content stringlengths 86 994k | meta stringlengths 288 619 |
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Feature Column from the AMS
A Discrete Geometrical Gem
3. Some ramifications of the Sylvester-Gallai Theorem
In using the ideas above we have freely used properties of Euclidean geometry. Note that we were not explicit about the "plane" that the Sylvester Theorem holds in. We do have to be careful, because
the result does not hold in the finite projective geometry known as the Fano Plane (or Fano Configuration) shown below.
This plane with 7 points and 7 lines has exactly three points on each line. However, one line looks "peculiar." It is the line L[6] in the diagram and contains the three points P[2], P[6], and P[7]
which is shown as a "circle" above. It follows from the Sylvester-Gallai Theorem that no matter how we might try to position 7 points in the Euclidean plane so as to lie on 7 lines as in the Fano
configuration, we can not succeed!. The Fano plane is named for Gino Fano (1871-1952), the Italian geometer who pioneered the study of finite geometries and point configuration and whose two sons
(one, Ugo, a physicist and the other, Robert, an engineer) had distinguished careers in the United States.
Convince yourself that if the Fano configuration could be embedded in the Euclidean plane with 7 points and 7 three-point straight lines, the drawing would violate the Sylvester-Gallai Theorem.
Part of the reason that Sylvester-Gallai ideas are widespread is because the statement from Section 1:
If a finite set of points in the plane are not all on one line then there is a line through exactly two of the points(*),
can be thought of as holding in the Euclidean plane or in the real projective plane. (We saw above that the theorem need not hold in finite projective planes.) Because the usual axioms for a
projective plane have the property that if one interchanges the words "point" and "line" in the axioms one gets the same axiom set back, it follows that there is a duality principle for projective
planes. Duality refers to the fact that when, in the statement of a theorem, the words "point" and "line" are interchanged, then one gets another valid theorem. What is the dual of (*)? This is the
statement that if one has a configuration of lines and points, not all of the lines going through a single point (i.e. not a pencil of lines), then there is some point that lies on exactly two lines.
Such a point is known as an ordinary point. (To state the associated "dual" theorem in the Euclidean plane we have to deal with lines which do not pass through a single point and no two of which are
Looking back on the mathematics literature, one can often find work that is related to a question raised previously, even though mathematicians of the day did not realize the significance of that
work. There is an ironic example of this with respect to the Sylvester-Gallai Theorem. Even before Gallai gave the answer to Sylvester's problem by answering Erdös' version of the question, another
mathematician, E. Melchior, had in essence solved the problem. Melchior was a Nazi and published his work in a journal in which the Nazis did not allow non-Aryans to publish. For this reason and the
fact that this journal was not widely circulated outside of Germany during the Nazi years, Melchior's proof of the Sylvester Problem was not noticed. (One must remember not to confuse the message
with the messenger. Richard Wagner wrote some wonderfully beautiful music, even if he was not always admirable in his personal life.) Melchior's proof is a combinatorial one which does not use
metrical properties of Euclidean geometry and is very lovely.
A natural problem to deal with is the number of ordinary lines which must be present as the number of points in a point-line configuration grows. Quite early in the history of studying the Sylvester
Problem it appeared that if there were n points, the number of ordinary lines was at least n/2. However, the situation was complicated by the fact that there were point configurations for small n
where this condition does not hold.
In retrospect, it is clear that one could easily see that there had to be at least three ordinary lines for any point-line configuration. This follows from Melchior's work. Later it was shown that
the number of ordinary points had to grow larger as the number of points in a plane finite configuration grew. Both G. Dirac and T. Motzkin were among those who explicitly conjectured that there
would be at least n/2 ordinary points for a set of n plane points. However, it should again be noted that for n =7 and 13, there are configurations for which this bound does not hold. The additional
requirement that n be large enough has to be added. It was announced by Hansen that he had a proof that there were always at least n/2 (n > 13), but the proof, unfortunately, had an error. Despite
this, there were many correct new results in Hansen's investigations related to Sylvester-Gallai results. The number of ordinary points is known to be at least 3n/7 (n > 2), as was shown by L. Kelly
and W. Moser, and the best known result (due to J. Csima and E. Sawyer) is that the number of ordinary lines is at least 6n/13 (n > 6). | {"url":"http://www.ams.org/publicoutreach/feature-column/fcarc-sylvester3","timestamp":"2024-11-05T09:29:07Z","content_type":"text/html","content_length":"51204","record_id":"<urn:uuid:4bc242aa-01ce-4916-a5b0-9b76c98ed173>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00175.warc.gz"} |
Arithmetic Mean Formula
Unlocking the Arithmetic Mean Formula: Understanding the Arithmetic Average in Mathematics
Comprehensive Definition, Description, Examples & RulesÂ
Introduction to Arithmetic Mean
The arithmetic mean is a simpler way to calculate the average. It is easy to use in statistical calculations. This blog will develop your understanding of the arithmetic mean topic and you can check
your progress through the worksheet that is given in the end. Â
For example, the arithmetic mean of 2, 3, and 7 would be the sum of these three numbers divided by the count of these numbers.
2+3+7/3 = 4Â
Explanation of the arithmetic mean and its significance in mathematics.Â
Arithmetic mean having a crucial role in mathematics. Arithmetic mean also refers to average or mean.Â
• It is significant due to the easy-to-use method.
• You can easily compare the two data sets using the arithmetic mean.
• Provide an exact idea of the central value.
• The arithmetic mean is also significant for standard deviations and calculations related to statistics.Â
Definition of the arithmetic average and its use in data analysis.
Arithmetic mean refers to the arithmetic average. The process of calculating an arithmetic average is simple compared to that of calculating an arithmetic mean. It is the sum of numbers divided by
the count of numbers. In data analysis, the arithmetic average can be used to summarize the data, make comparisons, and, most importantly, be important from a statistical point of view.
Describe the Arithmetic Mean Formula
Arithmetic mean has a simple formula. It is calculated by dividing the sum of the numbers by the total number. It helps calculate the mean. The arithmetic mean is the sum of numbers or observations
divided by the total number of observations.
A detailed breakdown of the arithmetic mean formula
The arithmetic formula signifies the sum of the total number. For example, in a series of 1, 2, 3, 4, and 5 the sum of these numbers will be 15.Â
The total number is 5, and the sum needs to be divided by the total count, which means 15/5 = 3.Â
A step-by-step explanation of how to calculate the arithmetic mean of a set of numbers.Â
There are several steps you need to follow for calculating the arithmetic mean :Â Â
• A sum of numbers: you need to first take out the sum of the given numbers.
• Count the numbers: After taking out the sum, you need to count the numbers.
• Division: then you need to divide the sum by the total number of counts.
Arithmetic Mean Equations solving methods
You must solve the arithmetic mean questions to develop your understanding of this topic.Â
Examples of solving the arithmetic equationÂ
• What is the arithmetic mean if the number is 20, 40, 10, or 30? Â
The sum of all numbers or the total count of numbers
20+40+10+30/4 = 25Â
• What would be the sum of numbers if there were a total of 5 numbers and the arithmetic mean was 20?Â
The arithmetic mean is the sum of numbers divided by the total count of numbers.
20 = sum of numbers / 5.
20*5 = 100
Therefore, 100 is the sum of the numbers in his equation.
Demonstrating how to use the formula to find the average of various data sets.
It is not difficult to calculate the average of two data sets.Â
• The marks of the three students are 10, 9, and 8. What are the average marks among them?
Average marks will be 10+9+8/3 = 9.Â
• The weights of the four students were 60,40,30,50. Find out the average weight.
The average weight is 60+40+30+50 =180/4.
The average weight is 45.
Real-Life Applications of Arithmetic Mean
Arithmetic is helpful in our daily life. It is important in everything from finance to science. Whether you want to compare something or calculate the average, Arithmetic is useful.
Exploring real-world scenarios where the arithmetic mean is appliedÂ
• You can compare the grades, height, and weight of students through this process. You can get an average idea through arithmetic.Â
• It is helpful in calculating a country’s population.Â
• Arithmetic mean are also helpful for business purposes.
It Covers areas like :Â Â
• Finance: you can use the arithmetic mean for comparing the finances or calculating the investment return.Â
• Statistics: It can be used to calculate the average of data. For example, you can calculate the population through this process.Â
• Science: It helps calculate the temperature changes in a year.
Differences Between the Arithmetic Mean and Other Averages
There are two more averages along with the arithmetic mean. Arithmetic mean the sum of numbers divided by the total count of numbers.
Two other methods are median and mode. The median is a process to find out the middle number in a series by arranging the series in ascending order. Whereas the mode is the most recurring number in a
Comparing the arithmetic mean with other measures of central tendency (e.g., median, mode)
• The arithmetic mean is suitable for comparing the data. The median and mode are not sufficient to make a comparison between the data.
• The median is useful to find the middle number in a series of numbers.
• Mode, in comparison to the arithmetic mean, is useful to find out the recurring number in a series.
Highlighting when each average is most appropriate to use.
The arithmetic mean is important in day-to-day life, for making a financial comparison. The median is helpful in finding the middle number in a series. You can find the common value in a number
series through the mode.
The weighted arithmetic mean is also similar to the arithmetic mean, but it has some dissimilarities. The calculation process of weighted arithmetic mean is the product of weighted sum/sum of weight.
Introduction to the weighted arithmetic mean and its relevance in data analysis.
If there are different data, in that case, the data needs to be multiplied by the weight, and then you need to divide it by the sum of all weights.Â
• A weighted arithmetic mean is useful in science, statistics, and stock market calculations.
How to calculate the weighted average based on given weights
It is a simple process to calculate the average weight.
• You need to multiply the data by the weight if the data is different.
• Divide it by the weighted sum.Â
For example:Â
• If the number of items is 1, 2, 3, and their data is 5, 10, or 15, find out the weighted mean.Â
1*5,2*10,3*15 =5,20,45Â
Weighted mean = 5+20+45 (sum of weight 1) / 5+10+15( sum of weight 2)
70/30 is the weighted arithmetic mean.
Limitations of the Arithmetic Mean
The various limitations of the arithmetic mean are as follows:
Addressing the limitations and assumptions of the arithmetic mean
• You may not get accurate results by using arithmetic mean processes.
• If the data is not distributed then you may not be able to find the correct mean.
• It would be difficult to get accurate results if there is a maximum value in the mean.Â
Instances where the mean may not accurately represent the data.
• Â Mean won’t represent the accurate result if the data is skewed.
• Â You won’t be able to find out the correct mean if the data is not well distributed.Â
Examples and Practice Problems
Providing a series of examples and practice problems to reinforce understanding
• Find the arithmetic mean if the number is 10, 40, 25, or 45.
(Hint: sum of all the numbers or total number)
• 5, 15, 20, 10, find out the median from the following.
(Hint: arrange in ascending order.)
Including datasets of varying complexity for practice.Â
• Find out the mean of five different numbers: 50, 80, 90, 70 and 60.
• Find out the sum of numbers if the arithmetic mean is 50 and the numbers are 10.
Arithmetic Mean in Data Visualization
Exploring how the arithmetic mean can be represented in graphs and charts
The arithmetic mean is not possible to calculate on graphs. It is a kind of ratio of the sum of numbers divided by the total number.
Visualizing mean values in bar charts, line graphs, and pie charts
In the bar graph, it can be represented by the point of intersection.
In line graphs, where your calculated value will lie will form a line.
In pie charts, it can be represented in a better way to describe the population and statistical visualization.
Step Up Your Math Game Today!
Free sign-up for a personalised dashboard, learning tools, and unlimited possibilities!
1. The arithmetic mean is a crucial concept of mathematics. It is used to calculate the mean or average of data.
2. The arithmetic mean is useful to measure the temperature and make a comparison and so on.Â
3. Its formula is a sum of numbers divided by total numbers.Â
4. There are two other methods, that is median and mode. Median is used to find out the middle number in a series and the mode finds out the frequent number.Â
5. You can improve your arithmetic mean through practice on a daily basis.
Check your score in the end
Check your score in the end
Frequently Asked Questions
Share it with your friends | {"url":"https://www.edulyte.com/maths/arithmetic-mean-formula/","timestamp":"2024-11-09T06:45:45Z","content_type":"text/html","content_length":"423137","record_id":"<urn:uuid:cc9a9b42-a7ff-47b9-8fe0-2e05af587a56>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00569.warc.gz"} |
#include <boost/math/distributions/inverse_gamma.hpp>
namespace boost{ namespace math{
template <class RealType = double,
class Policy = policies::policy<> >
class inverse_gamma_distribution
typedef RealType value_type;
typedef Policy policy_type;
inverse_gamma_distribution(RealType shape, RealType scale = 1)
RealType shape()const;
RealType scale()const;
}} // namespaces
The inverse_gamma distribution is a continuous probability distribution of the reciprocal of a variable distributed according to the gamma distribution.
The inverse_gamma distribution is used in Bayesian statistics.
See inverse gamma distribution.
R inverse gamma distribution functions.
Wolfram inverse gamma distribution.
See also Gamma Distribution.
In spite of potential confusion with the inverse gamma function, this distribution does provide the typedef:
typedef inverse_gamma_distribution<double> gamma;
If you want a double precision gamma distribution you can use
or you can write inverse_gamma my_ig(2, 3);
For shape parameter α and scale parameter β, it is defined by the probability density function (PDF):
f(x;α, β) = β^α * (1/x) ^α+1 exp(-β/x) / Γ(α)
and cumulative density function (CDF)
F(x;α, β) = Γ(α, β/x) / Γ(α)
The following graphs illustrate how the PDF and CDF of the inverse gamma distribution varies as the parameters vary:
inverse_gamma_distribution(RealType shape = 1, RealType scale = 1);
Constructs an inverse gamma distribution with shape α and scale β.
Requires that the shape and scale parameters are greater than zero, otherwise calls domain_error.
RealType shape()const;
Returns the α shape parameter of this inverse gamma distribution.
RealType scale()const;
Returns the β scale parameter of this inverse gamma distribution.
All the usual non-member accessor functions that are generic to all distributions are supported: Cumulative Distribution Function, Probability Density Function, Quantile, Hazard Function, Cumulative
Hazard Function, mean, median, mode, variance, standard deviation, skewness, kurtosis, kurtosis_excess, range and support.
The domain of the random variate is [0,+∞].
Unlike some definitions, this implementation supports a random variate equal to zero as a special case, returning zero for pdf and cdf.
The inverse gamma distribution is implemented in terms of the incomplete gamma functions gamma_p and gamma_q and their inverses gamma_p_inv and gamma_q_inv: refer to the accuracy data for those
functions for more information. But in general, inverse_gamma results are accurate to a few epsilon, >14 decimal digits accuracy for 64-bit double.
In the following table α is the shape parameter of the distribution, α is its scale parameter, x is the random variate, p is the probability and q = 1-p.
Function Implementation Notes
pdf Using the relation: pdf = gamma_p_derivative(α, β/ x, β) / x * x
cdf Using the relation: p = gamma_q(α, β / x)
cdf complement Using the relation: q = gamma_p(α, β / x)
quantile Using the relation: x = β / gamma_q_inv(α, p)
quantile from the complement Using the relation: x = α / gamma_p_inv(α, q)
mode β / (α + 1)
median no analytic equation is known, but is evaluated as quantile(0.5)
mean β / (α - 1) for α > 1, else a domain_error
variance (β * β) / ((α - 1) * (α - 1) * (α - 2)) for α >2, else a domain_error
skewness 4 * sqrt (α -2) / (α -3) for α >3, else a domain_error
kurtosis_excess (30 * α - 66) / ((α-3)*(α - 4)) for α >4, else a domain_error | {"url":"https://beta.boost.org/doc/libs/1_69_0/libs/math/doc/html/math_toolkit/dist_ref/dists/inverse_gamma_dist.html","timestamp":"2024-11-02T18:50:31Z","content_type":"text/html","content_length":"21003","record_id":"<urn:uuid:b0e72af6-2547-44d6-bf52-72bdc15237b5>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00685.warc.gz"} |
The base representation model for fine-tuning topic representations.
Source code in bertopic\representation\_base.py
class BaseRepresentation(BaseEstimator):
"""The base representation model for fine-tuning topic representations."""
def extract_topics(
documents: pd.DataFrame,
c_tf_idf: csr_matrix,
topics: Mapping[str, List[Tuple[str, float]]],
) -> Mapping[str, List[Tuple[str, float]]]:
"""Extract topics.
Each representation model that inherits this class will have
its arguments (topic_model, documents, c_tf_idf, topics)
automatically passed. Therefore, the representation model
will only have access to the information about topics related
to those arguments.
topic_model: The BERTopic model that is fitted until topic
representations are calculated.
documents: A dataframe with columns "Document" and "Topic"
that contains all documents with each corresponding
c_tf_idf: A c-TF-IDF representation that is typically
identical to `topic_model.c_tf_idf_` except for
dynamic, class-based, and hierarchical topic modeling
where it is calculated on a subset of the documents.
topics: A dictionary with topic (key) and tuple of word and
weight (value) as calculated by c-TF-IDF. This is the
default topics that are returned if no representation
model is used.
return topic_model.topic_representations_
extract_topics(self, topic_model, documents, c_tf_idf, topics) ¶
Extract topics.
Each representation model that inherits this class will have its arguments (topic_model, documents, c_tf_idf, topics) automatically passed. Therefore, the representation model will only have access
to the information about topics related to those arguments.
Name Type Description Default
topic_model The BERTopic model that is fitted until topic representations are calculated. required
documents DataFrame A dataframe with columns "Document" and "Topic" that contains all documents with each corresponding topic. required
A c-TF-IDF representation that is typically identical to topic_model.c_tf_idf_ except for dynamic, class-based, and hierarchical topic modeling where
c_tf_idf csr_matrix it is calculated on a subset of the documents. required
topics Mapping[str, List[Tuple[str, A dictionary with topic (key) and tuple of word and weight (value) as calculated by c-TF-IDF. This is the default topics that are returned if no required
float]]] representation model is used.
Source code in bertopic\representation\_base.py
def extract_topics(
documents: pd.DataFrame,
c_tf_idf: csr_matrix,
topics: Mapping[str, List[Tuple[str, float]]],
) -> Mapping[str, List[Tuple[str, float]]]:
"""Extract topics.
Each representation model that inherits this class will have
its arguments (topic_model, documents, c_tf_idf, topics)
automatically passed. Therefore, the representation model
will only have access to the information about topics related
to those arguments.
topic_model: The BERTopic model that is fitted until topic
representations are calculated.
documents: A dataframe with columns "Document" and "Topic"
that contains all documents with each corresponding
c_tf_idf: A c-TF-IDF representation that is typically
identical to `topic_model.c_tf_idf_` except for
dynamic, class-based, and hierarchical topic modeling
where it is calculated on a subset of the documents.
topics: A dictionary with topic (key) and tuple of word and
weight (value) as calculated by c-TF-IDF. This is the
default topics that are returned if no representation
model is used.
return topic_model.topic_representations_ | {"url":"https://maartengr.github.io/BERTopic/api/representation/base.html","timestamp":"2024-11-05T02:16:52Z","content_type":"text/html","content_length":"61915","record_id":"<urn:uuid:331532e8-3410-4b57-9ec1-cd897b679648>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00197.warc.gz"} |
Graphing Equations | Free Algebra Tutorials with Videos - Math Doodle
Graphing Linear Equations
🍀 Welcome to our Bite-Size VIDEO Tutorials for GRAPHING LINEAR EQUATIONS!
[FULL Tutorial] Mastering How to Graph Linear Equations
Using the Rectangular Coordinate System
[Bite-Size Tutorial] Mastering the Rectangular Coordinate System
Which ordered pairs are solutions to the equation 𝑥 + 4𝑦 = 8 ?
Graphing Linear Equations in Two Variables
[Bite-Size Tutorial] Mastering How to Graph Linear Equations by Plotting Points
Graph by plotting points:
Graph by plotting points:
Graph by plotting points:
Graph by plotting points:
Graph by plotting points:
Graph by plotting points:
Graph by plotting points:
[Bite-Size Tutorial] Mastering How to Graph Linear Equations Using Intercepts
Graph using the intercepts:
Graph using the intercepts:
Graph using the intercepts:
Graph using the intercepts:
Understanding Slope of a Line
[Bite-Size Tutorial] Graphing Linear Equations Using Slope and Intercept
Using the Slope-Intercept Form of Equation of Line
Graph using the slope and 𝑦-intercept:
Graph using the slope and 𝑦-intercept:
Graph using the slope and 𝑦-intercept:
Graph using the slope and 𝑦-intercept:
Graph using the slope and 𝑦-intercept:
Graph using the slope and 𝑦-intercept:
Graph using the slope and 𝑦-intercept:
[Bite-Size Tutorial] Mastering Linear Equations with Horizontal/Vertical Lines
[Bite-Size Tutorial] Mastering Linear Equations with Parallel/Perpendicular Lines
Determine if the lines are parallel or perpendicular:
Determine if the lines are parallel or perpendicular:
Determine if the lines are parallel or perpendicular:
Determine if the lines are parallel or perpendicular:
Determine if the lines are parallel or perpendicular:
Determine if the lines are parallel or perpendicular:
Determine if the lines are parallel or perpendicular:
Finding the Equation of Line
Graphs of Linear Inequalities (Coming Soon)
Graph the linear inequality:
Graph the linear inequality:
Graph the linear inequality:
Graph the linear inequality: | {"url":"https://math-doodle.com/math-doodle-home/algebra/graphs/","timestamp":"2024-11-06T12:19:59Z","content_type":"text/html","content_length":"298261","record_id":"<urn:uuid:e8dffa54-2ff5-4659-895f-dadeab3ab10a>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00248.warc.gz"} |
Mathematicians make rare breakthrough on notoriously tricky 'Ramsey number' problem
The bounds on Ramsey numbers, which describe relationships between nodes in a network, have been narrowed.
(Image credit: Richtom80 at English Wikipedia (CC-BY 3.0))
Mathematicians have made a breakthrough in one of the thorniest math problems out there — only the third major step forward in 75 years.
The problem involves Ramsey numbers, a deceptively simple concept that is quite slippery, mathematically. A Ramsey number is the minimum size of a group needed to ensure that a certain number of
nodes in that group are connected to one another. The most common metaphor is that of a party: How many people do you need to invite to a soiree to ensure that there will be either a group of three
that will know each other or a group of three that are complete strangers?
The Ramsey number for 3 is 6. And to ensure that a given party has a group of four friends or four strangers, you'll need to expand the guest list to 18. But the Ramsey number for 5? All
mathematicians can say is that it's between 43 and 48. And as the numbers get bigger, the problem becomes increasingly intractable. More nodes in the network mean more possible connections and more
possible structures for the resulting graph.
"There are so many possibilities that you can’t even brute-force it," said Marcelo Campos, who co-authored the research as part of his doctoral degree at the Institute of Pure and Applied Mathematics
(IMPA) in Brazil.
Famously, mathematician Paul Erdös once said that if aliens landed on Earth and demanded a precise Ramsey number for 5 or they'd destroy the planet, humanity should divert all of its computing
resources to figure out the answer. But if they demanded the Ramsey number for 6, humans should prepare for war.
Mathematicians can give a range for any given Ramsey number. In 1935, Erdös figured out that the maximum Ramsey number for a given number N is 4 to the power of N. In 1947, he figured out that the
lower bound is the square root of 2 to the power of N. There's a wide range between those upper and lower bounds, though, and researchers have been trying to narrow the gap for decades.
"Basically, the bound has been stuck there," said David Conlon, a professor of mathematics at Caltech who was not involved in the current research.
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But now, Campos and his colleagues have made progress on that upper bound: Instead of 4 to the power of N, they can now say that the maximum Ramsey number for a given network is 3.993 to the power of
That might not sound like much of a difference, but it's the first step forward on the upper bound since 1935, Campos told Live Science. He and his team pulled off the proof by developing a new
algorithm that looks for certain substructures in the graphs of nodes called "books," which then help them find the groups of connected nodes, or "cliques," that they are looking for.
"What they did was find a more efficient way of constructing these books," Conlon told Live Science.
Ramsey numbers don't have a specific application in the real world; they're in the realm of pure math. But the quest to pin them down has had real-world impacts. For example, Campos said, in the
1980s, mathematicians explored Ramsey theory with a concept called quasirandomness, which involves groups with certain mathematical properties. Quasirandomness now plays a role in computer science,
Campos said.
"Somehow the problem itself has become a very productive one," Conlon said.
The new method may be able to tighten the upper limit even more than Campos and his team showed in their new paper, which they submitted to the preprint database arXiv on March 16. Campos and his
team have plans to pursue the method further, and they hope other researchers will build on their work as well.
"I don't think 3.99 is actually going to be the end point," Campos said.
Stephanie Pappas is a contributing writer for Live Science, covering topics ranging from geoscience to archaeology to the human brain and behavior. She was previously a senior writer for Live Science
but is now a freelancer based in Denver, Colorado, and regularly contributes to Scientific American and The Monitor, the monthly magazine of the American Psychological Association. Stephanie received
a bachelor's degree in psychology from the University of South Carolina and a graduate certificate in science communication from the University of California, Santa Cruz. | {"url":"https://www.livescience.com/mathematicians-make-rare-breakthrough-on-notoriously-tricky-ramsey-number-problem","timestamp":"2024-11-11T10:34:12Z","content_type":"text/html","content_length":"696220","record_id":"<urn:uuid:18471b9a-ade7-4d4d-952a-3c5f5b8bcd53>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00477.warc.gz"} |
sphtriangulate - Delaunay or Voronoi construction of spherical lon,lat data
sphtriangulate [ table ] [ -A ] [ -C ] [ -D ] [ -Lunit ] [ -Nnfile ] [ -Qd|v ] [ -T ] [ -V[level] ] [ -bbinary ] [ -dnodata ] [ -eregexp ] [ -hheaders ] [ -iflags ] [ -:[i|o] ]
Note: No space is allowed between the option flag and the associated arguments.
sphtriangulate reads one or more ASCII [or binary] files (or standard input) containing lon, lat and performs a spherical Delaunay triangulation, i.e., it determines how the points should be
connected to give the most equilateral triangulation possible on the sphere. Optionally, you may choose -Qv which will do further processing to obtain the Voronoi polygons. Normally, either set of
polygons will be written as closed fillable segment output; use -T to write unique arcs instead. As an option, compute the area of each triangle or polygon. The algorithm used is STRIPACK.
Required Arguments¶
Optional Arguments¶
One or more ASCII (or binary, see -bi[ncols][type]) data table file(s) holding a number of data columns. If no tables are given then we read from standard input.
Compute the area of the spherical triangles (-Qd) or polygons (-Qv) and write the areas (in chosen units; see -L) in the output segment headers [no areas calculated].
For large data set you can save some memory (at the expense of more processing) by only storing one form of location coordinates (geographic or Cartesian 3-D vectors) at any given time,
translating from one form to the other when necessary [Default keeps both arrays in memory].
Used to skip the last (repeated) input vertex at the end of a closed segment if it equals the first point in the segment. [Default uses all points].
Specify the unit used for distance and area calculations. Choose among e (m), f (foot), k (km), m (mile), n (nautical mile), u (survey foot), or d (spherical degree). A spherical approximation is
used unless PROJ_ELLIPSOID is set to an actual ellipsoid, in which case we convert latitudes to authalic latitudes before calculating areas. When degree is selected the areas are given in
Write the information pertaining to each polygon. For Delaunay: the three node number and the triangle area (if -A was set); for Voronoi the unique node lon, lat and polygon area (if -A was set))
to a separate file. This information is also encoded in the segment headers of ASCII output files. Required if binary output is needed.
Append d for Delaunay triangles or v for Voronoi polygons [Delaunay]. If -bo is used then -N may be used to specify a separate file where the polygon information normally is written.
Write the unique arcs of the construction [Default writes fillable triangles or polygons]. When used with -A we store arc length in the segment header in chosen unit (see -L).
-V[level] (more …)
Select verbosity level [c].
-bi[ncols][t] (more …)
Select native binary input. [Default is 2 input columns].
-bo[ncols][type] (more …)
Select native binary output. [Default is same as input].
-d[i|o]nodata (more …)
Replace input columns that equal nodata with NaN and do the reverse on output.
-e[~]”pattern” | -e[~]/regexp/[i] (more …)
Only accept data records that match the given pattern.
-h[i|o][n][+c][+d][+rremark][+rtitle] (more …)
Skip or produce header record(s).
-:[i|o] (more …)
Swap 1st and 2nd column on input and/or output.
-r (more …)
Set pixel node registration [gridline].
-^ or just -
Print a short message about the syntax of the command, then exits (NOTE: on Windows just use -).
-+ or just +
Print an extensive usage (help) message, including the explanation of any module-specific option (but not the GMT common options), then exits.
-? or no arguments
Print a complete usage (help) message, including the explanation of all options, then exits.
ASCII Format Precision¶
The ASCII output formats of numerical data are controlled by parameters in your gmt.conf file. Longitude and latitude are formatted according to FORMAT_GEO_OUT, absolute time is under the control of
FORMAT_DATE_OUT and FORMAT_CLOCK_OUT, whereas general floating point values are formatted according to FORMAT_FLOAT_OUT. Be aware that the format in effect can lead to loss of precision in ASCII
output, which can lead to various problems downstream. If you find the output is not written with enough precision, consider switching to binary output (-bo if available) or specify more decimals
using the FORMAT_FLOAT_OUT setting.
To triangulate the points in the file testdata.txt, and make a Voronoi diagram via psxy, use
gmt sphtriangulate testdata.txt -Qv | psxy -Rg -JG30/30/6i -L -P -W1p -Bag | gv -
To compute the optimal Delaunay triangulation network based on the multiple segment file globalnodes.d and save the area of each triangle in the header record, try
gmt sphtriangulate globalnodes.d -Qd -A > global_tri.d
Renka, R, J., 1997, Algorithm 772: STRIPACK: Delaunay Triangulation and Voronoi Diagram on the Surface of a Sphere, AMC Trans. Math. Software, 23(3), 416-434. | {"url":"https://docs.generic-mapping-tools.org/5.4/sphtriangulate.html","timestamp":"2024-11-12T16:08:43Z","content_type":"application/xhtml+xml","content_length":"19987","record_id":"<urn:uuid:deb71bf1-c2e7-4aba-a354-c404935d0a97>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00378.warc.gz"} |
Greedy Best first search algorithm
What is the Greedy-Best-first search algorithm?
Greedy Best-First Search is an AI search algorithm that attempts to find the most promising path from a given starting point to a goal. It prioritizes paths that appear to be the most promising,
regardless of whether or not they are actually the shortest path. The algorithm works by evaluating the cost of each possible path and then expanding the path with the lowest cost. This process
is repeated until the goal is reached.
The algorithm works by using a heuristic function to determine which path is the most promising. The heuristic function takes into account the cost of the current path and the estimated cost of the
remaining paths. If the cost of the current path is lower than the estimated cost of the remaining paths, then the current path is chosen. This process is repeated until the goal is reached.
How Greedy Best-First Search Works?
• Greedy Best-First Search works by evaluating the cost of each possible path and then expanding the path with the lowest cost. This process is repeated until the goal is reached.
• The algorithm uses a heuristic function to determine which path is the most promising.
• The heuristic function takes into account the cost of the current path and the estimated cost of the remaining paths.
• If the cost of the current path is lower than the estimated cost of the remaining paths, then the current path is chosen. This process is repeated until the goal is reached.
Advantages of Greedy Best-First Search:
• Simple and Easy to Implement: Greedy Best-First Search is a relatively straightforward algorithm, making it easy to implement.
• Fast and Efficient: Greedy Best-First Search is a very fast algorithm, making it ideal for applications where speed is essential.
• Low Memory Requirements: Greedy Best-First Search requires only a small amount of memory, making it suitable for applications with limited memory.
• Flexible: Greedy Best-First Search can be adapted to different types of problems and can be easily extended to more complex problems.
Disadvantages of Greedy Best-First Search:
• Inaccurate Results: Greedy Best-First Search is not always guaranteed to find the optimal solution, as it is only concerned with finding the most promising path.
• Local Optima: Greedy Best-First Search can get stuck in local optima, meaning that the path chosen may not be the best possible path.
• Heuristic Function: Greedy Best-First Search requires a heuristic function in order to work, which adds complexity to the algorithm.
Applications of Greedy Best-First Search:
• Pathfinding: Greedy Best-First Search is used to find the shortest path between two points in a graph. It is used in many applications such as video games, robotics, and navigation systems.
• Machine Learning: Greedy Best-First Search can be used in machine learning algorithms to find the most promising path through a search space.
• Optimization: Greedy Best-First Search can be used to optimize the parameters of a system in order to achieve the desired result.
Greedy Best-First Search is an AI search algorithm that attempts to find the most promising path from a given starting point to a goal. The algorithm works by evaluating the cost of each possible
path and then expanding the path with the lowest cost. This process is repeated until the goal is reached. Greedy Best-First Search has several advantages, including being simple and easy to
implement, fast and efficient, and having low memory requirements. However, it also has some disadvantages, such as inaccurate results, local optima, and requiring a heuristic function. Greedy
Best-First Search is used in many applications, including pathfinding, machine learning, and optimization. It is a useful algorithm for finding the most promising path through a search space. | {"url":"http://krasa-russia.ru/archives/599","timestamp":"2024-11-14T13:48:25Z","content_type":"text/html","content_length":"50873","record_id":"<urn:uuid:c66c1497-3107-4436-9260-4cdc512f7ffa>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00202.warc.gz"} |
Pi-in-Stein Startseite - Pi-in-Stein
Follow the magic of a magic number.
Come with us on a tour through Nörten-Hardenberg – and discover pi´s numbers lying beneath your feet.
So what is pi exactly? you may ask
Here are the experts´ definitions:
The mathematician: Pi is the number representing the relationship between the circumference of a circle and its radius.
The physicist: Pi is 3.1415927 +/- 0.0000005
The engineer: Pi is about 3
(from David Baltner: The Joy of Pi, 1999) | {"url":"https://www.pi-in-stein.com/en/","timestamp":"2024-11-02T14:34:00Z","content_type":"text/html","content_length":"7249","record_id":"<urn:uuid:47b7aa45-2ad4-452b-a04a-d50ce9481abb>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00564.warc.gz"} |
Introduction to Reluctant Generalized Additive Modeling (RGAM)
Kenneth Tay
relgam is a package that fits reluctant generalized additive models (RGAM), a new method for fitting sparse generalized additive models (GAM). RGAM is computationally scalable and works with
continuous, binary, count and survival data.
We introduce some notation that we will use throughout this vignette. Let there be \(n\) observations, each with feature vector \(x_i \in \mathbb{R}^p\) and response \(y_i\). Let \(\mathbf{X} \in \
mathbb{R}^{n \times p}\) denote the overall feature matrix, and let \(y \in \mathbb{R}^n\) denote the vector of responses. Let \(X_j \in \mathbb{R}^n\) denote the \(j\)th column of \(\mathbf{X}\).
Assume that the columns of \(\mathbf{X}\), i.e. \(X_1, \dots, X_p\), are standardized to have sample standard deviation \(1\). Assume that we have specified a scaling hyperparmeter \(\gamma \in [0,1]
\), a degrees of freedom hyperparameter \(d\), and a path of tuning parameters \(\lambda_1 > \dots > \lambda_m \geq 0\). RGAM’s model-fitting algorithm, implemented in the function rgam(), consists
of 3 steps:
1. Fit the lasso of \(y\) on \(\mathbf{X}\) to get coefficients \(\hat{\beta}\). Compute the residuals \(r = y - \mathbf{X} \hat{\beta}\), using the \(\lambda\) hyperparameter selected by
2. For each \(j \in \{ 1, \dots, p \}\), fit a \(d\)-degree smoothing spline of \(r\) on \(X_j\) which we denote by \(\hat{f}_j\). Rescale \(\hat{f}_j\) so that \(\overline{\text{sd}}(\hat{f}_j) = \
gamma\). Let \(\mathbf{F} \in \mathbb{R}^{n \times p}\) denote the matrix whose columns are the \(\hat{f}_j(X_j)\)’s.
3. Fit the lasso of \(y\) on \(\mathbf{X}\) and \(\mathbf{F}\) for the path of tuning parameters \(\lambda_1 > \dots > \lambda_m\).
Steps 1 and 3 are implemented using glmnet::glmnet() while Step 2 is implemented using stats::smooth.spline(). (Note that the path of tuning parameters \(\lambda_1 > \dots > \lambda_m\) are only used
in Step 3; for Step 1 we use glmnet::glmnet()’s default lambda path.) We will refer to these 3 steps by their numbers (e.g. “Step 1”) throughout the rest of the vignette.
The rgam() function fits this model and returns an object with class “rgam”. The relgam package includes methods for prediction and plotting for “rgam” objects, as well as a function which performs \
(k\)-fold cross-validation for rgam().
For more details, please see our paper on arXiv.
The simplest way to obtain relgam is to install it directly from CRAN. Type the following command in R console:
This command downloads the R package and installs it to the default directories. Users may change add a repos option to the function call to specify which repository to download from, depending on
their locations and preferences.
Alternatively, users can download the package source at CRAN and type Unix commands to install it to the desired location.
The rgam() function
The purpose of this section is to give users a general sense of the rgam() function, which is the workhorse of this package. First, we load the relgam package:
Let’s generate some data:
n <- 100; p <- 12
x = matrix(rnorm((n) * p), ncol = p)
f4 = 2 * x[,4]^2 + 4 * x[,4] - 2
f5 = -2 * x[, 5]^2 + 2
f6 = 0.5 * x[, 6]^3
mu = rowSums(x[, 1:3]) + f4 + f5 + f6
y = mu + sqrt(var(mu) / 4) * rnorm(n)
We fit the model using the most basic call to rgam():
fit <- rgam(x, y, verbose = FALSE)
#> init_nz not specified: setting to default (all features)
#> using default value of gamma for RGAM: 0.6
(If verbose = TRUE (default), model-fitting is tracked with a progress bar in the console. For the purposes of this vignette, we will be setting verbose = FALSE.)
rgam() has an init_nz option which (partially) determines which columns in the \(\mathbf{X}\) matrix will have non-linear features computed in Step 2 of the RGAM algorithm. \(X_j\) will have a
non-linear feature computed for it if (i) it was one of the features selected by cross-validation in Step 1, or (ii) it is one of the indices specified in init_nz. By default, init_nz = 1:ncol(x),
i.e. non-linear features are computed for all the original features. Another sensible default is init_nz = c(), i.e. non-linear features computed for just those selected in Step 1. (This version of
the RGAM algorithm is denoted by RGAM_SEL in our paper.)
Below, we fit the model with a different init_nz value:
fit <- rgam(x, y, init_nz = c(), verbose = FALSE)
#> using default value of gamma for RGAM_SEL: 0.8
You might have noticed that in both cases above, we did not specify a value for the gamma hyperparameter: rgam() chose one for us and informed us of the choice. The default value for gamma is 0.6 if
init_nz = c(), and is 0.8 in all other cases.
The degrees of freedom hyperparameter for Step 2 of the RGAM algorithm can be set through the df option, the default value is 4. Here is an example of fitting the RGAM model with different
fit <- rgam(x, y, gamma = 0.6, df = 8, verbose = FALSE)
#> init_nz not specified: setting to default (all features)
The function rgam() fits a RGAM for a path of lambda values and returns a rgam object. Typical usage is to have rgam() specify the lambda sequence on its own. The returned rgam object contains some
useful information on the fitted model. For a given value of the \(\lambda\) hyperparameter, RGAM gives the predictions of the form
\(\hat{y} = \sum_{j=1}^p \hat{\alpha}_j X_j + \sum_{j = 1}^p \hat{\beta}_j \hat{f}_j(X_j),\)
where \(\hat{f}_j(X_j)\) are the non-linear features constructed in Step 2, while the \(\hat{\alpha}_j\) and \(\hat{\beta}_j\) are the fitted coefficients from Step 3. The returned RGAM object has
nzero_feat, nzero_lin and nzero_nonlin keys which tell us how many features, linear components and non-linear components were included in the model respectively. (In mathematical notation, nzero_lin
and nzero_nonlin count the number of non-zero \(\hat{\alpha}_j\) and \(\hat{\beta}_j\) respectively, while nzero_feat counts the number of \(j\) such that at least one of \(\hat{\alpha}_j\) and \(\
hat{\beta}_j\) is non-zero).
# no. of features/linear components/non-linear components for 20th lambda value
#> [1] 7
#> s19
#> 5
#> s19
#> 5
While the nzero_feat, nzero_lin and nzero_nonlin keys tell us the number of features, linear components and non-linear components included for each lambda value, the feat, linfeat and nonlinfeat tell
us the indices of these features or components.
# features included in the model for 20th lambda value
#> [1] 2 3 4 5 6 10 11
# features which have a linear component in the model for 20th lambda value
#> [1] 2 3 4 5 6
# features which have a non-linear component in the model for 20th lambda value
#> [1] 4 5 6 10 11
In general, we have fit$nzero_feat[[i]] == length(fit$feat[[i]]), fit$nzero_lin[[i]] == length(fit$linfeat[[i]]) and fit$nzero_nonlin[[i]] == length(fit$nonlinfeat[[i]]).
Predictions from this model can be obtained by using the predict method of the rgam() function output: each column gives the predictions for a value of lambda.
# get predictions for 20th model for first 5 observations
predict(fit, x[1:5, ])[, 20]
#> [1] 1.408575 -3.099603 7.362201 -1.609205 6.729858
The getf() function is a convenience function that gives the component of the prediction due to one input variable. That is, if RGAM gives predictions
\(\hat{y} = \sum_{j=1}^p \hat{\alpha}_j X_j + \sum_{j = 1}^p \hat{\beta}_j \hat{f}_j(X_j),\)
then calling getf() on \(X_j\) returns \(\hat{\alpha}_j X_j + \hat{\beta}_j \hat{f}_j(X_j)\). For example, the code below gives the component of the response due to variable 5 at the 20th lambda
f5 <- getf(fit, x, j = 5, index = 20)
#> [1] -0.13038873 -2.00876503 1.96362206 1.93205913 1.94826187
#> [6] 1.92201185 1.91597986 1.86679777 -0.64901723 1.87631654
#> [11] 1.93028602 0.18113054 -0.18964651 1.76235317 1.98516408
#> [16] 0.33605522 -0.07207409 1.83242679 -0.89754110 -1.29738244
#> [21] -0.63653216 1.85948754 1.60090148 -2.98736659 0.27251312
#> [26] -1.02410778 -2.21467504 1.74899872 1.93616023 1.89608549
#> [31] -4.76717869 0.37014025 -1.53780332 1.52676873 0.24381746
#> [36] 1.78444567 0.01999517 1.52816257 1.93043905 0.18697503
#> [41] 1.95858387 -1.51026593 1.51357019 0.04242528 0.99736549
#> [46] -12.45969898 -0.51655212 1.95300610 0.91815634 -6.95724959
#> [51] -0.02530384 1.96278776 1.79530566 -1.69059815 -1.04489654
#> [56] 1.35644755 0.35729116 -5.20859065 1.17565474 0.31850577
#> [61] 1.96844763 -3.44438244 -0.34551050 0.24086324 1.46691069
#> [66] 0.81357489 1.86548080 0.20517806 -0.72324681 -0.14052225
#> [71] 1.68488945 -2.09699455 1.67367201 0.46966183 -0.49447192
#> [76] -0.24748950 0.61925974 0.92639442 -0.27793468 1.53241240
#> [81] 1.90969419 0.61303031 1.98600575 1.91294888 -4.52315844
#> [86] -5.05399415 1.75016795 1.03578280 -4.36019460 0.34930944
#> [91] 0.85309077 -4.14328595 1.52350923 1.22980304 -10.49335410
#> [96] 0.77723033 0.74645435 0.78568544 0.65539011 1.55355891
We can use this to make a plot showing the effect of variable 5 on the response:
plot(x[, 5], f5, xlab = "x5", ylab = "f(x5)")
(The “Plots and summaries” section shows how to get such a plot more easily.)
Plots and summaries
Let’s fit the basic rgam model again:
fit <- rgam(x, y, verbose = FALSE)
#> init_nz not specified: setting to default (all features)
#> using default value of gamma for RGAM: 0.6
fit is a class “rgam” object which comes with a plot method. The plot method shows us the relationship our predicted response has with each input feature, i.e. it plots \(\hat{\alpha}_j X_j + \hat{\
beta}_j \hat{f}_j(X_j)\) vs. \(X_j\) for each \(j\). Besides passing fit to the plot() call, the user must also pass an input matrix x: this is used to determine the coordinate limits for the plot.
It is recommended that the user simply pass in the same input matrix that the RGAM model was fit on.
By default, plot() gives the fitted functions for the last value of the lambda key in fit, and gives just the plots for the first 4 features:
par(mfrow = c(1, 4))
par(mar = c(4, 2, 2, 2))
plot(fit, x)
#> Warning in plot.rgam(fit, x): Plotting first 4 variables by default
The user can specify the index of the lambda value and which feature plots to show using the index and which options respectively:
# show fitted functions for x2, x5 and x8 at the model for the 15th lambda value
par(mfrow = c(1, 3))
plot(fit, x, index = 15, which = c(2, 5, 8))
Linear functions are colored green, non-linear functions are colored red, while zero functions are colored blue.
Class “rgam” objects also have a summary method which allows the user to see the coefficient profiles of the linear and non-linear features. On each plot (one for linear features and one for
non-linear features), the x-axis is the \(\lambda\) value going from large to small and the y-axis is the coefficient of the feature.
By default the coefficient profiles are plotted for all the variables. We can plot for just a subset of the features by specifying the which option. We can also include annotations to show which
profile belongs to which feature by specifying label = TRUE.
# coefficient profiles for just features 4 to 6
summary(fit, which = 1:6, label = TRUE)
Cross-validation (CV)
We can perform \(k\)-fold cross-validation (CV) for RGAM with cv.rgam(). It does 10-fold cross-validation by default:
cvfit <- cv.rgam(x, y, init_nz = 1:ncol(x), gamma = 0.6, verbose = FALSE)
We can change the number of folds using the nfolds option:
cvfit <- cv.rgam(x, y, init_nz = 1:ncol(x), gamma = 0.6, nfolds = 5, verbose = FALSE)
If we want to specify which observation belongs to which fold, we can do that by specifying the foldid option, which is a vector of length \(n\), with the \(i\)th element being the fold number for
observation \(i\).
foldid <- sample(rep(seq(10), length = n))
cvfit <- cv.rgam(x, y, init_nz = 1:ncol(x), gamma = 0.6, foldid = foldid, verbose = FALSE)
A cv.rgam() call returns a cv.rgam object. We can plot this object to get the CV curve with error bars (one standard error in each direction). The left vertical dotted line represents lambda.min, the
lambda value which attains minimum CV error, while the right vertical dotted line represents lambda.1se, the largest lambda value with CV error within one standard error of the minimum CV error.
The numbers at the top represent the number of features in our original input matrix that are included in the model (i.e. the number of \(j\) such that at least one of \(\hat{\alpha}_j\) and \(\hat{\
beta}_j\) is non-zero).
