or the brake system will drag until the contact with the disc, for example, knocks the pads and pistons back from the rubbing surface. During this time, there can be significant brake drag. This brake drag can lead to significant parasitic power loss, thus impacting fuel economy and overall vehicle performance.
Early brake system
In the 1890s, Wooden block brakes became obsolete when Michelin brothers introduced rubber tires.[9]
During the 1960s, some car manufacturers replaced drum brakes with disc brakes.[10]
In 1966, the ABS was fitted in the Jensen FF grand tourer.[11]
In 1978, Bosch and Mercedes updated their 1936 anti-lock brake system for the Mercedes S-Class. That ABS is a fully electronic, four-wheel and multi-channel system that later became standard.[12]
In 2005, ESC — which automatically applies the brakes to avoid a loss of steering control — become compulsory for carriers of dangerous goods without data recorders in the Canadian province of Quebec.[13]
Since 2017, numerous United Nations Economic Commission for Europe (UNECE) countries use Brake Assist System (BAS) a function of the braking system that deduces an emergency braking event from a characteristic of the driver's brake demand and under such conditions assist the driver to improve barking.[14]
In July 2013[15] UNECE vehicle regulation 131 was enacted. This regulation defines Advanced Emergency Braking Systems (AEBS) for heavy vehicles to automatically detect a potential forward collision and activate the vehicle braking system.
On 23 January 2020[16] UNECE vehicle regulation 152 was enacted, defining Advanced Emergency Braking Systems for light vehicles.
From May 2022, in the European Union, by law, new vehicles will have advanced emergency-braking system.[17]
Air brake (rail)
Air brake (road vehicle)
Advanced Emergency Braking System
Band brake
Bicycle brake systems
Brake-by-wire (or electromechanical braking)
Brake tester
Brake wear indicator
Braking distance
Breeching (tack)
Bundy tube
Caster brake
Counter-pressure brake
Dynamic braking
Electromagnetic brake
Electronic Parking Brake
Emergency brake (train)
Hand brake
Line lock
Overrun brake
Railway brake
Threshold braking
Trail braking
Vacuum brake
Wagon brake
^ Bhandari, V.B. (2010). Design of machine elements. Tata McGraw-Hill. p. 472. ISBN 9780070681798. Retrieved 9 February 2016.
^ "Definition of brake". The Collins English Dictionary. Retrieved 9 February 2016.
^ "Foundation Brakes". ontario.ca. Retrieved 2017-07-22.
^ Nice, Karim (2000-08-22). "How Power Brakes Work". Howstuffworks.com. Retrieved 2011-03-12.
^ C.Michael Hogan, Analysis of highway noise, Journal of Water, Air, & Soil Pollution, Volume 2, Number 3, Biomedical and Life Sciences and Earth and Environmental Science Issue, Pages 387-392, September, 1973, Springer Verlag, Netherlands ISSN 0049-6979
^ David Hench (May 8, 2014). "Train-sparked fires cause explosions, destroy trailers, force evacuations". Portland Press Herald.
^ "Mercedes explains Hamilton brake fire on Mugello F1 grid". www.motorsport.com. Retrieved 2020-11-21.
^ "Sprinter 311 Rear Brakes on fire". Mercedes-Benz Owners' Forums. Retrieved 2020-11-21.
^ "The History of Brakes | Did You Know Cars". 28 August 2017.
^ Roll Stability Control system (RSC) Archived 2011-07-16 at the Wayback Machine
^ https://www.unece.org/fileadmin/DAM/trans/main/wp29/wp29regs/2020/ECE-TRANS-WP.29-343-Rev.28-Add.1.pdf[bare URL PDF]
^ "Parliament approves EU rules requiring life-saving technologies in vehicles | News | European Parliament". Europarl.europa.eu. 2019-04-16. Retrieved 2020-08-31.
How Stuff Works - Brakes
Automotive navigation system
Automotive night vision
Backup camera
Blind spot monitor
Boost gauge
Check engine light
Electronic instrument cluster
Fuel gauge
Head-up display
Parking sensor
Radar detector
Trip computer
Bowden cable
Cruise control
Electronic throttle control
Gear stick
Steering wheel
Automatic vehicle location
Power door locks
Remote keyless system
Smart key
VIN etching
Bench seat
Bucket seat
Child safety lock
Rear-view mirror
Rumble seat
Seat belt
Boot liner
Center console
Glove compartment
Molded carpet
Sun visor
Vehicle mat
Automobile auxiliary power outlet
Cup holder
Railway brakes
Countersteam brake
Dynamic brake
Eddy current brake
Exhaust brake
Heberlein brake
Kunze-Knorr brake
Railway air brake
Railway disc brake
Steam brake
Track brake
Faiveley Transport
Knorr-Bremse (New York Air Brake)
Westinghouse Air Brake Company
Westinghouse Brake and Signal Company
Brake van
Diesel brake tender
Diesel electric locomotive dynamic braking
Electronically controlled pneumatic brakes
Electro-pneumatic brake system on British railway trains
Dowty retarders
Air brake
Bicycle brake
Dead man's switch
Pearson's Coupling
Railroad Safety Appliance Act (United States)
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A scale-space approach with wavelets to singularity estimation
Bigot, Jérémie
This paper is concerned with the problem of determining the typical features of a curve when it is observed with noise. It has been shown that one can characterize the Lipschitz singularities of a signal by following the propagation across scales of the modulus maxima of its continuous wavelet transform. A nonparametric approach, based on appropriate thresholding of the empirical wavelet coefficients, is proposed to estimate the wavelet maxima of a signal observed with noise at various scales. In order to identify the singularities of the unknown signal, we introduce a new tool, “the structural intensity”, that computes the “density” of the location of the modulus maxima of a wavelet representation along various scales. This approach is shown to be an effective technique for detecting the significant singularities of a signal corrupted by noise and for removing spurious estimates. The asymptotic properties of the resulting estimators are studied and illustrated by simulations. An application to a real data set is also proposed.
Classification : 62G05, 62G08, 65Dxx
Mots clés : Lipschitz singularity, continuous wavelet transform, scale-space representation, zero-crossings, wavelet maxima, feature extraction, non parametric estimation, bagging, landmark-based matching
author = {Bigot, J\'er\'emie},
title = {A scale-space approach with wavelets to singularity estimation},
AU - Bigot, Jérémie
TI - A scale-space approach with wavelets to singularity estimation
Bigot, Jérémie. A scale-space approach with wavelets to singularity estimation. ESAIM: Probability and Statistics, Tome 9 (2005), pp. 143-164. doi : 10.1051/ps:2005007. http://www.numdam.org/articles/10.1051/ps:2005007/
[2] A. Antoniadis and I. Gijbels, Detecting abrupt changes by wavelet methods. J. Nonparam. Statist 14 (2001) 7-29. | Zbl 1017.62033
[3] A. Arneodo, E. Bacry, S. Jaffard and J.F. Muzy, Oscillating singularities and fractal functions, in Spline functions and the theory of wavelets (Montreal, PQ, 1996), Amer. Math. Soc., Providence, RI. CRM Proc. Lect. Notes 18 (1999) 315-329.. | Zbl 1003.28007
[4] A. Arneodo, E. Bacry, S. Jaffard and J.F. Muzy, Singularity spectrum of multifractal functions involving oscillating singularities. J. Fourier Anal. Appl. 4 (1998) 159-174. | Zbl 0914.28005
[5] A. Arneodo, E. Bacry, S. Jaffard and J.F. Muzy, Oscillating singularities on Cantor sets: a grand-canonical multifractal formalism. J. Statist. Phys. 87 (1997) 179-209. | Zbl 0917.28007
[6] A. Arneodo, E. Bacry and J.F. Muzy, The thermodynamics of fractals revisited with wavelets. Physica A 213 (1995) 232-275.
[7] E. Bacry, J.F. Muzy and A. Arneodo, Singularity spectrum of fractal signals: exact results. J. Statist. Phys. 70 (1993) 635-674. | Zbl 0943.37500
[8] J. Bigot, Automatic landmark registration of 1D curves, in Recent advances and trends in nonparametric statistics, M. Akritas and D.N. Politis Eds., Elsevier (2003) 479-496.
[9] J. Bigot, Landmark-based registration of 1D curves and functional analysis of variance with wavelets. Technical Report TR0333, IAP (Interuniversity Attraction Pole network) (2003).
[10] L. Breiman, Bagging Predictors. Machine Learning 24 (1996) 123-140. | Zbl 0858.68080
[11] L.D. Brown and M.G. Lo, Asymptotic equivalence of nonparametric regression and white noise. Ann. Statist. 3 (1996) 2384-2398. | MR 1425958 | Zbl 0867.62022
[12] P. Chaudhuri and J.S. Marron, SiZer for exploration of structures in curves. J. Am. Statist. Ass. 94 (1999) 807-823. | MR 1723347 | Zbl 1072.62556
[13] P. Chaudhuri and J.S. Marron Scale space view of curve estimation. Ann. Statist. 28 (2000) 408-428. | MR 1790003 | Zbl 1106.62318
[14] R.R. Coifman and D.L. Donoho, Translation-invariant de-noising, in Wavelets and Statistics, A. Antoniadis and G. Oppenheim, Eds., New York: Springer-Verlag. Lect. Notes Statist. 103 (1995) 125-150. | MR 1364669 | Zbl 0866.94008
[15] I. Daubechies, Ten Lectures on Wavelets. Philadelphia, SIAM (1992). | MR 1162107 | Zbl 0776.42018
[16] D.L. Donoho and I.M. Johnstone, Ideal spatial adaptation by wavelet shrinkage. Biometrika 81 (1994) 425-455. | Zbl 0815.62019
[17] D.L. Donoho and I.M. Johnstone, Adapting to unknown smoothness via wavelet shrinkage. J. Am. Statist. Ass. 90 (1995) 1200-1224. | MR 1379464 | Zbl 0869.62024
[19] D.L. Donoho and I.M. Johnstone, Asymptotic minimality of wavelet estimators with sampled data. Stat. Sinica 9 (1999) 1-32. | MR 1678879 | Zbl 1065.62518
[20] D.L. Donoho, I.M. Johnstone, G. Kerkyacharian and D. Picard, Wavelet shrinkage: Asymptotia? (with discussion). J. R. Statist. Soc. B 57 (1995) 301-337. | MR 1323344 | Zbl 0827.62035
[21] N.I. Fisher and J.S. Marron, Mode testing via the excess mass estimate. Biometrika 88 (2001) 499-517. | MR 1844848 | Zbl 0985.62034
[22] T. Gasser and A. Kneip, Searching for Structure in Curve Samples. J. Am. Statist. Ass. 90 (1995) 1179-1188. | Zbl 0864.62019
[23] B. Hummel and R. Moniot, Reconstruction from zero-crossings in scale-space. IEEE Trans. Acoust., Speech, and Signal Proc. 37 (1989) 2111-2130.
[24] S. Jaffard, Mathematical Tools for Multifractal Signal Processing. Signal Processing for Multimedia, J.S Byrnes Ed., IOS Press (1999) 111-128. | Zbl 0991.94012
[25] A. Kneip and T. Gasser, Statistical tools to analyze data representing a sample of curves. Ann. Statist. 20 (1992) 1266-1305. | MR 1186250 | Zbl 0785.62042
[26] M.R. Leadbetter, G. Lindgren and H. Rootzén, Extremes and Related Properties of Random Sequences and Processes. Springer-Verlag (1983). | MR 691492 | Zbl 0518.60021
[27] T. Lindeberg, Scale Space Theory in Computer Vision. Kluwer, Boston (1994). | Zbl 0812.68040
[28] S. Mallat, Zero-Crossings of a Wavelet Transform. IEEE Trans. Inform. Theory 37 (1991) 1019-1033. | MR 1111805
[30] S. Mallat and W.L. Hwang, Singularity Detection and Processing with Wavelets. IEEE Trans. Inform. Theory 38 (1992) 617-643. | MR 1162218 | Zbl 0745.93073
[31] S. Mallat and S. Zhong, Characterization of Signals From Multiscale Egde. IEEE Trans. Pattern Anal. Machine Intelligence 14 (1992) 710-732.
[32] S. Mallat and S. Zhong, Wavelet Transformation Maxima and Multiscale Edges, in Wavelets: A Tutorial in Theory and Applications, C.K. Chui Ed. Boston, Academic Press (1992) 66-104. | MR 1187338 | Zbl 0804.68158
[33] S. Mallat and S. Zhong, Wavelet Maxima Representation, in Wavelets and Applications, Y. Meyer Ed. New York, Springer-Verlag (1992) 207-284. | MR 1276526 | Zbl 0804.68157
[34] M.C. Minnotte and D.W. Scott, The mode tree: a tool for visualization of nonparametric density features. J. Computat. Graph. Statist. 2 (1993) 51-68.
[35] M.C. Minnotte, D.J. Marchette and E.J. Wegman, The bumpy road to the mode forest. J. Comput. Graph. Statist. 7 (1998) 239-251.
[36] M. Misiti, Y. Misiti, G. Oppenheim and J.-M. Poggi, Décomposition en ondelettes et méthodes comparatives : étude d'une courbe de charge éléctrique. Revue de Statistique Appliquée 17 (1994) 57-77. | EuDML 106352 | Numdam
[37] J.F. Muzy, E. Bacry and A. Arneodo, The multifractal formalism revisited with wavelets. Int. J. Bif. Chaos 4 (1994) 245-302. | MR 1287531 | Zbl 0807.58032
[38] D. Picard and K. Tribouley, Adaptive confidence interval for pointwise curve estimation. Ann. Statist. 28 (2000) 298-335. | Zbl 1106.62331
[39] M. Raimondo, Minimax estimation of sharp change points. Ann. Statist. 26 (1998) 1379-1397. | MR 1647673 | Zbl 0929.62039
[40] J.O. Ramsay and X. Li, Curve registration. J. R. Statist. Soc. B 60 (1998) 351-363. | MR 1616045 | Zbl 0909.62033
[41] J.O. Ramsay and B.W. Silverman, Functional data analysis. New York, Springer Verlag (1997). | MR 2168993 | Zbl 0882.62002
[42] Y. Raviv and N. Intrator, Bootstrapping with Noise: An Effective Regularization Technique. Connection Science, Special issue on Combining Estimator 8 (1996) 356-372.
[43] M. Unser, A. Aldroubi and M. Eden, On the Asymptotic Convergence of B-Spline Wavelets to Gabor Functions. IEEE Trans. Inform. Theory 38 (1992) 864-872. | MR 1162223 | Zbl 0757.41022
[44] Y. Wang, Jump and Sharp Cusp Detection by Wavelets. Biometrica 82 (1995) 385-397. | MR 1354236 | Zbl 0824.62031
[45] K. Wang and T. Gasser, Alignment of curves by dynamic time warping. Ann. Statist. 25 (1997) 1251-1276. | MR 1447750 | Zbl 0898.62051
[46] K. Wang and T. Gasser, Synchronizing sample curves nonparametrically. Ann. Statist. 27 (1999) 439-460. | MR 1714722 | Zbl 0942.62043
[47] Y.P. Wang and S.L. Lee, Scale-Space Derived From B-Splines. IEEE Trans. on Pattern Analysis and Machine Intelligence 20 (1998) 1040-1055.
[48] L. Younes, Deformations, Warping and Object Comparison. Tutorial (2000) http://www.cmla.ens-cachan.fr/
\stackrel{˜}{}
younes.
[49] A.L. Yuille and T.A. Poggio, Scaling Theorems for Zero Crossings. IEEE Trans. Pattern Anal. Machine Intelligence 8 (1986) 15-25. | Zbl 0575.94001
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Compare accuracies of two classification models by repeated cross-validation - MATLAB testckfold - MathWorks Benelux
Estimate the classification loss by comparing the two sets of estimated labels to the true labels. Denote
{e}_{crk}
as the classification loss when the test set is fold k in run r of classification model c.
{\stackrel{^}{\delta }}_{rk}={e}_{1rk}-{e}_{2rk}.
Estimate the within-fold averages of the differences and their average:
{\overline{\delta }}_{r}=\frac{1}{K}\sum _{k=1}^{K}{\stackrel{^}{\delta }}_{kr}.
Estimate the overall average of the differences:
\overline{\delta }=\frac{1}{KR}\sum _{r=1}^{R}\sum _{k=1}^{K}{\stackrel{^}{\delta }}_{rk}.
Estimate the within-fold variances of the differences:
{s}_{r}^{2}=\frac{1}{K}\sum _{k=1}^{K}{\left({\stackrel{^}{\delta }}_{rk}-{\overline{\delta }}_{r}\right)}^{2}.
Estimate the average of the within-fold differences:
{\overline{s}}^{2}=\frac{1}{R}\sum _{r=1}^{R}{s}_{r}^{2}.
Estimate the overall sample variance of the differences:
{S}^{2}=\frac{1}{KR-1}\sum _{r=1}^{R}\sum _{k=1}^{K}{\left({\stackrel{^}{\delta }}_{rk}-\overline{\delta }\right)}^{2}.
{t}_{paired}^{\ast }=\frac{{\stackrel{^}{\delta }}_{11}}{\sqrt{{\overline{s}}^{2}}}.
{t}_{paired}^{\ast }
has a t-distribution with R degrees of freedom under the null hypothesis.
To reduce the effects of correlation between the estimated differences, the quantity
{\stackrel{^}{\delta }}_{11}
occupies the numerator rather than
\overline{\delta }
{F}_{paired}^{\ast }=\frac{\frac{1}{RK}\sum _{r=1}^{R}\sum _{k=1}^{K}{\left({\stackrel{^}{\delta }}_{rk}\right)}^{2}}{{\overline{s}}^{2}}.
{F}_{paired}^{\ast }
has an F distribution with RK and R degrees of freedom.
{t}_{CV}^{\ast }=\frac{\overline{\delta }}{S/\sqrt{\nu +1}}.
{t}_{CV}^{\ast }
has a t distribution with ν degrees of freedom. If the differences were truly independent, then ν = RK – 1. In this case, the degrees of freedom parameter must be optimized.
{\stackrel{^}{p}}_{1j}
is the predicted class assignment of classification model 1 for observation j.
{e}_{1}=\frac{\sum _{j=1}^{{n}_{test}}{w}_{j}\mathrm{log}\left(1+\mathrm{exp}\left(-2{y}_{j}^{\prime }f\left({X}_{j}\right)\right)\right)}{\sum _{j=1}^{{n}_{test}}{w}_{j}}
f\left({X}_{j}\right)
is the classification score.
{e}_{1}=\frac{\sum _{j=1}^{{n}_{test}}{w}_{j}\mathrm{exp}\left(-{y}_{j}f\left({X}_{j}\right)\right)}{\sum _{j=1}^{{n}_{test}}{w}_{j}}.
yj and
f\left({X}_{j}\right)
take the same forms here as in the binomial deviance formula.
{e}_{1}=\frac{\sum _{j=1}^{n}{w}_{j}\mathrm{max}\left\{0,1-{y}_{j}\prime f\left({X}_{j}\right)\right\}}{\sum _{j=1}^{n}{w}_{j}},
f\left({X}_{j}\right)
{e}_{1}=\frac{\sum _{j=1}^{{n}_{test}}{w}_{j}I\left({\stackrel{^}{p}}_{1j}\ne {y}_{j}\right)}{\sum _{j=1}^{{n}_{test}}{w}_{j}}.
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For each topic, decide which type of association a scatterplot
For each topic, decide which type of association a scatterplot of the data would likely show.Outdoor temperature and layers of clothing
For each topic, decide which type of association a scatterplot of the data would likely show. Explain your choice.
Outdoor temperature and layers of clothing
The association is positive, when the pattern in the scatterplot slopes upwards and thus when one variable increases as the other increases.
The association is negative, when the pattern in the scatterplot slopes downwards and thus when one variable decreased as the other increases.
There is no association, when the pattern in the scatterplot does not slope upwards nor downwards and thus when one variable increases, the other variable is unaffected.
As the outdoor temperature increases, we expect the number of layers of clothing to decrense as the warmer it is, the less clothes you tend to wear.
This then implies that there is most likely a negative association between these two variables.
Tom is creating a scatterplot that depicts the perimeter and area of a square. (If s is the length of a side of a square, the perimeter is 4s and the area is s2). He sets up these coordinates in which the explanatory variable is the perimeter and the response variable is the area. (10, u) (12, v) (20, w) (24, x) (36, y) (z, 49) a. Find the values of u, v, w, x, y, and z. b. If the points are put on a scatterplot, do they depict a positive or negative correlation?
Ruthie did a survey among her classmates comparing the time spent playing video games to the time spent studying. The scatterplot of her data is shown. a. What association can you make from her data? b. Use an ordered pair (x, y) to identify any outliers.
Using the health records of ever student at a high school, the school nurse created a scatterplot relating
y=\text{ }\text{height (in centimeters) to}\text{ }x=\text{ }\text{age (in years).}
\text{After verifying that the conditions for the regression model were met, the nurse calculated the equation of the population regression line to be}\text{ }{\mu }_{0}=105\text{ }+\text{ }4.2x\text{ }\text{with}\text{ }\sigma =7\text{ }cm.
About what percent of 15-year-old students at this school are taller than 180 cm?
HOW can you use technology to describe associations in scatter plots?
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Gamma inverse cumulative distribution function - MATLAB gaminv - MathWorks Italia
Compute Gamma icdf
Confidence Interval of Gamma icdf Value
Gamma icdf
Gamma inverse cumulative distribution function
x = gaminv(p,a)
x = gaminv(p,a,b)
[x,xLo,xUp] = gaminv(p,a,b,pCov)
[x,xLo,xUp] = gaminv(p,a,b,pCov,alpha)
x = gaminv(p,a) returns the inverse cumulative distribution function (icdf) of the standard gamma distribution with the shape parameter a, evaluated at the values in p.
x = gaminv(p,a,b) returns the icdf of the gamma distribution with shape parameter a and the scale parameter b, evaluated at the values in p.
[x,xLo,xUp] = gaminv(p,a,b,pCov) also returns the 95% confidence interval [xLo,xUp] of x when a and b are estimates. pCov is the covariance matrix of the estimated parameters.
[x,xLo,xUp] = gaminv(p,a,b,pCov,alpha) specifies the confidence level for the confidence interval [xLo,xUp] to be 100(1–alpha)%.
Find the median of the gamma distribution with shape parameter 3 and scale parameter 5.
x = gaminv(0.5,3,5)
Find a confidence interval estimating the median using gamma distributed data.
Generate a sample of 500 gamma distributed random numbers with shape 2 and scale 5.
params = gamfit(x)
Store the estimates for the parameters as ahat and bhat.
Compute the covariance of the parameter estimates.
Create a confidence interval estimating x.
[x,xLo,xUp] = gaminv(0.50,ahat,bhat,nCov)
Probability values at which to evaluate the inverse cdf (icdf), specified as a scalar value or an array of scalar values, where each element is in the range [0,1].
If you specify pCov to compute the confidence interval [xLo,xUp], then p must be a scalar value (not an array).
To evaluate the icdfs of multiple distributions, specify a and b using arrays.
If one or more of the input arguments p, a, and b are arrays, then the array sizes must be the same. In this case, gaminv expands each scalar input into a constant array of the same size as the array inputs. Each element in x is the icdf value of the distribution specified by the corresponding elements in a and b, evaluated at the corresponding element in p.
If you specify pCov to compute the confidence interval [xLo,xUp], then p, a, and b must be scalar values.
You can estimate a and b by using gamfit or mle, and estimate the covariance of a and b by using gamlike. For an example, see Confidence Interval of Gamma icdf Value.
icdf values evaluated at the probability values in p, returned as a scalar value or an array of scalar values. x is the same size as p, a, and b, after any necessary scalar expansion. Each element in x is the icdf value of the distribution specified by the corresponding elements in a and b, evaluated at the corresponding element in p.
The gamma inverse function in terms of the gamma cdf is
x={F}^{-1}\left(p|a,b\right)=\left\{x:F\left(x|a,b\right)=p\right\},
p=F\left(x|a,b\right)=\frac{1}{{b}^{a}\Gamma \left(a\right)}\underset{0}{\overset{x}{\int }}{t}^{a-1}{e}^{\frac{-t}{b}}dt.
The result x is the value such that an observation from the gamma distribution with parameters a and b falls in [0,x] with probability p.
No known analytical solution exists for the integral equation shown in Gamma icdf. gaminv uses an iterative approach (Newton's method) to converge on the solution.
gaminv is a function specific to the gamma distribution. Statistics and Machine Learning Toolbox™ also offers the generic function icdf, which supports various probability distributions. To use icdf, create a GammaDistribution probability distribution object and pass the object as an input argument or specify the probability distribution name and its parameters. Note that the distribution-specific function gaminv is faster than the generic function icdf.
GammaDistribution | icdf | gamcdf | gampdf | gamstat | gamfit | gamlike | gamrnd
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Interpolate by a factor of two using polyphase IIR - MATLAB - MathWorks España
H\left(z\right)=0.5*\left[{A}_{1}\left({z}^{2}\right)+{z}^{-1}{A}_{2}\left({z}^{2}\right)\right]
{A}_{1}\left(z\right)=\prod _{k=1}^{{K}_{1}}\frac{{a}_{k}^{\left(1\right)}+{z}^{-1}}{1+{a}_{k}^{\left(1\right)}{z}^{-1}}
{A}_{2}\left(z\right)=\prod _{k=1}^{{K}_{2}}\frac{{a}_{k}^{\left(2\right)}+{z}^{-1}}{1+{a}_{k}^{\left(2\right)}{z}^{-1}}
{A}_{1}\left(z\right)={z}^{-k}
{A}_{2}\left(z\right)=\prod _{K=1}^{{K}_{2}^{\left(1\right)}}\frac{{a}_{k}+{z}^{-1}}{1+{a}_{k}{z}^{-1}}\prod _{K=1}^{{K}_{2}^{\left(2\right)}}\frac{{c}_{k}+{b}_{k}{z}^{-1}+{z}^{-2}}{1+{b}_{k}{z}^{-1}+{c}_{k}{z}^{-2}}
G\left(z\right)=0.5*\left[{A}_{1}\left({z}^{2}\right)-{z}^{-1}{A}_{2}\left({z}^{2}\right)\right]
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Numerical Simulation of Pre- and Postsurgical Flow in a Giant Basilar Aneurysm | J. Biomech Eng. | ASME Digital Collection
Vitaliy L. Rayz,
, VA Medical Center-San Francisco, 4150 Clement Street, San Francisco, CA 94121
e-mail: vlrayz@gmail.com
Michael T. Lawton,
Director of Cerebrovascular Disorders Program, Assistant Professor of Neurological Surgery
, 1001 Potrero Avenue, San Francisco, CA 94110
e-mail: lawtonm@neurosurg.ucsf.edu
Alastair J. Martin,
, 505 Parnassus Avenue, San Francisco, CA 94143
e-mail: amartin@radiology.ucsf.edu
William L. Young,
James P. Livingston Professor Vice-Chair of Department of Anesthesia and Perioperative Care Professor of Neurological Surgery and Neurology
e-mail: youngw@anesthesia.ucsf.edu
, San Francisco CA 9443; Vascular Imaging Research Center,
VA Medical Center-San Francisco
, 4150 Clement Street, San Francisco, CA 94121
e-mail: saloner@radmail.ucsf.edu
Michael T. Lawton Director of Cerebrovascular Disorders Program, Assistant Professor of Neurological Surgery
Alastair J. Martin Associate Adjunct Professor
David Saloner Professor
Rayz, V. L., Lawton, M. T., Martin, A. J., Young, W. L., and Saloner, D. (March 27, 2008). "Numerical Simulation of Pre- and Postsurgical Flow in a Giant Basilar Aneurysm." ASME. J Biomech Eng. April 2008; 130(2): 021004. https://doi.org/10.1115/1.2898833
Computational modeling of the flow in cerebral aneurysms is an evolving technique that may play an important role in surgical planning. In this study, we simulated the flow in a giant basilar aneurysm before and after surgical takedown of one vertebral artery. Patient-specific geometry and flowrates obtained from magnetic resonance (MR) angiography and velocimetry were used to simulate the flow prior to and after the surgery. Numerical solutions for steady and pulsatile flows were obtained. Highly three-dimensional flows, with strong secondary flows, were computed in the aneurysm in the presurgical and postsurgical conditions. The computational results predicted that occlusion of a vertebral artery would result in a significant increase of the slow flow region formed in the bulge of the aneurysm, where increased particle residence time and velocities lower than
2.5cm∕s
were computed. The region of slow flow was found to have filled with thrombus following surgery. Predictions of numerical simulation methods are consistent with the observed outcome following surgical treatment of an aneurysm. The study demonstrates that computational models may provide hypotheses to test in future studies, and might offer guidance for the interventional treatment of cerebral aneurysms.
biomedical NMR, computational fluid dynamics, haemodynamics, surgery, basilar aneurysm, computational fluid dynamics, medical imaging
Aneurysms, Flow (Dynamics), Surgery, Computational fluid dynamics, Computer simulation, Pulsatile flow
Surgical Strategies for Giant Intracranial Aneurysms
The Detection and Management of Unruptured Intracranial Aneurysms
1969, Intracranial Aneurysms and Subarachnoid Hemorrhage: A Cooperative Study,
Vertebrobasilar Occlusion Therapy of Giant Aneurysms. Significance of Angiographic Morphology of the Posterior Communicating Arteries
Computational Approach to Quantifying Hemodynamic Forces in Giant Cerebral Aneurysms
Am. J. Neuroadiol.
Predictive Medicine: Computational Techniques in Therapeutic Decision-Making
Analysis of Slipstream Flow in a Wide-Necked Basilar Artery Aneurysm: Evaluation of Potential Treatment Regimes
Elastodynamics and Arterial Wall Stress
In Vitro Study of Haemodynamics in a Giant Saccular Aneurysm Model: Influence of Flow Dynamics in the Parent Vessel and Effects of Coil Embolisation
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MATLAB Programming/Nichols Plot - Wikibooks, open books for an open world
MATLAB Programming/Nichols Plot
2 MATLAB's Nichols Command
2.1 Issues with the nichols command
This article is on the topic of creating Nichols plots in MATLAB. The quick answer is use the Nichols command. However, the Nichols command has several options and the plots generated by the Nichols command are not easily reformatted. The default formatting of most MATLAB plots is good for analysis but less than ideal for dropping into Word and PowerPoint documents or even this website. As a result this article presents an alternative that requires more lines of code but offers the full formatting flexibility of the generic plot command.
MATLAB's Nichols Command[edit | edit source]
The basic Nichols command is as follows
>> nichols(LTI_SYS)
The Nichols command will automatically call gcf which will put the Nichols plot on the current figure. If no figure exists then one is created by gcf.
>> nichols(LTI_SYS, freqVec * (2*pi))
{\displaystyle \pi }
and in this case it is used to convert freqVec to rad/sec as it is passed to the Nichols command
Issues with the nichols command[edit | edit source]
The main issue with the nichols command is reformatting of the plot. The nichols command appears to use a normal semilogx plot and then apply patches or something similar to the figure. This can lead to odd behavior when attempting to create multi-line titles, reformat line widths or font sizes, etc. The normal relationship of axes to figure is just not quite present.
More on the Nichols plot at ControlTheoryPro.com
Retrieved from "https://en.wikibooks.org/w/index.php?title=MATLAB_Programming/Nichols_Plot&oldid=1567581"
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Page 7-7: The CS559 Framework Code (GraphicsTown)
Box 1: Prelude to Graphics Town
For the remaining parts of the workbook (two exercises), you will work with some framework code we have created for this class. Learning to work with the Framework code is important practically (you will use it on all of the assignments from here on), but also pedagogically. In the real world, you often have to work with code that someone else wrote.
The framework code allows you to focus on creating graphics objects and defining their behavior (for animation). You don't need to worry about setting up a user interface, or the basic stuff of the world. It will give you "good enough" defaults that you can focus on the parts of the assignment you care about. For example, in this assignment, you can focus on making hierarchical objects. For Workbook 8 you can focus on making appearances (just wait).
The framework provides a "thin wrapper" around THREE.JS - you still program using THREE.JS, but you put your objects inside of wrapper objects, and define their behaviors as methods of these objects.
WARNING: The framework code will evolve over the next few assignments. In the next assignments you will be given newer versions that add new features.
The best documentation for the framework is the code itself, and the examples that we give you. The code is designed to be (reasonably) self-documenting. I am trying to structure the comments to automatically produce a documentation web (using jsdoc), but that only gets some of the information. The web of documentation is included in the workbook (start here).
For Spring 2019, the framework is being re-written from scratch. It is based on the version written in 2015 (which served graphics classes for 3 years).
The framework is really helpful for small exercises where we simply want you to make an object. It will let you put the object on the screen with simple lighting and a simple UI with very little code. However, the real point of the framework is to allow you make a world with lots of objects that are all defined independently. This will let you make much more complex worlds. It also lets you add objects that we give you (or even to share objects with other students).
The framework will allow us to build "Graphics Town" - where you will create a world with lots of "living" stuff in it. The last workbooks/assignments of the semester will ask you to do this. The new few assignments will ask you to make some objects and put them into simpler "worlds". However, the objects you create can be re-used later in later assignments.
If you're wondering where the "fancy features" of the framework are (shadows, multi-pass rendering, or other fancy things)... Just wait. As we learn things in class, the framework will be extended to support these kinds of things.
Box 2: Modern JavaScript
Learning about features of "modern" JavaScript is a learning goal of this assignment.
The framework code uses "modern" JavaScript. That includes modules and classes. We've actually had some of that in prior assignments, but this time you will see more of it. You will not be able to avoid writing a class or figuring out how to export from a module. Fortunately, we will provide a lot of examples, so you can always do things by modifying something we've given you to start with. However, we do recommend that you take some time to actually understand what is going on with the code.
If you want to learn about JavaScript classes (and other modern features), most up-to-date books discuss it. See the course JavaScript page for suggestions. One specific recommendation: Exploring JavaScript Chapter 15 has a nice introduction before it starts a deep-dive into some gory details and tricks to do fancy stuff.
Box 3: Getting Started with a simple example
Here's the spinning cube example (from page 3 and the previous workbook and lecture). This time in the framework:
Everything is pretty much the default - except that I made the groundplane gray, not its usual green, since the cube is green. We got some lighting, a camera, orbit controls, an animation loop, and some other stuff "for free". It may not be exactly what we want, but we get a reasonable default and can always change things.
The code is in 7-simplespin.js. I'll include the key parts here:
// define a special kind of cube that spins
class SpinCube extends GrCube {
super({color:"green"});
advance(ms,daytime) {
// this used to be .01 per step
// however, we want to advance things based on the frame rate
// if we get 60fps, that's 16 miliseconds
this.objects[0].rotation.x += 0.01 * ms / 16;
this.objects[0].rotation.y += 0.01 * ms / 16;
let world = new GrWorld({groundplanecolor:"gray",
where:document.getElementById("simplespin")});
let cube = new SpinCube();
world.add(cube);
// we need to place the cube above the ground
cube.objects[0].position.y = 1;
world.go();
onWindowOnload(go);
Walking through this...
The main thing we do is define a new object type, a spinning cube. We create a class SpinCube that subclasses the built-in Cube type. The super call in the constructor runs the constructor of the base class (giving it an option that tells us the color).
The main thing we do in the new object is to define its behavior by overriding the advance method. The advance method gets called every frame and tells the object to update itself. The first parameter (ms) gives an estimate of how long it has been since the last redraw. This way we can adapt the speed in case our frame rate drops, rather than assuming our computer keeps up with 60 frames per second (16 milliseconds per frame).
Inside the advance method we do the same angle increments we did in the original program. Two things to note:
First, since we need access to the THREE Object3D, we have to look inside the list of objects that are "owned" by the SpinCube GrObject. I took advantage of the fact that I know that a GrCube has only one Object3D.
Second, notice how I scale the spin rate by the time.
In the main program (the function go that is called on window load), all we need to do is:
create the world (making the new GrWorld)
make objects and add them to the world
start the animation loop
We could do more (add more objects, adjust things), but we don't have to. We get reasonable defaults.
When I create the GrWorld I tell it where to put the DOM element that THREE creates for us. Here, I find the simplespin html element (it's a div). If I don't tell GrWorld where to put the element, it will just stick it at the end of the page.
Since there is a ground, I needed to raise the cube above the ground. Again, this means accessing the internal Object3D.
For a simple example like this one, the framework isn't that much easier than just writing the code from scratch using THREE.js directly. It does save me from having to remember to turn the lights on and point the camera the right way.
In the simple example, I re-used a basic object (a cube). Just as when we use THREE directly, usually we will make more interesting objects. You'll see that on the next pages.
Box 4: Another Simple Example
The first example showed how to make a simple animation. Here is a different example, inspired by the EulerToy demo on page 2. This time, rather than animating the cube, we control it with sliders. The nice thing: the framework will make the sliders for us!
The code for this one is in 7-simpleslider.js. This defines a different subclass of cube:
class RotCube extends GrCube {
super({color:"green"},
[ ["X",-Math.PI/2,Math.PI/2,0],
["Y",-Math.PI/2,Math.PI/2,0],
["Z",-Math.PI/2,Math.PI/2,0]
update(vec) {
this.objects[0].rotation.x = vec[0];
this.objects[0].rotation.y = vec[1];
this.objects[0].rotation.z = vec[2];
The big difference here is that when I created the cube subclass, I defined a set of "parameters" (the X, Y, Z) for rotation. They each range from
-\frac{\pi}{2}
\frac{\pi}{2}
. Since these are defined (with a name, range, and default value), the AutoUI code can make a control panel for it. Also notice how rather than defining an advance method, I provided an update method that takes the slider values as its argument.
The main body of the code created the UI - but all this required was telling AutoUI to do its job.
Again, this is a really simple case where I extend the cube. Under normal circumstances, I would extend the base class GrObject and populate it with THREE objects myself.
Box 5: The Framework Code Ground Rules
Some ground rules about the Framework Code:
You are welcome to (required to!) read the framework code. The code is meant to be somewhat self-documenting (with comments and type declarations).
Please avoid changing the framework code. Do not edit the files in the Framework directory unless we give you explicit instructions to do so. If you feel like you want to make your own version of something, copy the code into your own file and make changes. (be sure to give the code proper attribution)
If you find a bug in the framework code, please notify the instructors by Piazza.
We appreciate suggestions on how to make it better - this includes things we could document better, or functionality that we might want to add. The framework code will most likely be used by many classes, so suggestions may help many future students.
You are strongly encouraged to use the framework code. Learning to work within someone else's (imperfect) code is part of the learning goals of the assignments. However, if you choose to implement things yourself, you will be responsible for providing all of the functionality that the framework provides.
Summary: The Graphics Town Framework Code
The main way to learn about the Framework code is to read the code and its documentation. But the most important thing to do is to try it out and do things with it. You'll get to do that in the next two exercises.
First, we'll go make Graphics Park on the next page.
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KIEAE Journal. 2022; 22(2):5
The International Journal of The Korea Institute of Ecological Architecture and Environment - Vol. 22, No. 2, pp.5-11
*PhD, Associate Professor, School of Architecture and Design Convergence, Hankyong National Univ., South Korea architectism@hknu.ac.kr
Thermal Dissatisfaction, Heating Energy, Control Process, Artificial Neural Network
{Q}_{loss}+{Q}_{gain}=\frac{du}{dt}
\begin{array}{c}{Q}_{loss}=\left({T}_{room}-{T}_{out}\right) \hfill \\ /\left\{\frac{1}{\left({h}_{out}A\right)}+\frac{D}{\left(kA\right)}+\frac{1}{\left({h}_{in}A\right)}\right\}\hfill \end{array}
\frac{du}{dt}={m}_{room}{C}_{v}\frac{{dT}_{room}}{dt}
\frac{{dT}_{room}}{dt}=\frac{1}{{m}_{room}{C}_{v}}\text{*}\left(\begin{array}{c}\left(\frac{{T}_{room}-{T}_{out}}{\frac{1}{{h}_{out}A}+\frac{D}{kA}+\frac{1}{{h}_{\in }A}}\right)\hfill \\ +\left({ṁ}_{ht}{C}_{p}\left({T}_{heater}-{T}_{room}\right)\right)\hfill \end{array}\right)
PMV=3.155\left(0.303{e}^{-0.114M}+0.028\right)L
\begin{array}{c}L={q}_{met,heat}-{f}_{cl}{h}_{c}\left({T}_{cl}-{T}_{a}\right) \hfill \\ -{f}_{cl}{h}_{r}\left({T}_{cl}-{T}_{r}\right)-156\left({W}_{sk,req}-{W}_{a}\right)\hfill \\ -0.42\left({q}_{met,heat}-18.43\right)\hfill \\ -0.00077M\left(93.2-{T}_{a}\right)\hfill \\ -2.78M\left(0.0365-{W}_{a}\right)\hfill \end{array}
PPD=100-95{e}^{\left(-0.03353{PMV}^{4}-0.2179{PMV}^{2\right)}}
E={T}_{set}-{T}_{room}
\Delta E=\frac{\left({E}_{n}-{E}_{n-1}\right)}{\Delta t}
\mu \left(x\right)=∆\left(x;{a}_{i}, {b}_{i}, {c}_{i}\right)= \left\{\begin{array}{c}x\le {a}_{i}\to 0\\ {a}_{i}\le x\le {b}_{i}\to \frac{\left(x-{a}_{i}\right)}{\left({b}_{i}-{a}_{i}\right)}\\ {b}_{i}\le x\le {c}_{i}\to \frac{\left({c}_{i}-x\right)}{\left({c}_{i}-{b}_{i}\right)}\\ {c}_{i}\le x\to 0\end{array}\right\
(Eq. 10)
C. Blasco et al., Modelling and PID Control of HVAC System According to Energy Efficiency and Comfort Criteria. Sustainability in Energy and Buildings 2012, 12, pp.365-374. [https://doi.org/10.1007/978-3-642-27509-8_31]
T. Kull, M. Thalfeldt and J. Kurnitski. PI Parameter Influence on Underfloor Heating Energy Consumption and Setpoint Tracking in nZEBs. Energies, 2020, 13. [https://doi.org/10.3390/en13082068]
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조혜운 외 4인, 시뮬레이션을 통한 융복합 히트펌프 시스템의 적응형 예측제어 알고리듬 성능평가, 한국: 한국생태환경건축학회 논문집, 제21권 제6호, 2021.12, pp.55-62.
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Numerical Simulation of Stochastic Kuramoto-Sivashinsky Equation
Numerical Simulation of Stochastic Kuramoto-Sivashinsky Equation
Ping Gao*, Chengjian Cai, Xiaoyi Liu
College of Mathematics and Information Science, Guangzhou University, Guangzhou, China
In this paper, the random Kuramoto-Sivashinsky equation with additive noise is studied numerically, using the finite difference method to simulate the effect of different amplitude of noise on the solitary wave. And numerical experiments show that the white noise does not affect the propagation of the solitary wave, but can increase the amplitude of the solitary wave.
Random Kuramoto-Sivashinsky Equation, Difference Scheme, White Noise, Wiener Process
In recent years, many scholars have studied deterministic k-s equations and made important achievements, but there are relatively few studies on stochastic Kuramoto-Sivashinsky equations, and studying the numerical solution of the equation is a new field. In general, there is no analytic solution to stochastic Kuramoto-Sivashinsky equation, so numerical analysis becomes an important tool to develop its properties. Moreover, it has high computational efficiency, low computational complexity and good reliability. In this paper, its accuracy can be seen by comparing the numerical solution with the exact solution. Moreover we can also discover some phenomena about solution properties directly by numerical analysis.
We consider the following form of nonlinear evolution equation
{u}_{t}+u{u}_{x}+\alpha {u}_{xx}+\beta {u}_{xxxx}=0
\alpha
\beta
are real constants, which are a number of important mathematical physics equations in many physical problems. The second and fourth order terms represent the dissipation and instability of the system respectively, and the second one represents the convective nonlinear effect. Equation (1.1) is called the Kuramoto-Sivashinsky equation, hereinafter referred to as k-s equation; it is independently obtained in the analysis of Kuramotol [1] in dissipative structure of reaction diffusion system and Sivashinsky [2] in flame combustion and fluid dynamics instability. However, in practical situations, we must consider the effect of small irregular random factors, for example, adding a random force term to the right of the equation.