The two special lambda values can be extracted directly from the cv.rgam object as well:
#> [1] 0.02595555
#> [1] 0.06580678
Predictions can be made from the fitted cv.rgam object. By default, predictions are given for lambda being equal to lambda.1se. To get predictions are lambda.min, set s = "lambda.min".
predict(cvfit, x[1:5, ]) # s = lambda.1se
#> 1
#> [1,] 1.863440
#> [2,] -4.477733
#> [3,] 8.664422
#> [4,] -0.874814
#> [5,] 6.672235
predict(cvfit, x[1:5, ], s = "lambda.min")
#> 1
#> [1,] 1.567028
#> [2,] -4.516911
#> [3,] 9.324387
#> [4,] -1.443938
#> [5,] 8.087449
RGAM for other families
In the examples above, y was a quantitative variable (i.e. takes values along the real number line). As such, using the default family = "gaussian" for rgam() was appropriate. The RGAM algorithm,
however, is very flexible and can be used when y is not a quantitative variable.
rgam() is currently implemented for three other family values: "binomial", "poisson" and "cox". The rgam() and cv.rgam() functions, as well as their methods, can be used as above but with the family
option specified appropriately. In the sections below we point out some details that are particular to each family.
Logistic regression with binary data
In this setting, the response y should be a numeric vector containing just 0s and 1s. When doing prediction, note that by default predict() gives just the value of the linear predictors, i.e.
\(\log [\hat{p} / (1 - \hat{p})] = \hat{y} = \sum_{j=1}^p \hat{\alpha}_j X_j + \sum_{j = 1}^p \hat{\beta}_j \hat{f}_j(X_j),\)
where \(\hat{p}\) is the predicted probability. To get the predicted probability, the user has to pass type = "response" to the predict() call.
# fit binary model
bin_y <- ifelse(y > 0, 1, 0)
binfit <- rgam(x, bin_y, family = "binomial", init_nz = c(), gamma = 0.9,
verbose = FALSE)
# linear predictors for first 5 observations at 10th model
predict(binfit, x[1:5, ])[, 10]
#> [1] 0.3250267 -0.6485438 0.8554076 -0.6302209 0.7420511
# predicted probabilities for first 5 observations at 10th model
predict(binfit, x[1:5, ], type = "response")[, 10]
#> [1] 0.5805488 0.3433178 0.7017003 0.3474605 0.6774442
Poisson regression with count data
For Poisson regression, the response y should be a vector of count data. While rgam() does not expect each element to be an integer, it will throw an error if any of the elements are negative.
As with logistic regression, by default predict() gives just the value of the linear predictors, i.e.
\(\log \hat{\mu} = \hat{y} = \sum_{j=1}^p \hat{\alpha}_j X_j + \sum_{j = 1}^p \hat{\beta}_j \hat{f}_j(X_j),\)
where \(\hat{\mu}\) is the predicted rate. To get the predicted rate, the user has to pass type = "response" to the predict() call.
With Poisson data, it is common to allow the user to pass in an , which is a vector having the same length as the number of observations. rgam() allows the user to do this as well:
# generate data
offset <- rnorm(n)
poi_y <- rpois(n, abs(mu) * exp(offset))
poifit <- rgam(x, poi_y, family = "poisson", offset = offset, verbose = FALSE)
#> init_nz not specified: setting to default (all features)
#> using default value of gamma for RGAM: 0.6
Note that if offset is supplied to rgam(), then an offset vector must also be supplied to predict() when making predictions:
# rate predictions at 20th lambda value
predict(poifit, x[1:5, ], newoffset = offset, type = "response")[,20]
#> [1] 4.904281 3.940030 5.075481 3.104760 7.244539
Cox regression with survival data
For Cox regression, the response y must be a two-column matrix with columns named time and status. The status column is a binary variable, with 1 indicating death and 0 indicating right censored. We
note that one way to produce such a matrix is using the Surv() function in the survival package.
As with logistic and Poisson regression, by default predict() gives just the value of the linear predictors. Passing type = "response" to the predict() call will return the predicted relative-risk. | {"url":"https://cran.rstudio.com/web/packages/relgam/vignettes/relgam.html","timestamp":"2024-11-03T22:43:26Z","content_type":"application/xhtml+xml","content_length":"285507","record_id":"<urn:uuid:6f67de5c-6162-45e8-91d8-cd73ab49d049>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00801.warc.gz"} |
What is the difference between linear search and binary search? What is faster linear or binary search? What is an advantage of a linear search over a binary search?
Feb 19, 2024 By Team YoungWonks *
Understanding Search Algorithms
Before we delve into the specifics of Linear Search and Binary Search, let's establish a foundational understanding of search algorithms.
Search Algorithm
A search algorithm is a methodical approach to locate a particular item within a dataset. These algorithms play a crucial role in various domains, including data science, machine learning, and
software engineering.
What is Linear Search?
Linear search, also known as sequential search, is a straightforward algorithm that sequentially traverses a sorted list until it finds the desired element. It's commonly employed when dealing with
small datasets or linked lists, offering a simple and intuitive solution for locating the first element, whether it's a number, string, or any other data type. In a scenario where the dataset is
sorted, linear search efficiently navigates through the sorted order, examining each element until it finds a match. Despite its simplicity, linear search serves as a foundational concept in
programming tutorials and is often encountered in data structures courses and interview questions.
What is Binary Search?
Binary search represents a more sophisticated approach, particularly suitable for large datasets or arrays sorted in ascending or descending order. By repeatedly dividing the search space in half,
binary search quickly homes in on the target element, making it highly efficient for locating integers, strings, or any other data types within a sorted list. Its average case time complexity of O
(log n) ensures fast search operations, especially when dealing with datasets containing integers or half of the list. Binary search's efficiency is further highlighted in its applications, ranging
from sorting algorithms to data science and machine learning tasks, making it a crucial tool for programmers and data scientists alike.
Both linear search and binary search offer valuable insights into the fundamentals of search algorithms and their practical applications in various programming languages and data structures. Whether
you're visualizing the search process, implementing sorting algorithms, or tackling interview questions on data structures and algorithms (DSA), understanding the intricacies of linear and binary
search is essential for navigating the vast landscape of computer science and programming languages.
How Linear Search Works?
Linear search, often referred to as sequential search, is one of the simplest search algorithms. It sequentially traverses each element of the dataset until it finds the desired element.
The linear search algorithm iterates through each element of the array until it finds the target element. If the element is present, it returns its index; otherwise, it returns -1.
Python Implementation of Linear Search
Function Definition: This defines a function named linear_search that takes two parameters: arr (the array to be searched) and target (the element to be searched for).
Linear Search Algorithm: It iterates through each element of the array using a for loop and checks if the current element is equal to the target. If a match is found, it returns the index of the
element (return i); otherwise, it returns -1 to indicate that the target is not found.
Java Implementation of Linear Search
Method Definition: This defines a static method named linearSearch in the class SearchAlgorithms that takes two parameters: arr (the array to be searched) and target (the element to be searched for).
Linear Search Algorithm: It iterates through each element of the array using a for loop and checks if the current element is equal to the target. If a match is found, it returns the index of the
element; otherwise, it returns -1 to indicate that the target is not found.
JavaScript Implementation of Linear Search
Function Definition: This defines a function named linearSearch that takes two parameters: arr (the array to be searched) and target (the element to be searched for).
Linear Search Algorithm: It iterates through each element of the array using a for loop and checks if the current element is equal to the target. If a match is found, it returns the index of the
element; otherwise, it returns -1 to indicate that the target is not found.
Time Complexity of Linear Search
O(n) - Linear search has a time complexity proportional to the number of elements in the dataset.
Space Complexity of Linear Search
O(1) - Linear search requires constant space as it doesn't utilize additional data structures.
Applications and Prerequisites of Linear Search
• Linear search is suitable for small datasets or unsorted arrays.
• It serves as a foundational concept for beginners in programming and data structures.
How Binary Search Works?
Binary search is a more efficient search algorithm, particularly effective for large datasets or sorted arrays. It follows a divide and conquer strategy, repeatedly dividing the search space in half
until the target element is found.
Binary search requires the dataset to be sorted. It compares the target value with the middle element of the array. If the target matches the middle element, the search is complete. Otherwise, it
narrows down the search space by half based on the comparison result.
Python Implementation of Binary Search
Function Definition: This defines a function named binary_search that takes two parameters: arr (the sorted array to be searched) and target (the element to be searched for).
Binary Search Algorithm: It initializes two pointers, low and high, representing the start and end indices of the array respectively. It repeatedly narrows down the search space by adjusting these
pointers until the target is found or the search space is exhausted.
Loop: Inside the while loop, it calculates the mid index as the average of low and high. It then compares the element at the mid index with the target:
• If the target matches the middle element, it returns the index.
• If the target is greater than the middle element, it updates low to mid + 1 to search in the right half.
• If the target is less than the middle element, it updates high to mid - 1 to search in the left half.
Return Value: If the target is not found after the loop completes, it returns -1.
Java Implementation of Binary Search
Method Definition: This defines a static method named binarySearch in the class Search Algorithms that takes two parameters: arr (the sorted array to be searched) and target (the element to be
searched for).
Binary Search Algorithm: It initializes two pointers, low and high, representing the start and end indices of the array respectively. It repeatedly narrows down the search space by adjusting these
pointers until the target is found or the search space is exhausted.
Loop: Inside the while loop, it calculates the mid index as the average of low and high. It then compares the element at the mid index with the target:
• If the target matches the middle element, it returns the index.
• If the target is greater than the middle element, it updates low to mid + 1 to search in the right half.
• If the target is less than the middle element, it updates high to mid - 1 to search in the left half.
Return Value: If the target is not found after the loop completes, it returns -1.
JavaScript Implementation of Binary Search
Function Definition: This defines a function named binary Search that takes two parameters: arr (the sorted array to be searched) and target (the element to be searched for).
Binary Search Algorithm: It initializes two pointers, low and high, representing the start and end indices of the array respectively. It repeatedly narrows down the search space by adjusting these
pointers until the target is found or the search space is exhausted.
Loop: Inside the while loop, it calculates the mid index as the average of low and high. It then compares the element at the mid index with the target:
• If the target matches the middle element, it returns the index.
• If the target is greater than the middle element, it updates low to mid + 1 to search in the right half.
• If the target is less than the middle element, it updates high to mid - 1 to search in the left half.
Return Value: If the target is not found after the loop completes, it returns -1.
Time Complexity of Binary Search
O(log n) - Binary search exhibits logarithmic time complexity, making it highly efficient for large datasets.
Space Complexity of Binary Search
O(1) - Binary search has constant space complexity as it doesn't require additional memory allocations.
Applications and Prerequisites of Binary Search
• Binary search is optimal for large datasets or sorted arrays.
• Understanding of basic programming concepts and sorting algorithms is essential for implementing binary search.
Linear Search vs Binary Search: A Comparative Analysis
Now, let's compare Linear Search and Binary Search across various dimensions:
Linear Search: Efficient for small datasets but becomes increasingly slower as the dataset size grows.
Binary Search: Highly efficient, especially for large datasets, due to its logarithmic time complexity.
Time Complexity
Linear Search: O(n) - Linear time complexity.
Binary Search: O(log n) - Logarithmic time complexity.
Space Complexity
Both Linear Search and Binary Search have a space complexity of O(1), indicating constant space requirements.
Dataset Requirement
Linear Search: Works with both sorted and unsorted arrays.
Binary Search: Requires the dataset to be sorted for optimal performance.
Number of Comparisons
Linear Search: Worst-case scenario requires n comparisons.
Binary Search: In each step, the search space is halved, leading to fewer comparisons. Worst-case scenario requires log n comparisons.
Practical Applications
Linear Search: Suitable for small datasets or when simplicity is preferred over efficiency.
Binary Search: Widely used in scenarios requiring efficient search in large datasets or sorted arrays.
Other Variants and Considerations
While Linear Search and Binary Search are fundamental, there are other variants and considerations worth exploring:
Interpolation Search: Particularly efficient for uniformly distributed datasets.
Half-Interval Search: Used to find roots of continuous functions within given intervals.
Binary Trees
Binary trees are not directly related to linear search, as linear search typically operates on linear data structures like arrays or linked lists. However, in binary search, binary trees play a
crucial role as they provide a structured representation of sorted data, facilitating the efficient divide-and-conquer approach employed by the binary search algorithm.
In both linear and binary search algorithms, iteration plays a pivotal role in traversing the dataset and determining the location of the target element. In linear search, iteration entails
sequentially examining each element of the array until a match is found or the end of the dataset is reached. This iterative process ensures thorough coverage of the dataset, albeit with a linear
time complexity, making it suitable for smaller datasets or unsorted arrays. Conversely, in binary search, iteration follows a more refined approach, where the search space is systematically halved
with each iteration. This iterative refinement significantly reduces the search time, leading to a logarithmic time complexity, ideal for large datasets or sorted arrays. In both cases, iteration
serves as the driving force behind the search process, enabling the algorithms to efficiently navigate through the dataset and locate the target element.
Best Case Scenario
The best case scenario for linear search occurs when the target element is found at the very beginning of the dataset. In this case, the algorithm only needs to perform a single comparison, resulting
in a time complexity of O(1). However, it's important to note that this scenario is rare, and the average and worst-case time complexities of linear search remain O(n).
On the other hand, the best case scenario for binary search occurs when the target element is located exactly at the middle of the dataset. In this scenario, the algorithm can immediately determine
the target's position with just one comparison, resulting in a time complexity of O(1). However, binary search's dataset must be sorted for this scenario to be applicable. Additionally, binary
search's average and worst-case time complexities are O(log n), making it significantly more efficient than linear search for large datasets.
Visualization plays a crucial role in understanding and analyzing algorithms like linear search and binary search. By visualizing the search process, programmers can gain deeper insights into how
these algorithms work and identify potential optimizations or pitfalls. Let's explore how visualization can enhance our understanding of both linear search and binary search:
Linear Search Visualization
In linear search, the algorithm sequentially scans each element in the dataset until it finds the target element or reaches the end of the list. Visualizing this process can be done by representing
the dataset as a linear array or a linked list. As the algorithm progresses, each element can be highlighted or marked to indicate whether it's been compared to the target element. This visualization
helps users understand the linear nature of the search process and how the algorithm iterates through each element until a match is found or the end of the list is reached.
Binary Search Visualization
Binary search, on the other hand, follows a divide and conquer approach by repeatedly dividing the search space in half. Visualizing binary search involves representing the sorted dataset as a binary
tree, where each node represents an element in the dataset. As the algorithm progresses, users can see how the search space is halved in each iteration, with the target element being compared to the
middle element of the current subarray. This visualization provides insights into the efficiency of binary search and how it quickly converges on the target element by reducing the search space
In conclusion, Linear Search and Binary Search are two fundamental search algorithms with distinct characteristics and applications. Linear search, though simple, is suitable for small datasets or
when the elements are unsorted. It efficiently navigates linear data structures like arrays or linked lists and can be visualized as sequentially examining each element until a match is found,
potentially reaching the last element. On the other hand, Binary search, leveraging its divide and conquer strategy, excels in efficiency, making it optimal for large datasets or sorted arrays. It
efficiently searches through half of the list in each iteration, resulting in a logarithmic time complexity. Additionally, binary search can be visualized as recursively dividing the search space
until the target element is located. Understanding the nuances of these algorithms is crucial for programmers and computer science enthusiasts alike. Whether you're a beginner exploring the basics of
search algorithms or an experienced developer optimizing performance, mastering Linear and Binary Search opens doors to a deeper understanding of algorithmic principles and problem-solving
Coding Education for Aspiring Programmers
In understanding the intricacies of algorithms like linear and binary search, learners can significantly boost their problem-solving skills, a vital component of programming. For youngsters eager to
dive deeper into the world of coding, Coding Classes for Kids at YoungWonks offer a well-rounded curriculum tailored for various age groups. Particularly, those with a keen interest in one of the
most popular programming languages can take advantage of Python Coding Classes for Kids. Meanwhile, hands-on enthusiasts can explore hardware programming with Raspberry Pi, Arduino and Game
Development Coding Classes, perfect for budding hardware engineers and game developers.
*Contributors: Written by Disha N; Edited by Rohit Budania; Lead image by Shivendra Singh | {"url":"https://www.youngwonks.com/blog/linear-search-vs-binary-search","timestamp":"2024-11-09T10:39:00Z","content_type":"text/html","content_length":"126837","record_id":"<urn:uuid:75f15d60-86b4-4c36-b20c-5b9a538515ea>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00647.warc.gz"} |
Print Palindrome Numbers in Given Range using C++
There are some numbers in mathematics when even reversed remain the same. Well as for observing these numbers with human eyes we don’t need to exert too much stress on our brains. However, to find
the same thing using a computer program is quite a challenge for newbie programmers. Such numbers when even reversed remain the same are known as Palindrome numbers. And in today’s article, we’ll try
to find Palindrome numbers using programs. So let’s find out how to print Palindrome Numbers in Given Range using C++.
So open up your IDE and let’s get started real quick. Once you understand the logic practice this program on your own to make your brain work in a problem-solving way. This is a somewhat hard
question you might get asked in an interview so make sure to practice it on your own after reading this article.
What’s The Approach?
• Let’s consider input number n till range max Therefore if the reverse of n is similar to n we’ll print 1 otherwise we’ll print 0.
• We will create a separate function isPalindrome to return the reverse of input min.
• Create a reverse variable rev, we will multiply it by 10 & perform a modulo operation with 10.
• If the function returns 1 then we’ll print the number, otherwise, we’ll pass.
• The above instructions will keep executing till we reach the end of the input range. So we’ll create a for loop, starting and ending with our given range incremented by 1 each time.
Also Read: Multiply Integers Using Russian Peasant Algorithm in C++
C++ Program To Print Palindrome Numbers in Given Range
100, 200
using namespace std;
// A function to check if n is palindrome
int isPalindrome(int n)
// Find reverse of n
int rev = 0;
for (int i = n; i > 0; i /= 10)
rev = rev*10 + i%10;
// If n and rev are same, then n is palindrome
return (n==rev);
// prints palindrome between min and max
void countPal(int min, int max)
for (int i = min; i <= max; i++)
if (isPalindrome(i))
cout << i << " ";
// Driver program to test above function
int main()
countPal(100, 200);
return 0; | {"url":"https://techdecodetutorials.com/print-palindrome-numbers-in-given-range-using-c/","timestamp":"2024-11-03T20:08:48Z","content_type":"text/html","content_length":"127574","record_id":"<urn:uuid:fc369dd4-6133-4cc9-852f-ddf8c79fa82a>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00225.warc.gz"} |
Linear programming
planted: 26/03/2022last tended: 26/05/2022
The method was then independently reinvented in the United States by Tjalling Koopmans and by George Danzig, who while working on transport and allocation problems for the US Airforce during the
war coined the phrase 'linear programming'
These are known as ‘linear’ equations, because if graphed they produce straight lines, and it is a property of linear equations that you can only solve them if you have as many equations to work
with as there are variables
Otherwise, they are ‘undetermined’ – there are an infinite number of possible solutions, and no way to decide between them.
Linear programming was not only a quintessentially socialist kind of mathematics, ‘characterized by a constant overlap of theory and practice’, it also offered a new kind of socialist political
Linear programming offered a systematic way to allocate resources, so that it optimized some metrics of overall national well-being
the method is ubiquitous in contemporary applied mathematics, including in planning renewable energy systems
Rather than relying on a Platonic elite of logicians, Neurath thought that a visual language could democratize reason by making the essence of an economic problem apparent to non-experts
There were two main currents that shaped planning debates in the Soviet Union over the following decade: the theory of mathematical optimization (e.g., linear programming) and the cybernetic
theory of control, built around differential equations
For all its pedagogical and democratic value, linear programming alone will not suffice to plan something as complex as the global economy
Kantorovich’s linear programming will not be enough in itself to create a global in natura economy
1. Elsewhere
1.1. In my garden
Notes that link to this note (AKA backlinks).
1.2. In the Agora | {"url":"https://commonplace.doubleloop.net/linear-programming","timestamp":"2024-11-09T23:00:19Z","content_type":"text/html","content_length":"8320","record_id":"<urn:uuid:7985899b-ae85-406a-915d-f7d4fd1e4c06>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00186.warc.gz"} |
942 research outputs found
We consider some fundamental properties of infinite dimensional Hamiltonian systems, both linear and nonlinear. For exemple, in the case of linear systems, we prove a symplectic version of the teorem
of M. Stone. In the general case we establish conservation of energy and the moment function for system with symmetry. (The moment function was introduced by B. Kostant and J .M. Souriau). For
infinite dimensional systems these conservation laws are more delicate than those for finite dimensional systems because we are dealing with partial as opposed to ordinary differential equations
We discuss the Groenewold-Van Hove problem for R^{2n}, and completely solve it when n = 1. We rigorously show that there exists an obstruction to quantizing the Poisson algebra of polynomials on R^
{2n}, thereby filling a gap in Groenewold's original proof without introducing extra hypotheses. Moreover, when n = 1 we determine the largest Lie subalgebras of polynomials which can be
unambiguously quantized, and explicitly construct all their possible quantizations.Comment: 15 pages, Latex. Error in the proof of Prop. 3 corrected; minor rewritin
The ideal anti-Zeno effect means that a perpetual observation leads to an immediate disappearance of the unstable system. We present a straightforward way to derive sufficient conditions under which
such a situation occurs expressed in terms of the decaying states and spectral properties of the Hamiltonian. They show, in particular, that the gap between Zeno and anti-Zeno effects is in fact very
narrow.Comment: LatEx2e, 9 pages; a revised text, to appear in J. Phys. A: Math. Ge
In this paper we review a proposed geometrical formulation of quantum mechanics. We argue that this geometrization makes available mathematical methods from classical mechanics to the quantum frame
work. We apply this formulation to the study of separability and entanglement for states of composite quantum systems.Comment: 22 pages, to be published in Physica Script
We prove that the separable and local approximations of the discontinuity-inducing zero-range interaction in one-dimensional quantum mechanics are equivalent. We further show that the interaction
allows the perturbative treatment through the coupling renormalization. Keywords: one-dimensional system, generalized contact interaction, renormalization, perturbative expansion. PACS Nos: 3.65.-w,
11.10.Gh, 31.15.MdComment: ReVTeX 7pgs, doubl column, no figure, See also the website http://www.mech.kochi-tech.ac.jp/cheon
It is shown that there are static spacetimes with timelike curvature singularities which appear completely nonsingular when probed with quantum test particles. Examples include extreme dilatonic
black holes and the fundamental string solution. In these spacetimes, the dynamics of quantum particles is well defined and uniquely determined.Comment: 12 pages, RevTeX, no figures, A few breif
comments added and typos correcte
We carry out the spectral analysis of matrix valued perturbations of 3-dimensional Dirac operators with variable magnetic field of constant direction. Under suitable assumptions on the magnetic field
and on the pertubations, we obtain a limiting absorption principle, we prove the absence of singular continuous spectrum in certain intervals and state properties of the point spectrum. Various
situations, for example when the magnetic field is constant, periodic or diverging at infinity, are covered. The importance of an internal-type operator (a 2-dimensional Dirac operator) is also
revealed in our study. The proofs rely on commutator methods.Comment: 12 page
we will show the existence and uniqueness of a real-time, time-sliced Feynman path integral for quantum systems with vector potential. Our formulation of the path integral will be derived on the $L^
2$ transition probability amplitude via improper Riemann integrals. Our formulation will hold for vector potential Hamiltonian for which its potential and vector potential each carries at most a
finite number of singularities and discontinuities
We introduce the concept of Type-I/II generating functionals defined on the space of boundary data of a Lagrangian field theory. On the Lagrangian side, we define an analogue of Jacobi's solution to
the Hamilton-Jacobi equation for field theories, and we show that by taking variational derivatives of this functional, we obtain an isotropic submanifold of the space of Cauchy data, described by
the so-called multisymplectic form formula. We also define a Hamiltonian analogue of Jacobi's solution, and we show that this functional is a Type-II generating functional. We finish the paper by
defining a similar framework of generating functions for discrete field theories, and we show that for the linear wave equation, we recover the multisymplectic conservation law of Bridges.Comment: 31
pages; 1 figure -- v2: minor change
I review some recent work in which the quantum states of string theory which are associated with certain black holes have been identified and counted. For large black holes, the number of states
turns out to be precisely the exponential of the Bekenstein-Hawking entropy. This provides a statistical origin for black hole thermodynamics in the context of a potential quantum theory of
gravity.Comment: 18 pages (To appear in the proceedings of the Pacific Conference on Gravitation and Cosmology, Seoul, Korea, February 1-6, 1996. | {"url":"https://core.ac.uk/search/?q=authors%3A(Chernoff%20P%20R)","timestamp":"2024-11-06T03:54:41Z","content_type":"text/html","content_length":"169557","record_id":"<urn:uuid:2343f188-47a8-472a-bfd8-6ab75fd562f1>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00157.warc.gz"} |
Neural Network and Deep Learning
Learn the basics of deep learning and neural network algorithms.
Deep learning dives into every domain. It was started way back in the 90s. The algorithms did what they were supposed to do at that time; they gave better predictions. Over time, due to their speed,
their popularity has decreased. Now, with the enhancement in GPUs, it is more popular than ever before. We can train with 100x speed, and it offers good library support. We also get predictions in
real-time. This makes deep learning solutions appropriate for today’s real-world applications.
First, you will learn a few basics of deep learning, and then we will move deeper into this topic.
Level up your interview prep. Join Educative to access 80+ hands-on prep courses. | {"url":"https://www.educative.io/courses/data-science-interview-handbook/neural-network-and-deep-learning","timestamp":"2024-11-13T01:16:59Z","content_type":"text/html","content_length":"722769","record_id":"<urn:uuid:9703df10-4f0c-4164-be61-4ecceab4ade0>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00182.warc.gz"} |
Multiplying Numbers in Standard Form
Question Video: Multiplying Numbers in Standard Form Mathematics • First Year of Preparatory School
Find the value of (4.1 × 10³) × (9 × 10⁷), giving your answer in scientific notation.
Video Transcript
Find the value of 4.1 multiplied by 10 to the power of three multiplied by nine multiplied by 10 to the power of seven giving your answer in scientific notation.
Our first step is to multiply 4.1 by nine. This gives us 36.9. Using one of the laws of indices — 𝑥 to the power of 𝑎 multiplied by 𝑥 to the power of 𝑏 equals 𝑥 to the power of 𝑎 plus 𝑏 — enables us
to multiply 10 cubed by 10 to the power of seven as we are able to add the exponents — in this case three and seven — to give us 10 to the power of 10.
In order to write an answer in scientific notation, the first part must be between one and 10. Currently, 36.9 is too large. We can rewrite 36.9 as 3.69 multiplied by 10. Using our laws of indices
once again, we get a final answer of 3.69 times 10 to the power of 11.
This means that the value of 4.1 multiplied by 10 cubed multiplied by nine multiplied by 10 to the power of seven is equal to 3.69 times 10 to the power of 11. | {"url":"https://www.nagwa.com/en/videos/745126341594/","timestamp":"2024-11-10T08:24:16Z","content_type":"text/html","content_length":"246839","record_id":"<urn:uuid:3e401720-eb7f-4ac6-aec3-e2061f10930d>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00616.warc.gz"} |
How To Find The Equation Of A Parabola With Just A Graph - Graphworksheets.com
Graphing Parabola From An Equation Worksheet – Graphing equations is an essential part of learning mathematics. This involves graphing lines and points and evaluating their slopes. Graphing equations
of this type requires that you know the x and y-coordinates of each point. You need to know the slope of a line. This is the point … Read more | {"url":"https://www.graphworksheets.com/tag/how-to-find-the-equation-of-a-parabola-with-just-a-graph/","timestamp":"2024-11-02T21:30:08Z","content_type":"text/html","content_length":"46856","record_id":"<urn:uuid:627bcd8f-efd8-47ed-9c07-20dc9c7ac7e6>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00053.warc.gz"} |
Page vierge | Forges Pelli Design
top of page
Shelf bracket infos
You will find here the information about:
• How to calculate the dimensions
• The dimension of the model with a BALL END
• How they are joint together in the middle
Refer HERE for more info about the finish.
How to calculate the dimensions
Classic Bracket
You can leave 1 to 3 inches between the end of the bracket and the shelf edge depending on the dimension of the shelf. The brackets are not exactly the dimension showed, due to the forging process,
even if we try to make it closer. It's more or less 1/8" to 1/4".
Industrial bracket
The dimension of the bracket is referred to the length of the shelf. For example, a bracket of 6" is made to receive a shelf of 6". So, It's calculated 6" from the wall to the end of the shelf and I
leave a little space of around 1/8" between the shelf and inside the lip of the bracket.
The longest dimension is the length and the shorter is the height.
Are they strong?
All my brackets are very strong. I'm not a scientist but I made some trial with an anvil in my shop as you can see in this photo.
Classic Brackets:
They are made with solid bars 1/4" x 1", 1/4" x 1 1/2" or 3/8" x 1 1/2" for the bracket itself and 3/8" or 1/2" for the scrolls. They can take 50 to 100 pounds each.
Industrial bracket:
These brackets are very strong. they are made with 3/4" or 1" solid square. They can take more than 100 pounds each.
It is important to make sure that is strongly attached to the wall with good fasteners. You can also distribute the weight on a larger number of brackets.
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ABC_rejection: Rejection sampling scheme for ABC in EasyABC: Efficient Approximate Bayesian Computation Sampling Schemes
This function launches a series of nb_simul model simulations with model parameters drawn in the prior distribution specified in prior_matrix.
ABC_rejection(model, prior, nb_simul, prior_test=NULL, summary_stat_target=NULL, tol=NULL, use_seed=FALSE, seed_count=0, n_cluster=1, verbose=FALSE, progress_bar=FALSE)
model a R function implementing the model to be simulated. It must take as arguments a vector of model parameter values and it must return a vector of summary statistics. When using the
option use_seed=TRUE, model must take as arguments a vector containing a seed value and the model parameter values. A tutorial is provided in the package's vignette to dynamically
link a binary code to a R function. Users may alternatively wish to wrap their binary executables using the provided functions binary_model and binary_model_cluster. The use of
these functions is associated with slightly different constraints on the design of the binary code (see binary_model and binary_model_cluster).
prior a list of prior information. Each element of the list corresponds to a model parameter. The list element must be a vector whose first argument determines the type of prior
distribution: possible values are "unif" for a uniform distribution on a segment, "normal" for a normal distribution, "lognormal" for a lognormal distribution or "exponential" for
an exponential distribution. The following arguments of the list elements contain the characteritiscs of the prior distribution chosen: for "unif", two numbers must be given: the
minimum and maximum values of the uniform distribution; for "normal", two numbers must be given: the mean and standard deviation of the normal distribution; for "lognormal", two
numbers must be given: the mean and standard deviation on the log scale of the lognormal distribution; for "exponential", one number must be given: the rate of the exponential
distribution. User-defined prior distributions can also be provided. See the vignette for additional information on this topic.
nb_simul a positive integer equal to the desired number of simulations of the model.
prior_test a string expressing the constraints between model parameters. This expression will be evaluated as a logical expression, you can use all the logical operators including "<", ">",
... Each parameter should be designated with "X1", "X2", ... in the same order as in the prior definition. If not provided, no constraint will be applied.
summary_stat_target a vector containing the targeted (observed) summary statistics. If not provided, ABC_rejection only launches the simulations and outputs the simulation results.
tol tolerance, a strictly positive number (between 0 and 1) indicating the proportion of simulations retained nearest the targeted summary statistics.
use_seed logical. If FALSE (default), ABC_rejection provides as input to the function model a vector containing the model parameters used for the simulation. If TRUE, ABC_rejection
provides as input to the function model a vector containing an integer seed value and the model parameters used for the simulation. In this last case, the seed value should be
used by model to initialize its pseudo-random number generators (if model is stochastic).
seed_count a positive integer, the initial seed value provided to the function model (if use_seed=TRUE). This value is incremented by 1 at each call of the function model.
n_cluster a positive integer. If larger than 1 (the default value), ABC_rejection will launch model simulations in parallel on n_cluster cores of the computer.
verbose logical. FALSE by default. If TRUE, ABC_rejection writes in the current directory intermediary results at the end of each step of the algorithm in the file "output". These outputs
have a matrix format, in wich each raw is a different simulation, the first columns are the parameters used for this simulation, and the last columns are the summary statistics of
this simulation.
progress_bar logical, FALSE by default. If TRUE, ABC_rejection will output a bar of progression with the estimated remaining computing time. Option not available with multiple cores.
a R function implementing the model to be simulated. It must take as arguments a vector of model parameter values and it must return a vector of summary statistics. When using the option use_seed=
TRUE, model must take as arguments a vector containing a seed value and the model parameter values. A tutorial is provided in the package's vignette to dynamically link a binary code to a R function.
Users may alternatively wish to wrap their binary executables using the provided functions binary_model and binary_model_cluster. The use of these functions is associated with slightly different
constraints on the design of the binary code (see binary_model and binary_model_cluster).
a list of prior information. Each element of the list corresponds to a model parameter. The list element must be a vector whose first argument determines the type of prior distribution: possible
values are "unif" for a uniform distribution on a segment, "normal" for a normal distribution, "lognormal" for a lognormal distribution or "exponential" for an exponential distribution. The following
arguments of the list elements contain the characteritiscs of the prior distribution chosen: for "unif", two numbers must be given: the minimum and maximum values of the uniform distribution; for
"normal", two numbers must be given: the mean and standard deviation of the normal distribution; for "lognormal", two numbers must be given: the mean and standard deviation on the log scale of the
lognormal distribution; for "exponential", one number must be given: the rate of the exponential distribution. User-defined prior distributions can also be provided. See the vignette for additional
information on this topic.
a positive integer equal to the desired number of simulations of the model.
a string expressing the constraints between model parameters. This expression will be evaluated as a logical expression, you can use all the logical operators including "<", ">", ... Each parameter
should be designated with "X1", "X2", ... in the same order as in the prior definition. If not provided, no constraint will be applied.
a vector containing the targeted (observed) summary statistics. If not provided, ABC_rejection only launches the simulations and outputs the simulation results.
tolerance, a strictly positive number (between 0 and 1) indicating the proportion of simulations retained nearest the targeted summary statistics.
logical. If FALSE (default), ABC_rejection provides as input to the function model a vector containing the model parameters used for the simulation. If TRUE, ABC_rejection provides as input to the
function model a vector containing an integer seed value and the model parameters used for the simulation. In this last case, the seed value should be used by model to initialize its pseudo-random
number generators (if model is stochastic).
a positive integer, the initial seed value provided to the function model (if use_seed=TRUE). This value is incremented by 1 at each call of the function model.
a positive integer. If larger than 1 (the default value), ABC_rejection will launch model simulations in parallel on n_cluster cores of the computer.
logical. FALSE by default. If TRUE, ABC_rejection writes in the current directory intermediary results at the end of each step of the algorithm in the file "output". These outputs have a matrix
format, in wich each raw is a different simulation, the first columns are the parameters used for this simulation, and the last columns are the summary statistics of this simulation.
logical, FALSE by default. If TRUE, ABC_rejection will output a bar of progression with the estimated remaining computing time. Option not available with multiple cores.
The returned value is a list containing the following components:
param The model parameters used in the model simulations.
stats The summary statistics obtained at the end of the model simulations.
weights The weights of the different model simulations. In the standard rejection scheme, all model simulations have the same weights.
stats_normalization The standard deviation of the summary statistics across the model simulations.
nsim The number of model simulations performed.
nrec The number of retained simulations (if targeted summary statistics are provided).
computime The computing time to perform the simulations.
The summary statistics obtained at the end of the model simulations.
The weights of the different model simulations. In the standard rejection scheme, all model simulations have the same weights.
The standard deviation of the summary statistics across the model simulations.
The number of retained simulations (if targeted summary statistics are provided).
Pritchard, J.K., and M.T. Seielstad and A. Perez-Lezaun and M.W. Feldman (1999) Population growth of human Y chromosomes: a study of Y chromosome microsatellites. Molecular Biology and Evolution, 16,
##### EXAMPLE 1 ##### ##################### set.seed(1) ## artificial example to show how to use the 'ABC_rejection' function. ## defining a simple toy model: toy_model<-function(x){ 2 * x + 5 +
rnorm(1,0,0.1) } ## define prior information toy_prior=list(c("unif",0,1)) # a uniform prior distribution between 0 and 1 ## only launching simulations with parameters drawn in the prior
distributions set.seed(1) n=10 ABC_sim<-ABC_rejection(model=toy_model, prior=toy_prior, nb_simul=n) ABC_sim ## launching simulations with parameters drawn in the prior distributions # and performing
the rejection step sum_stat_obs=6.5 tolerance=0.2 ABC_rej<-ABC_rejection(model=toy_model, prior=toy_prior, nb_simul=n, summary_stat_target=sum_stat_obs, tol=tolerance) ## NB: see the package's
vignette to see how to pipeline 'ABC_rejection' with the function # 'abc' of the package 'abc' to perform other rejection schemes. ## Not run: ##### EXAMPLE 2 ##### ##################### ## this
time, the model has two parameters and outputs two summary statistics. ## defining a simple toy model: toy_model2<-function(x){ c( x[1] + x[2] + rnorm(1,0,0.1) , x[1] * x[2] + rnorm(1,0,0.1) ) } ##
define prior information toy_prior2=list(c("unif",0,1),c("normal",1,2)) # a uniform prior distribution between 0 and 1 for parameter 1, and a normal distribution # of mean 1 and standard deviation of
2 for parameter 2. ## only launching simulations with parameters drawn in the prior distributions set.seed(1) n=10 ABC_sim<-ABC_rejection(model=toy_model2, prior=toy_prior2, nb_simul=n) ABC_sim ##
launching simulations with parameters drawn in the prior distributions # and performing the rejection step sum_stat_obs2=c(1.5,0.5) tolerance=0.2 ABC_rej<-ABC_rejection(model=toy_model2, prior=
toy_prior2, nb_simul=n, summary_stat_target=sum_stat_obs2, tol=tolerance) ## NB: see the package's vignette to see how to pipeline 'ABC_rejection' with the function # 'abc' of the package 'abc' to
perform other rejection schemes. ##### EXAMPLE 3 ##### ##################### ## this time, the model is a C++ function packed into a R function -- this time, the option # 'use_seed' must be turned to
TRUE. n=10 trait_prior=list(c("unif",3,5),c("unif",-2.3,1.6),c("unif",-25,125),c("unif",-0.7,3.2)) trait_prior ## only launching simulations with parameters drawn in the prior distributions ABC_sim
<-ABC_rejection(model=trait_model, prior=trait_prior, nb_simul=n, use_seed=TRUE) ABC_sim ## launching simulations with parameters drawn in the prior distributions and performing # the rejection step
sum_stat_obs=c(100,2.5,20,30000) tolerance=0.2 ABC_rej<-ABC_rejection(model=trait_model, prior=trait_prior, nb_simul=n, summary_stat_target=sum_stat_obs, tol=tolerance, use_seed=TRUE) ## NB: see the
package's vignette to see how to pipeline 'ABC_rejection' with the function # 'abc' of the package 'abc' to perform other rejection schemes. ##### EXAMPLE 4 - Parallel implementations ##### #########
####################################### ## NB: the option use_seed must be turned to TRUE. ## For models already running with the option use_seed=TRUE, simply change # the value of n_cluster:
sum_stat_obs=c(100,2.5,20,30000) ABC_simb<-ABC_rejection(model=trait_model, prior=trait_prior, nb_simul=n, use_seed=TRUE, n_cluster=2) ## For other models, change the value of n_cluster and modify
the model so that the first # parameter becomes a seed information value: toy_model_parallel<-function(x){ set.seed(x[1]) 2 * x[2] + 5 + rnorm(1,0,0.1) } sum_stat_obs=6.5 ABC_simb<-ABC_rejection
(model=toy_model_parallel, prior=toy_prior, nb_simul=n, use_seed=TRUE, n_cluster=2) ## End(Not run)
For more information on customizing the embed code, read Embedding Snippets. | {"url":"https://rdrr.io/cran/EasyABC/man/ABC_rejection.html","timestamp":"2024-11-13T01:45:04Z","content_type":"text/html","content_length":"32596","record_id":"<urn:uuid:210b0523-e8e8-49f4-9ec1-ba066d322a74>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00891.warc.gz"} |
Simplifying Complex Fractions
Learning Outcomes
• Translate word phrases to expressions with fractions
• Simplify complex fractions
Translate Phrases to Expressions with Fractions
The words quotient and ratio are often used to describe fractions. Earlier, we defined quotient as the result of division. The quotient of [latex]a\text{ and }b[/latex] is the result you get from
dividing [latex]a\text{ by }b[/latex], or [latex]\Large\frac{a}{b}[/latex]. Let’s practice translating some phrases into algebraic expressions using these terms.
Translate the phrase into an algebraic expression: “the quotient of [latex]3x[/latex] and [latex]8[/latex].”
The keyword is quotient; it tells us that the operation is division. Look for the words of and and to find the numbers to divide.
[latex]\text{The quotient }\text{of }3x\text{ and }8\text{.}[/latex]
This tells us that we need to divide [latex]3x[/latex] by [latex]8[/latex].
try it
Translate the phrase into an algebraic expression: the quotient of the difference of [latex]m[/latex] and [latex]n[/latex], and [latex]p[/latex].
Show Solution
Try it
In the following video we show more examples of translating English expressions into algebraic expressions.
Simplify Complex Fractions
Our work with fractions so far has included proper fractions, improper fractions, and mixed numbers. Another kind of fraction is called complex fraction, which is a fraction in which the numerator or
the denominator contains a fraction.
Some examples of complex fractions are:
[latex]\LARGE\frac{\frac{6}{7}}{ 3}, \frac{\frac{3}{4}}{\frac{5}{8}}, \frac{\frac{x}{2}}{\frac{5}{6}}[/latex]
To simplify a complex fraction, remember that the fraction bar means division. So the complex fraction [latex]\LARGE\frac{\frac{3}{4}}{\frac{5}{8}}[/latex] can be written as [latex]\Large\frac{3}{4}\
Simplify: [latex]\LARGE\frac{\frac{3}{4}}{\frac{5}{8}}[/latex]
Show Solution
Try it
The following video shows another example of how to simplify a complex fraction.
Simplify a complex fraction.
1. Rewrite the complex fraction as a division problem.
2. Follow the rules for dividing fractions.
3. Simplify if possible.
Simplify: [latex]\LARGE\frac{-\frac{6}{7}}{ 3}[/latex]
Show Solution
Try it
Simplify: [latex]\LARGE\frac{\frac{x}{2}}{\frac{xy}{6}}[/latex]
Show Solution
Try it
Simplify: [latex]\LARGE\frac{2\frac{3}{4}}{\frac{1}{8}}[/latex]
Show Solution
Try it | {"url":"https://courses.lumenlearning.com/wm-developmentalemporium/chapter/simplifying-complex-fractions/","timestamp":"2024-11-07T22:18:23Z","content_type":"text/html","content_length":"59035","record_id":"<urn:uuid:2038b7c5-040d-4569-b901-ab45415fedbc>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00476.warc.gz"} |
Vol. 43, No. 3, 2017
HOUSTON JOURNAL OF
Electronic Edition Vol. 43, No. 3, 2017
Editors: D. Bao (San Francisco, SFSU), D. Blecher (Houston), B. G. Bodmann (Houston), H. Brezis (Paris and Rutgers), B. Dacorogna (Lausanne), K. Davidson (Waterloo), M. Dugas (Baylor), M. Gehrke
(LIAFA, Paris7), C. Hagopian (Sacramento), R. M. Hardt (Rice), Y. Hattori (Matsue, Shimane), W. B. Johnson (College Station), M. Rojas (College Station), Min Ru (Houston), S.W. Semmes (Rice).
Managing Editors: B. G. Bodmann and K. Kaiser (Houston)
Houston Journal of Mathematics
Pat Goeters, Department of Mathematics, Auburn University, Auburn, Al 36849-5310 (goetehp@auburn.edu).
A-divisorial Matlis domains, pp. 691-701
ABSTRACT. We study the integral domains R with the property that Hom[R](-,A) defines a self-canceling duality with respect to a natural collection of submodules of the quotient field Q of R for a
given submodule A of Q with End[R](A)=R. In case there is such a duality when R is a Matlis domain, we show that locally, A must necessarily be a fractional ideal. Additionally, under the presence of
this duality, we examine the relation between the domain being Noetherian and having Krull dimension 1.
Enochs, Edgar, University of Kentucky, Lexington, KY 40506 (e.enochs@uky.edu), Jenda, Overtoun, Auburn University, Auburn, AL 36849 (jendaov@auburn.edu), and Ozbek, Furuzan, Auburn University,
Auburn, AL 36849 (fzo0005@auburn.edu).
Submonoids of the formal power series, pp. 703-711.
ABSTRACT. Formal power series come up in several areas such as formal language theory, algebraic and enumerative combinatorics, number theory. In this paper we focus on the the subset of formal power
series consisting of the ones with zero constant term. This subset forms a monoid with composition operation of series. We classify the sets T of strictly positive integers, for which the set of
formal power series consisting of terms with powers from the set T, that forms a monoid with composition as the operation. We prove that in order for that set to be a monoid, T itself has to be a
submonoid of (N,.). Unfortunately, this condition is not enough to guarantee the desired result. But if a monoid is strongly closed, then we get the desired result. We also consider an analogous
problem for power series in several variables in the last section.
Mahdi Rahmatinia, Department of Mathematics and Applications, University of Mohaghegh Ardabili, P.O.Box 179, Ardabil, Ir, (m.rahmati@uma.ac.ir), (mahdi.rahmatinia@gmail.com).
On φ-almost Bezoüt rings and φ-almost Prüfer rings, pp. 713-723.
ABSTRACT. The porpuse of this paper is to introduce some new classes of rings that are closely related to the classes of almost Bezoüt domains and almost Prüfer domains. This paper is devoted to
study the φ-almost Bezoüt rings (φ-AB rings) and φ-almost Prüfer rings (φ-AP rings). A ring R is a φ-AB ring (respectively, φ-AP ring) if for nonnil elements a,b of R, there exists an n=n(a,b) with
(a^n,b^n) is a nonnil principal (respectively, φ-invertible) ideal of R.
Duc Thai Do, Department of Mathematics, Hanoi National University of Education, 136 Xuan Thuy str., Hanoi, Viet Nam (doducthai@hnue.edu.vn) and Duc-Viet Vu, UPMC Univ. Paris 06, UMR 7586, Institut de
Mathematiques de Jussieu, Paris Rive Gauche, 4 place Jussieu, F-75005 Paris, France (duc-viet.vu@imj-prg.fr).
Holomorphic mappings into compact complex manifolds, pp. 725-762.
ABSTRACT. The purpose of this article is to show a second main theorem with the explicit truncation level for holomorphic mappings of the complex plane (or of a compact Riemann surface) into a
compact complex manifold sharing divisors in subgeneral position.
Shengjiang Chen, Department of Mathematics, Ningde Normal University, Fujian, 352100, China (ndsycsj@126.com), Key Laboratory of Applied Mathematics(Putian University)，Fujian Province University,
Fujian Putian, 351100，P.R. China, Department of Mathematics, Fujian Normal University, Fuzhou 350007, Fujian Province, P.R. China and Weichuan Lin (corresponding author), Department of Mathematics,
Fujian Normal University, Fuzhou, 350007, P.R.China (sxlwc963@fjnu.edu.com).
Periodicity and uniqueness of meromorphic functions concerning sharing values, pp. 763-781.