Let’s think about the k-s equation with a random term
\lambda
is the amplitude of the noise,
\stackrel{˙}{\xi }
is additive noise, and a real value gaussian process. Suppose that
u\left(x,t\right)
is defined in the region
u\left(0,x\right)={u}_{0}\left(x\right),-L<x<L;
and boundary condition
{u}_{x}\left(0,x\right)=0,t>0
The following is a mathematical definition of
\stackrel{˙}{\xi }
{\left(W\left(t\right)\right)}_{t\ge 0}
be a cylinder wiener process on
{L}^{2}\left({R}^{n},R\right)
, for the arbitrary orthogonal basis
{\left({e}_{i}\right)}_{i\in N}
{L}^{2}\left({R}^{n},R\right)
space, setting
{\beta }_{i}\left(t\right)=\left(W\left(t\right),{e}_{i}\right),i\in N,t\ge 0,
{\left({\beta }_{i}\right)}_{i\in N}
is a column of independent random Brownian motion, the column of Brownian motion
{\beta }_{i}\left(t,\omega \right),t\ge o,\omega \in \Omega
is stochastic process which is defined at random base
\left(\Omega ,Ϝ,P,{\left(Ϝ\left(t\right)\right)}_{t\ge 0}\right)
t\ge s
{Ϝ}_{t}
-measurable gaussian random independent variable of
{Ϝ}_{s}
for each i,
{\beta }_{i}\left(t\right)-{\beta }_{i}\left(s\right)
. Therefore, W can be written as:
W\left(t,x,\omega \right)=\underset{i\in N}{\sum }\text{ }{\beta }_{i}\left(t,\omega \right){e}_{i}\left(x\right),t\ge 0,x\in R,\omega \in \Omega .
Then the temporal and spatial white noise
\stackrel{˙}{\xi }
is the derivative of W to the time, that is:
\stackrel{˙}{\xi }=\frac{\text{d}W}{\text{d}t}={W}_{t}
In the same way, we can also define space related noise, giving a kernel k and a linear operator
\Phi
\Phi f\left(x\right)={\int }_{{R}^{n}}k\left(x,y\right)f\left(y\right)\text{d}y,f\in {L}^{2}\left({R}^{n}\right),
defining the Wiener process
\stackrel{˜}{W}=\Phi W
, then its time derivative
\stackrel{˙}{\stackrel{˜}{\xi }}
is the related noise of the time
{\delta }_{t-s}
, and its spatial correlation function c:
c\left(x,y\right)={\int }_{{R}^{n}}k\left(x,z\right)k\left(y,z\right).
In form, there are
E\left(\stackrel{˙}{\stackrel{˜}{\xi }}\left(x,t\right),\stackrel{˙}{\stackrel{˜}{\xi }}\left(y,s\right)\right)=c\left(x,y\right){\delta }_{t-s}.
k\left(x,y\right)=k\left(x-y\right)
is a convolution kernel, noise is uniform in space namely
c\left(x,y\right)=c\left(x-y\right)
and the noise is temporal and spatial white noise, then there are
k\left(x,y\right)={\delta }_{x-y}
\Phi ={I}_{d}
, then the Equation (1.2) can be written in the form of the following
\text{d}u+\left(u{u}_{t}+\alpha {u}_{xx}+\beta {u}_{xxxx}\right)\text{d}t=\lambda \text{d}\stackrel{˜}{W}
Literature [3] proves that the Equations (1.8), (1.3) and (1.4) have a unique solution. In this paper, the finite difference method is used to simulate the solutions of Equations (1.8), (1.3) and (1.4), and the results of the numerical analysis will be obtained.
2. Derivation of the Difference Scheme
u\left(x,t\right)
is defined on region , the following partition is made to R
-L={x}_{0}<{x}_{1}<\cdot \cdot \cdot <{x}_{J-1}<{x}_{J}=L,
0={t}_{0}<{t}_{1}<\cdot \cdot \cdot <{x}_{N-1}<{x}_{N}=t,
{u}_{j}^{n}=u\left(jh,n\tau \right)
, then in point
\left(jh,n\tau \right)
{\left[{u}_{t}\right]}_{j}^{n}+\frac{1}{2}{\left[{\left({u}^{2}\right)}_{x}\right]}_{j}^{n}+\alpha {\left[{u}_{xx}\right]}_{j}^{n}+\beta {\left[{u}_{xxxx}\right]}_{j}^{n}=\gamma {f}_{j}^{n+\frac{1}{2}}
First considering the above equation as the form of the K-S equation, replacing the
{\left[{u}_{t}\right]}_{j}^{n}
with the first order difference, and replace the
{\left[{\left({u}^{2}\right)}_{x}\right]}_{j}^{n}
{\left[{u}_{xx}\right]}_{j}^{n}
{\left[{u}_{xxxx}\right]}_{j}^{n}
with the center difference, so that
\begin{array}{l}{u}_{j}^{n+1}-{u}_{j}^{n}+\frac{\tau }{4}\left({\left[{\left({u}^{2}\right)}_{x}\right]}_{j}^{n+1}+{\left[{\left({u}^{2}\right)}_{x}\right]}_{j}^{n}\right)+\frac{\tau \alpha }{2}\left({\left[{u}_{xx}\right]}_{j}^{n+1}+{\left[{u}_{xx}\right]}_{j}^{n}\right)\\ +\frac{\tau \beta }{2}\left({\left[{u}_{xxxx}\right]}_{j}^{n+1}+{\left[{u}_{xxxx}\right]}_{j}^{n}\right)=0\end{array}
If the partial derivative of (2.2) with respect to x is simply substituted by the difference quotient, the problem of solving nonlinear equations will be encountered, in order to overcome this difficulty, we did Taylor expansion for nonlinear terms.
\begin{array}{c}{\left[{\left({u}^{2}\right)}_{x}\right]}_{j}^{n+1}={\left[{\left({u}^{2}\right)}_{x}\right]}_{j}^{n}+{\left[{\left({u}^{2}\right)}_{tx}\right]}_{j}^{n}\tau +o\left({\tau }^{2}\right)\\ ={\left[{\left({u}^{2}\right)}_{x}\right]}_{j}^{n}+{\left[{\left(2u{u}_{t}\right)}_{x}\right]}_{j}^{n}\tau +o\left({\tau }^{2}\right)\\ ={\left[{\left({u}^{2}\right)}_{x}\right]}_{j}^{n}+2{\left[\left({u}_{j}^{n}\right)\left(\frac{{u}_{j}^{n+1}-{u}_{j}^{n}}{\tau }+o\left(\tau \right)\right)\tau \right]}_{x}+o\left({\tau }^{2}\right)\\ ={\left[{\left({u}^{2}\right)}_{x}\right]}_{j}^{n}+2{\left[\left({u}_{j}^{n}\right)\left({u}_{j}^{n+1}-{u}_{j}^{n}\right)\right]}_{x}+o\left(\tau 2\right)\end{array}
\begin{array}{c}{\left[{\left({u}^{2}\right)}_{x}\right]}_{j}^{n}+{\left[{\left({u}^{2}\right)}_{x}\right]}_{j}^{n+1}=2{\left[\left({u}_{j}^{n}\right)\left({u}_{j}^{n+1}-{u}_{j}^{n}\right)\right]}_{x}+2{\left[{\left({u}^{2}\right)}_{x}\right]}_{j}^{n}+o\left({h}^{2}\right)\\ =2{\left[{u}_{j}^{n}{u}_{j}^{n+1}\right]}_{x}+o\left({h}^{2}\right)\\ =\frac{{u}_{j+1}^{n}{u}_{j+1}^{n+1}-{u}_{j-1}^{n}{u}_{j-1}^{n+1}}{h}+o\left({h}^{2}\right)\end{array}
You can get the following difference scheme
a{u}_{j-2}^{n+1}+{b}_{j}^{n}{u}_{j-1}^{n+1}+e{u}_{j}^{n+1}+{c}_{j}^{n}{u}_{j+1}^{n+1}+a{u}_{j+2}^{n+2}={d}_{j}^{n}
a=\frac{\tau \beta }{2{h}^{4}},{b}_{j}^{n}=\frac{\tau \alpha }{2{h}^{2}}-\frac{2\tau \beta }{{h}^{4}}-\frac{\tau }{4h}{u}_{j-1}^{n},
e=1+\frac{3\tau \beta }{{h}^{4}}-\frac{\tau \alpha }{{h}^{2}},{c}_{j}^{n}=\frac{\tau \alpha }{2{h}^{2}}-\frac{2\tau \beta }{{h}^{4}}+\frac{\tau }{4h}{u}_{j+1}^{n},
\begin{array}{l}{d}_{j}^{n}=-\frac{\tau \beta }{2{h}^{4}}{u}_{j-2}^{n}-\left(\frac{\tau \alpha }{2{h}^{2}}-\frac{2\tau \beta }{{h}^{4}}\right){u}_{j-1}^{n}+\left(1-\frac{3\tau \beta }{{h}^{4}}+\frac{\tau \alpha }{{h}^{2}}\right){u}_{j}^{n}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\left(\frac{\tau \alpha }{2{h}^{2}}-\frac{2\tau \beta }{{h}^{4}}\right){u}_{j+1}^{n}-\frac{\tau \beta }{2{h}^{4}}{u}_{j+2}^{n},\text{ }n=0,1,\cdots ,N,j=0,1,\cdots ,J.\end{array}
For the difference scheme (2.4), the value of each node is required, we need to solve a large linear system of linear equations with a matrix order of J at every step of time t, according to the supposition of the boundary conditions,
{u}_{-1}={u}_{0}={u}_{1}
{u}_{J+2}={u}_{J+1}={u}_{J}
{f}_{j}^{n+\frac{1}{2}}
of (2.1) can use the following formula to approximate
\frac{1}{h\tau }{\int }_{\left(j-\frac{1}{2}\right)h}^{\left(j+\frac{1}{2}\right)h}{\int }_{{t}_{n}}^{{t}_{n+1}}\stackrel{˙}{\xi }\text{d}s\text{d}x,j=0,\cdots ,J.
Substituting the previous (1.4) and (1.5) into the above equation, we can get
\begin{array}{c}{f}_{j}^{n+\frac{1}{2}}=\frac{1}{h\tau }{\int }_{\left(j-\frac{1}{2}\right)h}^{\left(j+\frac{1}{2}\right)h}{\int }_{{t}_{n}}^{{t}_{n+1}}\underset{i\in N}{\sum }{e}_{i}\left(x\right)\text{d}{\beta }_{i}\left(s\right)\text{d}x\\ =\frac{1}{h\tau }\underset{i\in N}{\sum }\left({\int }_{\left(j-\frac{1}{2}\right)h}^{\left(j+\frac{1}{2}\right)h}{\int }_{{t}_{n}}^{{t}_{n+1}}{e}_{i}\left(x\right)\text{d}x\right)\left({\beta }_{i}\left({t}_{n+1}\right)-{\beta }_{i}\left({t}_{n}\right)\right)\end{array}
if the orthogonal basis
{\left({e}_{i}\right)}_{i\in N}
{L}^{2}\left(-L,L\right)
is taken as the following form
{e}_{j}=\frac{1}{\sqrt{h}}{1}_{\left[\left(j-\frac{1}{2}\right)h,\left(j+\frac{1}{2}\right)h\right]},j=-J+1,\cdots ,J-1
{e}_{-J}=\frac{1}{\sqrt{h/2}}{1}_{\left[-Jh,\left(-J+\frac{1}{2}\right)h\right]},{e}_{J}=\frac{1}{\sqrt{h/2}}{1}_{\left[\left(J-\frac{1}{2}\right)h,Jh\right]},
then through orthogonalization,
{\int }_{\left(j-\frac{1}{2}\right)h}^{\left(j+\frac{1}{2}\right)h}{e}_{i}\left(x\right)\text{d}x=0
i\ne j,j=0,\cdots ,J,i\in N
{f}_{j}^{n+\frac{1}{2}}=\frac{1}{\tau \sqrt{h}}\left({\beta }_{i}\left({t}_{n+1}\right)-{\beta }_{i}\left({t}_{n}\right)\right),j=-J+1,\cdots ,J-1,
{f}_{-J}^{n+\frac{1}{2}}=\frac{1}{\tau \sqrt{h/2}}\left({\beta }_{-J}\left({t}_{n+1}\right)-{\beta }_{-J}\left({t}_{n}\right)\right),{f}_{J}^{n+\frac{1}{2}}=\frac{1}{\tau \sqrt{h/2}}\left({\beta }_{J}\left({t}_{n+1}\right)-{\beta }_{J}\left({t}_{n}\right)\right)
\left({\beta }_{j}\left({t}_{n+1}\right)-{\beta }_{j}\left({t}_{n}\right)\right)/\sqrt{\tau }
is independent random variable and obeys
the standard normal distribution
N\left(0,1\right)
, selecting
{\left({\chi }_{j}^{n+1/2}\right)}_{n\ge 0,j=-J,\cdots ,J}
random variable that obeys the standard normal distribution. So for each time increment,
{f}^{n+\frac{1}{2}}
can use the vector
\left({\chi }_{-J}^{n+1/2},\cdot \cdot \cdot ,{\chi }_{J}^{n+1/2}\right)
to simulate.
Although our purpose is to simulate the solution of K-S equation and study its properties, there is a very important problem that we need to verify whether the format described above is effective. First in the interval
I\times R=\left[\mathrm{0,1}\right]\times \left[-\mathrm{5,5}\right]
we simulate the initial value problem (1.1), and the initial condition is
{u}_{0}\left(x\right)=-\frac{c}{k}+\frac{60}{19}k\left(-38\beta {k}^{2}+\alpha \right)\mathrm{tanh}kx+120\beta {k}^{3}{\mathrm{tanh}}^{3}kx
k=\sqrt{\frac{11\alpha }{76\beta }}
, this problem has the following solitary wave solution [4]
u\left(x,t\right)=-\frac{c}{k}+\frac{60}{19}k\left(-38\beta {k}^{2}+\alpha \right)\mathrm{tanh}\left(kx+ct\right)+120\beta {k}^{3}{\mathrm{tanh}}^{3}\left(kx+ct\right)
\alpha =0.1,\beta =0.1
, space step
h=0.1
, and time step
\tau =0.01
, Figure 1 below shows an image of a numerical solution and an exact solution, Table 1 and Table 2 show the absolute error between different numerical solutions and exact solutions. The obtained numerical solution can well reflect the solution of the equation, indicating that the format described in this paper is valid.
Now we have the numerical simulation Equation (1.7), using the methods described above and the initial conditions.
When the amplitude of noise is small,
\lambda ={10}^{-3}
, as shown in Figure 2(a) made a time t in the interval
\left[0,3\right]
image, the other parameters are the same as before, it can be seen that the solitary wave is not destroyed, and the noise does not stop the propagation of the wave. We can see the same phenomenon by choosing different noise tracks.
In order to further study the stability of the solitary wave, we increase the amplitude of the noise,
\lambda =0.5\times {10}^{-2}
, the impact will be strengthened. The initial conditions are the same as before, As shown in Figure 2(b), it can be seen that the solitary wave is not destroyed, but it can be seen that the amplitude
Figure 1. The comparison of the numerical solution (a) and the exact solution (b), here
\alpha =0.1,\beta =0.1,c=1
\lambda =0.001
\lambda =0.005
\lambda =0.01
, the contour curves of
d\left(u\right)
\alpha =0.1,\beta =0.1,c=1
Table 1. The absolute error data table between the numerical solutions and the exact solutions, here
\alpha =0.1,\beta =0.1,c=1
Table 2. The absolute error data table between the numerical solutions and the exact solutions here
\alpha =0.1,\beta =0.1,c=1
increases through the propagation of the solitary wave, which makes it clear that the noise will enhance the amplitude of the wave.
Increasing the amplitude of the noise again,
\lambda =0.01
as shown in Figure2(c), in this case, the whole solitary wave is still intact, in order to further study this phenomenon, we use another representation method to represent the image of the solution, as shown in Figure 2(d), we draw the contour curves of solutions, we observed that although the amplitude of the noise is very high, we can clearly see that the amplitude of the wave increases during the propagation of solitary waves. We believe that this is the effect of the energy injected by the noise, which increases the amplitude of the solitary wave.
In this paper, the finite difference method is used to carry out numerical experiments on the solution of the random K-S equation. The results show that the noise does not affect the propagation of the solitary wave, but it can enhance the amplitude of the solitary wave. This is similar to the phenomenon observed in random Kdv equations [5] .
Gao, P., Cai, C.J. and Liu, X.Y. (2018) Numerical Simulation of Stochastic Kuramoto-Sivashinsky Equation. Journal of Applied Mathematics and Physics, 6, 2363-2369. https://doi.org/10.4236/jamp.2018.611198
1. Kuramoto, Y. and Tsuzuki, T. (1975) On the Formation of Dissipative Structures in Reaction-Diffusion Systems. Progress of Theoretical Physics, 54, 687-699. https://doi.org/10.1143/PTP.54.687
2. Sivashinsky, G.I. (1977) Nonlinear Analysis of Hydrodynamic Instability in Laminar Flames. Acta Astronautica, 4, 1177-1206. https://doi.org/10.1016/0094-5765(77)90096-0
3. Duan, J. and Ervin, V.J. (2001) On the Stochastic Kuramoto Sivashinsky Equation. Nonlinear Analysis, 44, 205-216. https://doi.org/10.1016/S0362-546X(99)00259-X
4. Khater, A.H. and Temsah, R.S. (2008) Numerical Solutions of the Generalized Kuramoto-Sivashinsky Equation by Chebyshev Spectral Collocation Methods. Computers and Mathematics with Applications, 56, 1465-1472. https://doi.org/10.1016/j.camwa.2008.03.013
5. Debussche, A. and Di Menza, L. (2002) Numerical Simulation of Focusing Stochastic Nonlinear Schrodinger Equations. Physica D, 162, 131-154. https://doi.org/10.1016/S0167-2789(01)00379-7
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Pictures of Julia and Mandelbrot Sets/Julia and Mandelbrot sets for non-complex functions - Wikibooks, open books for an open world
Pictures of Julia and Mandelbrot Sets/Julia and Mandelbrot sets for non-complex functions
{\displaystyle f(z)}
{\displaystyle f_{x}(x,y)}
{\displaystyle f_{y}(x,y)}
{\displaystyle {\frac {\partial }{\partial {x}}}f_{x}={\frac {\partial }{\partial {y}}}f_{y}}
{\displaystyle {\frac {\partial }{\partial {y}}}f_{x}=-{\frac {\partial }{\partial {x}}}f_{y}}
{\displaystyle f'(z)}
{\displaystyle z^{2}+c}
{\displaystyle (x,y)}
{\displaystyle (x^{2}-y^{2},2xy)+(u,v)}
2 The Julia sets
4 The formula for the distance function
5 Precise calculation of the critical point
The Mandelbrot setEdit
{\displaystyle f(x,y)}
{\displaystyle p(x,y)}
{\displaystyle q(x,y)}
{\displaystyle f(x,y)=p(x,y)+iq(x,y)}
{\displaystyle f'(z)=0}
{\displaystyle g(x,y)}
{\displaystyle g'(x,y)}
{\displaystyle {\frac {\partial }{\partial {x}}}g_{x}(x,y)}
{\displaystyle {\frac {\partial }{\partial {y}}}g_{x}(x,y)}
{\displaystyle {\frac {\partial }{\partial {x}}}g_{y}(x,y)}
{\displaystyle {\frac {\partial }{\partial {y}}}g_{y}(x,y)}
{\displaystyle g'(z)}
{\displaystyle g'(x,y)}
{\displaystyle |g'(x,y)|}
{\displaystyle (\partial g_{x}(x,y)/\partial {x})(\partial g_{y}(x,y)/\partial {y})-(\partial g_{x}(x,y)/\partial {y})(\partial g_{y}(x,y)/\partial {x})=0}
{\displaystyle f(x,y)}
{\displaystyle |g'(x,y)|}
{\displaystyle (zc_{x},zc_{y})}
{\displaystyle (zc_{x},zc_{y})}
{\displaystyle p(x,y)=xy}
{\displaystyle q(x,y)=y-x^{2}}
{\displaystyle y+2x^{2}}
{\displaystyle y=-2x^{2}}
{\displaystyle (zc_{x},zc_{y})=(0,0)}
The Julia setsEdit
{\displaystyle p(x,y)=x^{2}}
{\displaystyle q(x,y)=y^{2}}
{\displaystyle (p(x,y)+iq(x,y))/r(x^{2},y^{2})}
{\displaystyle p(x,y)=(x^{4}-y^{4})/10}
{\displaystyle q(x,y)=x-y}
{\displaystyle r(x,y)=1-(x-y)/5+((x+y)/10))^{2}}
{\displaystyle p(x,y)=1-xy^{3}}
{\displaystyle q(x,y)=x^{2}y^{2}+y^{4}}
{\displaystyle r(x,y)=1+x^{2}+xy+y^{2}}
{\displaystyle p(x,y)=x^{3}-y^{3}+x^{2}y^{2}}
{\displaystyle q(x,y)=x-y+x^{2}y^{2}}
{\displaystyle r(x,y)=1+x^{2}+y^{2}}
The formula for the distance functionEdit
{\displaystyle \delta (z)=\phi (z)/|\phi '(z)|}
{\displaystyle \phi (z)}
{\displaystyle \phi (z)}
{\displaystyle \phi '(z)}
{\displaystyle \partial \phi (z)/\partial {x}}
{\displaystyle \partial \phi (z)/\partial {y}}
{\displaystyle f(z)}
{\displaystyle z_{k}}
{\displaystyle z'_{k}}
{\displaystyle f'(z)}
{\displaystyle Df(z)}
{\displaystyle {\begin{bmatrix}\partial f_{x}/\partial {x}&\partial f_{x}/\partial {y}\\\partial f_{y}/\partial {x}&\partial f_{y}/\partial {y}\\\end{bmatrix}}}
{\displaystyle z'_{k}}
{\displaystyle {\begin{bmatrix}\partial x_{k}/\partial {x}&\partial x_{k}/\partial {y}\\\partial y_{k}/\partial {x}&\partial y_{k}/\partial {y}\\\end{bmatrix}}}
{\displaystyle {\begin{bmatrix}xx_{k}&xy_{k}\\yx_{k}&yy_{k}\\\end{bmatrix}}}
{\displaystyle z'_{k+1}=Df(z_{k})z'_{k}}
{\displaystyle z'_{0}=I}
{\displaystyle \phi '(z)}
{\displaystyle \phi '(z)=}
{\displaystyle x*-x_{kr},y*-y_{kr}}
{\displaystyle z'_{kr}/(|z_{kr}-z*|^{3}\alpha ^{k})}
{\displaystyle \phi '(z)=}
{\displaystyle x*-x_{kr},y*-y_{kr}}
{\displaystyle z'_{kr}/(|z_{kr}-z*|^{2}\alpha ^{k})}
{\displaystyle \phi '(z)=}
{\displaystyle x_{k},y_{k}}
{\displaystyle z'_{k}/(|z_{k}|^{2}d^{k})}
{\displaystyle x_{k},y_{k}}
{\displaystyle z'_{k}}
{\displaystyle x_{k}xx_{k}+y_{k}yx_{k}}
{\displaystyle x_{k}xy_{k}+y_{k}yy_{k}}
{\displaystyle \delta (z)=\phi (z)/|\phi '(z)|}
{\displaystyle \phi (z)}
{\displaystyle \phi '(z)}
{\displaystyle \delta (z)}
{\displaystyle |z'_{k}|}
{\displaystyle Df(z)}
{\displaystyle det(z'_{k+1})=det(Df(z_{k}))det(z'_{k})}
{\displaystyle det(z'_{k})}
{\displaystyle det(z'_{0})=1}
{\displaystyle z'_{k+1}=Df(z_{k})z'_{k}+I}
{\displaystyle z'_{0}=0}
Precise calculation of the critical pointEdit
{\displaystyle f(z)}
{\displaystyle z-f(z)Df(z)*/|Df(z)|^{2}}
{\displaystyle Df(z)}
{\displaystyle f(z)}
{\displaystyle \partial f(z)/\partial {x},\partial f(z)/\partial {y}}
{\displaystyle Df(z)*}
{\displaystyle f(z)=det(g'(x,y))=2x^{2}+y}
{\displaystyle (x+iy)-(2x^{2}+y)(4x+i)/(16x^{2}+1)}
Retrieved from "https://en.wikibooks.org/w/index.php?title=Pictures_of_Julia_and_Mandelbrot_Sets/Julia_and_Mandelbrot_sets_for_non-complex_functions&oldid=3547460"
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m (→Overview: Updated Intro)
WorldView-1, WorldView-2 and WorldView-3 are commercial earth observation satellites. Details about the sensors are provided at Digital Globe's collection of [http://www.digitalglobe.com/resources/satellite-information Spacecraft Data Sheets].
It is recommended -- actually it is a necessity! -- to have a look at the documentation explaining [http://www.digitalglobe.com/sites/default/files/Imagery_Support_Data_Documentation%20%281%29.pdf Digital Globe's products metadata]. Note, all of DigitalGlobe’s satellites collect data using an 11 bit dynamic range. For technical reasons, products are delivered as either 16-bit or 8-bit data.
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L_{\lambda {\text{Pixel, Band}}}={\frac {K_{\text{Band}}*q_{\text{Pixel, Band}}}{\Delta \lambda _{\text{Band}}}}}
{\displaystyle L_{\lambda {\text{Pixel,Band}}}}
{\displaystyle K_{\text{Band}}}
{\displaystyle q_{\text{Pixel,Band}}}
{\displaystyle \Delta _{\lambda _{\text{Band}}}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
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Basics of Linear Programming | Education Lessons
At the time of World War- II, G.B. Dantzing was working with US Air force and he was facing many problems such as allocation of weapons to different war locations, military logistics, etc. Since available resources are limited (restricted), it was very crucial to allocate optimum number of weapons or logistics in order to fulfil the objective of winning the war. At that time, he developed the technique to solve such problems, and that technique is known as Linear Programming (LP).
LP consists of two words – Linear and Programming. Linear word represents that relationship between variables is linear. Programming represents mathematical modelling and development of algorithms to solve a problem that involves optimum allocation of limited or restricted resources.
Linear Programming is a mathematical modelling for optimization of a function (Objective Function), subject to restricted resources of variables (represented ‘n’ form of linear equations and/or inequalities).
Objective function may be to maximize profit or to minimize loss or any other measures which to be obtained in the best optimal manner. Constraints are restrictions on available man hours, materials, money, machines, etc.
In general, if the resources are unlimited, then there will not need to define any kind of strategy. Since, resources are limited, there should be a well-defined strategy.
Pre-requirements of LP
Decision variables should be interrelated and non-negative. Decision variables are those variables which are decided at the starting of solving the sum.
There must be a well-defined objective function, which can be represented as a linear function of decision variables.
There must be constraints on amount of resources which can be expressed as linear inequalities in terms of decision variables.
Properties/Assumptions of LP
Tip: As suggested in video, properties of LP can be easily remembered as keyword
{PANC}^2
The amount of each resource consumed and its contribution to profit in objective function must be proportional to the value of each decision variable. For eg: If one egg can provide 8 g of protein, then 10 eggs can provide 80 g (10X8=80 g) of protein.
The total value of the objective function equals the sum of the contribution of each variable to the objective function. Similarly, total resource use is the sum of the resource use of each variable. For eg: Total profit by selling ‘n’ products must be equal to the profit earned by selling the items individually. On the other hand, the total cost of manufacturing ‘n’ products at a time, must be equal to the cost of manufacturing items individually.
It is assumed that values of decision variables are either positive or zero. It cannot be negative.
It is assumed that the solution value of decision variables and the amount of resources used are not be integer values. For eg: It is not possible to have 3.5 as the optimum number of products.
Certainty of coefficients:
In all LP models, the coefficients of objective function, R.H.S. coefficients of constraints and resources values are certainly and precisely known and measurable. Also, it is assumed that those values remain constant irrespective of time.
Optimum allocation of resources: LP shows the allocation of resources to fulfil the objective function and helps in making the optimum use of available resources.
Improvement in quality of decision: LP improves decision quality as it gives accurate values of decision variables to fulfil the objective function.
Exploring bottleneck machines: Bottleneck machines are those machines which cannot meet the demand and because of them other machines are idle. LP model identifies this bottleneck machines and explores the problem of low production capacity of the plant.
Re-evaluation of basic plan: LP re-evaluates the basic plan as per changing conditions. If condition change when the plan is partly carried out, they can be determined so as to adjust the remainder of the plan for best result.
Limitations of LP:
LP treats relationships between variables as Linear. However, always it is not possible to have linearity.
LP assumes that outcome of problem has decision variables as integers.
LP assumes the certainty of the coefficients, but in each case it is not possible to determine the values of coefficients.
LP deals with single objective functions, where as in real life situation business may have multiple objectives.
LP ignores the effect of time.
Applications of LP:
Now-a-days, LP has wide applications in different sectors like industrial, management, farming, financial and many other miscellaneous applications.
- Product Mix: Company produces number of products from same limited production resources. In such cases it is essential to determine the quantity of each product to be manufactured knowing the profit induced and amount of materials consumed by each of them. The objective is to maximize profit subject to all constraints. For example, wafer manufacturing companies like Balaji.
Image source: http://www.walkthroughindia.com/
- Blending problems: When a product can be made from a variety of raw materials, each of which has a particular composition and price. The objective here is to determine the minimum cost blend subject to availability of the raw materials and the minimum constraints on certain product constituents.
For example, we are producing some product-X, and for which we have four different raw materials- A, B, C & D. Now, product-X can be made of A blend with B, as well as B with D and A with C. Now, for each particular raw material has different price, so compositions have their different price too. LP helps in selecting optimum composition and its price, subjected to constraints.
Image illustrating Blending Problem
- Trim loss: When an item can be made of standard size (like glass, paper,sheet, etc.), the problem that arises is to determine which combination of requirements should be produced from standard materials in order to minimize the trim loss (wastage).
- Production Planning: This deals with the determination of minimum cost production plan over a planning period of an item with a fluctuating demand considering the initial number of units in inventory, production capacity,constraints on production, manpower and all relevant cost factors. The objective here is to minimize total operational costs.
- Media selection: LP techniques can helps in determining the advertising media mix so as to maximize the effective exposure, subject to limitation of budget, specified exposure rates to different market segments, specified minimum and maximum number of advertisements in various media.
- Travelling salesman problem: The problem of salesman is to find the shortest route starting from a given city, visiting each specified cities and then returning to the original point of departure, provided no city shall be visited twice during the tour.
- Physical distribution: LP determines the most economic and efficient manner of locating manufacturing plants and distribution centers for physical distribution.
3) Agricultural Applications:
- Farm management: LP can be applied in agricultural planning like allocation of limited resources such as available land, labour, water supply and working capital, etc. in such a way so as to maximize net revenue.
- Capital budgeting problems: This deals with the selection of specific investment activity among several other activities.
- Profit planning: This deals with the maximization of profit margin from investment in plant facilities and equipment, cash on hand and inventory.
5) Miscellaneous problems:
- Diet problems: Diet problem includes the optimization of food quantities in order to fulfil the nutrition proportions as well as to minimize the cost of the food. Dieticians can use LP for planning a balance diet for a particular patient.
- Inspection problems: These problems relates with the deciding the optimum number of inspectors in order to inspect the raw materials. Also, the cost of inspectors should be minimum and reliability requires being high. This helps in deciding the optimum time to move on machine inspection based on require output, so as to increase the profit by decreasing the manufacturing time of particular product.
- Military Applications: Military application includes the problem of allocation of limited tools and technologies to various locations.
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In this article, we learn about syntax-directed translation schemes, a complementary notation to syntax-directed definitions.
Postfix translation schemes.
Parser-stack implementation of postfix SDTs.
SDTs with actions inside productions.
Eliminating left-recursion.
SDTs for L-Attributed definitions.
A translation scheme is a context-free grammar whereby semantic rules are embedded within the right sides of productions.
A translation schema and a syntax-directed definition are close to being similar except that the order for evaluation of semantic rules is shown.
When we construct a parse tree for a translation scheme, we use an extra child that is connected by a dashed line to the node of the production to indicate an action.
In the parse tree, the order in which actions appear is the order in which they are executed.
These are SDTs that have all their actions at the right ends of the production body.
The following is an example of a postfix SDT implementation of a calculator;
Since the underlying grammar is LR and the SDD is S-attributed, actions can be correctly performed along with reduction steps of the parser.
Postfix SDTs can be implemented during LR parsing by executing actions when reductions occur. Attributes of each grammar symbol can be placed on the stack in a place they can be found during reduction. The best way is to place attributes along with grammar symbols in records on the stack itself.
The following is a parser that contains records with a field for a grammar symbol and below it a field for an attribute.
The grammar symbols XYZ are to be reduced according to production A
\to
XYZ.
The desk calculator in a bottom-up parsing stack;
Take for example a production B
\to
X{a} Y. Action a is performed after we recognize X(X is a terminal) or all terminals derived from X(X is a non-terminal)
We insert marker non-terminals so as to remove embedded actions and change the SDT into a postfix SDT then rewrite the product with marker non-terminal M into B
\to
XMY
\to
\epsilon
{a}.
Note that inserting marker non-terminals may introduce conflicts in the parse table.
Any SDT can be implemented as listed;
First we ignore actions, parse the input and return a parse tree as the result.
We then examine each interior node N, for a production A
\epsilon
\alpha
. Add more children to N for the actions in
\alpha
, therefore the children of N from left to right will have exactly symbols and actions of
\alpha
Finally we perform a pre-order traversal of the tree and when we visit a node labeled by an action, we perform the action.
The following is a parse tree for an expression 3 * 5 + 4;
No grammar with left-recursion can be parsed deterministically in a top-down manner.
The following principle guides us;
When the order in which actions in an SDT is considered, actions are treated like terminal symbols during grammar transformation.
This principle is based on the idea that grammar transformations preserve the order of the terminals in the generated string.
Actions are therefore executed in any left-to-right, top-down, or bottom-up ordering.
The key to eliminating left recursion is to take two productions;
\to
\alpha
\beta
That generate strings that consist of a
\beta
and any number of
\alpha
and replace them by productions that generate a similar string using a new non-terminal R of the first production;
\to
\beta
\to
\alpha
ϵ
\beta
does not start with A, A no longer has a left-recursive production.
We assume a grammar can be parsed top-down. Rules for converting an L-Attributed SDD to an SDT are as follows;
We embed the action that computes inherited attributed for a non-terminal A that comes immediately before the occurrence of A in the production body. If several inherited attributes for A depend on one another in an acyclic order, the evaluation of those attributes is computed first.
Secondly, we place actions that compute a synthesized attribute for the head of a production at the end of the production body.
Given the grammar;
\to
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
| (
{\mathrm{B}}_{1}
) | text
Although the above grammar is ambiguous we can use it to parse in a bottom-up manner, that is, if we make subscribing and juxtaposition right-associative with the former taking precedence over the latter.
The SDD for typesetting boxes;
The SDT for typesetting boxes;
Compilers Principles, Techniques, & Tools - Alfred V. Aho, Monica S. Lam
Applications for Syntax-Directed Translation
In this article, we learn about the main application of Syntax-directed Translation which is the construction of syntax trees.
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(→Bandlimitation and timing: Squares like LaTeX, more.)
Well, we can say it's a waveform that's
some positive value for half a cycle and then transitions
instantaneously to a negative value for the other half. But that doesn't
really tell us anything useful about how this input
becomes this output .
Then we remember that [any waveform is also [[WikiPedia:Fourier_series|the sum of discrete frequencies]],
\ x(t) = \begin{cases} 1, & |t| < T_1 \\ 0, & T_1 < |t| \leq {1 \over 2}T \end{cases}
and a square wave is particularly simple sum: a fundamental and an
infinite series of [[WikiPedia:Even_and_odd_functions#Harmonics|odd harmonics]]. Sum them all up, you get a
At first glance, that doesn't seem very useful either. You have to sum
up an infinite number of harmonics to get the answer. Ah, but we don't
have an infinite number of harmonics.
{\displaystyle \ x(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}x_{\mathrm {square} }(t)&{}={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+{\frac {4}{7\pi }}\sin(7\omega t)+\cdots \end{aligned}}}
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Radiation astronomy/Opticals/Quiz - Wikiversity
Radiation astronomy/Opticals/Quiz
< Radiation astronomy/Opticals
Optical astronomy is a lecture and an article as part of the astronomy department course on the principles of radiation astronomy.
You are free to take this quiz based on optical astronomy at any time.
2 True or False, The symbol
{\displaystyle \odot }
3 Astronomy is not purely mathematics because of which phenomena?
7 The atmosphere of Titan which has a natural orange color is composed largely of?
11 With respect to the zodiacal light which group of middle-easterners worshipped it?
12 True or False, When Venus is viewed in the ultraviolet, its color appears brownish.
21 Various optical radiation observatories occur at different altitudes and geographic locations due to what effect?
24 True or False, Optical telescopes use domes as a form of shelter.
33 Being outside during fair weather in the day light looking upward when the Sun is off to the East or West, you may see that the sky is what color.
35 Yes or No, Hydrogen has an emission line in the yellow.
Retrieved from "https://en.wikiversity.org/w/index.php?title=Radiation_astronomy/Opticals/Quiz&oldid=2146681"
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Cellular Automaton Practice Problems Online | Brilliant
In Classical Mechanics, you're going to learn how to predict the motion of classical systems like springs, rockets, and helicopters by thinking in terms of force and energy. But physics is more than an accounting system for forces that predicts how objects move.
At a deeper level, it is an optimistic belief that there are rules that take systems from one moment to the next, and a way of thinking that helps us find these rules.
A "glider" in the 2-dimensional cellular automata called Conway's Game of Life
In this quiz, we're going to look at systems called cellular automata that live and die according to simple rules on a lattice. By looking at these toy systems, we will form a clear picture of what the machinery of classical mechanics does and how it is able to describe so much of the world around us.
An important part of physics is deciding what information to keep track of and what to ignore. Consider, for example, air drag on a car. Billions of individual air molecules collide with a car's surface to produce a drag force, but modeling every collision would be mathematically overwhelming, even with the help of a modern computer.
The art of physics is distilling the essential qualities of a complex system into a simpler model without sacrificing predictive accuracy. Many of physics' most famous spokespersons have described this model-building process, but no one advocated for this doctrine as strongly as physicist Richard Feynman:
"Others... make guesses that are very complicated, and it sort of looks as if it is all right, but I know it is not true because the truth always turns out to be simpler than you thought."
In other words, no amount of calculation or mathematical sophistication is stand-in for developing physical intuition. In this course, we uphold Feynman's ideal as the light to guide us on our path.
Each time step, a square updates its color using the Update Rule according to the current state of its trio.
Cellular automata are a type of system that takes Feynman's doctrine to the extreme, unfolding in time according to a set of update rules that can be listed on a table. Each colored square simply looks at itself and each of its neighbors, and uses the information on the table to update its color. For simplicity, we're going to stick with
1
-dimensional automatons where each square has two neighbors.
We call each square, plus its two neighbors, a trio. If we want to define a "physics" for a cellular automaton, we need an update rule for each possible trio. For example, if a square's trio were
\{{\color{#66ff00}{\blacksquare}},{\color{#66ff00}{\blacksquare}},{\color{#333333}{\blacksquare}}\}
it would update to
\color{#333333}{\blacksquare}
according to the update rules in the diagram.
How many possible trios are there?
4
6
8
12
Once the update rule is complete with an entry for all possible trios, the dynamics of a cellular automaton are nearly unambiguous.
The last thing required is the starting color of each square. In physics, this is known as the initial state of a system. With an update rule and an initial state in hand, we have everything we need to predict any future state—the color of each square in our cellular automata.
Suppose the initial state of our system is a single green square:
What color will the green square be at the next time step using the update rule below?
\color{#66ff00}{\blacksquare}
\color{#333333}{\blacksquare}
The dynamics of a cellular automaton are fully specified by its initial state and an update rule, which encodes the state of any square one step into the future.
Nothing stops us from applying the update rule repeatedly to determine the state of the automaton at much later times. We can even string together all the states into a movie of the dynamics.
Can you figure out which of the movies below shows a cellular automaton that's following this update rule that we used in the previous question?
Note: the initial state of the three cellular automata below is all squares black, except for the middle square which is green.
\mathbf{A}
\mathbf{B}
\mathbf{C}
We just saw a particular update rule that resulted in squares streaming to the left, kind of like a particle. If we wanted, we could flip that update rule in a mirror to find squares that stream to the right.
But what if we wanted an update rule that could accommodate left and right moving squares, like the system below?
For a 1-dimensional cellular automaton that updates according to the current state of its trios, there are
256
possible update rules (
2^8
), which we can think of as
256
possible "physics".
Is there an update rule among the
256
that permits the rebound we see above? (try to make one!) If not, what modification would we need to make in order to see such phenomena?
Allow the update rule to keep track of previous time steps Allow the update rule to consider more than two neighbors Nothing new is needed, there's a one-step rule that can do it
The magic of cellular automata is that the update rule reflects all the information we know about the system, as well as any assumptions we've made about it. What's more is that the update rule encodes all the possible motions in the system.
But are there such update rules in "real" physics? Consider Newton's law of motion. It tells us that force is equal to mass times acceleration,
F = m a.
Is this an update rule that connects two states of a system? Remember that acceleration is simply the change in velocity divided by the change in time:
a = \Delta v / \Delta t.
Taking this into account, Newton's law becomes
\boxed{v_\text{final} = v_\text{initial} + \dfrac{F\Delta t}{m}},
a simple update rule!
Reduced to its essence, classical mechanics is a two-fold endeavor: understanding the lessons this update rule contains about the natural world, and learning to solve it for the forces we find in nature.
There is one caveat that we glossed over. In nature it seems that time and space aren't discrete like they are in cellular automata, they're continuous. This means we have to make the timestep
\Delta t
infinitesimally small, which changes the discrete update rule into a differential equation:
m\frac{dv}{dt} = F.
Although this will be our only encounter with cellular automata in this course, they are an archetype of how complexity arises from simple update rules. For instance, consider the cellular automaton better known as the Game of Life, which has the following update rule:
Any green cell with exactly 2 or 3 green neighbors stays green
Any other green cell turns black
Any black cell with exactly 3 green neighbors turns green
The simplicity of the update rule is superficial. Hidden beneath is a rich universe of lattice creatures that it encodes, which contains patterns that move themselves across the lattice, patterns that spit out other patterns, and patterns that make copies of themselves.
As but one example, the contraption below generates lattice gliders at regular intervals:
Despite the simplicity of their update rules, cellular automata can give rise to complex behavior.
In the final quiz, we'll use Newton's laws, the central update rule in Classical Mechanics, to illustrate complexity encoded by simple rules—the mysterious case of two clocks talking to each other through a wall.
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Iota and Jot - Wikipedia
Iota, Jot, Zot
Formal language, Turing tarpit, esoteric
2001 / 2001; 21 years ago (2001)
Scheme, JavaScript
Scheme interpreter, Web browser (JavaScript)
www.nyu.edu/projects/barker
In formal language theory and computer science, Iota and Jot (from Greek iota ι, Hebrew yodh י, the smallest letters in those two alphabets) are languages, extremely minimalist formal systems, designed to be even simpler than other more popular alternatives, such as the lambda calculus and SKI combinator calculus. Thus, they can also be considered minimalist computer programming languages, or Turing tarpits, esoteric programming languages designed to be as small as possible but still Turing-complete. Both systems use only two symbols and involve only two operations. Both were created by professor of linguistics Chris Barker in 2001. Zot (2002) is a successor to Iota that supports input and output.[1]
1 Universal iota
Universal iota[edit]
Chris Barker's universal iota combinator ι has the very simple λf.fSK structure defined here, using denotational semantics in terms of the lambda calculus,
{\displaystyle \iota :=\lambda f.((fS)K):=\lambda f.((f\lambda a.\lambda b.\lambda c.((ac)(bc)))\lambda d.\lambda e.d)}
From this, one can recover the usual SKI expressions, thus:
{\displaystyle I\,=\,(\iota \iota ),\;K\,=\,(\iota (\iota (\iota \iota ))),\;S\,=\,(\iota (\iota (\iota (\iota \iota ))))}
Because of its minimalism, it has influenced research concerning Chaitin's constant.[2]
Iota is the LL(1) language that prefix orders trees of the aforementioned Universal iota ι combinator leafs, consed by function application ε,
iota = "1" | "0" iota iota
so that for example 0011011 denotes
{\displaystyle ((\iota \iota )(\iota \iota ))}
, whereas 0101011 denotes
{\displaystyle (\iota (\iota (\iota \iota )))}
Jot[edit]
Jot is the regular language consisting of all sequences of 0 and 1,
jot = "" | jot "0" | jot "1"
The semantics is given by translation to SKI expressions. The empty string denotes
{\displaystyle I}
{\displaystyle w0}
{\displaystyle (([w]S)K)}
{\displaystyle [w]}
is the translation of
{\displaystyle w}
{\displaystyle w1}
{\displaystyle (S(K[w]))}
The point of the
{\displaystyle w1}
case is that the translation satisfies
{\displaystyle (([w1]A)B)=([w](AB))}
for arbitrary SKI terms
{\displaystyle A}
{\displaystyle B}
{\displaystyle [w11100]=(([w1110]S)K)=(((([w111]S)K)S)K)=((([w11](SK))S)K)=(([w1]((SK)S))K)=([w](((SK)S)K))=([w]K)}
holds for arbitrary strings
{\displaystyle w}
{\displaystyle [w11111000]=(((((([w11111]S)K)S)K)S)K)=([w](((((SK)S)K)S)K))=([w]S)}
holds as well. These two examples are the base cases of the translation of arbitrary SKI terms to Jot given by Barker, making Jot a natural Gödel numbering of all algorithms.
Jot is connected to Iota by the fact that
{\displaystyle [w0]=(\iota [w])}
and by using the same identities on SKI terms for obtaining the basic combinators
{\displaystyle K}
{\displaystyle S}
Zot[edit]
The Zot and Positive Zot languages command Iota computations, from inputs to outputs by continuation-passing style, in syntax resembling Jot,
zot = pot | ""
pot = iot | pot iot
iot = "0" | "1"
where 1 produces the continuation
{\displaystyle \lambda cL.L(\lambda lR.R(\lambda r.c(lr)))}
, and 0 produces the continuation
{\displaystyle \lambda c.c\iota }
, and wi consumes the final input digit i by continuing through the continuation w.
^ Barker, Chris. "Zot". The Esoteric Programming Languages Webring. Archived from the original on 12 March 2016. Retrieved 4 October 2016.
^ Stay, Michael (August 2005). "Very Simple Chaitin Machines for Concrete AIT". Fundamenta Informaticae. IOS Press. 68 (3): 231–247. arXiv:cs/0508056. Bibcode:2005cs........8056S. Retrieved 20 February 2011.
Barker, Chris. "Iota and Jot: the simplest languages?". The Esoteric Programming Languages Webring. Archived from the original on 7 May 2016. Retrieved 13 August 2004.
https://esolangs.org/wiki/Iota
https://esolangs.org/wiki/Jot
https://esolangs.org/wiki/Zot
Retrieved from "https://en.wikipedia.org/w/index.php?title=Iota_and_Jot&oldid=1066169439"
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One pass code generation.
Backpatching for boolean expressions.
Flow-control statements.
Break, continue and goto statements.
An issue with generating code for boolean expressions or flow control statements involves matching a jump instruction with the target of the jump.
The translation of a boolean expression (B) S contains a jump when B is false , to instruction following the the code for S.
During a one-pass translation, B is translated before S is examined therefore this begs the question, what is the target of the goto that jumps over the code for S?
This problem can be solved by passing labels as inherited attributes where relevant jump instructions were previously generated, although for this we need a separate pass, In the following sections, we shall look at a complementary approach referred to as backpatching.
In backpatching, lists of jumps are passed as synthesized attributes.
When a jump is generated, the target of the jump is left unspecified. Such jumps are placed in a list of jumps whose labels are to be filled when a label can be determined.
All jumps in the list will have the same target label
One-pass code generation.
Backpatching is used to generate code for boolean expressions and flow control statements.
With backpatching synthesized attributes truelist and falselist of a nonterminal B are used to manage labels in the jumping code for boolean expressions.
That is, B.truelist is a list of jump or conditional jump instructions into which we insert the label that control goes to if B is true and B.falselist is a list of instructions that gets the label to which control goes to when B is false.
During code generation for B, jumps to true and false exits stay incomplete with the label unfilled, these jumps are then placed in a list pointed to by B.truelist and B.falselist.
In a similar manner a statement S has a synthesized attribute s.nextlist that denotes a list of jumps to the instruction following the code for S.
To manipulate the lists we use the following three functions:
makelist(i) to create a new list with only an index i into the array of instructions. It returns a pointer to the new list.
merge(p1, p2), to concatenate the lists pointed to by p1 and p2 and return a pointer to this list.
backpatch(p, i) to insert i as the target label for each instruction on the list pointer to by p.
Backpatching Boolean expressions.
We construct a translation scheme for generating code for boolean expressions during bottom-up parsing.
We have the following grammar:
B →
{\mathrm{B}}_{1}
|| M
{\mathrm{B}}_{2}
{\mathrm{B}}_{1}
&& M
{\mathrm{B}}_{2}
| !
{\mathrm{B}}_{1}
| !(
{\mathrm{B}}_{1}
) |
{\mathrm{E}}_{1}
{\mathrm{E}}_{2}
| true | false M → ϵ
M is a marker nonterminal that invokes a semantic action to pick up the index of the next instruction that is to be generated at appropriate times.
The translation scheme for boolean expressions is as follows:
Consider a semantic action (1) for the production B →
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
{\mathrm{B}}_{1}
is true, B is also true therefore
{\mathrm{B}}_{1}
.truelist become part of B.truelist.
{\mathrm{B}}_{1}
is false, we test
{\mathrm{B}}_{2}
, the target for the jumps
{\mathrm{B}}_{1}
.falselist must be the beginning of the code that is generated for
{\mathrm{B}}_{2}
We use the marker nonterminal M to get this target. This nonterminal produces as a synthesized attribute M.instr, the index of the next instruction before
{\mathrm{B}}_{2}
We associate the semantic action {M.instr = nextinstr;} with the production M → ϵ to obtain the instruction index.
nextinstr variable holds this index.
Now, this value is backpatched onto
{\mathrm{B}}_{1}
.falselist when we see the remainder of the production B →
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
The semantic action (2) for B →
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
is similar to (1).
The action (3) for B → !B swaps the true and false lists and action (4) ignores parentheses.
For simplicity action (5) generates two instructions, a conditional and an unconditional goto neither of which has its target filled, therefore they are placed on new lists pointed to by B.truelist and B.falselist respectively.
The image is an annotated parse tree for x < 100 || x > 200 && x != y.
From the above tree we learn the following:
Attributes truelists and falselists are represented by t and f respectively.
Actions are performed during a depth first search traversal of the tree and since they all appear at the ends of the right sides, they are performed together with reductions during bottom-up parsing.
The following instructions are generated for the reduction of x < 100 to B by production (5);
100: if x < 100 goto _
100: goto _
The marker nonterminal M in the production B →
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
records nextinstr value which at this point is 102.
The reduction x > 200 to B by production (5) generates the following instructions:
102: if x > 200 goto _
The subexpression x > 200 corresponds to
{\mathrm{B}}_{1}
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
The maker nonterminal records the current value of nextinstr now 104.
Reducing x != y by production (5) yields;
104: if x != y goto _
Now to reduce
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
. Its semantic action calls backpatch(
{\mathrm{B}}_{1}
.truelist, M.instr) to bind the true exit of
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
which is the first instruction, since
{\mathrm{B}}_{1}
.truelist is {102} and M.instr 104, this call fills 104 in instruction 102.
Here are the generated instructions thus far;
102: if x > 200 goto 104
The semantic action that is associated with the final reduction by
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
calls backpatch({102}, {103}) which leaves the instructions as follows;
The expression is true only if the gotos of instructions 100 and 104 are arrived at.
It is false if gotos of instructions 103 and 105 are arrived at.
The instructions have their targets filled later in the compilation.
Now to translate flow-control statements using backpatching in a single pass.
We have the grammar;
S → if(B) S | if(B) S else S | while(B) S | {L} | A;
L → LS | S
S denotes a statement, L a list, A an argument and B a boolean.
The translation scheme is as follows;
The translations maintain a list of jumps filled in when their targets are found. Just like in the previous translation scheme, boolean expressions generated by nonterminal B have B.truelist and B.falselist jumps that correspond to the true and false exits from the code for B.
Statements generated by nonterminals S and L have a list of unfilled jumps that are given by nextlist attribute. These lists are completed during backpatching.
S.nextlist is a list of conditional and unconditional jumps to the instruction following the code statement S in execution order.
L.nextlist is defined similarly.
The two occurrences of the maker nonterminal M in production S → while
{\mathrm{M}}_{1}
{\mathrm{M}}_{2}
{\mathrm{S}}_{1}
record the instruction numbers of the beginning of the code for B and beginning of the code for
{\mathrm{S}}_{1}
The only production for M is M → ϵ, Action (6) in the translation scheme sets attribute M.instr to the number of the next instruction.
After the body of
{\mathrm{S}}_{1}
is executed, control flows to the beginning.
When we reduce while
{\mathrm{M}}_{1}
{\mathrm{M}}_{2}
{\mathrm{S}}_{1}
to S, we backpatch
{\mathrm{S}}_{1}
.nextlist to make all targets on the list
{\mathrm{M}}_{1}
.instr.
Since control may 'fall from the bottom', an explicit jump to the beginning of the code for B is appended after the code for
{\mathrm{S}}_{1}
B.truelist is backpatched to go to the start of S1 by making jumps on B.truelist go to
{\mathrm{M}}_{2}
We have a more compelling argument for using S.nextlist and L.nextlist when code is generated for the conditional statement if(B)
{\mathrm{S}}_{1}
{\mathrm{S}}_{2}
If control falls out from the bottom of S1, we use another marker nonterminal to generate this jump after S1.
Let N be this nonterminal maker with the production N → ϵ. N has N.nextlist attribute which is a list consisting of the instruction number of the jump goto_ generated by action(7) for N.
Action(2) deals with if-else-statements with the syntax;
S → if(B)
{\mathrm{M}}_{1}
{\mathrm{S}}_{1}
N else
{\mathrm{M}}_{2}
{\mathrm{S}}_{2}
We backpatch the jumps when B is true to the instruction
{\mathrm{M}}_{1}
.instr, similarly we backpatch jumps when B is false to the start of
{\mathrm{S}}_{2}
The list S.nextlist includes all jumps out of S1, S2 and the jump generated by N.
Actions(8) and (9) are responsible for handling statements.
In L →
{\mathrm{L}}_{1}
MS, the instruction following the code for L1 is the begining of S.
{\mathrm{L}}_{1}
.nextlist is backpatched to the start of the code for S given by M.instr.
In L → S, L.nextlist is similar to S.nextlist.
No new instructions are generated in these semantic rules except for rules (3) and (7). Semantic actions that are associated with assignment-statements and expressions generate the rest of the code.
Flow of control causes proper backpatching such that assignments and boolean expressions evaluated connect properly.
The goto statement is the most elementary programming language construct used to change the flow of control.
These statements are implemented by maintaining a list of unfilled jumps for each label then backpatching the target when it is known.
In Java, there are no goto statements, however, there are break statements that send control out of an enclosing construct and continue statements that trigger the next iteration of an enclosing loop.
1. for(; ; readch()){
2. if(peek == ' ' || peek == '\t') continue;
3. else if(peek == '\n') line = line + 1;
4. else break;
From the excerpt above, control jumps from the break statement at line 4 to the next statement after the for loop.
Control also jumps from continue statement at line 2 to evaluate readch() followed by the if statement on line 1.
If S is an enclosing loop, then a break statement is a jump to the first instruction after the loop.
We can generate code for the break statement by tracking the enclosing loop statement S, generating an unfilled jump for the break statement and putting the unfilled jump of S.nextlist.
In a two-pass front end, S.nextlist is implemented as a field in the node for S. To track S we use a symbol table. With this approach we can also handle labeled break statement in java since the table maps the label to a node in the syntax tree.
Alternatively, instead of using a symbol table we can use a pointer to S.nextlist in the symbol table and when we reach a break statement, we generate an unfilled jump, look up the nextlist in the symbol table and jump to the list where it is backpatched.
Continue statements are handled in an analogous manner compared to break statements. The main difference between them is that the target of the generated jump will be different.
Now we have an idea of how backpatching works for boolean expressions, flow control statements such as if-statements and loops and break, continue and goto statements.
We have looked at three main functions, makelist(i), merge(p1, p2) and backpatch(p, i) and their roles in the backpatching process.
Compilers Principles, Techniques, & Tools Alfred V. Aho, Jeffrey D. Ullman
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Which set of ordered pairs could be generated by an
Which set of ordered pairs could be generated by an exponential function?
\left(1,1\right)\left(2,\frac{1}{2}\right)\left(3,\frac{1}{3}\right)\left(4,\frac{1}{4}\right)
\left(1,1\right)\left(2,\frac{1}{4}\right)\left(3,\frac{1}{9}\right)\left(4,\frac{1}{16}\right)
\left(1,\frac{1}{2}\right)\left(2,\frac{1}{4}\right)\left(3,\frac{1}{8}\right)\left(4,\frac{1}{16}\right)
\left(1,\frac{1}{2}\right)\left(2,\frac{1}{4}\right)\left(3,\frac{1}{6}\right)\left(4,\frac{1}{8}\right)
A=1173.1{e}^{0.008t}
A=31.5{e}^{0.019t}
A=127.3{e}^{0.006t}
A=141.9{e}^{0.005t}
\begin{array}{|cc|}\hline x& y\\ 1& 2\\ 2& 4\\ 3& 8\\ 4& 16\\ 5& 32\\ \hline\end{array}
Write a linear
\left(y=mx+b\right)
, quadratic
\left(y=ax2\right)
, or exponential
\left(y=a\left(b\right)x\right)
function that models the data.
y=
Determine whether the function given by the table is linear, exponential, or neither. If the function is linear, find a linear function that models the data. If it is exponential, find an exponential function that models the data. x f(x) -1 8/7 0 8 1 56 2 392
Models that increase or decrease by a percentage will be exponential. Which of the following scenarios will use exponential modeling?
A) A person makes $50,000 per year and then has a salary increase of $6,000 for each year of experience.
B) A person makes $50,000 per year and their salary will increase by 2.5% each year.
Determine whether the given function is linear, exponential, or neither. For those that are linear functions, find a linear function that models the data; for those that are exponential, find an exponential function that models the data.
\begin{array}{}x& \text{g(x)}\\ \text{-1,-1}& \text{3}\\ \text{0}& \text{6}\\ \text{1}& \text{12}\\ \text{2}& \text{18}\\ \text{3}& \text{30}\end{array}
CBU Tuition
Let C(t)C(t) be the cost of tuition at CBU in dollars for t years since 1990.
A quadratic model for the data is
C(t)C(t) = .
Use three decimals in your answer.
Estimate the tuition in 2016. $
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On a trip, the total distance (in miles) you travel
On a trip, the total distance (in miles) you travel in x hours is represented by
On a trip, the total distance (in miles) you travel in x hours is represented by the piccewise function.
d\left(x\right)=\left\{\begin{array}{ll}55x,& if\text{ }0\le x\le 2\\ 65x-20,& if\text{ }2<x\le 5\end{array}
How far do you travel in 4 hours?
Calculation: The given function:
d\left(x\right)=\left\{\begin{array}{ll}55x,& if\text{ }0\le x\le 2\\ 65x-20,& if\text{ }2<x\le 5\end{array}
Use the second equation because
2<4<5
d\left(x\right)=65x-20
d\left(4\right)=65\left(4\right)-20
d\left(4\right)=260-20
d\left(4\right)=240
Therefore the distance travel in 4 hours is 240 miles.
y=3.5x+2.8
Discuss the data modeling concepts in DynamoDB.
Use the strategy for solving word problems modeling the verbal conditions of the problem with a linear inequality. On two examinations, you have grades of 86 and 88. There is an optional final examination, which counts as one grade. You decide to take the final in order to get a course grade of A, meaning a final average of at least 90. What must you get on the final to earn an A in the course?
You and your friend drive toward each other. The equation
50h=190-45h
represents the number h of hours until you and your friend meet. When will you meet?
For Questions
1-2
, use the following. Scooters are often used in European and Asian cities because of their ability to negotiate crowded city streets. The number of scooters (in thousands) sold each year in India can be approximated by the function
N=61.86{t}^{2}—237.43t+943.51
where f is the number of years since 1990. 1. Find the vertical intercept. What is the practical meaning of the vertical intercept in this situation? 2. Use a numerical method to find the year when the number of scooters sold reaches 1 million. (Note that 1 million is 1,000 thousand, so
N=1000
) Show three rows of the table you used to support your answer and write a clear answer to the problem.
The value of the variable bysolving the exponential equation.
The given exponential equation is
4{e}^{x}=5
|
(→Bandlimitation and timing: move graphics around)
* Panel 0: This panel presents buttons that allow the sound card to be configured in several sampling rates and bit depths. Samples read from the audio inputs are sent to the output pipes and audio outputs for playback without modification.
{\displaystyle \ x(t)={\begin{cases}1,&|t|<T_{1}\\0,&T_{1}<|t|\leq {1 \over 2}T\end{cases}}}
{\displaystyle {\begin{aligned}x_{\mathrm {square} }(t)={\frac {4}{\pi }}\sin(\omega t)+{\frac {4}{3\pi }}\sin(3\omega t)+{\frac {4}{5\pi }}\sin(5\omega t)+\\{\frac {4}{7\pi }}\sin(7\omega t)+{\frac {4}{9\pi }}\sin(9\omega t)+{\frac {4}{11\pi }}\sin(11\omega t)+\\{\frac {4}{13\pi }}\sin(13\omega t)+{\frac {4}{15\pi }}\sin(15\omega t)+{\frac {4}{17\pi }}\sin(17\omega t)+\\{\frac {4}{19\pi }}\sin(19\omega t)+{\frac {4}{21\pi }}\sin(21\omega t)+{\frac {4}{23\pi }}\sin(23\omega t)+\\{\frac {4}{25\pi }}\sin(25\omega t)+{\frac {4}{27\pi }}\sin(27\omega t)+{\frac {4}{29\pi }}\sin(29\omega t)+\\{\frac {4}{31\pi }}\sin(31\omega t)+{\frac {4}{33\pi }}\sin(33\omega t)+\cdots \end{aligned}}}
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Conic Sections/Identifying Conics - Wikibooks, open books for an open world
Conic Sections/Identifying Conics
< Conic Sections
1 Identifying Conics
1.1.1 Example 1: (Ellipse) '"`UNIQ--postMath-00000006-QINU`"'
1.1.2 Example 2: '"`UNIQ--postMath-00000019-QINU`"'
1.1.3 Example 3: '"`UNIQ--postMath-0000001A-QINU`"'
1.1.4 Example 4: '"`UNIQ--postMath-0000001B-QINU`"'
1.1.5 Example 5: (Rotated Ellipse) '"`UNIQ--postMath-0000001C-QINU`"'
Identifying Conics[edit | edit source]
Conics are reasonably easy to graph if they are given in their standard form. Unfortunately, this is often not the case. Conics can be represented by polynomials of the form:
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}
, and to graph these it is necessary to convert them back to their standard form. Before you do this, though, the formula:
{\displaystyle B^{2}-4AC}
can be used to determine the type of conic from the original equation before you start graphing:
{\displaystyle B^{2}-4AC=0}
: Parabola
{\displaystyle B^{2}-4AC<0}
{\displaystyle B^{2}-4AC>0}
: Hyperbola
Firstly, if B is not zero then the graph represents a rotated conic. Follow the instructions given on Rotation of Axes to determine how to convert this to a non-rotated form. Then convert this to the standard form of a conic as detailed later in this section, and graph it on a set of axes at the appropriate angle to the x- and y- axes.
If B is equal to zero, then use completing the square and algebraic manipulation to obtain standard forms of the conics.
Example 1: (Ellipse)
{\displaystyle {\frac {1}{9}}x^{2}+{\frac {1}{25}}y^{2}-{\frac {2}{3}}x-{\frac {8}{25}}y+{\frac {16}{25}}=0}
It is first necessary to complete the square for the x and y terms. First we will start with x. While it appears that in this case b is
{\displaystyle -{\frac {2}{3}}}
, you must remember that the coefficient of
{\displaystyle x^{2}}
must be equal to 1. Therefore:
{\displaystyle {\frac {1}{9}}x^{2}-{\frac {2}{3}}x}
{\displaystyle ={\frac {1}{9}}[x^{2}-6x]}
Now, b = -6.
{\displaystyle \left({\frac {b}{2}}\right)^{2}=\left({\frac {-6}{2}}\right)^{2}=9}
{\displaystyle ={\frac {1}{9}}[x^{2}-6x+9-9]}
{\displaystyle ={\frac {1}{9}}[(x-3)^{2}-9]}
{\displaystyle ={\frac {1}{9}}(x-3)^{2}-1}
Doing the same for the y:
{\displaystyle {\frac {1}{25}}y^{2}-{\frac {8}{25}}y}
{\displaystyle ={\frac {1}{25}}[y^{2}-8y]}
{\displaystyle \left({\frac {b}{2}}\right)^{2}=\left({\frac {-8}{2}}\right)^{2}=16}
{\displaystyle ={\frac {1}{25}}[y^{2}-8y+16-16]}
{\displaystyle ={\frac {1}{25}}[(y-4)^{2}-16]}
{\displaystyle ={\frac {1}{25}}(y-4)^{2}-{\frac {16}{25}}}
These can then be placed back into the original equation:
{\displaystyle \left({\frac {1}{9}}(x-3)^{2}-1\right)+\left({\frac {1}{25}}(y-4)^{2}-{\frac {16}{25}}\right)+{\frac {16}{25}}=0}
{\displaystyle {\frac {(x-3)^{2}}{9}}+{\frac {(y-4)^{2}}{25}}=1}
Which is the standard form for an ellipse. To check our answer, we can use the formula
{\displaystyle B^{2}-4AC}
getting the values from the original equation. Substituting these in gives:
{\displaystyle (0)^{2}-4\left({\frac {1}{9}}\right)\left({\frac {1}{25}}\right)=-{\frac {4}{225}}}
which is less than zero. Therefore the conic is an ellipse, confirming our previous answer.
{\displaystyle Circle}
{\displaystyle Parabola}
{\displaystyle Hyperbola}
Example 5: (Rotated Ellipse)
{\displaystyle 6x^{2}+{\sqrt {3}}xy+7y^{2}-36=0}
The fact that it has an xy term means that this is a rotated conic. The angle that it has been rotated is given by:
{\displaystyle tan2\theta ={\frac {B}{A-C}}={\frac {({\sqrt {3}})}{(6)-(7)}}}
{\displaystyle 2\theta =atan\left({\frac {({\sqrt {3}})}{(6)-(7)}}\right)}
{\displaystyle 2\theta =atan(-{\sqrt {3}})}
{\displaystyle 2\theta ={\frac {-\pi }{3}}}
{\displaystyle \theta ={\frac {-\pi }{6}}}
This tells you that the conic has been rotated
{\displaystyle -{\frac {\pi }{6}}}
radians, or -30°, anticlockwise, which is the same as
{\displaystyle {\frac {\pi }{6}}}
radians or 30° clockwise about the origin.
You can then make the following substitutions to obtain the polynomial for an identical, non-rotated conic:
{\displaystyle x=Xcos\theta -Ysin\theta }
{\displaystyle y=Xsin\theta +Ycos\theta }
{\displaystyle \theta }
{\displaystyle -{\frac {\pi }{6}}}
. Make sure you do not confuse the sign as this will make further calculations invalid. the value for the angle used is always the amount anti-clockwise. Therefore, substituting
{\displaystyle \theta =-{\frac {\pi }{6}}}
into the previous will give you:
{\displaystyle x=Xcos{\frac {-\pi }{6}}-Ysin{\frac {-\pi }{6}}}
{\displaystyle y=Xsin{\frac {-\pi }{6}}+Ycos{\frac {-\pi }{6}}}
{\displaystyle x={\frac {\sqrt {3}}{2}}X-{\frac {1}{2}}Y}
{\displaystyle y=-{\frac {1}{2}}X+{\frac {\sqrt {3}}{2}}Y}
When these are substituted into the rotated equation, you get:
{\displaystyle 6\left({\frac {\sqrt {3}}{2}}X-{\frac {1}{2}}Y\right)^{2}+{\sqrt {3}}\left({\frac {\sqrt {3}}{2}}X-{\frac {1}{2}}Y\right)\left(-{\frac {1}{2}}X+{\frac {\sqrt {3}}{2}}Y\right)+7\left(-{\frac {1}{2}}X+{\frac {\sqrt {3}}{2}}Y\right)^{2}-36=0}
{\displaystyle 6\left({\frac {3}{4}}X^{2}+{\frac {\sqrt {3}}{2}}XY+{\frac {1}{4}}Y^{2}\right)+{\sqrt {3}}\left(-{\frac {\sqrt {3}}{4}}X^{2}+{\frac {1}{2}}XY+{\frac {\sqrt {3}}{4}}Y\right)+7\left({\frac {1}{4}}X^{2}-{\frac {\sqrt {3}}{2}}XY+{\frac {3}{4}}Y^{2}\right)-36=0}
When you expand the brackets and collect like terms, you end up with:
{\displaystyle \left(6\times {\frac {3}{4}}+{\sqrt {3}}\times {\frac {-{\sqrt {3}}}{4}}+7\times {\frac {1}{4}}\right)X^{2}+\left(6\times {\frac {\sqrt {3}}{2}}+{\sqrt {3}}\times {\frac {1}{2}}+7\times {\frac {-{\sqrt {3}}}{2}}\right)XY+\left(6\times {\frac {1}{4}}+{\sqrt {3}}\times {\frac {\sqrt {3}}{4}}+7\times {\frac {3}{4}}\right)Y^{2}-36=0}
{\displaystyle {\frac {11}{2}}X^{2}+0XY+{\frac {15}{2}}Y^{2}-36=0}
The fact that there is now no XY term means that this is an equation for a non-rotated conic. Now all that remains is a little algebraic manipulation to get it into its standard form:
{\displaystyle {\frac {11}{2}}X^{2}+{\frac {15}{2}}Y^{2}=36}
{\displaystyle {\frac {X^{2}}{\frac {2}{11}}}+{\frac {Y^{2}}{\frac {2}{15}}}=36}
{\displaystyle {\frac {X^{2}}{{\frac {2}{11}}\times 36}}+{\frac {Y^{2}}{{\frac {2}{15}}\times 36}}=1}
{\displaystyle {\frac {X^{2}}{\frac {72}{11}}}+{\frac {Y^{2}}{\frac {24}{5}}}=1}
Which is the standard form of an ellipse which you can use to find the axis lengths, eccentricity, foci and anything else. Remember though that this must be graphed at the angle found earlier of
{\displaystyle {\frac {-\pi }{6}}}
to the x-axis. You can use the same formulae from above to find the rotated location of points from the non-rotated ellipse:
The point (x,y) when rotated by an angle of
{\displaystyle \theta }
about the origin becomes (X,Y) where:
{\displaystyle X=xcos\theta +ysin\theta }
{\displaystyle Y=-xsin\theta +ycos\theta }
Retrieved from "https://en.wikibooks.org/w/index.php?title=Conic_Sections/Identifying_Conics&oldid=3240968"
Book:Conic Sections
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Range of 3 sin theta+4 cos theta+2= - Maths - Relations and Functions - 9247467 | Meritnation.com
Range of 3 sin theta+4 cos theta+2=?
\mathrm{Consider} \mathrm{the} \mathrm{expression}.\phantom{\rule{0ex}{0ex}} 3\mathrm{sin\theta }+4\mathrm{cos\theta }+2\phantom{\rule{0ex}{0ex}}\mathrm{Note} \mathrm{that} \mathrm{the} \mathrm{range} \mathrm{of} \mathrm{cosine} \mathrm{function} \left(\mathrm{cos\theta }\right) \mathrm{is} \left[-1,1\right]\phantom{\rule{0ex}{0ex}}\mathrm{Also}, \mathrm{the} \mathrm{range} \mathrm{of} \mathrm{sine} \mathrm{function} \left(\mathrm{sin\theta }\right) \mathrm{is} \left[-1,1\right]\phantom{\rule{0ex}{0ex}}\mathrm{This} \mathrm{implies} \mathrm{that},\phantom{\rule{0ex}{0ex}} -1\le \mathrm{sin\theta }\le 1 \mathrm{and} -1\le \mathrm{cos\theta }\le 1\phantom{\rule{0ex}{0ex}}\mathrm{This} \mathrm{further} \mathrm{implies} \mathrm{that},\phantom{\rule{0ex}{0ex}} -3\le 3\mathrm{sin\theta }\le 3 \mathrm{and} -4\le 4\mathrm{cos\theta }\le 4\phantom{\rule{0ex}{0ex}}\mathrm{So} \mathrm{the} \mathrm{minimum} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{expression} \mathrm{will} \mathrm{be}\phantom{\rule{0ex}{0ex}} -3-4+2=-5\phantom{\rule{0ex}{0ex}}\mathrm{and} \mathrm{the} \mathrm{maximum} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{expression} \mathrm{will} \mathrm{be}\phantom{\rule{0ex}{0ex}} 3+4+2=9\phantom{\rule{0ex}{0ex}}\mathrm{So} \mathrm{the} \mathrm{range} \mathrm{of} 3\mathrm{sin\theta }+4\mathrm{cos\theta }+2 \mathrm{is} \left[-5,9\right]
let theta = x (for typing comforts)
Meethi, there is a formula .... for f(x) = a sinx + b cosx + c ..... range of f(x) is from "-1 * [root(a^2 + b^2)] + c" to "root(a^2 + b^2) + c" (closed intervals)
so using that ..... ans is [-3,7]
Hope it helps u!!! .... if it does .... do thumbs up
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LMIs in Control/Matrix and LMI Properties and Tools/Passivity and Positive Realness - Wikibooks, open books for an open world
LMIs in Control/Matrix and LMI Properties and Tools/Passivity and Positive Realness
This section deals with passivity of a system.
{\displaystyle {\begin{aligned}\ {\dot {x}}=Ax+Bu\\\ y=Cx+Du\\\end{aligned}}}
{\displaystyle x\in \mathbb {R} ^{n},y\in \mathbb {R} ^{m},u\in \mathbb {R} ^{r}}
{\displaystyle A,B,C,D}
The linear system with the same number of input and output variables is called passive if
{\displaystyle {\begin{aligned}\int \limits _{0}^{T}u^{T}y(t)dt\geq 0\\\end{aligned}}}
{\displaystyle T\geq 0}
{\displaystyle u(t)}
{\displaystyle y(t)}
{\displaystyle x(0)=0}
{\displaystyle {\begin{aligned}G(s)&=C(sI-A)^{-1}B+D\\\end{aligned}}}
{\displaystyle {\begin{aligned}\ G^{H}(s)+G(s)\geq 0\forall s\in \mathbb {C} ,Re(s)>0\\\end{aligned}}}
Let the linear system be controllable. Then, the system is passive if an only if there exists
{\displaystyle P>0}
{\displaystyle {\begin{aligned}\ {\begin{bmatrix}A^{T}P+PA&PB-C^{T}\\B^{T}P-C&-D^{T}-D\end{bmatrix}}\leq 0\\\end{aligned}}}
https://github.com/ShenoyVaradaraya/LMI--master/blob/main/Passivity.m
Retrieved from "https://en.wikibooks.org/w/index.php?title=LMIs_in_Control/Matrix_and_LMI_Properties_and_Tools/Passivity_and_Positive_Realness&oldid=3969277"
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First-to-Default Swaps - MATLAB & Simulink - MathWorks France
Formally, if the time to default of a particular issuer is denoted by
\tau
, and we know its default probability function
P\left(t\right)
, a latent variable
A
and corresponding thresholds
C\left(t\right)
Pr\left(\tau \le t\right)=P\left(t\right)=Pr\left(A\le C\left(t\right)\right)
Pr\left(s\le \tau \le t\right)=P\left(t\right)-P\left(s\right)=Pr\left(C\left(s\right)\le A\le C\left(t\right)\right)
These relationships make latent variable approaches convenient for both simulations and analytical derivations. Both
P\left(t\right)
C\left(t\right)
are functions of time.
The choice of a distribution for the variable
A
determines the thresholds
C\left(t\right)
. In the standard latent variable model, the variable
A
is chosen to follow a standard normal distribution, from which
C\left(t\right)={\Phi }^{-1}\left(P\left(t\right)\right)
\Phi
Given parameters
{\beta }_{i}
for each issuer
i
, and given independent standard normal variables
Z
{ϵ}_{i}
, the one-factor latent variable model assumes that the latent variable
{A}_{i}
associated to issuer
i
{A}_{i}={\beta }_{i}*Z+\sqrt{1-{\beta }_{i}^{2}}*{ϵ}_{i}
This induces a correlation between issuers
i
j
{\beta }_{i}{\beta }_{j}
. All latent variables
{A}_{i}
share the common factor
Z
as a source of uncertainty, but each latent variable also has an idiosyncratic source of uncertainty
{ϵ}_{i}
. The larger the coefficient
{\beta }_{i}
, the more the latent variable resembles the common factor
Z
Using the latent variable model, you can derive an analytic formula for the survival probability of the basket. The probability that issuer
i
survives past time
{t}_{j}
, in other words, that its default time
{\tau }_{i}
{t}_{j}
Pr\left({\tau }_{i}>{t}_{j}\right)=1-Pr\left({A}_{i}\le {C}_{i}\left({t}_{j}\right)\right)
{C}_{i}\left({t}_{j}\right)
is the default threshold computed above for issuer
i
, for the
j
-th date in the discretization grid. Conditional on the value of the one-factor
Z
, the probability that all issuers survive past time
{t}_{j}
Pr\left(\text{No}\text{ }\text{defaults}\text{ }\text{by}\text{ }\text{time}\text{ }{t}_{j}|Z\right)
=Pr\left({\tau }_{i}>{t}_{j}\text{ }\text{for}\text{ }\text{all}\text{ }i|Z\right)
=\prod _{i}\left[1-Pr\left({A}_{i}\le {C}_{i}\left({t}_{j}\right)|Z\right)\right]
where the product is justified because all the
{ϵ}_{i}
's are independent. Therefore, conditional on
Z
{A}_{i}
's are independent. The unconditional probability of no defaults by time
{t}_{j}
is the integral over all values of
Z
of the previous conditional probability
Pr\left(\text{No}\text{ }\text{defaults}\text{ }\text{by}\text{ }\text{time}\text{ }{t}_{j}\right)
={\int }_{Z}\prod _{i}\left[1-Pr\left({A}_{i}\le {C}_{i}\left({t}_{j}\right)|Z\right)\right]\varphi \left(Z\right)dZ
\varphi \left(Z\right)
the standard normal density.
By evaluating this one-dimensional integral for each point
{t}_{j}
in the grid, you get a discretization of the survival curve for the whole basket, which is the FTD survival curve.
|
Oxygen Transport in Brain Tissue | J. Biomech Eng. | ASME Digital Collection
Education and Research Center for Frontier Science and Engineering,
, 1-5-1 Chofugaoka, Chofu-shi, Tokyo 182-8585, Japan
e-mail: masamoto@mce.uec.ac.jp
Department of System Design Engineering,
, 3-14-1 Hiyoshi, Kohoku-ku, Yokohama 223-8522, Japan
e-mail: tanishita@sd.keio.ac.jp
Masamoto, K., and Tanishita, K. (July 27, 2009). "Oxygen Transport in Brain Tissue." ASME. J Biomech Eng. July 2009; 131(7): 074002. https://doi.org/10.1115/1.3184694
Oxygen is essential to maintaining normal brain function. A large body of evidence suggests that the partial pressure of oxygen
(pO2)
in brain tissue is physiologically maintained within a narrow range in accordance with region-specific brain activity. Since the transportation of oxygen in the brain tissue is mainly driven by a diffusion process caused by a concentration gradient of oxygen from blood to cells, the spatial organization of the vascular system, in which the oxygen content is higher than in tissue, is a key factor for maintaining effective transportation. In addition, a local mechanism that controls energy demand and blood flow supply plays a critical role in moment-to-moment adjustment of tissue
pO2
in response to dynamically varying brain activity. In this review, we discuss the spatiotemporal structures of brain tissue oxygen transport in relation to local brain activity based on recent reports of tissue
pO2
measurements with polarographic oxygen microsensors in combination with simultaneous recordings of neural activity and local cerebral blood flow in anesthetized animal models. Although a physiological mechanism of oxygen level sensing and control of oxygen transport remains largely unknown, theoretical models of oxygen transport are a powerful tool for better understanding the short-term and long-term effects of local changes in oxygen demand and supply. Finally, emerging new techniques for three-dimensional imaging of the spatiotemporal dynamics of
pO2
map may enable us to provide a whole picture of how the physiological system controls the balance between demand and supply of oxygen during both normal and pathological brain activity.
biological tissues, biotransport, brain, diffusion, haemodynamics, oxygen, gas transport, neural activity, energy metabolism, blood flow regulation
Biological tissues, Blood flow, Brain, Oxygen
Selective Neocortical and Thalamic Cell Death in the Gerbil After Transient Ischemia
Dynamic Observation of Oxygenation-Induced Contraction of and Transient Fiber-Network Formation-Disassembly in Cultured Human Brain Microvascular Endothelial Cells
The Supply of Oxygen to the Tissues and the Regulation of the Capillary Circulation
Brain Tissue Oxygen Concentration Measurements
Tissue Oxygen Tension and Brain Sensitivity to Hypoxia
Heterogeneities and Profiles of Oxygen Pressure in Brain and Kidney as Examples of the pO2 Distribution in the Living Tissue
Oxygen in Mammalian Tissue: Methods of Measurement and Affinities of Various Reactions
Heterogeneity of Brain Oxidative Metabolism and Hypoxia Response. Mammalian Systems and Nature's Solutions
Distribution of Oxygen Tension on the Surface of Arterioles, Capillaries and Venules of Brain Cortex and in Tissue in Normoxia: An Experimental Study on Rats
Local Oxygen Tension and Spike Activity of the Cerebral Grey Matter of the Rat and Its Response to Short Intervals of O2 Deficiency or CO2 Excess
Oxygenation of the Cat Primary Visual Cortex
The Architecture of the Cerebral Capillary Bed
Biol. Rev. Cambridge Philos. Soc.
Microangioarchitecture of Rat Parietal Cortex With Special Reference to Vascular ‘Sphincters.’ Scanning Electron Microscopic and Dark Field Microscopic Study
Non-Random Distribution of Blood Vessels in the Posterior Region of the Rat Somatosensory Cortex
Three-Dimensional Reconstruction of Brain Capillaries From Frozen Serial Sections
Blood Supply of the Cerebral Cortex
The Human Brain Surface, Three-Dimensional Sectional Anatomy With MRI, and Blood Supply
Springer-Wien
Reina-De La Torre
Sahuquillo-Barris
Morphological Characteristics and Distribution Pattern of the Arterial Vessels in Human Cerebral Cortex: A Scanning Electron Microscope Study
Staining of PO2 Measuring Points Demonstrated for the Rat Brain Cortex
Local Cerebral Glucose Utilization in the Neocortical Areas of the Rat Brain
Temporal Dynamics of the Partial Pressure of Brain Tissue Oxygen During Functional Forepaw Stimulation in Rats
Single-Neuron Activity and Tissue Oxygenation in the Cerebral Cortex
Separate Spatial Scales Determine Neural Activity-Dependent Changes in Tissue Oxygen Within Central Visual Pathways
Apparent Diffusion Time of Oxygen From Blood to Tissue in Rat Cerebral Cortex: Implication for Tissue Oxygen Dynamics During Brain Functions
Increased Cortical Oxidative Metabolism Due to Sensory Stimulation: Implications for Functional Brain Imaging
Kohl-Bareis
No Evidence for Early Decrease in Blood Oxygenation in Rat Whisker Cortex in Response to Functional Activation
NADH Fluorescence, [K+]0 and Oxygen Consumption in Cat Cerebral Cortex During Direct Cortical Stimulation
Neural Activity Triggers Neuronal Oxidative Metabolism Followed by Astrocytic Glycolysis
Interaction Between Tissue Oxygen Tension and NADH Imaging During Synaptic Stimulation and Hypoxia in Rat Hippocampal Slices
Comparisons of Oxygen Metabolism and Tissue PO2 in Cortex and Hippocampus of Gerbil Brain
PtiO2 and CMRO2 Changes in Cortex and Hippocampus of Aging Gerbil Brain
Metabolism and Function in the Cerebral Cortex Under Local Perfusion, With the Aid of an Oxygen Cathode for Surface Measurement of Cortical Oxygen Consumption
The Oxygen Pressure Field of the Brain and Its Significance for the Normal and Critical Oxygen Supply of the Brain
Oxygen Transport in Blood and Tissue
Witaleb
Leniger-Follert
Direct Determination of Local Oxygen Consumption of the Brain Cortex In Vivo
Increases in Oxygen Consumption Without Cerebral Blood Volume Change During Visual Stimulation Under Hypotension Condition
Spatial Specificity of the Enhanced Dip Inherently Induced by Prolonged Oxygen Consumption in Cat Visual Cortex: Implication for Columnar Resolution Functional MRI
Trial-by-Trial Relationship Between Neural Activity, Oxygen Consumption, and Blood Flow Responses
Changes in Evoked Brain Oxygen During Sensory Stimulation and Conditioning
Regional Control of Cerebral Vascular Reactivity and Oxygen Supply in Man
Oxygen Tension Changes Evoked in the Brain by Visual Stimulation
Behavior of Microflow and Local pO2 of the Brain Cortex During and After Direct Electrical Stimulation. A Contribution to the Problem of Metabolic Regulation of Microcirculation in the Brain
Cellular Microenvironment in Relation to Local Blood Flow
Measurement of Brain Tissue Oxygen at a Carbon Past Electrode Can Serve as an Index of Increases in Regional Cerebral Blood Flow
Activity-Induced Tissue Oxygenation Changes in Rat Cerebellar Cortex: Interplay of Postsynaptic Activation and Blood Flow
Gamma-Aminobutyric Acid Modulates Local Brain Oxygen Consumption and Blood Flow in Rat Cerebellar Cortex
Cerebral Regional Oxygen Consumption and Supply in Anesthetized Cat
Tracer Oxygen Distribution is Barrier-Limited in the Cerebral Microcirculation
A Multiparametric Assessment of Oxygen Efflux From the Brain
Brain Tissue Oxygen Consumption and Supply Induced by Neural Activation: Determined Under Suppressed Hemodynamic Response Conditions in the Anesthetized Rat Cerebral Cortex
Brain Energy Metabolism and Catecholaminergic Activity in Hypoxia, Hypercapnia and Ischemia
J. Neural Transm., Suppl.
Wylezinska
Oxygen Dependency of Cerebral Oxidative Phosphorylation in Newborn Piglets
Mathematical Model of Oxygen Transport in the Cerebral Cortex
Theoretical Simulation of Oxygen Transport to Brain by Networks of Microvessels: Effects of Oxygen Supply and Demand on Tissue Hypoxia
Vlassenko
Blood Flow and Oxygen Delivery to Human Brain During Functional Activity: Theoretical Modeling and Experimental Data
Relation Between Cerebral Blood Flow and Metabolism Explained by a Model of Oxygen Exchange
Dynamics of Oxygen Delivery and Consumption During Evoked Neural Stimulation Using a Compartment Model and CBF and Tissue P(O2) Measurements
A Model for the Regulation of Cerebral Oxygen Delivery
On the Oxygenation of Hemoglobin in the Human Brain
Dependence of Oxygen Delivery on Blood Flow in Rat Brain: A 7 Tesla Nuclear Magnetic Resonance Study
Model of Blood-Brain Transfer of Oxygen Explains Nonlinear Flow-Metabolism Coupling During Stimulation of Visual Cortex
A Theoretical Model of Oxygen Delivery and Metabolism for Physiologic Interpretation of Quantitative Cerebral Blood Flow and Metabolic Rate of Oxygen
Dominant Events that Modulate Mass Transfer Coefficient of Oxygen in Cerebral Cortex
Increased Oxygen Consumption Following Activation of Brain: Theoretical Footnotes Using Spectroscopic Data From Barrel Cortex
A Model of the Hemodynamic Response and Oxygen Delivery to Brain
Oxygen Transport in the Microvessel Network
A Three-Compartment Model of the Hemodynamic Response and Oxygen Delivery to Brain
Kouuchi
Dynamic Imaging of Somatosensory Cortical Activity in the Rat Visualized by Flavoprotein Autofluorescence
Optical Imaging of the Spatiotemporal Dynamics of Cerebral Blood Flow and Oxidative Metabolism in the Rat Barrel Cortex
Flavoprotein Autofluorescence Imaging of Neuronal Activation in the Cerebellar Cortex In Vivo
Quantitative Determination of Localized Tissue Oxygen Concentration In Vivo by Two-Photon Excitation Phosphorescence Lifetime Measurements
Oxygen Microscopy by Two-Photon-Excited Phosphorescence
Oxyphor R2 and G2: Phosphors for Measuring Oxygen by Oxygen-Dependent Quenching of Phosphorescence
Benrashid
Applications of Quantum Dots in Optical Fiber Luminescent Oxygen Sensors
A Passive Model of the Heat, Oxygen and Carbon Dioxide Transport in the Human Body
|
Numerical on PERT (Program Evaluation and Review Technique) | Education Lessons
For the given activities determine:
1. Critical path using PERT.
2. Calculate variance and standard deviation for each activity.
3. Calculate the probability of completing the project in 26 days.
Solution: First of all draw the network diagram for given data as shown below:
Here the time for completion of activities are probabilistic. So, using given values of time we will find the expected time to completion the activities and variance.