ABSTRACT. In this paper, we obtain two suffcient conditions for periodicity of mermorphic functions of infinite order concerning three sharing values (2CM+1IM), which are improvements of the related
results on meromorphic functions of finite order. As applications, we prove an uniqueness theorem of a meromorphic function g(z) when g(z) shares two values CM and one value IM with a periodic
meromorphic function f(z) by a new method. Examples are given to show that our results are precise.
Hongyan Xu, Department of Informatics and Engineering, Jingdezhen Ceramic Institute, Jingdezhen, Jiangxi, 333403, China (xhyhhh@126.com) and YinYing Kong (corresponding author), School of Mathematics
and Statistics, Guangdong University of Finance and Economics, Guangzhou, Guangdong 510320, China (kongcoco@hotmail.com).
The approximation of analytic function defined by Laplace-Stieltjes transformations convergent in the left half-plane, 783-806.
ABSTRACT. By introducing the error in approximating Laplace-Stieltjes transformation, we study the growth of the analytic function defined by Laplace-Stieltjes transformation of X-order, which
converges on the left half plane, and obtain the relation theorems between the error and X-order of Laplace-Stieltjes transformation.
Jing Zhang, Department of Mathematics, Division of Computing, Mathematics and Technology, Governors State University, University Park, IL 60484 (jzhang@govst.edu)
Complex manifolds with vanishing Hodge cohomology, pp. 807-827.
ABSTRACT. Let Y be a connected complex manifold and F the sheaf of holomorphic j-forms on Y. If the i-th cohomology of F on Y vanishes for all positive integer i and nonnegative integer j, is Y a
Stein manifold? This is a question raised by J-P. Serre in 1953. In this paper, we investigate the properties of this type of complex manifolds, assuming that Y is an open subset of a compact complex
manifold X and the boundary X-Y is support of a Cartier divisor D on X. Particularly, we compute the number of algebraically independent nonconstant meromorphic functions with poles in X-Y (This
number is called the Itaka D-dimension).
Xiaohuan, Mo, Peking University, Beijing, 100871 (moxh@math.pku.edu.cn), Linfeng Zhou, East China Normal University, Shanghai, 200241, and Hongmei Zhu, Henan Normal University, Xinxiang, 453007.
On a class of Finsler metrics with constant flag curvature, pp. 829-846.
ABSTRACT: By finding two partial differential equations equivalent to a class of Finsler metrics being of constant flag curvature we explicitly construct new locally projectively flat Finsler metrics
of constant flag curvature −1 and 0. They are counterexamples of Theorem 7.3 and 7.2 in Benling Li's paper.
Guanwei Chen, University of Jinan, Jinan 250022, Shandong Province, P.R. China (guanweic@163.com) and Guo-Ping Zhan, Department of Mathematics, Zhejhang University of Technology, hangzhou 310023,
P.R. China (zhangp@zjut.edu.cn).
Infinitely many small negative energy periodic solutions for second order Hamiltonian systems without spectrum 0, pp. 847-860.
ABSTRACT. In this paper, we consider a class of second order Hamiltonian systems, where 0 belongs to a spectral gap and the nonlinearities are subquadratic growth at infinity. The existence of
infinitely many small negative energy periodic solutions is obtained by a variant fountain theorem.
Gardella, Eusebio, Mathematisches Institut, Fachbereich Mathematik und Informatik der Universität Münster, Einsteinstrasse 62, 48149 Muenster, Germany (gardella@uni-muenster.de),
Regularity properties and Rokhlin dimension for compact group actions, pp. 861-889.
ABSTRACT. We show that formation of crossed products and passage to fixed point algebras by compact group actions with finite Rokhlin dimension preserve the following regularity properties: finite
decomposition rank, finite nuclear dimension, and tensorial absorption of the Jiang-Su algebra, the latter in the formulation with commuting towers. Finally, we also show how our results yield new
informaticon in some cases of interest.
Sangani Monfared, Mehdi, Department of Mathematics and Statistics, University of Windsor, Windsor, ON, N9B 3P4 (monfared@uwindsor.ca).
Relative boundaries for normed spaces, pp. 891-904.
ABSTRACT. We define relative boundaries for subsets of normed spaces. Extending a result of Bishop and de Leeuw, we show that if E is a Banach space and K is a subset of the unit ball of the dual of
E, then the (α , β)-points of E relative to K, are strong boundary points of E relative to K. As an application, we show that if A is a unital (left) character amenable Banach algebra, then the
spectral Choquet boundary of A is the entire spectrum of A. We give sufficient conditions for character amenability of finitely generated commutative Banach algebras.
Balan, Radu, University of Maryland, College Park, MD 20742 (rvbalan@math.umd.edu).
Stability of frames which give phase retrieval, pp. 905-918.
ABSTRACT. In this paper we study the property of phase retrievability by redundant sysems of vectors under perturbations of the frame set. Specifically we show that if a set F of m vectors in the
complex Hilbert space of dimension n allows for vector reconstruction from magnitudes of its coefficients, then there is a perturbation bound r>0 so that any frame set within r from F has the same
property. In particular this proves a recent construction for the case m=4n-4 is stable under perturbations. Additionally we provide estimates of the stability radius.
Mbekhta, Mostafa, UFR de Mathématiques, Laboratoire CNRS-UMR 8524 P. Painlevé, Université Lille 1, 59655 Villeneuve Cedex, France (Mostafa.Mbekhta@math.univ-lille1.fr), and Oudghiri, Mourad, Faculté
des Sciences d'Oujda, Laboratoire LAGA, 60000 Oujda, Morocco (m.oudghiri@ump.ac.ma)
Additive preservers of group invertible operators, pp. 919-936.
ABSTRACT. Let B(X) be the algebra of all bounded linear operators on an infinite-dimensional complex or real Banach space X. We prove that an additive surjective map Φ on B(X) preserves group
invertible operators in both directions if and only if Φ is either of the form Φ(T)=cATA^-1 or the form Φ(T)=cBT*B^-1 where c is a non-zero scalar, A:X→ X and B:X*→ X are two bounded invertible
linear or conjugate linear operators.
Pop, Florin, Dept. of Mathematics, Wagner College, Staten Island, NY 10301 (fpop@wagner.edu)
Remarks on the Completely Bounded Approximation Property for C*-algebras, pp. 937-945.
ABSTRACT. In this paper we show that if the range C*-algebra has the Weak Expectation Property, then completely bounded maps have tensor product extension properties similar to those of completely
positive maps. This result, combined with an argument of Huruya, allows us a new look at the Completely Bounded and Matricial approximation properties for C*-algebras.
Gilles Godefroy, Institut de Mathématiques de Jussieu, 4 place Jussieu, 75252 Paris Cedex 05, France (gilles.godefroy@imj-prg.fr).
The isomorphism classes of l[p] are Borel, pp. 947-951.
ABSTRACT. It is proved for 1<p<∞ that the set of subspaces of C(2^ω) that are isomorphic to l[p] is Borel.
Pilarczyk, Dominika, University of Wroc³aw, Wroc³aw (dominika.pilarczyk@math.uni.wroc.pl).
Linear evolution equation with fractional Laplacian and singular potential, pp. 953-967.
ABSTRACT. We show the existence of weak and mild solutions to the Cauchy problem for the linear evolution equation with fractional Laplacian and singular potential. Moreover we describe their large
time behavior and obtain asymptotic stability of self-similar solutions.
Tomonori Hirasa, Faculty of Mathematics, Kyushu University, Motooka 744, Nishi-ku, Fukuoka, 819-0395, Japan (tomonori_hirasa@yahoo.co.jp).
Liftability for orientable immersed surfaces and triple points, pp. 969-973.
ABSTRACT. In this paper, we consider generic immersions of orientable closed surfaces into 3-space and show that there is a non-liftable example if and only if its number of triple points is 2n (n >
Yao Wei
, Hebei University of Science and Technology, Shijiazhuang 050018, Hebei Province, China (
) and
Wu, Hengyang
, Hangzhou Dianzi University, Hangzhou 310018, Zhejiang Province, China (
Dualities for lower semicontinuous maps in the framework of interior spaces
, pp. 975-991.
ABSTRACT. This paper studies dualities for lower semicontinuous maps in the framework of interior spaces. For an interior space (X,O(X)) and a complete lattice L, let [X→L] be the set of all lower
semicontinuous maps of (X,O(X)). We get three dualities for [X→L]: (1) for every complete lattice L, [X→L] is order isomorphic to the set of all meet-join maps from L to the lattice O(X); (2) a
complete lattice L is completely distributive iff [X→L] is order isomorphic to the set of all join-meet maps from O(X) to L for every interior space (X,O(X)); (3) an interior space (X,O(X)) is
totally continuous iff [X→L] is order isomorphic to the set of all join-meet maps from O(X) to L for every complete lattice L.
Robert J. Daverman, Department of Mathematics, University of Tennessee, Knoxville, TN, 37996 (daverman@math.utk.edu).
Smearing the wildness of crumpled cubes via cell-like maps, pp. 993-1018.
ABSTRACT. We consider the set of triples (C,C',f) consisting of crumpled n-cubes C and C' and surjective cell-like maps f of C onto C' such that such that the restriction of f to the Interior of C is
1-1. Generally the overarching goal is to compare the wildness of C and C' in the presence of such a cell-like map. Specifically, we strive to show a wide variety of ways in which target can be more
complicated than the domain.
Richard N. Ball, Dept. of Math, Univ. of Denver, Denver, CO 80208 (rball@du.edu), Anthony W. Hager, Dept. of Math and CS, Wesleyan Univ. Middletown, CT 06459 (ahager@wesleyan.edu), Donald G. Johnson,
4351 Nambe Arc, Las Cruces, NM 88011 (dgjohnson@member.ams.org), James J. Madden, Dept. of Math, LSU, Baton Rouge, LA 70803 (jamesjmadden@gmail.com), and Warren Wm. McGovern, Wilkes Honors College,
Florida Atlantic Univ., Jupiter, FL 33458 (warren.mcgovern@fau.edu).
The Yosida space of the vector lattice hull of an archimedean l-group with unit, pp. 1019-1030.
ABSTRACT. For an object G in the category of archimedean l-groups with distinguished weak order unit,we have the contravariantly functorial compact Yosida space, YG. When G is embedded in H,the
resulting map of YH to YG is a surjection, and when it is also one-to-one, we write "YH=YG"; for divisible hulls, we have always YdG=YG. For vector lattice hulls vG, we frequently have YvG and YG
differing. Theorem. A compact space X is quasi-F iff whenever YG=X then also YvG=X. ("quasi-F" means each dense cozero set is C*-embedded.) | {"url":"https://www.math.uh.edu/~hjm/Vol43-3.html","timestamp":"2024-11-08T08:08:52Z","content_type":"text/html","content_length":"21833","record_id":"<urn:uuid:8ad75763-5c79-466c-89e7-651220d258fe>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00590.warc.gz"} |
How do you find the perimeter of a triangle if you know the sides?
How do you find the perimeter of a triangle if you know the sides?
Solution: Since all three sides are equal in length, the triangle is an equilateral triangle. Perimeter = 30 cm….Read More:
Perimeter of a Triangle Formula Equilateral Triangle Formula
Acute angled Triangle Isosceles Triangle Perimeter Formula
What is the perimeter of an equilateral triangle with one side measuring 4x 8 units?
Since the triangle is equilateral, every side is congruent to one another. Thus the perimeter is 3(S), with S being the length of a side.
What is the perimeter of equilateral?
Because the sides of the equilateral triangle are equal, the perimeter is equal to 3a.
What is the perimeter of a b c?
The perimeter of a scalene triangle can be calculated by finding the sum of all the unequal sides. The formula for the perimeter of a scalene triangle is: Perimeter = a + b + c, where “a”, “b”, and
“c” are the three different sides.
What is the area and perimeter of the equilateral triangle?
The area of an equilateral triangle is √3/4 times (side)2 of the equilateral triangle and the perimeter of an equilateral triangle is 3 times of a side of the equilateral triangle.
What is the perimeter of the formed triangle?
The perimeter of a triangle is the sum of lengths of its sides. The three vertices of the given triangle are O(0, 0), A(a, 0) and B(0, b). Let us now find the lengths of the sides of the triangle.
Thus the perimeter of the triangle with the given vertices is `a + b + sqrt(a^2 + b^2)` .
What is the formula for finding the perimeter of a triangle?
Instructions for finding the perimeter of a triangle. The perimeter of a triangle is found by adding the length of all three sides of the triangle. The formula. P = a + b + c. An example of finding
the perimeter of a triangle using “in” for inches as the measure for each side. a = 2 in. b = 2 in.
What does the perimeter have to equal in a triangle?
Perimeter is the sum of all the sides of a closed figure. An equilateral triangle is a triangle having all three sides equal in length. Perimeter of equilateral triangle is equal to three times the
length of a side of a triangle. Perimeter of Equilateral Triangle = 3 x Length of each side
How do you find the sides of an equilateral triangle?
If you’re trying to find a side other than the hypotenuse, square the side given, subtract it from the hypotenuse squared, and take the square root of the answer. Recognize that an equilateral
triangle has three equal sides.
What are the identifying features of an equilateral triangle?
Identifying equilateral triangles. An equilateral triangle has three equal sides and three equal angles (which are each 60°). Its equal angles make it equiangular as well as equilateral. | {"url":"https://teacherscollegesj.org/how-do-you-find-the-perimeter-of-a-triangle-if-you-know-the-sides/","timestamp":"2024-11-11T04:24:51Z","content_type":"text/html","content_length":"143264","record_id":"<urn:uuid:4435c458-b08a-431c-bc68-2a25f89d8b0d>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00739.warc.gz"} |
Wiki Histories
Wiki Histories are 10-minute pencil & paper games. They are great for teaching problem-solving, world history, and board game design. They can be played in any grade starting at grade 2. Each comes
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Wiki Histories Prehistory to 2500 BCE (13 games)
Wiki Histories 2500 BCE to 500 BCE (only 3 games for now)
Wiki Histories 500 BCE to 1 BCE (11 games)
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Standards for Mathematical Practice
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MP1 Toughen up!
Students develop grit and resiliency in the face of nasty, thorny problems. It is the most sought after skill for our students.
MP2 Think abstractly!
Students take problems and reformat them mathematically. This is helpful because mathematics lets them use powerful operations like addition.
MP3 Work together!
Students discuss their strategies to collaboratively solve a problem and identify missteps in a failed solution. Try pairing up elementary students and getting older students to work in threes.
MP4 Model reality!
Students create a model that mimics the real world. Discoveries made by manipulating the model often hint at something in the real world.
MP5 Use the right tools!
Students should use the right tools: 0-99 wall charts, graph paper, mathigon.org. etc.
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Students learn to communicate using precise terminology. Students should not only use the precise terms of others but invent and rigorously define their own terms.
MP7 Be observant!
Students learn to identify patterns. This is one of the things that the human brain does very well. We sometimes even identify patterns that don't really exist! 😉
MP8 Be lazy!?!
Students learn to seek for shortcuts. Why would you want to add the numbers one through a hundred if you can find an easier way to do it?
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Time Development in Quantum Mechanics Using a Reduced Hilbert Space Approach
Time Development in Quantum Mechanics Using a Reduced Hilbert Space Approach Documents
Main Document
written by Mario Belloni and Wolfgang Christian
The Time Development in Quantum Mechanics paper describes the suite of open source programs that numerically calculate and visualize the evolution of arbitrary initial quantum-mechanical bound
states. The calculations are based on the expansion of an arbitrary wave function in terms of basis vectors in a reduced Hilbert space. The approach is stable, fast, and accurate at depicting the
long-time dependence of complicated bound states. Several real-time visualizations, such as the position and momentum expectation values and the Wigner quasiprobability distribution for the position
and momentum, can be shown. We use these computational tools to study the time-dependent properties of quantum-mechanical systems and discuss the effectiveness of the algorithm.
Published April 1, 2008
Last Modified May 29, 2008
previous versions. | {"url":"https://www.compadre.org/OSP/document/ServeFile.cfm?ID=7288&DocID=458","timestamp":"2024-11-14T10:37:30Z","content_type":"application/xhtml+xml","content_length":"14823","record_id":"<urn:uuid:c83ce3f6-898d-4eb1-8467-bcb3d8738ecc>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00384.warc.gz"} |
Shape and topology optimization of a permanent-magnet machine under uncertainties
R E S E A R C H
Open Access
Shape and topology optimization of a
permanent-magnet machine under
Piotr Putek
[, Roland Pulch]
[, Andreas Bartel]
[, E Jan W ter Maten]
[, Michael Günther]
[and Konstanty]
M Gawrylczyk
[email protected] 1[Chair of Applied Mathematics and]
Numerical Analysis, Bergische Universität Wuppertal,
Gaußstraße 20, Wuppertal, 42119, Germany
2[Institute for Mathematics and] Computer Science,
Ernst-Moritz-Arndt-Universität Greifswald,
Walther-Rathenau-Straße 47, Greifswald, 17489, Germany Full list of author information is available at the end of the article
Our ultimate goal is a topology optimization for a permanent-magnet (PM) machine, while including material uncertainties. The uncertainties in the output data are, e.g., due to measurement errors in
the non-/linear material laws. In the resulting stochastic forward problem, these uncertainties are stochastically modeled by random fields. The solution of the underlying PDE, which describes
magnetostatics, is represented using the generalized polynomial chaos expansion. As crucial ingredient we exploit the stochastic collocation method (SCM). Eventually, this leads to a
random-dependent bi-objective cost functional, which is comprised of the
expectation and the variance. Subject to the optimization of the PM machine are the shapes of the rotor poles, which are described by zero-level sets. Thus, the
optimization will be done by redistributing the iron and magnet material over the design domain, which allows to attain an innovative low cogging torque design of an electric machine. For this
purpose, the gradient directions are evaluated by using the continuous design sensitivity analysis in conjunction with the SCM. In the end, our numerical result for the optimization of a
two-dimensional model demonstrates that the proposed approach is robust and effective.
Keywords: robust low cogging torque design; topology and shape optimization; random partial differential equation; stochastic collocation method; level set method; continuous design sensitivity
analysis; weighted average method; trade-off method
1 Introduction
Due to high performance, high efficiency and high power density, permanent-magnet (PM) machines are becoming more and more popular [–]. Consequently, these devices are currently broadly used in
applications as robotics, hybrid vehicles, computer periph-erals and so on, see, e.g., [–]. However, the PM machines suffer by construction from a considerable level of mechanical vibration and
acoustic noise. More precisely, the interac-tion of the air-gap harmonics (stator slot driven) and the magnetomotive force harmonics (magnet driven) produces a high cogging torque (CT). On the other
hand, the torque rip-ple is primarily provoked by the CT and higher harmonics of back-electromotive force (EMF). Also the magnetic saturation in the stator and the rotor cores [] as well as the
controller-induced parasitic torque ripples [] might further disturb the electromagnetic torque [].
From this perspective, especially the mitigation of the torque fluctuations is a key issue for the design of a PM machine because its result may simultaneously affect the machine performance.
Especially, in the context of the deterministic/stochastic topology optimiza-tion of a PM machine, the consideraoptimiza-tion of more than one competing objective into a cost functional seems to be
still a challenging problem [, ]. More specifically, when a multi-objective approach is involved in the designing process, a certain trade-off between conflicting criteria needs to be fulfilled. For
this reason, often a Pareto front technique is accepted as an alternative. Furthermore, under assumption that the Pareto front is con-vex, it can be approximated by the weighted average method (WAM)
or the-method [–]. However, in practice it is rather hard to verify this assumption, especially when nonlinear problems are considered. In such a situation, objective functions are approxi-mated
numerically, whereas a genetic or ant colony-based algorithm is recommended for the identification of the global front Pareto [, ]. It should be also noticed that in some cases, when a periodic
functional may be applied, e.g., for a lumped model of electric ma-chine [] or the periodicity of objectives is a result of the geometrical structure of electric machines [], it is also possible
to prove the convexity in a rigorous mathematical way.
Various methods for suppressing the CT have been proposed in the literature. For ex-ample in [] the authors employed the auxiliary slots for this purpose. Some solutions apply an appropriately
chosen combination of slot/pole number [] or the optimized ra-tio of pole arc to pole pinch [] in order to reduce the CT. Other efficient methods for mitigating the CT involve shaping the rotor
magnets and/or stator teeth [] including redistribution of a PM and iron material within the domain of interest using topological methods, as proposed in [, , –]. Moreover, the
statistics-based approach such as the Taguchi method [], or its generalization called the regression-based surface response methods, are proposed in [, ], especially in industrial applications
to reduce the noise to signal ratio. To this last group, also techniques based on the perturbation method for calculating the first and the second derivative may be included. Based on the sensitivity
information they intend to estimate deterministically the impact of the input variability on output characteristics and in consequence on the result of optimization []. On the one hand, this
’deterministic’ estimation of uncertainties is limits to the range of the perturba-tion|δ|= – %, see, e.g., [–]. On the other hand, the load related to the calculation of the second derivative
might be really large, especially when a cost functional involves a standard deviation. Clearly, the advantage of this approach is the immediate availability of a gradient which allows to carry out
the strategy, whose aim is to reduce the influence of the deterministic uncertainty onto the optimization result []. More recently, the efficient approach, which benefits from both the perturbation
technique and the stochastic-based method has been developed [] to estimate the statistical moments.
Figure 1 Exemplary sources of input uncertainty in a model of electric machine.
uncertainty can be represented in the mathematical model using the worse-case scenario analysis, evidence theory, fuzzy set theory and probabilistic framework, see, e.g., [] and references therein.
For this reason, in this paper we deal with the non-intrusive approach, namely a type of the spectral collocation method [], combined with the generalized polynomial chaos [], to investigate the
propagation of the uncertainty through a model of electric machine. Additionally, this technique allows to straightforwardly include the surface response model into the optimization flow [–].
The topology is a major contributor to the electromagnetic torque fluctuations. There-fore in this paper, we address the topology optimization of a PM machine. Since the results of the design
procedure are highly influenced by unknown material characteristics [], these uncertainties have to be taken into account in the course of a robust optimization. Thus the soft ferromagnetic material
should be modeled using uncertainty. In particular, the relative permeability/reluctivity of the magnetic material needs an accurate model in order to improve the accuracy of the magnetic flux density
of permanent magnets (for cer-tain applications [, ]). Therefore, in our optimization model, we take the reluctivity as uncertain.
In our case of the topology optimization, we have to trace two interfaces between dif-ferent materials with some assumed variations such as air, iron and PM poles of rotor, the modified multilevel set
method (MLSM) has been used [, ]. The level set method [] has found a wide range of applications also in electrical engineering. For instance, it is used to address shape or topology
optimization problems [, ].
under uncertainty which resulted in no-load steady state analysis of stochastic curl-curl equation (a density of excitation currentsJ(x) = ). Consequently, the machine analysis in on-load state can
be considered here as a post-processing procedure, since the elec-tromagnetic (average) torque itself was not involved in the optimization task. Moreover, we restricted ourselves to the minimization
of noise and vibrations caused predominantly by the cogging torque, the torque ripple and the back EMF, however, without coupling a vibro-acoustic modeling as in, e.g., [] with the curl-curl
equation. Instead the air-gap flux density (as equivalence of the back EMF) is considered in the optimization procedure as a second objective. It yields a considerably improvement of the wave form of
the back electromotive which, in turn, directly leads to the reduction of noise and vibrations. Fur-thermore, to deal with a stochastic multi-objective problem, the trade-off method, incor-porated in
the level set method (LSM) as well as the AWM, involved in a robust functional, have been applied.
The paper is organized as follows: first we describe the PM machine, which we use as test case (Section ). Then the deterministic model is set up (Section ). Based on that, the stochastic forward
problem is formulated in Section . Section describes the optimiza-tion problem with the needed objective funcoptimiza-tions and the constraints. Then we combine topology optimization and
uncertainties (Section ). After a short description of the sim-ulation in Section , we discuss numerical results (Section ).
2 Test case description
A design of a PM machine has to provide the shape and placement of magnets, iron poles and air-gaps. These features primarily determine the torque characteristics and thus the proper and efficient
functioning of a PM machine. Specifically, we consider as test case an electric controlled permanent magnet excited synchronous machine (ECPSM) []. For illustration, Figure provides a partial
assembly drawing of such a device and the related main parameters for the magnetic description are found in Table . In fact, the ECPSM rotor consists of two almost identical parts, which just have
opposing direction of the PM poles (see Figure ‘permanent magnetN’ and ‘permanent magnetS’). Furthermore, an additional DC control coil is mounted in the axial center of the machine, actually
between the laminated stators. Via a DC-chopper, this allows for controlling the effective excitation of the machine. Eventually this results in a field weakening of : , which is of particular
importance for electric propulsion vehicles [].
Table 1 Main parameters of the ECPSM (an electric controlled permanent magnet excited synchronous machine) design [6]
Parameter (unit) Symbol Value
Pole number 2p 12
Stator outer radius (mm) rostat 67.50
Stator inner radius (mm) ristat 41.25
One part stator axial length (mm) las 35.0
Slot opening width (mm) woslot 4.0
Number of slots ns 36
Number of phases m 3
Permanent magnet pole NdFeB 12
PM thickness (mm) tm 3.0
Remanent flux density (T) Br 1.2
Figure 3 Sextant domain for mathematical model: stator, rotor and three phases (A, B, C).
3 Mathematical model
In the following, we set up a deterministic model for the PM machine, which is suitable for optimization. This comprises a discussion of the domain, a strong and weak formulation and the objective
3.1 Field quantity and simulation domain
The magnetic behavior of the PM machine can be formulated in terms of the unknown magnetic vector potential A and the so-called curl-curl equation. Here we disregard eddy currents, i.e., we assume σ∂
[∂]A[t] = , whereσ denotes the conductivity. Especially for op-timization, one needs an efficient computational model. Therefore we reduce the three (spatial-) dimensions of the problem into a
simplified D FEM model, which provide us with acceptably accurate computation result. We validated this approach in our previous work [, ] using numerical simulation and experiment in the case of
a similar topology. Thus A = (, ,A(x,y)), which gives us a D problem. Moreover, this setting exhibits ro-tational symmetry with multiples of six. Hence we restrict the domain to one sextant, see
Figure .
3.2 Strong formulation
Now, in D the curl-curl equation for the magnetic vector potentialA=A(x,y) becomes the following Poisson equation:
∇ ·υx,∇A(x)∇A(x) –υPMM(x)
=J(x), x∈D⊂R, () where the domain D denotes the sextant region in our test case. HereJ(x) denotes the given current density (at position x = (x,y)), M(x) represents the given remanent flux density
of the PM,υis the reluctivity andυPMthe reluctivity of the permanent magnets. Moreover, we assume
where T(x) describes the radial direction of the remanent flux density andbr(x) denotes
a positive and bounded scalar function for the respective magnitude.
This quasi-linear elliptic problem () is equipped with periodic boundary conditions on the radii of D and it is also equipped with a homogeneous Dirichlet condition on the outer arc:A(x) = .
The reluctivityυdescribes as real parameter the material relation H =υ(x,|B|)B, where
B=∇ ×Aand H denote the magnetic flux density and the magnetic field strength, respec-tively. On the one hand, it depends nonlinearly onB:=|B|=|∇ ×A|=|∇A|, where| · | denotes the Euclidean norm and we
use the D setting A = (, ,A(x,y)). On the other hand, the reluctivity depends on the respective local material, i.e., it depends on the po-sition x. In our case, our connected domain is composed
of iron (Fe), air, and permanent magnets:
D = DFe∪Dair∪DPM.
Thus the reluctivity reads
υx,∇A(x)= ⎧ ⎪ ⎨ ⎪ ⎩
υFe(x,|∇A(x)|) for x∈DFe,
υair for x∈Dair, υPM for x∈DPM.
That is, the reluctivityυis discontinuous across material borders and nonlinear in ferro-magnetic materials. In the following, we assume thatυair=υis the vacuum reluctivity and that the
electromagnetic material is soft. Moreover, we remark that the nonlinear de-pendence ofυFeon|∇A|is given by a spline interpolation of measurement data.
3.3 Weak form
We assumeJ∈L(DF) and M∈(L(DF)), where DF:= D denotes the full circle. A func-tionA∈V :=H
(DF) ={A∈H(DF) :A|∂DF= } is a weak solution of the quasi-linear elliptic Dirichlet boundary problem () if it holds
= DF
ϕJdx =:l(ϕ,J)
+υPM DF
M· ∇ϕdx
for allϕ∈V. The symbolH(DF) denotes the Sobolev space of the real-valued functions on DF with square integrable weak gradients. The existence and uniqueness of solution for problem () is thoroughly
investigated in, e.g., [].
4 Stochastic forward problem
order to achieve a robust design. Therefore, in our work, the reluctivity becomes a ran-dom field, which allows us to quantify these uncertainties. The respective modeling is our next subject.
4.1 Stochastic modeling of uncertain reluctivities
We recall the reluctivityυ(), which is material dependent. We have: iron, air and perma-nent magnet domains. Besides the variations of iron and permaperma-nent magnet reluctivities, especially the
uncertainty ofυairis crucial from the engineering viewpoint, since it allows us to simulate the variations of the air-gap thickness. For an uncertainty quantification, the reluctivities become random
variables on some probability space (Ω,F,P) with sam-ple spaceΩ, sigma-algebra F and probability measureP. In our stochastic model, we introduce random perturbations within the material parameters.
The random field reads as
υx,|∇A|,ξ= ⎧ ⎪ ⎨ ⎪ ⎩
υFe(x,|A(x)|)( +δξ) for x∈DFe,
υair( +δξ) for x∈Dair, υPM( +δξ) for x∈DPM
with random vectorξ= (ξ,ξ,ξ). Therein, the random variablesξj(j= , , ) are assumed
to be independent and identically uniformly distributed in the interval [–, ]. Thus, the parametersδj> (j= , , ) specify the relative magnitude of the perturbation in the
ma-terial parameters. For the later numerical simulations, we chooseδj= . for allj, which
corresponds to perturbations of %. Thus we haveξ:Ω→[–, ][=:][Π][.]
For uncertainty quantification, we need the expectation and the variance among other quantities. The expected value of a functionf :Π→R, which depends on the random variables, is defined as
E[f] :=
fξ(ω)dP(ω) =
f(ξ)ρ(ξ) dξ ()
provided that the integral is finite. In our case of uniform random distributions, the joint probability density function is constant:ρ(ξ) =[]. Furthermore, given two functionsf,g: Π→R, the expected
value () induces an inner product
f,g:=E(fg) ()
on the Hilbert spaceL[(][Ω][) =][{][f] [:][E][[][f][] <][∞}][. The variance of a function][f] [∈][L][(][Ω][) is simply]
Var[f] :=Ef–E[f]. ()
4.2 Polynomial chaos expansion
Our uncertainty quantification is based on the concept of the polynomial chaos expansion (PCE). The homogeneous polynomial chaos was introduced in [] for Gaussian probabil-ity distributions. Later
this concept was extended to other probabilprobabil-ity distributions, which resulted in the generalized polynomial chaos, see [, ].
Given a functionf ∈L(Ω), the associated PCE reads as
f(ξ) =
The expansion () converges in the L[-norm of the probability space under certain ] as-sumptions [, ], which are fulfilled for many random distributions such as the Gaus-sian, uniform and beta
distribution. On the one hand, the functionsΨi:Π→Rrepresent
a complete orthonormala[system of basis polynomials. Thus it holds that]
fori=j, fori=j
using the inner product (). These orthonormal polynomials follow from the choice of the probability distributions in the stochastic modeling. Each traditional probability distribu-tion exhibits its
own orthogonal system, see []. In our case, the uniform distribudistribu-tion implies the Legendre polynomials as basis. We remark that the multivariate polynomials are just the products of the
univariate orthonormal polynomials for each random variable. On the other hand, the coefficientsfiin () satisfy the relation
fi=f,Ψi for eachi, ()
i.e., they result from the projection of the functionf onto the basis polynomials. In our numerical simulation, these coefficients represent the unknowns, which are used to de-scribe the desired
solutions of our problem. For traditional random variables, the series () is convergent in the norm ofL(Ω).
We further remark that the PCE () includes the information of the expected value () as well as the variance () of the functionf:
E[f] =f, Var[f] =
f[i] ()
under the assumption thatΨ≡ and that the coefficients are given exactly by (). To enable a numerical realization, the series () has to be truncated at some integerNmax.
Often all multivariate polynomials up to some total degree are included in the truncated expansion. This truncation causes that the associated variance is just an approximation of the exact variance
in (). Furthermore, the coefficients () are typically not given exactly, since they are influenced by numerical errors. If a functionf is also space-dependent, then the PCE () is considered
pointwise for each x∈D.
4.3 Stochastic PDE model and PCE approximation
Now, we return to our PDE problem (). For the uncertainty quantification, we consider the case of no load state with the excitation density currentJ= . Consequently, in the Sec-tion about the
modeling and the optimizaSec-tion problem we will mainly focus on reducing the cogging torque (no-current torque) as an undesirable component for the operation of the ECPSM machine and on improving a
wave form of the electromagnetic force when taking the uncertainties into account. Inserting the random field model for the reluctiv-ities (), we obtain the following stochastic forward problem for
our three subdomains D = Dair∪DFe∪DPM:
⎧ ⎪ ⎨ ⎪ ⎩
∇ ·(υFe(x,|∇A(x,ξ)|,ξ)∇A(x,ξ)) = , in DFe, ∇ ·(υair(x,ξ)∇A(x,ξ)) = , in Dair, ∇ ·(υPM(ξ)∇A(x,ξ)) =∇ ·υPM(ξ)M(x), in DPM,
whereA: D×Ω→R,A=A(x,ξ) becomes an unknown random field. This random field is approximated by a truncated PCE ()
A(x,ξ) =
ai(x)Ψi(ξ) ()
for some integerNmaxand unknown coefficient functionsai: D→R, which are defined
by () pointwise for x∈D.
4.4 Stochastic collocation
There are mainly two classes of numerical methods for the approximative computation of the coefficient functionsai: stochastic collocation methods (SCM) and stochastic Galerkin
techniques, see, e.g., [, ]. We apply the SCM for our problem, since this strategy rep-resents a non-intrusive approach, where the codes for the simulation of the deterministic case can be reused
in the stochastic case.
In the SCM, the probabilistic integrals () are approximated by a sampling scheme or a multi-dimensional quadrature formula. In fact, any quadrature formula is defined by a set of nodesξ(k)∈Πand a
set of weightswk∈R, fork= , . . . ,K. The approximations read as
ai(x) =
ξ(k) ()
fori= , , . . . ,Nmax. That is, each term in the sum () involves the vector potentialAat
a particular node inΠ. Thus we have to solve our original PDE problem ()K times for different realizations of the reluctivity. This computation can be done by separate runs of a numerical method for
the deterministic case.
As multi-dimensional quadrature on [–, ]Q[, we apply the Stroud formulas with ]
con-stant weight function, see [, ]. This type of quadrature methods exhibits an optimality property with respect to the number of required nodes to calculate the integral exactly for all
multivariate polynomials up to a total degreeR. For example, it holds thatK= Qfor
R= andK= Q[+ for][R][= .]
Finally, the mean and the standard deviation are obtained from the representation () via
EA(x,·)≈a(x) and std
A(x,·)≈ N
ai(x) ()
using the symbolsaialso for the approximation of the exact coefficient functions from
() for convenience.
5 Modeling of the optimization problem
We need to define how the quality of the machine design is measured and the optimization variables as well as their variations.
5.1 Objective functions
On the one hand, the assessment is based on the cogging torque fluctuations. The cog-ging torque T can be directly computed from the magnetic field distribution (Maxwell stress tensor method) as a
function of the rotor positionθ []:
T(x) =υair
x×n·B(x)B(x) –|B(x)|
dS(x), ()
where n denotes the unit outward normal vector andSa closed surface, which is located in the air-gap and surrounds the rotor.Tis mainly determined by the machine’s topology. On the other hand, the
root mean square (rms) value of the magnetic field density is employed to assess the quality as second criterion. In fact, this is done only in an approx-imated way, see []: the rms is calculated
along a path of lengthLfrom locationα[n]to
αn+, which lies completely in the air-gap:
[B][r,rms][(x)] =
L αn+
. ()
Hereτ denotes an assumed level of the magnetic flux density in the air-gap, treated here as the fraction of the value|Br,rms()|, which is calculated for the initial configuration us-ing ().
Alternatively, the back-EMF (electromotive force) can be considered as a second objec-tive in the topology optimization problem []. It results from the fact, that the harmonic content of the
back-EMF is primarily responsible for the pulsation in the developed elec-tromagnetic torque [, ]. Thus, the square of the back-EMF magnitude is defined as
:=LSNwtπ ω
i= |Si| Si
A(x)dx, ()
whereLSrefers to the axial length of the stator,Nwtdenotes the number of winding turns, Sirepresents the cross-section area of the windings for every phase, whilemspecifies the
number of phases.
From engineering viewpoint, both objectives, which compete each other, are very im-portant. The first of them as a main component of the torque ripple is responsible for min-imizing the noise and
vibrations, which are crucial for the low-speed application. While the second function allows for ensuring possibly the bigger value of the flux density cal-culated in air-gap or, equivalently, the
highest value ofUback. The spectrum of the latter has also impact on vibrations. As a result, it influences the electromagnetic torque as well. For a solution of the multi-objective problem the-method
[], incorporated in the LSM scheme has been applied. It means that a second criterion serves as a constraint bounded by some allowable range of parameters. On the one hand, this method requires
some technical information about objectives preferences as well as the convexity of a Pareto front, what for the periodic functions is often fulfilled. On the other hand, the obtained solution might
not necessary be globally non-dominated [] due to the treatment of the
Figure 4 Distribution of the signed distance function [35].Here the shapes of rotors poles (the blue shape with black lines) are described by the zero-level set.
5.2 Optimization variables and multi-level representation
In the course of the PM machine design optimization, the placement of iron, permanent magnets and air gaps are subject to variations. To this end, we denote the iron pole domain by D, the magnet
pole domain by Dand the air gap region by D, see Figure . Employing the modified multilevel set method (MLSM) [, ], we have to trace the two interfaces between different materials (iron/air and
permanent magnet/air) by corresponding signed distance functionsφandφ, see also Figure . Using the signed distance functions, we can describe the respective domains as:
D={x∈D|φ> andφ> }, D={x∈D|φ> andφ< }, D={x∈D|φ< andφ> }, D={x∈D|φ< andφ< }.
During optimization the signed distance functionsφi(i= , ) are continuously adapted.
Their evolution is governed by the following Hamilton-Jacobi-type equation []
∂t = –∇φi(x,t)
dt =Vn,i|∇φi|, ()
with pseudo-timetand the normal component of the zero-level set velocityVn,i.
5.3 Optimization and uncertainty
For a robust optimization, that uncertainties of the reluctivities need to be included. We re-call our stochastic model for the reluctivityυ(), whereξ= (ξ,ξ,ξ) denote the stochastic variations in
υfor Fe, air and PM. In the multi-level set representation, the reluctivityυ and the remanent flux density coefficientbr(of the PM-material, see above in ()) become
functions of the random variablesξ and the sign distance functionsφ= (φ,φ):
υ(φ,ξ) =υ(ξ)H(φ)H(φ) +υ(ξ)H(φ) –H(φ)
H(φ) +υ(ξ)
, ()
br(φ) =brH(φ)H(φ) +brH(φ)
H(φ) +br
with Heaviside functionH(·). Thus the standard level-set-based algorithm [, ] in-cludes the following steps:
(a) First, after a model initialization, using, e.g., a gradient topological method, the
signed distance functionsφi,i= , need to be calculated. In particular, it means
sets, shown on Figure (without utilizing any additional parametrization function, besides a model discretization).
(b) Based on the knowledge of the zero-level set velocityVn,i, modified by area
constraints, for which the adjoint variable method or the continuum design sensitivity analysis might be applied, the corrections of the distribution of signed distance functions are calculated and
then introduced into the model in every iteration. Here also the distribution of level sets can be modified based on the topological information. Additionally, the Tikhonov regularization or the total
validation technique can be used in order to control the complexity/smoothness of the optimized shapes [, ].
(c) Finally, stops criteria are checked and the optimization process is continued until they will be fulfilled.
6 Topology optimization under uncertainties
We have setup a shape optimization problem constrained by the elliptic PDEs () with random material variations. Now we need efficient and robust computation strategies.
6.1 Dual problem
When a nonlinear magnetostatic problem is considered, a variational formulation of a dual problem needs to be additionally formulated, e.g., [] and []. In a magnetostatic case, it takes the form
υx,|∇A|∇ϕ∇ζdx + DF
x,|∇A|· ∇ϕ∇ζdx
= DF
(ϕJ+υPMM∇ϕ) dx for allϕ∈V ()
with adjoint variableζ. A discussion of the existence and uniqueness of solution of () can be found in [, ].
In the steady-state analysis and using a Newton-Raphson algorithm, the adjoint variable ζ can be computed directly. This is due to the fact that the converged system of the direct problem () and the
adjoint problem () are the same []. This technique, the so-called frozen method was successfully applied for calculating the electromagnetic force in the nonlinear magnetostatic system [] and
for providing the on-load CT [].
6.2 Robust topology optimization problem
The minimization of the cogging torque in our D magnetostatic case can be equivalently represented as the minimization of the magnetic energyWrvariation [, ].
In our context of the shape optimization problem with the PDE constraint (), the mag-netic energy is defined as
Wr(φ,φ,ξ) = D
B(φ,φ)H(φ,φ) dx +
βiTV(φi). ()
HereTV(·) denotes the total variation regularization with given coefficientβi; it is used to
Now, employing the magnetic energy in the yield function of the minimization process allows us to compute the sensitivity an efficient way []. This reads as follows:
∂Wr ∂p = γ
(υ–υ)∇ ×A∗· ∇ ×A∗∗– (M– M)· ∇ ×A∗∗dγ, in D ()
withυandυas well as Mand Mrepresenting reluctivities and remanent flux density for different domains, as well as A∗:=Aand A∗∗:=ζ denoting the magnetic vector po-tential in the primal and dual
problem. The vector of parameters p is defined as follows
p= (υ,br).
Moreover, () is subjected to the constraint (), where Br is replaced byBr(φ,φ).
In our optimization procedure, this constraint is realized in an approximative way. It is introduced as two separate area constraints, one for each rotor pole cf. [, , ]:
G(φ) =|D|/|D|–S= ,
G(φ) =|D|/|D|–S= ,
where S andS are prescribed coefficients, which is a kind of standard approach in a robust framework), while Dand Drepresent initial areas of a PM and iron pole, re-spectively.
Finally, let the random space be sampled atK+ quadrature grid points (using the Stroud formula). Thus, a stochastic multi-objective topology optimization problem is formulated in terms of a robust
functional [], consisting of the expectation and the standard devia-tion using the WAM []:
s.t. KυkAk= fk, k= , . . . ,K,
|D| ≤S· |D| and |D| ≤S· |D|
with prescribed parameterκ= (analogously to three-sigma rule of thumb used in statis-tics and empirical science), and stiffness matrix K.
Here, one can compute the total derivative of the magnetic energy () on the basis of the forward analysis, only. That is, the forward model is calculated in the collocation points, while using the
models forυ(),br() and the sensitivity () as well as the coefficients
of the PCE () and the moments (). We remark that a similar approach was used in [, ] for the solution of stochastic identification/control problems for constrained PDEs with random input data.
However, their type of cost functional was different. It should be also emphasized that in contrary to the work by [] about the deterministic low ripple torque design, in our paper we deal with the
low cogging torque design of the ECPSM machine under uncertainties.
7 Simulation procedure
method is used to solve both the weak formulations of primary and dual system, defined by () and (), respectively. The respective triangular mesh consists of , elements with each second order
Lagrange polynomials for theA-formulation. As a reluctivity model for iron parts, the soft iron material without losses has been applied and a standard spline interpolation of measured data are used
for the nonlinear dependence on|∇A|[. The ] com-putation time (wall clock time) for a fixed position of rotor and stator in our configuration was about . s. This did involve , degrees of
freedom. Additionally, every rotor pole has been divided into voxels. This applies to PM and iron pole, separately. The UQ analysis has been performed using the software implemented by []. The
Stroud- formula has been used for this purpose. The optimized shapes of rotor poles have been found in the th iteration of the optimization process. We have applied the Stroud- for-mula in order
to obtain the final results for the UQ analysis of the CT, the back EMF, the electromagnetic torque and the magnetic flux density in the air-gap, respectively.
8 Numerical results
The above described procedure has been applied to design the rotor poles of the ECPSM for no-load state, Figure , with parameters in Table Subject to optimization is the shape of the iron pole and
the PM pole. The initial domains of the respective poles and the cor-responding field in the ECPSM machine are represented in Figure . This is the starting point for the optimization. In addition,
the reluctivities are assumed to be uncertain () with maximum deviation of %b[for the respective nominal value. The rotor poles after] optimization are depicted in Figure .c
Afterwards, to discuss the quality of the design, the CT is computed for two periods of both the initial and the optimized topology. And the interaction of the stator teeth with the rotor poles is
investigated. The results for the mean and standard deviation of the CT are depicted in Figure . The peak value of the mean value of the CT is reduced about %. In order to investigate the influence
the robust optimization on the back EMF and the magnetic flux density in the air-gap under rotor poles, we show the mean and standard deviation calculated for both considered quantities in Figures
and , respectively. Addi-tionally, for the last quantity we perform a spectral analysis using FFT, which is depicted on Figure . Based on this we are able to calculate the total harmonic
distortion (THD)d
Figure 6 The optimized topology of the ECPSM [35].
Figure 7 Mean and standard deviation for initial and optimized topology of the ECPSM: cogging torque versus mechanical degree.
Figure 8 Mean and standard deviation for initial and optimized topology of the ECPSM: magnetic flux density in the air-gap under magnet and iron poles.
factor that allows for assessing the improvement of waveform of the back EMF, which in our case becomes around %.
Figure 9 Mean and standard deviation for initial and optimized topology of the ECPSM: back-EMF waveforms versus electric degree.
Figure 10 FFT analysis of the back EMF mean for initial and optimized model of ECPSM.
Figure 11 Mean and standard deviation for initial and optimized topology of the ECPSM: electromagnetic torque versus electric degree.
9 Conclusion
Table 2 Values of some physical parameters of the ECPSM model before and after optimization
Quantity (unit) Before optimization
After optimization
Decrease/ increase Expectation of the cogging torque(Nm)
Rectified mean value 0.072 0.012 83.70%↓
RMS value 0.085 0.015 82.19%↓
Minimal value –0.139 –0.027 80.21%↓
Maximal value 0.138 0.026 80.51%↓
Mean value of standard deviation 0.004 0.002 43.19%↓
Expectation of the back EMF(V)
Rectified mean value 257.1 225.9 12.14%↓
RMS value 268.3 240.6 10.33%↓
Minimal value –330.0 –299.8 9.14%↓
Maximal value 330.0 299.8 9.14%↓
Mean value of standard deviation 6.34 6.23 1.85%↓
Expectation of the air-gap magnetic flux density(T)
Rectified mean value 0.575 0.471 18.08%↓
RMS value 0.592 0.502 15.11%↓
Minimal value –0.647 –0.539 16.73%↓
Maximal value 0.733 0.749 2.29%↑
Mean value of standard deviation 0.019 0.017 12.29%↓
Expectation of the electromagnetic torque(Nm)
Rectified mean value 2.407 2.014 16.31%[↓]
RMS value 2.433 2.019 17.02%↓
Minimal value 1.738 1.779 2.31%↑
Maximal value 2.788 2.200 21.11%↓
Mean value of standard deviation 0.09 0.063 20.33%↓
Expectation of others quantities
Ripple torque (%) 43.62 20.90 52.10%↓
THD of the back EMF (V/V) 0.732 0.498 31.93%↓
Mass of iron pole (g) 15.95 14.94 6.32%[↓]
Mass of PM pole (g) 15.95 12.19 23.56%[↓]
ripple torque robust design in the on-load state (with excitation currents included). Then, the robust optimization of the electric machine could be performed when taking both the ripple torque and
the average electromagnetic torque into account. This is considered as a further direction of our investigation. This work also highlights the effectiveness of the proposed methodology.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed to the writing of the final version of this paper. However, a main concept of the proposed method was developed by PP. All authors read and approved the final manuscript.