\text {Expected time } \quad t_e = {t_o + 4t_m + t_p \over 6}
\text {Variance } V = \Big({t_p - t_o \over 6}\Big)^2
For each given activity we will calculate the expected time as follows:
t_e = {t_o + 4t_m + t_p \over 6}
{6+4 \times 9+12 \over 6} = 9
({12-6 \over 6})^2 = 1.000
1-3 3 4 11 5 1.778
3-4 4 6 8 6 0.444
3-5 1 1.5 5 2 0.444
Now based on estimate time, we calculate the EST, EFT, LST and LFT for each activity to find out critical path of project as shown below. (Click here to know about calculation of EST, EFT, LST and LFT from CPM numerical)
2-4 6 9 15 9 15 0
2-6 6 9 15 18 24 9
Here the critical path is along the activities 1-2, 2-4, 4-6. So the critical path is 1-2-4-6. Following diagram is prepared to show critical path along with EST and LFT.
\therefore
The critical path = 1-2-4-6 with time duration
(t_{cp})
of 24 days.
Here standard deviation is calculated for activities of critical path. So we get
\begin{aligned} \sigma&= \sqrt{V_1 + V_2 + V_3} \\ \sigma&= \sqrt{1 + 4 + 1.778} \\ &= 2.6034 \end{aligned}
Now the probability of completion of project in that given time (t) of 26 days, can be calculate by below formula,
Z = {t - t_{cp} \over \sigma} = {26 - 24 \over 2.6034} = 0.7682
Using table in Appendix-B, we get probability
= 77.8 \%
As you can see below in Appendix-B the first column Z that is the probability we find out in the example through formula. The second column
\psi(z)
represent the probability in percentage
(\%)
As we have value of
Z=0.7682
now you can see in table we have
Z=0.76
\psi(z)=0.7764
Z=0.77
\psi(z)=0.7794
we take average of both we get
\psi(z)=0.7779=77.8\%
As you can see for
Z=0 \space \space
\psi(z)=0.5000 \space \space
50\%
Z=1 \space \space
\psi(z)=0.8413 \space \space
84.13\%
|
Revision as of 01:20, 29 July 2013 by NikosA (talk | contribs) (→Pan Sharpening: one approach creating a color-balanced pan-sharpened composite from 11-bit bands)
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L\lambda ={\frac {10^{4}*DN\lambda }{CalCoef\lambda *Bandwidth\lambda }}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
{\displaystyle [0,255]}
{\displaystyle [0,2047]}
|
Triple DES - Wikipedia
(Redirected from TripleDES)
48 DES-equivalent rounds
Lucks: 232 known plaintexts, 2113 operations including 290 DES encryptions, 288 memory; Biham: find one of 228 target keys with a handful of chosen plaintexts per key and 284 encryptions
In cryptography, Triple DES (3DES or TDES), officially the Triple Data Encryption Algorithm (TDEA or Triple DEA), is a symmetric-key block cipher, which applies the DES cipher algorithm three times to each data block. The Data Encryption Standard's (DES) 56-bit key is no longer considered adequate in the face of modern cryptanalytic techniques and supercomputing power. A CVE released in 2016, CVE-2016-2183 disclosed a major security vulnerability in DES and 3DES encryption algorithms. This CVE, combined with the inadequate key size of DES and 3DES, NIST has deprecated DES and 3DES for new applications in 2017, and for all applications by 2023. It has been replaced with the more secure, more robust AES.
While the government and industry standards abbreviate the algorithm's name as TDES (Triple DES) and TDEA (Triple Data Encryption Algorithm),[1] RFC 1851 referred to it as 3DES from the time it first promulgated the idea, and this namesake has since come into wide use by most vendors, users, and cryptographers.[2][3][4][5]
5 Encryption of more than one block
In 1978, a triple encryption method using DES with two 56-bit keys was proposed by Walter Tuchman; in 1981 Merkle and Hellman proposed a more secure triple key version of 3DES with 112 bits of security.[6]
The Triple Data Encryption Algorithm is variously defined in several standards documents:
RFC 1851, The ESP Triple DES Transform[7] (approved in 1995)
ANSI ANS X9.52-1998 Triple Data Encryption Algorithm Modes of Operation[8] (approved in 1998, withdrawn in 2008[9])
FIPS PUB 46-3 Data Encryption Standard (DES)[10] (approved in 1999, withdrawn in 2005[11])
NIST Special Publication 800-67 Revision 2 Recommendation for the Triple Data Encryption Algorithm (TDEA) Block Cipher[12] (approved in 2017)
ISO/IEC 18033-3:2010: Part 3: Block ciphers[13] (approved in 2005)
The original DES cipher's key size of 56 bits was generally sufficient when that algorithm was designed, but the availability of increasing computational power made brute-force attacks feasible. Triple DES provides a relatively simple method of increasing the key size of DES to protect against such attacks, without the need to design a completely new block cipher algorithm.
A naive approach to increase strength of a block encryption algorithm with short key length (like DES) would be to use two keys
{\displaystyle (K1,K2)}
instead of one, and encrypt each block twice:
{\displaystyle E_{K2}(E_{K1}({\textrm {plaintext}}))}
. If the original key length is
{\displaystyle n}
bits, one would hope this scheme provides security equivalent to using key
{\displaystyle 2n}
bits long. Unfortunately, this approach is vulnerable to meet-in-the-middle attack: given a known plaintext pair
{\displaystyle (x,y)}
{\displaystyle y=E_{K2}(E_{K1}(x))}
, one can recover the key pair
{\displaystyle (K1,K2)}
{\displaystyle 2^{n+1}}
steps, instead of the
{\displaystyle 2^{2n}}
steps one would expect from an ideally secure algorithm with
{\displaystyle 2n}
bits of key.
Therefore, Triple DES uses a "key bundle" that comprises three DES keys,
{\displaystyle K1}
{\displaystyle K2}
{\displaystyle K3}
, each of 56 bits (excluding parity bits). The encryption algorithm is:
{\displaystyle {\textrm {ciphertext}}=E_{K3}(D_{K2}(E_{K1}({\textrm {plaintext}}))).}
That is, DES encrypt with
{\displaystyle K1}
, DES decrypt with
{\displaystyle K2}
, then DES encrypt with
{\displaystyle K3}
{\displaystyle {\textrm {plaintext}}=D_{K1}(E_{K2}(D_{K3}({\textrm {ciphertext}}))).}
That is, decrypt with
{\displaystyle K3}
, encrypt with
{\displaystyle K2}
, then decrypt with
{\displaystyle K1}
In each case the middle operation is the reverse of the first and last. This improves the strength of the algorithm when using keying option 2 and provides backward compatibility with DES with keying option 3.
Keying options[edit]
The standards define three keying options:
Keying option 1
All three keys are independent. Sometimes known as 3TDEA[14] or triple-length keys.[15]
This is the strongest, with 3 × 56 = 168 independent key bits. It is still vulnerable to meet-in-the-middle attack, but the attack requires 22 × 56 steps.
K1 and K2 are independent, and K3 = K1. Sometimes known as 2TDEA[14] or double-length keys.[15]
This provides a shorter key length of 112 bits and a reasonable compromise between DES and Keying option 1, with the same caveat as above.[16] This is an improvement over "double DES" which only requires 256 steps to attack. NIST has deprecated this option.[14]
All three keys are identical, i.e. K1 = K2 = K3.
This is backward compatible with DES, since two operations cancel out. ISO/IEC 18033-3 never allowed this option, and NIST no longer allows K1 = K2 or K2 = K3.[14][12]
Each DES key is 8 odd-parity bytes, with 56 bits of key and 8 bits of error-detection.[8] A key bundle requires 24 bytes for option 1, 16 for option 2, or 8 for option 3.
NIST (and the current TCG specifications version 2.0 of approved algorithms for Trusted Platform Module) also disallows using any one of the 64 following 64-bit values in any keys (note that 32 of them are the binary complement of the 32 others; and that 32 of these keys are also the reverse permutation of bytes of the 32 others), listed here in hexadecimal (in each byte, the least significant bit is an odd-parity generated bit, it is discarded when forming the effective 56-bit keys):
01.01.01.01.01.01.01.01, FE.FE.FE.FE.FE.FE.FE.FE, E0.FE.FE.E0.F1.FE.FE.F1, 1F.01.01.1F.0E.01.01.0E,
01.01.FE.FE.01.01.FE.FE, FE.FE.01.01.FE.FE.01.01, E0.FE.01.1F.F1.FE.01.0E, 1F.01.FE.E0.0E.01.FE.F1,
01.01.E0.E0.01.01.F1.F1, FE.FE.1F.1F.FE.FE.0E.0E, E0.FE.1F.01.F1.FE.0E.01, 1F.01.E0.FE.0E.01.F1.FE,
01.01.1F.1F.01.01.0E.0E, FE.FE.E0.E0.FE.FE.F1.F1, E0.FE.E0.FE.F1.FE.F1.FE, 1F.01.1F.01.0E.01.0E.01,
01.FE.01.FE.01.FE.01.FE, FE.01.FE.01.FE.01.FE.01, E0.01.FE.1F.F1.01.FE.0E, 1F.FE.01.E0.0E.FE.01.F1,
01.FE.FE.01.01.FE.FE.01, FE.01.01.FE.FE.01.01.FE, E0.01.01.E0.F1.01.01.F1, 1F.FE.FE.1F.0E.FE.FE.0E,
01.FE.E0.1F.01.FE.F1.0E, FE.01.1F.E0.FE.01.0E.F1, E0.01.1F.FE.F1.01.0E.FE, 1F.FE.E0.01.0E.FE.F1.01,
01.FE.1F.E0.01.FE.0E.F1, FE.01.E0.1F.FE.01.F1.0E, E0.01.E0.01.F1.01.F1.01, 1F.FE.1F.FE.0E.FE.0E.FE,
01.E0.01.E0.01.F1.01.F1, FE.1F.FE.1F.FE.0E.FE.0E, E0.1F.FE.01.F1.0E.FE.01, 1F.E0.01.FE.0E.F1.01.FE,
01.E0.FE.1F.01.F1.FE.0E, FE.1F.01.E0.FE.0E.01.F1, E0.1F.01.FE.F1.0E.01.FE, 1F.E0.FE.01.0E.F1.FE.01,
01.E0.E0.01.01.F1.F1.01, FE.1F.1F.FE.FE.0E.0E.FE, E0.1F.1F.E0.F1.0E.0E.F1, 1F.E0.E0.1F.0E.F1.F1.0E,
01.E0.1F.FE.01.F1.0E.FE, FE.1F.E0.01.FE.0E.F1.01, E0.1F.E0.1F.F1.0E.F1.0E, 1F.E0.1F.E0.0E.F1.0E.F1,
01.1F.01.1F.01.0E.01.0E, FE.E0.FE.E0.FE.F1.FE.F1, E0.E0.FE.FE.F1.F1.FE.FE, 1F.1F.01.01.0E.0E.01.01,
01.1F.FE.E0.01.0E.FE.F1, FE.E0.01.1F.FE.F1.01.0E, E0.E0.01.01.F1.F1.01.01, 1F.1F.FE.FE.0E.0E.FE.FE,
01.1F.E0.FE.01.0E.F1.FE, FE.E0.1F.01.FE.F1.0E.01, E0.E0.1F.1F.F1.F1.0E.0E, 1F.1F.E0.E0.0E.0E.F1.F1,
01.1F.1F.01.01.0E.0E.01, FE.E0.E0.FE.FE.F1.F1.FE, E0.E0.E0.E0.F1.F1.F1.F1, 1F.1F.1F.1F.0E.0E.0E.0E,
With these restrictions on allowed keys, Triple DES has been reapproved with keying options 1 and 2 only. Generally the three keys are generated by taking 24 bytes from a strong random generator and only keying option 1 should be used (option 2 needs only 16 random bytes, but strong random generators are hard to assert and it's considered best practice to use only option 1).
Encryption of more than one block[edit]
As with all block ciphers, encryption and decryption of multiple blocks of data may be performed using a variety of modes of operation, which can generally be defined independently of the block cipher algorithm. However, ANS X9.52 specifies directly, and NIST SP 800-67 specifies via SP 800-38A[17] that some modes shall only be used with certain constraints on them that do not necessarily apply to general specifications of those modes. For example, ANS X9.52 specifies that for cipher block chaining, the initialization vector shall be different each time, whereas ISO/IEC 10116[18] does not. FIPS PUB 46-3 and ISO/IEC 18033-3 define only the single block algorithm, and do not place any restrictions on the modes of operation for multiple blocks.
In general, Triple DES with three independent keys (keying option 1) has a key length of 168 bits (three 56-bit DES keys), but due to the meet-in-the-middle attack, the effective security it provides is only 112 bits.[14] Keying option 2 reduces the effective key size to 112 bits (because the third key is the same as the first). However, this option is susceptible to certain chosen-plaintext or known-plaintext attacks,[19][20] and thus it is designated by NIST to have only 80 bits of security.[14] This can be considered insecure, and, as consequence Triple DES has been deprecated by NIST in 2017.[21]
Logo of the Sweet32 attack
The short block size of 64 bits makes 3DES vulnerable to block collision attacks if it is used to encrypt large amounts of data with the same key. The Sweet32 attack shows how this can be exploited in TLS and OpenVPN.[22] Practical Sweet32 attack on 3DES-based cipher-suites in TLS required
{\displaystyle 2^{36.6}}
blocks (785 GB) for a full attack, but researchers were lucky to get a collision just after around
{\displaystyle 2^{20}}
blocks, which took only 25 minutes.
The security of TDEA is affected by the number of blocks processed with one key bundle. One key bundle shall not be used to apply cryptographic protection (e.g., encrypt) more than
{\displaystyle 2^{20}}
64-bit data blocks.
— Recommendation for Triple Data Encryption Algorithm (TDEA) Block Cipher (SP 800-67 Rev2)[12]
OpenSSL does not include 3DES by default since version 1.1.0 (August 2016) and considers it a "weak cipher".[23]
The electronic payment industry uses Triple DES and continues to develop and promulgate standards based upon it, such as EMV.[24]
Earlier versions of Microsoft OneNote,[25] Microsoft Outlook 2007[26] and Microsoft System Center Configuration Manager 2012[27] use Triple DES to password-protect user content and system data. However, in December 2018, Microsoft announced the retirement of 3DES throughout their Office 365 service.[28]
Firefox and Mozilla Thunderbird[29] use Triple DES in CBC mode to encrypt website authentication login credentials when using a master password.
Below is a list of cryptography libraries that support Triple DES:
Trusted Platform Module (alias TPM, hardware implementation)
Some implementations above may not include 3DES in the default build, in later or more recent versions.
^ "Triple DES Encryption". IBM. Retrieved 2010-05-17.
^ Alanazi, Hamdan. O.; Zaidan, B. B.; Zaidan, A. A.; Jalab, Hamid A.; Shabbir, M.; Al-Nabhani, Y. (March 2010). "New Comparative Study Between DES, 3DES and AES within Nine Factors". Journal of Computing. 2 (3). arXiv:1003.4085. Bibcode:2010arXiv1003.4085A. ISSN 2151-9617.
^ "Cisco PIX 515E Security Appliance Getting Started Guide: Obtaining a DES License or a 3DES-AES License" (PDF). Cisco. 2006. Retrieved 2017-09-05.
^ "3DES Update: Most Banks Are Done, But..." ATM & Debit News. 2007-03-29. Archived from the original on 2013-05-10. Retrieved 2017-09-05.
^ RFC 2828 and RFC 4949
^ Merkle, R. and M. Hellman, “On the Security of Multiple Encryption”, Communications of the ACM, vol. 24, no. 7, pp. 465–467, July 1981.
^ Karn, P.; Metzger, P.; Simpson, W. (September 1995). The ESP Triple DES Transform. doi:10.17487/RFC1851. RFC 1851.
^ a b "ANSI X9.52-1998 Triple Data Encryption Algorithm Modes of Operation". Retrieved 2017-09-05. Extends ANSI X3.92-1981 Data Encryption Algorithm.
^ "ANSI Standards Action" (PDF). Vol. 39, no. 46. ANSI. 2008-11-14. Retrieved 2017-09-05. {{cite magazine}}: Cite magazine requires |magazine= (help)
^ "FIPS PUB 46-3: Data Encryption Standard (DES)" (PDF). United States Department of Commerce. Oct 25, 1999. Retrieved 2017-09-05.
^ "Announcing Approval of the Withdrawal of Federal Information Processing Standard (FIPS) 46–3..." (PDF). Federal Register. 70 (96). 2005-05-19. Retrieved 2017-09-05.
^ a b c Barker, Elaine; Mouha, Nicky (November 2017). "NIST Special Publication 800-67 Revision 2: Recommendation for the Triple Data Encryption Algorithm (TDEA) Block Cipher". NIST. doi:10.6028/NIST.SP.800-67r2. {{cite journal}}: Cite journal requires |journal= (help)
^ "ISO/IEC 18033-3:2010 Information technology -- Security techniques -- Encryption algorithms -- Part 3: Block ciphers". ISO. December 2010. Retrieved 2017-09-05.
^ a b c d e f Barker, Elaine (January 2016). "NIST Special Publication 800-57: Recommendation for Key Management Part 1: General" (PDF) (4 ed.). NIST. Retrieved 2017-09-05.
^ a b "The Cryptography Guide: Triple DES". Cryptography World. Archived from the original on 2017-03-12. Retrieved 2017-09-05.
^ Katz, Jonathan; Lindell, Yehuda (2015). Introduction to Modern Cryptography. Chapman and Hall/CRC. p. 223. ISBN 9781466570269.
^ NIST Special Publication 800-38A, Recommendation for Block Cipher Modes of Operation, Methods and Techniques, 2001 Edition (PDF)
^ "ISO/IEC 10116:2006 Information technology -- Security techniques -- Modes of operation for an n-bit block cipher" (3 ed.). February 2006. Retrieved 2017-09-05.
^ Merkle, Ralph; Hellman, Martin (July 1981). "On the Security of Multiple Encryption" (PDF). Communications of the ACM. 24 (7): 465–467. CiteSeerX 10.1.1.164.251. doi:10.1145/358699.358718. S2CID 11583508.
^ van Oorschot, Paul; Wiener, Michael J. (1990). A known-plaintext attack on two-key triple encryption. EUROCRYPT'90, LNCS 473. pp. 318–325. CiteSeerX 10.1.1.66.6575.
^ "Update to Current Use and Deprecation of TDEA". nist.gov. 11 July 2017. Retrieved 2 August 2019.
^ "Sweet32: Birthday attacks on 64-bit block ciphers in TLS and OpenVPN". sweet32.info. Retrieved 2017-09-05.
^ Salz, Rich (2016-08-24). "The SWEET32 Issue, CVE-2016-2183". OpenSSL. Retrieved 2017-09-05.
^ "Annex B Approved Cryptographic Algorithms – B1.1 Data Encryption Standard (DES)". EMV 4.2: Book 2 – Security and Key Management (4.2 ed.). EMVCo. June 2008. p. 137. The double-length key triple DES encipherment algorithm (see ISO/IEC 18033-3) is the approved cryptographic algorithm to be used in the encipherment and MAC mechanisms specified in Annex A1. The algorithm is based on the (single) DES algorithm standardised in ISO 16609.
^ Daniel Escapa's OneNote Blog, Encryption for Password Protected Sections, November 2006.
^ "Encrypt e-mail messages – Outlook – Microsoft Office Online". office.microsoft.com. Archived from the original on 2008-12-25. Applies to: Microsoft Office Outlook 2007
^ Microsoft TechNet product documentation, Technical Reference for Cryptographic Controls Used in Configuration Manager, October 2012.
^ https://portal.office.com/AdminPortal/home?switchtomodern=true#/MessageCenter?id=MC171089
^ Mozilla NSS source code. See Explanation of directory structure (especially the introductory and "security" sections) for background information.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Triple_DES&oldid=1078804116"
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François Viète - Wikiquote
François Viète (1540 – 23 February 1603), Seigneur de la Bigotière, was a French mathematician, also known as Franciscus Vieta, Francois Vieta or Francois Viete, whose new algebra was an important step towards modern algebra, with innovations such as the use of letters as parameters in equations. He was a lawyer serving as a privy councillor to kings of France, Henry III and Henry IV.
1.1 In artem analyticem Isagoge (1591)
2 Quotes about Viète
2.1 Introduction to the Literature of Europe in the Fifteenth, Sixteenth, and Seventeenth Centuries (1866)
2.3 History of Mathematics (1925) Vol. 2
2.4 The Development of Mathematics (1940)
In artem analyticem Isagoge (1591)[edit]
f unless otherwise noted
9.If the quantities in the same proportion are subtracted likewise from amounts in the same proportion, the differences are in proportion. [a:b::c:d => (a-c):(b-d)::c:d]
15. If we have three or four magnitudes and the product of the extremes is equal to the product means, they are in proportion.[ad=bc => a:b::c:d OR ac=b2 => a:b::b:c]
Ch. 1 as quoted by Jacob Klein, Greek Mathematical Thought and the Origin of Algebra (1934-1936) Appendix.
There is a certain way of searching for the truth in mathematics that Plato is said first to have discovered; Theon named it analysis, and defined it as the assumption of that which is sought as if it were admitted and working through its consequences to what is admitted to be true. This is opposed to synthesis, which is the assuming what is admitted and working through its consequences to arrive at and to understand that which is sought.
Ch. 1 as quoted by Douglas M. Jesseph, Squaring the Circle: The War Between Hobbes and Wallis (1999) p. 225
Quotes about Viète[edit]
Vieta's innovation contains three interrelated and interdependent aspects. ...methodical ...making calculation possible with both known and unknown indeterminate (and therefore 'general') numbers. ...cognitive ...resolving mathematical problems in this general mode, such that its indeterminate solution allows arbitrarily many determinate solutions based on numbers assumed at will. ...analytic ...being applicable indifferently to the numbers of traditional arithmetic and the magnitudes of traditional geometry.
Burt C. Hopkins, "Nastalgia and Phenomenon: Hussel and Patočka on the End of the Ancient Cosmos," (2015) ibid.
A major advance in notation with far-reaching consequences was François Viète's idea, put forward in his "Introduction to the Analytic Art"... of designating by letters all quantities, known or unknown, occurring in a problem. ...for the first time it was possible to replace various numerical examples by a single "generic" example, from which all others could be deduced by assigning values to the letters. ...by using symbols as his primary means of expression and showing how to calculate with those symbols, Viète initiated a completely formal treatment of algebraic expressions, which he called logistice speciosa (as opposed to logistice numerosa, which deals with numbers). This "symbolic logistic" gave some substance, some legitimacy to algebraic calculations, which allowed Viète to free himself from the geometric diagrams used... as justifications.
Jean-Pierre Tignol, Galois' Theory of Algebraic Equations (2001)
Introduction to the Literature of Europe in the Fifteenth, Sixteenth, and Seventeenth Centuries (1866)[edit]
Henry Hallam, Vol. 1
Ars Magna, published in 1545... contains many valuable discoveries; but that which has been most celebrated is the rule for the solution of cubic equations, generally known by Cardan's name, though he had obtained it from a man of equal genius in algebraic science, Nicolas Tartaglia. ...Cossali has ingeniously attempted to trace the process by which Tartaglia arrived at this discovery; one which, when compared with the other leading rules of algebra, where the invention... has generally lain much nearer the surface, seems an astonishing effort of sagacity. Even Harriott's beautiful generalization of the composition of equations was prepared by what Cardan and Vieta had done before, or might have been suggested by observation in the less complex cases.
Cardan, though not entitled to the honor of this discovery, nor even equal, perhaps, in mathematical genius to Tartaglia, made a great epoch in the science of algebra; and according to Cossali and Hutton, has a claim to much that Montucla has unfairly or carelessly attributed to his favorite, Vieta.
Cossali has given the larger part of a quarto volume to the algebra of Cardan; his object being to establish the priority of the Italian's claim to most of the discoveries ascribed by Montucla to others, and especially to Vieta. Cardan knew how to transform a complete cubic equation into one wanting the second term; one of the flowers which Montucla has placed on the head of Vieta; and this he explains so fully, that Cossali charges the French historian of mathematics with having never read the Ars Magna.
Rhaeticus was not a ready calculator only... Up to his time, the trigonometric functions had been considered always with relation to the arc; he was the first to construct the right triangle and to make them depend directly upon its angles. It was from the right triangle that Rhæticus got his idea of calculating the hypotenuse; i.e., he was the first to plan a table of secants. Good work in trigonometry was done also by Vieta and Romanus.
Cardan applied the Hindoo rule of "false position" (called by him regula aurea) to the cubic, but this mode of approximating was exceedingly rough. An incomparably better method was invented by Franciscus Vieta... whose transcendent genius enriched mathematics with several important innovations... For this process, Vieta was greatly admired by his contemporaries. It was employed by Harriot, Oughtred, Pell, and others. Its principle is identical with the main principle involved in the methods of approximation of Newton and Horner. The only change lies in the arrangement of the work. This alteration was made to afford facility and security in the process of evolution of the root.
Vieta [was] the most eminent French mathematician of the sixteenth century.
He was employed throughout life in the service of the state, under Henry III and Henry IV. He was, therefore, not a mathematician by profession, but his love for the science was so great that he remained in his chamber studying, sometimes several days in succession, without eating and sleeping more than was necessary to sustain himself. So great devotion to abstract science is the more remarkable because he lived at a time of incessant political and religious turmoil.
During the war against Spain, Vieta rendered service to Henry IV by deciphering intercepted letters written in a species of cipher, and addressed by the Spanish Court to their governor of Netherlands. The Spaniards attributed the discovery of the key to magic.
An ambassador from Netherlands once told Henry IV that France did not possess a single geometer capable of solving a problem propounded to geometers by a Belgian mathematician, Adrianus Romanus. It was the solution of the equation of the forty fifth degree:—
{\displaystyle 45y-3795y^{3}+95634y^{3}-\ldots +945y^{41}-45y^{43}+y^{45}=C}
...Vieta, who, having already pursued similar investigations, saw at once that this awe-inspiring problem was simply the equation by which C = 2 sin φ was expressed in terms of y = 2 sin 1⁄45 φ that since 45 = 3·3·5, it was necessary only to divide an angle once into 5 equal parts, and then twice into 3,—a division which could be effected by corresponding equations of the fifth and third degrees. Brilliant was the discovery by Vieta of 23 roots to this equation, instead of only one. The reason why he did not find 45 solutions, is that the remaining ones involve negative sines, which were unintelligible to him.
Detailed investigations on the famous old problem of the section of an angle into an odd number of equal parts, led Vieta to the discovery of a trigonometrical solution of Cardan's irreducible case in cubics. He applied the equation (2 cos 1⁄3 φ)3 - 3 (2 cos 1⁄3 cos φ) = 2 cos φ to the solution of x3 - 3 a2x = a2b, when a > ½ b, by placing x = 2 a cos 1⁄3 φ, and determining φ from b = 2a cos φ.
The main principle employed by him in the solution of equations is that of reduction. He solves the quadratic by making a suitable substitution which will remove the term containing x to the first degree. Like Cardan, he reduces the general expression of the cubic to the form x3 + mx + n = 0; then assuming x = (1⁄3 a - z2)÷z and substituting, he gets z6 - bz3 - 1⁄27 a3 = 0. Putting z3 = y, he has a quadratic. In the solution of bi-quadratics, Vieta still remains true to his principle of reduction. This gives him the well-known cubic resolvent. He thus adheres throughout to his favourite principle, and thereby introduces into algebra a uniformity of method which claims our lively admiration.
In Vieta's algebra we discover a partial knowledge of the relations existing between the coefficients and the roots of an equation. He shows that if the coefficient of the second term in an equation of the second degree is minus the sum of two numbers whose product is the third term, then the two numbers are roots of the equation. Vieta rejected all except positive roots; hence it was impossible for him to fully perceive the relations in question.
The most epoch making innovation in algebra due to Vieta is the denoting of general or indefinite quantities by letters of the alphabet. To be sure, Regiomontanus and Stifel in Germany, and Cardan in Italy, used letters before him, but Vieta extended the idea and first made it an essential part of algebra. The new algebra was called by him logistica speciosa in distinction to the old logistica numerosa.
Vieta's formalism differed considerably from that of to-day. The equation a3 + 3a2b + 3ab2 + b3 = (a + b)3 was written by him "a cubus + b in a quadr. 3 + a in b quadr. 3 + b cubo æqualia a+b cubo."
In numerical equations the unknown quantity was denoted by N, its square by Q, and its cube by C. Thus the equation x3 - 8 x2 + 16 x = 40 was written 1 C - 8 Q - 16 N œqual. 40.
Exponents and our symbol (=) for equality were not yet in use; but... Vieta employed the Maltese cross (+) as the short-hand symbol for addition, and the (-) for subtraction. These two characters had not been in general use before the time of Vieta.
History of Mathematics (1925) Vol. 2[edit]
{\displaystyle x^{6}-15x^{4}+85x^{3}-225x^{2}+274x=120}
{\displaystyle x^{2}+ax+b=0}
{\displaystyle u+z}
{\displaystyle x}
{\displaystyle u^{2}+(2z+a)u+(z^{2}+az+b)=0.}
{\displaystyle 2z+a=0,}
{\displaystyle z=-{\frac {1}{2}}a,}
{\displaystyle u^{2}-{\frac {1}{4}}(a^{2}-4b)=0.}
{\displaystyle u=\pm {\frac {1}{2}}{\sqrt {a^{2}-4b}}.}
{\displaystyle x=u+z=-{\frac {1}{2}}a\pm {\sqrt {a^{2}-4b}}.}
{\displaystyle x^{2}}
{\displaystyle x^{3}+px^{2}+qx+r=0}
{\displaystyle x=y-{\frac {1}{3}}p,}
{\displaystyle y^{3}+3by=2c.}
{\displaystyle z^{3}+yz=b,}
{\displaystyle y={\frac {b-z^{2}}{z}},}
{\displaystyle z^{6}+2cz^{2}=b^{2},}
{\displaystyle x^{4}+2gx^{2}+bx=c,}
{\displaystyle x^{4}+2gx^{2}=c-bx,}
{\displaystyle gx^{2}+{\frac {1}{4}}y^{2}+yx^{2}+gy}
The Development of Mathematics (1940)[edit]
Letters had been used before Vieta to denote numbers, but he introduced the practice for both given and unknown numbers as a general procedure. He thus fully recognized that algebra is on a higher level of abstraction than arithmetic. This advance in generality was one of the most important steps ever taken in mathematics. The complete divorce of algebra and arithmetic was consummated only in the nineteenth century, when the postulational method freed the symbols of algebra from any necessary arithmetical connotation.
Improving on the devices of his European predecessors, Vieta gave a uniform method for the numerical solution of algebraic equations. ...it was essentially the same as Newton's (1669)... Although Vieta's method has been displaced by others... The method applies to transcendental equations as readily as to algebraic when combined with expansions to a few terms by Taylor's or Maclaurin's series.
An algebraic equation of degree 45 which Vieta attacked in reply to a challenge indicates the quality of his work in trigonometry. Consistently seeking the generality underlying particulars, Vieta had found how to express sin nθ (n a positive integer) as a polynomial in sin θ, cos θ. He saw at once that the formidable equation of his rival had manufactured from an equivalent of dividing the circumference of the unit circle into 45 equal parts. ...More important than this spectacular feat was Vieta's suggestion that cubics can be solved trigonometrically.
Vieta's principle advance in trigonometry was his systematic application of algebra. ...he worked freely with all six of the usual functions, and... obtained many of the fundamental identities algebraically. With Vieta, elementary (non-analytic) trigonometry was practically completed except on the computational side. All computation was greatly simplified early in the seventeenth century by the invention of logarithms.
Introduction à l'art Analytique (1868) French Tr. Frédéric Louis Ritter
Retrieved from "https://en.wikiquote.org/w/index.php?title=François_Viète&oldid=2628047"
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Star | Brilliant Math & Science Wiki
Infinity Mathematics, S V, and Jimin Khim contributed
For further reference: A light year is the distance travelled by light in a vacuum in one year (approximately 9,460,730,472,580.8 kilometers or 5,878,625,373,183.6 miles). In this article, distances will be in light years unless otherwise specified.
Stars viewed from the Hubble Space Telescope. Retrieved from the U.S. Public Domain.
A star is an astronomical object that consists of a sphere of luminous plasma held together by its own gravity. The Sun is the closest star to Earth, but numerous others are visible to the naked eye from Earth during the night, appearing as fixed, luminous points. However, although thousands of stars can be viewed from Earth, most of the stars in the universe, including all stars outside our galaxy, are invisible even with the most powerful telescopes.[1][2][3]
For the majority of the life of an average star, thermonuclear fusion of hydrogen and helium occurs within the star's core, releasing energy that transverses the outer layers of the star and radiates into outer space, causing the star to shine and emit various types of electromagnetic radiation. During a star's lifetime, almost all of the 92 naturally occurring elements are produced within the star by stellar nucleosynthesis and, for high-mass stars, by supernova nucleosynthesis when the star explodes.[1]
Astral Measurement
Stars within the Universe
Stellar Classification and Evolution
Characteristics and Structure of Stars
Although astral parameters can be expressed using SI units, it is most convenient to represent characteristics of stars, including mass, luminosity, and radii, based upon solar units, which are based upon characteristics of the Sun. In 2015, the International Astronomical Union defined the set of nominal solar values, which can be used to describe astral parameters:[1]
nominal solar luminosity:
L_{\bigodot} = 3.828 \times 10^{26} \text{ Watts}
nominal solar radius:
R_{\bigodot} = 6.957 \times 10^8 \text{ meters} = 6.957 \times 10^5 \text{ kilometers}
nominal solar mass:
GM_{\bigodot} = 1.3271244 \times 10^{20}\text{ m}^3\text{/s}^{2}.
As described by Everipedia,[1]
Large lengths, such as the radius of a giant star or the semi-major axis of a binary star system, are often expressed in terms of the astronomical unit, approximately equal to the mean distance between Earth and the Sun (150 million km or approximately 93 million miles). In 2012, the IAU defined the astronomical constant to be an exact length in meters: 149,597,870,700 meters.
Astronomical Objects Within the Observable Universe (approximate):
and 30 sextillion (30 billion trillion) stars
within the universe.[4]
Closest Stars to the Solar System:[5][6][7]
Star Name(s) Distance from Earth Star Name(s) Distance from Earth
1 Proxima Centauri 4.24 light years 21 DX Cancri 11.83 light years
2 α Centauri A and α Centauri B 4.37 light years 22 Tau Ceti 11.89 light years
3 Barnard's Star 5.96 light years 23 GJ 1061 11.99 light years
4 Luhman 16A and Luhman 16B 6.59 light years 24 WISE 0350−5658 12.07 light years
5 WISE 0855−0714 7.27 light years 25 YZ Ceti 12.13 light years
6 Lalande 21185 8.29 light years 26 Luyten's Star 12.37 light years
7 Sirius A and Sirius B 8.58 light years 27 Teegarden's Star 12.51 light years
\hspace{5mm}
Luyten 726-8 A and Luyten 726-8 B
\hspace{10mm}
8.73 light years
\hspace{30mm}
\hspace{5mm}
SCR 1845-6357 A/SCR 1845-6357 B
\hspace{10mm}
12.57 light years
9 Ross 154 9.68 light years 29 Kpteyn's Star 12.78 light years
10 Ross 248 10.32 light years 30 Lacaille 8760 12.87 light years
11 Epsilon Eridani 10.52 light years 31 WISE 0535−7500 13.00 light years
12 Lacaille 9352 10.74 light years 32 Kruger 60 A/Kruger 60 B 13.15 light years
13 Ross 128 10.92 light years 33 DEN 1048-3956 13.17 light years
14 WISE 1506+7027 11.01 light years 34 UGPS 0722-05 13.26 light years
15 EZ Aquarii A/EZ Aquarii B/EZ Aquarii C 11.27 light years 35 Ross 614A/Ross 614B 13.35 light years
16 Procyon A/Procyon B 11.40(2) light years 36 Wolf 1061 13.82 light years
17 61 Cygni A/61 Cygni B 11.40(3) light years 37 Van Maanen's Star 14.07 light years
18 Struve 2398 A/Struve 2398 B 11.53 light years 38 Gliese 1 14.23 light years
19 Groombridge 34 A/Groombridge 34 B 11.62 light years 39 Wolf 424 A/Wolf 424 B 14.31 light years
20 Epsilon Indi A/Ba/Bb 11.82 light years 40 2MASS J154043.42-510135.7 14.40 light years
Stellar Formation within Nebulas
Mass-dependent Evolutionary Stages
The Hertzsprung-Russell diagram and the specific regions where certain star types (e.g. giants, supergiants, etc.) are found. Retrieved from the U.S. Public Domain.
Various color spectrums relating to certain stars (labeled). Retrieved from the U.S. Public Domain.
The Hertzsprung-Russell diagram is a 2-dimensional scatterplot showing the relationship between stars' absolute magnitudes (also known as luminosities or brightnesses) and their effective temperatures (which correlates to the stars' colors). The diagram was created independently by Ejnar Hertzsprung and Henry Norris Russell (hence the name "The Hertzsprung-Russell diagram") in 1910 and represented a major step towards the complete understanding of stellar evolution.[6][8]
The horizontal axis of the Hertzsprung-Russell diagram represents a star's temperature, which directly correlates to the spectral type or spectral class (color) of the star. Astronomers recognize 7 different spectral types or temperature divisions of stars, which are also used in the Hertzsprung-Russell diagram:[9]
Spectral Type Color Surface Temperature Mass (Sun = 1) Luminosity (Sun = 1) Examples
O Blue Over 25,000 K 60 1,400,000 10 Lacertra
B Blue 11,000 - 25,000 K 18 20,000 Rigel
A Blue 7,500 - 11,000 K 3.2 80 Sirius
F White 6,000 - 7,500 K 1.7 6 Canopus
G Yellow 5,000 - 6,000 K 1.1 1.2 Sun, Capella
K Orange 3,500 - 5,000 K 0.8 0.4 Arcturus
M Red Under 3,500 K 0.3 0.04 Antares
Everipedia, E. Star. Retrieved April 7, 2018, from www.everipedia.org/wiki/Star/
The Gale Group, I. Star - UXL Encyclopedia of Science, Encyclopedia.com. Retrieved from www.encyclopedia.com/science-and-technology/astronomy-and-space-exploration/astronomy-general/star
Fernie, J. Star - Encyclopædia Britannica. Retrieved from www.britannica.com/science/star-astronomy
Atlas of the Universe, I. The Universe within 14 Billion Light Years - The Visible Universe. Retrieved from www.atlasoftheuniverse.com/universe.html
Tate, K. The Nearest Stars to Earth (Infographic) - Space.com. Retrieved from www.space.com/18964-the-nearest-stars-to-earth-infographic.html
University of Wisconsin-Madison, I. The Closest Stars to the Earth - The 26 Nearest Stars. Retrieved from www.astro.wisc.edu/~dolan/constellations/extra/nearest.html
Atlas of the Universe, I. The Universe within 12.5 Light Years - The Nearest Stars. Retrieved from www.atlasoftheuniverse.com/12lys.html
Wikipedia, I. Hertzsprung-Russell diagram - Wikipedia. Retrieved from https://en.wikipedia.org/wiki/Hertzsprung–Russell_diagram
Enchanted Learning, I. Star Classification - Zoom Astronomy. Retrieved from http://www.enchantedlearning.com/subjects/astronomy/stars/startypes.shtml
Cite as: Star. Brilliant.org. Retrieved from https://brilliant.org/wiki/star/
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Vapor Pressure and Raoult's Law | Brilliant Math & Science Wiki
Vapor Pressure and Raoult's Law
The molecules at the surface of a liquid are weakly bonded compared to the molecules beneath the surface. For this reason, the molecules at the surface easily vaporize at temperatures lower than the boiling point. This process is called evaporation.
All liquids undergo evaporation, but not forever. They evaporate until the partial pressure of their gas form reaches a certain level, which is known as vapor pressure. Vapor pressure is defined as the partial pressure of a vapor in dynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.
To understand vapor pressure, we must get familiar with the concept of dynamic equilibrium. Let's discuss this through an example. Normally if you leave a cup of water outside, the water will completely vaporize if you give it enough time. This is because "outside" is not a closed system, and the pressure of water vapor will never reach the vapor pressure as the evaporated vapor molecules will fly away. However if you put a cup of water in a small closed area, the surface level will decrease at first but will stay still afterwards. At this point, we say that the water and vapor have reached a dynamic equilibrium state, where the rate of condensation and rate of vaporization are equal. In other words, the rate at which vapor condenses into water is equal to the rate at which water vaporizes, and therefore it looks like nothing is happening. The partial pressure of water vapor will stay at some constant value, which is the vapor pressure.
Vapor pressure can be a measure of intermolecular forces. Molecules that bond weakly to each other usually have high vapor pressures, and molecules that interact weakly with each other generally have low vapor pressures.
Vapor pressure tends to increase as temperature. This is because as temperature increases, the molecules have more energy, which means they become more labile. The graph below shows the vapor pressures of some liquids according to temperature. Note that mercury and ethylene glycol have relatively low vapor pressures as they have strong intermolecular attractions.
Also observe from the graph the transverse line labeled "normal boiling point (1 atm)". Another interesting fact about vapor pressure is that the boiling point is equal to the temperature at which vapor pressure equals atmospheric pressure.
Raoult's law states that the partial vapor pressure of a component of an ideal mixture is the vapor pressure of the pure component multiplied by its mole fraction. An ideal mixture presumes that the intermolecular interactions are equal between all molecules inside the mixture. If there is a mixture of of ethanol and water, the partial vapor pressure that water vapor exerts in this situation is
p_{\text{water}}=p^*_{\text{water}}x_{\text{water}},
p^*_{\text{water}}
is the vapor pressure of pure water and
x_{\text{water}}
is the mole fraction of water in this mixture.
Suppose you have an ideal mixture of 1 mol each of water and ethanol. If the temperature of the solution is
50^\circ\text{ C}
and the partial vapor pressure exerted by water vapor is
50\text{ mmHg},
what is the vapor pressure of pure water at
50^\circ\text{ C}?
According to Raoult's law we have
\begin{aligned} p_{\text{water}}&=p^*_{\text{water}}x_{\text{water}}\\ 50\text{ mmHg}&=p^*_{\text{water}}\times0.5\\ \Rightarrow p^*_{\text{water}}&=100\text{ mmHg}.\ _\square \end{aligned}
Cite as: Vapor Pressure and Raoult's Law. Brilliant.org. Retrieved from https://brilliant.org/wiki/vapor-pressure-and-raoults-law/
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Ramanujan Master Theorem | Brilliant Math & Science Wiki
Ramanujan Master Theorem
Rishabh Deep Singh, Tapas Mazumdar, and Jimin Khim contributed
In mathematics, Ramanujan's master theorem (named after mathematician Srinivasa Ramanujan) is a technique that provides an analytic expression for the Mellin transform of a function.
The result is stated as follows:
Assume function
f(x)
has an expansion of the form
f\left( x \right) =\sum _{ k=0 }^{ \infty }{ \left( \frac { \phi (k) }{ k! } \right) { \left( -x \right) }^{ k } }
for some function (say analytic or integrable)
\phi(k)
, then the Mellin transform of
f(x)
\left\{ \mathcal{M} f(x) \right\} (s) = \int _{ 0 }^{ \infty }{ { \left( x \right) }^{ s-1 } } f\left( x \right)\, dx=\Gamma (s)\phi (-s),
\Gamma(s)
denotes the gamma function. It was widely used by Ramanujan to calculate definite integrals and infinite series.
Multi-dimensional versions of this theorem also appear in quantum physics (through Feynman diagrams).
Alternative Formalism
Application to Bernoulli Polynomials
Application to the Gamma Function
Evaluation of Quartic Integral
An alternative formulation of Ramanujan's master theorem is as follows:
\int _{ 0 }^{ \infty }{ { \left( x \right) }^{ s-1 }\left( \lambda (0)-x\lambda (1)+{ x }^{ 2 }\lambda (2)-.... \right) } dx =\frac { \pi }{ \sin { \left( \pi s \right) } } \lambda (-s)
0 < \mathfrak{R} (s) < 1
This gets converted to the result expressed before when a substitution of
\lambda(n) = \dfrac{\phi(n)}{\Gamma(n+1)}
is made and the functional equation for the gamma function is further used.
Note: Some authors denote
\phi(n) \to \lambda(n)
\lambda(n) \to \phi(n)
The generating function of the Bernoulli polynomials
B_k(x)
\frac { z{ e }^{ xz } }{ { e }^{ z }-1 } =\sum _{ k=0 }^{ \infty }{ { B }_{ k }\left( x \right) \frac { { \left( z \right) }^{ k } }{ k! } }.
These polynomials are given in terms of Hurwitz zeta function:
\zeta \left( s,a \right) =\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( n+a \right) }^{ s } } }
\zeta \left( 1-n,a \right) =-\frac { { B }_{ n }\left( a \right) }{ n } \quad\text{ for } n\ge 1.
By means of Ramanujan master theorem and generating function of Bernoulli polynomials, one will have the following integral representation:
\int _{ 0 }^{ \infty }{ { x }^{ s-1 }\left( \frac { { e }^{ -ax } }{ 1-{ e }^{ -x } } -\frac { 1 }{ x } \right) dx } =\Gamma (s)\zeta \left( s,a \right)
0<\mathfrak{R}(s)<1
Weierstrass's definition of the gamma function is
\Gamma (x)=\frac { { e }^{ -\gamma x } }{ x } \prod _{ n=1 }^{ \infty }{ { \left( 1+\frac { x }{ n } \right) }^{ -1 } } { e }^{ \left( \frac { x }{ n } \right) },
\gamma
denotes the Euler-Mascheroni constant.
This is equivalent to the expression
\log { \big( \Gamma \left( 1+x \right) \big) } =-\gamma x+\sum _{ k=2 }^{ \infty }{ \frac { \zeta \left( k \right) }{ k } { \left( -x \right) }^{ k } },
\zeta(k)
Then, applying Ramanujan master theorem, we have
\int _{ 0 }^{ \infty }{ { \left( x \right) }^{ s-1 }\left( \frac { \gamma x+\log { \big( \Gamma \left( 1+x \right) \big) } }{ { x }^{ 2 } } \right) dx } =\frac { \pi }{ \sin { \left( \pi s \right) } } \frac { \zeta \left( 2-s \right) }{ (2-s) }
0<\mathfrak{R}(s)<1
Special cases of
s=\frac{1}{2}
s=\frac{3}{4}
\begin{aligned} \int_0^\infty \frac{\gamma x+\log\Gamma(1+x)}{x^{5/2}} \, dx &=\frac{2\pi}{3} \zeta\left( \frac{3}{2} \right) \\\\ \int_0^\infty \frac{\gamma x+\log\Gamma(1+x)}{x^{9/4}} \,dx &= \sqrt{2} \frac{4\pi}{5} \zeta\left(\frac 5 4\right). \end{aligned}
It is well known for the evaluation of
F(a,m)=\int_0^\infty \frac{dx}{(x^4+2ax^2+1)^{m+1}},
which is a quartic integral.
Cite as: Ramanujan Master Theorem. Brilliant.org. Retrieved from https://brilliant.org/wiki/ramanujan-mastered-theorem/
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High School Calculus/The Length of a Plane Curve - Wikibooks, open books for an open world
High School Calculus/The Length of a Plane Curve
Length of a Plane Curve[edit | edit source]
{\displaystyle y=x^{\frac {3}{2}}}
is a curve in the x-y plane. How long is that curve? A definite integral needs endpoints and we specify x = 0 and x = 4. The first problem is to know what "length function" to integrate.
Here is the unofficial reasoning that gives the length of the curve. A straight piece has
{\displaystyle (\Delta x)^{2}+(\Delta y)^{2}}
. Within that right triangle, the height
{\displaystyle \Delta y}
is the slope
{\displaystyle \left({\frac {\Delta y}{\Delta x}}\right)}
{\displaystyle \Delta x}
. This secant slope is close to the slope of the curve. Thus
{\displaystyle \Delta y}
{\displaystyle \left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)\Delta x}
{\displaystyle \Delta s\approx {\sqrt {(\Delta x)^{2}+\left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)^{2}(\Delta x)^{2}}}={\sqrt {1+\left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)^{2}}}\Delta x}
Now add these pieces and make them smaller. The infinitesimal triangle has
{\displaystyle (\operatorname {d} s)^{2}=(\operatorname {d} x)^{2}+(\operatorname {d} y)^{2}}
. Think of
{\displaystyle \operatorname {d} s}
{\displaystyle {\sqrt {1+\left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)^{2}}}\operatorname {d} x}
and integrate:
length of curve =
{\displaystyle \int \operatorname {d} s=\int {\sqrt {1+\left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)^{2}}}\operatorname {d} x}
Retrieved from "https://en.wikibooks.org/w/index.php?title=High_School_Calculus/The_Length_of_a_Plane_Curve&oldid=2178801"
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Health Overview - Mango Markets
Collateral, borrows and perp position values are summed up to give the Health metric
Each asset is weighted differently (Read: Asset Specs)
Health is used to determine liquidations and new position eligibility
The health of an account is used to determine if one can open a new position or if one can be liquidated. There are two types of health, initial health used for opening new positions and maintenance health used for liquidations. They are both calculated as a weighted sum of the assets minus the liabilities but the maint. health uses slightly larger weights for assets and slightly smaller weights for the liabilities. Zero is used as the bright line for both i.e. if your init health falls below zero, you cannot open new positions and if your maint. health falls below zero you will be liquidated. They are calculated as follows:
health = \sum\limits_{i} a_i \cdot p_i \cdot w^a_i - l_i \cdot p_i \cdot w^l_i \newline where \newline a_i-quantity\space asset \space i \newline p_i - price \space of \space i \newline w^a_i - asset \space weight \newline w^l_i - liability \space weight \newline l_i - quantity \space liability \space i
If the health calculation is for determining liquidations, liability weight and asset weight will use the maintenance version. The quote currency, in the typical case USD, will always have all weights and prices equal to 1.
Example - BTC-PERP
As an example, one asset might be BTC-PERP with
init_asset_weight = 0.9
init_liab_weight = 1.1
maint_asset_weight = 0.95
maint_liab_weight = 1.05
These correspond to a 10x initial leverage and 20x maintenance leverage.
Suppose a user deposits 10k USDC and goes long 10 BTC-PERP at 10k each. Then the user has 10 BTC-PERP in assets, and a net 90k in USD liabilities. The init health would be exactly 0 and the maint health would be: maint_health = 10 * 10000 * maint_asset_weight - 90000 = 5000
Suppose the BTC-PERP mark price moves to $9400, then:
maint_health = 10 * 9400 * maint_asset_weight - 90000 = -700
Since the maint_health is now below zero, this account can be liquidated until init health is above zero.
A short position would have negative contract sizes and use maint_liab_weight instead of maint_asset_weight.
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(Redirected from Statistical power)
{\displaystyle H_{0}}
{\displaystyle H_{1}}
{\displaystyle 1-\beta }
{\displaystyle \beta }
10.1 Bayesian power
10.2 Predictive probability of success
11 Software for power and sample size calculations
Main article: Type I and type II errors
{\displaystyle H_{1}}
{\displaystyle {\text{power}}=\Pr {\big (}{\text{reject }}H_{0}\mid H_{1}{\text{ is true}}{\big )}.}
{\displaystyle H_{1}}
{\displaystyle H_{0}}
{\displaystyle H_{0}:\mu =0}
{\displaystyle \mu ,}
{\displaystyle H_{1}:\mu \neq 0}
{\displaystyle n}
{\displaystyle \beta =0.2}
{\displaystyle \alpha =0.05}
{\displaystyle n=16{\frac {s^{2}}{d^{2}}},}
{\displaystyle s^{2}}
{\displaystyle d=\mu _{1}-\mu _{2}}
{\displaystyle d}
{\displaystyle {\bar {Y}}-{\bar {X}}}
{\displaystyle ({\bar {Y}}-{\bar {X}})/\sigma }
{\displaystyle \sigma }
{\displaystyle A_{i}}
{\displaystyle B_{i}}
{\displaystyle i}
{\displaystyle D_{i}=B_{i}-A_{i},}
{\displaystyle H_{0}:\mu _{D}=0.}
{\displaystyle H_{1}:\mu _{D}>0.}
{\displaystyle T_{n}={\frac {{\bar {D}}_{n}-0}{{\hat {\sigma }}_{D}/{\sqrt {n}}}},}
{\displaystyle {\bar {D}}_{n}={\frac {1}{n}}\sum _{i=1}^{n}D_{i},}
{\displaystyle {\hat {\sigma }}_{D}/{\sqrt {n}}}
{\displaystyle N(\mu _{D},\sigma _{D}^{2})}
{\displaystyle \alpha =0.05\,.}
{\displaystyle \Phi ^{-1}}
{\displaystyle T_{n}>1.64\,.}
{\displaystyle \mu _{D}=\theta }
{\displaystyle {\begin{aligned}B(\theta )&=\Pr \left(T_{n}>1.64~{\big |}~\mu _{D}=\theta \right)\\&=\Pr \left({\frac {{\bar {D}}_{n}-0}{{\hat {\sigma }}_{D}/{\sqrt {n}}}}>1.64~{\Big |}~\mu _{D}=\theta \right)\\&=\Pr \left({\frac {{\bar {D}}_{n}-\theta +\theta }{{\hat {\sigma }}_{D}/{\sqrt {n}}}}>1.64~{\Big |}~\mu _{D}=\theta \right)\\&=\Pr \left({\frac {{\bar {D}}_{n}-\theta }{{\hat {\sigma }}_{D}/{\sqrt {n}}}}>1.64-{\frac {\theta }{{\hat {\sigma }}_{D}/{\sqrt {n}}}}~{\Big |}~\mu _{D}=\theta \right)\\&=1-\Pr \left({\frac {{\bar {D}}_{n}-\theta }{{\hat {\sigma }}_{D}/{\sqrt {n}}}}<1.64-{\frac {\theta }{{\hat {\sigma }}_{D}/{\sqrt {n}}}}~{\Big |}~\mu _{D}=\theta \right)\\\end{aligned}}}
{\displaystyle {\frac {{\bar {D}}_{n}-\theta }{{\hat {\sigma }}_{D}/{\sqrt {n}}}}}
{\displaystyle B(\theta )\approx 1-\Phi \left(1.64-{\frac {\theta }{{\hat {\sigma }}_{D}/{\sqrt {n}}}}\right).}
{\displaystyle \theta .}
{\displaystyle \theta }
{\displaystyle \theta ,}
{\displaystyle \theta }
{\displaystyle \alpha ,}
{\displaystyle \theta =0}
{\displaystyle \theta >1,}
{\displaystyle B(1)\approx 1-\Phi \left(1.64-{\frac {\sqrt {n}}{{\hat {\sigma }}_{D}}}\right)>0.90,}
{\displaystyle \Phi \left(1.64-{\frac {\sqrt {n}}{{\hat {\sigma }}_{D}}}\right)<0.10\,.}
{\displaystyle {\frac {\sqrt {n}}{{\hat {\sigma }}_{D}}}>1.64-z_{0.10}=1.64+1.28\approx 2.92\qquad {\text{or}}\qquad n>8.56{\hat {\sigma }}_{D}^{2},}
{\displaystyle z_{0.10}}
{\displaystyle \Phi }
Retrieved from "https://en.wikipedia.org/w/index.php?title=Power_of_a_test&oldid=1078810186"
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Time Delays in Linear Systems - MATLAB & Simulink - MathWorks Benelux
First Order Plus Dead Time Model
Input and Output Delay in State-Space Model
Transport Delay in MIMO Transfer Function
Discrete-Time Transfer Function with Time Delay
Use the following model properties to represent time delays in linear systems.
InputDelay, OutputDelay — Time delays at system inputs or outputs
ioDelay, InternalDelay — Time delays that are internal to the system
In discrete-time models, these properties are constrained to integer values that represent delays expressed as integer multiples of the sample time. To approximate discrete-time models with delays that are a fractional multiple of the sample time, use thiran.
This example shows how to create a first order plus dead time model using the InputDelay or OutputDelay properties of tf.
To create the following first-order transfer function with a 2.1 s time delay:
G\left(s\right)={e}^{-2.1s}\frac{1}{s+10},
G = tf(1,[1 10],'InputDelay',2.1)
where InputDelay specifies the delay at the input of the transfer function.
You can use InputDelay with zpk the same way as with tf:
G = zpk([],-10,1,'InputDelay',2.1)
For SISO transfer functions, a delay at the input is equivalent to a delay at the output. Therefore, the following command creates the same transfer function:
G = tf(1,[1 10],'OutputDelay',2.1)
Use dot notation to examine or change the value of a time delay. For example, change the time delay to 3.2 as follows:
G.OutputDelay = 3.2;
To see the current value, enter:
G.OutputDelay
An alternative way to create a model with a time delay is to specify the transfer function with the delay as an expression in s:
Specify G(s) as an expression in s.
G = exp(-2.1*s)/(s+10);
This example shows how to create state-space models with delays at the inputs and outputs, using the InputDelay or OutputDelay properties of ss.
Create a state-space model describing the following one-input, two-output system:
\begin{array}{l}\frac{dx\left(t\right)}{dt}=-2x\left(t\right)+3u\left(t-1.5\right)\\ y\left(t\right)=\left[\begin{array}{c}x\left(t-0.7\right)\\ -x\left(t\right)\end{array}\right].\end{array}
This system has an input delay of 1.5. The first output has an output delay of 0.7, and the second output is not delayed.
In contrast to SISO transfer functions, input delays are not equivalent to output delays for state-space models. Shifting a delay from input to output in a state-space model requires introducing a time shift in the model states. For example, in the model of this example, defining T = t – 1.5 and X(T) = x(T + 1.5) results in the following equivalent system:
\begin{array}{l}\frac{dX\left(T\right)}{dT}=-2X\left(T\right)+3u\left(T\right)\\ y\left(T\right)=\left[\begin{array}{c}X\left(T-2.2\right)\\ -X\left(T-1.5\right)\end{array}\right].\end{array}
All of the time delays are on the outputs, but the new state variable X is time-shifted relative to the original state variable x. Therefore, if your states have physical meaning, or if you have known state initial conditions, consider carefully before shifting time delays between inputs and outputs.
Define the state-space matrices.
G = ss(A,B,C,D,'InputDelay',1.5,'OutputDelay',[0.7;0])
G is a ss model.
Use delayss to create state-space models with more general combinations of input, output, and state delays, of the form:
\begin{array}{l}\frac{dx}{dt}=Ax\left(t\right)+Bu\left(t\right)+\sum _{j=1}^{N}\left(Ajx\left(t-tj\right)+Bju\left(t-tj\right)\right)\\ y\left(t\right)=Cx\left(t\right)+Du\left(t\right)+\sum _{j=1}^{N}\left(Cjx\left(t-tj\right)+Dju\left(t-tj\right)\right)\end{array}
This example shows how to create a MIMO transfer function with different transport delays for each input-output (I/O) pair.
Create the MIMO transfer function:
H\left(s\right)=\left[\begin{array}{cc}{e}^{-0.1}\frac{2}{s}& {e}^{-0.3}\frac{s+1}{s+10}\\ 10& {e}^{-0.2}\frac{s-1}{s+5}\end{array}\right].
Time delays in MIMO systems can be specific to each I/O pair, as in this example. You cannot use InputDelay and OutputDelay to model I/O-specific transport delays. Instead, use ioDelay to specify the transport delay across each I/O pair.
To create this MIMO transfer function:
Use the variable s to specify the transfer functions of H without the time delays.
H = [2/s (s+1)/(s+10); 10 (s-1)/(s+5)];
Specify the ioDelay property of H as an array of values corresponding to the transport delay for each I/O pair.
H.IODelay = [0.1 0.3; 0 0.2];
H is a two-input, two-output tf model. Each I/O pair in H has the time delay specified by the corresponding entry in tau.
This example shows how to create a discrete-time transfer function with a time delay.
In discrete-time models, a delay of one sampling period corresponds to a factor of
{z}^{-1}
in the transfer function. For example, the following transfer function represents a discrete-time SISO system with a delay of 25 sampling periods.
H\left(z\right)={z}^{-25}\frac{2}{z-0.95}.
To represent integer delays in discrete-time systems in MATLAB, set the 'InputDelay' property of the model object to an integer value. For example, the following command creates a tf model representing
H\left(z\right)
with a sampling time of 0.1 s.
If system has a time delay that is not an integer multiple of the sampling time, you can use the thiran command to approximate the fractional portion of the time delay with an all-pass filter. See Time-Delay Approximation.
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Condense the following log expression. -2\log_3(xy)+\frac{4}{5}\log_3z
Condense the following log expression.-2\log_3(xy)+\frac{4}{5}\log_3z^{1
Condense the following log expression.
-2{\mathrm{log}}_{3}\left(xy\right)+\frac{4}{5}{\mathrm{log}}_{3}{z}^{10}+{\mathrm{log}}_{3}\left(x+1\right)
x{\mathrm{log}}_{a}b={\mathrm{log}}_{a}{b}^{x}
Condensing the logarithmic expression.
-2{\mathrm{log}}_{3}\left(xy\right)+\frac{4}{5}{\mathrm{log}}_{3}{z}^{10}+{\mathrm{log}}_{3}\left(x+1\right)
x{\mathrm{log}}_{a}b={\mathrm{log}}_{a}{b}^{x}
Apply log rule:
x{\mathrm{log}}_{a}b={\mathrm{log}}_{a}{b}^{x}
-{\mathrm{log}}_{3}{\left(xy\right)}^{2}+{\mathrm{log}}_{3}{\left({z}^{10}\right)}^{\frac{4}{5}}+{\mathrm{log}}_{3}\left(x+1\right)
By the property of logarithms.
{\mathrm{log}}_{a}\left(cd\right)={\mathrm{log}}_{a}c+{\mathrm{log}}_{a}d
⇒-{\mathrm{log}}_{3}{\left(xy\right)}^{2}+{\mathrm{log}}_{3}\left({z}^{8}\right)+{\mathrm{log}}_{3}\left(x+1\right)
⇒-{\mathrm{log}}_{3}{\left(xy\right)}^{2}+{\mathrm{log}}_{3}\left({z}^{8}\left(x+1\right)\right)
{\mathrm{log}}_{a}\left(\frac{c}{d}\right)={\mathrm{log}}_{a}c-{\mathrm{log}}_{a}d
⇒{\mathrm{log}}_{3}\left({z}^{8}\left(x+1\right)\right)-{\mathrm{log}}_{3}{\left(xy\right)}^{2}
⇒{\mathrm{log}}_{3}\left(\frac{{z}^{8}\left(x+1\right)}{{\left(xy\right)}^{2}}\right)
\mathrm{ln}\left(\frac{1}{\sqrt{e}}\right)
{10}^{-9}\left[2×{10}^{6}+{3}^{1000}\right]
2×{10}^{-3}+{3}^{1000}×{10}^{-9}
\mathrm{log}\left[2×{10}^{-3}+{3}^{1000}×{10}^{-9}\right]
\mathrm{log}\left(2×{10}^{-3}\right)+\mathrm{log}\left({3}^{1000}×{10}^{-9}\right)
\left(\mathrm{log}2+{\mathrm{log}10}^{-3}\right)+\left({\mathrm{log}3}^{1000}+{\mathrm{log}10}^{-9}\right)
\left(\mathrm{log}2-3\right)+\left(1000×\mathrm{log}3-9\right)
\mathrm{ln}\left(-4-x\right)+\mathrm{ln}3=\mathrm{ln}\left(2-x\right)
Write the following expression as a sum and/or difference of logarithms. Express powers as factors.
\mathrm{ln}\left(\frac{{e}^{4}}{8}\right)
{\mathrm{ln}\left(x+1\right)}^{2}=2
Is it standard to say
-i\mathrm{log}\left(-1\right)
\pi
\underset{x\to 0}{lim}\frac{\mathrm{log}\left(\mathrm{cos}\left(x\right)\right)}{\mathrm{log}\left(\mathrm{cos}\left(3x\right)\right)}
without l'Hopital?
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On the minimizing point of the incorrectly centered empirical process and its limit distribution in nonregular experiments
{F}_{n}
be the empirical distribution function (df) pertaining to independent random variables with continuous df
F
. We investigate the minimizing point
{\stackrel{^}{\tau }}_{n}
of the empirical process
{F}_{n}-{F}_{0}
{F}_{0}
is another df which differs from
F
F
{F}_{0}
are locally Hölder-continuous of order
\alpha
\tau
our main result states that
{n}^{1/\alpha }\left({\stackrel{^}{\tau }}_{n}-\tau \right)
converges in distribution. The limit variable is the almost sure unique minimizing point of a two-sided time-transformed homogeneous Poisson-process with a drift. The time-transformation and the drift-function are of the type
{|t|}^{\alpha }
Classification : 60E15, 60F05, 60F17, 62E20
Mots clés : rescaled empirical process, argmin-CMT, Poisson-process, weak convergence in
D\left(ℝ\right)
author = {Ferger, Dietmar},
title = {On the minimizing point of the incorrectly centered empirical process and its limit distribution in nonregular experiments},
AU - Ferger, Dietmar
TI - On the minimizing point of the incorrectly centered empirical process and its limit distribution in nonregular experiments
Ferger, Dietmar. On the minimizing point of the incorrectly centered empirical process and its limit distribution in nonregular experiments. ESAIM: Probability and Statistics, Tome 9 (2005), pp. 307-322. doi : 10.1051/ps:2005014. http://www.numdam.org/articles/10.1051/ps:2005014/
[1] P. Billingsley, Convergence of probability measures. Wiley, New York (1968). | MR 233396 | Zbl 0172.21201
[2] Z.W. Birnbaum and R. Pyke, On some distributions related to the statistic
{D}_{n}^{+}
. Ann. Math. Statist. 29 (1958) 179-187. | Zbl 0089.14803
[3] Z.W. Birnbaum and F.H. Tingey, One-sided confidence contours for probability distribution functions. Ann. Math. Statist. 22 (1951) 592-596. | Zbl 0044.14601
[4] F.P. Cantelli, Considerazioni sulla legge uniforme dei grandi numeri e sulla generalizzazione di un fondamentale teorema del sig. Paul Levy. Giorn. Ist. Ital. Attuari 4 (1933) 327-350. | Zbl 0007.21802
[5] J. Donsker, Justification and extension of Doob's heuristic approach to the Kolmogorov-Smirnov theorems. Ann. Math. Statist. 23 (1952) 277-281. | Zbl 0046.35103
[6] R.M. Dudley, Weak convergence of probabilities on nonseparable metric spaces and empirical measures on Euclidean spaces. Illinois J. Math. 10 (1966) 109-126. | Zbl 0178.52502
[7] R.M. Dudley, Measures on nonseparable metric spaces. Illinois J. Math. 11 (1967) 449-453. | Zbl 0152.24501
[8] R.M. Dudley, Uniform central limit theorems. Cambridge University Press, New York (1999). | MR 1720712 | Zbl 0951.60033
[9] M. Dwass, On several statistics related to empirical distribution functions. Ann. Math. Statist. 29 (1958) 188-191. | Zbl 0089.14804
[10] R. Dykstra and Ch. Carolan, The distribution of the argmax of two-sided Brownian motion with parabolic drift. J. Statist. Comput. Simul. 63 (1999) 47-58. | Zbl 0946.65001
[11] D. Ferger, The Birnbaum-Pyke-Dwass theorem as a consequence of a simple rectangle probability. Theor. Probab. Math. Statist. 51 (1995) 155-157. | Zbl 0934.62017
[12] D. Ferger, Analysis of change-point estimators under the null hypothesis. Bernoulli 7 (2001) 487-506. | Zbl 1006.62022
[13] D. Ferger, A continuous mapping theorem for the argmax-functional in the non-unique case. Statistica Neerlandica 58 (2004) 83-96. | Zbl 1090.60032
[14] D. Ferger, Cube root asymptotics for argmin-estimators. Unpublished manuscript, Technische Universität Dresden (2005).
[15] V. Glivenko, Sulla determinazione empirica delle leggi die probabilita. Giorn. Ist. Ital. Attuari 4 (1933) 92-99. | Zbl 0006.17403
[16] P. Groneboom, Brownian motion with a parabolic drift and Airy Functions. Probab. Th. Rel. Fields 81 (1989) 79-109.
[17] P. Groneboom and J.A. Wellner, Computing Chernov's distribution. J. Comput. Graphical Statist. 10 (2001) 388-400.
[18] J. Hoffman-Jørgensen, Stochastic processes on Polish spaces. (Published (1991): Various Publication Series No. 39, Matematisk Institut, Aarhus Universitet) (1984). | MR 1217966 | Zbl 0919.60003
[19] I.A. Ibragimov and R.Z. Has'Minskii, Statistical Estimation: Asymptotic Theory. Springer-Verlag, New York (1981). | Zbl 0467.62026
[20] O. Kallenberg, Foundations of Modern Probability. Springer-Verlag, New York (1999). | MR 1876169 | Zbl 0892.60001
[21] K. Knight, Epi-convergence in distribution and stochastic equi-semicontinuity. Technical Report, University of Toronto (1999) 1-22.
[22] A.N. Kolmogorov, Sulla determinazione empirica di una legge di distribuzione. Giorn. Ist. Ital. Attuari 4 (1933) 83-91. | Zbl 0006.17402
[23] N.H. Kuiper, Alternative proof of a theorem of Birnbaum and Pyke. Ann. Math. Statist. 30 (1959) 251-252. | Zbl 0119.15003
[24] T. Lindvall, Weak convergence of probability measures and random functions in the function space
D\left[0,\infty \right)
. J. Appl. Prob. 10 (1973) 109-121. | Zbl 0258.60008
[25] P. Massart, The tight constant in the Dvoretzky-Kiefer-Wolfowitz inequality. Ann. Probab. 18 (1990) 1269-1283. | Zbl 0713.62021
[26] G.Ch. Pflug, On an argmax-distribution connected to the Poisson process, in Proc. of the fifth Prague Conference on asymptotic statistics, P. Mandl, H. Husková Eds. (1993) 123-130.
[28] N.V. Smirnov, Näherungsgesetze der Verteilung von Zufallsveränderlichen von empirischen Daten. Usp. Mat. Nauk. 10 (1944) 179-206. | Zbl 0063.07087
[29] L. Takács, Combinatorial Methods in the theory of stochastic processes. Robert E. Krieger Publishing Company, Huntingtun, New York (1967). | Zbl 0376.60016
[30] A.W. Van Der Vaart and J.A. Wellner, Weak convergence of empirical processes. Springer-Verlag, New York (1996). | MR 1385671
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Mode | Brilliant Math & Science Wiki
Mode is that observation in a frequency distribution which occurs the maximum times in the frequency distribution, or technically speaking, which has the highest frequency. Mode is derived from the French word La Mode which means fashion. A frequency distribution may have one or more than one modes. The distribution having only a single mode is called unimodal frequency distribution and the distribution having two modes is called bi-modal frequency distribution. Here are some formulas for calculating mode:
Individual Series: Just check out the maximum number of times an individual observation occurs.
Discrete Series: Just check out the highest frequency of the observations.
Continuous Series: The formula is
l+\frac{f_{1}-f_{0}}{2f_{1}-f_{2}-f_{0}}h,
l
represents the lower limit of the modal class;
f_{1}
represents the frequency of the modal class;
f_{2}
represents the frequency of the class interval succeeding the modal class;
f_{0}
represents the frequency of the class interval preceding the modal class;
h
represents the width of the class interval.
For example, if the class interval is
30-40
, then its
Lower Limit:
30
40-30=10
Upper Limit:
40
For more information on different types of series, see Statistical Series.
Merits: Mode is easy to calculate and understand. In some cases, it can be located merely by inspection. It can also be estimated graphically from a histogram. Mode is not at all affected by extreme observations. It can be conveniently obtained in the case of open end classes.
Demerits: Mode is not rigidly defined. From the modal values and the sizes of two or more series, we cannot find the mode of the combines series.
What is the mode of the following distribution:
1,2,3,3,4,4,4,5,5,6,6,7,7,7,7,8,8,9?
Mode is the observation that occurs the most in a frequency distribution. Since
7
occurs the most number of times in this distribution, thus the mode is
7
_\square
Cite as: Mode. Brilliant.org. Retrieved from https://brilliant.org/wiki/data-mode/
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Let’s use the random vector
x
to represent an uncertain state, and the random vector
z
to represent an uncertain measurement. Even before making any actual measurements, we should have prior idea of the likelihoods of different values of the combined vector
\begin{bmatrix} x\\ z \end{bmatrix}
. These are subjective assessments of the following sort.
x
is probably close to the known vector
a
z
will probably turn out to be close to the known vector
b
z
is probably close to
F x
F
is a known matrix
To encode these prior subjective beliefs numerically, we can say that
\begin{bmatrix} x\\ z \end{bmatrix}
is distributed as a Gaussian random variable.
\begin{bmatrix} x\\ z \end{bmatrix} \sim N(\begin{bmatrix} \mu_x\\ \mu_z \end{bmatrix}, \begin{bmatrix} \Sigma_{xx} & \Sigma_{xz}\\ \Sigma_{zx} & \Sigma_{zz} \end{bmatrix})
\Sigma_{xx}
describes how close we believe
x
\mu_x
\Sigma_{zz}
z
will be to
\mu_z
\Sigma_{xz} = \Sigma_{zx}^T
describes how correlated we think
z
x
The Kalman Filter can be viewed as a principled way to choose
\mu_x \, \mu_z \, \Sigma_{xx} \, \Sigma_{xz} \, \Sigma_{zx} \, \Sigma_{zz}
. There are also other ways to choose these priors, but suppose for now that we have chosen them sensibly.
Now that we have a prior
p(x,z)
, we can incorporate any measurement
z_0
into the state by simply taking the posterior estimate
p(x|z=z_0)
. We will find in the next section that the posterior is
x|z=z_0
distributed as a Gaussian with the following parameters.
\mu_{x|z=z_0} = \mu_x + \Sigma_{xx}\Sigma_{xz}^{-1}(z_0 - \mu_z)
\Sigma_{x|z=z_0} =\Sigma_{xx} - \Sigma_{xz}\Sigma_{zz}^{-1}\Sigma_{zx}
I will call these the equations the Bayes Inference equations.
Deriving the Bayes Inference Equations
In this section I’ll derive the Bayes inference equations.
\mu_{x|z=z_0} = \mu_x + \Sigma_{xx}\Sigma_{xz}^{-1}(z_0 - \mu_z)
\Sigma_{x|z=z_0} =\Sigma_{xx} - \Sigma_{xz}\Sigma_{zz}^{-1}\Sigma_{zx}
Feel free to come back to this section later if you’re willing to take these equations on faith for now.
We have the proportionality relationship
p(x|z=z_0) = \frac {p(x, z_0)} {p(z_0)}\propto p(x,z_0)
. This means only have to evaluate the right hand side
p(x,z_0)
in order to know the distribution
p(x|z=z_0)
Remember the Gaussian density, where
K
represents a constant of integration that we don’t care about.
p(x,z) = K\exp(-\frac{1}{2}\begin{bmatrix} x - \mu_x\\ z - \mu_z\end{bmatrix} ^T\begin{bmatrix} \Sigma_{xx} & \Sigma_{xz}\\ \Sigma_{zx} & \Sigma_{zz} \end{bmatrix}^{-1}\begin{bmatrix} x - \mu_x\\ z - \mu_z \end{bmatrix})
It will be convenient to use the inverse covariance matrix, also known as the information matrix.
\begin{bmatrix} \Lambda_{xx} & \Lambda_{xz}\\ \Lambda_{zx} & \Lambda_{zz} \end{bmatrix}\equiv \begin{bmatrix} \Sigma_{xx} & \Sigma_{xz}\\ \Sigma_{zx} & \Sigma_{zz} \end{bmatrix}^{-1}
We can substitute the information matrix and expand.
p(x,z) = K\exp(-\frac{1}{2}\begin{bmatrix} x\\ z \end{bmatrix}^T\begin{bmatrix} \Lambda_{xx} & \Lambda_{xz}\\ \Lambda_{zx} & \Lambda_{zz} \end{bmatrix}\begin{bmatrix} x\\ z \end{bmatrix} + \begin{bmatrix} x\\ z \end{bmatrix}^T \begin{bmatrix} \Lambda_{xx} & \Lambda_{xz}\\ \Lambda_{zx} & \Lambda_{zz} \end{bmatrix}\begin{bmatrix} \mu_x\\ \mu_z \end{bmatrix} )
z = z_0
and expand more. We can collect any terms that are not multiplied by
x
into a constant
C
p(x,z_0) = K\exp(-\frac{1}{2}x^T\Lambda_{xx}x - x^T\Lambda_{xy}z_0 + x^T \Lambda_{xx} \mu_x + x^T\Lambda_{xz}\mu_z + C)
C
K
both drop out as scaling constants.
p(x,z_0) \propto \exp(-\frac{1}{2}x^T\Lambda_{xx}x - x^T\Lambda_{xy}z_0 + x^T \Lambda_{xx} \mu_x + x^T\Lambda_{xz}\mu_z)
p(x,z_0) \propto \exp(-\frac{1}{2}x^T\Lambda_{xx}x + x^T(\Lambda_{xx} \mu_x - \Lambda_{xz}(z_0 - \mu_z))
Complete the square by first by rewriting
\Lambda_{xx} \mu_x - \Lambda_{zz}(z_0 - \mu_z) \rightarrow \Lambda_{xx} (\mu_x - \Lambda_{xx}^{-1}\Lambda_{xz}(z_0 - \mu_z))
p(x,z_0) \propto \exp(-\frac{1}{2}x^T\Lambda_{xx}x + x^T\Lambda_{xx} (\mu_x - \Lambda_{xx}^{-1}\Lambda_{xz}(z_0 - \mu_z)))
p(x,z_0) \propto \exp(-\frac{1}{2} (x - (\mu_x - \Lambda_{xx}^{-1}\Lambda_{zz}(z_0 - \mu_z)))^T\Lambda_{xx}(x - (\mu_x - \Lambda_{xx}^{-1}\Lambda_{xz}(z_0 - \mu_z))))
Note that this is the probability density of a Gaussian with mean
\mu_x - \Lambda_{xx}^{-1}\Lambda_{xz}(z_0 - \mu_z)
\Lambda_{xx}^{-1}
\mu_{x|z=z_0} = \mu_x - \Lambda_{xx}^{-1}\Lambda_{xz}(z_0 - \mu_z)
\Sigma_{x|z=z_0} = \Lambda_{xx}^{-1}
This formula is written in terms of the information matrix, but in many cases it is more convenient to write it in terms of the covariance matrix. To accomplish this, we can use the block-matrix inversion formula where
\Sigma/\Sigma_{zz}
is the Schur complement
\Sigma_{xx} - \Sigma_{xz}\Sigma_{zz}^{-1}\Sigma_{zx}
\begin{bmatrix} \Lambda_{xx} & \Lambda_{xz}\\ \Lambda_{zx} & \Lambda_{zz} \end{bmatrix} = \begin{bmatrix} \Sigma_{xx} & \Sigma_{xz}\\ \Sigma_{zx} & \Sigma_{zz} \end{bmatrix}^{-1} = \begin{bmatrix} (\Sigma/\Sigma_{zz})^{-1} & - (\Sigma/\Sigma_{zz})^{-1}\Sigma_{xz}\Sigma_{zz}^{-1} \\ \Sigma_{zz}^{-1}\Sigma_{zx}(\Sigma/\Sigma_{zz})^{-1} & \Sigma_{zz}^{-1} + \Sigma_{zz}^{-1}\Sigma_{zx} (\Sigma/\Sigma_{zz})^{-1}\Sigma_{xz}\Sigma_{zz}^{-1}\end{bmatrix}
\Lambda_{xx}^{-1} = \Sigma/\Sigma_{zz}
-\Lambda_{xx}^{-1} \Lambda_{xz} = \Sigma_{xx}\Sigma_{xz}^{-1}
. Therefore we can write the distribution of
x|_{z=z_0}
in terms of the covariance matrix.
\mu_{x|z=z_0} = \mu_x + \Sigma_{xz}\Sigma_{zz}^{-1}(z_0 - \mu_z)
\Sigma_{x|z=z_0} =\Sigma_{xx} - \Sigma_{xz}\Sigma_{zz}^{-1}\Sigma_{zx}
In the first section, I mentioned that the Kalman Filter can be seen as a principled way to establish the priors
\mu_x, \, \mu_z, \, \Sigma_{xx}, \, \Sigma_{zz},\, \Sigma_{xz}
Remember we wanted these priors so that, given an actual measurement
z_0
, we could apply the Bayes inference equations.
\mu_{x|z=z_0} = \mu_x + \Sigma_{xz}\Sigma_{zz}^{-1}(z_0 - \mu_z)
\Sigma_{x|z=z_0} =\Sigma_{xx} - \Sigma_{xz}\Sigma_{zz}^{-1}\Sigma_{zx}
The Kalman Filter sets up
\mu_x, \, \mu_z, \, \Sigma_{xx}, \, \Sigma_{zz},\, \Sigma_{xz}
by supposing that the state variable
x
and the measurement variable
z
are both caused by a single prior variable
x_0 \sim N(\mu_{x_0}, \Sigma_{x_0})
, via a state-update matrix
F
, and a measurement matrix
H
w \sim N(0, \Sigma_w)
as independent process noise, we assume our state
x
arises from
x_0
x = F x_0 + w
v \sim N(0, \Sigma_v)
as independent measurement error, we assume our measurement
z
x
(and ultimately from
x_0
z = Hx + v
These two equations are enough to generate the list
\mu_x, \, \mu_z, \, \Sigma_{xx}, \, \Sigma_{zz},\, \Sigma_{xz}
via straightforward computations. See the next section for those derivations in detail.
We will end up with the following.
\mu_x = F\mu_{x_0}
\mu_z = HF\mu_{x_0}
\Sigma_{xx} =F\Sigma_{x_0}F^T + \Sigma_w
\Sigma_{xz} = \Sigma_{xx}H^T
\Sigma_{zz} = H\Sigma_{xx}H^T + \Sigma_v
That’s it! Now plug those values into the Bayes update rule and you have a Kalman Filter!
\mu_{x|z=z_0} = F\mu_{x_0}+ \Sigma_{xz}\Sigma_{zz}^{-1}(z_0 - \mu_z)
\Sigma_{x|z=z_0} =\Sigma_{xx} - \Sigma_{xz}\Sigma_{zz}^{-1}\Sigma_{zx}
A note on terminology for comparison to the Wikipedia article on Kalman Filter:
\mu_x
is called the predicted mean
\Sigma_{xx}
is called the predicted covariance
\Sigma_{zz}
is called the innovation, or pre-fit residual covariance
\Sigma_{xz} \Sigma_{zz}^{-1} = \Sigma_{xx}H^T\Sigma_{zz}^{-1}
is called the optimal Kalman Gain
Deriving the Kalman Filter In Detail
In this section, I’ll show theses equalities.
\mu_x = F\mu_{x_0}
\mu_z = HF\mu_{x_0}
\Sigma_{xx} =F\Sigma_{x_0}F^T + \Sigma_w
\Sigma_{xz} = \Sigma_{xx}H^T
\Sigma_{zz} = H\Sigma_{xx}H^T + \Sigma_v
Here are the means.
\mu_x = E[x] =E[Fx_0 + w] = FE[x_0] + E[w] = F\mu_{x_0} + 0 = F\mu_{x_0}
\mu_z = E[z] = E[Hx + v] = HE[x] + E[v] = H\mu_x + 0 = HF\mu_{x_0}
Here are the covariances and cross covariance. It will be convenient to define the delta operator
\Delta
\Delta y = y - E[y]
. Also, for zero-mean variables like
v, \Delta v = v
\Sigma_{xx} = E[\Delta x \Delta x^T]
= E[ (F \Delta x_0 + w)(F \Delta x_0 + w)^T]
= E[ (F \Delta x_0 \Delta x_0^T F^T)] + E[ F\Delta x w^T] + E[ w \Delta x^T F^T] + E[ w w^T]
Use independence to distribute expectation in the second and third terms.
= F E[\Delta x_0 \Delta x_0^T] F^T + E[ F\Delta x]E[ w^T] + E[ w]E[ \Delta x^T F^T] + E[ w w^T]
=F\Sigma_{x_0}F^T + 0 +0 + \Sigma_w
=F\Sigma_{x_0}F^T + \Sigma_w
\Sigma_{xz} = E[\Delta x \Delta z^T]
= E[\Delta x\Delta(Hx + v)^T]
= E[\Delta x (H \Delta x + v)^T]
= E[\Delta x \Delta x^T]H^T + E[\Delta x ]E[v^T]
=\Sigma_{xx} H^T + 0
=\Sigma_{xx} H^T
\Sigma_{zz} = E[\Delta z\Delta z^T]
= E[\Delta (Hx + v)\Delta (Hx + v)^T]
= E[(H\Delta x + v)(H\Delta x + v)^T]
= HE[\Delta x \Delta x^T]H^T + E[v]E[(H\Delta x+ v)^T] +E[H\Delta x+ v]E[v^T] + E[vv^T]
= H\Sigma_{xx} H^T + 0 + 0 + \Sigma_v
= H\Sigma_{xx} H^T + \Sigma_v
We Learned the Wrong Way to Matrix Multiply →
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5.2 Ledger State | IOTA Wiki
The introduction of a voting-based consensus requires a fast and easy way to determine a node's initial opinion for every received transaction. This includes the ability to both detect double spends and transactions that try to spend non-existing funds. These conditions are fulfilled by the introduction of an Unspent Transaction Output (UTXO) model for record-keeping, which enables the validation of transactions in real time, see also the section on UTXO.
The Ledger State depends on:
UTXO: see the Section on UTXO DAG as well as 5.1 UTXO.
Tangle: the Tangle maps the approval relations between messages as well as transactions, see 4.1 The Tangle.
Solidification: Secures that all non-conflicting transactions converge to the same ledger state, see 4.4 Solidification.
5.2.3 Realities Ledger State
its extension to the corresponding branches and the branch DAG,
the Tangle, which maps the parent relations between messages and thus also transactions.
5.2.4 The UTXO DAG
The UTXO DAG models the relationship between transactions, by tracking which outputs have been spent by what transaction, see also the section on UTXO. Since outputs can only be spent once, we use this property to detect double spends.
We allow for different versions of the ledger to coexist temporarily. This is enabled by extending the UTXO DAG by the introduction of branches (see the following section). We can then determine which conflicting versions of the ledger state exist in the presence of conflicts. Thus, we allow for different versions of the ledger to coexist temporarily.
5.2.4.1 Conflict Sets and Detection of Double Spends
For every output we maintain a list of consumers consumerList, and where the consumers have the unique identifier consumerID. For a given output this list keeps track of which transactions have spent that particular output. For every spending transaction we add an element with consumerID=transactionID. Outputs without consumers are considered to be unspent outputs. Transactions that consume an output that have more than one consumer are considered to be double spends.
When there are more than one consumer in the consumer list we shall create a conflict set list conflictSet, whose elements have a unique identifier conflictID each. The conflictSet is uniquely identified by the unique identifier conflictSetID. Since the outputID is directly and uniquely linked to the conflict set, we set conflictSetID=outputID. For every transaction that shall be added to the conflict set we add an element with conflictID=transactionID.