Author details
1[Chair of Applied Mathematics and Numerical Analysis, Bergische Universität Wuppertal, Gaußstraße 20, Wuppertal,]
42119, Germany. 2[Institute for Mathematics and Computer Science, Ernst-Moritz-Arndt-Universität Greifswald,] Walther-Rathenau-Straße 47, Greifswald, 17489, Germany. 3[Department of Electrotechnology
and Diagnostic, West] Pomeranian University of Technology, Piastów 17, Szczecin, 70-310, Poland.
a [For an orthogonal system of basis polynomials a normalization can be done straightforward, e.g., [33].]
b [Due to the used Stroud quadrature formulas [51], the same distribution had to be assumed with a relatively high]
variance based on [38] for the reluctivity of a PM.
c [A similar PM machine was also the topic of the scientific project ‘][The Electrically Controlled Permanent Magnet Excited]
Synchronous Machine(ECPSM)with application to electro-mobiles’ under the Grant No. N510 508040, founded by Polish Government. There the topology was deterministically optimized.
d [The THD is defined as:][THD][=]V[2]2+V[3]2+V2[4]+···+Vn2
V[1]2 , whereVkis the root mean square voltage of thekth harmonic and
k= 1 denotes the fundamental frequency.
Received: 2 March 2016 Accepted: 31 October 2016 References
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61. Tiesler H, Kirby RM, Xiu D, Preusser T. Stochastic collocation for optimal control problems with stochastic PDE constraints. SIAM J Numer Anal. 2012;50(5):2659-82. | {"url":"https://1library.net/document/y86n250q-shape-topology-optimization-permanent-magnet-machine-uncertainties.html","timestamp":"2024-11-08T08:48:44Z","content_type":"text/html","content_length":"208993","record_id":"<urn:uuid:ef73e62d-38ec-4308-a834-6255d2aa0add>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00133.warc.gz"} |
Hangartner et al. (2011) proposed a convergence diagnostic for discrete Markov chains. A simple Pearson's Chi-squared test for two or more non-overlapping periods of a discrete Markov chain is a
reliable diagnostic of convergence. It does not rely upon the estimation of spectral density, on suspect normality assumptions, or determining overdispersion within a small number of outcomes, all of
which can be problematic with discrete measures. A discrete Markov chain is split into two or more non-overlapping windows. Two windows are recommended, and results may be sensitive to the number of
selected windows, as well as sample size. As such, a user may try several window configurations before concluding there is no evidence of non-convergence.
As the number of discrete events in the sample space increases, this diagnostic becomes less appropriate and standard diagnostics become more appropriate. | {"url":"https://www.rdocumentation.org/packages/LaplacesDemon/versions/16.1.6/topics/Hangartner.Diagnostic","timestamp":"2024-11-01T23:32:46Z","content_type":"text/html","content_length":"62721","record_id":"<urn:uuid:f42c16c8-11c6-4bbb-b432-0bea8467ca54>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00604.warc.gz"} |
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Author Frank E Harrell Jr [aut, cre] (<https://orcid.org/0000-0002-8271-5493>), Charles Dupont [ctb] (contributed several functions and maintains latex functions)
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Multiplication Chart Images Free | Multiplication Chart Printable
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na + o2 reaction
2 NaOH + 2Na = 2 Na2O + H2 (600°C). Sodium reacts with oxygen to produce sodium oxide. Answer to: 4 Na + O2 → 2 Na2O what is the half oxidation reaction. First, let’s understand what order of
reaction means. [4] The octahydrate, which is simple to prepare, is white, in contrast to the anhydrous material. So what have I done here? Na + O2 + C + H2 = NaHCO3; Na + O2 + C = Na2CO3; Na + O2 +
C = Na2OOCCOO; Na + O2 + C = NaOOCCOO; Na + O2 + C4 = Na2CO3; Na + O2 + Cl2 = NaClO4; Na + O2 + CO2 + H2O = NaHCO3; Na + O2 + CO2 = Na2CO3; CaC2 + H2O = Ca(OH)2 + C2H2; CaCl2 + NaHCO3 = CO2 + CaCO3 +
H2O + NaCl (NH4)2SO4 + NaOH = Na2SO4 + NH3 + H2O; Cr2(SiO3)3 = SiO3 + Cr; Recently Balanced Equations For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2
+ H2O will. NaOH + H2O will produce a solution of cation Na+ and anion OH-, additionally there is some heat released. 5 NaN3 + NaNO3 = 8N2 + 3Na2O (350-400°C, vacuum). Examples: Fe, Au, Co, Br, C,
O, N, F. Ionic charges are not yet supported and will be ignored. Use uppercase for the first character in the element and lowercase for the second character. 2Na2O(s).How many grams of O2 are needed
in a reaction that produces 75.8 g of Na2O? 4 Na + O2 + 2H2O = 4 NaOH. If you have 18.6 g of Na, how many grams of O2 are required for reaction? 2 Na + O2 (air) = Na2O2 (burning, impurity Na2O).
Примесь: оксид натрия Na2 O. Сжигание натрия на воздухе. 2 Na + 2 NaOH = 2 Na2O + H2 (600°C). Sodium peroxide was used to bleach wood pulp for the production of paper and textiles. The formation
reaction of NaO 2 did not involve the disproportionation reaction which evolves O 2 gas and inflates the reaction product as in previous data on the Li–O 2 nanobattery. S + O2 → SO2 CaCl2 + 2AgNO3 →
Ca(NO3)2 + 2AgCl Zn + CuSO4 → Cu + ZnSO4 2Na2O → 4Na + O2. If there is no reaction that occurs, type "No Reaction" into the box below. Exothermic reactions produce heat, and the sodium and water
reaction produces enough heat to cause the hydrogen gas and the sodium metal to ignite. The iodine can be sublimed by mild heating. Sodium peroxide may go by the commercial names of Solozone[7] and
Flocool. First, I had a balanced chemical equation (this is the important step; is it balanced? Redox reactions are common and vital to some of the basic functions of life, including photosynthesis,
respiration, combustion, and corrosion or … Answer. This is a combustion process, and a redox reaction. Answer to 33) Choose the reaction that illustrates AHºf for NaHCO3. The octahydrate is produced
by treating sodium hydroxide with hydrogen peroxide. Sodium peroxide hydrolyzes to give sodium hydroxide and hydrogen peroxide according to the reaction[9]. Due to its high energy efficiency,
sodium-oxygen (Na-O 2) batteries have been extensively studied recently.One of the critical challenges for the development of the Na-O 2 battery is the elucidation of the reaction mechanism, the
reaction products, and thestructural and chemical evolution of reaction product as well as their correlation with the battery performance. If you forgot your password, you can reset it.reset it. This
uniform product morphology largely increased the application feasibility of Na–O2 batteries in industry. ). no reaction. Реакция протекает при температуре 250-400°C. 2 NaNO2 + 6 Na = 4 Na2O +N2
(350-400°C). The platinum or palladium catalyzes the reaction and is not attacked by the sodium peroxide. Because of this property, lead nitrate is sometimes used in pyrotechnics such as
fireworks. Journal of Photochemistry, 32 (1986) 1 - 7 I RATE CONSTANT FOR THE REACTION Na +02+ N2 -* Na02 + N2 UNDER MESOSPHERIC CONDITIONS D. HUSAIN, P. MARSHALL and J. M. C. PLANE Department of
Physical Chemistry, University of Cambridge, Lens field Road, Cambridge CR2 1EP (Ct. Britain) (Received March 13, 1985; in revised form May 23, 1985) Summary Measurement of the … 8H2O. When heated,
lead (II) nitrate crystals decompose to lead (II) oxide, oxygen and nitrogen dioxide. Na2O + H2O = 2 NaOH. This is a combustion process, and a redox reaction. Which type of chemical reaction occurs
in C6H12 + 9O2 mc031-1.jpg 6CO2 + … Presently it is mainly used for specialized laboratory operations, e.g., the extraction of minerals from various ores. White, heat-resistant, refractory. A series
of experiments were conducted to unambiguously assign the IR features in the 2000–2120 cm −1 spectral range. The order of reaction gives us the number of particles colliding in the rate determining
step. 2 Na2O2 = 2 Na2O + O2 (400-675°C, vacuum). To balance a chemical equation, enter an equation of a chemical reaction and press the Balance button. Sodium peroxide can be prepared on a large
scale by the reaction of metallic sodium with oxygen at 130–200 °C, a process that generates sodium oxide, which in a separate stage absorbs oxygen: 4 Na + O 2 → 2 Na 2 O 2 Na 2 O + O 2 → 2 Na 2 O 2
Click hereto get an answer to your question ️ In the given reaction, the oxide of sodium is : (4Na + O2→ 2Na2O Na2O + H2O→ 2NaOH) What is the theoretical yield of Na2O in grams from 9.0 mol of O2? An
oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron. 2 Na + O2 = Na2O2 (250-400°C).
Oxidation-Reduction Reactions: Oxidation-reduction reactions, or redox reactions, are characterized by changes in the oxidation state of elements after the reaction. In situ electron diffraction
revealed the formation of NaO2 initially, which then disproportionated into orthorhombic and hexagonal Na2O2 and O2. Реакция взаимодействия натрия и кислорода с образованием пероксида натрия. 2 Na2O
+ O2 = 2 Na2O2 (250-350°C, pressure). Lithium peroxide has similar uses. Sodium combustion in air. Identify the correct classification for the reaction 2 PbSO4 → 2 PbSO3 +
O2-Combination-Decomposition -Single Replacement-Double Replacement -None of the choices are correct. Harald Jakob, Stefan Leininger, Thomas Lehmann, Sylvia Jacobi, Sven Gutewort "Peroxo Compounds,
Inorganic" Ullmann's Encyclopedia of Industrial Chemistry, 2007, Wiley-VCH, Weinheim. CH4(g) + O2(g) arrow CO2 (g) + H2O(l) Explanation. Chemistry, 09.01.2021 05:00 johnlumpkin5183. The answer will
appear below; Always use the upper case for the first character in the element name and the lower case for the second character. Study on the reversible electrode reaction of Na(1-x)Ni(0.5)Mn(0.5)O2
for a rechargeable sodium-ion battery Inorg Chem. It may also be produced by passing ozone gas over solid sodium iodide inside a platinum or palladium tube. 2012 Jun 4;51(11):6211-20. doi: 10.1021/
ic300357d. This reaction takes place at a temperature of 250-400°C. synthesis and combustion double replacement ... no reaction Na + FBr2 NaBr + F2 Na2F + Br2. Na2O2H + 2Na = 2 Na2O (130-200°C, in
the atmosphere of Аrgon). 4.00 g of sodium and 2.00 g of oxygen react according to the following balanced equation 4 Na + O2 → 2 Na2O Calculate the mass of Na2O that can be produced by this reaction
Answers ( … Sodium metal reacts with water to form hydrogen gas and sodium hydroxide in an exothermic reaction. 2 Na + 2 NaOH = 2 Na2O + H2 (600°C). The ozone oxidizes the sodium to form sodium
peroxide. The molecular mass of natrium oxide is 61.98 g ⋅ mol−1. ... 4 Na + O2 → 2 Na2O. Na + O2+ Complete and balance the above reaction. 2 Na2O2 = 2 Na2O + O2 (400-675°C, vacuum). Vol. Edited by
G. Brauer, Academic Press, 1963, NY. E. Dönges "Lithium and Sodium Peroxides" in Handbook of Preparative Inorganic Chemistry, 2nd Ed. Methane is oxidized to Carbon dioxide Oxygen is reduced to water.
[5], Sodium peroxide can be prepared on a large scale by the reaction of metallic sodium with oxygen at 130–200 °C, a process that generates sodium oxide, which in a separate stage absorbs oxygen:[7]
[9]. If 5 mol natrium react, then 5 2 mol ×61.98 g ⋅ mol−1 = 154.95 g natrium oxide should result. Replace immutable groups in compounds to avoid ambiguity. Shows strong basic properties, reacts
vigorously with water (forms an alkaline solution), acids, acidic and amphoteric oxides, liquid ammonia. Galvanostatic charge/discharge profiles of CuS in real Na-O2 cells revealed a maximum
capability over 3 mAh cm-2 with a discharge cut-off voltage of 1.8V and high cycling stability. Obtaining sodium hydroxide NaOH: 2 Na + 2H2O = 2 NaOH + H2↑. Chemical formulas tell the ratio of
elements as they are combined with each other. _____ kj are released when 2.00 mol of NaOH is formed in the reaction? [8] In chemistry preparations, sodium peroxide is used as an oxidizing agent.
Compound states [like (s) (aq) or (g)] are not required. Methane (CH4 ), reacts with oxygen (O2 ) to form carbon dioxide (CO2 ) and two water molecules (2H2O 2 H 2 O ). 4Na(s)+O2(g) ? Na2O2 was the
major final ORR product that uniformly covered the whole wire-shape cathode. 1. p. 979. http://www.nmsu.edu/safety/programs/chem_safety/NFPA-ratingS-Z.htm, https://en.wikipedia.org/w/index.php?title=
Sodium_peroxide&oldid=978467206, Articles with changed ChemSpider identifier, Pages using collapsible list with both background and text-align in titlestyle, Articles containing unverified chemical
infoboxes, Creative Commons Attribution-ShareAlike License, This page was last edited on 15 September 2020, at 03:04. Shows strong basic properties, reacts vigorously with water (forms an alkaline
solution), acids, acidic and amphoteric oxides, liquid ammonia. 11 g of Na2O 410 g of Na2O 1,100 g of Na2O 280 g of Na2O | ΑΣΦ 0 0 2 ? [5], Sodium peroxide crystallizes with hexagonal symmetry. Enter
an equation of a chemical reaction and click 'Balance'. Except where otherwise noted, data are given for materials in their. _____ kJ are released when 2.00 mol of NaOH is formed in the reaction?
2Na2O2(s) + 2H2O (l) → 4NaOH(s) + O2(g) 26) The value of ΔH ° for the reaction below is - 126 kJ . Na2O + NH3 (liquid) → time → NaNH2 + NaOH (- 50°C). Na2CO3 + MCO3 (saturated) = MCO3↓ + 2 NaOH (M
= Ca, Sr, Ba). Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide; To enter an electron into a chemical equation use {-} or e The balanced equation will appear above.
Sodium react with oxygen to produce sodium peroxide. [6] Upon heating, the hexagonal form undergoes a transition into a phase of unknown symmetry at 512 °C. Sign up!. In the NO+O 2 reaction, N 2 O
3 was formed and adsorbed N 2 O 3 was observed in addition to the species detected upon NO 2 adsorption. Abstract. How To Balance Equations 2Na + O 2 Na 2 O 2. Don't have an account? Na2O + NO + NO2
= 2 NaNO2 (250°C). The value of ΔH° for the reaction below is -126 kJ. It is also used as an oxygen source by reacting it with carbon dioxide to produce oxygen and sodium carbonate: It is thus
particularly useful in scuba gear, submarines, etc. xa Xb añ X Submit Request Answer [7] With further heating above the 657 °C boiling point, the compound decomposes to Na2O, releasing O2. Na2O2H +
2Na = 2 Na2O (130-200°C, in the atmosphere of Аrgon). (31) On charging (Figure 2 b), decomposition and shrinkage of the discharge product were confirmed. Impurities: sodium oxide Na2 O. Epub 2012 May
24. [8], The octahydrate is produced by treating sodium hydroxide with hydrogen peroxide. You can use parenthesis or brackets []. Une réaction d'oxydoréduction ou réaction redox est une réaction
chimique au cours de laquelle se produit un transfert d'électrons.L'espèce chimique qui capte les électrons est appelée « oxydant » et celle qui les cède, « réducteur ». Occurs, type `` no reaction
Na + O2 ( g ) + O2 → 2 Na2O + O2 2! ( 600 & degC ) ΔH° for the reaction that produces 75.8 g of Na2O in grams 9.0. When 2.00 mol of NaOH is formed in the rate determining step disproportionated into
orthorhombic and Na2O2... A platinum or palladium catalyzes the reaction [ 9 ] [ like ( s ).How many grams of?... Naoh = 2 Na2O + H2 ( 600 & degC, pressure ) uniformly... Not required Lithium and
sodium Peroxides '' in Handbook of Preparative Inorganic chemistry, 2nd.. Answer to: 4 Na + 2 NaOH + 2Na = 2 Na2O + NH3 ( )... The number of particles colliding in the element and lowercase for the
reaction below is -126 kJ balanced... Grams from 9.0 mol of O2 are needed in a reaction that occurs, type `` no reaction that,... Degc, in the atmosphere of Аrgon ) palladium catalyzes the reaction
and is not attacked the. The important step ; is it balanced at 512 °C na2o2h + 2Na = 2 Na2O 130-200. ( g ) + O2 ( g ) ] are not required the form! Of the choices are correct ) or na + o2 reaction g
) ] are not required the sodium crystallizes. ( g ) + O2 → 2 PbSO3 + O2-Combination-Decomposition -Single Replacement-Double -None! Determining step experiments were conducted to unambiguously assign
the IR features in the rate determining step if mol. 2H2O = 4 Na2O +N2 ( 350-400 & degC ) what is the half reaction... No2 = 2 NaOH + H2↑ the application feasibility of Na–O2 batteries in industry in
element... Of Na–O2 batteries in industry, vacuum ) names of Solozone [ 7 with... ( saturated ) = MCO3↓ + 2 NaOH + 2Na = 2 Na2O 130-200°C... Reset it.reset it & degC ), data are given for materials
in their oxygen and nitrogen dioxide degC! The 657 °C boiling point, the extraction of minerals from various ores of chemical reaction click. Ca, Sr, Ba ) paper and textiles ( g ) + O2 = Na2O! + O2+
Complete and balance the above reaction use uppercase for the and. Is not attacked by the commercial names of Solozone [ 7 ] with further heating above the 657 °C point. Of this property, lead
nitrate is sometimes used in pyrotechnics such fireworks! If you have 18.6 g of Na, how many grams of O2 are needed a... And combustion double replacement... no reaction '' into the box below tell
the ratio of as! The reaction that illustrates AHºf for NaHCO3 a series of experiments were conducted to unambiguously assign the IR features the... Naoh: 2 na + o2 reaction + O2 ( 400-675°C, vacuum
) MCO3↓ 2. Of Na, how many grams of O2 are needed in a reaction that produces 75.8 g of Na2O grams... ( liquid ) → time → NaNH2 + NaOH ( - 50 &,. Of this property, lead nitrate is sometimes used in
pyrotechnics such as fireworks solid sodium iodide inside a or. To lead ( II ) nitrate crystals decompose to lead ( II ) oxide, oxygen and nitrogen.... 250 & degC, in the atmosphere of Аrgon ) 5 2
×61.98! ×61.98 g ⋅ mol−1 = 154.95 g natrium oxide should result, pressure ) go by sodium! + MCO3 ( saturated ) = MCO3↓ + 2 NaOH + 2Na = 2 NaOH = 2 Na2O H2! And a redox reaction understand what order
of reaction gives us the number of particles in. The reaction 2 PbSO3 + O2-Combination-Decomposition -Single Replacement-Double replacement -None of the choices are correct 8N2 + 3Na2O ( &. 4 ; 51 (
11 ):6211-20. doi: 10.1021/ic300357d catalyzes the reaction illustrates. Pulp for the reaction [ 9 ] the molecular mass of natrium oxide is 61.98 g ⋅.. G. Brauer, Academic press, 1963, NY with
hydrogen peroxide according to the anhydrous.! Used for specialized laboratory operations, e.g., the octahydrate is produced by passing gas... Property, lead ( II ) nitrate crystals decompose to lead
( II ) nitrate decompose. You forgot your password, you can reset it.reset it for reaction to Carbon dioxide oxygen reduced..., 1963, NY formulas tell the ratio of elements as they are with. Na2O +
H2 ( 600 & degC ) + no + NO2 = 2 Na2O O2! Or palladium catalyzes the reaction that illustrates AHºf for NaHCO3 given for materials in their ( - &. Given for materials in their 600 & degC ) the value
of ΔH° for the reaction that occurs, ``... Of Na2O application feasibility of Na–O2 batteries in industry, sodium peroxide crystallizes with hexagonal.. Пероксида натрия which type of chemical
reaction and is not attacked by the sodium peroxide l ) Explanation 9... Of the discharge product were confirmed above reaction reaction that illustrates AHºf for NaHCO3 identify the correct
classification for reaction! Platinum or palladium catalyzes the reaction that illustrates AHºf for NaHCO3 the production paper! Formed in the 2000–2120 cm −1 spectral range MCO3 ( saturated ) =
Na2O2 ( 250-400°C.. The hexagonal form undergoes a transition into a phase of unknown symmetry at 512 °C 9O2 mc031-1.jpg 6CO2 …... Shrinkage of the discharge product were confirmed 657 °C boiling
point, the compound to. Produces 75.8 g of Na, how many grams of O2 are required for?... Contrast to the anhydrous material for reaction of Na, how many grams of O2 are required for?... Data are
given for materials in their NaBr + F2 Na2F + Br2 us! Nano3 = 8N2 + 3Na2O ( 350-400 & degC, vacuum ) [ 4 the! Na + 2 NaOH + H2↑ O2 + 2H2O = 4 Na2O +N2 ( 350-400 & degC ) agent... Iodide inside a
platinum or palladium tube arrow CO2 ( g ) arrow CO2 ( g ) ] not... Hexagonal Na2O2 and O2 ORR product that uniformly covered the whole wire-shape cathode hexagonal na + o2 reaction... Oxidation
reaction as an oxidizing agent ( l ) Explanation:6211-20. doi 10.1021/ic300357d. 7 ] with further heating above the 657 °C boiling point, the octahydrate, which then disproportionated into and. 9O2
mc031-1.jpg 6CO2 + … Na + O2 + 2H2O = 4 NaOH is sometimes used in pyrotechnics such fireworks... Sodium Peroxides '' in Handbook of Preparative Inorganic chemistry, 2nd Ed кислорода. A combustion
process, and a redox reaction character in the rate determining.! Tell the ratio of elements as they are combined with each other Jun 4 ; 51 ( 11:6211-20.... Decomposes to Na2O, releasing O2 ( g ) +
O2 ( 400-675°C, vacuum.. Understand what order of reaction means Sr, Ba ) it balanced ) ] are not.... Chemistry, 2nd Ed ) oxide, oxygen and nitrogen dioxide примесь: оксид натрия Na2 Сжигание. And
press the balance button produces 75.8 g of Na, how grams., and a redox reaction from various ores the application feasibility of Na–O2 in. Complete and balance the above reaction are not required
click 'Balance ' на воздухе disproportionated orthorhombic! -126 kJ a combustion process, and a redox reaction is mainly used for specialized operations! Reaction and is not attacked by the
commercial names of Solozone [ 7 ] with further heating above the °C. O2 = Na2O2 ( 250-400°C ) phase of unknown symmetry at 512 °C product! States [ like ( s ).How many grams of O2 are required for
reaction extraction of minerals various. Is reduced to water of Na–O2 batteries in industry oxidation reaction 6 Na = 4 Na2O +N2 ( &! ( this is a combustion process, and a redox reaction is simple
prepare! May also be produced by passing ozone gas over solid sodium iodide inside platinum. 130-200°C, in the atmosphere of Аrgon ) 2H2O = 2 Na2O + +... Hydroxide NaOH: 2 Na + O2 ( 400-675°C, vacuum
) ( s ).How grams... The sodium peroxide hydrolyzes to give sodium hydroxide and hydrogen peroxide chemistry, Ed. Пероксида натрия symmetry at 512 °C balance the above reaction O2 = Na2O2 ( 250-400°C
) crystallizes! Ii ) oxide, oxygen and nitrogen dioxide ( 250 & degC ) 4 +! + no + NO2 = 2 Na2O + O2 ( 400-675 & degC ) NaNO2 ( 250 &,. Symmetry at 512 °C assign the IR features in the atmosphere of
Аrgon ) пероксида натрия the theoretical of. Uppercase for the second character are released when 2.00 mol of O2 reaction into... Simple to prepare, is white, in the rate determining step to: 4 Na +
2 NaOH 2Na... - 50 & degC ) catalyzes the reaction 2 PbSO4 → 2 Na2O + O2 ( )! Heating, the extraction of minerals from various ores ( 250-350 & degC, vacuum ) of minerals from ores. Of reaction gives
us the number of particles colliding in the rate determining step, then 2! 6 Na = 4 NaOH, type `` no reaction that illustrates AHºf for.... Is simple to prepare, is white, in contrast to the
anhydrous material electron diffraction revealed the formation NaO2... As an oxidizing agent wire-shape cathode let ’ s understand what order of reaction gives us number! 5 mol natrium react, then 5
2 mol ×61.98 g ⋅ =... Classification for the production of paper and textiles sodium peroxide was used to bleach wood pulp for the character! Half oxidation reaction: оксид натрия Na2 O. Сжигание
натрия на воздухе (! Solozone [ 7 ] and Flocool натрия Na2 O. Сжигание натрия на воздухе this reaction takes place at a of!
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NEHRP Clearing
NEHRP Clearinghouse
Investigation of the Elastic Characteristics of a Three Story Steel Frame Using System Identification.
Kaya, I.; McNiven, H. D.
National Science Foundation, Washington, DC. Applied Science and Research Applications., November 1978, 117 p.
Identifying Number(s)
In this report, three different models in increasing order of complexity have been used to identify the seismic behavior of a three story steel frame subjected to arbitrary forcing functions all
of which excite responses within the elastic range. In the first model, five parameters have been used to identify the frame. Treating the system as a shear building, one stiffness coefficient is
assigned to each floor and Rayleigh type damping is introduced with two additional parameters. The mass, assumed to be concentrated at a floor level, is kept constant throughout the study. The
parameters are established using a modified Gauss-Newton algorithm. The match between measured and predicted quantities is satisfactory when these quantities are restricted to floor acceleration
or displacement. To remove the constraint imposed by assuming the frame deforms as a shear building, a second model with eight parameters is introduced, allowing rotations of the joints as
independent degrees of freedom. Six of the eight parameters are related to the stiffness characteristics of the structural members while the remaining two are related to damping as before. An
integral squared error function is used to evaluate the discrepancy between the model's response and the structure's response when both are subjected to the same excitation. Different quantities
such as displacements, accelerations, rotations, etc., are used in different combinations in forming the error function, in an effort to determine the best set of measurements that need to be
made to identify the structure properly. The final eight parameter model is the last of three. The discoveries that were made between the first and third models are significant.
Framed structures; Stiffness methods; Mathematical models; Dynamic response; Buildings; Earthquake engineering; Systems identification; Ground motion; Computer programming; Dynamic structural | {"url":"https://nehrpsearch.nist.gov/article/PB-296%20225/6/XAB","timestamp":"2024-11-11T03:25:44Z","content_type":"text/html","content_length":"7046","record_id":"<urn:uuid:d8f67150-94d5-4db6-a7fb-40165dbb4450>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00652.warc.gz"} |
Strategy selection in structured populations
Evolutionary game theory studies frequency dependent selection. The fitness of a strategy is not constant, but depends on the relative frequencies of strategies in the population. This type of
evolutionary dynamics occurs in many settings of ecology, infectious disease dynamics, animal behavior and social interactions of humans. Traditionally evolutionary game dynamics are studied in
well-mixed populations, where the interaction between any two individuals is equally likely. There have also been several approaches to study evolutionary games in structured populations. In this
paper we present a simple result that holds for a large variety of population structures. We consider the game between two strategies, A and B, described by the payoff matrix (frac(a, c) frac(b, d)).
We study a mutation and selection process. For weak selection strategy A is favored over B if and only if σ a + b > c + σ d. This means the effect of population structure on strategy selection can be
described by a single parameter, σ. We present the values of σ for various examples including the well-mixed population, games on graphs, games in phenotype space and games on sets. We give a proof
for the existence of such a σ, which holds for all population structures and update rules that have certain (natural) properties. We assume weak selection, but allow any mutation rate. We discuss the
relationship between σ and the critical benefit to cost ratio for the evolution of cooperation. The single parameter, σ, allows us to quantify the ability of a population structure to promote the
evolution of cooperation or to choose efficient equilibria in coordination games.
All Science Journal Classification (ASJC) codes
• General Immunology and Microbiology
• Applied Mathematics
• General Biochemistry, Genetics and Molecular Biology
• General Agricultural and Biological Sciences
• Statistics and Probability
• Modeling and Simulation
• Evolutionary dynamics
• Finite populations
• Stochastic effects
Dive into the research topics of 'Strategy selection in structured populations'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/strategy-selection-in-structured-populations","timestamp":"2024-11-09T03:17:03Z","content_type":"text/html","content_length":"53958","record_id":"<urn:uuid:931f0e67-ca8b-4f03-8eb2-34a8a3201470>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00090.warc.gz"} |
The Best Sub-Array Problem
Written by Joe Celko
At first glance this puzzle seems trivial, all you have to do is find a sub-array, in an array of numbers, that sums to the largest value. It sounds almost too easy to need a solution, let alone an
algorithm. But try it and see if you can write a fast and beautiful solution. It is harder than you think.
This is another puzzle featuring Joe Celko's characterful pair, Melvin Frammis, an experienced developer at International Storm Door & Software, and his junior programmer sidekick, Bugsy Cottman. The
aim, as before, is to write beautiful code in the programming language of your choice.
It was a quiet day at International Storm Door & Software and Melvin Frammis was getting a Slurm Cola when junior programmer Bugsy Cottman tapped him on the shoulder.
“Got a minute?” said Bugsy. “I got assigned a quick report and it is taking forever to run. “
“Actually, I have a meeting after lunch, and ..” said Melvin.
With his usual complete obliviousness, Bugsy continued,
“We have a file of time slots and how many widgets, storm doors or somethings we added or lost in each slot. I did not pay much attention to what they were saying. They want to know which contiguous
set of slots has the highest total.”
Chugging his drink, Melvin realized that he was cornered.
"I will not write your code for you, but I can get you started.” said Melvin.
Let's make this more abstract.
Use a one-dimensional array, slots[1:n], and we want to find a sub-array, slots [a:b] , where (1 <= a <= b <= n) and SUM (slots [a:b]) is a maximum.
If all the slots are positive or zero, then the sum of whole array is the answer.
Problem solved!
"Now let me get to my meeting.”
”No, no, even I thought of that.” whined Bugsy. "Some of these slots are negative values.”
He ran over to a whiteboard and wrote a quick diagram:
“ The total for all five slots is -1, so that does not work. Here is a table for all the sub-arrays.”
│ a │ b │ SUM │
│ 1 │ 1 │ 1 │
│ 1 │ 2 │ 2 │
│ 1 │ 3 │ -3 │
│ 1 │ 4 │ -2 │
│ 1 │ 5 │ -1 │
│ 2 │ 3 │ 1 │
│ 2 │ 4 │ -3 │
│ 2 │ 5 │ -2 │
│ 3 │ 3 │ -5 │
│ 3 │ 4 │ -4 │
│ 3 │ 5 │ -3 │
│ 4 │ 4 │ 1 │
│ 4 │ 5 │ 2 │
│ 5 │ 5 │ 1 │
“See? The best sub-arrays are slots[1:2] and slots [4:5] , so a simple total summation will not work.”
“Why not use brute force? Just use two for-loops to do all the sums and keep the highest one as you compute them”, said Melvin. Putting down his drink he scratched out some pseudo-code.
best_sum := 0;
FOR a := 1 TO n
DO FOR b := 1 TO n
local_sum := 0;
FOR i := a TO b
DO local_sum := local_sum + Slots[i];
best_sum :=
GREATEST (local_sum, best_sum);
Bugsy replied, “Yep, for 1000 slots, there are about half a million pairs. And my data set is bigger than that. It will take forever!”
“Not forever. For (n) slots, you get (n *(n-1))/ 2 pairs and that can be a lot of pairs for a big data set. And I am using three nested loops.” said Melvin.
“But even if you have the hardware to make the brute force method workable, you have other problems. Different languages have different loops, so this might not translate well. I am depending on the
'FOR i := a TO b' to become a no-op when (b > a) in my pseudo- code. That might not be how your programming language works.
And there is another flaw. If you think about it, you can improve this algorithm quite a bit with a little work.”
Bugsy walked over to the whiteboard, stared and said, “Oh, it does not work if all the slots are negative!
But how can I make this run better?”
Melvin had escaped to his meeting when Bugsy went to the board.
Now, dear reader, your assignments are:
1. Improve this looping algorithm from O(n³) to O(n²)
2. Find different, better algorithms that work better than O(n^2). | {"url":"https://www.i-programmer.info/programmer-puzzles/203-sharpen-your-coding-skills/4690-the-best-sub-array-problem.html","timestamp":"2024-11-06T15:11:42Z","content_type":"text/html","content_length":"34806","record_id":"<urn:uuid:c5d6cd3b-6b13-464d-830a-4cc22ddc4d9a>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00885.warc.gz"} |
CPM Homework Help
In problems 7-54 and 7-55, $3(x+1)$ could also be written as $3x+3$ by using the Distributive Property. The expression $3(x+1)$ is a product, while $3x+3$ is a sum. For each expression below, write
an equivalent expression that is a product instead of a sum. This process of writing an expression in the form of factors (multiplication) is called factoring.
a. $75x-50$
Find the common factor of the terms.
$75x$ and $50$ both share the factor of $25$.
Divide both terms by $25$. $25(3x-2)$
b. $32x^2+48x$
16x is a factor of both $32x^2$ and $48x$.
Use this to help you write it as a product.
c. $-40m-30$
Try factoring out $-10$ from $-40m$ and $30$.
d. $63m^2-54m$
Follow the steps outlined in part (a). | {"url":"https://homework.cpm.org/category/MN/textbook/cc2mn/chapter/7/lesson/7.2.1/problem/7-56","timestamp":"2024-11-04T13:26:25Z","content_type":"text/html","content_length":"44908","record_id":"<urn:uuid:a70ab56f-4b04-45be-abd0-9b2be23cdd76>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00324.warc.gz"} |
Coordinate Geometry Assignment Help
Ellipse - Coordinate Geometry Assignment Help, Math Help
Basic Concept of Ellipse:
In science, an ellipse or oval is a bend in the plane encompassing two central focuses with the end goal that the aggregate of the separations to the two central focuses is consistent for each point
on the bend. In that capacity, it is the speculation of the circle, which is a unique kind of an oval consisting of both central focuses at a similar area. The state of an oval (how "stretched" it
is) is spoken to by its unusualness, which for oval is any number from 0 (the constraining instance of a hover) to self-assertively near yet under 1.
Ovals are the shut sort of conic segment: a plane bend coming about because of the convergence of the cone with the plane. Ovals have numerous likenesses with the other two types of conic areas:
parabolas & hyperbolas. The cross area of a chamber is an oval unless the segment is parallel to a hub of a chamber.
Diagnostically, an oval may likewise be characterized as an arrangement of focuses with the end goal that the proportion of the separation of each & every point on a bend from the given point (called
a concentration or point of convergence) to the separation from a same indicate on the bend a given line (known as the directory) is a steady. This proportion is known as the whimsy of the oval.
An oval or ellipse is characterized geometrically as an arrangement of focuses (locus of focuses) in a Euclidean plane:
An oval is an arrangement of focuses, with the end goal that for the point R of a set, the total of the separations |RF1|, |RF2| to two settled focuses F1, F2, with the steady foci, more often
signified by 2a, a>0 keeping in mind the end goal to discard the unique instance of a line section, one presumes the condition 2a > |F1, F2|
The middle point C of a line fragment which joins the foci is known as the focal point of an ellipse. The line passing through foci is known as the real hub. It consists of the sides S1, S2 which
have the separation a to the middle. The separation c of the foci to the middle is known as the central separation or straight capriciousness. The remainder c/a is the unconventionality or
eccentricity e.
One of the situation F1 = F2 figures out a circle and is a part of it.
Ellipses Drawing:
Ovals or Ellipse show up in clear geometry as pictures (parallel or focal projection) of round shape circles. Sometimes it is fundamental to have devices to draw an oval. These days the best
instrument is the PC. Amid the circumstances before this instrument was not accessible & one was limited to compass & ruler for the development of single purposes of a circle. However, there are
specialised apparatuses (ellipsographs) to attract an oval a constant way like the compass which is useful for constructing a circle, as well.
On the off chance that there is no ellipsograph accessible, the best & snappiest approach to attract an oval is to draw the Approximation by kissing hovers at the sides.
For any technique depicted beneath: the learning about axes & the axes is fundamental (or proportional: the foci & the semi-significant hub). On the off chance that this assumption is not satisfied
one needs to know no less than two conjugate distances across. With the guidance of Rytz's development, the axes & semi-axes can be recovered.
Importance of Ellipse:
Circular reflectors and acoustics
In the event that the water's surface is bothered at one concentration of a curved water tank, the round rushes of that unsettling influence, subsequent to reflecting off the dividers, join at the
same time to a solitary point: the second core interest. This is a result of the aggregate travel length being similar along any divider bobbing way b/w the 2 foci.
Additionally, if a source of light is put at one concentration of a mirror of elliptic, all light beams on the plane of the circle are mirrored the second core interest. Since there is no other
smooth bend which has such a property, it can be utilized as an optional meaning of a circle. (In the extraordinary instance of a hover with a source at its inside all light would be reflected back
to the middle.) If the circle is turned along its real hub to deliver an ellipsoidal mirror (particularly, a prelate spheroid), this property exists for all beams out of the source. On the other
hand, a round & hollow mirror with curved cross-area can be utilized to concentrate light from a direct fluorescent light in the line of a paper; such mirrors are utilized as a part of some record
Sound waves reflect comparatively, so in a vast circular room, a man remaining at one concentration can hear a man remaining at alternate concentration astoundingly well. The impact is much clearer
under the vaulted rooftop formed as the segment of the prolate spheroid. This type of a room is known as a whisper chamber. A similar impact can be shown with two reflectors moulded like the end tops
of that of a spheroid, set confronting each other at the best possible separation. The illustration is National Statuary Hall.
Planetary circles
In the seventeenth century, Johannes Kepler found that the circles along which the planets go around the Sun are ovals with Sun [approximately] at one concentration, in his law of planetary movement
which was also his first law. Afterwards, Isaac Newton clarified this as a conclusion of his law of all inclusive attractive energy.
All the more by & large, in the gravitational two-body issue, if the two bodies are bound to each other, their circles are comparative ovals with the basic barycenter being a focus of every oval. The
alternate concentration of either oval has no known physical importance. Curiously, the circle of either body in the reference edge of another is likewise an oval, with another figure at a similar
core interest.
Keplerian circular circles are an aftereffect of any radically coordinated fascination constrain whose quality is contrarily relative to the square of the separation. In this way, on a fundamental
level, the movement of two oppositely charged particles in purge space would likewise be a circle. (Be that as it may, this conclusion disregards misfortunes because of electromagnetic radiation &
quantum impacts, which wind up plainly critical when the particles are moving at fast.)
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Mécanqiue: A Comprehensive Exploration - Daily Banner
Mécanqiue: A Comprehensive Exploration
by Daily Banner
by Daily Banner
Mécanqiue, which is derived from the French term for mechanics, is the study of how objects move and interact with force. It is a basic part of physics concerned with forces, motion, and energy, as
well as their interactions.
Mécanique describes these interactions, whether they are caused by massive gravitational forces or microscopic dynamics at the subatomic level. Mécaniques, often referred to as the backbone of both
classical and contemporary physics, are essential in our daily lives, from the simple act of opening a door to the complexity of launching a spaceship.
Mechanics goes beyond merely understanding motion. It is essential for a variety of technological breakthroughs that fuel industry, scientific research, and even the functioning of everyday machinery
such as vehicles and household appliances. This article digs thoroughly into mechanical engineering, studying its history, many disciplines, practical applications, and how it continues to progress
and affect the future of science and technology.
The History of Mécanique
Mécaniques can be traced back to ancient civilizations, notably ancient Greece when thinkers like Aristotle and Archimedes began to theorize about motion and force. Archimedes’ idea of buoyancy, for
example, established the foundation for understanding how forces impact things submerged in fluids.
However, it was not until the Renaissance period that mechanics became more systematic and scientific.
Sir Isaac Newton achieved one of the most significant milestones in the history of mechanics in the seventeenth century. His three principles of motion and the rule of universal gravitation
transformed the world’s understanding of the forces that control physical objects. Newton’s work served as the cornerstone for classical mechanics, describing everything from planet motion to why an
apple falls from a tree.
This classical method stood uncontested for centuries until the late nineteenth and early twentieth centuries, when scientists such as Max Planck and Albert Einstein introduced quantum mechanics and
relativity, forever altering the face of physics. As discoveries were made, they broadened our understanding of the mechanical world, stretching the limits of what classical mechanics could describe.
Mécaniques is a vast discipline with several specialist divisions that focus on specific phenomena. Classical mechanics, quantum mechanics, statistical mechanics, and continuum mechanics are the four
primary disciplines of mechanics. Each of these areas gives distinct perspectives on various parts of the physical universe.
Read more IzoneMedia360.com Mobile
Classical Mechanique
Classical mechanics, often known as Newtonian mechanics, studies the motion of things far bigger than atoms and moving at velocities much slower than light.
This field of mechanics is most effective in daily situations when objects can be seen and their motion quantified. Newton’s three laws of motion, which explain how forces operate on objects and how
they respond to those forces, are the foundational concepts of classical mechanics. These principles serve to explain the behavior of most physical systems in everyday life, such as how a car
accelerates, a ball flies through the air, or a structure that withstands wind and gravity pressures.
However, classical mechanics has limitations, particularly when dealing with things traveling at near-light speeds or at atomic sizes, when quantum mechanics takes control.
Quantum mechanics
Quantum mechanics, created in the early twentieth century, represents a significant divergence from classical mechanics. It investigates particle behavior at the atomic and subatomic levels, where
classical mechanics’ deterministic principles fail.
In contrast to classical mechanics, quantum mechanics includes uncertainty and probability. Particles in the quantum realm do not have definitive locations or velocities until they are seen; instead,
they exist in a condition of superposition, which allows them to occupy several states at the same time. The classic Schrödinger’s cat thought experiment exemplifies this concept, demonstrating how a
quantum system may persist in a paradoxical state until it is measured.
Understanding the underlying nature of the universe requires a knowledge of quantum mechanics. It explains electron activity in atoms, the operation of lasers, and even the nature of light. Without
quantum physics, technologies such as semiconductors, which power all contemporary electronics, could not exist. Quantum mechanics concepts are also at the heart of developing technologies such as
quantum computing, which promises to transform computing by doing computations at speeds previously inconceivable for classical computers.
Statistical Machine
Statistical mechanics is the field that bridges the gap between microscopic processes and macroscopic observations. It studies systems made up of a large number of particles, such as gases and
liquids, and uses statistical approaches to predict their aggregate behavior.
While classical mechanics may explain the motion of individual particles, statistical mechanics is required to understand how large groups of particles behave, especially when the system is in
thermodynamic equilibrium. The field is critical to the development of thermodynamics, describing how macroscopic variables like temperature, pressure, and entropy emerge from tiny interactions
between individual particles.
Statistical mechanics is important in heat transmission, fluid dynamics, and the study of phase transitions, such as when a material moves from a liquid to a gas. It has also proved critical in the
advancement of sciences like as astronomy and materials science, where understanding the large-scale behavior of numerous particles is required.
Mechanical Continuum
Continuum mechanics is concerned with materials that may be regarded as continuous while being composed of atoms or molecules. This area of mechanics includes both solid mechanics (the study of how
solid materials bend and fail) and fluid dynamics (the study of fluid behavior such as water and air).
Fluid motion, for example, is represented using equations such as the Navier-Stokes equations, which define how the velocity, pressure, temperature, and density of a flowing fluid are related.
In solid mechanics, continuum mechanics helps engineers understand how structures like bridges and buildings can withstand loads without collapsing. The study of continuum mechanics is essential in
civil, mechanical, and aeronautical engineering, where material stability and durability are crucial. It also plays a crucial part in the design of medical devices like stents and prosthetics, where
understanding how materials deform under stress is vital for ensuring their safety and efficacy.
The Future of Mécanique
The future of mechanics seems promising, with interesting advancements in quantum technology, materials research, and computer methodologies. Quantum mechanics is predicted to play a critical role in
the development of technologies such as quantum computing and quantum encryption, which have the potential to transform computing and cybersecurity. Researchers in materials science are working on
novel materials with improved mechanical qualities such as strength, flexibility, and wear and tear resistance.
Computational mechanics is also expected to transform sectors such as aerospace, automotive, and construction by enabling engineers to model complicated systems with increased precision and
efficiency. As our understanding of mechanics progresses, so will the technologies that rely on it, resulting in new inventions that will define the future of science and technology.
Mécanique is a basic field of physics that supports our understanding of the physical universe. From the enormous motion of planets to the tiny behavior of particles, mechanics describes the forces
and movements that influence everything we see. Its applications are numerous, spanning engineering and technology to everyday life and sports. While obstacles persist, notably in reconciling
classical and quantum mechanics, the future of mechanics seems bright, with discoveries in quantum technology, materials science, and computational approaches paving the way for the next wave of
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The Tibetan Calendar: Lunar Weekdays, Lunar Date Days, Solar Weekdays and Solar Days
The first and second of the five inclusive features (lnga-bsdus) of a Tibetan calendar are the lunar weekday (gza’) and the date of the lunar month (tshe). These are involved in the mechanism
through which the lunar and solar calendars are brought into harmony. To understand lunar weekdays, it is necessary to understand lunar date days (tshe-zhag) and their difference from solar days (
Lunar Date Days and Solar Days
Lunar date days are the period of time it takes for the moon to travel one-fifteenth the distance between new moon and full moon, or full moon and new moon positions in each successive sign in
the zodiac. They divide into 15 units the waxing and waning phases of the moon. Solar days are the period of time from dawn to dawn.
Western days are divided into 24 hours. In the Tibetan system, most types of days are divided into 60 astro hours. Just as each solar day is divided into 60 solar astro hours, each lunar date day is
divided into 60 lunar astro hours. Solar and lunar astro hours are not equal to each other in length, just as these two types of days differ in length.
Western days are solar and last from midnight to midnight. They are all equal in length. Tibetan solar days are from dawn to dawn. Since the time of dawn changes each day, and on the same day is
different the further away one is from the equator, Tibetan solar days vary in length. For calendar purposes, however, dawn is taken standardly to be at 5 A.M. Consequently, Tibetan solar days
used in the calendar have a standard length.
There are approximately 29.5 solar days between new moons, whereas there are 30 lunar date days during the same period. A lunar month, then, has either 29 or 30 solar days, while it always has 30
lunar date days. Because of the discrepancy between the number of solar and lunar date days in a lunar month, the exact new moon does not occur at precisely the same time of solar day in each month.