5.2.5 Branches
The UTXO model and the concept of solidification, see section 4.4 Solidification, makes all non-conflicting transactions converge to the same ledger state no matter in which order the transactions are received. Messages containing these transactions could always reference each other in the Tangle without limitations.
However, every double spend creates a new possible version of the ledger state that will no longer converge. Whenever a double spend is detected (see the previous section), we track the outputs created by the conflicting transactions and all the transactions that spend these outputs, by creating a container for them in the ledger which we call a branch.
More specifically a container branch shall be created for each transaction that double spends one or several outputs, or if messages aggregate those branches. Every transaction that spends directly or indirectly from a transaction that created a branch, i.e. double spent funds, is also contained in this branch or one of its child branches. Note that a branch that was created by a transaction that spends multiple outputs can be part of multiple conflict sets.
In other words, a branch is a downward closed, conflict free collection of conflicts.
5.2.5.1 Conflict Branches
On solidification of a message, we shall store the corresponding branchID together with every output, as well as the transaction metadata to enable instant lookups of this information. Thus, on solidification, a transaction can be immediately associated with a branch.
5.2.5.2 Aggregated Branches
A transaction that does not create a double spend inherits the branches of the input's branches. In the simplest case, where there is only one input branch the transaction inherits that branch. If outputs from multiple non-conflicting branches are spent in the same transaction, then the transaction and its resulting outputs are part of an aggregated branch. This type of branch is not part of any conflict set. Rather it simply combines the perception that the individual conflict branches associated to the transaction's inputs are the ones that will be accepted by the network.
Furthermore, since a message inherits the branches from its parents, it also can create aggregated branches.
Each aggregated branch shall have a unique identifier branchID, which is the same type as for conflict branches. Furthermore, the container for an aggregated branch is also of type branch.
To calculate the unique identifier of a new aggregated branch, we take the identifiers of the branches that were aggregated, sort them lexicographically and hash the concatenated identifiers once:
# AggregatedBranchID returns the identifier for an aggregated branch.
FUNCTION aggregatedBranchID = GetAggregatedBranchID(branchIDs)
sortedBranchIDs = Sort(branchIDs)
RETURN Hash(sortedBranchIDs)
IF NOT (branch IN reducedBranches)
5.2.5.3 The Branch DAG
A new branch is created for each transaction that is part of a conflict set, or if a transaction aggregates branches.
In the branch DAG, branches constitute the vertices of the DAG. A branch that is created by a transaction that is spending outputs from other branches has edges pointing to those branches. The branch DAG maps the UTXO DAG to a simpler structure that ignores details about relations between transactions inside the branches and instead retains only details about the interrelations of conflicts.
The set of all non-conflicting transactions form the master branch. Thus, at its root the branch DAG has the master branch, which consists of non-conflicting transaction and resolved transactions. From this root of the branch DAG the various branches emerge.
In other words the conflict branches and the aggregated branches appear as the children of the master branch.
5.2.5.4 Detecting Conflicting Branches
Branches are conflicting if they, or any of their ancestors, are part of the same conflict set.
The branch DAG can be used to check if branches are conflicting, by applying an operation called normalization, to a set of input branches.
From this information we can identify messages or transactions that are trying to combine branches belonging to conflicting double spends, and thus introduce an invalid perception of the ledger state.
Since branches represent the ledger state associated with a double spend and sub-branches implicitly share the perception of their parents, we define a function NormalizeBranches() to normalize a list of branches and that gets rid of all branches that are referenced by other branches in that list. The function returns NULL if the branches are conflicting and can not be merged.
In order to explain this function in pseudo code we require the following global variables
seenConflictSets = map[]conflictSetID
traversedBranches = map[]branch
parentsToCheck = map[]branch
as well as a function BranchCheck() that performs certain checks and returns TRUE when the branch is conflicting with a previously seen branch. However, we note that this is an implementation detail that must not match the implementation.
# reduce list of branches to normalized branches, and return NULL when detecting conflicting branches
FUNCTION normalizedBranches = NormalizeBranches(initialBranches)
IF Len(initialBranches) == 0
RETURN masterBranch
RETURN initialBranches
# check original set of branches
normalizedBranches = ReduceBranches(initialBranches)
FOR branch IN normalizedBranches
BranchCheck(branch)
# check every ancestor
WHILE Len(parentsToCheck) != 0
branch = parentsToCheck[0]
Delete(parentsToCheck,branch) # delete this branch from the list
# remove this ancestor
IF branch IN normalizedBranches
Delete(normalizedBranches,branch)
# if branch check fails, i.e. a conflict set was seen twice, return a null list
IF BranchCheck(branch)
RETURN normalizedBranches
The branch check function BranchCheck() checks if the branch was already traversed, i.e. we have handled this branch already. Then it checks if the branch's conflict set has been already seen, which proofs that the current branch conflicts with an already traversed branch. Lastly it adds new branches to the queue of branches that should be traversed.
FUNCTION isConflicting = BranchCheck(branch)
# abort if branch was traversed already
IF branch IN traversedBranches
Append(traversedBranches,branch.ID)
# check if conflict set was seen twice
IF branch.conflictSetID IN seenConflictSets
Append(seenConflictSets,branch.conflictSetID)
# queue parents to be checked when traversing ancestors
IF branch NOT IN parentsToCheck
Append(parentsToCheck,parentBranch)
5.2.5.5 Merging of Branches
A branch gains approval weight when messages from (previously non-attached) nodeIDs attach to messages in the future cone of that branch. Once the approval weight exceeds a certain threshold we consider the branch as confirmed, see also section 6.4 Finality.
However, there are two special cases of branches:
First the branch that is created by the genesis transaction is called master branch and has the identifier masterBranchID. The masterBranchID is confirmed on creation and thus it is the "correct" reality by definition. Once a conflict branch is confirmed, it can be merged into the master branch. Since the approval weight is monotonically increasing for branches from the past to the future, branches are only merged into the master branch.
Second, a branch rejectedBranch is created that is rejected by definition, and it has the identifier rejectedBranchID. Messages that are contained in a rejected branch or in one its child branches are booked into the rejectedBranch.
5.2.6 Relation to the Tangle
Since messages in the Tangle are dependent on the fate of the messages they approve, we shall create dependencies between payloads, messages, and branches. The branch ID of a message or of a transaction represents all the conflicts upon which that object depends. Specifically, we associate a branch to a payload and to a message in the following way.
The branch of a non-value payload is always the master branch.
The branch of a transaction is assigned in one of two ways:
If the transaction is a conflict, then a new branch is created whose branchID is that transactionID. The transaction gets assigned to this new branch.
Otherwise, the transaction is assigned to the aggregated branch of all its inputs.
The branch of a message is the aggregate of
The branch of its payload
The branches of each strong parent
The branches of the payloads of the weak parents.
This assignments captures the essence of weak and strong parents, see 4.3 Tip Selection Specification. Strong arrows pick up the dependencies of the whole past cone, where as the weak arrows only penetrate to the paylaod of the parent, ignoring the history of the parent.
We that a message
M
(resp. transaction
X
) belongs to a branch
B
if the branch
A
M
X
) is in the branch past of
B
. Thus, branches, represent certain coherent sections of the Tangle which are then ordered by inclusion.
After a message is solidified, it and its payload are both assigned to their branch. During this check, the message is flagged as invalid if:
The payload is a transaction, and the node cannot aggregate branchIDs of the transaction's input into a valid branchID.
The branchIF of the message cannot be aggregated. If these branchID's cannot be computed, then the message contains a pair of its history, and thus does not support a coherent view of the ledger.
5.2.3 Realities Ledger State
5.2.4 The UTXO DAG
5.2.4.1 Conflict Sets and Detection of Double Spends
5.2.5.1 Conflict Branches
5.2.5.2 Aggregated Branches
5.2.5.3 The Branch DAG
5.2.5.4 Detecting Conflicting Branches
5.2.5.5 Merging of Branches
5.2.6 Relation to the Tangle
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Double Chooz - Wikipedia
The Double Chooz logo
Double Chooz was a short-baseline neutrino oscillation experiment in Chooz, France. Its goal was to measure or set a limit on the θ13 mixing angle, a neutrino oscillation parameter responsible for changing electron neutrinos into other neutrinos. The experiment uses reactors of the Chooz Nuclear Power Plant as a neutrino source and measures the flux of neutrinos they receive. To accomplish this, Double Chooz has a set of two detectors situated 400 meters and 1050 meters from the reactors. Double Chooz was a successor to the Chooz experiment; one of its detectors occupies the same site as its predecessor. Until January 2015 all data has been collected using only the far detector. The near detector was completed in September 2014, after construction delays,[1] and started taking data at the beginning of 2015. Both detectors stopped taking data in late December 2017.
1 Detector design
1.1 Neutrino target and γ-catcher
1.2 Buffer vessel and PMTs
1.3 Inner and outer vetos
2.1 Neutrino mixing
3.1 Mixing angle
Detector design[edit]
Double Chooz used two identical gadolinium-doped liquid scintillator detectors[2] placed in vicinity of two 4.25 GW thermal power reactors to measure antineutrino disappearance. The two detectors are aptly referred to as "near", 400 meters from the reactor; and "far", 1,050 meters from the reactor. The far detector is placed inside a hill such that there is a 300 meters of water equivalent of shielding from cosmic muons. The detector itself is a calorimetric liquid scintillator consisting of four concentric cylindrical vessels.[3][4]
Neutrino target and γ-catcher[edit]
The innermost vessel is made of acrylic plastic and has a diameter of 230 cm, a height of 245.8 cm, and a thickness of 0.8 cm. This chamber is filled with 10,000 liters of gadolinium (Gd) loaded (1 gram/liter) liquid scintillator; it is the neutrino target. The next layer out is the γ-catcher. It surrounds the neutrino target with a 55 cm thick layer of Gd-free liquid scintillator. The casing for the γ-catcher is 12 cm thick and made of the same material as the neutrino catcher. The materials are chosen so that both of these vessels are transparent to photons with a wavelength greater than 400 nm.[3][4]
Buffer vessel and PMTs[edit]
The buffer vessel is made of stainless steel 304L with dimensions of 552.2 cm wide by 568.0 cm tall and 0.3 cm thick. The remainder of the interior space that isn't occupied by the acrylic double vessel is filled with a non-scintillating mineral oil. On the inner surface of the buffer vessel are 390 10-inch photomultiplier tubes. The purpose of the buffer layer is to shield from radioactivity in the PMTs and the surrounding rock. These to layers in addition to the neutrino target and γ-catcher are collectively referred to as the "inner detector."[3][4]
Inner and outer vetos[edit]
The inner veto surrounds the buffer vessel with a 50 cm thick layer of scintillating mineral oil. In addition, it has 78 8-inch PMTs distributed on the top, bottom and sides. This inner veto layer serves as an active veto layer for muons and fast neutrons. The surrounding 15 cm thick steel casing further serves to shield against external γ-rays. The outer veto covers the top of the detector tank. It consists of strips with a 5 cm x 1 cm cross section laid in orthogonal directions.[3][4]
Signals from the inner detector and the inner veto are recorded by 8-bit flash ADC electronics with a sampling rate of 500 MHz. The trigger threshold for the detectors is set to 350 keV, much lower than the 1.02 MeV expected of the electron anti-neutrinos.[3][4]
For several years Double Chooz has operated with only the far detector and has used models such as Bugey4 to calculate the expected flux. The completed near detector will allow increased precision in the next years of data taking.
Neutrino mixing[edit]
Neutrinos are electrically neutral, extremely light particles that only interact weakly, meaning they can travel vast distances without ever being noticed. One of the properties of neutrinos is that as the propagate they have a chance to oscillate from one flavor (
{\displaystyle e,\mu ,\tau }
) to another, and this is the principle under which the experiment operates. The goal of Double Chooz is to more tightly constrain the value for the
{\displaystyle \theta _{13}}
mixing angle.
The Chooz experiment, performed in the 1990s, found that the
{\displaystyle \theta _{13}}
mixing angle is constrained by
{\displaystyle \sin ^{2}(2\theta _{13})<0.2}
which was the best experimental upper limit for over a decade. The goal of the Double Chooz experiment is to continue to explore the
{\displaystyle \theta _{13}}
angle by probing an even smaller region
{\displaystyle 0.03<\sin ^{2}(2\theta _{13})<0.2}
Observations of the mixing angle are accomplished by observing the
{\displaystyle {\bar {\nu }}_{e}}
flux that comes off of the reactors during their fission reactions. The expected
{\displaystyle {\bar {\nu }}_{e}}
flux from the reactors is about 50 per day. Because one of the neutrino mass-squared differences is much smaller than the other, the Double Chooz experiment only needs to consider a two-flavor oscillation. In the two-flavor model the survival probability of any given neutrino is modelled by
{\displaystyle P=1-\sin ^{2}(2\theta _{13})\,\sin ^{2}\left({\frac {1.27\Delta m_{31}^{2}L}{E_{\nu }}}\right)\quad \mathrm {(in\;natural\;units).} }
{\displaystyle L}
is the length in meters the neutrino has travelled and
{\displaystyle E_{\nu }}
is the energy of the
{\displaystyle {\bar {\nu }}_{e}}
particle. From this the value of the mixing angle can be measured from the oscillation amplitude in reactor neutrino oscillations.[4]
The neutrinos from the reactor are observed via the inverse beta decay (IBD) process
{\displaystyle {\bar {\nu }}_{e}+p\to e^{+}+n.}
Since there are backgrounds to consider, candidates for (IBD) are determined by the following: visible energy from the prompt signal must be between 0.5 and 20 MeV; the delayed signal must have an energy between 4 and 10 MeV; the time difference between those two signals must be between 0.5 and 150 microseconds; the distance between the vertices of the two signals should be less than 100 cm; and no other signals (except for the delayed signal) are found 200 microseconds before or 600 microseconds after the prompt signal. Detection of the prompt signal has reached nearly 100% efficiency, however it is not as easy to detect the delayed signal due to issues such as Gd-concentration and neutron scattering models.[4]
Mixing angle[edit]
In November 2011, first results of the experiment, using 228 days of data, were presented at the LowNu conference in Seoul, hinting at a non-zero value of θ13,[5] followed by an article submitted to arXiv in December 2011.[6] In the PRL article[7] (published in 2012), the zero θ13 oscillation hypothesis was excluded at 2.9 sigma by combining the Double Chooz experiment disappearance data and the T2K experiment appearance data, that had been released only some months before. This result became both the most important evidence at the time and the first accurate measurement of the amplitude of θ13. Only some months later, the Daya Bay experiment provided its confirming measurement and the ultimate discovery (i.e ≥5σ significance) evidence. The central values of both Double Chooz and Daya Bay experiments were in excellent agreement and has remained so (within ≤2σ) so far. A similar analysis combination technique as done by the Double Chooz experiment in 2012 has been employed by the T2K experiment to yield the first constraints on the non-zero CP-violation phase in 2020.
Neutron capture on hydrogen was used to produce independent data, which was analysed to yield a separate measurement in 2013:[8]
{\displaystyle \sin ^{2}(2\theta _{13})=0.097\pm 0.034\,\mathrm {(stat)} \pm 0.034\,\mathrm {(syst)} .}
Using reactor-off data, a background-independent measurement[9] was published July 2014 in Physics Letters B:
{\displaystyle \sin ^{2}(2\theta _{13})=0.102\pm 0.028\,\mathrm {(stat)} \pm 0.033\,\mathrm {(syst)} .}
An improved measurement with reduced background and systematic uncertainties after 467.90 days of data was published in the Journal of High Energy Physics in 2014:[4]
{\displaystyle \sin ^{2}(2\theta _{13})=0.090_{-0.029}^{+0.032}.}
Double Chooz was able to identify positronium formation in their detector, which delays positron annihilation and distorts the scintillation signal.[10] A tagging algorithm was developed that could be used in neutrino detectors for improved background rejection, which was similarly done by Borexino for cosmogenic 11C background. An ortho-positronium lifetime of 3.68±0.15 ns was measured, compatible with other dedicated setups.
Limits on Lorentz violation parameters were also set.[11]
Apollonio, M.; et al. (2003). "Search for neutrino oscillations on a long base-line at the CHOOZ nuclear power station". The European Physical Journal C. 27 (3): 331–374. arXiv:hep-ex/0301017. Bibcode:2003EPJC...27..331A. doi:10.1140/epjc/s2002-01127-9. S2CID 14226312.
Ardellier, F.; et al. (2006). "Double Chooz: A Search for the Neutrino Mixing Angle θ13". arXiv:hep-ex/0606025.
Huber, P.; et al. (2006). "From Double Chooz to Triple Chooz — Neutrino Physics at the Chooz Reactor Complex". Journal of High Energy Physics. 0605 (72): 072. arXiv:hep-ph/0601266. Bibcode:2006JHEP...05..072H. doi:10.1088/1126-6708/2006/05/072. S2CID 13576581.
^ "Inauguration of second neutrino detector for Double Chooz experiment". 25 September 2014.
^ L, Mikaelyan and; V, Sinev (2000). "Neutrino Oscillations at Reactors: What Is Next?". Physics of Atomic Nuclei. 63 (6): 1002. arXiv:hep-ex/9908047. Bibcode:2000PAN....63.1002M. doi:10.1134/1.855739. S2CID 15221390.
^ a b c d e Ardellier, F.; et al. (2006). "Double Chooz: A Search for the Neutrino Mixing Angle θ13". arXiv:hep-ex/0606025. Bibcode:2006hep.ex....6025G. {{cite journal}}: Cite journal requires |journal= (help)
^ a b c d e f g h i Abe, Y.; et al. (Double Chooz Collaboration) (2014). "Improved measurements of the neutrino mixing angle θ13 with the Double Chooz detector". Journal of High Energy Physics. 2014 (10): 86. arXiv:1406.7763. Bibcode:2014JHEP...10..086A. doi:10.1007/JHEP10(2014)086. S2CID 53849018.
^ Herve de Kerret, "First results from the Double Chooz experiment", Talk at the LowNu conference, Seoul, November 2011, via: "First Results from Double Chooz". Archived from the original on 2011-11-12. Retrieved 2011-11-10.
^ Y, Abe; et al. (Double Chooz collaboration) (28 March 2012). "Indication for the disappearance of reactor electron antineutrinos in the Double Chooz experiment". Physical Review Letters. 108 (19): 131801. arXiv:1112.6353. Bibcode:2012PhRvL.108m1801A. doi:10.1103/PhysRevLett.108.131801. PMID 22540693. S2CID 19008791.
^ Abe, Y.; et al. (Double Chooz Collaboration) (18 September 2012). "Reactor ν¯e disappearance in the Double Chooz experiment". Physical Review D. 86 (5): 052008. arXiv:1207.6632. Bibcode:2012PhRvD..86e2008A. doi:10.1103/PhysRevD.86.052008. S2CID 30891902.
^ Abe, Y.; et al. (Double Chooz Collaboration) (2012). "First Measurement of θ13 from Delayed Neutron Capture on Hydrogen in the Double Chooz Experiment". Physics Letters B. 723 (1–3): 66–70. arXiv:1301.2948. Bibcode:2013PhLB..723...66A. doi:10.1016/j.physletb.2013.04.050. S2CID 59462260.
^ Abe, Y.; et al. (Double Chooz Collaboration) (2014). "Background-independent measurement of θ13 in Double Chooz". Physics Letters B. 735: 51–56. arXiv:1401.5981. Bibcode:2014PhLB..735...51A. doi:10.1016/j.physletb.2014.04.045. S2CID 27219821.
^ Abe, Y.; et al. (Double Chooz Collaboration) (October 2014). "Ortho-positronium observation in the Double Chooz experiment". Journal of High Energy Physics. 2014 (10): 32. arXiv:1407.6913. Bibcode:2014JHEP...10..032A. doi:10.1007/JHEP10(2014)032. hdl:1721.1/92880.
^ Abe, Y.; et al. (Double Chooz Collaboration) (December 2012). "First test of Lorentz violation with a reactor-based antineutrino experiment". Physical Review D. 86 (11): 112009. arXiv:1209.5810. Bibcode:2012PhRvD..86k2009A. doi:10.1103/PhysRevD.86.112009. hdl:1721.1/76809. S2CID 3282231.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Double_Chooz&oldid=1081193649"
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energy - Maple Help
Home : Support : Online Help : Science and Engineering : Units : Known Units : energy
Energy has the dimension mass length squared per time squared. The SI unit of energy is the joule, which is defined as a kilogram meter squared per second squared.
Work and heat are physical quantities with the same dimension as energy.
Maple knows the units of energy listed in the following table. The context IT indicates International Table.
watt_hour
watt_hours
thermochemical *
calory, calories
`15degC`
British_thermal_unit
British_thermal_units
`39degF`
Q_units
Celsius_heat_unit
Celsius_heat_units
planck_energy
planck_energies
An asterisk ( * ) indicates the default context, an at sign (@) indicates an abbreviation, and under the prefixes column, SI indicates that the unit takes all SI prefixes, IEC indicates that the unit takes IEC prefixes, and SI+ and SI- indicate that the unit takes only positive and negative SI prefixes, respectively. Refer to a unit in the Units package by indexing the name or symbol with the context, for example, joule[SI] or Cal[thermodynamic]; or, if the context is indicated as the default, by using only the unit name or symbol, for example, joule or Cal.
The units of energy are defined as follows.
An electron volt is defined as
1.60217733×{10}^{-19}
joule, that is, the product of the charge of an electron in coulombs with a joule per coulomb.
An erg is defined as
1.×{10}^{-7}
A watt hour is defined as
3600
A ton of nuclear equivalent TNT is equal to
1.×{10}^{9}
thermochemical calories.
An EU therm is defined as
105506000
A US therm is defined as
105480400
A planck energy is defined as a planck mass times planck length squared per planck time squared.
For each context of the calorie, there exists a unit, the Calorie, that is by definition
1000
times the value. Therefore, a Calorie is a kilocalorie.
A 15 degree Celsius or 20 degree Celsius calorie is approximately the amount of energy required to raise the temperature of
1
gram of water by
1
degree Celsius to 15.5 or 20.5 degrees Celsius, respectively.
A thermochemical calorie is defined as
4.184
A International Table calorie is defined as
4.1868
A mean calorie is
\frac{1}{100}
the energy required to raise the temperature of
1
gram of water from
0
degrees Celsius to
100
degrees Celsius, and is approximately
4.19002
A 39 degree Fahrenheit, 59 degree Fahrenheit, or 60 degree Fahrenheit calorie is the approximate amount of energy required to raise the temperature of
1
pound of water by
1
degree Fahrenheit to 39.5, 59.5, or 60.5 degrees Fahrenheit, respectively.
A thermochemical British thermal unit is defined by the relationship:
1
thermochemical British thermal unit per pound degree Fahrenheit equals
1
thermochemical kilocalorie per kilogram kelvin.
A International Table British thermal unit is defined by the relationship:
1
International Table British thermal unit per pound degree Fahrenheit equals
1
International Table kilocalorie per kilogram kelvin.
A mean British thermal unit is
\frac{1}{180}
1
pound of water from
32
degrees Fahrenheit to
212
degrees Fahrenheit.
For the thermochemical, International Table, and mean contexts, there are the following associated units.
A therm is defined as
10000.
British thermal units.
A quad is defined as
1.×{10}^{15}
A Q unit is defined as
1.×{10}^{18}
Celsius Heat Units
A 15 degree Celsius or 20 degree Celsius heat unit is the approximate amount of energy required to raise the temperature of
1
1
A International Table Celsius heat unit is defined by the relationship:
1
International Table Celsius heat unit per pound equals
1
International Table kilocalorie per kilogram.
A thermochemical Celsius heat unit is defined by the relationship:
1
thermochemical Celsius heat unit per pound equals
1
thermochemical kilocalorie per kilogram.
A mean Celsius heat unit is
\frac{1}{100}
1
0
100
degrees Celsius.
\mathrm{convert}\left('J','\mathrm{dimensions}','\mathrm{base}'=\mathrm{true}\right)
\frac{\textcolor[rgb]{0,0,1}{\mathrm{mass}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{length}}}^{\textcolor[rgb]{0,0,1}{2}}}{{\textcolor[rgb]{0,0,1}{\mathrm{time}}}^{\textcolor[rgb]{0,0,1}{2}}}
\mathrm{convert}\left(1,'\mathrm{units}','J','\mathrm{kilowatt}''\mathrm{hour}'\right)
\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3600000}}
\mathrm{convert}\left(1,'\mathrm{units}','J','\mathrm{Btu}'\right)
\frac{\textcolor[rgb]{0,0,1}{22500000}}{\textcolor[rgb]{0,0,1}{23722880951}}
\mathrm{convert}\left(1.0,'\mathrm{units}','\mathrm{cal}','\mathrm{Btu}[\mathrm{IT}]'\right)
\textcolor[rgb]{0,0,1}{0.003965666831}
\mathrm{convert}\left(1.0,'\mathrm{units}','\mathrm{cal}[\mathrm{IT}]','\mathrm{Btu}[\mathrm{IT}]'\right)
\textcolor[rgb]{0,0,1}{0.003968320719}
\mathrm{convert}\left(325,'\mathrm{units}','J','\mathrm{therm}[\mathrm{US}]'\right)
\frac{\textcolor[rgb]{0,0,1}{13}}{\textcolor[rgb]{0,0,1}{4219216}}
\mathrm{convert}\left(325,'\mathrm{units}','J','\mathrm{therm}[\mathrm{EU}]'\right)
\frac{\textcolor[rgb]{0,0,1}{13}}{\textcolor[rgb]{0,0,1}{4220240}}
\mathrm{convert}\left(22000,'\mathrm{units}','\mathrm{Q_unit}[\mathrm{thermochemical}]','\mathrm{ton}[\mathrm{TNT}]'\right)
\frac{\textcolor[rgb]{0,0,1}{49895160700000000}}{\textcolor[rgb]{0,0,1}{9}}
|
The solution for the equation. Use the change of base
The solution for the equation. Use the change of base formula to approximate axa
The solution for the equation. Use the change of base formula to approximate axact answer to the nearest hundredth when approximate.
2×{10}^{x}=66
doplovif
Change of base formula for logarithm is given by
{\mathrm{log}}_{a}x=\frac{{\mathrm{log}}_{b}x}{{\mathrm{log}}_{b}a}
a>0
x>0
2×{10}^{x}=66
{10}^{x}=\frac{66}{2}
{10}^{x}=33
x={\mathrm{log}}_{10}\left(33\right)
Change to an logarithmic equation
x=\frac{\mathrm{log}\left(33\right)}{\mathrm{log}\left(10\right)}
Using the Change of Base Formula
x=\frac{1.5185}{1}\mathrm{log}\left(13\right)=1.5185
\mathrm{log}\left(10\right)=1
x\approx 1.52
The solution for the given equation is
x\approx 1.52
y=3.5x+2.8
Whether the given data is linear or exponential and find the corresponding model.
\begin{array}{|cccccc|}\hline x& 0& 1& 2& 3& 4\\ y& 2& 8& 32& 128& 512\\ \hline\end{array}
An equation that expresses a relationship between two or more variables, such as
H=\frac{9}{10}\left(220-a\right)
\frac{a}{an}?.
The process of finding such equations to describe real-world phenomena is called mathematical ? .
Such equations, together with the meaning assigned to the variables, are called mathematical ? .
The speed of each of the runners by modeling the system of linear equations.
Consider a 3000 lb car whose speed is increased by 30 mph. Modeling the car as a particle and assuming that the car is traveling on a rectilinear and horizontal stretch of road, determine the amount of work done on the car throughout the acceleration process if the car starts from rest.
To calculate: Percentage of gene combinations that result in albino coloring, further model a polynomial representing possible gene combinations.
|
The Rational(f, k) command computes a closed form of the indefinite sum of
k
f\left(k\right)
s\left(k\right)
t\left(k\right)
f\left(k\right)=s\left(k+1\right)-s\left(k\right)+t\left(k\right)
t\left(k\right)
k
\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{\sum }_{k}t\left(k\right)
g,[p,q]
g
is the closed form of the indefinite sum of
k
p
is a list containing the integer poles of
q
s
that are not poles of
\mathrm{with}\left(\mathrm{SumTools}[\mathrm{IndefiniteSum}]\right):
f≔\frac{1}{{n}^{2}+\mathrm{sqrt}\left(5\right)n-1}
\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{1}}{{\textcolor[rgb]{0,0,1}{n}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\sqrt{\textcolor[rgb]{0,0,1}{5}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}
g≔\mathrm{Rational}\left(f,n\right)
\textcolor[rgb]{0,0,1}{g}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\frac{\sqrt{\textcolor[rgb]{0,0,1}{5}}}{\textcolor[rgb]{0,0,1}{2}}\right)}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\frac{\sqrt{\textcolor[rgb]{0,0,1}{5}}}{\textcolor[rgb]{0,0,1}{2}}\right)}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\frac{\sqrt{\textcolor[rgb]{0,0,1}{5}}}{\textcolor[rgb]{0,0,1}{2}}\right)}
\mathrm{evala}\left(\mathrm{Normal}\left(\mathrm{eval}\left(g,n=n+1\right)-g\right),\mathrm{expanded}\right)
\frac{\textcolor[rgb]{0,0,1}{1}}{{\textcolor[rgb]{0,0,1}{n}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\sqrt{\textcolor[rgb]{0,0,1}{5}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}
f≔\frac{13-57x+2y+20{x}^{2}-18xy+10{y}^{2}}{15+10x-26y-25{x}^{2}+10xy+8{y}^{2}}
\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{20}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{18}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{10}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{57}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{13}}{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{25}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{10}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{8}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{10}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{26}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{15}}
g≔\mathrm{Rational}\left(f,x\right)
\textcolor[rgb]{0,0,1}{g}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}}{\textcolor[rgb]{0,0,1}{5}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}}{\textcolor[rgb]{0,0,1}{25}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{34}}{\textcolor[rgb]{0,0,1}{25}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{\Psi }}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}}{\textcolor[rgb]{0,0,1}{5}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0,0,1}{5}}\right)\textcolor[rgb]{0,0,1}{+}\left(\frac{\textcolor[rgb]{0,0,1}{17}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}}{\textcolor[rgb]{0,0,1}{25}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0,0,1}{5}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{\Psi }}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}}{\textcolor[rgb]{0,0,1}{5}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)
\mathrm{simplify}\left(\mathrm{combine}\left(f-\left(\mathrm{eval}\left(g,x=x+1\right)-g\right),\mathrm{\Psi }\right)\right)
\textcolor[rgb]{0,0,1}{0}
f≔\frac{1}{n}-\frac{2}{n-3}+\frac{1}{n-5}
\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{n}}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{2}}{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{5}}
g,\mathrm{fp}≔\mathrm{Rational}\left(f,n,'\mathrm{failpoints}'\right)
\textcolor[rgb]{0,0,1}{g}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{fp}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{5}}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{n}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{..}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{..}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{..}\textcolor[rgb]{0,0,1}{5}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}]]
f
n=0,3,5
g
n=1,2,4
|
Game semantics - Wikipedia
Game semantics (German: dialogische Logik, translated as dialogical logic) is an approach to formal semantics that grounds the concepts of truth or validity on game-theoretic concepts, such as the existence of a winning strategy for a player, somewhat resembling Socratic dialogues or medieval theory of Obligationes.
3 Intuitionistic logic, denotational semantics, linear logic, logical pluralism
5 Computability logic
In the late 1950s Paul Lorenzen was the first to introduce a game semantics for logic, and it was further developed by Kuno Lorenz. At almost the same time as Lorenzen, Jaakko Hintikka developed a model-theoretical approach known in the literature as GTS (game-theoretical semantics). Since then, a number of different game semantics have been studied in logic.
Shahid Rahman (Lille) and collaborators developed dialogical logic into a general framework for the study of logical and philosophical issues related to logical pluralism. Beginning 1994 this triggered a kind of renaissance with lasting consequences. This new philosophical impulse experienced a parallel renewal in the fields of theoretical computer science, computational linguistics, artificial intelligence, and the formal semantics of programming languages, for instance the work of Johan van Benthem and collaborators in Amsterdam who looked thoroughly at the interface between logic and games, and Hanno Nickau who addressed the full abstraction problem in programming languages by means of games. New results in linear logic by Jean-Yves Girard in the interfaces between mathematical game theory and logic on one hand and argumentation theory and logic on the other hand resulted in the work of many others, including S. Abramsky, J. van Benthem, A. Blass, D. Gabbay, M. Hyland, W. Hodges, R. Jagadeesan, G. Japaridze, E. Krabbe, L. Ong, H. Prakken, G. Sandu, D. Walton, and J. Woods, who placed game semantics at the center of a new concept in logic in which logic is understood as a dynamic instrument of inference. There has also been an alternative perspective on proof theory and meaning theory, advocating that Wittgenstein's "meaning as use" paradigm as understood in the context of proof theory, where the so-called reduction rules (showing the effect of elimination rules on the result of introduction rules) should be seen as appropriate to formalise the explanation of the (immediate) consequences one can draw from a proposition, thus showing the function/purpose/usefulness of its main connective in the calculus of language.(de Queiroz (1988), de Queiroz (1991), de Queiroz (1994), de Queiroz (2001), de Queiroz (2008))
If the formula contains negations or implications, other, more complicated, techniques may be used. For example, a negation should be true if the thing negated is false, so it must have the effect of interchanging the roles of the two players.
More generally, game semantics may be applied to predicate logic; the new rules allow a dominant quantifier to be removed by its "owner" (the Verifier for existential quantifiers and the Falsifier for universal quantifiers) and its bound variable replaced at all occurrences by an object of the owner's choosing, drawn from the domain of quantification. Note that a single counterexample falsifies a universally quantified statement, and a single example suffices to verify an existentially quantified one. Assuming the axiom of choice, the game-theoretical semantics for classical first-order logic agree with the usual model-based (Tarskian) semantics. For classical first-order logic the winning strategy for the Verifier essentially consists of finding adequate Skolem functions and witnesses. For example, if S denotes
{\displaystyle \forall x\exists y\,\phi (x,y)}
then an equisatisfiable statement for S is
{\displaystyle \exists f\forall x\,\phi (x,f(x))}
. The Skolem function f (if it exists) actually codifies a winning strategy for the Verifier of S by returning a witness for the existential sub-formula for every choice of x the Falsifier might make.[1]
The above definition was first formulated by Jaakko Hintikka as part of his GTS interpretation. The original version of game semantics for classical (and intuitionistic) logic due to Paul Lorenzen and Kuno Lorenz was not defined in terms of models but of winning strategies over formal dialogues (P. Lorenzen, K. Lorenz 1978, S. Rahman and L. Keiff 2005). Shahid Rahman and Tero Tulenheimo developed an algorithm to convert GTS-winning strategies for classical logic into the dialogical winning strategies and vice versa.
Formal dialogues and GTS games may be infinite and use end-of-play rules rather than letting players decide when stop playing. Reaching this decision by standard means for strategic inferences (iterated elimination of dominated strategies or IEDS) would, in GTS and formal dialogues, be equivalent to solving the Halting Problem and exceeds the reasoning abilities of human agents. GTS avoids this with a rule to test formulas against an underlying model; logical dialogues, with a non-repetition rule (similar to Threefold Repetition in Chess). Genot and Jacot (2017)[2] proved that players with severely bounded rationality can reason to terminate a play without IEDS.
For most common logics, including the ones above, the games that arise from them have perfect information—that is, the two players always know the truth values of each primitive, and are aware of all preceding moves in the game. However, with the advent of game semantics, logics, such as the independence-friendly logic of Hintikka and Sandu, with a natural semantics in terms of games of imperfect information have been proposed.
Intuitionistic logic, denotational semantics, linear logic, logical pluralism[edit]
The primary motivation for Lorenzen and Kuno Lorenz was to find a game-theoretic (their term was dialogical, in German Dialogische Logik [de]) semantics for intuitionistic logic. Andreas Blass[3] was the first to point out connections between game semantics and linear logic. This line was further developed by Samson Abramsky, Radhakrishnan Jagadeesan, Pasquale Malacaria and independently Martin Hyland and Luke Ong, who placed special emphasis on compositionality, i.e. the definition of strategies inductively on the syntax. Using game semantics, the authors mentioned above have solved the long-standing problem of defining a fully abstract model for the programming language PCF. Consequently, game semantics has led to fully abstract semantic models for a variety of programming languages, and to new semantic-directed methods of software verification by software model checking.
Shahid Rahman [fr] and Helge Rückert extended the dialogical approach to the study of several non-classical logics such as modal logic, relevance logic, free logic and connexive logic. Recently, Rahman and collaborators developed the dialogical approach into a general framework aimed at the discussion of logical pluralism.
Quantifiers[edit]
Foundational considerations of game semantics have been more emphasised by Jaakko Hintikka and Gabriel Sandu, especially for independence-friendly logic (IF logic, more recently information-friendly logic), a logic with branching quantifiers. It was thought that the principle of compositionality fails for these logics, so that a Tarskian truth definition could not provide a suitable semantics. To get around this problem, the quantifiers were given a game-theoretic meaning. Specifically, the approach is the same as in classical propositional logic, except that the players do not always have perfect information about previous moves by the other player. Wilfrid Hodges has proposed a compositional semantics and proved it equivalent to game semantics for IF-logics.
More recently Shahid Rahman [fr] and the team of dialogical logic in Lille implemented dependences and independences within a dialogical framework by means of a dialogical approach to intuitionistic type theory called immanent reasoning.[4]
Computability logic[edit]
Japaridze’s computability logic is a game-semantical approach to logic in an extreme sense, treating games as targets to be serviced by logic rather than as technical or foundational means for studying or justifying logic. Its starting philosophical point is that logic is meant to be a universal, general-utility intellectual tool for ‘navigating the real world’ and, as such, it should be construed semantically rather than syntactically, because it is semantics that serves as a bridge between real world and otherwise meaningless formal systems (syntax). Syntax is thus secondary, interesting only as much as it services the underlying semantics. From this standpoint, Japaridze has repeatedly criticized the often followed practice of adjusting semantics to some already existing target syntactic constructions, with Lorenzen’s approach to intuitionistic logic being an example. This line of thought then proceeds to argue that the semantics, in turn, should be a game semantics, because games “offer the most comprehensive, coherent, natural, adequate and convenient mathematical models for the very essence of all ‘navigational’ activities of agents: their interactions with the surrounding world”.[5] Accordingly, the logic-building paradigm adopted by computability logic is to identify the most natural and basic operations on games, treat those operators as logical operations, and then look for sound and complete axiomatizations of the sets of game-semantically valid formulas. On this path a host of familiar or unfamiliar logical operators have emerged in the open-ended language of computability logic, with several sorts of negations, conjunctions, disjunctions, implications, quantifiers and modalities.
Games are played between two agents: a machine and its environment, where the machine is required to follow only effective strategies. This way, games are seen as interactive computational problems, and the machine's winning strategies for them as solutions to those problems. It has been established that computability logic is robust with respect to reasonable variations in the complexity of allowed strategies, which can be brought down as low as logarithmic space and polynomial time (one does not imply the other in interactive computations) without affecting the logic. All this explains the name “computability logic” and determines applicability in various areas of computer science. Classical logic, independence-friendly logic and certain extensions of linear and intuitionistic logics turn out to be special fragments of computability logic, obtained merely by disallowing certain groups of operators or atoms.
^ J. Hintikka and G. Sandu, 2009, "Game-Theoretical Semantics" in Keith Allan (ed.) Concise Encyclopedia of Semantics, Elsevier, ISBN 0-08095-968-7, pp. 341–343
^ Genot, Emmanuel J.; Jacot, Justine (2017-09-01). "Logical Dialogues with Explicit Preference Profiles and Strategy Selection". Journal of Logic, Language and Information. 26 (3): 261–291. doi:10.1007/s10849-017-9252-4. ISSN 1572-9583. S2CID 37033818.
^ Andreas R. Blass
^ S. Rahman, Z. McConaughey, A. Klev, N. Clerbout: Immanent Reasoning or Equality in Action. A Plaidoyer for the Play level. Springer (2018). https://www.springer.com/gp/book/9783319911489.
For an application of the dialogical approach to intuitionistic type theory to the axiom of choice see S. Rahman and N. Clerbout: Linking Games and Constructive Type Theory: Dialogical Strategies, CTT-Demonstrations and the Axiom of Choice. Springer-Briefs (2015). https://www.springer.com/gp/book/9783319190624.
^ G. Japaridze, “In the beginning was game semantics”. In: Games: Unifying Logic, Language and Philosophy. O. Majer, A.-V. Pietarinen and T. Tulenheimo, eds. Springer 2009, pp.249-350. [1]
J. van Benthem, G. Heinzmann, M. Rebuschi and H. Visser (eds.) The Age of Alternative Logics. Springer (2006).ISBN 978-1-4020-5011-4.
L. Keiff Le Pluralisme Dialogique. Thesis Université de Lille 3 (2007).
S. Rahman and N. Clerbout: Linking Games and Constructive Type Theory: Dialogical Strategies, CTT-Demonstrations and the Axiom of Choice. Springer-Briefs (2015). https://www.springer.com/gp/book/9783319190624.
S. Rahman, Z. McConaughey, A. Klev, N. Clerbout: Immanent Reasoning or Equality in Action. A Plaidoyer for the Play levell. Springer (2018). https://www.springer.com/gp/book/9783319911489.
J. Redmond & M. Fontaine, How to play dialogues. An introduction to Dialogical Logic. London, College Publications (Col. Dialogues and the Games of Logic. A Philosophical Perspective N° 1). (ISBN 978-1-84890-046-2)
S. Abramsky and R. Jagadeesan, Games and full completeness for multiplicative linear logic. Journal of Symbolic Logic 59 (1994): 543-574.
J.M.E.Hyland and H.L.Ong On Full Abstraction for PCF: I, II, and III. Information and computation, 163(2), 285-408.
E.J. Genot and J. Jacot, Logical Dialogues with Explicit Preference Profiles and Strategy Selection, Journal of Logic, Language and Information 26, 261–291 (2017). doi.org/10.1007/s10849-017-9252-4
D.R. Ghica, Applications of Game Semantics: From Program Analysis to Hardware Synthesis. 2009 24th Annual IEEE Symposium on Logic In Computer Science: 17-26. ISBN 978-0-7695-3746-7.
G. Japaridze, In the beginning was game semantics. In Ondrej Majer, Ahti-Veikko Pietarinen and Tero Tulenheimo (editors), Games: Unifying logic, Language and Philosophy. Springer (2009).
Krabbe, E. C. W., 2001. "Dialogue Foundations: Dialogue Logic Restituted [title has been misprinted as "...Revisited"]," Supplement to the Proceedings of the Aristotelian Society 75: 33-49.
H. Nickau (1994). "Hereditarily Sequential Functionals". In A. Nerode; Yu.V. Matiyasevich (eds.). Proc. Symp. Logical Foundations of Computer Science: Logic at St. Petersburg. Lecture Notes in Computer Science. Vol. 813. Springer-Verlag. pp. 253–264. doi:10.1007/3-540-58140-5_25.
de Queiroz, R. (1988). "A Proof‐Theoretic Account of Programming and the Role of Reduction Rules". Dialectica. 42 (4): 265–282. doi:10.1111/j.1746-8361.1988.tb00919.x.
de Queiroz, R. (1991). "Meaning as Grammar plus Consequences". Dialectica. 45 (1): 83–86. doi:10.1111/j.1746-8361.1991.tb00979.x.
de Queiroz, R. (1994). "Normalisation and Language Games". Dialectica. 48 (2): 83–123. doi:10.1111/j.1746-8361.1994.tb00107.x.
de Queiroz, R. (2001). "Meaning, Function, Purpose, Usefulness, Consequences – Interconnected Concepts". Logic Journal of the IGPL. 9 (5): 693–734. doi:10.1093/jigpal/9.5.693.
de Queiroz, R. (2008). "On Reduction Rules, Meaning-as-use, and Proof-theoretic Semantics". Studia Logica. 90 (2): 211–247. doi:10.1007/s11225-008-9150-5. S2CID 11321602.
S. Rahman and T. Tulenheimo, From Games to Dialogues and Back: Towards a General Frame for Validity. In Ondrej Majer, Ahti-Veikko Pietarinen and Tero Tulenheimo (editors), Games: Unifying logic, Language and Philosophy. Springer (2009).
Johan van Benthem (2003). "Logic and Game Theory: Close Encounters of the Third Kind". In G. E. Mints; Reinhard Muskens (eds.). Games, logic, and constructive sets. CSLI Publications. ISBN 978-1-57586-449-5.
Thomas Piecha. "Dialogical Logic". Internet Encyclopedia of Philosophy.
"Logic and Games" entry by Wilfrid Hodges in the Stanford Encyclopedia of Philosophy
"Dialogical Logic" entry by Laurent Keiff in the Stanford Encyclopedia of Philosophy
Retrieved from "https://en.wikipedia.org/w/index.php?title=Game_semantics&oldid=1077244706"
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We can place a point in a plane by polar coordinates
(r, \ \theta).
e^{i\theta} = \cos{\theta} + i \sin{\theta}.
Employing this formula, we have
r e^{i\theta} = r \cos{\theta} + i r \sin{\theta} = x + iy,
so we have Cartesian coordinates
(x, \ y)
from the polar coordinates. Therefore, a point in a plane can be represented by polar coordinates
(r, \ \theta),
which have their equivalent complex numbers
r e^{i\theta}.
The multiplication of two points
(r_1, \ \theta_1)
(r_2, \ \theta_2),
which have their equivalent complex numbers, can be easily calculated. Given two equivalent complex numbers
r_1 e^{i\theta_1}
r_2 e^{i\theta_2},
the multiplication is
\begin{aligned} re^{i\theta} =& \left(r_1 e^{i \theta _1}\right)\left(r_2 e^{i \theta _2}\right) \\ =& (r_1 r_2)\left( e^{i\theta _1} e^{i\theta _2}\right)\\ =& r_1 r_2 e^{i (\theta _1 + \theta _2)}, \end{aligned}
where the rule of addition of exponents over real numbers is extended to that over complex numbers in the same way. Therefore, the multiplication of two points is
(r, \ \theta),
r = r_1 r_2
\theta = \theta _1 + \theta _2.
What is the multiplication of two points
(3, 5)
(2, 4)
in polar coordinates?
The first coordinate of the multiplication is the product of the two first coordinates. The second coordinate of the multiplication is the sum of the two second coordinates. Therefore, we have
(r, \theta) = (3 \times 2, 5 + 4) = (6, 9). \ _\square
(5, 2)
in polar coordinates and
(4, 3)
in Cartesian coordinates? Approximate the angles in radians up to two digits below the decimal point.
(4, 3)
in Cartesian coordinates to a representation in polar coordinates:
\left(\sqrt{4^2 + 3^2}, \arctan{\frac{3}{4}}\right) = (5, 0.64).
In polar coordinates, the first coordinate of the multiplication is the product of the two first coordinates, and the second coordinate of the multiplication is the sum of the two second coordinates. Therefore, we have
(r, \theta) \approx (5 \times 5, 2 +0.64) = (25, 2.64). \ _\square
(5, -12)
(6, 8)
(5, -12)
(6, 8)
in Cartesian coordinates to representations in polar coordinates:
\left(\sqrt{5^2 + (-12)^2}, \arctan{\frac{-12}{5}}\right) = (13, -1.176)
\left(\sqrt{6^2 + 8^2}, \arctan{\frac{8}{6}}\right) = (10, 0.927 ).
(r, \theta) = (13 \times 10, -1.176 \ldots + 0.927 \ldots) \approx (130, -0.25). \ _\square
z = x + \imath y
is represented by the point
(x,y)
in the complex plane. From the properties of complex numbers, we can write
\begin{aligned} x &= \Re(z) = |z| \cos(\theta)\\ y &= \Im(z) = |z| \sin(\theta), \end{aligned}
|z| = \sqrt{x^2 + y^2}
. This is shown in the picture below:
(x,y)
pair can equivalently be described by trigonometric functions of another pair
(|z|, \theta)
(r, \theta)
. These are referred to as the polar coordinates of the complex number
z
r
is a non-negative number denoting the magnitude of the complex number (the radius of the circle) and is represented on the radial axis that extends outward from the origin at
(0,0)
. An expression for
\theta
can be obtained by dividing
y = |z| \sin (\theta)
x = |z| \cos (\theta)
\theta = \arctan\left(\frac{y}{x}\right)
and is called the argument of the complex number.
What is the polar form of the complex number
z = 1 + 1i?
We can see from the image that the triangle has two equal sides; from this we know that it must also have two equal angles. Since one of the angles is
90
degrees, we can safely conclude that
\theta
45
degrees or
\frac{\pi}{4}
radians. We can then find
|z|
by computing the hypotenuse of the triangle using the Pythagoras theorem:
x^2 + y^2 = z^2 \implies |z| = \sqrt{1^2 + 1^2} = \sqrt{2}.
Thus we can write our complex number as
z = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right ).\ _\square
1 + \sqrt{3} i
in its polar form.
Applying our formula,
\begin{aligned} \theta &= \arg(z)\\ &= \arctan \frac{y}{x}\\ &= \arctan \sqrt{3}= \frac{\pi}{3} \\\\ r &= |z| \\&= \sqrt{ 1^2 + (\sqrt{3}) ^2} = 2\\\\ z &= |z|(\cos \theta + i \sin \theta ) \\ z &= 2\left(\cos \frac{\pi}{3} + i \sin\frac{\pi}{3} \right).\ _\square \end{aligned}
6\left(\cos \frac{5\pi}{3} +i \sin \frac{5\pi}{3} \right)
in its rectangular form.
|z| = 6,
and we evaluate
\sin \frac{5\pi}{3} = \frac{-\sqrt{3}}{2}, \quad \cos \frac{5\pi}{3} = \frac{1}{2}.
Thus our complex number
z
\begin{aligned} z &= |z|(\cos \theta + i\sin \theta )\\ &= 6\left( \frac{1}{2} + \frac{-\sqrt{3}}{2} i\right)\\ &= 3\left(1 - \sqrt{3} i\right) \\ &= 3 -3\sqrt{3}i.\ _\square \end{aligned}
\left(1 + \sqrt{2} i\right)\cdot\left(1-\sqrt{2} i\right)
\begin{aligned} (1 + \sqrt{2} i)\cdot(1-\sqrt{2} i) &= 1\cdot1 - \sqrt{2}i +\sqrt{2}i - \sqrt{2}i \cdot \sqrt{2}i \\ &= 3 + 0\cdot i. \end{aligned}
Then from our formula, we have
\theta = arg(z)= \arctan \frac{y}{x}= \arctan \frac{0}{3}= 0.
r=\lvert z \rvert = \sqrt{3^2 +0^2 } = 3,
\begin{aligned} z &= |z|(\cos \theta + i \sin \theta ) \\ &= 3(\cos 0 + i \sin 0 ).\ _\square \end{aligned}
i\left(1-\sqrt{3}i \right)
\begin{aligned} i(1-\sqrt{3}i ) = \sqrt{3}+i. \end{aligned}
\theta = arg(z)= \arctan \frac{y}{x}= \arctan \frac{1}{\sqrt{3}}= \frac{\pi}{6},
r = |z| = \sqrt{ 1^2 + (\sqrt{3}) ^2} = 2.
Thus, the polar form
z
\begin{aligned} z &= |z|(\cos \theta + i \sin \theta ) \\ &= 2\left(\cos \frac{\pi}{6} + i \sin\frac{\pi}{6} \right).\ _\square \end{aligned}
b \left ( \cos \frac{\pi}{a} + \sin \frac{\pi}{a}i \right )
is the polar form of
3+ 3\sqrt{3}i ,
and
b?
3+ 3\sqrt{3} i
c \left ( \frac{3}{c} + \frac{3\sqrt{3}}{c} i \right ),
c
is a positive integer. Then since
\sin^2 \theta + cos^2 \theta =1,
c
\begin{aligned} \left ( \frac{3}{c} \right )^2 + \left ( \frac{3\sqrt{3}}{c} \right )^2 = \frac{ 3^2 + (3\sqrt{3})^2}{c^2} = \frac{36}{c^2} &=1 \\ \Rightarrow c&=6. \end{aligned}
c=6,
3+ 3\sqrt{3} i
c\left( \frac{3}{c} + \frac{3\sqrt{3}}{c} i \right ) = 6\left ( \frac{3}{6} + \frac{3\sqrt{3}}{6}i \right ) = 6\left ( \frac{1}{2} + \frac{\sqrt{3}}{2}i\right ).
\cos \frac{\pi}{a} = \frac{1}{2}
\sin \frac{\pi}{a} = \frac{\sqrt{3}}{2},
from trigonometry we can conclude that
a=3.
a=3, b=6. \ _\square
\dfrac{-2\pi}{\sqrt3}
\dfrac{\pi}{3}
\dfrac{-2\pi}{3}
\dfrac{\pi}{\sqrt3}
\large \ln\left(\dfrac{\omega^\omega}{\omega^{\omega^2}}\right)
\omega
is a non-real complex cube root of unity, then what is the value of the expression above?
z_1 = 6+i
z_2 = 4-3i
z
\text{arg}\left(\dfrac{z-z_1}{z_2-z}\right)=\dfrac{\pi}{2}
|z-(5-i)|
\sqrt{m}
m.
\text{arg}(x)
x
z_1,z_2,
z
\LARGE { i }^{ { i }^{ i^{.^{.^{.}}} } }
i = \sqrt{-1}
, the value of the infinitely nested exponent above is equal to
A + Bi
A
B
{ A }^{ 2 }+{ B }^{ 2 }
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High School Calculus/Evaluating Definite Integrals - Wikibooks, open books for an open world
High School Calculus/Evaluating Definite Integrals
Evaluating a Definite Integral[edit | edit source]
Let's say you have the parabola
{\displaystyle x^{2}}
and you want to find the area from x=2 to x=4
{\displaystyle 2\leq A\leq 4}
{\displaystyle \int _{2}^{4}x^{2}\,dx}
In order to take the integral of the function you have to do the opposite that of the derivative
The power of the variable x will have a number added to it. So,
{\displaystyle x^{(a+1)}}
then the number gets inverted and brought out front.
{\displaystyle {\frac {1}{a+1}}*x^{(a+1)}}
{\displaystyle \int _{2}^{4}x^{2}\,dx}
From here we integrate and plug (b) into the indefinite integral and subtract the integral from (a) plugged into the indefinite integral.
{\displaystyle [{\frac {1}{3}}*4^{3}]-[{\frac {1}{3}}*2^{3}]}
Now we evaluate the integral
{\displaystyle [{\frac {1}{3}}*64]-[{\frac {1}{3}}*8]}
{\displaystyle [{\frac {64}{3}}]-[{\frac {8}{3}}]}
{\displaystyle {\frac {56}{3}}}
{\displaystyle {\frac {56}{3}}}
is the area underneath the curve from 2 to 4. In other words
{\displaystyle 2\leq A\leq 4}
Retrieved from "https://en.wikibooks.org/w/index.php?title=High_School_Calculus/Evaluating_Definite_Integrals&oldid=2063013"
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Network Analysis - Dealing with Network Construction Basics | Education Lessons
Many of you are engineering students, and you will be Project Managers soon in your particular field. Also, many of you are working as Project managers.
So, this topic of network analysis is related all such management work.
Why Network Diagram ?
Project managers have to deal with all limited human and non-human resources like, man, machine, space, raw-materials, transport vehicles, etc.
Project manager have to do most important basic thing as follows:
Project Planning (How to?)
Project Scheduling (When to?)
Project Controlling (All right?)
Let me brief these important activities of project management:
→ Project Planning mainly includes indentification of different activities to be performed for project completion, resources requirement and project completion time.
→ Resources here involves man, machine, materials, vehicles, etc.
→ Project Scheduling includes the squencing of the project activities in the given time frame.
→ This also includes indentification of critical and special tasks and also, evalution of resources required at particular time.
→ Project controlling involves - to identify the work status after planning and scheduling.
→ Controlling is to control the squence of processes which are already planned to be completed in a particular time frame.
→ It is a kind of analysis process too. By controlling a particular project, one can identify the deviation between the planned work and the work in progress.
→ By indentifying this, one can re-scheduled the project activities and resource requirement to complete project in given project completion time.
To manage the project efficiently, different tools are required to be implemented. A network is one the most important and widely used tool.
You can see here many arrows, circles(which are called as nodes) and some alphabets like A, B, C, etc. and some numbers like 1, 2, 3, etc.
This all together generates a Network as such shown above. We generally called this as Network Diagram.
A network is symbolic representation of essential characteristics of the project.
Critical Path Method (CPM) and Project Evalution and Review Technique (PERT) are two most widely used network techniques.
Network Diagram is a graphical representation of logically and sequentially connected arrows and nodes (i.e., activities and events) of the project.
We will understand the whole concept of Network Diagram (Project management technique) and different terms used in this by using a following example throughout:
We want to manufacture some item-A. To manufacture it, we have to fabricate and assemble as following sequence:
Cutting → Welding → Machining → Assembly → Packaging
Let us now understand different terms used in network diagram:
Activities in the network diagram are identifiable parts of a project, which consumes time as well as resources for its execution.
It is represented by an arrow as here;
The tail of this arrow indicates the start of the activity and head of the arrow indicates the finish of the activity.
Try to keep this arrow straight while drawing network diagrams. Don't use curves in the networks.
As per example above presented, the tasks mentioned here like cutting, welding, machining, etc. are activities, and that all are physically identifiable.
2) Event:
Event is representation of the beginning and finishing points of an activity.
Event doesn't consume time as it is just a notation.
It is represented by a node (circle):
As illustrated above, Tail event is the starting event of an activity, and is denoted as "
i^{th}
" event. Whereas, Head event represents finishing of the an activity, and is donted as "
j^{th}
" event.
As per example provided above, we can draw an activity as illustrated in (1) for each of the activities mentioned in the problem with its starting and end points, and that denotes the events.
3) Path:
A continous chain of activity arrows connecting the initial event to some other event is called as path.
As we can see in the following figure, the arrows are forming a kind of chain. This chain of arrows shows the flow of project require to complete the project on time.
4) Predecessor Activities:
In group of many activities which are required to complete a particular job, the activity which is required to be completed before starting of a particular acivity is called as predecessor activity.
In a constructed network diagram, the activity which is required to be completed before starting of a particular activity is called precedessor activity.
Remember: Predecessor means "previous"
As per example provided above, Welding can only be done after completion of cutting.
So, we can say that Cutting is the predecessor activity of Welding.
5) Successor Activities:
In group of many activities which are required to complete a particular job, the activity which must be followed after completion of a particular acivity is called as successor activity.
In a constructed network diagram, the activity which must follow any particular activity is called successor activity.
Remember: Successor means "next"
As per example provided above, after cutting of components, welding must be executed before machining.
So, we can say that Welding is the successor activity of Cutting.
6) Dummy Activities:
An activity which only shows dependency of one activity on the other is known as Dummy activity.
Dummy activities are very important activity in network diagram, as it will help you to correct the network with given precedence.
We will understand dummy activities in detail later in this note.
Steps for Construction of Network Diagram:
Following steps to be performed for the preparation of network diagram.
Identify numbers of activities.
Decide the logical order in which activities to be performed or executed.
Fix predecessor and successor activites.
Find out parallel activities. (Dummy activities)
Rules for Construction of Network Diagram:
We have to keep in mind the following rules while drawing the network diagram.
Arrows should not cross each other. If crossing is not possible to avoid, bridging should be done.
No two or more activities have same tail and head events. (Very improtant)
An event is not said to be completed, untill all the activities flowing into it are completed.
No subsequent activity can begin until its tail event is completed.
Only one initial and one end event is to be there in a network diagram.
Solving problem using Network Diagram:
Predecessor Activity
As we are trying to solve this problem using Network Diagram as already mentioned above.
As per steps, we have total of 7 activites and we will be following the order as starting from top row to the bottom, as you can see activities A & B have no predecessor activities, and so these are two starting activities.
Also, as we have well defined precedecessor activities, so we can easily identify the successor & predecessor activities.
Let's move to best part now, i.e., construction/drawing of Network diagram.
But wait, let me introduce first, two different models of Network Diagram as following:
1) Activity on Arrow (AOA):
→ In this model of Network Diagram, each activity starts with a node(circle) and also ends with a node(cicle).
→ Here, arrow itself indicates the time duration of activity and annotation of activity.
→ This type of diagrams starts with a single node and then follows from left to right along with the a single node, where no followers come together.
→ Dummy activities follows the same pattern as real activities, as they are treated as real activities only.
Example for this AOA model of Network Diagram:
Consider if activity B & C must follow activity A, then following AOA diagram can be constructed
You may wonder, what this numbers in the event nodes means???
Actually, numbering Network Diagram is very important to identify the logical order followed for its construction.
Numbering a Network Diagram follows the Fulkerson's Rule, which are proposed by D.R. Fulkerson:
Fulkerson's Rule:
The initial event which has no incoming events and all outgoing arrows is numbered '1'.
Delete all the arrows coming out from node '1', which will result into more starting nodes. Provide 2, 3, 4, ... numbers to these new nodes.
No two events can have same numbering.
Head event must have higher number compared to tail event.
Follow the same above steps untill last or final node is reached. Final node must have all incoming arrows and no outgoing arrows.
Final event must have highest number, since its end event.
2) Activity on Node (AON):
→ As we know now, Network Diagrams are all arrow diagrams. So, if we have Dummy activities in the network, which are used to show precedence of an activity which depends on more than one activity, then we have to number them also.
→ So, this results in increased number of activities, which makes networks lengthy and cumbersome, and which will also consume more time and effort for analysis.
→ To avoid such difficulty, activity is represented on node connected to precedent activity. This kind of diagrams are called as Acitivity on Node (AON) diagram.
→ In this type of diagrams, the tail of each arrow is on predecessor activity, while the head is on successor activity.
→ The activity is indicated within the node, while the arrows only show the sequencing.
Example for this AON model of Network Diagram:
Consider if activity B & C must follow activity A, then following AON diagram can be constructed (same example is considered as that was in AOA for better understanding)
Now, solving the problem we have mentioned above with AOA:
→ As we don't have any predecessor activity associated with activities A & B, we will draw a node and two starting activities as A & B.
→ Now, as for activity C, we have predecessor activity as A. This means that, activity A must be completed before the start of activity C. So, the end event of activity A will be the start event for activity C. Draw activity C as shown in the figure.
→ Similarly we have predecessor activity B for activity D and E. Drawing the arrows representing activities D & E with same logic used in the above step.
→ Repeat the same procedure and complete the network diagram such that, all activities ends in a single event.
→ Use Fulkerson's Rule for numbering the events.
Why Dummy Activity?
Dummy activity is an improtant activity when there is starting of one particular activity is depended on completion of more than one activity.
C A & B
We will try to draw network diagram of this example as follows:
As you can see in the above image, we tried to draw a network and it almost satisfies the given example, but at activity C, we fail to show that it has two predecessor activities.
So, what do we do now???
As rules of Network Diagram says that: No two or more activities have same tail and head events.
We have to consider a dummy activity will help us to show the the dependency of activity C on both activities A & B.
We will be representing dummy activities with dotted line arrows. We will draw something like this, which will serve the purpose.
Now you can easily understand from the above diagram, that activity C can be started only if both the activities A & B are completed.
Remember that, Dummy activities only to show dependency of activities and so they does not consume any resources like time and cost for their execution.
Finally, we can define Dummy activity as:
An activity which only shows dependency of one activity on the other but does not consume any resource is called as Dummy Activity.
Errors in Network Diagram:
1) Looping of Activities:
→ Sometimes due to some error while planning of different activities (project planning), it may happen that there will be loop of activities generated in the network diagram.
→ Such looping of activities must be avoided. Looping looks like following:
2) Dangling:
→ Same as looping, if there's mistake in project planning, dangling occurs.
→ In dangling, any activity may be disconnected before the completion of all activities. As shown in following figure, activity 'B' is disconnected from the flow of the project.
→ New notes on numerical with dummy activities will be uploaded soon on the website.
→ Get latest updates of EL Website by subscribing to new notes notification.
Related Notes on Network Diagrams:
Crashing Special Case- Multiple Critical Paths
Crashing Special Case- Indirect cost less than Crash cost
Check out videos on:
Dummy acitivy in networks
Crashing of Project Network
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Measurement and Memory Practice Problems Online | Brilliant
In the last quiz, we encountered head-on the measurement problem in quantum mechanics: we could not devise an experiment capable of measuring the neutron spin components without destroying previously collected information along the way. It appears that the experimental apparatus we’re using must disturb the neutron spins while observing them, making it impossible to precisely determine the direction of the spin vector. This interesting consequence of quantum objects is not limited to spin, as we’ll see in later chapters.
How might we be disturbing the neutrons? Since this quantum measurement problem is generally only observed with microscopic systems, we might guess that our macroscopic magnets generating the magnetic field
\mathbf{B}
are simply too strong: it's easy to imagine measuring the speed of a baseball with a radar gun, but difficult to imagine measuring the speed of a neutron, whether or not quantum mechanics governs its motion.
Let’s investigate this measurement problem a bit more. Can we determine what step in the SG experiment is disturbing the state of our neutrons?
Measurement and Memory
Let’s revisit the experimental setup with three SG analyzers. The first SG-
z
analyzer filters the unknown state
\ket{\psi}
into two possibilities:
\ket{\uparrow}
\ket{\downarrow}
, corresponding to “up” and “down” spin neutrons. We’ve now set up our apparatus so that 100
\ket{\uparrow}
neutrons are prepared by the first analyzer.
The output from this first analyzer is directed towards the input of an SG-
x
analyzer. Measuring the
x
-component of spin is incompatible with the
z
measurement we’ve already performed, so this analyzer resets the neutrons into two states:
\ket{\rightarrow}
\ket{\leftarrow}
, aligned parallel and anti-parallel with the
x
The state that an incident
\ket{\uparrow}
neutron collapses into is probabilistic, and so with
100
neutrons input into the analyzer, we would expect
50
neutrons in the
\ket{\rightarrow}
channel, and
50
\ket{\leftarrow}
Finally, we direct the
\ket{\leftarrow}
channel into a third SG-
z
analyzer. Even though these neutrons have been measured previously as
\ket{\uparrow}
, that knowledge has been permanently erased by measuring an incompatible observable. Instead of confirming the
\ket{\uparrow}
spin of all
50
of these neutrons, we again observe a reset:
25
\ket{\uparrow}
25
\ket{\downarrow}
We’ve made a slight variation to our experimental apparatus.
Instead of measuring the
\ket{\leftarrow}
channel, we've selected the
\ket{\rightarrow}
channel and used that as input for the third analyzer.
Given what we've learned about incompatible measurements destroying previously collected information, what do we expect to output from the third SG device?
The neutrons are split evenly between both channels All neutrons emerge from the
\ket{\downarrow}
channel All neutrons emerge from the
\ket{\uparrow}
Measuring the
x
-component of the spin destroys any previously collected knowledge of the spin state. So whether or not you analyze the
\ket{\rightarrow}
\ket{\leftarrow}
channels, you still have no information about the
z
-component, so you get a random
50:50
Let’s now combine the
\ket{\rightarrow}
\ket{\leftarrow}
beams of neutrons which emerge from the second SG analyzer. The exact experimental method we use to perform this combination isn’t important; we use an arbitrary set of magnets to deflect the beams back into alignment.
Can we tease out what the composition of this mixed state is based on the two experiments we just looked at?
We can assume that since it went through the
x
-aligned SG analyzer, we’ve already disturbed the system and presumably lost any information about the
z
-alignment of the neutrons. The neutrons leaving the
x
-aligned analyzer have been reset into
\ket{\rightarrow}
\ket{\leftarrow}
states. We must assume that the set of magnets used to realign the two deflected beams did not further disturb the alignment of the neutrons.
Based on the two previous experiments, what do you expect the result of the third SG analyzer to be?
\ket{\downarrow}
\ket{\uparrow}
This is not what we observe. The result is classically absurd, and contradicts our expectations: all of the neutrons emerge from the
\ket{\uparrow}
exit of the third analyzer.
What could explain this observation?
The state resets are completely random The
x
-component was never actually observed, so the state wasn't reset The
z
x
magnetic fields added to bias the
\ket{\uparrow}
The measurement problem is more sinister than a practical disturbance caused by incident magnetic fields. What we thought might be a shortcoming of our experimental device turned out to be the reflection of something much deeper: it's the act of measurement itself that is changing the state of our neutrons.
Though the neutrons went through the SG-
x
analyzer in the experiment above, their
x
-components were never actually observed, and were recombined so that the channels could not be distinguished again. From the point of view of the neutrons, they were never observed, and no incompatible measurements have been performed that would reset their states. The role of observers in quantum mechanical measurement is completely unique and has no classical equivalent. This fundamental quantum mechanical property is the most important we've learned so far:
Observers can change the state of quantum objects. Quantum states are reset by observation itself, though the point in the experiment when this occurs is still the subject of active debate.
This counterintuitive result of combining spin channels also reveals an important caveat to keep in mind when analyzing quantum systems:
The probabilistic behavior of quantum objects is not limited to SG analyzers, and it's often tempting to break a quantum system down into simpler independent paths and combine their probabilities.
We can try to understand the combined channel experiment by analyzing two single channel experiments, and note the probability of a neutron emerging in each state:
But when we put these two results together and see what they predict for each channel of the combined experiment, we reach a glaring contradiction:
The two states
\ket{\uparrow}
\ket{\downarrow}
are identical Stern Gerlach analyzers only output from one channel at a time Quantum measurements can't be treated as independent events
In the single channel experiments,
50\%
of the neutrons are blocked after the second analyzer and
25\%
of the neutrons exit the
\ket{\downarrow}
channel of the third analyzer. In the combined channel experiment,
100\%
of the neutrons pass from the second analyzer to the third analyzer, yet fewer neutrons come out in the
\ket{\downarrow}
channel. In fact, there are no neutrons in that channel at all!
This combined experiment allowed the neutrons more paths to reach the final
\ket{\downarrow}
channel but ends up with fewer neutrons making it. It's as though we opened a second window to a room and part of the room got darker: classical probability can't hope to explain this part of quantum mechanics known as interference, in which combining two effects can lead to cancellation, rather than enhancement. Interference will play an essential role in our later adventures with quantum objects.
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Dioptre Knowpia
A dioptre (British spelling) or diopter (American spelling) is a unit of measurement with dimension of reciprocal length, equivalent to one reciprocal metre, 1 dioptre = 1 m−1. It is normally used to express the optical power of a lens or curved mirror, which is a physical quantity equal to the reciprocal of the focal length, expressed in metres. For example, a 3-dioptre lens brings parallel rays of light to focus at 1⁄3 metre. A flat window has an optical power of zero dioptres, as it does not cause light to converge or diverge. Dioptres are also sometimes used for other reciprocals of distance, particularly radii of curvature and the vergence of optical beams.
The main benefit of using optical power rather than focal length is that the thin lens formula has the object distance, image distance, and focal length all as reciprocals. Additionally, when relatively thin lenses are placed close together their powers approximately add. Thus, a thin 2.0-dioptre lens placed close to a thin 0.5-dioptre lens yields almost the same focal length as a single 2.5-dioptre lens.
Though the dioptre is based on the SI-metric system, it has not been included in the standard, so that there is no international name or symbol for this unit of measurement—within the international system of units, this unit for optical power would need to be specified explicitly as the inverse metre (m−1). However most languages have borrowed the original name and some national standardization bodies like DIN specify a unit name (dioptrie, dioptria, etc.) and unit symbol dpt. In vision care the symbol D is frequently used.
The idea of numbering lenses based on the reciprocal of their focal length in metres was first suggested by Albrecht Nagel in 1866.[1][2] The term dioptre was proposed by French ophthalmologist Ferdinand Monoyer in 1872, based on earlier use of the term dioptrice by Johannes Kepler.[3][4][5]
In vision correctionEdit
In humans, the total optical power of the relaxed eye is approximately 60 dioptres.[6][7] The cornea accounts for approximately two-thirds of this refractive power (about 40 dioptres) and the crystalline lens contributes the remaining one-third (about 20 dioptres).[6] In focusing, the ciliary muscle contracts to reduce the tension or stress transferred to the lens by the suspensory ligaments. This results in increased convexity of the lens which in turn increases the optical power of the eye. The amplitude of accommodation is about 11 to 16 dioptres at age 15, decreasing to about 10 dioptres at age 25, and to around 1 dioptre above age 60.
Convex lenses have positive dioptric value and are generally used to correct hyperopia (farsightedness) or to allow people with presbyopia (the limited accommodation of advancing age) to read at close range. Concave lenses have negative dioptric value and generally correct myopia (nearsightedness). Typical glasses for mild myopia have a power of −0.50 to −3.00 dioptres, while over the counter reading glasses are rated at +1.00 to +4.00 dioptres. Optometrists usually measure refractive error using lenses graded in steps of 0.25 dioptres.
The dioptre can also be used as a measurement of curvature equal to the reciprocal of the radius measured in metres. For example, a circle with a radius of 1/2 metre has a curvature of 2 dioptres. If the curvature of a surface of a lens is C and the index of refraction is n, the optical power is φ = (n − 1)C. If both surfaces of the lens are curved, consider their curvatures as positive toward the lens and add them. This gives approximately the right result, as long as the thickness of the lens is much less than the radius of curvature of one of the surfaces. For a mirror the optical power is φ = 2C.
Relation to magnifying powerEdit
{\displaystyle V=0.25\ \mathrm {m} \times \varphi +1}
^ Rosenthal, J. William (1996). Spectacles and Other Vision Aids: A History and Guide to Collecting. Norman. p. 32. ISBN 9780930405717.
^ Collins, Edward Treacher (1929). The history & traditions of the Moorfields Eye Hospital: one hundred years of ophthalmic discovery & development. London: H.K. Lewis. p. 116.
^ Monoyer, F. (1872). "Sur l'introduction du système métrique dans le numérotage des verres de lunettes et sur le choix d'une unité de réfraction". Annales d'Oculistiques (in French). Paris. 68: 101.
^ Thomas, C. "Monoyer, Ferdinand". La médecine à Nancy depuis 1872 (in French). Retrieved 2011-04-26.
^ Colenbrander, August. "Measuring Vision and Vision Loss" (PDF). Smith-Kettlewell Institute. Archived from the original (PDF) on 2014-12-04. Retrieved 2009-07-10.
^ a b Najjar, Dany. "Clinical optics and refraction". Eyeweb. Archived from the original on 2008-03-23. Retrieved 2008-03-25.
^ Palanker, Daniel (October 28, 2013). "Optical Properties of the Eye". American Academy of Ophthalmology. Retrieved 2017-10-16.
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Solving the Linear Oscillatory Problem without Damping with Random Loading Condition Using the Decomposition Method
Solving the Linear Oscillatory Problem without Damping with Random Loading Condition Using the Decomposition Method
Amnah S. Al-Juhani*, Aleh A. Al-Shammari
Faculty of Science, Tabuk University, Tabuk, KSA
Received: April 2, 2018; Accepted: March 10, 2019; Published: March 13, 2019
In this paper we study the solution of random linear oscillatory equation
\stackrel{¨}{x}+{w}^{2}x=F\left(t;\omega \right)
without damping and with random leading condition using the method. Finally, the time evolution of the mean, variance and standard deviation has been plotted for a range of values of the natural frequency w.
Linear Stochastic Differential Equations, Adomian Decompositions, Linear Oscillatory, Mathematica
The Adomian decomposition technique was firstly introduced by Adomian in 1975. This technique can be used to solve differential, integral, algebraic and many other equations (linear or nonlinear) [1] - [12] . The method is based on a suggestion by Adomian G. that the solution can be decomposed into components. In the coming sections we will see that the Adomian decomposition method is also very convenient computationally and offers some significant advantages [13] - [20] . The Adomian decomposition method is not a perturbation procedure, so no assumption concerning the size of randomness is necessary, where each term from the decomposed solution depends only on the preceding terms. A little work in the convergence of the procedure had been done [21] [22] [23] [24] [25] .
In this paper, we focus on solving the following Solving the linear oscillatory problem
\stackrel{¨}{x}+{w}^{2}x=F\left(t;\omega \right)
F\left(t;\omega \right)=e\left(t\right)\left[1+\epsilon n\left(t;\omega \right)\right]
under stochastic excitation
F\left(t;\omega \right)
with the deterministic initial conditions
x\left(0\right)={x}_{0},\text{ }\stackrel{˙}{x}\left(0\right)={\stackrel{˙}{x}}_{0}
w: frequency of oscillation,
\epsilon
: deterministic nonlinearity scale,
\omega \in \left(\Omega ,\sigma ,P\right)
: a triple probability space with
\Omega
as the sample space, where σ is a σ-algebra on event in
\Omega
and P is a probability measure, and
n\left(t;\omega \right)
is a white noise with the following properties:
En\left(t;\omega \right)=0
En\left({t}_{1};\omega \right)\cdot n\left({t}_{2};\omega \right)=\mathrm{cov}\left[n\left({t}_{1}\right),n\left({t}_{2}\right)\right]=\delta \left({t}_{1}-{t}_{2}\right)
By obtaining the P.d.f. of
x\left(t\right)
, the average and variance of the solution process in terms of t: time, the general solution is
x\left(t\right)={x}_{0}\mathrm{cos}wt+\frac{{\stackrel{˙}{x}}_{0}}{w}\mathrm{sin}wt+\frac{1}{\omega }\underset{0}{\overset{t}{\int }}\mathrm{sin}w\left(t-s\right)F\left(s;q\right)\text{d}s
The ensemble average is given by
\begin{array}{c}Ex\left(t\right)={\mu }_{x\left(t\right)}={x}_{0}\mathrm{cos}wt+\frac{{\stackrel{˙}{x}}_{0}}{w}\mathrm{sin}wt+\frac{1}{w}\underset{0}{\overset{t}{\int }}\mathrm{sin}w\left(t-s\right)EF\left(s;q\right)\text{d}s\\ ={x}_{0}\mathrm{cos}wt+\frac{{\stackrel{˙}{x}}_{0}}{w}\mathrm{sin}wt+\frac{1}{\omega }\underset{0}{\overset{t}{\int }}\mathrm{sin}w\left(t-s\right)e\left(s\right)\text{d}s\end{array}
The covariance takes the form
\begin{array}{c}\mathrm{cov}\left(x\left({t}_{1}\right),x\left({t}_{2}\right)\right)=E\left(x\left({t}_{1}\right)-{\mu }_{x\left({t}_{1}\right)}\right)\cdot \left(x\left({t}_{2}\right)-{\mu }_{x\left({t}_{2}\right)}\right)\\ =\frac{{\epsilon }^{2}}{{w}^{2}}\underset{0}{\overset{{t}_{1}}{\int }}\mathrm{sin}w\left({t}_{1}-s\right)\mathrm{sin}w\left({t}_{2}-s\right){e}^{2}\left(s\right)\text{d}s\end{array}
{\sigma }_{x}^{2}\left(t\right)=\frac{{\epsilon }^{2}}{{w}^{2}}\underset{0}{\overset{t}{\int }}{\mathrm{sin}}^{2}w\left(t-s\right){e}^{2}\left(s\right)\text{d}s
Due to linearity and the deterministic properties of and the frequency w we obtain a Gaussian solution process:
{f}_{x\left(t\right)}=\frac{1}{{\sigma }_{x\left(t\right)}\sqrt{2\text{π}}}{\text{e}}^{-\frac{1}{2}{\left(\frac{x\left(t\right)-{\mu }_{x\left(t\right)}}{{\sigma }_{x\left(t\right)}}\right)}^{2}}
{\sigma }_{x\left(t\right)}^{2}=\frac{{\epsilon }^{2}}{{w}^{2}}\underset{0}{\overset{t}{\int }}{\mathrm{sin}}^{2}w\left(t-s\right){e}^{2}\left(s\right)\text{d}s
Equation (9) represents a closed form solution of problem (1) with random loading condition.
F\left(t;\omega \right)={\text{e}}^{-t}+\epsilon n\left(t;\omega \right)
In the Adomian decomposition method, differential operators are decomposed. Thus Equation (1) is rewritten in the following form:
\left(L+R\right)x=F\left(t;q\right)
L=\frac{{\text{d}}^{2}}{\text{d}{t}^{2}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}R={\omega }^{2}
Lx=F\left(t;q\right)-Rx
Solving for x we obtain
x={L}^{-1}F\left(t;q\right)-{L}^{-1}Rx+\varphi \left(t\right)
\varphi \left(t\right)
is the solution of
Lx=0
Lx=0⇒\frac{{\text{d}}^{2}x}{\text{d}{t}^{2}}=0⇒x=at+c
\varphi \left(t\right)={x}_{0}+{\stackrel{˙}{x}}_{0}t
Thus, the solution of equation takes the form:
x={x}_{0}+{\stackrel{˙}{x}}_{0}t+\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}F\left(t;q\right)\text{d}t\text{d}t-{w}^{2}\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}x\left(t\right)\text{d}t\text{d}t
We now assume that the solution can be written in the following form:
x\left(t\right)={x}^{\left(0\right)}\left(t\right)+{x}^{\left(1\right)}\left(t\right)+\cdots =\underset{i=0}{\overset{\infty }{\sum }}{x}^{\left(i\right)}\left(t\right)
Substituting (17) in (16) we obtain:
\underset{i=0}{\overset{\infty }{\sum }}{x}^{\left(i\right)}={x}_{0}+{\stackrel{˙}{x}}_{0}t+\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}F\left(t;q\right)\text{d}t\text{d}t-{\omega }^{2}\underset{i=0}{\overset{\infty }{\sum }}\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{x}^{\left(i\right)}\left(t\right)\text{d}t\text{d}t
By matching the boundaries, we obtain:
{x}^{\left(0\right)}\left(t\right)={x}_{0}+{\stackrel{˙}{x}}_{0}t+\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}F\left(t;q\right)\text{d}t\text{d}t
{x}^{\left(1\right)}\left(t\right)=-{w}^{2}\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{x}^{\left(0\right)}\text{d}t\text{d}t
{x}^{\left(2\right)}\left(t\right)=-{w}^{2}\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{x}^{\left(1\right)}\left(t\right)\text{d}t\text{d}t
And the nth term will be:
{x}^{\left(n\right)}\left(t\right)=-{w}^{2}\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{x}^{\left(n-1\right)}\left(t\right)\text{d}t\text{d}t,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ge 1
By applying this procedure to equation, we obtain:
{x}^{\left(1\right)}\left(t\right)=-{w}^{2}{x}_{o}\frac{{t}^{2}}{2!}-{w}^{2}{\stackrel{˙}{x}}_{0}\frac{{t}^{3}}{3!}-{w}^{2}{L}^{-1}{L}^{-1}F\left(t;q\right)
{x}^{\left(2\right)}\left(t\right)={w}^{4}{x}_{0}\frac{{t}^{4}}{4!}+{w}^{4}{\stackrel{˙}{x}}_{0}\frac{{t}^{5}}{5!}+{w}^{4}{L}^{-1}{L}^{-1}{L}^{-1}F\left(t;q\right)
{x}^{\left(3\right)}\left(t\right)=-{w}^{6}{x}_{0}\frac{{t}^{6}}{6!}-{w}^{6}{\stackrel{˙}{x}}_{0}\frac{{t}^{7}}{7!}-{w}^{6}{\left({L}^{-1}\right)}^{4}F\left(t;q\right)
{x}^{\left(4\right)}\left(t\right)={w}^{8}{x}_{0}\frac{{t}^{8}}{8!}+{w}^{8}{\stackrel{˙}{x}}_{0}\frac{{t}^{9}}{9!}+{w}^{8}{\left({L}^{-1}\right)}^{5}F\left(t;q\right)
The nth term is:
{x}^{\left(n\right)}\left(t\right)={w}^{2n}{x}_{0}\frac{{t}^{2n}}{2n!}+{w}^{2n}{\stackrel{˙}{x}}_{0}\frac{{t}^{2n+1}}{\left(2n+1\right)!}+{w}^{2n}{\left({L}^{-1}\right)}^{n+1}F\left(t;q\right)
\begin{array}{l}x\left(t\right)={x}^{\left(0\right)}+{x}^{\left(1\right)}+{x}^{\left(2\right)}+\cdots \\ ={x}_{0}\left[1-\frac{{\left(wt\right)}^{2}}{2!}+\frac{{\left(wt\right)}^{4}}{4!}+\cdots \right]+\frac{{\stackrel{˙}{x}}_{0}}{\omega }\left[\left(wt\right)-\frac{{\left(wt\right)}^{3}}{3!}+\frac{{\left(wt\right)}^{5}}{5!}+\cdots \right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+\frac{1}{w}\left[w{L}^{-1}-{w}^{3}{\left({L}^{-1}\right)}^{2}+{w}^{5}{\left({L}^{-1}\right)}^{3}-{w}^{7}{\left({L}^{-1}\right)}^{4}+{w}^{9}{\left({L}^{-1}\right)}^{5}+\cdots \right]F\left(t;q\right)\\ ={x}_{0}\mathrm{cos}\omega t+\frac{{\stackrel{˙}{x}}_{0}}{\omega }\mathrm{sin}\omega t+\frac{1}{\omega }\left[\omega {L}^{-1}-{\omega }^{2}{\left({L}^{-1}\right)}^{2}\right]F\left(t;q\right)\end{array}
\underset{0}{\overset{t}{\int }}\cdots \underset{0}{\overset{t}{\int }}F\left(u\right)\text{d}{u}^{n}=\underset{0}{\overset{t}{\int }}\frac{{\left(t-u\right)}^{n-1}}{\left(n-1\right)!}F\left(u\right)\text{d}u
{L}^{-1}F\left(t;q\right)=\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}F\left(t;q\right)\text{d}{t}^{2}=\underset{0}{\overset{t}{\int }}\left(t-u\right)F\left(u;q\right)\text{d}u
{L}^{-1}{L}^{-1}F\left(t;q\right)=\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}F\left(t;q\right)\text{d}{t}^{4}=\underset{0}{\overset{t}{\int }}\frac{{\left(t-u\right)}^{3}}{3!}F\left(u;q\right)\text{d}u
{L}^{-1}{L}^{-1}{L}^{-1}F\left(t;q\right)=\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}F\left(t;q\right)\text{d}{t}^{6}=\underset{0}{\overset{t}{\int }}\frac{{\left(t-u\right)}^{5}}{5!}F\left(u;q\right)\text{d}u
Figure 1. The mean of
x\left(t\right)
\omega =1
Figure 2. The variance of
x\left(t\right)
\omega =1
Figure 3. The covariance of
x\left(t\right)
\epsilon =0.1,\omega =1
x\left(t\right)
\epsilon =0.3,\omega =1
x\left(t\right)
\omega =0.5
x\left(t\right)
\omega =0.5
x\left(t\right)
\epsilon =0.1,\omega =0.5
x\left(t\right)
\epsilon =0.3,\omega =0.5.
\begin{array}{c}x\left(t\right)={x}_{0}\mathrm{cos}wt+\frac{{\stackrel{˙}{x}}_{0}}{\omega }\mathrm{sin}wt+\frac{1}{w}\left[w\underset{0}{\overset{t}{\int }}\left(t-u\right)F\left(u;q\right)\text{d}u\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{w}^{3}\underset{0}{\overset{t}{\int }}\frac{{\left(t-u\right)}^{3}}{3!}F\left(u;q\right)\text{d}u+{w}^{5}\underset{0}{\overset{t}{\int }}\frac{{\left(t-u\right)}^{5}}{5!}F\left(u;q\right)\text{d}u\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{w}^{7}\underset{0}{\overset{t}{\int }}\frac{{\left(t-u\right)}^{7}}{7!}F\left(u;q\right)\text{d}u+\cdots \right]\\ ={x}_{0}\mathrm{cos}wt+\frac{{\stackrel{˙}{x}}_{0}}{\omega }\mathrm{sin}wt+\frac{1}{w}\underset{0}{\overset{t}{\int }}\left[w\left(t-u\right)-\frac{{\left[w\left(t-u\right)\right]}^{3}}{3!}+\cdots \right]F\left(u;q\right)\text{d}u\\ ={x}_{0}\mathrm{cos}wt+\frac{{\stackrel{˙}{x}}_{0}}{w}\mathrm{sin}wt+\frac{1}{w}\underset{0}{\overset{t}{\int }}\mathrm{sin}w\left(t-u\right)F\left(u;q\right)\text{d}u\end{array}
F\left(t;\omega \right)=e\left(t\right)\left[1+\epsilon n\left(t;q\right)\right]
in the previous case-study. By using the decomposition method, the following results are obtained (Figures 1-8).