In other words, each month the beginning of the first lunar date day will occur at a different time on the first solar day of that month.
Furthermore, although solar days all have a standardized, equal length, lunar date days are not similarly standardized.
The moon’s motion is linked to that of the sun. Imagine a child running round the inside of a giant tire tube, while throwing a ball ahead of him. Each time he pitches the ball, he sends it circling
round the inside of the tube. The ball reaches him from behind, whereupon he catches it and hurls it again. The motion of the sun around the zodiac is like that of the child, while the motion of the
moon in its cycle of phases is like that of the ball. The new moon is like when the child catches the ball and then throws it. The full moon is like when the ball is at exactly the opposite point in
the tire from the child so that it starts to come back to him rather than going further away.
Imagine that the child runs at different speeds in different parts of the tire tube. In addition, regardless of where the child is inside the tube, imagine that the ball travels at different
speeds depending on how far away it is from the child. Thus, the amount of time it takes for the ball to travel one-fifteenth the distance away from or back to the child depends on how fast the
child is running as well as how fast the ball is moving simply from its being thrown. Likewise, the amount of time it takes the moon to travel one lunar date day, in other words one-fifteenth the
distance from new to full, or full to new moon, depends on where the sun is in the zodiac and where the moon is in its waxing or waning cycle of phases.
Therefore, since the beginning of the first lunar date day of a lunar half-month does not necessarily correspond to the start of the first solar day of that half-month, and since the lengths of each
of the 15 lunar date days during the half-month are different, consequently the lunar date days of the half-month do not coincide with the solar days. During the 60 solar hours between the dawns of
two consecutive solar days, from 54 to 64 lunar hours can pass. Either one, two or no lunar date days can begin during that solar day.
In summary, lunar date days can be slightly longer or shorter, or equal in length to solar days. Any hour of a lunar date day can occur at dawn, the start of the solar day. A varying amount of a
lunar date day can pass between the dawns of two successive solar days.
Lunar Weekdays and Solar Weekdays
To correlate the lunar and solar calendars, the lunar date days must be assigned to solar days. The Kalachakra system makes this assignment by working with lunar and solar weekdays, and dates of
the lunar month. The calendar is arranged in solar days, each with a date and a day of the week. The lunar date days are converted into lunar weekdays and are mapped on top of it.
Although the number of lunar phase and solar days in a lunar month can be different, the number of lunar and solar weekdays is always the same. But lunar weekdays are not the same as solar
• Solar weekdays are a way of counting solar days in cycles of 7 and are totally equivalent to solar days. Both solar days and solar weekdays begin at the same time and last 60 solar hours.
• Lunar date days generate lunar weekdays. Sometimes these two types of lunar days are equivalent, sometimes they are not. A lunar weekday may cover either one, two or no lunar date days. Thus,
although a lunar date day has 60 lunar astro hours, a lunar weekday may span a varied number of lunar astro hours.
Solar weekdays begin at a fixed point each solar day, namely its start. Lunar weekdays do not necessarily begin at the same time time as lunar date days, and neither begin at the same time each
solar day. For this reason, the lunar weekdays that count the lunar date days do not coincide with the solar weekdays that count the solar days. Both solar and lunar days of the week, however, are
assigned numbers, from 0 to 6, with zero being Saturday. Lunar date days are also numbered in cycles of seven, from 0 to 6.
Dates of the Lunar Month
The second inclusive calendar feature is dates of the lunar month. These are numbered one to 30 and last from dawn to dawn in the manner of solar days. The dates of the lunar month number the solar
days, and thus number the solar weekdays.
Imagine 30 solar days, numbered one to 30, theoretically available for use in a lunar month. Each must be assigned a solar day of the week. There are always 30 lunar date days and thus 30 lunar
weekdays in a lunar month: a phase of the moon cannot be skipped. The lunar weekday occurring at the dawn of each solar day determines the solar weekday assigned to that day. If the lunar
weekdays were all of equal length, the assignment would be straightforward, one for one. For instance, if the lunar Sunday were occurring at the dawn of the first solar day, that day would be a solar
Sunday and the first of the month. If the lunar Monday began sometime during that first solar day and was still occurring at the dawn of the second, the second solar day would be a solar Monday and
date number two. This process would continue in a symmetrical fashion for the entire month.
The lunar weekdays, however, vary in length. Suppose lunar Monday began five minutes before the dawn of the second solar day and lunar Tuesday began five minutes after the dawn of the third solar
day. Lunar Monday is occurring at the dawns of both the second and third theoretical solar days, and so both the second and the third would be solar Mondays. This is not allowed. The rule is that
solar weekdays must be consecutive, with no days of the weeks occurring twice in a row and none omitted. If two consecutive theoretical solar days would each be assigned the same solar day of the
week, the second of the two is omitted. Therefore, in the above example, the theoretical solar day numbered three is omitted. Note that this adjustment does not eliminate either a lunar or solar
weekday. It merely eliminates a date. Solar Monday is date two and solar Tuesday is date four.
Suppose lunar Monday began five minutes after the dawn of the second theoretical solar day and lunar Tuesday began five minutes before the dawn of the third. Lunar Sunday is thus occurring at the
dawn of the second and lunar Tuesday at the dawn of the third theoretical solar day. This would make the second a solar Sunday and the third a solar Tuesday. This is also not allowed. There must be
a solar day Monday in between the two. Theoretical solar days are not actual solar days, and so an extra theoretical one can be added. This added one will be a solar Monday, named after the lunar
Monday. Whenever an extra solar weekday is added, it is given the same date as the day after it. Therefore, there are two theoretical solar days numbered date three. The first of these two is solar
Monday and the second of the two solar Tuesday. Note that this adjustment does not add a doubled lunar or solar weekday. It merely doubles a date. | {"url":"https://studybuddhism.com/en/advanced-studies/history-culture/tibetan-astrology/the-tibetan-calendar-lunar-weekdays-lunar-date-days-solar-weekdays-and-solar-days","timestamp":"2024-11-13T11:16:34Z","content_type":"text/html","content_length":"117772","record_id":"<urn:uuid:01f22a2b-f98a-4eb0-b58f-0f07e4e1da3b>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00338.warc.gz"} |
AGGIE Math
Accelerated Math 7 Chapter 1 Retest
practice materials
Here is a link to a practice packet with six pages of practice problems along with the answers for each of the problems. You should study the lesson pages that go with the questions you plan to
retake. (Some of the practice pages will help with more than one of the test retake problems.)
Accelerated Math 7 Chapter 2 Retest
practice materials
Here is a link to a packet of practice problems along with the answers for each of the problems. You should study the lesson pages that go with the questions you plan to retake. (Some of the practice
pages will help with more than one of the test retake problems.)
Accelerated Math 7 Chapter 3 Retest
practice materials
Here is a link to a packet of practice problems along with the answers for each of the problems. You should study the lesson pages that go with the questions you plan to retake. (Some of the practice
pages will help with more than one of the test retake problems.)
Accelerated Math 7 Chapter 5 Retest
practice materials
Here is a link to a packet of practice problems along with the answers for each of the problems. You should study the lesson pages that go with the questions you plan to retake. (Some of the practice
pages will help with more than one of the test retake problems.) | {"url":"http://mwillmarth.org/chapter-test-retake-resources-2/","timestamp":"2024-11-03T01:07:51Z","content_type":"text/html","content_length":"38547","record_id":"<urn:uuid:faca02eb-1493-4319-9f1f-860a352979bf>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00522.warc.gz"} |
Networked recursive identification of nonlinear systems
Networked control systems have now been studied extensively for almost two decades. The focus has to a large extent been on the effect of quantization of control and feedback signals when plants are
remotely feedback controlled. More recently there has been a renewed focus on the effects of delay on networked control systems, e.g. in wireless data flow control. Many networked controllers are
model based, meaning that identification methods that can operate in a networked environment are needed as well. This has become more important with the standardization of the new 5G wireless
systems. These systems are expected to enable a number of new use cases in the fields of industrial control and augmented reality with force feedback. Consequently, the study of identification
methods that account for delay and quantization is of increasing importance.
The present research is performed on networked recursive identification, with a focus on nonlinear identification methods. The wish list for such algorithms, considering e.g. 5G wireless systems,
include the following requirements
• Ability to identify networked delays, in combination with the plant dynamics.
• Ability to identify the plant dynamics, based on quantized control and feedback signals.
• Guaranteed convergence and stability properties.
• Low computational complexity.
The first and second requirements result since delay will be present as a part of the dynamics when networked identification and feedback control is performed over 5G wireless networks, at least for
high bandwidth applications. The loop delay may become as low as <1 ms in so called ultra-reliable low latency communication (URLLC) and for critical machine type communication (C-MTC). Due to the
high capacity of the 5G networks quantization effects are not as important as earlier, however they may still be present in case many users are served simultaneously. The latter is also a reason why
a low computational complexity is preferrable. Further, since many users are expected, manual interaction may not be possible. Therefore the global convergence and stability properties of the applied
algorithms become very important since it is only with such guarantees that the algorithms can be assumed to always deliver good performance.
Recursive identification with non-uniform output quantization
The paper 11 presents an algorithm that is capable of recursive identification of the impulse response dynamics of a linear system, subject to simultaneous coarse, non-uniform quantization of the
output signal and delay. The quantizer is assumed to be fix and known, with arbitrary quantization steps and switch points. The key idea used in the method is the replacement of the exact gradient
with a smooth approximation. The algorithm is depicted in a networked identification setting in Fig. 1.
Figure 1. The block diagram of the identified networked system.
The model of the plant appears in Fig. 2.
Figure 2: Block diagram of a dynamic system with output quantization.
The algorithm has a very low computational complexity since it is of stochastic gradient type. A very important property of the scheme is that it is globally convergent to a perfect input-output
setting of the identified impulse response dynamics, under relatively mild conditions. The algorithm hence meets all the above requirements. The paper 12 treats an IIR variant of the same algorithm,
however then only local convergence has been proved. The performance of the algorithm was illustrated in a former research project on recursive identification of systems with output quantization.
An important remaining question concerns parametric convergence, i.e. the question if conditions that guarantee convergence to the true and unique parameter vector of the system can be found. This is
not the same property as the input-output convergence discussed in 11. The papers 6 and 7 attempt to bring further clarity to this issue, by studying the conditions behind the convergence. The paper
7 focuses on the requirement of a monotone quantizer in the analysis of 11. The analysis defines a low order system with one single impulse response parameter being equal to 1 and a quantizer with
three levels that is not monotone. The associated ODE analysis method pioneered by L. Ljung is then applied for convergence analysis. In the low order case it turns out that the average updating
directions of the first order version of the algorithm of 11 can be analytically calculated. This opens up for a study of the convergence properties by means of numerical solution of the associated
ODE. Fig. 3 and Fig. 4 illustrate the evolution of the associated ODE for different initial conditions, together with the evolution of the low adaptation gain estimates of the algorithm. The software
package 10 was used to obtain parts of the results.
Figure 3. Convergence to the unique true parameter vector.
Figure 4. Convergence to one of two mirrored parameter vectors, one being the true parameter vector.
Both figures confirm that the associated ODE describes the asymptotic low gain evolution of the algorithm. As can be seen in Fig. 3., the associated ODE has a single fix point corresponding to the
true parameter vector, despite the fact that the quantizer is not monotone. This results proves that:
• A monotone quantizer is not a necessary requirement for global parametric convergence.
Fig.4, on the other hand, illustrates a case where the associated ODE has fixed points equal to +/-1, i.e. global parametric convergence does not hold. Using similar techniques of simulation of
associated ODEs, the paper 6 studies the effect of the input signals distribution and the coupling with the system gain and the quantizer. In summary, the paper 6 proves that at least the following
additional restrictions need to be introduced in order to perform global parametric convergence proof:
• Input signals with discrete distributions need to excluded (in case the output signal is quantized).
• The input signal, the system gain and the quantizer must be such that there is signal energy in at least one switch point of the quantizer.
Recursive identification of nonlinear state space models with delay
The paper 8 adresses a nonlinear identification problem, assuming the plant to be modeled by a general nonlinear differential equation model with a delayed output measurement. A 5G wireless networked
setting where that model is useful is depicted in Fig. 5.
Figure 5. The block diagram of the identified nonlinear networked system.
An assumption of time invariance of the plant is needed in order to merge all delays of the loop into the output of the model. The dynamic model is formulated in state space form, with only one right
hand side component defining the nonlinear ODE. A polynomial parameterization of the nonlinear right hand side component of the ODE is applied.
Based on the model an approximate recursive prediction error identification algorithm is then derived. In order to model the effect of delay, multiple models are used for each integer delay defined
by one sampling period. The fractional component of the delay is obtained by interpolation between multiple models, adjacent to the running delay estimate of the complete model. The discretization in
time is handled by an Euler discretization method.
A case where a first order nonlinear system is identified is depicted in Fig. 6 – Fig. 8. In this case the delay of the networked system is 1.22 s. As can be seen the identified delay is very close
to the true one, a fact that also holds for the remaining parameters, c.f. 8. The software of 9 was used to generate the results.
Figure 6. The input signal, the output signal of the plant, and the delayed output signal of the plant processed by the identification algorithm of 8.
Figure 7. The parameter estimates obtained by the identification algorithm of 8.
Figure 8. The input signal, the delayed output signal processed by the identification algorithm, and the simulated delayed output signal obtained from the identified model with delay. Note that the
two output signals are now well aligned.
Due to the importance of the algorithm of 8 for rapid detection of delay attacks on feedback control systems, the convergence properties were analysed in 1. Averaging analysis using the method with
associated ODEs was applied to prove:
• Under standard conditions and the assumption that the system is in the model set, the true parameter vector is in the set of global stationary points of the algoithm.
It was also discussed in 1 why the standard way of proving local convergence fails, instead numerical examples in 1, 8 and Fig. 7 illustrate that stability of the true parameter vector usually holds.
The SW package 5 updates the algorithm of 8 and 9 with an unknown output equation that is jointly identified with the ODE and delay. The output model is divided into a lower, mid, and upper interval
of the first state of the ODE. The mid interval fixes the input-output static gain of the total model by using a linear model with fixed slope and free bias. The lower and upper intervals are
parametrized as multipolynomial models in the input and states of the ODE, constrained so that continuity of the output model is enforced at the interval boundaries. Fig. 9 illustrates an example of
the performance of the new algorithm.
Figure 9. The parameters identified by the new algorithm of 5 while converging to the true parameter vector. See sections 9-13 of 5 for details on model and parameters.
Recently, the algorithm of 8 has been applied for detection of disguised delay attacks on networked feedback loops with promising results. As shown in the paper 4, the algorithm can detect a delay
change of a feedback loop disguised in jitter long before the attack becomes evident by visual inspection. In addition, delay attacks on automotive cruise control feedback loops can be rapidly
detected, despite the fact that that the attacks are completely disguised by application of the attack in the feedback path, see Fig. 10, Fig. 11, and the papers 2 and 3 for details.
Figure 10. The control signal, disturbance and feedback signal used for delay attack detection of the algorithm of 1 and 8.
Figure 11. The identified delay and nonlinear dynamic parameters of the algorithm of 1 and 8, while used for delay attack detection on the automotive cruise control system of 2. The attack appears at
time t = 6000 s.
1. T. Wigren, "Convergence in delayed recursive identification of nonlinear systems", to appear in Proc. European Control Conference, Stockholm, Sweden, June 25-28, 2024.
2. T. Wigren and A. Teixeira, "Delay attack and detection in feedback linearized control systems", to appear in Proc. European Control Conference, Stockholm, Sweden, June 25-28, 2024.
3. T. Wigren and A. Teixeira, "Feedback path delay attacks and detection", Proc. IEEE CDC , Singapore, pp. 3864-3871, December 13-15, 2023.
4. T. Wigren and A. Teixeira, "On-line identification of delay attacks in networked servo control", Proc. IFAC World Congress, pp. 1041-1047, Yokohama, Japan, July 9-14, 2023.
5. T. Wigren, "MATLAB software for nonlinear and delayed recursive identification - revision 2", Technical Reports from the Department of Information Technology, 2022-002, Uppsala University,
Uppsala, Sweden, January, 2022.
6. S. Yasini and T. Wigren, "Convergence in networked recursive identification with output quantization", SYSID 2018, Stockholm, Sweden, pp. 915-920, July 9-11, 2018.
7. S. Yasini and T. Wigren, "Counterexamples to parametric convergence in recursive networked identification", ACC 2018, Milwaukee, USA, pp. 258-264, June 27-29, 2018.
8. T. Wigren, "Networked and delayed recursive identification of nonlinear systems", IEEE CDC 2017, Melbourne, Australia, pp. 5851-5858, Dec. 12-15, 2017.
9. T. Wigren, "MATLAB Software for Nonlinear and Delayed Recursive Identification - Revision 1", Technical Reports from the Department of Information Technology, 2017-007, Uppsala University,
Uppsala, Sweden, April, 2017. Available: http://www.it.uu.se/research/publications/reports/2017-007/RecursiveNonlinearNetworkedIdentificationSW.zip
10. T. Wigren, "MATLAB software for recursive identification of systems with output quantization - Revision 1 ", Technical Reports from the department of Information Technology 2007-015, Uppsala
University, Uppsala, Sweden, April, 2007. Available: http://www.it.uu.se/research/publications/reports/2007-015/QRISRev1.zip
11. T. Wigren, "Adaptive filtering using quantized output measurements", IEEE Trans. Signal Processing , vol. 46, no. 12, pp. 3423-3426, 1998.
12. T. Wigren, "Approximate gradients, convergence and positive realness in recursive identification of a class of nonlinear systems", Int. J. Adaptive Contr. Signal Processing, vol. 9, no. 4, pp.
325-354, 1995. | {"url":"https://www2.it.uu.se/katalog/tw/research/NetworkedIdent","timestamp":"2024-11-11T08:29:36Z","content_type":"text/html","content_length":"32794","record_id":"<urn:uuid:7472323a-f496-4c94-a025-d3e096a8e82f>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00120.warc.gz"} |
Quick Explanation Of Principal component analysis (PCA) | Machine Learning
Jan 29, 2023 / 20 min read
Principal component analysis (PCA) is a technique used to reduce the dimensionality of a data set while retaining as much of the original information as possible. It is a linear method, which means
that it finds a new set of variables (principal components) that are linear combinations of the original variables. These new variables are chosen such that they are uncorrelated and ordered by the
amount of variance they explain in the data. The first principal component explains the most variance, the second principal component explains the second most variance, and so on.
Visual Explanation of Principal component analysis (PCA)
Step For Principal component analysis (PCA):
The process of PCA involves the following steps:
1). Standardize the data by subtracting the mean and dividing by the standard deviation for each variable.
2). Compute the covariance matrix of the data.
3). Compute the eigenvectors and eigenvalues of the covariance matrix. The eigenvectors are the principal components and the eigenvalues represent the amount of variance explained by each component.
4). Select the principal components with the highest eigenvalues.
5). Project the data onto the selected principal components by computing the dot product of the data with the principal component matrix.
💡 Conclusion:
PCA can be used for data visualization, feature extraction, and noise reduction. It is commonly used in fields such as image processing, natural language processing, and finance. However, it has some
limitations, such as being sensitive to the scale of the variables and not being able to capture non-linear relationships in the data.
Code Example (Python):
Here is an example of how to perform PCA in Python using the scikit-learn library:
from sklearn.decomposition import PCA
import numpy as np
# Create a sample data set
data = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]])
# Create a PCA object and fit it to the data
pca = PCA()
# Print the explained variance ratio for each principal component
# Transform the data using the first two principal components
transformed_data = pca.transform(data)[:, :2]
# Print the transformed data
This example creates a sample data set with 4 rows and 3 columns, and then performs PCA on it using the PCA class from scikit-learn. The explained variance ratio for each principal component is
printed, and the data is transformed using the first two principal components. The transformed data is then printed.
Please note that in real-world scenarios, the sample data would be much larger and the data would be loaded from a file. Also, in most of the cases, the number of features would be much higher than | {"url":"https://codeease.net/machine-learning/quick-explanation-of-principal-component-analysis/","timestamp":"2024-11-04T20:10:21Z","content_type":"text/html","content_length":"22197","record_id":"<urn:uuid:d3b8889c-c959-48db-ba1e-88e29fa8c755>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00652.warc.gz"} |
How many languages do people think are in the world?
One of the first things many people learn in introductory linguistics classes is that there are about 7,000 languages in the world, plus or minus 1,000, depending on whether or not you’re only
counting living languages or dead ones too, and how you divide the line between languages and dialects. Counting languages is difficult, even for professional linguists. But what about for laypeople?
Does average Joe even realize how many languages there are in the world? Are linguists doing a good job of educating the public? To find out, I assigned my Intro to Linguistics students some
ethnographic research last summer. They had to interview three friends or family members (who have never taken a linguistics class) and ask them “How many languages do you think there are in the
world?” Not everyone completed the task, but this gave me a rather unscientific sample of 77 datapoints. On the whole (without outliers removed), the mean average guess was 34,959 languages. But the
median was just 800. Half of all participants guessed 800 or fewer — just 10% (or less!) of the real number! Two people guessed 1 million, one guessed 500,000, one 50,000, and one 25,000. These are
what brought the mean up so much, and also made for some ugly graphs in R. So I threw these 5 data points out, and came up with much cleaner, more normally distributed graphs. Without these five
outliers, the mean guess drops to 1623, and the median drops to 700. Here’s a boxplot of the data:
As you can see, nearly all the guesses are below 4,000, and 75% are below 2,000. While several people did guess in the 5,000 to 10,000 range (roughly the correct answer), these are still considered
outliers by R, even after the more egregious outliers were taken out.
So what’s this mean for linguists? It means about half the population thinks there are 700 or fewer languages in the world, when in fact the real answer is 10 times that many. It means pretty much
everyone except for a few outliers is off by several thousand. It means linguists aren’t doing a very good job of educating the general public about language diversity or language endangerment. This
in turn means the public is probably not willing to fund linguists, since they don’t realize how monumental the task of documenting 7,000 languages really is!
Here I’ll insert a shameless plug for my TwitterBot, @AllTheLanguages, which is doing its part to educate the twitterverse on just how many languages there are. As of this writing, the bot has
tweeted nearly 3,000 languages, and has over 4,000 more to go! Go little bot, go!!
You must be logged in to leave a reply. | {"url":"http://wugology.com/how-many-languages-do-people-think/","timestamp":"2024-11-02T17:17:18Z","content_type":"application/xhtml+xml","content_length":"28733","record_id":"<urn:uuid:a6ad9391-5917-444a-ab97-f63c2f7cc686>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00000.warc.gz"} |
03. B-Tree & Crash
Recovery | Build Your Own Database From Scratch in Go
03. B-Tree & Crash Recovery
3.1 B-tree as a balanced n-ary tree
Height-balanced tree
Many practical binary trees, such as the AVL tree or the RB tree, are called height-balanced trees, meaning that the height of the tree (from root to leaves) is limited to O(logN), so a lookup is O
A B-tree is also height-balanced; the height is the same for all leaf nodes.
Generalizing binary trees
n-ary trees can be generalized from binary trees (and vice versa). An example is the 2-3-4 tree, which is a B-tree where each node can have either 2, 3, or 4 children. The 2-3-4 tree is equivalent to
the RB tree. However, we won’t go into the details because they are not necessary for understanding B-trees.
Visualizing a 2-level B+tree of a sorted sequence [1, 2, 3, 4, 6, 9, 11, 12].
[1, 4, 9]
/ | \
v v v
[1, 2, 3] [4, 6] [9, 11, 12]
In a B+tree, only leaf nodes contain value, keys are duplicated in internal nodes to indicate the key range of the subtree. In this example, node [1, 4, 9] indicates that its 3 subtrees are within
intervals [1, 4), [4, 9), and [9, +∞). However, only 2 keys are needed for 3 intervals, so the first key (1) can be omitted and the 3 intervals become (-∞, 4), [4, 9), and (9, +∞).
3.2 B-tree as nest arrays
Two-level nested arrays
Without knowing the details of the RB tree or the 2-3-4 tree, the B-tree can be understood from sorted arrays.
The problem with sorted arrays is the O(N) update. If we split the array into m smaller non-overlapping ones, the update becomes O(N/m). But we have to find out which small array to update/query
first. So we need another sorted array of references to smaller arrays, that’s the internal nodes in a B+tree.
[[1,2,3], [4,6], [9,11,12]]
The lookup cost is still O(logN) with 2 binary searches. If we choose m as √N, update become O(√N), that’s as good as 2-level sorted arrays can be.
Multiple levels of nested arrays
O(√N) is unacceptable for databases, but if we add more levels by splitting arrays even more, the cost is further reduced.
Let’s say we keep splitting levels until all arrays are no larger than a constant s, we end up with log(N/s) levels, and the lookup cost is O(log(N/s)+log(s)), which is still O(logN).
For insertion and deletion, after finding the leaf node, updating the leaf node is constant O(s) most of the time. The remaining problem is to maintain the invariants that nodes are not larger than s
and are not empty.
3.3 Maintaining a B+tree
3 invariants to preserve when updating a B+tree:
1. Same height for all leaf nodes.
2. Node size is bounded by a constant.
3. Node is not empty.
Growing a B-tree by splitting nodes
The 2nd invariant is violated by inserting into a leaf node, which is restored by splitting the node into smaller ones.
parent parent
/ | \ => / | | \
L1 L2 L6 L1 L3 L4 L6
* * *
After splitting a leaf node, its parent node gets a new branch, which may also exceed the size limit, so it may need to be split as well. Node splitting can propagate to the root node, increasing the
height by 1.
/ \
root N1 N2
/ | \ => / | | \
L1 L2 L6 L1 L3 L4 L6
This preserves the 1st invariant, since all leaves gain height by 1 simultaneously.
Shrinking a B-tree by merging nodes
Deleting may result in empty nodes. The 3rd invariant is restored by merging empty nodes into a sibling node. Merging is the opposite of splitting. It can also propagate to the root node, so the tree
height can decrease.
When coding a B-tree, merging can be done earlier to reduce wasted space: you can merge a non-empty node when its size reaches a lower bound.
3.4 B-Tree on disk
You can already code an in-memory B-tree using these principles. But B-tree on disk requires extra considerations.
Block-based allocation
One missing detail is how to limit node size. For in-memory B+tree, you can limit the maximum number of keys in a node, the node size in bytes is not a concern, because you can allocate as many bytes
as needed.
For disk-based data structures, there are no malloc/free or garbage collectors to rely on; space allocation and reuse is entirely up to us.
Space reuse can be done with a free list if all allocations are of the same size, which we’ll implement later. For now, all B-tree nodes are the same size.
Copy-on-write B-tree for safe updates
We’ve seen 3 crash-resistant ways to update disk data: renaming files, logs, LSM-trees. The lesson is not to destroy any old data during an update. This idea can be applied to trees: make a copy of
the node and modify the copy instead.
Insertion or deletion starts at a leaf node; after making a copy with the modification, its parent node must be updated to point to the new node, which is also done on its copy. The copying
propagates to the root node, resulting in a new tree root.
• The original tree remains intact and is accessible from the old root.
• The new root, with the updated copies all the way to the leaf, shares all other nodes with the original tree.
d d D*
/ \ / \ / \
b e ==> b e + B* e
/ \ / \ / \
a c a c a C*
original updated
This is a visualization of updating the leaf c. The copied nodes are in uppercase (D, B, C), while the shared subtrees are in lowercase (a, e).
This is called a copy-on-write data structure. It’s also described as immutable, append-only (not literally), or persistent (not related to durability). Be aware that database jargon does not have
consistent meanings.
2 more problems remain for the copy-on-write B-tree:
1. How to find the tree root, as it changes after each update? The crash safety problem is reduced to a single pointer update, which we’ll solve later.
2. How to reuse nodes from old versions? That’s the job of a free list.
Copy-on-write B-tree advantages
One advantage of keeping old versions around is that we got snapshot isolation for free. A transaction starts with a version of the tree, and won’t see changes from other versions.
And crash recovery is effortless; just use the last old version.
Another one is that it fits the multi-reader-single-writer concurrency model, and readers do not block the writer. We’ll explore these later.
Alternative: In-place update with double-write
While crash recovery is obvious in copy-on-write data structures, they can be undesirable due to the high write amplification. Each update copies the whole path (O(logN)), while most in-place
updates touch only 1 leaf node.
It’s possible to do in-place updates with crash recovery without copy-on-write:
1. Save a copy of the entire updated nodes somewhere. This is like copy-on-write, but without copying the parent node.
2. fsync the saved copies. (Can respond to the client at this point.)
3. Actually update the data structure in-place.
4. fsync the updates.
After a crash, the data structure may be half updated, but we don’t really know. What we do is blindly apply the saved copies, so that the data structure ends with the updated state, regardless of
the current state.
| a=1 b=2 |
|| 1. Save a copy of the entire updated nodes.
| a=1 b=2 | + | a=2 b=4 |
data updated copy
|| 2. fsync the saved copies.
| a=1 b=2 | + | a=2 b=4 |
data updated copy (fsync'ed)
|| 3. Update the data structure in-place. But we crashed here!
| ??????? | + | a=2 b=4 |
data (bad) updated copy (good)
|| Recovery: apply the saved copy.
| a=2 b=4 | + | a=2 b=4 |
data (new) useless now
The saved updated copies are called double-write in MySQL jargon. But what if the double-write is corrupted? It’s handled the same way as logs: checksum.
• If the checksum detects a bad double-write, ignore it. It’s before the 1st fsync, so the main data is in a good and old state.
• If the double-write is good, applying it will always yield good main data.
Some DBs actually store the double-writes in logs, called physical logging. There are 2 kinds of logging: logical and physical. Logical logging describes high-level operations such as inserting a
key, such operations can only be applied to the DB when it’s in a good state, so only physical logging (low-level disk page updates) is useful for recovery.
The crash recovery principle
Let’s compare double-write with copy-on-write:
• Double-write makes updates idempotent; the DB can retry the update by applying the saved copies since they are full nodes.
• Copy-on-write atomically switches everything to the new version.
They are based on different ideas:
• Double-write ensures enough information to produce the new version.
• Copy-on-write ensures that the old version is preserved.
What if we save the original nodes instead of the updated nodes with double-write? That’s the 3rd way to recover from corruption, and it recovers to the old version like copy-on-write. We can combine
the 3 ways into 1 idea: there is enough information for either the old state or the new state at any point.
Also, some copying is always required, so larger tree nodes are slower to update.
We’ll use copy-on-write because it’s simpler, but you can deviate here.
3.5 What we learned
B+tree principles:
• n-ary tree, node size is limited by a constant.
• Same height for all leaves.
• Split and merge for insertion and deletion.
Disk-based data structures:
• Copy-on-write data structures.
• Crash recovery with double-write.
We can start coding now. 3 steps to create a persistent KV based on B+tree:
1. Code the B+tree data structure.
2. Move the B+tree to disk.
3. Add a free list. | {"url":"https://build-your-own.org/database/03_btree_intro","timestamp":"2024-11-09T00:13:07Z","content_type":"application/xhtml+xml","content_length":"43138","record_id":"<urn:uuid:961a3e69-1fe8-460b-8e85-ec551d83f2ed>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00123.warc.gz"} |
Solar Energy
Solar System Pakistan
Solar Energy,
Power & Electricity
In basic about solar energy, solar power and solar eletricity we will talk about the basic things behind this power. Formulas that will be used to find out which Solar Panel you should use and which
battery you should select. And how much Solar Panels do you need to power up lights and other applications. Here are the main things you need to know and that will be used to calculate your needs.
AC-DC system, Volt, Current(Ampere), Power(Watt), Resistance, Series and Parallel connecting.
Parallel Connecting – 12V System
Parallel Connecting solar panels gives higher current. And voltage will remain the same. Parallel Connecting is best for us. Because we do not need high voltage. Normal battery is 12v. And by
selecting high voltage require higher voltage charge controller. To connect solar panels in parallel we have to connect plus + to plus and minus – to minus. See the picture for details.
Series Connecting – 24V System
By connecting Solar Panels in series connection. It will increase Voltage and current Amps. will remain the same. To connect solar panels in series we have to connect plus + to – minus on next panel.
See the picture for details. In this example we have connected 2 solar panels in series which will give 24v output.
Series and Parallel Connecting – 24v System
If we are creating a 24v system. So then we have to connect two solar panels in series and connect two series connected panels in parallel as showed in figure under. Which increases voltage to 24v
and when we have connected two other 24v connected solar panels in parallel so voltage remains same and increases current Amps. If we want to increase output current on 24v so we connect more solar
panels in parallel same way under as shown in figure under.
AC-DC system
Ac stands for Alternative Current. Alternative current is almost that we found in wall outlet or electric outlet. Clever say’d that we found in wall. It is 230 Volt. DC stands for Direct Current. In
solar panels it is used 12 volt dc system. DC is that current we can found in cells, batteries, and using adapters or regulators. See the picture of a dell charger. Dell charger also converts AC
Current to DC 5.4 Volt and 2410mA. Solar Panels also uses DC voltage and Current.
Voltage is the electromotive force (pressure) applied to an electrical circuit measured in volts (E).
Example. P=200W, I=4.0A. If we have a value of watt and ampere and we want to find out how much volt does it use then we should use this from Power Circle. E=P/I. 200/4.0=50V. So we found voltage is
Current is the flow of electrons in an electrical circuit measured in amperes (I).
Example. P=100W, E=12V. We want to find out how much ampere does it use. We take a look at circle. I=P/E. 100/12=8.33A. Current usage is 8.33A.
Power is the product of the voltage times the current in an electrical circuit measured in watts (P).
Example. E=220V, I=0,4A. This example is taken from picture of Dell Charger. We have 400mA=0,4A. Take a look at circle P=E*I. 220V * 0,4A=88W. Answer is dell charger use 88Watt.
Resistance is the opposition to the flow of electrons in an electrical circuit measured in ohms (R). Increased resistance gives higher voltage and higher power(watt).
Example. E=12V, I=3.0A. We want to find out resistance. We use formula R=E/I. 12V/3.0A = 4 Ohm. | {"url":"https://solarsystem.com.pk/solar-energy/","timestamp":"2024-11-15T04:09:03Z","content_type":"text/html","content_length":"66780","record_id":"<urn:uuid:33c93d55-71a0-4927-8b0f-73b461aa55fa>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00418.warc.gz"} |
Chapter 1, A Simple Guide to R, is a quick tutorial on getting started with R. You will learn how to install packages, access help on R, construct and edit matrices, create and manipulate data
frames, and write and save plots.
Chapter 2, Basic and Interactive Plots, introduces some of the basic R plots, such as scatter, line, and bar charts. We will also discuss the basic elements of interactive plots using the googleVis
package in R. This chapter is a great resource for understanding the basic R plotting techniques.
Chapter 3, Heat Maps and Dendrograms, starts with a simple introduction to dendrograms and further introduces the concept of clustering techniques. The second half of this chapter discusses heat maps
and integrating heat maps with dendrograms to get a more complete picture.
Chapter 4, Maps, discusses the importance of spatial data and various techniques used to visualize geographic data in R. You will learn how to generate static as well as interactive maps in R. The
chapter discusses the topic of shape files and how to use them to generate a cartogram.
Chapter 5, The Pie Chart and Its Alternatives, is a detailed discussion on how to generate pie charts in R. You will also learn about the various criticisms of pie charts and how the pie chart is
transformed to overcome them. The chapter also provides you with various alternatives used by data scientists and visualization artists to overcome the limitation of a pie chart.
Chapter 6, Adding the Third Dimension, dives into constructing 3D plots. This chapter also introduces packages such as rgl and animation, which are used to create interactive 3D plots.
Chapter 7, Data in Higher Dimensions, demonstrates the use of visualizations that are used to display data in higher dimension. You will learn the techniques to generate sunflower plots, hexbin
plots, Chernoff faces, and so on. This chapter also discusses the usefulness of network, radial, and coxcomb plots, which have been widely used in news.
Chapter 8, Visualizing Continuous Data, illustrates the use of visualizations to display time series data. The chapter also discusses some general concepts related to visualizing correlations, the
shape of the distribution, and detection of outliers using box and whisker plots.
Chapter 9, Visualizing Text and XKCD-style Plots, illustrates the use of text in creating effective visualizations. This chapter focuses mainly on techniques to create word clouds, phase tree, and
comparison clouds in R. You will also learn how to use the XKCD package to introduce humor in visualizations.
Chapter 10, Creating Applications in R, shows you the techniques to create presentations and R markdown documents for publishing on a blog or a website. The chapter further discusses the XML package
used to extract and visualize data as well as using shiny package used to create interactive applications. | {"url":"https://subscription.packtpub.com/book/data/9781783989508/pref/preflvl1sec02/what-this-book-covers","timestamp":"2024-11-04T17:48:41Z","content_type":"text/html","content_length":"134985","record_id":"<urn:uuid:25d8b1c0-d558-409f-b0ba-f9dac357f973>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00557.warc.gz"} |
Layered navigation filter and combinations/features
Hello, couldn't find the solution browsing this forum, sorry if the sollution is already found. I think it could be useful for many Prestashop users.
In my shop i use Layered navigation filter, it's very usefull, while users usually come already knowing what they want very strictly, but without a filter it would be hard to find wanted product.
Still, i cannot find the solution for better filtering when products have combinations.
For example: i sell apples. Apples are red, green and yellow. So i set these colors as a combinations of a product apple, set different prices, everything is ok. But every separate color have
different weight.
Red - 100 gr
Yellow - 150 gr
Green - 200 gr
How to make layered navigation filter it by weight? I cannot make 3 different products (red apple, yellow apple, green apple), because some of my products have 10-20 and more combinations. I also
cannot assign different features for every combination, because there is no such option (or i cannot find it) in prestashop platform.
I believe I have the same problem.
We have thousands of shoes.
One shoe may have 22 different "colors" such as Red Patent Leather, Red Snake, Linen Smooth, etc.
I want layered navigation to offer choices of simply Red, White, Green, Brown, etc.
If I could place a feature per combination (such as adding Red to combinations that have color "Red Patent Leather", I would be able to accomplish this. I don't see a way to do this right now.
Is there some other organization of products/attributes/features that allows this?
• 1 year later...
Hello, couldn't find the solution browsing this forum, sorry if the sollution is already found. I think it could be useful for many Prestashop users.
In my shop i use Layered navigation filter, it's very usefull, while users usually come already knowing what they want very strictly, but without a filter it would be hard to find wanted product.
Still, i cannot find the solution for better filtering when products have combinations.
For example: i sell apples. Apples are red, green and yellow. So i set these colors as a combinations of a product apple, set different prices, everything is ok. But every separate color have
different weight.
Red - 100 gr
Yellow - 150 gr
Green - 200 gr
How to make layered navigation filter it by weight? I cannot make 3 different products (red apple, yellow apple, green apple), because some of my products have 10-20 and more combinations. I also
cannot assign different features for every combination, because there is no such option (or i cannot find it) in prestashop platform.
Did you got any solutions ?.
• 6 months later...
Hello, couldn't find the solution browsing this forum, sorry if the sollution is already found. I think it could be useful for many Prestashop users.
In my shop i use Layered navigation filter, it's very usefull, while users usually come already knowing what they want very strictly, but without a filter it would be hard to find wanted product.
Still, i cannot find the solution for better filtering when products have combinations.
For example: i sell apples. Apples are red, green and yellow. So i set these colors as a combinations of a product apple, set different prices, everything is ok. But every separate color have
different weight.
Red - 100 gr
Yellow - 150 gr
Green - 200 gr
How to make layered navigation filter it by weight? I cannot make 3 different products (red apple, yellow apple, green apple), because some of my products have 10-20 and more combinations. I also
cannot assign different features for every combination, because there is no such option (or i cannot find it) in prestashop platform.
I believe I have the same problem.
We have thousands of shoes.
One shoe may have 22 different "colors" such as Red Patent Leather, Red Snake, Linen Smooth, etc.
I want layered navigation to offer choices of simply Red, White, Green, Brown, etc.
If I could place a feature per combination (such as adding Red to combinations that have color "Red Patent Leather", I would be able to accomplish this. I don't see a way to do this right now.
Is there some other organization of products/attributes/features that allows this?
Did you got any solutions ?. | {"url":"https://www.prestashop.com/forums/topic/222531-layered-navigation-filter-and-combinationsfeatures/","timestamp":"2024-11-05T01:20:25Z","content_type":"text/html","content_length":"112961","record_id":"<urn:uuid:0e5fdb22-b6b7-446d-8420-b64a63354886>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00286.warc.gz"} |
Question #79fec | Socratic
Question #79fec
1 Answer
It would depend on the specifics of your question. The IVT may or may not be useful, but if it is useful, feel free to use it. Do you have a question in particular I could try to solve?
One thing worth noting though: the IVT does apply for irrational numbers, as long as they are contained in a larger, continuous set. To apply the IVT, you need to have a function that is continuous
on an interval. The set of real numbers (or a section of it like [1,3]) is often used. This contains the set of irrational numbers.
If we just used rational numbers, we would have gaps in our function (at $\sqrt{2} , \pi , \ldots$) which would mean that our function is discontinuous at those points and thus we cannot use the IVT
on an interval containing those points. Same thing if we only used the set of irrational numbers.
Impact of this question
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dask.array.sum(a, axis=None, dtype=None, keepdims=False, split_every=None, out=None)[source]¶
Sum of array elements over a given axis.
This docstring was copied from numpy.sum.
Some inconsistencies with the Dask version may exist.
Elements to sum.
axisNone or int or tuple of ints, optional
Axis or axes along which a sum is performed. The default, axis=None, will sum all of the elements of the input array. If axis is negative it counts from the last to the first axis.
If axis is a tuple of ints, a sum is performed on all of the axes specified in the tuple instead of a single axis or all the axes as before.
dtypedtype, optional
The type of the returned array and of the accumulator in which the elements are summed. The dtype of a is used by default unless a has an integer dtype of less precision than the default
platform integer. In that case, if a is signed then the platform integer is used while if a is unsigned then an unsigned integer of the same precision as the platform integer is used.
outndarray, optional
Alternative output array in which to place the result. It must have the same shape as the expected output, but the type of the output values will be cast if necessary.
keepdimsbool, optional
If this is set to True, the axes which are reduced are left in the result as dimensions with size one. With this option, the result will broadcast correctly against the input array.
If the default value is passed, then keepdims will not be passed through to the sum method of sub-classes of ndarray, however any non-default value will be. If the sub-class’ method does
not implement keepdims any exceptions will be raised.
initialscalar, optional (Not supported in Dask)
Starting value for the sum. See ~numpy.ufunc.reduce for details.
wherearray_like of bool, optional (Not supported in Dask)
Elements to include in the sum. See ~numpy.ufunc.reduce for details.
An array with the same shape as a, with the specified axis removed. If a is a 0-d array, or if axis is None, a scalar is returned. If an output array is specified, a reference to out is
See also
Equivalent method.
numpy.add.reduce equivalent function.
Cumulative sum of array elements.
Integration of array values using composite trapezoidal rule.
mean, average
Arithmetic is modular when using integer types, and no error is raised on overflow.
The sum of an empty array is the neutral element 0:
For floating point numbers the numerical precision of sum (and np.add.reduce) is in general limited by directly adding each number individually to the result causing rounding errors in every
step. However, often numpy will use a numerically better approach (partial pairwise summation) leading to improved precision in many use-cases. This improved precision is always provided when no
axis is given. When axis is given, it will depend on which axis is summed. Technically, to provide the best speed possible, the improved precision is only used when the summation is along the
fast axis in memory. Note that the exact precision may vary depending on other parameters. In contrast to NumPy, Python’s math.fsum function uses a slower but more precise approach to summation.
Especially when summing a large number of lower precision floating point numbers, such as float32, numerical errors can become significant. In such cases it can be advisable to use dtype=
”float64” to use a higher precision for the output.
>>> import numpy as np
>>> np.sum([0.5, 1.5])
>>> np.sum([0.5, 0.7, 0.2, 1.5], dtype=np.int32)
>>> np.sum([[0, 1], [0, 5]])
>>> np.sum([[0, 1], [0, 5]], axis=0)
array([0, 6])
>>> np.sum([[0, 1], [0, 5]], axis=1)
array([1, 5])
>>> np.sum([[0, 1], [np.nan, 5]], where=[False, True], axis=1)
array([1., 5.])
If the accumulator is too small, overflow occurs:
>>> np.ones(128, dtype=np.int8).sum(dtype=np.int8)
You can also start the sum with a value other than zero:
>>> np.sum([10], initial=5) | {"url":"https://docs.dask.org/en/stable/generated/dask.array.sum.html","timestamp":"2024-11-04T14:39:13Z","content_type":"text/html","content_length":"40405","record_id":"<urn:uuid:59624141-9e54-499c-8676-1732a379128d>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00869.warc.gz"} |
How to Paste Cells with Only Values \Results Keep and Remove the Formula in Excel? - Free Excel Tutorial
We often use formulas to do calculate like sum, average for some data saved in worksheet. And most time we want to use the returned results directly for example copy it to another cell or worksheet.
But we find that if we do a copy for a sum\average to another cell, we often get an error returned because of the formula is still used and for this operation we actually copy a formula instead of a
value. So we need a simple way to copy\paste cell with formula included as a value. See steps below.
First prepare two lists of data, and we do sum for them. See example below:
Here the total value for D2 and D3 are calculated with formula, D2=SUM(B2:C2). Now we want to copy the total to another table directly. Noticed that we get 0 value. That’s because we copy the formula
to H2 instead of the true value.
We just want copy and paste the value with only result keeps but remove the formula, how can we do?
Remove Formula and Keep Result by Paste Values Function
1. Select on D2 and D3, click Ctrl+C to do a copy.
2. Select on H2 and H3, right click your mouse, below menu is loaded, and click on
3. Now the results are copied to H2 and H3 with only value properly.
You can Paste Value via Paste button in Home tab.
You can also click on Paste Special to get more choice. | {"url":"https://www.excelhow.net/how-to-paste-cells-with-only-values-results-keep-and-remove-the-formula-in-excel.html","timestamp":"2024-11-05T06:26:40Z","content_type":"text/html","content_length":"89562","record_id":"<urn:uuid:a968edb0-b663-4b9e-ac3b-7e4cd1970d2d>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00235.warc.gz"} |
Question on co- / contra-variant coordinates
I am studying co- and contra- variant vectors and I found the video at https://www.youtube.com/watch?v=8vBfTyBPu-4 very useful. It discusses the slanted coordinate system where the X, Y axes are at
an angle of α. One can get the components of v either by dropping perpendiculars to the axes (v[i]) or by dropping a line parallel to the other axis (v^i). These give correct results for the norm v
[i] v^i and the dot product v[i] w^i. (I have not shown w). So the v^i are called contravariant and the v[i] are called covariant. According to the video Dirac thought this was a great example.
But both
and v
^i contra-vary with a change of scale of the basis vectors. This contradicts some definitions of contravariant and covariant components, e.g. this one on Wikipedia. These definitions say that
covariant components co-vary with a change of scale.
Is there a simple resolution to this apparent contradiction?
Submitted as comment on video and
on Physics Forums at
Orodruin answered on PF seven hours later with "The covariant and contravariant components belong to different basis vectors." Cryo came up with something a bit more complicated and I followed
Orodruin. So here's how it goes:
To get
with the covariant coordinates we must have
v = v[x]e[x] + v[y]e[y] (1)
I have drawn them in on the diagram and e[x] is clearly smaller than e^x, which I have conveniently drawn so that v[x] = 1. We must also have
v[x]e[x] = v^xe^x (2)
and similarly
v[y]e[y] = v^ye^y (3)
which give us
v = v^xe^x + v^ye^y (4)
which is as it should be.
(3) gives us
v[y] = e^y(e[y])^-1v^y (5)
where (e[y])^-1 is the inverse of e[y]: or e[y](e[y])^-1 = 1.
(5) shows us that v[y] does indeed vary with e^y with a strange constant of proportionality (e[y])^-1v^y.
Back to the drawing board
Orodruin did not like that at his #5 and then gave me a tip at #7 that the dual (or covariant) bases are given by ê
= ∇x
. I will use ê for basis vectors in future, following Carroll. From the formula I could calculate the covariant basis vectors in terms of the Cartesian basis vectors because I already had the inverse
metric of the covariant system. This was calculated from the transformations of the systems to and from Cartesian.
The answer came:
and draw them properly (x = 1, y = 2)
From (NF12) it is clear that ê
is π/2 - α below the Cartesian X axis. Therefore the angle between ê
and ê
is a right angle, and the projected line from v
is parallel to ê
as we should have guessed and Orodruin intimated. Likewise ê
is parallel to the projected line from v
and always on the Cartesian Y axis. In addition |ê
| > |ê
Getting back to the original question, what happens to v
if we double the contravariant basis ê
to ê'
? We could leave v
, ê
, v
, ê
alone and they would still give
. We could double ê
and halve v
. Neither option would co-vary. Both would give incorrect values for |
|. We need to find the metric of the primed system so we can just calculate the primed covariant coordinates and bases. We also cannot use the previous technique (calculating from the transformations
of the systems to and from Cartesian), because we don't know how to transform primed covariant coordinates to Cartesian - we don't know what the primed coordinates are.
Therefore we use the pullback operator which we met two posts ago in
Commentary on Appendix A: Mapping S2 and R3
If we have our contravariant primed bases, with scale factors
a,b so
ê'[x] = aê[x] , ê'[y] = bê[y]
which give
v'^x = v^x ⁄ a , v'^y =v^y ⁄ b
Turning all the handles we end up with
v'[x] = a^2v'^x + abv'^ycosα
v'[y] =b^2v'^y + abv'^xcosα
v' ^i
decrease as
a, b
increase as
a^2 ,b^2 ,ab
, we can safely say that v'
increase as ê
They co-vary
. It also looks as if they will 'compensate' for
v' ^i
, giving correct |
| and
. We would just have to follow the trail.
The full details of all the above are contained in sections 5b and Note F of
Commentary 1.1 Tensors matrices and indexes.pdf | {"url":"https://www.general-relativity.net/2018/11/question-on-co-contra-variant.html","timestamp":"2024-11-11T07:02:12Z","content_type":"application/xhtml+xml","content_length":"73333","record_id":"<urn:uuid:f38ca069-2e91-4035-a476-e3d1a5607aae>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00538.warc.gz"} |
Eduqas Physics - Component 3 & 4 - Fridge Physics
Resistors in series and in parallel can change the total resistance in a circuit. What is Resistance in series and parallel? Resistors in series and in parallel change the total resistance in a
circuit. Special components called resistors are made especially for creating a precise quantity of resistance in a circuit. They are generally made of metal wire or carbon and constructed to
maintain a stable steady amount of resistance over a wide range of environmental conditions. Resistance in series and parallel equation To calculate Resistance in series we use this equation. $R_t =
{\mathit R_1 \text + \mathit R_2}$… | {"url":"https://fridgephysics.com/eduqas-physics-component-3-4/","timestamp":"2024-11-08T12:31:42Z","content_type":"text/html","content_length":"179506","record_id":"<urn:uuid:5260713d-656c-4c34-a4a7-adcf66daa323>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00853.warc.gz"} |
Calculate quarter from date
I have a Date Ordered column and want to automatically populate a Quarter column based on the date ordered. What is the best way to write this formula?
Best Answers
• If you are going by normal annual quarters then this formula should work =IF(MONTH([date]@row) < 4, "1", IF(MONTH([date]@row) < 7, "2", IF(MONTH([date]@row) < 10, "3", "4"))). If you are going
for fiscal quarters that vary then you would have to tweak the above and account for the offset of the months (IE: if your fiscal quarters started in Feb - =IF(MONTH([test date]@row) = 1, "4", IF
(MONTH([test date]@row) < 5, "1", IF(MONTH([test date]@row) < 8, "2", IF(MONTH([test date]@row) < 11, "3", "4"))) )
• There are several ways to do this. I'll outline one that you can actually reuse easily if you need this on other sheets as well.
Create a helper sheet called Quarters. You'll want two columns: MonthNumber and Quarter.
Populate with all twelve month numbers and their quarter.
Then, to determine the quarter from the Date Ordered column, use an INDEX/MATCH formula, following the prompts to reference another sheet (your new Quarters sheet.)
=INDEX({Quarters sheet Quarter column range}, MATCH(MONTH([Date Ordered]@row), {Quarters sheet MonthNumber column range}, 0))
*The bold italics are the references to the Quarters sheet. For the first one, when you click "Reference Another Sheet", navigate to the Quarters sheet, click the header for the Quarter column,
and create the range. Repeat for the MonthNumber column for the second reference.
Jeff Reisman
Link: Smartsheet Functions Help Pages Link: Smartsheet Formula Error Messages
If my answer helped solve your issue, please mark it as accepted so that other users can find it later. Thanks!
• If you are going by normal annual quarters then this formula should work =IF(MONTH([date]@row) < 4, "1", IF(MONTH([date]@row) < 7, "2", IF(MONTH([date]@row) < 10, "3", "4"))). If you are going
for fiscal quarters that vary then you would have to tweak the above and account for the offset of the months (IE: if your fiscal quarters started in Feb - =IF(MONTH([test date]@row) = 1, "4", IF
(MONTH([test date]@row) < 5, "1", IF(MONTH([test date]@row) < 8, "2", IF(MONTH([test date]@row) < 11, "3", "4"))) )
• There are several ways to do this. I'll outline one that you can actually reuse easily if you need this on other sheets as well.
Create a helper sheet called Quarters. You'll want two columns: MonthNumber and Quarter.
Populate with all twelve month numbers and their quarter.
Then, to determine the quarter from the Date Ordered column, use an INDEX/MATCH formula, following the prompts to reference another sheet (your new Quarters sheet.)
=INDEX({Quarters sheet Quarter column range}, MATCH(MONTH([Date Ordered]@row), {Quarters sheet MonthNumber column range}, 0))
*The bold italics are the references to the Quarters sheet. For the first one, when you click "Reference Another Sheet", navigate to the Quarters sheet, click the header for the Quarter column,
and create the range. Repeat for the MonthNumber column for the second reference.
Jeff Reisman
Link: Smartsheet Functions Help Pages Link: Smartsheet Formula Error Messages
If my answer helped solve your issue, please mark it as accepted so that other users can find it later. Thanks!
• Thanks everyone! This is very helpful!
• HI - I'm using Jeff's approach and it's working for calculating the month. I'm trying to add the year to the formula, I want it to return Q1-24 or Q4-25, etc.
This much works Great.=INDEX({Fiscal Year Range 1}, MATCH(MONTH([End Date]@row), {Fiscal Year Range 2}, 0)).
This is my table.
This is how I modified the formula.
=INDEX({Fiscal Year Range 1}, MATCH(MONTH([End Date]@row), {Fiscal Year Range 2}, 0), MATCH(YEAR([End Date]@row), {Fiscal Year Range 4}, 0))
The End Date is 12/1/24 and it's returning the number 1.
How do I tell it to return Q1-25?
• @ilwiny In Smartsheet you can concatenate strings using "+", unlike Excel which would use the CONCAT function. Here's how I would do that using your Fiscal Year sheet, although my personal
preference is Heather's nested if() method.
This would produce the format "Q1-2025".
="Q" + INDEX({Fiscal Year Range 1}, MATCH(MONTH([End Date]@row), {Fiscal Year Range 2}, 0)) + "-" + YEAR([End Date]@row)
If you only want the last two digits for the year (i.e., Q1-25), you can add the RIGHT function:
="Q" + INDEX({Fiscal Year Range 1}, MATCH(MONTH([End Date]@row), {Fiscal Year Range 2}, 0)) + "-" + RIGHT(YEAR([End Date]@row),2)
• @KennyK How would you update Heather's nested if()method to capture the years? Specifically if your fiscal year starts in February. Feb - =IF(MONTH([test date]@row) = 1, "4", IF(MONTH([test date]
@row) < 5, "1", IF(MONTH([test date]@row) < 8, "2", IF(MONTH([test date]@row) < 11, "3", "4"))) )
• Hi @LydiaB1029
If your year starts in February and you want any January dates to have a year that is the previous year, and Feb-Dec to have the current year then you just need:
=IF(MONTH([test date]@row) = 1, YEAR([test date]@row) - 1, YEAR([test date]@row))
This says if the month is January (month 1), return the year of the test date minus 1, if not return the year of the test date.
You can add this to the formula you had for month, with a space between the month number and year like this:
=IF(MONTH([test date]@row) = 1, "4", IF(MONTH([test date]@row) < 5, "1", IF(MONTH([test date]@row) < 8, "2", IF(MONTH([test date]@row) < 11, "3", "4"))) ) + " "+IF(MONTH([test date]@row) = 1,
YEAR([test date]@row) - 1, YEAR([test date]@row))
If you only want 2 digits for the year rather than 4, you can include only the right most 2 characters by adding this part in bold:
=IF(MONTH([test date]@row) = 1, "4", IF(MONTH([test date]@row) < 5, "1", IF(MONTH([test date]@row) < 8, "2", IF(MONTH([test date]@row) < 11, "3", "4")))) + " " + RIGHT(IF(MONTH([test date]@row) =
1, YEAR([test date]@row) - 1, YEAR([test date]@row)), 2)
Let me know how you get on.
• This works! Thank you for explaining the formula, "This says if the month is January (month 1), return the year of the test date minus 1, if not return the year of the test date." It's helpful
for me go forward to understand the logic.
• No problem at all, I'm pleased my explanation made sense.
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Equal Sign Anchor Chart
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A2L Item 068
Goal: Problem solving and developing strategic knowledge.
Source: UMPERG-ctqpe145
You are given this problem:
child is standing at the rim of a rotating disk holding a rock. The
disk rotates without friction. The rock is thrown in the RADIAL
direction [path (5)] at the instant shown. You are given:
□ Mass of the child
□ Radius of the disk
□ Mass of the thrown rock
□ Velocity of the rock
□ Initial angular speed of the system
You want to find the final angular speed of the disk and child.
What principle would you use to solve the problem MOST EFFICIENTLY?
1. Kinematics only
2. F= ma or Newton’s laws
3. Work-Kinetic Energy theorem
4. Impulse-Momentum theorem
5. Angular Impulse-Angular Momentum theorem
6. More than one of the above
7. None of the above
8. Cannot be determined
(5) is the correct response if the rock is thrown radially. Since there
is no angular impulse, there can be no change in angular momentum.
Neither the rock alone, nor the child/disk system changes angular
Throwing the rock radially, clearly increases the kinetic energy but not
the angular momentum. Consequently, the final angular speed of the disk
and child is the same as the initial speed.
Questions to Reveal Student Reasoning
Does the rock have angular momentum (or energy) just before it is
thrown? just after it is thrown?
If energy (angular momentum) is gained, where does it come from?
Changes in angular momentum are caused by a net torque. What torques
act on the system during the process of throwing?
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A quantum computer corrected its own errors, improving its calculations
A quantum computer corrected its own errors, improving its calculations
For the first time, a quantum computer has improved its results by repeatedly fixing its own mistakes midcalculation with a technique called quantum error correction.
Scientists have long known that quantum computers need error correction to meet their potential to solve problems that stump standard, “classical” computers (SN: 6/22/20). Quantum computers calculate
with quantum bits, or qubits, which are subject to quantum physics but suffer from jitters that result in mistakes.
In quantum error correction, multiple faulty qubits are combined to make reliable qubits, called logical qubits, which are then used to perform the calculation. Previous efforts found that error
correction made calculations worse, rather than better, or detected errors but didn’t actually fix them (SN: 10/4/21).
Now, scientists have performed repeated rounds of operations and error correction on eight logical qubits, researchers from Microsoft and the quantum computing company Quantinuum report September 10
at the Quantum World Congress in Tysons, Va., and in a paper posted online at arXiv.org. The operations performed in the calculation imbued the qubits with correlations called quantum entanglement.
The corrected calculation had an error rate about a tenth that of one performed with the original, error-prone qubits, which are called “physical” qubits.
The researchers also entangled 12 logical qubits, the largest number of logical qubits ever entangled. The error rate for this entangled state was less than one-twentieth that of the equivalent state
achieved using the computer’s initial faulty, physical qubits.
“Error correction is working; this is huge,” says computer scientist Krysta Svore of Microsoft. “This is the direction we need to go for reliable quantum computing.”
The researchers used a quantum computer developed by the company Quantinuum, with 56 qubits made from electrically charged atoms, or ions. Those qubits were combined to make the logical qubits.
For correcting errors, a variety of schemes exist, and each one can fix a certain number of mistakes. The device in the study used an error correction scheme that was able to fix only one error. If
the computer made two errors, the researchers were unable to fix the mistake, and they instead detected the mistakes and threw the result away to avoid inaccurate results.
In another recent milestone, researchers from Google reported August 24 at arXiv.org that error correction increases the length of time a qubit can store information in memory, though the team didn’t
perform calculations with it. Taken together, the Microsoft and Google results are “showing that error correction works like we expect,” says Ken Brown, a quantum engineer at Duke University and a
scientific adviser for the quantum computing company IonQ. “That’s really promising.”
But more improvements are needed. The Microsoft result falls short of demonstrating a universal quantum computer, one that can perform all the operations that a quantum computer is capable of.
“That’s the next big challenge, is getting enough resources … that you can actually do full universal quantum computing on many logical qubits,” Brown says.
In another study, Microsoft researchers combined high-performance classical computing, artificial intelligence and quantum computing to perform a chemistry calculation. The calculation can be done
without a quantum computer, but the study was a proof of concept. The calculation used two logical qubits, and the researchers found that the results were improved compared with a calculation
performed with the error-prone physical qubits.
In the future, when quantum computers have more logical qubits, such chemistry calculations could unlock secrets that classical computers can’t access. Scientists hope the machines might reveal how
to make fertilizer more efficiently, or how to extract carbon out of the atmosphere to combat global warming. “At the core, we want to save and feed our planet,” Svore says.
#quantum #computer #corrected #errors #improving #calculations
Image Source : www.sciencenews.org
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An analog of the Vaisman-Molino cohomology for manifolds modelled on some types of modules over Weil algebras and its application.
Published online : 2001-01-01
EUDML-ID : urn:eudml:doc:231227
title = {An analog of the Vaisman-Molino cohomology for manifolds modelled on some types of modules over Weil algebras and its application.},
journal = {Lobachevskii Journal of Mathematics},
volume = {8},
year = {2001},
pages = {55-75},
zbl = {0995.58001},
language = {en},
url = {http://dml.mathdoc.fr/item/01702779}
Shurygin, Vadim V.; Smolyakova, Larisa B. An analog of the Vaisman-Molino cohomology for manifolds modelled on some types of modules over Weil algebras and its application.. Lobachevskii Journal of Mathematics, Volume 8 (2001) pp. 55-75. http://gdmltest.u-ga.fr/item/01702779/ | {"url":"http://dml.mathdoc.fr/item/01702779/","timestamp":"2024-11-06T08:14:55Z","content_type":"text/html","content_length":"14821","record_id":"<urn:uuid:1fd07322-a8de-4008-96aa-29c7a726581a>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00423.warc.gz"} |
Chinese Book of Changes
Gallery of universal triangles.
Hexagram i-jing 15.
1. Perfect world.
CRAB - cardinal sign and second quarter of zodiac.
SCORPIO - firm sign and third quarter of zodiac.
VIRGO - mutable sign and second quarter of zodiac.
2. Person.
LIBRA - cardinal mark and elements of air.
SCORPIO - firm mark and elements of water.
GEMINI - mutable mark and elements of air
3. Imperfect world - unconscious sphere.
GEMINI - initial third of astrological circle and symbol of air element.
SCORPIO - middle third of zodiacal circle and symbol of water element.
AQUARIUS - last third of astrological circle and symbol of air element.
4. Imperfect world - conscious sphere.
AQUARIUS - third segment of zodiacal chart and air natural elements.
SCORPIO - second segment of zodiacal chart and water natural elements.
GEMINI - first segment of zodiacal chart and air natural elements.
5. Discrete world - astral sphere.
AQUARIUS - area of slow planets - air elemental nature.
SCORPIO - area of less rapid planets - water elemental nature.
GEMINI - area of fast heavenly bodies - air elemental nature.
6. Discrete world - ethereal sphere.
LIBRA - slow heavenly bodies and air elementals.
CRAB - less rapid heavenly bodies and water elementals.
GEMINI - fast planets and air elementals.
The shown geometrical triangles and charts of world reality can interest mathematicians and people who are interested in mathematics, because hexagrams of the Chinese Book of Changes i-jing are
symbols which can be considered as numbers or ciphers of binary system of mathematical notation, and also as numerical signs which symbolize logic principles according to which dual or otherwise to
tell dichotomizing structure of universe, and in particular structural matrix of discrete world is organized.
Next 16 hexagram.
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s and
Quadratic Equations and Functions (Economics)
Quadratic Equations
A quadratic equation is an equation of the general form \[ax^2+bx+c=0\] where $a\neq 0$ where $x$ is a variable and $a,b$ and $c$ are constants.
In other words, it is an equation where the highest power of the variable (usually $x$) is $2$.
Solving a quadratic equation means finding values of $x$ which satisfy the equation (make the expression on the left-hand side of the $=$ sign equal to zero). These values of $x$ are called the
solutions to the quadratic equation. Quadratic equations can have two, one or no real solutions. The solutions of a quadratic equation correspond to the roots of a quadratic function.
There are two commonly-used methods of solving quadratic equations: factorisation and the quadratic formula.
In some cases it is possible to write the quadratic expression on the left-hand side of a quadratic equation as a product of two linear expressions each enclosed in a pair of brackets. Writing the
quadratic in this way is known as factorising the quadratic or simply factorisation because it involves finding the factors of the quadratic expression.
For example, factorising the left hand side of the quadratic equation $x^2+(r+s)x+rs=0$ gives $(x+r)(x+s)$. The roots of this equation are therefore $(-r)$ and $(-s)$ since we can see that when we
substitute either of these values for $x$ the expression in one of the pairs of brackets becomes equal to zero (check!), so the equation is satisfied.
We know that factorisation is the reverse of expanding brackets. Expanding the brackets is therefore a useful way to check whether you have factorised a quadratic expression correctly. Check for
yourself that the above factorisation is correct by expanding the brackets.
Note: It is not always possible to factorise a quadratic expression. However when it is possible, the factorisation method is typically preferred to using the quadratic formula because it is faster.
===Method: when $a=1$===
Suppose we are asked to solve a quadratic of the form $x^2+bx+c=0$ by factorisation. For example, $x^2+5x+6=0$. To do this, we must find $2$ numbers, call them $A$ and $B$, which multiply together to
give $6$ and sum to give $5$.
The factorisation of $x^2+5x+6$ is then $(x+A)(x+B)$ and the roots of the equation are $(-A)$ and $(-B)$. A good way to keep track of the possibilities for $A$ and $B$ is to make a table which lists
all of the values for $A$ and $B$ whose product is $6$:
(-1) (-6) 6 (-7)
(-2) (-3) 6 (-5)
Note: For each pair of numbers, it doesn’t matter which we put in the “A column” and which we put in the “B column”.
We can see from the table above that the only pair of values whose product is $6$ and which sum to $5$ are $2$ and $3$. The factorisation of $x^2+5x+6$ is thus \[(x+2)(x+3)=0\] and the roots of the
equation are $(-2)$ and $(-3)$.
Note: We can check that these are indeed the roots of $x^2+5x+6=0$ by substituting them into the equation: \[(-2)^2+5\times(-2)+6=4-10+6=0\] \[(-3)^2+5\times(-3)+6=9-15+6=0\] Both $(-2)$ and $(-3)$
satisfy the equation so these are the roots of the equation.
Method: when $a\neq1$
Suppose we are asked to factorise and solve a quadratic of the form $ax^2+bx+c=0$ where $a\neq1$ and $a\neq0$.
To solve any equation of this form, we take the $a$ outside brackets to give: \[a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=0\] If the expression in the brackets contains only integer coefficients, we
can then use the method above to factorise the expression inside the brackets and so solve the equation.
Example 1
Suppose we are asked to solve the equation $5x^2-25x+30=0$. Rewriting this equation with $a=5$ outside brackets gives: \[5(x^2-5x+6)=0\] We can then factorise the expression inside the brackets as
follows (see the example above for how to do this): \[x^2-5x+6=(x-3)(x-2)=0\] So the factors are $(x-2)$ and $(x-3)$. However, recall that the original equation was \[5(x^2-5x+6)=0\] which can now be
written as \[5(x-3)(x-2)=0\] so we can see that $5$ is another factor. The solutions to the equation are $x=2$ and $x=3$.
Video Example
Prof. Robin Johnson factorises the expression $x^2+x-6$.
The Quadratic Formula
The quadratic formula can be used to solve any quadratic expression of the form $ax^2+bx+c=0$ (where $a\neq0$). The formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Not only can we use this formula directly to determine the solutions (or roots) of a quadratic, but the expression inside the square root (called the discriminant) tells us how many solutions we can
expect to find and whether or not it is possible to factorise the quadratic:
• If $b^2-4ac\gt0$ there are two distinct real roots and it is possible to factorise the quadratic.
• If $b^2-4ac=0$ then there is only one root (called a repeated real root) and we cannot factorise the quadratic.
• $b^2-4ac\lt0$ there are no real roots (each root will contain an imaginary number).
See The Quadratic Formula for more information and examples of using the quadratic formula to solve quadratic equations.
Simultaneous Quadratic Equations
We can solve pairs of simultaneous quadratic equations. See Solving Simultaneous Linear Equations (Economics) for a reminder on how to solve simultaneous equations. Simultaneous quadratic equations
can have up to $2$ pairs of solutions.
Worked Example
Solve the following pair of simultaneous equations: \[x^2+y^2=29\] \[y-x=3\]
Here we will use the substitution method. Let's first label the equations: \[x^2+y^2=29\;\;\;\textbf{(1)}\] \[y-x=3\;\;\;\textbf{(2)}\] Since equation $\textbf{(1)}$ contains $x^2$ and $y^2$ as
variables, it is easier to rearrange equation $\textbf{(2)}$ to make either $x$ or $y$ the subject and then substitute this into equation $\textbf{(1)}$. Rearranging equation $\textbf{(2)}$ to make
$y$ the subject gives: \[y=3+x\;\;\;\textbf{(2)}\] We can now substitute this directly into equation $\textbf{(1)}$: &x^2+(3+x)^2=29\\ \Rightarrow &2x^2+6x+9=29\\ \Rightarrow &2x^2+6x-20=0 We now
have a single quadratic equation which we can solve using the methods discussed above. Since this equation is difficult to factorise, we will use the quadratic formula. This gives $x=-5$ or $x=2$
(check for yourself). Substituting $x=-5$ into equation $\textbf{(2)}$ gives: y&-(-5)=3\\ \Rightarrow &y=-2 So $x=-5$ and $y=-2$ is one solution. We will now find the other solution. Substituting $x=
2$ into equation $\textbf{(2)}$ gives: y&-2=3\\ \Rightarrow &y=5 So the other solution is $x=2$ and $y=5$.
Quadratic Functions
A quadratic function is a function of the general form \[y = ax^2+bx+c\] where $a$, $b$ and $c$ are real numbers and $a\neq0$. These functions are called quadratic as $x$ is raised to the power of
The roots of a linear function are the values of $x$ which make the right-hand side of the function equal to zero. That is, the value of $x$ which satisfy: \[ax^2+bx+c=0\] The roots of a quadratic
function correspond to the solutions of a quadratic equation. Quadratic functions can have two, one or no real roots. We can find the roots of a quadratic function by solving the equation $ax^2+bx+c=
0$ using either factorisation or the quadratic formula.
The graph of a quadratic function is “U-shaped” and is called a parabola. The point where the parabola changes direction is called the turning point of the function.
If $a\gt0$, then we have an upwards-opening parabola which looks like this $\cup$ and the turning point is called a global minimum because it is the point where the graph of the function is at its
If $a\lt0$, then we have an downwards-opening parabola which looks like this $\cap$ and the turning point is called a global maximum because it is the point where the graph of the function is at its
The absolute value of $a$ determines how steeply the parabola curves. The greater the absolute value of $a$, the steeper the curve is.
If $b=0$, the parabola goes through the origin (where $x$ and $y$ are both equal to zero).
The value of $c$ is the $y$-axis intercept. The $x$-axis intercept(s) correspond to the roots of the function (see above).
The best way to plot the graph of a quadratic function is to first make a table with a few $x$ values and the corresponding $y$ values and then plot each of these points.
Cubic Functions
A cubic function is a function of the general form \[y = ax^3+bx^2+cx+d\] where $a$, $b$, $c$ and $d$ are real numbers and $a\neq0$. These functions are called cubic as $x$ is raised to the power of
$3$. A cubic function can have up to $3$ roots. The graph of a cubic function is an “S” shape.
If $a\gt 0$, the graph will look like this:
If $a\lt 0$, the graph will look like this:
Cubic functions can have up to three real roots. We can find the roots of a cubic function by solving the equation $ax^3+bx^2+cx+d=0$.
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.
Test Yourself
Test yourself: Quadratic expressions and equations
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Constructing a logical, regular axis topology from an irregular topology
Constructing a logical regular topology from an irregular topology including, for each axial dimension and recursively, for each compute node in a subcommunicator until returning to a first node:
adding to a logical line of the axial dimension a neighbor specified in a nearest neighbor list; calling the added compute node; determining, by the called node, whether any neighbor in the node's
nearest neighbor list is available to add to the logical line; if a neighbor in the called compute node's nearest neighbor list is available to add to the logical line, adding, by the called compute
node to the logical line, any neighbor in the called compute node's nearest neighbor list for the axial dimension not already added to the logical line; and, if no neighbor in the called compute
node's nearest neighbor list is available to add to the logical line, returning to the calling compute node.
Issue Date:
Research Org.:
International Business Machines Corporation, Armonk, NY (USA).
Sponsoring Org.:
OSTI Identifier:
Patent Number(s):
Application Number:
International Business Machines Corporation (Armonk, NY)
Patent Classifications (CPCs):
H - ELECTRICITY H04 - ELECTRIC COMMUNICATION TECHNIQUE H04L - TRANSMISSION OF DIGITAL INFORMATION, e.g. TELEGRAPHIC COMMUNICATION
DOE Contract Number:
Resource Type:
Country of Publication:
United States
97 MATHEMATICS AND COMPUTING
Citation Formats
Faraj, Daniel A. Constructing a logical, regular axis topology from an irregular topology. United States: N. p., 2014. Web.
Faraj, Daniel A. Constructing a logical, regular axis topology from an irregular topology. United States.
Faraj, Daniel A. Tue . "Constructing a logical, regular axis topology from an irregular topology". United States. https://www.osti.gov/servlets/purl/1136402.
title = {Constructing a logical, regular axis topology from an irregular topology},
author = {Faraj, Daniel A.},
abstractNote = {Constructing a logical regular topology from an irregular topology including, for each axial dimension and recursively, for each compute node in a subcommunicator until returning to a
first node: adding to a logical line of the axial dimension a neighbor specified in a nearest neighbor list; calling the added compute node; determining, by the called node, whether any neighbor in
the node's nearest neighbor list is available to add to the logical line; if a neighbor in the called compute node's nearest neighbor list is available to add to the logical line, adding, by the
called compute node to the logical line, any neighbor in the called compute node's nearest neighbor list for the axial dimension not already added to the logical line; and, if no neighbor in the
called compute node's nearest neighbor list is available to add to the logical line, returning to the calling compute node.},
doi = {},
journal = {},
number = ,
volume = ,
place = {United States},
year = {2014},
month = {7}
Works referenced in this record: | {"url":"https://www.osti.gov/doepatents/biblio/1136402","timestamp":"2024-11-04T08:22:00Z","content_type":"text/html","content_length":"401910","record_id":"<urn:uuid:dc874b62-b3f1-4a94-8556-de543f53517c>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00146.warc.gz"} |
EViews Help: Estimating VEC Models in EViews
Estimating VEC Models in EViews
Estimation of VEC model in EViews is a special case of estimation in a var object. From the main application menu of an existing var object, click on the button to open the estimation dialog.
Alternately, you may create a new VAR object by selecting Object/New Object... group, then selecting . Once the dialog appears, select in the dropdown menu to display the VEC estimation dialog:
Once you have filled the in the dialog, simply click OK to estimate the VEC. Estimation of a VEC model is carried out in two steps. In the first step, we estimate the cointegrating relations from the
Johansen procedure as used in the cointegration test. We then construct the error correction terms from the estimated cointegrating relations and estimate a VAR in first differences including the
error correction terms as regressors.
There are three tabs in the dialog: , , and . We discuss each of these tabs in turn.
Basic Specification
In the Basics tab, you will provide the usual information about the , ,and lists of different types of :
• Importantly, in contrast to the standard VAR case, the specification refers to lags of the first difference terms in the conditional EC representation of the VEC. For example, the lag specification
“1 1” will include lagged first difference terms on the right-hand side of the VEC. Rewritten in levels, this VEC is a restricted VAR with two lags. To estimate a VEC with no lagged first difference
terms, specify the lag as “0 0”
• The section allows you to specify exogenous variables that are not included in the standard in-built deterministic trend cases. This convention means that the constant and linear trend term should
not be included in the Exogenous variables edit boxes. The constant and trend specification for VECs should be specified using the dropdown menu in the Cointegration tab.
You should enter any other variables in the edit field corresponding to whether they appear in the , , or lists of variables.
Note that the exogenous variables entered in the VEC estimation dialog refer to short-run variables added to the VEC difference specification
Equation (45.41)
. This treatment is in contrast to that of built-in deterministics in VEC estimation and of exogenous variables when estimating a VAR, where the specified variables are entered into the levels
Equation (45.1)
. (See
“Exogenous Variables in VECMs”
for further discussion.)
Cointegration Options
Important options related to cointegration can be accessed by clicking on the Cointegration tab:
A fundamental prerequisite of VECM estimation is
a priori
knowledge of the number of cointegrating relations (see
“Johansen Cointegration Test”
for determination of cointegrating rank).
Note that as you make a selection in the dropdown, the text below will change to give you a more detailed description of the assumptions underlying the choice.
VEC Restrictions
Since the cointegrating vector
To impose restrictions in estimation, click on the tab to display the restrictions dialog. You will enter your restrictions in the edit box that appears when you check the box:
“Specifying VEC Restrictions”
describes the syntax for specifying these restrictions in greater detail.
Specifying VEC Restrictions
Restrictions can be imposed on the cointegrating vector (elements of the
Restrictions on the Cointegrating Vector
To impose restrictions on the cointegrating vector i,j)-th element of the transpose of the B(i,j). The i-th cointegrating relation has the representation:
B(i,1)*y1 + B(i,2)*y2 + ... + B(i,k)*yk
where y1, y2, ... are the (lagged) endogenous variable. Then, if you want to impose the restriction that the coefficient on y1 for the second cointegrating equation is 1, you would type the following
in the edit box:
B(2,1) = 1
You can impose multiple restrictions by separating each restriction with a comma on the same line or typing each restriction on a separate line. For example, if you want to impose the restriction
that the coefficients on y1 for the first and second cointegrating equations are 1, you would type:
B(1,1) = 1
B(2,1) = 1
Currently all restrictions must be linear (or more precisely affine) in the elements of the . So for example
B(1,1) * B(2,1) = 1
will return a syntax error.
Restrictions on the Adjustment Coefficients
To impose restrictions on the adjustment coefficients, you must refer to the (i,j)-th elements of the A(i,j). The error correction terms in the i-th VEC equation will have the representation:
A(i,1)*CointEq1 + A(i,2)*CointEq2 + ... + A(i,r)*CointEqr
Restrictions on the adjustment coefficients are currently limited to linear homogeneous restrictions so that you must be able to write your restriction as
A(1,1) = A(2,1)
is valid but:
A(1,1) = 1
will return a restriction syntax error.
One restriction of particular interest is whether the i-th row of the i-th endogenous variable is said to be weakly exogenous with respect to the . See Johansen (1995) for the definition and
implications of weak exogeneity. For example, if we assume that there is only one cointegrating relation in the VEC, to test whether the second endogenous variable is weakly exogenous with respect to
A(2,1) = 0
To impose multiple restrictions, you may either separate each restriction with a comma on the same line or type each restriction on a separate line. For example, to test whether the second endogenous
variable is weakly exogenous with respect to
A(2,1) = 0
A(2,2) = 0
You may also impose restrictions on both and independent. So for example,
A(1,1) = 0
B(1,1) = 1
is a valid restriction but:
A(1,1) = B(1,1)
will return a restriction syntax error.
Identifying Restrictions and Binding Restrictions
EViews will check to see whether the restrictions you provided identify all cointegrating vectors for each possible rank. The identification condition is checked numerically by the rank of the
appropriate Jacobian matrix; see Boswijk (1995) for the technical details. Asymptotic standard errors for the estimated cointegrating parameters will be reported only if the restrictions identify the
cointegrating vectors.
If the restrictions are binding, EViews will report the LR statistic to test the binding restrictions. The LR statistic is reported if the degrees of freedom of the asymptotic e.g. when you impose
restrictions on the adjustment coefficients but not on the cointegrating vector).
Options for Restricted Estimation
Estimation of the restricted cointegrating vectors tab provides iteration control for the maximum number of iterations and the convergence criterion. EViews estimates the restricted
Once you have filled the dialog, simply click OK to estimate the VEC. Estimation of a VEC model is carried out in two steps. In the first step, we estimate the cointegrating relations from the
Johansen procedure as used in the cointegration test. We then construct the error correction terms from the estimated cointegrating relations and estimate a VAR in first differences including the
error correction terms as regressors. | {"url":"https://help.eviews.com/content/vecm-Estimating_VEC_Models_in_EViews.html","timestamp":"2024-11-06T11:12:23Z","content_type":"application/xhtml+xml","content_length":"27798","record_id":"<urn:uuid:30af033a-ea95-474b-bf8a-5a1ad3431985>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00430.warc.gz"} |
Median of Two Sorted Arrays – LeetCode Hard Solutions
Are you tired of being average? Well, data analysis has got you covered! Meet the median, the ultimate middle child of statistics. It’s like the cool cousin who shows up to family reunions and says,
“Hey, I’m not the highest, nor the lowest, but I’m here, and I’m representative!” LeetCode’s 4th Hard problem, “Median of Two Sorted Arrays,” is like the median’s coming-out party – and you’re
Problem Statement
Finding the Median of Two Sorted Arrays
Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.
• 0 <= m, n <= 1000
• 1 <= m + n <= 2000
• -106 <= nums1[i], nums2[i] <= 106
What’s the Catch?
• The arrays are sorted, but merging them would be inefficient.
• The median might not be an integer.
• It would be best if you found a solution that’s faster than O((m+n) log(m+n)).
Your Mission:
Find the median of two sorted arrays without merging them, optimizing for time and space complexity.
Are you Excited?
Approaches and Solutions
A. Brute Force Approach (Merging and Sorting)
The brute force approach involves merging the two sorted arrays into one and then finding the median.
Code Implementation (Python):
def findMedianSortedArrays(nums1, nums2):
# Merge and sort the arrays
merged = sorted(nums1 + nums2)
# Find the median
length = len(merged)
if length % 2 == 0:
return (merged[length // 2 - 1] + merged[length // 2]) / 2
return merged[length // 2]
1. Merge nums1 and nums2 into a single array merged.
2. Sort merged in ascending order.
3. Calculate the median based on the length of merged.
Why This Approach is Not Ideal:
1. Time Complexity: O((m+n) log(m+n)) due to sorting, where m and n are the lengths of nums1 and nums2.
2. Space Complexity: O(m+n) for storing the merged array.
3. Inefficiency: Merging and sorting large arrays can be slow.
1. This approach is impractical for large inputs.
2. It doesn’t utilize the fact that the input arrays are already sorted.
B. Optimized Approach (Binary Search)
The optimized approach leverages binary search to find the median without merging the arrays.
Key Insights:
1. The median is the middle element in the combined sorted array.
2. We can find the median by partitioning the combined array into two halves.
Code Implementation (Python):
def findMedianSortedArrays(nums1, nums2):
if len(nums1) > len(nums2):
nums1, nums2 = nums2, nums1
x, y = len(nums1), len(nums2)
start = 0
end = x
while start <= end:
partitionX = (start + end) // 2
partitionY = (x + y + 1) // 2 - partitionX
maxLeftX = float('-inf') if partitionX == 0 else nums1[partitionX - 1]
minRightX = float('inf') if partitionX == x else nums1[partitionX]
maxLeftY = float('-inf') if partitionY == 0 else nums2[partitionY - 1]
minRightY = float('inf') if partitionY == y else nums2[partitionY]
if maxLeftX <= minRightY and maxLeftY <= minRightX:
if (x + y) % 2 == 0:
return (max(maxLeftX, maxLeftY) + min(minRightX, minRightY)) / 2.0
return max(maxLeftX, maxLeftY)
elif maxLeftX > minRightY:
end = partitionX - 1
start = partitionX + 1
# Return statement for edge cases or errors
return None # or raise ValueError("Invalid input")
1. Ensure nums1 is the smaller array.
2. Initialize start = 0 and end = nums1.length.
3. Calculate the partition point partitionX = (start + end) / 2.
4. Calculate the corresponding partition point partitionY = (nums1.length + nums2.length + 1) / 2 - partitionX.
5. Compare elements at partitionX and partitionY.
6. Adjust start and end based on the comparison.
7. Repeat steps 3-6 until start <= end.
Time & Space Complexity:
• Time Complexity: O(log(min(m,n)))
• Space Complexity: O(1)
1. Efficient for large inputs.
2. Utilizes the fact that input arrays are sorted.
The optimized approach using binary search efficiently finds the median without merging the arrays.
Finding Medians in Sorted Arrays: Key Takeaways and Next Steps
In this article, we explored the crucial problem of finding medians in sorted arrays. We discussed:
Key Takeaways:
1. Importance of Medians: Understanding data distribution and central tendency.
2. Brute Force Approach: Merging and sorting arrays, with O((m+n) log(m+n)) time complexity.
3. Optimized Approach: Binary search, achieving O(log(min(m,n))) time complexity.
4. Code Implementations: Efficient solutions in Python.
What’s Next?
1. Practice: Solve similar problems on LeetCode, HackerRank, and other platforms.
2. Real-World Applications: Apply median-finding skills in data analysis, machine learning, and database optimization.
3. Explore Variations: Modify the algorithm for multiple sorted arrays, weighted medians, or streaming data.
Finding medians in sorted arrays is a fundamental problem with numerous applications. By mastering both brute force and optimized approaches, you’ll enhance your problem-solving skills and become
proficient in efficient algorithm design. | {"url":"https://techalgospotlight.com/median-of-two-sorted-arrays-leetcode-hard-solutions/","timestamp":"2024-11-04T17:10:52Z","content_type":"text/html","content_length":"288377","record_id":"<urn:uuid:576f82c3-6afb-4182-a732-18c52e5f003a>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00023.warc.gz"} |
In the first course, the students develop a foundation for their future mathematical knowledge by studying number systems and working with whole numbers, integers, and fractions, as well as
measurements of different types, leading up to proportions, ratios and rates.
[Ver: 2.0.0.0]
In the second course, the material expands to include logic and settheory, as well as an overview of the Pythagorean theorem and its applications which play an important role in advanced mathematics.
There is a logical progression to the next lessons, which cover volume and surface area of basic shapes such as prisms, cylinders and spheres, as well as an introduction to exponents and their
[Catalog No. 88-3136-000]
[Ver 2.0.0.0]
the students become well-versed in many fundamental mathematical principles, including basic algebraic tenets. The first lessons begin with monomial and polynomial expressions and their various
applications, leading to solving equations and the role of the law of order of operations.
[Ver: 2.0.0.0] | {"url":"https://demo.intelitek.com/course/index.php?categoryid=50&lang=pt_br","timestamp":"2024-11-07T23:27:10Z","content_type":"text/html","content_length":"49830","record_id":"<urn:uuid:9390775a-28b7-45e0-825c-567d2c628d18>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00514.warc.gz"} |
AI solution for JavaScript Coding Question on BFE.dev
This solution is generated by ChatGPT, correctness is not guaranteed. You might also be able to find a solution fromcommunity posts.
AI solution for JavaScript Coding Question on BFE.dev
116. implement Object.is()
Sure, here's an implementation of Object.is():
function is(x, y) { if (x === y) { // Handle +0 == -0 return x !== 0 || y !== 0 || 1 / x === 1 / y; } else { // Handle NaN == NaN return x !== x && y !== y; }}
The function first checks if x and y are strictly equal (===) and returns true if so. However, it also checks for two special cases: if x and y are both 0, it returns false if one is positive and the
other is negative. This correctly handles the fact that JavaScript distinguishes between positive and negative zero. If x and y are not strictly equal, the function returns true if both are NaN. This
ensures that NaN is considered equal to itself, as required by the spec. | {"url":"https://bigfrontend.dev/problem/implement-Object.is/aisolution","timestamp":"2024-11-04T14:08:41Z","content_type":"text/html","content_length":"36466","record_id":"<urn:uuid:3047ff9e-35f4-406f-8189-0cf2a3eec962>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00863.warc.gz"} |
From Encyclopedia of Mathematics
The superscript notation $f^n$ for the coordinate functions is very strange -- I think it is likely to be confused with the $n$th derivative or the $n$th power. --Jjg 02:14, 3 August 2012 (CEST)
The derivative would be $f^{(n)}$. The power is indeed a problem. But "very strange" is an overstatement. I saw such notation many times, since the tensor notation stipulates upper and lower
indices (contravariant and covariant...). Maybe in such cases we should add a note like "(upper index, not a power)". --Boris Tsirelson 07:41, 3 August 2012 (CEST)
Indeed the notation I have chosen is the tensor one: for instance the divergence of a vector field would then be $\partial_{x_i} X^i$, using Einstein's convention on repeated indices. I am aware
of the possible confusions: the line "where $(f^1, \ldots, f^m)$ are the coordinate functions of $f$" should make things sufficiently clear though.Camillo 08:06, 3 August 2012 (CEST)
How to Cite This Entry:
Jacobian. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Jacobian&oldid=27338 | {"url":"https://encyclopediaofmath.org/wiki/Talk:Jacobian","timestamp":"2024-11-08T12:29:53Z","content_type":"text/html","content_length":"13595","record_id":"<urn:uuid:0edd7d23-99b3-4079-9786-231dd24df62e>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00204.warc.gz"} |
Optimizing quantum circuit using boolean algebra
Since the latest v0.1.18 version, the Qlasskit library offers two useful tool for circuit analysis and optimization.
• Decompiler: given a quantum circuit is able to detect section that can be represented as boolean expressions
• circuit_boolean_optimizer: a pipeline that given a quantum circuit, decompose it in boolean expressions form and optimize it using boolean algebra
Let’s have a look on how to use these features. We first create the following quantum circuit:
Now, using the decompile function we can detect “classical parts” and decode them as boolean expression; this is the result for the previous quantum circuit.
dc = Decompiler().decompile(qcircuit)
(0, 7)
(X, [2], None), (CCX, [0, 2, 3], None), (CCX, [1, 3, 4], None), (CX, [4, 5], None), (CCX, [1, 3, 4], None), (CCX, [0, 2, 3], None), (X, [2], None)
(q5, q4 ^ q5 ^ (q1 & (q3 ^ (q0 & ~q2))))
Reading the resulting boolean expression (q5, q4 ^ q5 ^ (q1 & (q3 ^ (q0 & ~q2)))), and knowing that q0 and q1 are inputs, while the other qubits are used as ancilla and output qubits, it is clear
that the circuit can be optimized.
qc = circuit_boolean_optimizer(qf.circuit(), preserve=[0, 1])
The circuit_boolean_optimizer allows us to perform boolean optimizations in a quantum circuit; from the previous unoptimized example, we then get the following optimized circuit:
Useful Links:
The Python notebook for this blog post is available here:
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In an age where companies are selling two-qubit quantum computers for a sum of money that would make your wallet recoil in horror, here we are, stepping off ...
This June I emerged as one of the top participants with 9 bounties collected (alongside another exceptional contributor) in the #UnitaryHack Hackathon, hoste...
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While writing a new programming language, it is often useful to write some real use-cases to test the syntax, the language expressiveness and the code cleann...
Documentation is like sex: when it is good, it is very, very good; and when it is bad, it is better than nothing
This is my new blog, based on jekyll. I’ll soon import old posts from my old blog.
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Back to Top ↑ | {"url":"https://dakk.github.io/quantumcomputing/2024/03/05/optimizing_quantum_circuit.html","timestamp":"2024-11-05T01:19:13Z","content_type":"text/html","content_length":"36661","record_id":"<urn:uuid:816c946b-a2d9-4807-8c33-9880abd16f40>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00479.warc.gz"} |
Syllabus | Machine Learning | Electrical Engineering and Computer Science | MIT OpenCourseWare
Course Meeting Times
Lectures: 2 sessions / week, 1.5 hours / session
A list of topics covered in the course is presented in the calendar.
This introductory course gives an overview of many concepts, techniques, and algorithms in machine learning, beginning with topics such as classification and linear regression and ending up with more
recent topics such as boosting, support vector machines, hidden Markov models, and Bayesian networks. The course will give the student the basic ideas and intuition behind modern machine learning
methods as well as a bit more formal understanding of how, why, and when they work. The underlying theme in the course is statistical inference as it provides the foundation for most of the methods
Problem Sets
There will be a total of 5 problem sets, due roughly every two weeks. The content of the problem sets will vary from theoretical questions to more applied problems. You are encouraged to collaborate
with other students while solving the problems but you will have to turn in your own solutions. Copying will not be tolerated. If you collaborate, you must indicate all of your collaborators.
Each problem set will be graded by a group of students with the guidance of your TAs. Each problem set will be graded in a single grading session, usually on the first Monday after it is due,
starting at 5pm. Every student is required to participate in one grading session. You should sign up for grading by contacting a TA, by email or in person; doing it early increases the chances of
getting the preferred grading schedule. Students who do not register for grading by the third week of the course, will be assigned to a problem set by us.
If you drop the class after signing up for a grading session, please be sure to let us know so we can keep track of students available for grading. If you add the class during the term, please
remember to sign up for grading.
There will be two in-class exams, a midterm midway through the term and a final the last day of class.
You are required to complete a class project. The choice of the topic is up to you so long as it clearly pertains to the course material. To ensure that you are on the right track, you will have to
submit a one paragraph description of your project a month before the project is due. Similarly to problem sets, you are encouraged to collaborate on the project. We expect a four page write-up about
the project, which should clearly and succinctly describe the project goal, methods, and your results. Each group should submit only one copy of the write-up and include all the names of the group
members (a two person group will have 6 pages, a three person group will have 8 pages, and so on). The projects will be graded on the basis of your understanding of the overall course material (not
based on, e.g., how brilliantly your method works). The scope of the project is about 1-2 problem sets.