Al-Juhani, A.S. and Al-Shammari, A.A. (2019) Solving the Linear Oscillatory Problem without Damping with Random Loading Condition Using the Decomposition Method. Journal of Applied Mathematics and Physics, 7, 527-535. https://doi.org/10.4236/jamp.2019.73038
1. Rubinstein, R. and Choudhari, M. (2005) Uncertainty Quantification for Systems with Random Initial Conditions Using Wiener-Hermite Expansions. Studies in Applied Mathematics, 114, 167-188. https://doi.org/10.1111/j.0022-2526.2005.01543.x
2. He J.H. (1999) Homotopy Perturbation Technique. Computer Methods in Applied Mechanics and Engineering, 178, 257-292. https://doi.org/10.1016/S0045-7825(99)00018-3
3. Nayfeh, A. (1993) Problems in Perturbation. Wiley, New York.
4. Tamura, Y. and Nakayama, J. (2005) Enhanced Scattering from a Thin Film with One-Dimensional Disorder. Waves in Random and Complex Media, 15, 269-295. https://doi.org/10.1080/17455030500053336
5. Jahedi, A. and Ahmadi, G. (1983) Application of Wiener-Hermite Expansion to Non-Stationary Random Vibration of a Duffing Oscillator. Journal of Applied Mechanics, Transactions ASME, 50, 436-442. https://doi.org/10.1115/1.3167056
6. Orabi and Ismail, I. (1988) Response of the Duffing Oscillator to a Non-Gaussian Random Excitation. Journal of Applied Mechanics, Transaction of ASME, 55, 740-743. https://doi.org/10.1115/1.3125861
7. Kayanuma, Y. and Noba, K. (2001) Wiener-Hermite Expansion Formalism for the Stochastic Model of a Driven Quantum System. Chemical Physics, 268, 177-188. https://doi.org/10.1016/S0301-0104(01)00305-6
8. Kenny, O. and Nelson, D. (1997) Time-Frequency Methods for Enhancing Speech Proceedings of SPIE—The International Society for Optical Engineering, 3162, 48-57.
9. De Feriet, K. (1955) Random Solutions of Partial Differential Equations. 3rd Berkeley Symposium on Mathematical Statistics and Probability, Vol. III, 199-208.
10. El Tawil, M. (1990) On Stochastic Partial Differential Equations. AMSE Review, 14, 1-8.
11. El-Tawil, M. and Saleh, M. (1998) The Stochastic Diffusion Equation with a Random Diffusion Coefficient. Ain Shams Engineering Journal, 33, 605-613.
12. Mckean (1969) Stochastic Integrals. Academic Press, New York.
13. Kloeden, P.E. and Platen, E. (1992) Numerical Solution of Stochastic Differential Equations. Springer-Verlag, Berlin. https://doi.org/10.1007/978-3-662-12616-5
14. Arnold, L. (1974) Stochastic Differential Equation Theory and Applications. John Wiley, New York.
15. Adomian, G. (1988) A Review of the Decomposition in Applied Mathematics. Mathematical Analysis and Applications, 135, 501-544. https://doi.org/10.1016/0022-247X(88)90170-9
16. Adomian, G. (1992) A Review of the Decomposition Method and Some Recent Results for Nonlinear Equations. Computers & Mathematics with Applications, 21, 101-127.
17. Ahmed, A. (2008) Adomian Decomposition Method: Convergence Analysis and Numerical Approximations. Msc, McMaster University, Hamilton.
18. Johan, G.J. and Snell, I. (1976) Finite Markov Chains. Springer-Verlag, New York.
19. Harold, J. (1977) Probability Methods for Approximation in Stochastic Control for Elliptic Equation. Academic Press, New York.
21. Robert, B. and Melvin, F. (1975) Topics in Stochastic Processes. Academic Press, New York.
22. Yong, Y. (1995) Convergence of Adomian Method and a Lgorithm for Adomian Polynomials. Journal of Mathematical Analysis and Applications, 33, 442-449.
23. El-Tawil, M., et al. (2002) Decomposition Solution of Stochastic Nonlinear Oscillator. International Journal of Differential Equations, 6, 441-422.
24. El-Tawil, M. and Al-Jihany, A. (2009) Approximate Solution of Mixed Nonlinear Stochastic Oscillator. Computers & Mathematics with Applications, 58, 2236-2259. https://doi.org/10.1016/j.camwa.2009.03.057
25. Al-Jihany, A. (2010) Comparisons between WHEP and Homotopy Perturbation Techniques in Solving Stochastic Cubic Oscillatory Problems. Computers & Mathematics with Applications, 58, 19-25.
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Bon, Jean-Louis ; Păltănea, Eugen
The paper is motivated by the stochastic comparison of the reliability of non-repairable
k
-out-of-
n
systems. The lifetime of such a system with nonidentical components is compared with the lifetime of a system with identical components. Formally the problem is as follows. Let
{U}_{i},\phantom{\rule{0.166667em}{0ex}}i=1,...,n,
be positive independent random variables with common distribution
F
{\lambda }_{i}>0
\mu >0
, let consider
{X}_{i}={U}_{i}/{\lambda }_{i}
{Y}_{i}={U}_{i}/\mu ,\phantom{\rule{4pt}{0ex}}i=1,...,n
. Remark that this is no more than a change of scale for each term. For
k\in \left\{1,2,...,n\right\},
{X}_{k:n}
k
th order statistics of the random variables
{X}_{1},...,{X}_{n}
{Y}_{k:n}
k
th order statistics of
{Y}_{1},...,{Y}_{n}.
{X}_{i},\phantom{\rule{4pt}{0ex}}i=1,...,n,
are the lifetimes of the components of a
n
1
k
non-repairable system, then
{X}_{k:n}
is the lifetime of the system. In this paper, we give for a fixed
k
a sufficient condition for
{X}_{k:n}{\ge }_{st}{Y}_{k:n}
st
is the usual ordering for distributions. In the markovian case (all components have an exponential lifetime), we give a necessary and sufficient condition. We prove that
{X}_{k:n}
is greater that
{Y}_{k:n}
according to the usual stochastic ordering if and only if
\left(\begin{array}{c}n\\ k\end{array}\right){\mu }^{k}\ge \sum _{1\le {i}_{1}<{i}_{2}<...<{i}_{k}\le n}{\lambda }_{{i}_{1}}{\lambda }_{{i}_{2}}...{\lambda }_{{i}_{k}}.
Classification : 60E15, 62N05, 62G30, 90B25, 60J27
Mots clés : stochastic ordering, Markov system, order statistics,
k
n
author = {Bon, Jean-Louis and P\u{a}lt\u{a}nea, Eugen},
title = {Comparison of order statistics in a random sequence to the same statistics with {I.I.D.} variables},
AU - Bon, Jean-Louis
AU - Păltănea, Eugen
TI - Comparison of order statistics in a random sequence to the same statistics with I.I.D. variables
Bon, Jean-Louis; Păltănea, Eugen. Comparison of order statistics in a random sequence to the same statistics with I.I.D. variables. ESAIM: Probability and Statistics, Tome 10 (2006), pp. 1-10. doi : 10.1051/ps:2005020. http://www.numdam.org/articles/10.1051/ps:2005020/
[1] J.-L. Bon and E. Păltănea, Ordering properties of convolutions of exponential random variables. Lifetime Data Anal. 5 (1999) 185-192. | Zbl 0967.60017
[2] G.H. Hardy, J.E. Littlewood and G. Pólya, Inequalities. Cambridge University Press, Cambridge (1934). | JFM 60.0169.01 | Zbl 0010.10703
[3] B.-E. Khaledi and S. Kochar, Some new results on stochastic comparisons of parallel systems. J. Appl. Probab. 37 (2000) 1123-1128. | Zbl 0995.62104
[4] A.W. Marshall and I. Olkin, Inequalities: Theory of Majorization and Its Applications. Academic Press, New York (1979). | MR 552278 | Zbl 0437.26007
[5] E. Păltănea. A note of stochastic comparison of fail-safe Markov systems2003) 179-182.
[6] P. Pledger and F. Proschan, Comparisons of order statistics and spacing from heterogeneous distributions, in Optimizing Methods in Statistics. Academic Press, New York (1971) 89-113. | Zbl 0263.62062
[7] M. Shaked and J.G. Shanthikumar, Stochastic Orders and Their Applications. Academic Press, New York (1994). | MR 1278322 | Zbl 0806.62009
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SAT Data - Graphs and Charts | Brilliant Math & Science Wiki
SAT Data - Graphs and Charts
To successfully solve problems about graphs and charts on the SAT, you need to know how to create and read:
SAT Tips for Data - Graphs and Charts
The graph above shows the cost of manufacturing of two products, A and B, per year. According to the graph, during which year was the cost of manufacturing both products the same?
\ \ 2000
\ \ 2001
\ \ 2002
\ \ 2003
\ \ 2004
f
g
(x,y),
f(x) = y = g(x)
The two lines intersect at (2002, 40). So, in the year 2002, the two products had the same manufacturing cost, $40,000.
The two lines do not intersect in any of these years. Therefore, the manufacturing costs for the two products were different in these choices.
The graph above shows the cost of manufacturing of two products, A and B, per year. In which year was the difference in the cost of manufacturing the two products the greatest?
\ \ 2001
\ \ 2002
\ \ 2003
\ \ 2007
\ \ 2008
The bar graph above shows the distribution of recipes in a certain cookbook. Which of the following pie charts most accurately displays the same data?
In the bar graph, each cookbook item is represented by a rectangle with a certain area, a percentage, and a different pattern. In the pie chart, the areas of the sectors should correspond to the percentages in the bar graph. In other words, desserts should take up 13% of the area of the circle, main entrees should take up 64% of the area of the circle, and so on. Also, that corresponding shapes should have matching patterns. We analyze each of the answer choices.
(A) This seems to be the correct answer. According to the bar graph, the desserts (dots) and the salads (solid black) have the same percentage. The dotted and the solid black sectors in the pie chart indeed have about the same area. There are about half as many snacks as salads, and the area taken up by the snacks (little squares) in the pie chart is about half the area taken up by the salads (solid black). Only 3% of the cookbook is devoted to soups, and the sector with the slanted stripes in the pie chart is the smallest, as expected. 64% of the cookbook is devoted to main entrees, and they seem to take up about two thirds of the circle, as is expected.
(B) Here, the dessert section is hardly discernible, but it is supposed to take up 13% of the area of the circle. And the main section takes up about 80%, instead of 64%. We eliminate this choice.
(C) This pie chart only has 4 categories, but we are given five. We eliminate this choice.
(D) Here, the desserts, snacks, salads, and soups sections seem to have equal areas. But we're given that only the desserts and the salads have the same percentage. This choice is wrong.
(E) Here, the soups and the salads sections have the same area. But we're told that the desserts and the salads have the same percentage, not the soups and the salads. This choice is wrong.
Only the pie chart in choice (A) corresponds to the given bar graph. Therefore, it is the correct answer.
Cite as: SAT Data - Graphs and Charts. Brilliant.org. Retrieved from https://brilliant.org/wiki/sat-data-graphs-and-charts/
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Multisignal 1-D wavelet packet transform - MATLAB dwpt - MathWorks España
Multichannel Discrete Wavelet Packet Transform
packetlevels
Multisignal 1-D wavelet packet transform
wpt = dwpt(X)
wpt = dwpt(X,wname)
wpt = dwpt(X,LoD,HiD)
[wpt,l] = dwpt(___)
[wpt,l,packetlevels] = dwpt(___)
[wpt,l,packetlevels,f] = dwpt(___)
[wpt,l,packetlevels,f,re] = dwpt(___)
[___] = dwpt(___,Name,Value)
wpt = dwpt(X) returns the terminal (final-level) nodes of the discrete wavelet packet transform (DWPT) of X. The input X is a real-valued vector, matrix, or timetable. By default, the fk18 wavelet is used, and the decomposition level is floor(log2(Ns)), where Ns is the number of data samples. The wavelet packet transform wpt is a 1-by-N cell array, where N = 2^floor(log2(Ns)).
wpt = dwpt(X,wname) uses the wavelet specified by wname for the DWPT. wname must be recognized by wavemngr.
wpt = dwpt(X,LoD,HiD) uses the scaling (lowpass) filter, LoD, and wavelet (highpass) filter, HiD.
[wpt,l] = dwpt(___) also returns the bookkeeping vector using any of the previous syntaxes. The vector l contains the length of the input signal and the number of coefficients by level. The bookkeeping vector is required for perfect reconstruction.
[wpt,l,packetlevels] = dwpt(___) also returns the transform levels of the nodes of wpt using any of the previous syntaxes.
[wpt,l,packetlevels,f] = dwpt(___) also returns the center frequencies of the approximate passbands in cycles per sample using any of the previous syntaxes.
[wpt,l,packetlevels,f,re] = dwpt(___) also returns the relative energy for the wavelet packets in wpt using any of the previous syntaxes. The relative energy is the proportion of energy contained in each wavelet packet by level.
[___] = dwpt(___,Name,Value) specifies options using name-value pair arguments in addition to the input arguments in the previous syntaxes. For example, 'Level',4 specifies the decomposition level.
Load the 23-channel EEG data Espiga3 [3]. The data is sampled at 200 Hz.
Compute the 1-D DWPT of the data using the sym3 wavelet down to level 4. Obtain the terminal wavelet packet nodes, bookkeeping vector, and center frequencies of the approximate passbands.
[wpt,bk,~,f] = dwpt(Espiga3,'sym3','Level',4);
The output wpt is a 1-by-
{2}^{4}
cell array. Every element of wpt is a matrix. Choose any terminal node, and confirm the size of the matrix is 23-by-M, where M is the last element of the bookkeeping vector bk.
size(wpt{nd})
bk(end)
Extract the final-level coefficients of the fifth channel.
p5 = cell2mat(cellfun(@(x) x(5,:).',wpt,'UniformOutput',false));
size(p5)
The terminal nodes are sequency-ordered. Plot the center frequencies of the approximate passbands in hertz, and confirm they are in order of increasing frequency.
plot(200*f,'x')
title('Center Frequencies')
{H}_{0}\left(z\right)=1/8\left(-{z}^{2}+2z+6+2{z}^{-1}-{z}^{-2}\right)
{H}_{1}\left(z\right)=1/2\left(z+2+{z}^{-1}\right)
\sqrt{2}
real-valued vector | real-valued matrix | timetable
Input data, specified as a real-valued vector, matrix, or timetable. If X is a matrix, the transform is applied to each column of X. If X is a timetable, X must either contain a matrix in a single variable or column vectors in separate variables. X must be uniformly sampled.
Wavelet to use in the DWPT, specified as a character vector or string scalar. wname must be recognized by wavemngr.
Example: wpt = dwpt(data,"sym4") specifies the sym4 wavelet.
LoD,HiD — Wavelet analysis filters
Wavelet analysis (decomposition) filters to use in the DWPT, specified as a pair of real-valued vectors. LoD is the scaling (lowpass) analysis filter, and HiD is the wavelet (highpass) analysis filter. You cannot specify both wname and a filter pair, LoD and HiD. See wfilters for additional information.
dwpt does not check that LoD and HiD satisfy the requirements for a perfect reconstruction wavelet packet filter bank. See PR Biorthogonal Filters for an example of how to take a published biorthogonal filter and ensure that the analysis and synthesis filters produce a perfect reconstruction wavelet packet filter bank using dwpt.
Example: wpt = dwpt(x,'sym4','Level',4) specifies a level 4 decomposition using the sym4 wavelet.
floor(log2(Ns)) (default) | positive integer
Wavelet decomposition level, specified as a positive integer less than or equal to floor(log2(Ns)), where Ns is the number of samples in the data. If unspecified, Level defaults to floor(log2(Ns)).
FullTree — Wavelet packet tree handling
Wavelet packet tree handling, specified as a numeric or logical 1 (true) or 0 (false). When set to true, wpt contains the full packet tree. When set to false, wpt contains only the terminal nodes. If unspecified, FullTree defaults to false.
Boundary — Wavelet packet transform boundary handling
Wavelet packet transform boundary handling, specified as 'reflection' or 'periodic'. Setting to 'reflection' or 'periodic', the wavelet packet coefficients are extended at each level based on the 'sym' or 'per' mode in dwtmode, respectively. If unspecified, Boundary defaults to 'reflection'.
Wavelet packet transform, returned as a 1-by-M cell array. If taking the DWPT of one signal, each element of wpt is a vector. Otherwise, each element is a matrix. The coefficients in the jth row of the matrix correspond to the signal in the jth column of X. The packets are sequency-ordered.
If returning the terminal nodes of a level N decomposition, wpt is a 1-by-2N cell array. If returning the full wavelet packet tree, wpt is a 1-by-(2N+1−2) cell array.
Bookkeeping vector, returned as a vector of positive integers. The vector l contains the length of the input signal and the number of coefficients by level, and is required for perfect reconstruction.
packetlevels — Transform levels
Transform levels, returned as a vector of positive integers. The ith element of packetlevels corresponds to the ith element of wpt. If wpt contains only the terminal nodes, packetlevels is a vector with each element equal to the terminal level. If wpt contains the full wavelet packet tree, then packetlevels is a vector with 2j elements for each level j.
f — Center frequencies
Center frequencies of the approximate passbands in cycles per sample, returned as a real-valued vector. The jthe element of f corresponds to the jth wavelet packet node of wpt. You can multiply the elements in f by a sampling frequency to convert to cycles per unit time.
re — Relative energy
Relative energy for the wavelet packets in wpt, returned as a cell array. The relative energy is the proportion of energy contained in each wavelet packet by level. The jth element of re corresponds to the jth wavelet packet node of wpt.
Each element of re is a scalar when taking the DWPT of one signal. Otherwise, when taking the DWPT of M signals, each element of re is a M-by-1 vector, where the ith element is the relative energy of the ith signal channel. For each channel, the sum of relative energies in the wavelet packets at a given level is equal to 1.
The dwpt function performs a discrete wavelet packet transform and produces a sequency-ordered wavelet packet tree. Compare the sequency-ordered and normal (Paley)-ordered trees.
\stackrel{˜}{G}\left(f\right)
is the scaling (lowpass) analysis filter, and
\stackrel{˜}{H}\left(f\right)
represents the wavelet (highpass) analysis filter. The labels at the bottom show the partition of the frequency axis [0, ½].
modwpt | idwpt
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Revision as of 20:39, 4 August 2013 by NikosA (talk | contribs) (→Deriving Physical Quantities: script added)
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L_{\lambda {\text{Pixel, Band}}}={\frac {K_{\text{Band}}*q_{\text{Pixel, Band}}}{\Delta \lambda _{\text{Band}}}}}
{\displaystyle L_{\lambda {\text{Pixel,Band}}}}
{\displaystyle K_{\text{Band}}}
{\displaystyle q_{\text{Pixel,Band}}}
{\displaystyle \Delta _{\lambda _{\text{Band}}}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
Converting Digital Numbers to Radiance/Reflectance requires knowledge about the sensor's specific spectral band parameters. In the following listed paramters (see below), the Effective Bandwidths and the Band-Averaged Solar Spectral Irradiances are extracted from the document Radiometric Use of WorldView-2 Imagery, Technical Note (2010) by Todd Updike and Chris Comp. The Absolute Calibration Factors are extracted from a WorldView-2 product, specifically from an image metadata file (extension .IMD).
The absolute radiometric calibration factor is dependent on the specific band, as well as the TDI exposure level, line rate, pixel aggregration, and bit depth of the product. Based on these parameters, the appropriate value is provided in the .IMD file. For this reason, care should be taken not to mix absolute radiometric calibration factors between products that might have different collection conditions.
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
eval K_BAND="K_${BAND}"
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WRITE A RULE FOR THE nth TERM OF THE ARITHMETIC SEQUENCE THEN FIND a_25: {1.6,4,
WRITE A RULE FOR THE nth TERM OF THE ARITHMETIC SEQUENCE THEN FIND a_25: {1.6,4,6.4,8.8,11.2,…} *
WRITE A RULE FOR THE nth TERM OF THE ARITHMETIC SEQUENCE THEN FIND
{a}_{25}
: {1.6,4,6.4,8.8,11.2,…} *
tabuordg
{a}_{n}=2.4n-0.8,{a}_{25}=59.2
We know an arithmetic sequence has the same difference for every two consecutive terms. So, the difference is
4-1.6=2.4
Then, note that the nth term of an arithmetic sequence is the first term plus
\left(n-1\right)×
the difference. Thus,
{a}_{n}=1.6+2.4×\left(n-1\right)=2.4n-0.8
{a}_{25}=2.4×25-0.8=60-0.8=59.2
Find the z-transform for the sequences
x\left(n\right)=10u\left(n\right)
x\left(n\right)=10\mathrm{sin}\left(0.25\pi n\right)u\left(n\right)
Give an example of such
f\left(x\right)
and prove, why it works
A non constant function
f:\left[0,1\right]\to R
f\left(x\right)\ge 0
x\in \left[0,1\right]
{\int }_{0}^{1}f\left(x\right)dx=0
Make the first 10 terms of the sequence. Logs are to base 2.
Indicate whether the sequence is increasing, decreasing, non-increasing, or non-decreasing. The sequence may have more than one of those properties.
{n}^{th}
⌈\mathrm{log}⌉
The 4th term of an AP is 4 and the 8th term is 22 what is the 1st term?
Find the z-transform f(z)
f\left(n\right)={\left(0.5\right)}^{\nu }\left(n\right)+{\left(1.5\right)}^{\nu }\left(-n-1\right)-2\delta \left(n\right)
Give the first six terms of the following sequences.
{g}_{1}=2
{g}_{2}=1
{c}_{1}=4,{c}_{2}=5
{c}_{n}={c}_{n-1}×{c}_{n-2}
n\ge 3
{b}_{1}=1,{b}_{2}=3
{b}_{n}={b}_{n-1}-{b}_{n-2}
n\ge 3
{d}_{1}=1,{d}_{2}=1
{d}_{n}={\left({d}_{n-1}\right)}^{2}+{d}_{n-2}
n\ge 3
{a}_{1},{a}_{2},{a}_{3}\dots \text{ }\text{and}\text{ }{b}_{1},{b}_{2},{b}_{3}\dots
{a}_{1}{b}_{1},{a}_{2}{b}_{2},{a}_{3}{b}_{3},\dots
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Equations - EXODIA
deposit = withdrawal
Swaps between EXOD and sEXOD during staking and unstaking are always honored 1:1. The amount of EXOD deposited into the staking contract will always result in the same amount of sEXOD. And the amount of sEXOD withdrawn from the staking contract will always result in the same amount of EXOD.
rebase = 1 - ( exodDeposits / sEXODOutstanding )
The treasury deposits EXOD into the distributor. The distributor then deposits EXOD into the staking contract, creating an imbalance between EXOD and sEXOD. sEXOD is rebased to correct this imbalance between EXOD deposited and sEXOD outstanding. The rebase brings sEXOD outstanding back up to parity so that 1 sEXOD equals 1 staked EXOD.
bond Price = 1 + Premium
EXOD has an intrinsic value of 1 DAI, which is roughly equivalent to $1. In order to make a profit from bonding, EXODIA charges a premium for each bond.
Premium = debt Ratio * BCV
The premium determines profit due to the protocol and in turn, stakers. This is because new EXOD is minted from the profit and subsequently distributed among all stakers.
debt Ratio = bondsOutstanding/exodSupply
The debt ratio is the total of all EXOD promised to bonders divided by the total supply of EXOD. This allows us to measure the debt of the system.
bondPayout_{reserveBond} = marketValue_{asset}\ /\ bondPrice
Bond payout determines the number of EXOD sold to a bonder. For reserve bonds, the market value of the assets supplied by the bonder is used to determine the bond payout. For example, if a user supplies 1000 DAI and the bond price is 250 DAI, the user will be entitled 4 EXOD.
bondPayout_{lpBond} = marketValue_{lpToken}\ /\ bondPrice
For liquidity bonds, the market value of the LP tokens supplied by the bonder is used to determine the bond payout. For example, if a user supplies 0.001 EXOD-DAI LP token which is valued at 1000 DAI at the time of bonding, and the bond price is 250 DAI, the user will be entitled 4 EXOD.
EXOD Supply
EXOD_{supplyGrowth} = EXOD_{stakers} + EXOD_{bonders} + EXOD_{DAO}
EXOD supply does not have a hard cap. Its supply increases when:
EXOD is minted and distributed to the stakers.
EXOD is minted for the bonder. This happens whenever someone purchases a bond.
EXOD is minted for the DAO. This happens whenever someone purchases a bond. The DAO gets the same number of EXOD as the bonder.
EXOD_{stakers} = EXOD_{totalSupply} * rewardRate
At the end of each epoch, the treasury mints EXOD at a set reward rate. These EXOD will be distributed to all the stakers in the protocol.
EXOD_{bonders} = bondPayout
Whenever someone purchases a bond, a set number of EXOD is minted. These EXOD will not be released to the bonder all at once - they are vested to the bonder linearly over time. The bond payout uses a different formula for different types of bonds. Check the bonding section above to see how it is calculated.
EXOD_{DAO} = EXOD_{bonders}
The DAO receives the same amount of EXOD as the bonder. This represents the DAO profit.
Backing per EXOD
EXOD_{backing} = treasuryBalance_{stablecoin} + treasuryBalance_{otherAssets}
Every EXOD in circulation is backed by the EXODIA treasury. The assets in the treasury can be divided into two categories: stablecoin and non-stablecoin.
treasuryBalance_{stablecoin} = RFV_{reserveBond} + RFV_{lpBond}
RFV_{reserveBond} = assetSupplied
For reserve bonds such as DAI bond, the RFV simply equals to the amount of the underlying asset supplied by the bonder.
RFV_{lpBond} = 2sqrt(constantProduct) * (\%\ ownership\ of\ the\ pool)
For LP bonds such as EXOD-DAI bond, the RFV is calculated differently because the protocol needs to mark down its value. Why? The LP token pair consists of EXOD, and each EXOD in circulation will be backed by these LP tokens - there is a cyclical dependency. To safely guarantee all circulating EXOD are backed, the protocol marks down the value of these LP tokens, hence the name risk-free value (RFV).
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High School Calculus/Implicit Differentiation - Wikibooks, open books for an open world
High School Calculus/Implicit Differentiation
Implicit Differentiation[edit | edit source]
When a functional relation between x and y cannot be readily solved for y, the preceding rules may be applied directly to the implicit function.
The derivative will usually contain both x and y. Thus the derivative of an algebraic function, defined by setting the polynomial of x and y to zero.
Given the function y of x
{\displaystyle x^{5}+y^{5}-5xy+1=0}
{\displaystyle {\operatorname {d} y \over \operatorname {d} x}}
{\displaystyle {\operatorname {d} \over \operatorname {d} x}(x^{5}+y^{5}-5xy+1)=0}
{\displaystyle =5x^{4}+5y^{4}{\operatorname {d} y \over \operatorname {d} x}-5y-5x{\operatorname {d} y \over \operatorname {d} x}=0}
{\displaystyle {\operatorname {d} y \over \operatorname {d} x}}
we must first factor the differentiation problem
In doing this we get
{\displaystyle {\operatorname {d} y \over \operatorname {d} x}(5y^{4}-5x)+(5x^{4}-5y)=0}
From here we subtract the
{\displaystyle {\operatorname {d} y \over \operatorname {d} x}}
to one side
Thus giving us
{\displaystyle 5x^{4}-5y=-{\operatorname {d} y \over \operatorname {d} x}(-5x+5y^{4})}
Here I am going to skip a step in solving this implicit differentiation problem. I am going to skip the step where I divide the -1 over to the other side.
From here we divide the polynomial from the
{\displaystyle \operatorname {d} y \over \operatorname {d} x}
over to the other side. Giving us
{\displaystyle \left({\frac {-5x^{4}+5y}{-5x+5y^{4}}}\right)={\operatorname {d} y \over \operatorname {d} x}}
Now we simplify and get
{\displaystyle {\operatorname {d} y \over \operatorname {d} x}=\left({\frac {x^{4}-y}{x-y^{4}}}\right)}
Other problems to work on
{\displaystyle {\operatorname {d} y \over \operatorname {d} x}}
{\displaystyle xy^{2}+x^{2}y=1}
{\displaystyle {\operatorname {d} y \over \operatorname {d} x}}
{\displaystyle x+y+(x-y)^{2}+(2x-3y)^{3}=0}
Retrieved from "https://en.wikibooks.org/w/index.php?title=High_School_Calculus/Implicit_Differentiation&oldid=2178797"
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Solve the equation. \frac{2x}{2x+3}+\frac{6}{4x+6}=5
\frac{2x}{2x+3}+\frac{6}{4x+6}=5
\frac{2x}{2x+3}+\frac{6}{4x+6}=5
To solve this equation:
A=\left[\begin{array}{cc}3& 1\\ 1& 1\\ 1& 4\end{array}\right],b\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]
\stackrel{―}{x}=
Convert the given polar equation to a Cartesian equation. Write in the standard form of a conic if possible, and identify the conic section represented.
r=2\mathrm{sec}0
Write the system of equations in the image in matrix form
{x}_{1}^{\prime }\left(t\right)=3{x}_{1}\left(t\right)-2{x}_{2}\left(t\right)+{e}^{t}{x}_{3}\left(t\right)
{x}_{2}^{\prime }\left(t\right)=\mathrm{sin}\left(t\right){x}_{1}\left(t\right)+\mathrm{cos}\left(t\right){x}_{3}\left(t\right)
{x}_{3}^{\prime }\left(t\right)={t}^{2}{x}_{1}\left(t\right)+t{x}^{2}\left(t\right)+{x}_{3}\left(t\right)
Solve the equation 23v+16=-10.
3.(-1,7)
\left\{\begin{array}{c}4x+y=3\\ 2x-3=y\end{array}\right\}
{8}^{2x}=32
\mathrm{ln}\left(-3x+2\right)=\mathrm{ln}\left(14\right)
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â ‹Please solve in a diagrammed manner: - Maths - Playing with Numbers - 12489521 | Meritnation.com
Please solve in a diagrammed manner:
Find the value of A, B and C in :\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}} 8 A\phantom{\rule{0ex}{0ex}}+8 B\phantom{\rule{0ex}{0ex}}_________\phantom{\rule{0ex}{0ex}}C B 3\phantom{\rule{0ex}{0ex}}_________
Since A, B and C are digits, they can only take values 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9.
So, the minimum value of the sum above is 80 + 80 = 160 and the maximum value is 89 + 89 = 178. Thus, C = 1.
Now, B can either be equal to 6 or 7.
Let us assume that B = 6. Then the final sum is 163. Here, the only two numbers which add up to 163 are 82 and 81. So, 82 + 81 = 163, which gives B = 2 or 1, which is a contradiction since we had assumed B to be 6. Thus, the final sum must be 173. This gives B = 7.
Now, 87 + 86 = 173, which gives A = 6.
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Position-Time Graph | Brilliant Math & Science Wiki
Sravanth C., Tim O'Brien, and Jimin Khim contributed
Position-time graphs are the most basic form of graphs in kinematics, which allow us to describe the motion of objects. In these graphs, the vertical axis represents the position of the object while the horizontal axis represents the time elapsed: the dependent variable, position, depends on the independent variable, time. In this way, the graph tells us where the particle can be found after some amount of time. Graphs such as these help us visualize the trajectory of objects. An amazing amount can be learned by studying a position-time graph for an object, as long as we know how to properly analyze them.
Slope of a Position-time Graph
We know that the slope of a function is its derivative, and that the derivative of the displacement function is the velocity function, and thus the slope of a position-time graph gives us the velocity of the object:
(\text{Slope of a position time graph})=v=\dfrac{s_2-s_1}{t_2-t_1}.
(1,2)
We have seen that the slope of the position-time graph gives us the velocity over the time period. Thus,
v=\dfrac{s_2-s_1}{t_2-t_1}=\dfrac{10-7.5}{2-1}=2.5\text{ (m/s)}.
Note: We can see that the displacement of this object is given by the function
s(t)=2.5t+5
. Thus, we can find the velocity of this function which is nothing but its derivative:
v(t)=\dfrac{d}{dt}(2.5t+5)=2.5\text{ (m/s)}.
(2,4)
Cite as: Position-Time Graph. Brilliant.org. Retrieved from https://brilliant.org/wiki/position-time-graph/
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student(deprecated)/completesquare - Maple Help
Home : Support : Online Help : student(deprecated)/completesquare
completesquare(f)
completesquare(f, x)
one or more of the indeterminates occurring in f
Important: The student package has been deprecated. Use the superseding command Student[Precalculus][CompleteSquare] instead.
This function completes the square of polynomials of degree 2 in x by re-writing such polynomials as perfect squares plus a remainder. If more than one variable appears in f, then x must be specified. x can be a name, list or set.
The polynomial may appear as a subexpression of f, so that equations, reciprocals, and in general, parts of expressions, may be simplified. Since the polynomial may occur as an argument to Int this function can be used to help simplify unevaluated integrals.
Terms not involving the indicated variable are treated as constants.
The command with(student,completesquare) allows the use of the abbreviated form of this command.
\mathrm{with}\left(\mathrm{student}\right):
\mathrm{completesquare}\left(9{x}^{2}+24x+16\right)
\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{}{\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{4}}{\textcolor[rgb]{0,0,1}{3}}\right)}^{\textcolor[rgb]{0,0,1}{2}}
\mathrm{completesquare}\left({x}^{2}-2xa+{a}^{2}+{y}^{2}-2yb+{b}^{2}=23,x\right)
{\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{a}\right)}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{b}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{23}
\mathrm{completesquare}\left(,y\right)
{\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{b}\right)}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{a}\right)}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{23}
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Distance, Rate, and Time | Brilliant Math & Science Wiki
Janae Pritchett, Brilliant Mathematics, Johnny Gérard, and
A rate of change is the ratio between the change in one quantity to the change in another quantity. A very common rate is speed, the ratio between distance and time.
While traveling in your car, your rate of speed might be 180 miles per 3 hours. We often simplify rates to express them as unit rates, which is how much of something there is per one unit of something else. Using the previous example, your unit rate of speed while traveling in the car is 60 miles per one hour.
If Starr walks 50 meters in 8 seconds, what is her rate of speed?
Starr's rate is 50 meters per 8 seconds, or
50 \div 8 = 6.25
_\square
Alice is running at a constant speed of
4
meters per second. How long (in seconds) will it take her to run
52
If an object moves at a constant rate of speed, we can determine how far the object travels by multiplying its rate by the time it has been traveling:
\text{distance} = \text{rate} \times \text{ time}.
Likewise, we can find rate of speed by dividing distance by time
\text{rate}=\frac{\text{distance}}{\text{time}}
and determine time by dividing distance traveled by rate of speed
\text{time}=\frac{\text{distance}}{\text{rate}}.
An ant is traveling at an average of 8 millimeters per second. How far will the ant travel in one minute?
We can find the ant's total distance by multiplying
\text{rate} \times \text{ time}: 8 \text{ mm/s} \times 60 \text{ seconds} = 480 \text { mm}.\ _\square
If a car travels at 80 kilometers per hour, how long will it take the car to travel 500 kilometers?
We can find the time by dividing the distance by the rate:
500 \text{ kilometers} \div 80 \text{ kilometers/hour} = 6.25 \text{ hours}.\ _\square
Calvin can run on a treadmill at a steady
6
miles per hour pace. How long (in minutes) will it take for him to run 13 miles?
It took a train
20
seconds to completely pass a
500
-m iron bridge. When the train was passing through a
1900
-m tunnel, it was invisible from outside of the tunnel for
30
seconds. If the speed of the train was constant, what was the length of the train (in meters)?
The train passes the iron bridge when no part of the train is on the bridge.
The train is invisible from outside of the tunnel when no part of the train is outside of the tunnel.
In order to determine average speed, we need to divide the total distance traveled by the total time.
Charlie bikes 500 meters to work in 6 minutes and returns home in 4 minutes. What is his average speed for the day?
Charlie's total distance is 1000 meters and his total time is 10 minutes. Therefore, Charlie's average speed is
1000 \div 10 = 100
meters per minute.
_\square
Lokman bikes from her home to her office at a constant speed of 20 km/h, and she bikes back home at a constant speed of 40 km/h. What is closest to the average speed of her bike over time in km/h?
Cite as: Distance, Rate, and Time. Brilliant.org. Retrieved from https://brilliant.org/wiki/distance-rate-and-time/
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Revision as of 19:28, 31 July 2013 by NikosA (talk | contribs) (→Post-Processing: some re-structuring)
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L\lambda ={\frac {10^{4}*DN\lambda }{CalCoef\lambda *Bandwidth\lambda }}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
Having beforehand satellite image data expressed in physical quantities (radiance or reflectance) is preferred to follow-up with digital image analysis techniques. A few common post-processing practices are Contrast-Enhancing, Pan-Sharpening and creating Pseudo- or True-Color Composites. Other well known enhancing manipualtions to support the analyses of satellite imagery, include deriving Vegetation Indices, transforming multi-spectral data based on PCA and Segmenting images.
{\displaystyle [0,255]}
{\displaystyle [0,2047]}
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High School Calculus/The Chain Rule - Wikibooks, open books for an open world
High School Calculus/The Chain Rule
When differentiating a square root function or a quantity raised to a power the chain rule is used.
{\displaystyle (ax^{n}+b)^{n}}
Using the chain rule we take the derivative of the entire quantity to the power, and multiply that by the derivative of the interior quantity:
{\displaystyle n(ax^{n}+b)^{n-1}*(anx^{n-1})}
{\displaystyle =an^{2}x^{n-1}(ax^{n}+b)^{n-1}}
What we did was take the power of the quantity and moved it out front. From there we took the derivative of the inside and then multiplied our previous steps by the quantity with n-1 for the power of the quantity.
{\displaystyle f(x)=(2x+4)^{3}}
In order to differentiate this properly we must use the chain rule. First thing we do is take the power number and derive it.
{\displaystyle 3(2x+4)^{2}}
From here we take the derivative of the inside
{\displaystyle 3(2)(2x+4)^{2}}
All we have to do now is some minor simplification
{\displaystyle f^{\prime }(x)=6(2x+4)^{2}}
And that is the derivative of
{\displaystyle f(x)=(2x+4)^{3}}
Other examples to work on
{\displaystyle {\sqrt {4x^{5}+x}}}
Remember that a square root is just something raised to the one half power
{\displaystyle 2(x^{3}+x^{2})^{5}}
{\displaystyle 2x(x^{2}+1)^{3}}
Hint: You will have to use the product rule as well as the chain rule
An alternate way of completing a chain rule is considering the problem as a composite function.
In this instance you consider simply taking the derivative of the outside multiplied by the derivative of the inside.
Chain Rule becomes very useful when dealing with a function to a power, or a square root of a function.
The following proof outlines this idea.
{\displaystyle h(x)=f(g(x)).}
{\displaystyle x=c}
and that as
{\displaystyle g(x)\rightarrow g(c)}
,that
{\displaystyle x\rightarrow c}
{\displaystyle h'(c)=\lim _{x\rightarrow c}{\frac {f(g(x))-f(g(c))}{x-c}}}
{\displaystyle =\lim _{x\rightarrow c}[{\frac {f(g(x))-f(g(c))}{g(x)-g(c)}}*{\frac {g(x)-g(c)}{x-c}}]}
{\displaystyle g(x)\neq g(c)}
{\displaystyle =[\lim _{x\rightarrow c}{\frac {f(g(x))-f(g(c))}{g(x)-g(c)}}][\lim _{x\rightarrow c}{\frac {g(x)-g(c)}{x-c}}]}
{\displaystyle =f'(g(c))g'(c){\square }.}
In order to find the derivative, you can take that proof into consideration.
You can use any method you deem viable to find the derivative.
{\displaystyle h(x)=(3x-1)^{2}.}
{\displaystyle g(x)=3x-1}
{\displaystyle f(x)=x^{2}.}
According to chain rule,
{\displaystyle f'(x)=2x.}
In the problem,
{\displaystyle g(x)}
is inside of
{\displaystyle f(x).}
So, then
{\displaystyle f'(g(x))=2g(x).}
But because of the chain rule, we must also multiply by
{\displaystyle g'(x).}
Thus giving us,
{\displaystyle 2g(x)g'(x)}
We also see that,
{\displaystyle g'(x)=3.}
So if we put this all together we get,
{\displaystyle h'(x)=2(3x-1)(3)}
{\displaystyle h'(x)=6(3x-1).}
You may have noticed that in some cases, a chain rule may be multiplied out to create a simple power rule.
For the previous example, if you multiply the power out, you are left with a simpler equation.
In some cases, chain rule is easier.
{\displaystyle h(x)=(3x-1)^{2}}
{\displaystyle h(x)=(3x-1)(3x-1)}
{\displaystyle h(x)=9x^{2}-6x+1}
{\displaystyle h'(x)=18x-6}
{\displaystyle h'(x)=6(3x-1).}
Another reason the chain rule becomes very effective is when you are dealing with trigonometric functions.
For example, the derivative of
{\displaystyle f(x)=sin(2x)}
{\displaystyle f'(x)=cos(2x)}
Instead, by applying chain rule,
{\displaystyle f'(x)=2cos(2x).}
This shows that chain rule is useful whenever a trigonometric function is applied to a function other than x.
Sometimes, you may need to apply chain rule more than once.
{\displaystyle f(x)=sin^{3}(3x)}
{\displaystyle f'(x)=3sin^{2}(3x)cos(3x)(3)}
{\displaystyle f'(x)=9sin^{2}(3x)cos(3x).}
As you can see, first chain rule was applied to find the derivative of the outside, which was the power on the sine function.
Next, the derivative of the sine function itself.
The last part was finding the derivative of the inside of the sine function.
This method required the chain rule twice. Other problems may require more uses of the chain rule.
Retrieved from "https://en.wikibooks.org/w/index.php?title=High_School_Calculus/The_Chain_Rule&oldid=3820947"
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Torque in the rotational form of Newton's second law | Brilliant Math & Science Wiki
Torque in the rotational form of Newton's second law
Dale Gray, Nihar Mahajan, Josh Silverman, and
In the consideration of the motion of composite systems, it is often necessary to consider the effects of forces applied at different points within the system and rotations which may be produced as a result. When rotations about a fixed axis are the relevant motion produced, the concept of torque is the relevant quantity for producing changes in the rotation rate. Torque is also referred to as the "moment of a force."
Consider a rigid body free to rotate around a fixed axis as a force
\mathbf{F}
is applied at a distance of
r
from the axis of rotation. Because the body is "rigid," the effect of the force would be the same if the force were applied anywhere along a line in the body called the line of action. What is relevant is the perpendicular distance from the axis of rotation and the line of action. This perpendicular distance is called the lever arm,
l
. See the figure below for a description of the quantities in the definition of torque: The torque vector,
\mathbf{\tau}
, produced by the force,
\mathbf{F}
, has a magnitude given by
\tau
lF
rF \sin \theta
. The direction of the torque vector is determined from the right hand rule for the cross product,
\mathbf{\tau}
\mathbf{r} \times \mathbf{F}
The response of a rigid body is dependent on how the mass of the body is distributed relative to the axis of rotation. The quantity that describes the distribution of mass, as it relates to the effectiveness of a torque, is called the moment of inertia,
I
, about the axis of rotation: The integral is of the appropriate type, a double integral for a body of negligible thickness or a triple integral in the general case, and extends over the surface area or volume
V
of the body. The quantity
\rho
occurring in the integral is the mass density.
For a point mass in an orbit of radius
r
about an axis
I
mr^2
For a disc of uniform density, radius
R
, and mass
M
I
\frac{1}{2}
MR^2
For an ellipsoid of revolution of constant density, with semi-minor axis
b
, rotating around the major axis,
I
\frac{2}{5}
Mb^2
M
is the mass of the ellipsoid.
When a torque is applied to a rigid body constrained to rotate around a fixed axis, the magnitude of the torque is related to the moment of inertia by
\tau
I \alpha
\alpha
is the angular acceleration of the body about the axis of rotation in radians per second squared. This equation is called the rotational form of Newton's second law.
For those who would like more information on moments of inertia in general, please see the following: Newton's Second Law for Composite Systems
Cite as: Torque in the rotational form of Newton's second law. Brilliant.org. Retrieved from https://brilliant.org/wiki/torque-rotational-form-newtons-second-law/
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