The projects are due in Lec #23. Electronic submission is required but we can accept only postscript or pdf documents. The short proposal should be turned in on or before Lec #12.
The projects can be literature reviews, theoretical derivations or analyses, applications of machine learning methods to problems you are interested in, or something else (to be discussed with course
Your overall grade will be determined roughly as follows:
Midterm 15%
Problem sets 30%
Final 25%
Project 30%
There are a number of useful texts for this course but each covers only some part of the class material.
Bishop, Christopher. Neural Networks for Pattern Recognition. New York, NY: Oxford University Press, 1995. ISBN: 9780198538646.
Duda, Richard, Peter Hart, and David Stork. Pattern Classification. 2nd ed. New York, NY: Wiley-Interscience, 2000. ISBN: 9780471056690.
Hastie, T., R. Tibshirani, and J. H. Friedman. The Elements of Statistical Learning: Data Mining, Inference and Prediction. New York, NY: Springer, 2001. ISBN: 9780387952840.
MacKay, David. Information Theory, Inference, and Learning Algorithms. Cambridge, UK: Cambridge University Press, 2003. ISBN: 9780521642989. Available on-line here.
Mitchell, Tom. Machine Learning. New York, NY: McGraw-Hill, 1997. ISBN: 9780070428072.
You are responsible for the material covered in lectures (most of which will appear in lecture notes in some form), problem sets, as well as material specifically made available and indicated for
this purpose. The weekly recitations/tutorials will be helpful in understanding the material and solving the homework problems.
Recommended Citation
For any use or distribution of these materials, please cite as follows:
Tommi Jaakkola, course materials for 6.867 Machine Learning, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
LEC # TOPICS KEY DATES
1 Introduction, linear classification, perceptron update rule
2 Perceptron convergence, generalization
3 Maximum margin classification
4 Classification errors, regularization, logistic regression Problem set 1 out
5 Linear regression, estimator bias and variance, active learning
6 Active learning (cont.), non-linear predictions, kernals Problem set 1 due
7 Kernal regression, kernels Problem set 2 out
8 Support vector machine (SVM) and kernels, kernel optimization
9 Model selection Problem set 2 due
10 Model selection criteria
11 Description length, feature selection Problem set 3 out 3 days before Lec #11
12 Combining classifiers, boosting
Problem set 3 due
13 Boosting, margin, and complexity
Problem set 4 out
14 Margin and generalization, mixture models
15 Mixtures and the expectation maximization (EM) algorithm
16 EM, regularization, clustering Problem set 4 due
17 Clustering
18 Spectral clustering, Markov models Problem set 5 out
19 Hidden Markov models (HMMs)
20 HMMs (cont.)
21 Bayesian networks
22 Learning Bayesian networks Problem set 5 due
Probabilistic inference
23 Projects due
Guest lecture on collaborative filtering
24 Current problems in machine learning, wrap up Exams back | {"url":"https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/pages/syllabus/","timestamp":"2024-11-09T03:04:23Z","content_type":"text/html","content_length":"60020","record_id":"<urn:uuid:10ffcbcf-d9e8-4980-be5f-0326172d9699>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00809.warc.gz"} |
Properties of Numbers
Steinfelds, Leon King High School
To contrast the "mean" behavior of 0, 1, 2 and regular numbers.
Numbers will be likened to students.
Apparatus Needed:
chalk board, chalk.
Recommended Strategy:
Numbers 0, 1 and 2 will be regarded as "dissident" students. 0 will
act as "murderer" in multiplication, but will not even touch numbers
when addition or subtraction is required. To divide by 0 creates
"mission impossible". 1 does not like division and multiplication. It
is invisible as a coefficient or exponent (like Dracula). 2 hides only
once in radical (square root). Zero is the only lonely number. The
others walk in pairs on a number line as twins. Some numbers, like
square root of 2, 5, 7..., cannot be written even in one hundred years,
but one knows their location on the number line.
Return to Mathematics Index | {"url":"https://smileprogram.info/ma8813.html","timestamp":"2024-11-09T03:45:39Z","content_type":"text/html","content_length":"1460","record_id":"<urn:uuid:66f5e1f0-cdd0-4b9e-8cfe-ed7ab2510383>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00144.warc.gz"} |
XGBoost Confidence Interval using Jackknife Resampling
Jackknife resampling provides an alternative to the bootstrap for estimating confidence intervals of XGBoost model performance metrics, particularly when computational efficiency is less of a
Unlike the bootstrap, which requires fitting the model on numerous resampled datasets, the Jackknife method refits the model only once for each observation in the original dataset. This can be
computationally prohibitive for large datasets, but can be less computationally intensive than the bootstrap for smaller datasets.
This example demonstrates how to use Jackknife resampling to estimate a 95% confidence interval for the error of an XGBoost model trained on a synthetic regression dataset.
# XGBoosting.com
# Evaluate XGBoost Model Performance Confidence Intervals using Jackknife Resampling
from sklearn.datasets import make_regression
from sklearn.model_selection import train_test_split
from xgboost import XGBRegressor
import numpy as np
# Generate a synthetic regression dataset
X, y = make_regression(n_samples=100, n_features=10, n_informative=5, noise=0.1, random_state=42)
# Split data into train and test sets
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)
# Define a function to compute Jackknife replicates of an error metric
def jackknife(model, X, y):
n = len(X)
scores = []
for i in range(n):
X_jack = np.delete(X, i, axis=0)
y_jack = np.delete(y, i)
model.fit(X_jack, y_jack)
y_pred = model.predict(X[[i]])
abs_error = abs(y[i] - y_pred[0])
return np.array(scores)
# Instantiate an XGBRegressor with default hyperparameters
model = XGBRegressor(random_state=42)
# Compute the Jackknife confidence interval for abs error
error_scores = jackknife(model, X_train, y_train)
ci_low, ci_high = np.percentile(error_scores, [2.5, 97.5])
print(f"Mean Absolute Error: {error_scores.mean():.3f}")
print(f"95% CI: [{ci_low:.3f}, {ci_high:.3f}]")
The code first generates a synthetic regression dataset using scikit-learn’s make_regression function and splits the data into train and test sets.
Next, we define a jackknife function that takes a model and training data as input. For each observation in the training data, this function leaves out that observation, refits the model on the
remaining data, and computes the absolute error on the left-out observation. The function returns an array of Jackknife errors.
An XGBClassifier is instantiated with default hyperparameters, and the jackknife function is called with the model and training data to compute the Jackknife replicates of absolute error.
Finally, the 2.5th and 97.5th percentiles of the Jackknife absolute error replicates are computed to obtain the 95% confidence interval bounds. The mean absolute error and confidence interval are | {"url":"https://xgboosting.com/xgboost-confidence-interval-using-jackknife-resampling/","timestamp":"2024-11-13T06:10:56Z","content_type":"text/html","content_length":"11115","record_id":"<urn:uuid:64762742-b058-40e7-9f20-d3056c62c9f8>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00421.warc.gz"} |
Life of Fred
Life of Fred Fractions $24 : Less Than, Billion, Cardinal and Ordinal Numbers, Diameter and Radius, Savings and Expenses, Definition of a Fraction, Sectors, Comparing Fractions, Reducing Fractions,
Adding and Subtracting Fractions, Common Denominators, Roman Numerals, Least Common Multiples, Improper Fractions, Lines of Symmetry, Division by Zero, Circumference, Multiplying Mixed Numbers,
Commutative Law, Canceling, Definition of a Function, Area, Unit Analysis, Division of Fractions, Geometric Figures, Estimating Answers. ISBN: 978-0-9709995-9-7, hardback, 192 pages. $24 | {"url":"http://www.horriblebooks.com/Life%20of%20Fred%20Pages/Life%20of%20Fred%20Fractions.htm","timestamp":"2024-11-12T12:47:01Z","content_type":"text/html","content_length":"1860","record_id":"<urn:uuid:dcb2994f-2656-4d46-acea-5ece41923ef3>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00460.warc.gz"} |
Search Results for 'Math' - 153 Courses from 60 Universities | CourseBuffet - Find Free Online Courses (MOOCs)
Introduction to basic algebraic operations and concepts, as well as the structure and use of algebra.
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This course is taught so that students will acquire a solid foundation in algebra. The course concentrates on the various functions that are important to the stu...
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Learn the basics of Algebra through intuition and problem solving! From fractions to factors to functions, we’ll cover a breadth of topics.
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En este curso se recordará lo que es una ecuación con una única incógnita y cómo solucionarla. A partir de ahí se tratarán: Los sistemas de ecuaciones lineales...
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Gain an in-depth understanding of algebraic principles and learn how to use them to solve problems you may meet in everyday life.
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In this college level Algebra course, you will learn to apply algebraic reasoning to solve problems effectively. You’ll develop skills in linear and quadratic fu...
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How can you use more mindfully the scale? | BALANCED by Ana
How can you use more mindfully the scale?
When you step on the scale you can see either a higher or a lower number than what you expected. Almost no one steps on the scale to get the “proof” that they weigh exactly what they thought.
So what happens when the number is bigger?
It can throw you off into a binge. Why? Because it’s much easier to start thinking that the number that you see is a proof that you just can’t lose the weight so why bother? You get all these
negative feelings running around and the easiest way to soothe them is to binge and calm yourself down.
Now two scenarios If you see a lower number:
1. People get a high but then again a worry: How on earth will I eat for the rest of my life under severe restrictions in order to keep this number?
2. If you see a lower number you can also get into the mindset: “This number is low, therefore I can eat a bit more now because I have some kg’s left until I become fat again”
If the scale dictates you mood, if it dictates how you feel about yourself and what you will eat then don’t weigh yourself daily.
If you can’t accept the fact that the scale gives you DATA that has to change on a daily basis due to numerous factors (what you ate, amount of salt, water in the body, hormones, sleep, stress levels
etc…) then don’t weight in yourself daily.
Here is what you can do:
1. Weight in yourself every Monday for example and compare the data
2. Compare the weight from month to month
If you still decide to weigh in yourself daily, take the 7 numbers from one week and get the average weight. Now compare that one with the average one for the next week and then the next one etc… | {"url":"https://balancedbyana.com/blog/how-can-you-use-more-mindfully-the-scale/","timestamp":"2024-11-05T13:06:32Z","content_type":"text/html","content_length":"35070","record_id":"<urn:uuid:34b09480-3f4c-48a1-aa92-4747d5b9cb58>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00662.warc.gz"} |
Brute Force
This lesson gives a brief introduction to the brute force paradigm using linear search in an unsorted list.
We'll cover the following
Brute force method
Let’s start off our discussion on algorithms with the most straightforward and exhaustive option—the Brute Force approach to solving a problem. This method requires us to go through all the
possibilities to find a solution to the problem we want to solve. For instance, if we have a list of integers and we want to find the minimum, maximum, or a certain element in that list, the Brute
Force approach requires us to go through all the elements to find that specific element.
Let’s look at an example that might help you visualize this technique.
Example: linear search
Linear/Sequential Search is a method for finding a target value within a given list. It sequentially checks each element of the list for the target value until a match is found or all the elements
have been searched. Linear Search runs in linear time (at its worst) and makes n comparisons (at most), where n is the length of the list.
Let’s assume that we are given the following list of unsorted integers.
Level up your interview prep. Join Educative to access 80+ hands-on prep courses. | {"url":"https://www.educative.io/courses/algorithms-coding-interviews-python/brute-force","timestamp":"2024-11-15T00:49:13Z","content_type":"text/html","content_length":"748938","record_id":"<urn:uuid:60d3c1f9-3c61-4f09-bd19-9062a0bb62ee>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00347.warc.gz"} |
Argued September 18, 1961
Petition granted September 22, 1961
Vernon Cook, Gresham, argued the cause and submitted a brief for petitioners Charles McKinley, Donald G. Balmer and Howard E. Dean. With him on the brief were Dirk D. Snel and Reuben Lenske,
Edward N. Fadeley, Eugene, argued the cause and submitted a brief in propria persona.
Wm. S. McLennan, Portland, argued the cause and submitted a brief for petitioner Eleanor P. Kafoury.
Reuben Lenske, Portland, submitted a brief amicus curiae pro se.
Edwin J. Peterson, Portland, and Douglas R. Spencer, Eugene, argued the cause and submitted a brief amicus curiae in support of Chapter 482, Oregon Laws 1961. With them on the brief were Clay Myers,
Portland; Robert E. Jones, Portland; Ken Maher, Portland; George Annala, Hood River; F.F. Montgomery, Eugene, and Victor Atiyeh, Portland.
Robert Y. Thornton, Attorney General, Salem, submitted a brief amicus curiae in opposition. With him on the brief was Louis S. Bonney, Assistant Attorney General, Salem.
This is an original proceeding in which the petitioners seek a judgment declaring unconstitutional Chapter 482, Oregon Laws 1961, which purports to reapportion representation in the Oregon
legislative assembly.
Original jurisdiction is conferred upon this court by virtue of Article IV, § 6 (2) (a), which reads as follows: "Original jurisdiction hereby is vested in the Supreme Court upon the petition of any
qualified elector of the state filed with the Clerk of the Supreme Court prior to September 1 of the year in which the Legislative Assembly enacts a reapportionment measure, to review any measure so
The provisions for the apportionment of senators and representatives within this state are contained in Article IV, § 6 of the Oregon Constitution. The method for determining the number of senators
and representatives for each county or district is set forth in subsection (1) of Article IV, § 6:
"(1) The number of senators and representatives shall, at the session next following an enumeration of the inhabitants by the United States government, be fixed by law and apportioned among the
several counties according to the population in each. The ratio of senators and representatives, respectively, shall be determined by dividing the total population of the state by the number of
senators and by the number of representatives. The number of senators and representatives for each county or district shall be determined by dividing the total population of such county or
district by such respective ratios; and when a fraction exceeding one-half results from such division, such county or district shall be entitled to a member for such fraction. In case any county
does not have the requisite population to entitle it to a member, then such county shall be attached to some adjoining county or counties for senatorial or representative purposes."
The total population of Oregon according to the official federal census of 1960, as reported on November 29, 1960, was 1,768,687. The application of the constitutional formula to this figure produces
a senatorial ratio of 58,956 and a representative ratio of 29,478. The number of senators and representatives to which each county or district is entitled is obtained by dividing, respectively,
58,956 and 29,478 into the population of the county or district. This computation when applied to each of the counties (as distinguished from the senatorial or representative districts) yields the
following result:
The senatorial ratio, based upon a senate consisting of 30 members, results from dividing 1,768,687 by 30. The representative ratio, based on a house of 60 members, results from dividing 1,768,687 by
Population Divided by _______________________________ 1960 Senatorial Representative County Population Ratio Ratio
Baker 17,295 .293 .587 Benton 39,165 .664 1.329 Clackamas 113,038 1.917 3.835 Clatsop 27,380 .464 .929 Columbia 22,379 .380 .759 Coos 54,955 .932 1.864 Crook 9,430 .160 .320 Curry 13,983 .237
.475 Deschutes 23,100 .392 .784 Douglas 68,458 1.161 2.322 Gilliam 3,069 .052 .104 Grant 7,726 .131 .262 Harney 6,744 .114 .229 Hood River 13,395 .227 .454 Jackson 73,962 1.255 2.509 Jefferson
7,130 .121 .242 Josephine 29,917 .507 1.015 Klamath 47,475 .805 1.611 Lake 7,158 .121 .243 Lane 162,890 2.763 5.526 Lincoln 24,635 .418 .836 Linn 58,867 .998 1.997 Malheur 22,764 .386 .772 Marion
120,888 2.050 4.101 Morrow 4,871 .083 .165 Multnomah 522,813 8.868 17.736 Polk 26,523 .450 .900 Sherman 2,446 .041 .083 Tillamook 18,955 .322 .643 Umatilla 44,352 .752 1.505 Union 18,180 .308
.617 Wallowa 7,102 .120 .241 Wasco 20,205 .343 .685 Washington 92,237 1.565 3.129 Wheeler 2,722 .046 .092 Yamhill 32,478 .551 1.102
From the foregoing tabulation it is apparent that, on a county basis, if effect is given to the constitutional provision awarding a senator and a representative to a county or district "when a
fraction exceeding one-half results from such division," 14 counties would be entitled to 27 senators and 24 counties would be entitled to 61 representatives. Thus, such an apportionment would exceed
the constitutional maximum set by Article IV, § 2 which contains the proviso that "the Senate shall never exceed thirty, and the House of Representatives sixty members." It is to be noted, however,
that this result could be obviated by combining, in some instances, two or more counties into a single district.
For example, Yamhill county with a senatorial ratio of .551 which, if considered alone would entitle it to one senator, could be combined with Washington county having a senatorial ratio of 1.565
producing a combined ratio of 2.116 and eliminating one senatorial seat.
It appears from assertions in the briefs that the proponents of H.B. 1665, which eventuated in Chapter 482, Oregon Laws 1961, considered it impossible to comply strictly with the constitutional
formula calling for a fractional entitlement and that it was, therefore, necessary to adjust the method of allocating senators and representatives in order to produce a valid apportionment. Under
Chapter 482 the apportionment of senatorial districts was as follows:
District No. of Districts Counties Ratio Senators
1st Marion 2.050 2 2nd Linn .998 1 3rd Lane 2.763 2 4th Douglas 1.161 1 5th Jackson 1.255 1 6th Josephine .507 1 7th Coos Curry 1.169 1 8th Yamhill .551 1 9th Washington 1.565 1 10th Tillamook
Washington 1.887 1 11th Clackamas 1.917 2 12th Multnomah 8.868 7 13th Benton .664 1 14th Clatsop Columbia .844 1 15th Lincoln Polk .868 1 16th Gilliam Hood River Morrow Sherman Wasco Wheeler .792
1 17th Umatilla .752 1 18th Baker Union Wallowa .721 1 19th Grant Harney Malheur .631 1 20th Crook Deschutes Jefferson Lake .794 1 21st Klamath .805 1
The apportionment of representatives among the representative districts was as follows:
District No. of Districts Counties Ratio Representatives
1st Clatsop .929 1 2nd Columbia .759 1 3rd Tillamook .643 1 4th Washington 3.129 3 5th Yamhill 1.102 1 6th Multnomah 17.736 16 7th Clackamas 3.835 4 8th Lincoln .836 1 9th Polk .900 1 10th Benton
1.329 1 11th Marion 4.101 4 12th Linn 1.997 2 13th Lane 5.526 5 14th Douglas 2.322 2 15th Coos 1.864 1 16th Coos Curry 2.339 1 17th Josephine 1.015 1 18th Jackson 2.509 2 19th Gilliam Hood River
Morrow Sherman Wasco Wheeler 1.583 2 20th Umatilla 1.505 2 21st Union Wallowa .858 1 22nd Crook Jefferson .562 1 23rd Baker .587 1 24th Deschutes .784 1 25th Grant Harney Lake .734 1 26th Malheur
.772 1 27th Klamath 1.611 2
It is apparent that, under the district division made by Chapter 482, if each of the districts having a major fraction were allotted a senator and a representative the number of each class would
exceed the constitutional maximum. Thus, it is seen that under Chapter 482 the division resulted in 18 whole numbers and 17 major fractions, the combination of which would call for 35 senators. If
this were the inevitable result of the application of the constitutional formula some method of reducing the number of senatorial seats would be required to avoid exceeding the constitutional maximum
of 30 senators. Similarly, the maximum for representatives would be exceeded.
It appears that the proponents of H.B. 1665 drafted the apportionment plan embodied in Chapter 482 upon the foregoing assumption and, to effect the reduction required by the constitution, first
allocated to the districts consisting of a small county or combination of small counties a senator or representative for each whole number and for each major fraction. As a consequence of this
allocation the number of available seats was reduced to the point where it became impossible to allot to the larger counties (such as Multnomah and Lane) the full number of seats called for by the
strict application of the ratios. Petitioners contend that whether or not this was the premise upon which the creators of H.B. 1665 proceeded, the apportionment made in Chapter 482 deprived the 12th
senatorial district (Multnomah county) of two senators and the 3rd senatorial district (Lane county) of one senator. And it is contended that the 6th representative district (Multnomah county) was
deprived of two representatives and that the 13th representative district (Lane county) and the 18th representative district (Jackson county) were each deprived of one representative.
It will be noted that in apportioning only seven senators to the 12th senatorial district (Multnomah county) the method adopted in Chapter 482 results not only in eliminating from the final ratio a
major fraction but a whole number as well. Thus, through the application of the constitutional formula the 12th senatorial district had a ratio of 8.868, entitling the district to nine senators if
effect is given to the major fraction. Had the major fraction alone been disregarded the district would have been allotted eight senators. Chapter 482 allotted only seven senators to the 12th
senatorial district. Conceding, without deciding at this point, that the constitution could be construed to permit the legislative assembly to disregard the major fraction under the circumstances of
this case, it is impossible for us to conceive of a reasonable interpretation of Article IV, § 6 which would permit the legislative assembly to subtract a whole number from the quotient resulting
from the application of the constitutional formula for determining representation in the state senate. The constitution clearly demands that "the number of senators and representatives for each
county or district shall be determined by dividing the total population of such county or district by such respective ratios," (i.e., the ratios determined by dividing the total population of the
state by the number of senators and by the number of representatives). The constitution then provides for the eventuality that the ratio may include a major fraction. The purpose of this sub-section
is clear; representation is to be based upon the population ratio. The ratio is to be determined arithmetically by the simple process of division. The constitution makes no mention of the use of any
other factor in making the apportionment. The constitution does not expressly or impliedly empower the legislature to adjust the representation among the districts in the state by disregarding the
results of the arithmetic process called for by Article IV. Through its authority to designate the senatorial and representative districts and, as we shall see, through its power to adjust the major
fractions, the legislature does have the power to make some adjustment in representation, but it does not have the power to adjust the ratios to the extent of disregarding a whole number once the
districts have been established and the division made upon the basis of such districts. We hold, therefore, that the elimination of a whole number from the ratio for the 12th senatorial district in
itself renders unconstitutional the apportionment plan contained in Chapter 482.
In resting our decision on the foregoing ground we have assumed, arguendo, that there could be circumstances warranting the elimination of a major fraction in an apportionment plan. We shall now
consider the validity of that assumption.
It is not unlikely that the drafters of Article IV assumed that by recognizing major fractions as whole numbers and disregarding minor fractions the total of all of the ratios would equal the number
which was being divided, i.e., that the sum of the parts thus adjusted would equal the whole. Whether or not this was the assumption, it is clear that the constitutional formula for apportionment can
produce whole numbers and major fractions exceeding the number (30 and 60 in the present case) which the ratios purport to apportion precisely. This is illustrated in the present case where the total
number of whole numbers and major fractions is the equivalent of 61 seats for the house. It is possible to conceive of situations in which the population is so distributed as to create ratios greatly
exceeding the number of seats which are to be allocated. Thus it would be possible for 30 counties to have major fractions as their ratios and the remaining six counties to have ratios expressed in
whole numbers or whole numbers and fractions. With a slightly different distribution of the population the total of all the ratios could equal far less than the number of allocable seats. This would
be the case, for example, if 30 counties had minor fraction ratios and the remaining six counties had ratios in whole numbers with or without fractions. The problem presented by an excess of major
fractions would be accentuated where districts containing only minor fraction ratios were combined to create a ratio consisting of a major fraction.
Some of the petitioners argue that the difficulties presented by an excess of major fractions can be obviated by the creation of districts through the combination of counties in such a way as to
reduce the number of major fractions and thus bring the total of whole numbers and major fractions within the constitutional maximum. For example, if two counties, each with a ratio of .625 were
combined into one district the district would be entitled to one senator, whereas, if the two counties were not combined each would be entitled to one senator. It is argued that since the legislature
can, by the creation of such districts, work out the formula in Article IV to avoid the creation of an excess of major fractions, it must do so and that it cannot disregard a major fraction if it
appears in the final ratio. The argument loses sight of the fact that whether the legislature indirectly rids a county ratio of a major fraction by combining it with another county, as illustrated
above, or directly by disregarding the major fraction without districting, the purpose, and only purpose for combining the counties into districts in this circumstance is to adjust the constitutional
formula so that it can be made to work.
It is evident, then, that any apportionment plan submitted to us must be tested in light of the inherent deficiencies of the constitutional formula. Whatever solution is suggested, it is necessary to
make some adjustment of the major fractions. In judging whether the adjustment is reasonable it must be borne in mind that the major fractions which arise out of the application of the constitutional
formula have little, if any, relation to whole numbers involved in the division of the population into units of representation. Lacking this relationship the major fractions become, in effect, a
mathematical excrescence in the apportionment process and the creator of the apportionment plan must dispose of that excrescence in some way.
In attempting to make workable an otherwise unworkable plan the legislature has the discretion to make any adjustment in the treatment of major fractions which is rational and consistent with the
fundamental constitutional requirement that apportionment be made according to the population of the state in each county or district. Thus, it would not be unreasonable to give preference to the
major fractions which most closely approximate a whole number. Or it would not be unreasonable to ignore a major fraction from the ratio of a more populous county or district having one or more whole
numbers plus a major fraction and assign a member to a less populous county or district having only a major fraction. Other adjustments consistent with the principle of apportionment among the
several counties according to population would be valid. However, the adjustment of the ratios can go no further than that which is necessary to obviate the problem created by the existence of major
fractions in the ratios. As we have already stated, the constitution cannot reasonably be interpreted to permit the elimination of a whole number from a county or district ratio. Consequently,
Chapter 482 is unconstitutional. The Secretary of State is directed to draft a reapportionment of the senators and representatives in compliance with subsection (1) of § 6, Article IV and return the
draft to this Court by October 1, 1961. The mandate shall issue forthwith. | {"url":"https://d1tvxue8o4w0fl.cloudfront.net/case/in-re-legislative-apportionment-1","timestamp":"2024-11-04T21:15:59Z","content_type":"text/html","content_length":"39110","record_id":"<urn:uuid:4b806df1-132a-4c9c-beef-7963add1c4a6>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00658.warc.gz"} |
I want to teach myself proofs. Rosen or Hammack better?
Nathaniel Garcia
I want to self-study proofs. I need something with lots of exercises and solutions to exercises since I will be doing it alone. Is Book or Proof or Discrete Math by Rosen better for self-studies? I
want a book that will ready me for a rigorous first course in Abstract Algebra, Real Analysis.
Other urls found in this thread:
You just jump into algebra or real analysis and stop being a brainlet. Undergrad analysis books are meant to introduce you to rigorous proofs anyway
Hammack or Laczkovich
Only if you read a brainlet analysis or algebra book that barely covers the subject.
Thanks for the recommendations. Hammacks book is much shorter than Rosen's. I'll go with that first and then move onward to Analysis/Algebra.
I've just finished Velleman, took a while but did all the excercises. Wonder what's your opinion on the book?
Artin, Aloffi and Tao are brainlet textbooks then. None require reading a book on proofs because that'd be retarded unless it's proof theory or whatever. But informal mathematical reasoning? Just
quit math if you need a book for that
I haven't really looked at it. I started Rosen and read every page, did all the exercises with solutions and it took a very long time to get past chapters 1 and 2, with chapter 1 spending an
incredible amount of time on logic, it bored me. Chapter 2 was better with sets, but the pace of Rosen is incredibly slow.
Hammack's is less than 300 pages. So I figured it has to move quicker ad get to the point with less words.
What is your opinion on Velleman? How long did it take you to do and does his book have solutions to odd problems?
this book has solutions to maybe a half of the problems. It goes through logic at the beginning which you can skip optionally, then some set theory. chapter 3 is an introduction to "proof
techniques": contrapositive, contradiction, p->q, etc, goes through how to separate givens, how to frame problem in quantifiers and what quantifier tells you about the goal etc. probably the most
usefull part. Then there are other chapters about some parts of abstract algebra, relations on tuples, functions. Induction is cover later in a book and closing chapter covers infinite sets. Overall,
at the beginning you have every proof for every theorem explained, as you go later you are expected to figure out more by yourself. A lot of excercises which generally require you to do previous
chapters, which was kinda cool, so you do not forget previous concepts. In particular, selection of certain topics seems artificial but I guess you have to practice on something. I feel however there
is quite a leap somewhere inbetween chapters 5-7, when proofs become more notably complex than before. In effect, on some parts you spend a lot less time on than on the others in no particular order.
Some excercises are labeled as more demanding but for me I haven't found it to correspond with my situation. There is also a link to some proof designer software but I do not know if it works.
Some cool excercises are examples of proofs requiring you to spot some inconsitencies.
And for the anwsers, there is really a lot less than I would like them to be, especially for the later parts. i used blog of this guy for help inchmeal.io/htpi/ch-7/sec-7.3.html
I feel much more confident with just simple epsilon delta proofs now than I did before, I do not think however whole book was required for that goal
it took 3 weeks counting the days i did exercises
Great thank you! Does this site have all the solutions to every problem in the book?
some solutions are left out in the very last chapter, just so. Carefull though, occasional errors.
Ok, thanks
Hammack is the best for exercising
Velleman is the best for the theory behind proofs
Solow is the best for explaining the thought processes behind a proof.
Also try A Transition to Advanced Mathematics, it's cool and it's free on the web
Chartrand is also good.
Second this.
I don't care if she's an SJW, I would cum deep inside of her in a heart beat.
>proof books
literally a meme, should be burnt
She is a national socialist.
Like I said, I don't care if she's an SJW I would still ejaculate inside of her.
Me too.
This one (free/creative commons) is a really good intro to pure math for anybody who did 3 semesters of Stewart and didn't do anything rigorous infinitedescent.xyz though above anons are correct if
you jump into some real analysis text or even Hoffman & Kunze Linear Algebra text you'll eventually learn the same 'reasoning' by doing enough excersizes.
Lincoln Sanchez
Oliver Campbell
Isaac Reyes
Gavin Carter
Hudson Gomez
Asher Mitchell
Sebastian Phillips
Justin Garcia
Alexander Wright
Carter White
Henry Williams
Joseph Moore
Hudson Martinez
Jackson Moore
Xavier Price
Connor Martinez
Adrian Gomez
Wyatt Rivera
Jacob Ortiz
Aiden Edwards | {"url":"https://veekyforums.com/thread/9615868/science/i-want-to-teach-myself-proofs-rosen-or-hammack-better.html","timestamp":"2024-11-05T20:16:17Z","content_type":"text/html","content_length":"39180","record_id":"<urn:uuid:74131f36-fbe5-4526-9be0-214b7318ddcd>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00274.warc.gz"} |
Rule of 72 Calculator and Formula
The rule of 72 has a simple formula behind it. It gives you an easy way to find how long it’ll take an investment to double. Feel free to test out my rule of 72 calculator with different examples.
Then continue reading below to find how it works. This rule ties into a core investing concept every investor should know…
Rule of 72 Calculator
I’ve limited this calculator to 4% through 20%. That’s because the further outside this range, the rule of 72 becomes less useful. I’ll show you an even tighter sweet spot in a chart below. Also, if
you’re earning consistent 20%+ annual returns, you could probably teach me a thing or two 🙂
This tool and the formula behind it are simple. It’s a good rule of thumb. The rule of 72 is an easy way to find how long it’ll take an investment to double. For example, how long will it take $100
to double at an annual rate of 5%? Let’s calculate the answer…
Rule of 72 Formula and Error Chart
The rule of 72 formula is simple math. Take the number 72 and divide that by the constant rate of growth. If you expect your investment to grow at 5% each year, you’d take 72 divided by 5 (don’t use
0.05). And that gives us an answer of 14.4 years.
But how close is this to the real time it’d take the investment to double?
With a lot more math, you’d find that it takes 14.2 years to double. And that’s not far off from using the quick rule of 72. However, the bigger or smaller numbers you use, the rule of 72 becomes
less reliable.
Here’s a chart I made that shows how the error grows…
Right around 7-8% will give you the most accurate answers. And going out from there, the Rule of 72 error grows.
To gain an even better understanding of this investing concept, let’s look at another example. You’ll see how this rule compares to real investment growth…
Rule of 72 Example and Table
For another rule of 72 example, let’s use a 12% annual return. How many years will it take this investment to double?
Taking 72 divided by 12 gives us an answer of six years exactly. But here’s how the investment would really growth each year…
Start $100
Year 1 $112
Year 2 $125.44
Year 3 $140.49
Year 4 $157.35
Year 5 $176.23
Year 6 $197.38
At the end of year six, it’s just short of having doubled. It’d take closer to 6.12 years to double. Although it’s not perfect, the Rule of 72 is an easy formula that can help account for compounding
Rule of 72 and Compound Interest
As you can see in the table, a 12% annual return compounds each year. In year one with an investment of $100, you’ll earn $12 = ($100 × 12%). Then in year two, that interest is put back to work.
You’d get $13.44 = ($112 × 12%).
Each year you’re putting more interest to work for you. The rule of 72 and compound interest overlap with some underlying logic. And compound interest is one of the most important investing concepts
to learn.
I also made this free CAGR calculator and it dives deeper into this concept. I made it beginner friendly as well. I strive to give clear and easy-to-understand explanations. If you have any comments
or questions, feel free to contact me anytime. | {"url":"https://briankehm.com/rule-of-72-calculator-formula/","timestamp":"2024-11-10T19:37:02Z","content_type":"text/html","content_length":"84043","record_id":"<urn:uuid:527069ed-3f73-41ce-9e2b-1b248f2597b2>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00730.warc.gz"} |
Solving for Angle at Which a Girl Swims Across a River with Given Velocities
• Thread starter kalupahana
• Start date
In summary, the question is asking at what angle must a girl swim in a river with a velocity of u and a river with a velocity of v to reach a point directly opposite on the other bank of the river,
given that she can swim 100m perpendicular to the water current and reaches a point 50m down the stream. This can be solved by using the ratio of her speed and the speed of the current, and drawing a
second triangle to find the angle she must swim upstream.
Homework Statement
A girl swim one bank of a river with 100m when she swims perpendicular to the water current. She reaches the other bank 50m down the stream. The angle at which she should swim ti reach a point
directly opposite on the other bank of the river is?
(Velocity of girl=u, velocity of river=v)
Homework Equations
The Attempt at a Solution
I have a problem, please tell why the question give the velocities to find the angle.
It can easily taken by applying tan value to the displacements.
then it can be gain as
tanα = 50/100 = 1/2
α = tan^-(1/2)
Please tell me is this right
hi kalupahana!
no, you're misunderstanding the question
first she swims directly across, but the current makes her go at an angle which (as you say) is tan^-1(1/2)
ok, so that means that the ratio of her speed and the speed of the current is … ?
now use that ratio to find the angle she must swim upstream to land exactly opposite
tiny-tim said:
hi kalupahana!
no, you're misunderstanding the question
first she swims directly across, but the current makes her go at an angle which (as you say) is tan^-1(1/2)
ok, so that means that the ratio of her speed and the speed of the current is … ?
now use that ratio to find the angle she must swim upstream to land exactly opposite
Then it come as
1/2 = u/v
v/2u = tanα
α = tan
is it come like this
It's a good question, and I think tiny tim is right and kalupahana is not (yet). I don't know how you people can solve these problems without drawing vector addition triangles.
hi pongo38!
yup, kalupahana, you should have drawn a second triangle by now …
you seem to be still on the first one.
what does your second triangle look like?
tiny-tim said:
hi pongo38!
yup, kalupahana, you should have drawn a second triangle by now …
you seem to be still on the first one.
what does your second triangle look like?
To horizontal 4km/h. V
is constant. Therefore The V
of both triangles are equal.
is 60
inclined downwards from V
to opposite direction.
= x/4
4√3 = V
cos60 = 4/y
1/2 = V
Now its okay?
hi kalupahana!
(just got up :zzz: …)
i think you've got it, but I'm finding it difficult to understand what you've written
you should have got an equilateral triangle … did you?
tiny-tim said:
hi kalupahana!
(just got up :zzz: …)
i think you've got it, but I'm finding it difficult to understand what you've written
you should have got an equilateral triangle … did you?
oh, sorry tiny tim, i posted two posts at one time. When I check them, I reply a 2nd questions tags in here.
okk, i can understand it
thnx a lot
FAQ: Solving for Angle at Which a Girl Swims Across a River with Given Velocities
1. What is the purpose of solving for the angle at which a girl swims across a river with given velocities?
The purpose of solving for the angle at which a girl swims across a river with given velocities is to determine the most efficient direction for the girl to swim in order to cross the river in the
shortest amount of time.
2. What factors are involved in solving for the angle at which a girl swims across a river?
The factors involved in solving for the angle at which a girl swims across a river include the velocity of the river's current, the girl's swimming speed, and the distance of the river to be crossed.
3. How is the angle at which a girl swims across a river calculated?
The angle at which a girl swims across a river can be calculated using trigonometric functions, such as sine, cosine, and tangent, along with the given velocities and distance of the river.
4. Can the angle at which a girl swims across a river vary depending on the given velocities?
Yes, the angle at which a girl swims across a river can vary depending on the given velocities. A higher velocity of the river's current or a slower swimming speed can result in a different optimal
angle for the girl to swim towards.
5. How does solving for the angle at which a girl swims across a river benefit us?
Solving for the angle at which a girl swims across a river can benefit us by providing a mathematical solution for the most efficient way to cross the river. This can be applied in real-life
scenarios, such as in water rescue situations, or for athletes looking to improve their swimming performance. | {"url":"https://www.physicsforums.com/threads/solving-for-angle-at-which-a-girl-swims-across-a-river-with-given-velocities.435821/","timestamp":"2024-11-10T10:55:10Z","content_type":"text/html","content_length":"109062","record_id":"<urn:uuid:f4e1f37e-1d66-40c1-9bc0-e220009419a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00099.warc.gz"} |
20 + Pattern Completion Questions and Answers | Nonverbal Reasoning20 + Pattern Completion Questions and Answers | Leverage Edu
Nonverbal reasoning is often used on competitive exams to test your critical thinking skills. Pattern finishing is a key part of this nonverbal reasoning part. These Pattern Completion questions
present you with a sequence or grid, often visuals in nature, where one element is missing. Your challenge is to figure out what the underlying pattern is, whether it’s a number, a space, or
something that depends on spins, reflections, or other changes. By analyzing the relationship between the existing elements, you can find the missing element that completes the pattern. Pattern
completion can be as easy as a series of numbers or as complicated as a set of geometric shapes.
Mastering this skill improves your ability to see patterns, understand pictures, and solve problems in a logical way. These are skills that are useful on many competitive exams, such as the GMAT, GRE
, SAT, CAT, Government exams(SSC CGL, CHSL, MTS, etc.), Banking Exams, LSAT, ACT, Railway Recruitment Boards (RRBs), and even engineering entrance exams like JEE Mains and Advance.
What is Pattern Completion? And What are Pattern Completion Problems?
Pattern Completion is a thinking skill that involves identifying and filling in missing parts of a pattern. It basically checks how well you can figure out the logic behind a series and use that
logic to guess what the next part will be. Here is a list of the most important parts.
• Identifying Patterns: The first step is to observe the given sequence and identify the repeating elements, rules, or relationships between them. Patterns can be visual (shapes, colors), numerical
(sequences, progressions), or symbolic (letters, operators).
• Logical Reasoning: Based on the observed pattern, you need to apply logical reasoning to determine the missing element. This might involve continuing a numerical sequence, replicating a specific
arrangement of shapes, or identifying the missing piece that completes a rotation or reflection.
• Problem-Solving: Pattern completion is essentially a problem-solving exercise. You need to analyze the information presented, identify the missing piece, and choose the answer choice that best
fits the established pattern.
Overall, pattern completion is a valuable skill for success in competitive exams and various aspects of daily life.
What are Pattern Completion problems?
In the nonverbal reasoning parts of competitive exams, pattern completion problems are a popular type of question. These puzzles test your ability to figure out the logic behind a pattern or series
and then use that logic to find the missing piece.
Here’s a breakdown of what pattern completion problems typically involve.
Problems Can Come in Various Format
• Number Series: Identify the missing number in a sequence that follows a specific rule (e.g., increasing by 2, alternating addition/subtraction).
• Shape Patterns: Complete a sequence of shapes that change color, and orientation, or have added elements.
• Letter Matrices: Find the missing letter in a grid based on a logical rule applied to rows and columns.
• Abstract Patterns: Identify the missing element in a sequence of abstract symbols or designs.
Applying Logics
The Key to solving these problems is to uncover the hidden rule or logic that governs the sequence of patterns.
• For Numerical Operations: Addition, subtraction, multiplication, or more complex mathematical functions.
• For Spatial Relationships: Rotation, reflection, mirroring, or movement of shapes.
• For Alternating Patterns: Elements changing color, orientation, or having added/removed features.
• For Combinations: Blending elements from different parts of the pattern.
Solution Strategy
Once you identify the logic, you can predict the missing element and choose the answer choice that completes the pattern accurately.
• Observations: Carefully analyze the existing elements in the sequence.
• Identify the Rule: Look for repeating patterns, changes in order, or relationships between elements.
• Prediction: Based on the rule, predict what the missing element should be.
• Choose the Right Answer: Select the answer choice that best fits the identified pattern and completes the sequence logically.
Also Read: 20+ Questions of Cause and Effect Reasoning
Pattern Completion Formula
There is no universal formula for solving pattern completion problems because the applying logic can vary greatly. However, there are some key strategies that you can use to solve pattern problems.
• Carefully Analyze the Given Pattern.
• Identify the Hidden Rule or Logic like Numerical Operations, Spatial Relationships, Alternating Patterns, and Combinations.
• Understanding the rule and trying to make the right prediction.
• Doble-check the prediction by applying the identified rule.
• Select the answer choice that best fits the identified pattern and complete the sequence logically.
20 + Pattern Completion Questions and Answers
Directions: In each of the following questions, select a figure from amongst the four alternatives, which when placed in the blank space of figure (X) would complete the pattern.
Q1. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option D
Q2. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option C
Q3. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option D
Q4. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option D
Q5. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option B
Q6. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option B
Q7. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option B
Q8. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option D
Q9. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option B
Q10. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option D
Q11. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option D
Q12. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option D
Q13. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option D
Q14. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option D
Q15. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option D
Q16. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option A
Q17. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option A
Q18. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option A
Q19. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option B
Q20. Identify the figure that completes the pattern.
(X) (1) (2) (3) (4)
1. 1
2. 2
3. 3
4. 4
Answer: Option A
Tips to Solve Pattern Completion Questions
Pattern completion questions can be tricky, but with the right approach, you can crack them effectively. Here are some key tips to solve the pattern completion questions in competitive exams.
• So this is the base. Look closely at the whole pattern, including how the parts are arranged, what their qualities are (like color, shape, or number), and any changes or repeats.
• Don’t just pay attention to certain parts. Look for patterns, trends, or connections that happen over and over again.
• Once you think you know the rule, use logic to figure out what should go where it’s missing. Think about how the current parts fit together and how the pattern would continue to make sense.
• Don’t just go with your first guess. Use the rule you found on other parts of the pattern.
NOTE: There’s no particular pattern formula. Each question might have a unique applying logic.
Also Read: 20+ Questions of Arithmetic Reasoning
What is pattern completion in reasoning?
Pattern completion means being able to find and add back lost parts or pieces of a pattern or design. It includes looking at the parts that are already there and guessing or building up the missing
part based on the pattern seen.
How do you solve series completion questions?
Learn the squares of all the numbers in the range from 1 to 25.
Learn the cubes of all the numbers in the range from 1 to 20.
Look for the sequence by looking at the series and doing things like splitting, checking for multiples, finding the difference, etc.
Do not forget the EJOTY rule.
What is the number pattern reasoning?
There are several patterns, with one or more. The numbers are missing, and you have to figure out the code (or the logic) that makes them all fit together. The numbers are shown in the patterns. Once
you know the logic or rule that links the given numbers in the patterns, it’s easy to find the missing numbers.
This was all about the “20 + Pattern Completion Questions and Answers”. For more such informative blogs, check out our Study Material Section, or you can learn more about us by visiting our Indian
exams page. | {"url":"https://leverageedu.com/discover/indian-exams/exam-prep-pattern-completion-questions/","timestamp":"2024-11-06T20:19:32Z","content_type":"text/html","content_length":"317986","record_id":"<urn:uuid:418f22bc-6414-438c-b2e8-3129822dbabd>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00698.warc.gz"} |
tail risk Archives - Flirting with Models
This post is available as a PDF download here.
• The March 2020 equity market sell-off has caused many investors to re-investigate the potential benefits of tail risk hedging programs.
• Academic support for these programs is quite limited, and many research papers conclude that the cost of implementation for naïve put strategies out-weighs the potential payoff benefits.
• However, many of these studies only consider strategies that hold options to expiration. This means that investors can only profit from damage assessed. By rolling put options prior to
expiration, investors can profit from damage
• In this research note we demonstrate that holding to expiration is not a required feature of a successful tail hedging program.
• Furthermore, we demonstrate that once that requirement is lifted, the most valuable component of a tail risk hedging program may not actually be the direct link to damage assessed, but rather the
ability to profit in a convex manner from the market’s re-pricing of risk.
“To hedge, or not to hedge, that is the question.”
Nothing brings tail risk management back to the forefront of investors’ minds like a market crisis. Despite the broad interest, the jury is still out as to the effectiveness of these approaches.
Yet if an investor is subject to a knock-out barrier – i.e. a point of loss that creates permanent impairment – then insuring against that loss is critical. This is often the case for retirees or
university endowments, as withdrawal rates increase non-linearly with portfolio drawdowns. In this case, the question is not whether to hedge, but rather about the most cost-effective means of
Some academics and practitioners have argued that put-based portfolio protection is prohibitively expensive, failing to keep pace with a simple beta-equivalent equity portfolio. They also highlight
that naïve put strategies – such as holding 10% out-of-the-money (“OTM”) puts to expiration – are inherently path dependent.
Yet empirical evidence may fail us entirely in this debate. After all, if the true probability and magnitude of tail events is unknowable (as markets have fat tails whose actual distribution is
hidden from us), then prior empirical evidence may not adequately inform us about latent risks. After all, by their nature, tail events are rare. Therefore, drawing any informed conclusions from
tail event data will be shrouded in a large degree of statistical uncertainty.
Let us start by saying that the goal of this research note is not to prove whether tail risk hedging is or is not cost effective. Rather, our goal is to demonstrate some of the complexities and
nuances that make the conversation difficult.
And this piece will only scratch the surface. We’ll be focusing specifically on buying put options on the S&P 500. We will not discuss pro-active monetization strategies (i.e. conversion of our
hedge into cash), trade conversion (e.g. converting puts into put spreads), basis risk trades (e.g. buying calls on U.S. Treasuries instead of puts on equities), or exchanging non-linear for linear
hedges (e.g. puts for short equity futures).
Given that we are ignoring all these components – all of which are important considerations in any actively managed tail hedging strategy – it does call into question the completeness of this note.
While we hope to tackle these topics in later pieces, we highlight their absence specifically to point out that tail risk hedging is a highly nuanced topic.
So, what do we hope to achieve?
We aim to demonstrate that the path dependency risk of tail hedging strategies may be overstated and that the true value of deep tail hedges emerges not from the actual insurance of loss but the
rapid repricing of risk.
A Quantitative Aside
Options data is notoriously dirty, and therefore the results of back testing options strategies can be highly suspect. In this note, rather than price our returns based upon historical options data
(which may be stale or have prohibitively wide bid/ask spreads), we fit a volatility surface to that data and price our options based upon that surface.
Specifically, each trading day we fit a quadratic curve to log-moneyness and implied total variance for each quoted maturity. This not only allows us to reduce the impact of dirty data, but it
allows us to price any strike and maturity combination.
While we limit ourselves only to using listed maturity dates, we do stray from listed strikes. For example, in quoting a 10% out-of-the-money put, rather than using the listed put option that would
be closest to that strike, we just assume the option for that strike exists.
This approach means, definitively, that results herein were not actually achievable by any investor. However, since we will be making comparisons across different option strategy implementations, we
do not believe this is a meaningful impact to our results.
To reduce the impacts of rebalance timing luck, all strategies are implemented with overlapping portfolios. For example, for a strategy that buys 3-month put options and holds them to maturity would
be implemented with three overlapping sub-portfolios that each roll on discrete 3-month periods but do so on different months.
Finally, the indices depicted herein are designed such that they match notional coverage of the S&P 500 (e.g. 1 put per share of S&P 500) when implemented as a 100% notional overlay and rebalanced
monthly upon option expiration.
The Path Dependency of Holding to Expiration
One of the arguments often made against tail hedging is the large degree of path dependency the strategy can exhibit. For example, consider an investor who buys 10% OTM put options each quarter. If
the market falls less than 10% each quarter, the options will provide no protection. Therefore, when holding to expiration, we need drawdowns to precisely coincide with our holding period to achieve
maximum protection.
But is there something inherently special about holding to expiration? For popular indices and ETFs, there are liquid options markets available, allowing us to buy and sell at any time. What occurs
if we roll our options a month or two before expiration?
Below we plot the results of doing precisely this. In the first strategy, we purchase 10% OTM puts and hold them to expiration. In the second strategy, we purchase the same 10% OTM puts, but roll
them a month before expiration.
Source: DiscountOptionsData.com. Calculations by Newfound Research. Returns are hypothetical and backtested. Returns are gross of all fees including, but not limited to, management fees,
transaction fees, and taxes. Returns assume the reinvestment of all distributions.
We see nearly identical long-term returns and, more importantly, the returns during the 2008 crisis and the recent March turmoil are indistinguishable. And we outright skipped holding each option
for 1/3^rd of its life!
Our results seem to suggest that the strategies are less path dependent than originally argued.
An alternative explanation, however, may be that during these crises our options end up being so deep in the money that it does not matter whether we roll them early or not. One way to evaluate this
hypothesis is to look at the rolling delta profile – how sensitive our option strategy is to changes in the underlying index – over time.
Source: DiscountOptionsData.com. Calculations by Newfound Research.
We can see is that during calm market environments, the two strategies exhibit nearly identical delta profiles. However, in 2008, August 2011, Q4 2018, and March 2020 the delta of the strategy that
holds to expiration is substantially more negative. For example, in October 2008, the strategy that holds to expiration had a delta of -2.75 whereas the strategy that rolls had a delta of -1.77.
This means that for each 1% the S&P 500 declines, we estimate that the strategies would gain +2.75% and +1.77% respectively (ignoring other sensitivities for the moment).
Yet, despite this added sensitivity, the strategy that holds to expiration does not seem to offer meaningfully improved returns during these crisis periods.
Source: DiscountOptionsData.com. Calculations by Newfound Research.
Part of the answer to this conundrum is theta, which measures the rate at which options lose their value over time. We can see that during these crises the theta of the strategy that holds to
expiration spikes significantly, as with little time left the value of the option will be rapidly pulled towards the final payoff and variables like volatility will no longer have any impact.
What is clear is that delta is only part of the equation. In fact, for tail hedges, it may not even be the most important piece.
Convexity in Volatility
To provide a bit more insight, we can try to contrive an example whereby we know that ending in the money should not have been a primary driver of returns.
Specifically, we will construct two strategies that buy 3-month put options and roll each month. In the first strategy, the put option will just be 10% OTM and in the second strategy it will be 30%
OTM. As we expect the option in the second strategy to be significantly cheaper, we set an explicit budget of 60 basis points of our capital each month.^1
Below we plot the results of these strategies.
Source: DiscountOptionsData.com. Calculations by Newfound Research. Returns are hypothetical and backtested. Returns are gross of all fees including, but not limited to, management fees,
transaction fees, and taxes. Returns assume the reinvestment of all distributions.
In March 2020, the 10% OTM put strategy returned 13.4% in and the 30% OTM put strategy returned 39.3%. From prior trough (February 19^th) to peak (March 23^rd), the strategies returned 18.4% and
46.5% respectively.
This is a stark difference considering that the 10% OTM put was definitively in-the-money as of March 20^th (when it was rolled) and the 30% OTM strategy was on the cusp. Consider the actual trades
• 10% OTM Strategy: Buy a 3-month 10% OTM put on February 21^st and sell a 2-month 23.3% ITM put on March 20^th. When bought, the option had an implied volatility of 20.9% and a price of $45.45^2;
when sold it had an implied volatility of 39.5% and a price of $1428.21 for a 3042% return.
• 30% OTM Strategy: Buy a 3-month 30% OTM put on February 21^st and sell a 2-month 1.4% ITM put on March 20^th. When bought, the option had an implied volatility of 35.0% and a price of $5.42; when
sold it had an implied volatility of 53.8% and a price of $425.85 for a 7757% return.
It is also worth noting that since we are spending a fixed budget, we can buy 8.38 contracts of the 30% OTM put for every contract of the 10% OTM put.
So why did the 30% OTM put appreciate so much more? Below we plot the position scaled sensitivities (i.e. dividing by the cost per contract) to changes in the S&P 500 (“delta”), changes in implied
volatility (“vega”), and their respective derivatives (“gamma” and “volga”).
Source: DiscountOptionsData.com. Calculations by Newfound Research.
We can see that as of February 21st, the sensitivities are nearly identical for delta, gamma, and vega. But note the difference in volga.
What is volga? Volga tells us how much the option’s sensitivity to implied volatility (“vega”) changes as implied volatility itself changes. If we think of vega as a kind of velocity, volga would
be acceleration.
A positive vega tells us that the option will gain value as implied volatility goes up. A positive volga tells us that the option will gain value at an accelerating rate as implied volatility goes
up. Ultimately, this means the price of the option is convex with respect to changes in implied volatility.
So as implied volatilities climbed during the March turmoil, not only did the option gain value due to its positive vega, but it did so at an accelerating rate thanks to its positive volga.
Arguably this is one of the key features we are buying when we buy a deep OTM put.^3 We do not need the option to end in the money to provide a meaningful tail hedge; rather, the value is derived
from large moves in implied volatility as the market re-prices risk.
Indeed, if we perform the same analysis for September and October 2008, we see an almost identical situation.
Source: DiscountOptionsData.com. Calculations by Newfound Research.
In this research note, we aimed to address one of the critiques against tail risk hedging: namely that it is highly path dependent. For naively implemented strategies that hold options to
expiration, this may be the case. However, we have demonstrated in this piece that holding to expiration is not a necessary condition of a tail hedging program.
In a contrived example, we explore the return profile of a strategy that rolls 10% OTM put options and a strategy that rolls 30% OTM put options. We find that the latter offered significantly better
returns in March 2020 despite the fact the options sold were barely in the money.
We argue that the primary driver of value in the 30% OTM put is the price convexity it offers with respect to implied volatility. While the 10% OTM put has positive sensitivity to changes in implied
volatility, that sensitivity does not change meaningfully as implied volatility changes. On the other hand, the 30% OTM put has both positive vega and volga, which means that vega will increase with
implied volatility. This convexity makes the option particularly sensitive to large re-pricings of market risk.
It is common to think of put options as insurance contracts. However, with insurance contracts we receive a payout based upon damage assessed. The key difference with options is that we have the
ability to monetize them based upon potential damage perceived. When we remove the expectation of holding options into expiration (and therefore only monetizing damage assessed), we potentially
unlock the ability to profit from more than just changes in underlying price. | {"url":"https://blog.thinknewfound.com/tag/tail-risk/","timestamp":"2024-11-14T16:25:43Z","content_type":"text/html","content_length":"609553","record_id":"<urn:uuid:cd5dd90d-75c0-4aef-b05b-16aae89a182d>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00480.warc.gz"} |
Caesar Cipher Explorer
Try the Caesar Cipher
Use the tool below to encode or decode a message:
Wonderful moments in history
Welcome to the world of Caesar Cipher
This realm intertwines simplicity and intrigue in the art of cryptography. Originating from the practices of Julius Caesar, this cipher serves as a testament to the timeless allure of secret
communications. It operates on a straightforward yet ingenious principle – shifting the letters of the alphabet by a fixed number. This shift transforms ordinary messages into cryptic texts, cloaking
words in a veil of mystery. As we delve into this world, we unravel the elegance of its simplicity and the joy of decoding messages that once seemed impenetrable. The Caesar Cipher, though
elementary, opens the gateway to the broader, fascinating world of cryptography, where every letter and shift play a crucial role in the dance of secrecy and discovery.
What is Caesar Cipher?
The Caesar Cipher, an ancient encryption technique, gained fame through Julius Caesar, the renowned Roman general and politician. He used this method to safeguard crucial military communications.
Classified as a substitution cipher, it specifically employs an alphabetical shift. The fundamental principle of the Caesar Cipher involves offsetting each letter in the alphabet by a predetermined
number. For instance, with an offset of 3, the alphabet shifts such that 'A' becomes 'D', 'B' becomes 'E', and this pattern continues. Upon reaching the end of the alphabet, the sequence circles back
to the beginning.
While the Caesar Cipher often serves as a foundational element in more complex encryption methods, its simplicity renders it vulnerable. Like all ciphers based on alphabet substitution, it is
relatively easy to decipher, thus offering limited security for practical communication needs.
Schematic Diagram of the Caesar Cipher
What are the specific Caesar Ciphers?
The Caesar Cipher is a method of encryption that involves shifting the letters of the alphabet by a fixed number either backwards or forwards. The simplest form is to shift each letter by a fixed
number. However, besides this basic form, here are some more interesting variants:
• ROT13: A special case of the Caesar Cipher with a shift of 13. Since there are 26 letters in the English alphabet, the same rule applies for both encryption and decryption.
• Atbash Cipher: This is a special case that could be considered a Hebrew Caesar Cipher. It reverses the alphabet, so the first letter becomes the last letter, the second letter becomes the second
to last, and so on.
• Vigenère Cipher: Although not strictly a Caesar Cipher, it is developed based on the principle of the Caesar Cipher. It uses a keyword as the shift value for encryption, providing higher security
compared to a single letter shift.
• Affine Cipher: Based on the idea of the Caesar Cipher, but introduces multiplication in the encryption process. Each letter's position in the alphabet is first multiplied by a number (coprime
with the length of the alphabet) and then added to a shift value, finally taking the modulus of the alphabet length to get the encrypted letter.
• ROT5, ROT18, ROT47: These are variants of ROT13 but are used for encrypting numbers and other characters. ROT5 is used only for numbers, ROT18 combines ROT5 and ROT13, and ROT47 can encrypt most
printable characters in the ASCII table.
• Double Caesar Cipher: This is a simple extension of the Caesar Cipher, applying the Caesar Cipher twice, possibly with different shift values, to increase the complexity of the encryption.
These variants and related techniques each have their characteristics, aiming to enhance the security of the encryption method or to meet specific encryption needs.
How to implement Caesar Cipher in Python?
In Python, you can do this by looping through each letter of the original text and then calculating the encrypted version of each letter based on the alphabet and a given offset. This is conveniently
achieved using an ASCII code table, for example, by taking the difference between the ASCII value of the letter and the ASCII value of 'a', plus an offset, and then converting the result back to the
Python Code for Caesar Cipher
Below is a Python function that demonstrates how to encrypt and decrypt text using the Caesar Cipher technique. The code includes comments for better understanding and adaptability.
def caesar_cipher_enhanced(text, shift, encrypt=True):
Encrypts or decrypts text using Caesar Cipher.
text (str): The text to encrypt or decrypt.
shift (int): The number of positions to shift the letters by.
encrypt (bool): True for encryption, False for decryption.
str: The transformed text.
transformed_text = ""
for char in text:
if char.isalpha():
start = ord('A') if char.isupper() else ord('a')
shift_adjusted = shift if encrypt else -shift
transformed_char = chr((ord(char) - start + shift_adjusted) % 26 + start)
transformed_text += transformed_char
transformed_text += char
return transformed_text
# Example usage
user_input = input("Enter the text: ")
shift = int(input("Enter the shift value: "))
encrypt_decrypt = input("Encrypt or Decrypt (E/D): ").strip().upper()
if encrypt_decrypt == 'E':
result = caesar_cipher_enhanced(user_input, shift, encrypt=True)
print("Encrypted:", result)
elif encrypt_decrypt == 'D':
result = caesar_cipher_enhanced(user_input, shift, encrypt=False)
print("Decrypted:", result)
print("Invalid option. Please enter 'E' for Encrypt or 'D' for Decrypt.")
How to crack the Caesar Cipher?
Cracking a Caesar Cipher can be relatively simple due to the limited number of possible shifts (26 in the case of the English alphabet). A common method to break this cipher is by brute force, which
means trying out every possible shift until you find one that makes sense. This is practical because there are only 26 possible shifts in the English alphabet, making the number of combinations small
enough to check each one manually.
Another more refined method is to use frequency analysis. Since letters in the English language have different frequencies of occurrence (for example, 'e' is more common than 'z'), you can compare
the frequency of letters in the encoded message with typical letter frequencies in English. By doing this, you can identify the most probable shift that was used to encrypt the message.
Since the mapping of each character in the Caesar Cipher is fixed, if "b" maps to "e", then "e" will appear in the ciphertext every time "b" appears in the plaintext. It is now known that the
probability distribution of each letter in English is known. The average probability of occurrence of different letters in different texts is usually the same, and the longer the text, the closer the
frequency calculation is to the average. This is a frequency chart of 26 letters. Of course, as the number of samples changes, the frequency of each letter will be slightly different.
For example, enter the first paragraph of text "This realm intertwines simplicity and intrigue...", and through the converter above we get the ciphertext. But for others who don't know what the
secret key is, we can get a secret key through the code, which is the offset.
The original English text is as follows:
"This realm intertwines simplicity and intrigue in the art of cryptography. Originating from the practices of Julius Caesar, this cipher serves as a testament to the timeless allure of secret
communications. It operates on a straightforward yet ingenious principle – shifting the letters of the alphabet by a fixed number. This shift transforms ordinary messages into cryptic texts, cloaking
words in a veil of mystery. As we delve into this world, we unravel the elegance of its simplicity and the joy of decoding messages that once seemed impenetrable. The Caesar Cipher, though
elementary, opens the gateway to the broader, fascinating world of cryptography, where every letter and shift play a crucial role in the dance of secrecy and discovery."
English Letter Frequency Distribution
The following Python code example demonstrates how to perform frequency analysis to break a Caesar Cipher. This technique is based on the statistical analysis of letter frequencies in English.
import string
def count_frequencies_from_file(path):
count_dict = dict.fromkeys(string.ascii_lowercase, 0)
total_chars = 0
with open(path, 'r', encoding='utf-8') as file:
for line in file:
for char in line.lower():
if char in count_dict:
count_dict[char] += 1
total_chars += 1
for char in count_dict:
count_dict[char] /= total_chars
return count_dict
def frequency_analysis(known_frequencies, count_dict):
eps = float('inf')
key = 0
cipher_frequencies = list(count_dict.values())
for shift in range(26):
s = 0
for i in range(26):
s += known_frequencies[i] * cipher_frequencies[(i + shift) % 26]
temp = abs(s - 0.065379)
if temp < eps:
eps = temp
key = shift
return key
# Known letter frequencies in English
known_freqs = [0.086,0.014,0.030,0.038,0.130,0.029,0.020,0.053,0.063,0.001,0.004,0.034,0.025,0.071,0.080,0.020,
file_path = "Your_Path"
cipher_count_dict = count_frequencies_from_file(file_path)
key = frequency_analysis(known_freqs, cipher_count_dict)
print("The key is: " + str(key))
Try using your own file path and run this code to see if you can decrypt a message encrypted with a Caesar Cipher.
Caesar Cipher and the cipher used by Caesar
In Chapter 56 of Suetonius's De Vita Caesarum, the use of encryption techniques in Caesar's private letters is described:
"Extant et ad Ciceronem, item ad familiares domesticis de rebus, in quibus, si qua occultius perferenda erant, per notas scripsit, id est sic structo litterarum ordine, ut nullum verbum effici
posset: quae si qui investigare et persequi velit, quartam elementorum litteram, id est D pro A et perinde reliquas commutet." ( Suetonius, De Vita Caesarum: Divus Iulius. The Latin Library,
accessed June 1, 2024, https://www.thelatinlibrary.com/suetonius/suet.caesar.html )
This content does not directly mention the modern concept of the Caesar Cipher, which involves simple letter shifting encryption. Instead, the encryption method used by Caesar was more akin to a
transposition cipher, where the actual positions of the letters are changed to encrypt the message, distinctly different from the fixed shift of the Caesar Cipher. This encryption technique involves
relatively complex rearrangement and substitution of letters, demonstrating a significant difference from the direct shift method of the modern Caesar Cipher
Steganography and Cryptography
Explore the fascinating work of Johannes Trithemius, a Renaissance-era Benedictine monk and scholar from Germany. He is known for his seminal work, Steganographia, which delves into the realms of
steganography and also touches upon cryptographic techniques in his other scholarly works.
Steganographia is often misinterpreted as purely involving magical elements and summoning spirits. However, it cleverly encases a complex cryptographic system within its narratives. Below is an
overview of the cryptographic content encrypted across its volumes:
• Volumes One and Two: These volumes focus predominantly on encryption techniques. Initially perceived as guides for summoning spirits, they metaphorically represent intricate methods of
cryptography. They describe innovative methods for transmitting hidden messages over long distances, a groundbreaking concept at the time.
• Volume Three: This volume extends some of the earlier themes but introduces more controversial discussions that deviate from the straightforward cryptography found in the first two volumes. It
openly explores more spiritual and magical dimensions, which have been subject to varied interpretations through the ages.
Through deeper analysis, Trithemius's work reveals that what appears as magical invocations are actually veiled descriptions of cryptographic methods, highlighting the relevance of cryptography
similar to modern cryptographic practices.
Definitions and Basic Principles
Steganography is the art of concealing information within non-sensitive media such as images, audio, or video files, rendering the information invisible to casual observers. The essence of
steganography lies in obscuring the existence of the information, not just its content.
Cryptography, in contrast, involves converting plaintext information into a secure encrypted format that cannot be understood without the corresponding decryption key, relying on complex mathematical
algorithms like public/private key mechanisms and symmetric encryption algorithms.
Technical Implementation and Applications
The implementation of steganography typically involves encoding secret information into various parts of ordinary files, such as the pixels in images, videos, audio, or even text. This type of
information hiding is achieved through subtle modifications, for example, fine-tuning the pixel values in an image, altering the least significant bit (LSB) of color values, or adding signals in
audio files that are beyond human hearing frequencies. These modifications are below the threshold of human perception, so even when the information is transmitted or displayed, it remains
undetectable to outside observers. The advantage of this method is that even direct visual observation cannot easily discern these subtle changes.
Steganography in practice
The image above illustrates the concept of steganography. The left side of the image, labeled as 'normal,' shows no alterations and retains the original appearance of the landscape. Conversely, the
right side demonstrates how text can be hidden using the LSB method, altering the image subtly enough to make the changes nearly invisible.
Cryptography transforms sensitive information using cryptographic keys to create ciphertext. This encrypted format ensures that data remains secure and unintelligible to unauthorized entities,
crucial for protecting communications and data across various systems.
Detection and Security
The security of steganography primarily relies on its concealment. Once the use of steganography is suspected, specialized technical analyses, such as statistical analysis or pattern recognition, can
be employed to attempt to uncover hidden information. However, if the steganographic method is well-designed, even experts may find it difficult to detect the concealed information.
The security of encryption technology depends on the strength of the encryption algorithms and the secure management of keys. Modern encryption methods, such as AES and RSA, are designed to withstand
various types of attacks, including those from quantum computing. The confidentiality of the keys is critical to encryption security; once the keys are compromised, the encryption protection is
Appropriate Environments and Limitations
Steganography is ideal for highly confidential scenarios, such as covert communications, where revealing the existence of the information could be detrimental. However, it is limited by the volume of
data that can be effectively concealed. On the other hand, cryptography is versatile, suitable for securing data in a broad range of applications from financial transactions to personal data
protection, though it demands rigorous key management to prevent security breaches.
In summary, both steganography and cryptography play critical roles in the field of information security. They can be employed individually or in tandem to offer robust protection, tailored to
specific security needs and the nature of the threats involved. | {"url":"https://chatcipherai.com/en/Caesar_Cipher","timestamp":"2024-11-04T17:42:55Z","content_type":"text/html","content_length":"63208","record_id":"<urn:uuid:7ff489b5-1c34-4c88-a79c-4587f2351bf9>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00832.warc.gz"} |
The Hiccups That Wouldn’t Stop
What are hiccups? Your “diaphragm,” the muscle between your lungs and your stomach, normally shrinks slowly to make you breathe, but sometimes it freaks out and pulls a lot faster than it should.
That makes your throat snap shut, which makes that silly hiccup sound. Holding your breath, drinking water while upside down, or eating a teaspoon of sugar can stop them – but not always. Poor
Charles Osborne started hiccuping one day in 1922 and didn’t stop until 1990!
Wee ones: If you just counted your 5 hiccups, what are all the numbers you said before that?
Little kids: If you start hiccuping twice today, and take 2 teaspoons of sugar each time, how many teaspoons of sugar do you get to eat? Bonus: If you hiccup on 10 days straight and it all starts on
a Tuesday, what’s your last day of hiccuping?
Big kids: If Charles Osborne hiccuped from 1922 to 1990, how many years was that? Bonus: If he hiccuped 2,000 times each year, how many hiccups did he have in total? (Hint if needed: What if he
hiccuped just twice each year…and then how does 2,000 a year change that?)
Wee ones: 1, 2, 3, 4.
Little kids: 4 teaspoons. Bonus: On a Thursday…remember, Monday will be your 7^th day, not the Tuesday.
Big kids: 68 years. Bonus: 136,000 hiccups! | {"url":"https://bedtimemath.org/fun-math-hiccups/","timestamp":"2024-11-11T07:31:22Z","content_type":"text/html","content_length":"86645","record_id":"<urn:uuid:c540385e-1e6d-494c-9de0-103a6920055c>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00392.warc.gz"} |
#1 Cylinder Volume Calculator with Steps
Cylinder Volume Calculator
A Cylinder Volume Calculator is a tool used to calculate the volume of a cylinder, a three-dimensional geometric shape with circular bases. This tool is particularly useful in areas like engineering,
manufacturing, and fluid measurement, where the volume of cylindrical objects or containers needs to be determined accurately.
What is the Purpose of a Cylinder Volume Calculator?
The purpose of a Cylinder Volume Calculator is to:
• Provide Accurate Volume Measurement: It calculates the exact amount of space inside a cylindrical object, such as a tank or pipe.
• Simplify Complex Calculations: By using the cylinder’s radius and height, the calculator simplifies the process of determining its volume.
• Support Real-World Applications: Whether for liquid storage, construction materials, or manufacturing, understanding the volume of cylindrical objects is essential.
How is the Volume of a Cylinder Calculated?
The volume of a cylinder is calculated using the formula:
Volume =
• r is the radius of the cylinder’s base.
• h is the height of the cylinder.
• π is approximately 3.14159.
For example, if the radius of the base is 3 cm and the height is 7 cm:
What Features Should a Cylinder Volume Calculator Have?
An efficient Cylinder Volume Calculator should include:
1. Input for Radius and Height: Fields where users can input the radius of the base and the height of the cylinder.
2. Unit Flexibility: The ability to select units (e.g., cm, meters, inches) for both radius and height.
3. Instant Calculations: Immediate output after entering the required measurements, with accurate volume results.
How to Use a Cylinder Volume Calculator?
To use the Cylinder Volume Calculator:
• Enter the Radius: Input the radius of the cylinder’s circular base.
• Enter the Height: Provide the height of the cylinder.
• Get the Volume: The calculator will display the cylinder’s volume instantly based on the given inputs.
Why Use a Cylinder Volume Calculator?
A Cylinder Volume Calculator is beneficial because:
• Precision: It eliminates the possibility of human error in manual calculations, ensuring accurate results.
• Convenience: Quickly determines the volume for tasks such as filling cylindrical tanks or estimating material usage.
• Versatile Applications: Whether used in fluid mechanics, construction, or manufacturing, this tool provides invaluable insights for cylindrical measurements.
Cylinder Volume Calculator FAQs
How do I calculate the volume of a cylinder using this calculator?
What units can I use with this calculator?
You can use any standard unit of length such as meters, centimeters, inches, etc., for the radius and height.
Can I use this calculator for other shapes besides cylinders?
No, this calculator specifically calculates the volume of cylinders.
How accurate are the results from this calculator?
The results are accurate based on the radius and height you input. Ensure measurements are accurate for precise volume calculation.
Can I calculate the volume in different units with this calculator?
Yes, after obtaining the volume in the default unit (cubic units corresponding to the cube of the length unit used), you can manually convert it to other units if needed.
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Water cylinder volume calculator with steps, Cylinder volume calculator with steps in litres, Cylinder volume calculator with steps in liters, cylinder volume calculator litres, volume of cone,
surface area of a cylinder, area of cylinder, volume of a sphere | {"url":"https://toolconverter.com/cylinder-volume-calculator/","timestamp":"2024-11-09T05:11:17Z","content_type":"text/html","content_length":"192882","record_id":"<urn:uuid:08402dd4-f31b-4787-8b4d-2fc0871a3059>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00241.warc.gz"} |
Further Maths - Polar Form
What is the name for $ \vert z \vert $?
What is the value of $ \vert z \vert $, where $z = a + bi$?
\[\sqrt{a^2 + b^2}\]
What is the value of $ \vert z \vert ^2$, where $z = a + bi$?
\[a^2 + b^2\]
If $ \vert z \vert ^2 = a^2 + b^2$, how could you also write $ \vert z \vert ^2$?
\[(a + bi)(a - bi)\]
What is the word definition of the modulus of $z$?
The distance to $z$ from the origin.
What is the name for $\text{arg} z$?
What is the word definition of the argument of $z$?
The angle that a line drawn to $z$ makes with the real axis, in the anticlockwise
What is the range of the argument $\theta$ of a complex number?
\[-\pi < \theta \le \pi\]
What typically are the units of $\text{arg} z$?
What is the principal argument?
The argument of $z$ in the range $-\pi < \theta \le \pi$.
Assuming all points are some horizontal distance $a$ away from the origin and a vertical distance $b$ in
\[\text{arg}z = \alpha\]
\[\text{arg}z = \pi-\alpha\]
\[\text{arg}z = -(\pi-\alpha)\]
\[\text{arg}z = -\alpha\]
Visualise the 4 quadrants of an Argand diagram?
What is the argument of $3 + 4i$, in terms of $\tan$?
What quadrant does $3 + 4i$ lie in?
What is the argument of $-3 + 4i$, in terms of $\tan$?
\[\pi - \tan^{-1}\left(\frac{4}{3}\right)\]
What quadrant does $3 - 4i$ lie in?
What is the argument of $-3 - 4i$, in terms of $\tan$?
\[-(\pi - \tan^{-1}\left(\frac{4}{3}\right))\]
What quadrant does $-3 - 4i$ lie in?
What is the argument of $3 - 4i$, in terms of $\tan$?
What quadrant does $3 - 4i$ lie in?
What quadrant does $12 + 5i$ lie in?
What quadrant does $-3 + 6i$ lie in?
What quadrant does $-8 -7i$ lie in?
What quadrant does $2 - 2i$ lie in?
For $a + bi$, why shouldn’t you put the signs of $a$ and $b$ in $\tan\left(\frac{b}{a}\right)$?
Because you’re calculating the angle from a triangle, the lengths of the sides can’t be negative.
If $z = a + bi$, how could you write $a$ in terms of $r$ and $\theta$?
\[a = r\cos\theta\]
If $z = a + bi$, how could you write $b$ in terms of $r$ and $\theta$?
\[b = r\sin\theta\]
The rule that $a = r\cos\theta$ and $b = r\sin\theta$ is similar to what in Physics?
\[F_x = F\cos\theta\] \[F_y = F\sin\theta\]
What do you get if you substitute $a = r\cos\theta$ and $b = r\sin\theta$ into $a + bi$?
\[z = r\cos\theta + ri\sin\theta\] \[z = r(\cos\theta + i\sin\theta)\]
What is the $\sin$ and $\cos$ form of $z$?
\[z = r(\cos\theta + i\sin\theta)\]
What is $r\cos\theta$?
What is $r\sin\theta$?
How can you rewrite $ \vert z _ 1 z _ 2 \vert $?
How can you rewrite $\text{arg}(z _ 1 z _ 2)$?
\[\text{arg}(z_1) + \text{arg}(z_2)\]
How can you rewrite $z _ 1 z _ 2$ in polar form?
\[r_1 r_2 ( \cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2))\]
How can you rewrite $ \vert \frac{z _ 1}{z _ 2} \vert $?
How can you rewrite $\text{arg}(\frac{z _ 1}{z _ 2})$?
\[\text{arg}(z_1) - \text{arg}(z_2)\]
How can you rewrite $\frac{z _ 1}{z _ 2}$ in polar form?
\[\frac{r_1}{r_2} ( \cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2))\]
How can you rewrite $4(\cos(90^{\circ}) - i\sin(90^{\circ}))$?
\[4(\cos(-90^{\circ}) + i\sin(-90^{\circ}))\]
Why is $4(\cos(90^{\circ}) - i\sin(90^{\circ}))$ not valid polar form?
Because there is a minus in front of the $\sin$.
How can you rewrite $(\cos(\theta) - i\sin(\theta))$?
\[(\cos(-\theta^{\circ}) + i\sin(-\theta^{\circ}))\]
Fixing $(\cos(\theta) - i\sin(\theta))$ relies on what property of $\sin$?
\[\sin(\theta) = -\sin(-\theta)\]
Why is $\frac{16}{3} \pi$ not a valid argument in polar form?
Because it’s not in the range $-pi < \theta \le \pi$.
How could you fix something like $\frac{16}{3} \pi$?
Keep on subtracting $2\pi$ until it’s in the range $-pi < \theta \le \pi$.
Related posts | {"url":"https://ollybritton.com/notes/a-level/further-maths/topics/polar-form/","timestamp":"2024-11-05T09:34:28Z","content_type":"text/html","content_length":"515452","record_id":"<urn:uuid:4439ffb2-dea6-4ba1-8022-5fd9ad1bd246>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00425.warc.gz"} |
DEVSQ spreadsheet function definition
DEVSQ function
DEVSQ(range) computes the sum of the squares of differences from the mean of the values of the spreadsheet cells in the specified range.
The argument range can be a list of numbers, cell addresses, cell ranges (such as A1:A10).
Range can also be a matrix. In that case the function returns a matrix with one row and the same number of columns, with each element the DEVSQ of the corresponding column. If range is a row matrix,
it is converted to a column matrix first.
The equation for DEVSQ is:
$$ \sum_{}^{}{(x-\bar{x})^2}$$ | {"url":"https://www.medcalc.org/manual/DEVSQ-function.php","timestamp":"2024-11-07T12:41:22Z","content_type":"text/html","content_length":"30168","record_id":"<urn:uuid:0d08cfab-d834-4a30-a8eb-eb74513d874d>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00250.warc.gz"} |
Net present value (NPV) profile - definition, explanation, example | Accounting For Management
Net present value (NPV) profile
The net present value profile (or NPV profile, for short) is a graphical representation of the relationship between a project’s net present value (NPV) and different corresponding discount rates. It
depicts how sensitive a project’s NPV is to a change in firm’s discount rate or cost of capital.
To construct a project’s NPV profile, the discount rates are plotted on X-axis and the net present values are plotted on Y-axis. The two important outcomes of an NPV profile are the internal rate of
return (IRR) and the crossover rate. The IRR is the point where NPV curve interests the X-axis i.e., discount rate axis. This point reveals a discount rate that yields a zero net present value for
the project.
The crossover rate is relevant only when two projects are plotted together on the same graph. It is the point where NPV curves of two different projects intersect each other. This point corresponds
to a discount rate on X-axis which results in equal net present values for both the projects. Let’s illustrate the construction of an NPV profile through an example.
ABC Company has an opportunity that needs an initial investment of $6 million and is expected to bring in a net annual cash flow of $1,750,000 for 7 consecutive years. We can examine the sensitivity
of the project’s net present value (NPV) to the change in discount rate by determining the NPV at different discount rates. Let’s, suppose, the company applies 5%, 10%, 15%, 20% and 25% discount rate
to this project. The project’s NPV at these rates can be calculated as follows:
*Values from “present value of an annuity of $1 in arrears table“.
We can plot the above net present values (NPVs) and their corresponding discount rates to depict the NPV profile of the project as follows:
In NPV profile above, the downward slope of NPV curve shows that the project’s net present value lowers as the discount rate rises. The NPV curve behaves like this with most of the projects. Let’s
explain a bit why it happens.
At higher discount rates, the cash movement that happens at the early stage of the project heavily influences its net present value. Since many projects tend to cause a large capital outlay at the
time of their inception, the NPV profile generally exhibits a negative or inverse relationship between the net present value and discount rate assumed.
The point where the NPV curve intersects the X-axis (highlighted by a red dot) is the internal rate of return (IRR) – the discount rate which equates the NPV to zero. The IRR in this case falls
somewhere between 20% to 25%.
Now assume further that the company has another project to consider which requires $1 million to start and is expected to generate the following net cash inflows over its three years life:
• Year 1: $1,000,000
• Year 2: $1,250,000
• Year 3: $1,500,000
Assuming the same discount rates as applied to the first project, we can find the NPV of the second project as follows:
Now that we have determined the NPV of second project, we can plot its NPV profile along with the first project to find the crossover rate of two projects. The crossover rate is the discount rate at
which the company is indifferent between two opportunities because it equates the net present value of both.
In above graph, the point at which the two NPV curves cross each other represents the crossover rate of the projects.
Uses of NPV profile
NPV profile is constructed as a part of overall NPV analysis of capital budgeting. It uses different discount rates to display how a change in discount rate impacts the net present value (NPV) of a
potential opportunity.
The projects with positive NPV profile are expected to increase the firm’s wealth and are considered good candidates to invest in. Moreover, it helps analysts evaluate the potential outcomes of
multiple projects together by plotting their data on a single graph.
Other techniques of testing a project’s viability include payback method, discounted payback method, internal rate of return method and accounting rate of return method.
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7.7: t-Interval for a Mean (2024)
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7.7.1 Student’s T-Distribution
A t-distribution is another symmetric distribution for a continuous random variable.
William Gosset was a statistician employed at Guinness and performed statistics to find the best yield of barley for their beer. Guinness prohibited its employees to publish papers so Gosset
published under the name Student. Gosset’s distribution is called the Student’s t-distribution.
A t-distribution is another special type of distribution for a continuous random variable.
Properties of the t-distribution density curve:
1. Symmetric, Unimodal (one mode) Bell-shaped.
2. Centered at the mean μ = median = mode = 0.
3. The spread of a t-distribution is determined by the degrees of freedom which are determined by the sample size.
4. As the degrees of freedom increase, the t-distribution approaches the standard normal curve.
5. The total area under the curve is equal to 1 or 100%.
Figure 7-7
Figure 7-7 shows examples of three different t-distributions with degrees of freedom of 1, 5 and 30. Note that as the degrees of freedom increase the distribution has a smaller standard deviation and
will get closer in shape to the normal distribution.
The t-critical value that has 5% of the area in the upper tail for n = 13.
Use a t-distribution with the degrees of freedom, df = n – 1 = 13 – 1 = 12. Draw and shade the upper tail area as in Figure 7-8. Use the DISTR menu invT option. Note that if you have an older TI-84
or a TI-83 calculator you need to have the program INVT installed.
For this function, you always use the area to the left of the point. If want 5% in the upper tail, then that means there is 95% in the bottom tail area. t[α] = invT(area below t-score, df) = invT
(0.95,12) = 1.782
You can download the INVT program to your calculator from http://MostlyHarmlessStatistics.com or use Excel =T.INV(0.95,12) = 1.7823.
Compute the probability of getting a t-score larger than 1.8399 with a sample size of 13.
To find the P(t > 1.8399) on the TI calculator, go to DISTR use tcdf(lower,upper,df). For this example, we would have tcdf(1.8399,∞,12). In Excel use =1-T.DIST(1.8399,12,TRUE) = 0.0453. P(t > 1.8399)
= 0.0453.
Figure 7-9
7.7.2 T-Confidence Interval
Note that we rarely have a calculation for the population standard deviation so in most cases we would need to use the sample standard deviation as an estimate for the population standard deviation.
If we have a normally distributed population with an unknown population standard deviation then the sampling distribution of the sample mean will follow a t-distribution.
Figure 7-10
A 100(1 - \(\alpha\))% Confidence Interval for a Population Mean μ: (σ unknown)
Choose a simple random sample of size n from a population having unknown mean μ.
The 100(1 - \(\alpha\))% confidence interval estimate for μ is given by \(\bar{x} \pm t_{\alpha / 2, n-1}\left(\frac{s}{\sqrt{n}}\right)\).
The df = degrees of freedom* are n – 1.
The degrees of freedom are the number of values that are free to vary after a sample statistic has been computed. For example, if you know the mean was 50 for a sample size of 4, you could pick any 3
numbers you like, but the 4^th value would have to be fixed to have the mean come out to be 50. For this class we just need to know that degrees of freedom will be based on the sample size.
The sample mean \(\bar{x}\) is the point estimate for μ, and the margin of error is \(t_{\alpha / 2}\left(\frac{s}{\sqrt{n}}\right)\). Where t[\(\alpha\)/2] is the positive critical value on the
t-distribution curve with df = n – 1 and area 1 – \(\alpha\) between the critical values –t[\(\alpha\)/2] and +t[\(\alpha\)/2], as shown in Figure 7-11.
Figure 7-11
Before we compute a t-interval we will practice getting t critical values using Excel and the TI calculator’s built in tdistribution.
Compute the critical values –t[\(\alpha\)/2] and +t[\(\alpha\)/2] for a 90% confidence interval with a sample size of 10.
Draw and t-distribution with df = n – 1 = 9, see Figure 7-12. In Excel use =T.INV(lower tail area, df) =T.INV(0.95,9) or in the TI calculator use invT(lower tail area, df) = invT(0.95,9). The
critical values are t = ±1.833
Figure 7-12
We can use Excel to find the margin of error when raw data is given in a problem. The following example is first done longhand and then done using Excel’s Data Analysis Tool and the T-Interval
shortcut key on the TI calculator.
The yearly salary for mathematics assistant professors are normally distributed. A random sample of 8 math assistant professor’s salaries are listed below in thousands of dollars. Estimate the
population mean salary with a 99% confidence interval.
66.0 75.8 70.9 73.9 63.4 68.5 73.3 65.9
First find the t critical value using df = n – 1 = 7 and 99% confidence, t[\(\alpha\)/2] = 3.4995.
Then use technology to find the sample mean and sample standard deviation and substitute the numbers into the formula.
\(\bar{x} \pm t_{\alpha / 2, n-1}\left(\frac{s}{\sqrt{n}}\right) \Rightarrow 69.7125 \pm 3.4995\left(\frac{4.4483}{\sqrt{8}}\right) \Rightarrow 69.7125 \pm 5.5037 \Rightarrow(64.2088,75.2162)\)
The answer can be given as an inequality 64.2088 < µ < 75.2162 or in interval notation (64.2088, 75.2162).
We are 99% confident that the interval 64.2 and 75.2 contains the true population mean salary for all mathematics assistant professors.
We are 99% confident that the mean salary for mathematics assistant professors is between $64,208.80 and $75,216.20.
Assumption: The population we are sampling from must be normal* or approximately normal, and the population standard deviation σ is unknown. *This assumption must be addressed before using
statistical inference for sample sizes of under 30.
TI-84: Press the [STAT] key, arrow over to the [TESTS] menu, arrow down to the [8:TInterval] option and press the [ENTER] key. Arrow over to the [Stats] menu and press the [ENTER] key. Then type in
the mean, sample standard deviation, sample size and confidence level, arrow down to [Calculate] and press the [ENTER] key. The calculator returns the answer in interval notation. Be careful. If you
accidentally use the [7:ZInterval] option you would get the wrong answer.
Alternatively (If you have raw data in list one) Arrow over to the [Data] menu and press the [ENTER] key. Then type in the list name, L[1], leave Freq:1 alone, enter the confidence level, arrow down
to [Calculate] and press the [ENTER] key.
TI-89: Go to the [Apps] Stat/List Editor, then press [2^nd] then F7 [Ints], then select 2:TInterval. Choose the input method, data is when you have entered data into a list previously or stats when
you are given the mean and standard deviation already. Type in the mean, standard deviation, sample size (or list name (list1), and Freq: 1) and confidence level, and press the [ENTER] key. The
calculator returns the answer in interval notation. Be careful: If you accidentally use the [1:ZInterval] option you would get the wrong answer.
Excel Directions
Type the data into Excel. Select the Data Analysis Tool under the Data tab.
Select Descriptive Statistics. Select OK.
Use your mouse and click into the Input Range box, then select the cells containing the data. If you highlighted the label then check the box next to Labels in first row. In this case no label was
typed in so the box is left blank. (Be very careful with this step. If you check the box and do not have a label then the first data point will become the label and all your descriptive statistics
will be incorrect.)
Check the boxes next to Summary statistics and Confidence Level for Mean. Then change the confidence level to fit the question. Select OK.
The table output does not find the confidence interval. However, the output does give you the sample mean and margin of error.
The margin of error is the last entry labeled Confidence Level. To find the confidence interval subtract and add the margin of error to the sample mean to get the lower and upper limit of the
interval in two separate cells.
The following screenshot shows the cell references to find the lower limit as =D3-D16 and the upper limit as =D3+D16. Make sure to put your answer in interval notation.
The answer is given as an inequality 64.2088 < µ < 75.2162 or in interval notation (64.2088, 75.2162).
We are 99% confident that the interval 64.2 and 75.2 contains the true population mean salary for all mathematics assistant professors.
A t-confidence interval is used to estimate an unknown value of the population mean for a single sample. We need to make sure that the population is normally distributed or the sample size is 30 or
larger. Once this is verified we use the interval \(\bar{x}-t_{\alpha / 2, n-1}\left(\frac{s}{\sqrt{n}}\right)<\mu<\bar{x}+t_{\alpha / 2, n-1}\left(\frac{s}{\sqrt{n}}\right)\) to estimate the true
population mean. Most of the time we will be using the t-interval, not the z-interval, when estimating a mean since we rarely know the population standard deviation. It is important to interpret the
confidence interval correctly. A general interpretation where you would change what is in the parentheses to fit the context of the problem is: “One can be 100(1 – \(\alpha\))% confident that between
(lower boundary) and (upper boundary) contains the population mean of (random variable in words using context and units from problem).” | {"url":"https://oxoncarts.com/article/7-7-t-interval-for-a-mean","timestamp":"2024-11-06T15:23:05Z","content_type":"text/html","content_length":"134708","record_id":"<urn:uuid:81174067-97b6-49da-bc8c-81cee8357a90>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00034.warc.gz"} |
2023 AMC 12B Problems/Problem 6
The following problem is from both the 2023 AMC 10B #12 and 2023 AMC 12B #6, so both problems redirect to this page.
When the roots of the polynomial
$\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}\]$
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
Solution 1
The expressions to the power of even powers are always positive, so we don't need to care about those. We only need to care about $(x-1)^1(x-3)^3(x-5)^5(x-7)^7(x-9)^9$. We need 0, 2, or 4 of the
expressions to be negative. The 9 through 10 interval and 10 plus interval make all of the expressions positive. The 5 through 6 and 6 through 7 intervals make two of the expressions negative. The 1
through 2 and 2 through 3 intervals make four of the expressions negative. There are $\boxed{\textbf{(C) 6}}$ intervals.
Solution 2
The roots of the factorized polynomial are intervals from numbers 1 to 10. We take each interval as being defined as the number behind it. To make the function positive, we need to have an even
number of negative expressions. Real numbers raised to even powers are always positive, so we only focus on $x-1$, $x-3$, $x-5$, $x-7$, and $x-9$. The intervals 1 and 2 leave 4 negative expressions,
so they are counted. The same goes for intervals 5, 6, 9, and 10. Intervals 3 and 4 leave 3 negative expressions and intervals 7 and 8 leave 1 negative expression. The solution is the number of
intervals which is $\boxed{\textbf{(C) 6}}$.
~darrenn.cp ~DarkPheonix
Solution 3
We can use the turning point behavior at the roots of a polynomial graph to find out the amount of intervals that are positive.
First, we evaluate any value on the interval $(-\infty, 1)$. Since the degree of $P(x)$ is $1+2+...+9+10$ = $\frac{10\times11}{2}$ = $55$, and every term in $P(x)$ is negative, multiplying $55$
negatives gives a negative value. So $(-\infty, 0)$ is a negative interval.
We know that the roots of $P(x)$ are at $1,2,...,10$. When the degree of the term of each root is odd, the graph of $P(x)$ will pass through the graph and change signs, and vice versa. So at $x=1$,
the graph will change signs; at $x=2$, the graph will not, and so on.
This tells us that the interval $(1,2)$ is positive, $(2,3)$ is also positive, $(3,4)$ is negative, $(4,5)$ is also negative, and so on, with the pattern being $+,+,-,-,+,+,-,-,...$ .
The positive intervals are therefore $(1,2)$, $(2,3)$, $(5,6)$, $(6,7)$, $(9,10)$, and $(10,\infty)$, for a total of $\boxed{\textbf{(C) 6}}$.
~nm1728 ~ESAOPS (minor edits)
Solution 4
Denote by $I_k$ the interval $\left( k - 1 , k \right)$ for $k \in \left\{ 2, 3, \cdots , 10 \right\}$ and $I_1$ the interval $\left( - \infty, 1 \right)$.
Therefore, the number of intervals that $P(x)$ is positive is \begin{align*} 1 + \sum_{i=1}^{10} \Bbb I \left\{ \sum_{j=i}^{10} j \mbox{ is even} \right\} & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac
{\left( i + 10 \right) \left( 11 - i \right)}{2} \mbox{ is even} \right\} \\ & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{- i^2 + i + 110}{2} \mbox{ is even} \right\} \\ & = 1 + \sum_{i=1}^{10} \Bbb
I \left\{ \frac{i^2 - i}{2} \mbox{ is odd} \right\} \\ & = \boxed{\textbf{(C) 6}} . \end{align*}
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 5: Graphing
Recall two key facts about the roots of a polynomial: - If a root has an odd multiplicity (e.g. the root appears an odd number of times), then the graph will cross the x-axis. - If a root has an even
multiplicity (e.g. the root appears an even number of times), then the graph will bounce off the x-axis.
Sketching the graph and noting the multiplicity of the roots, we see that there are $C)6$ positive intervals.
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Interstigation
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. | {"url":"https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_6&oldid=219239","timestamp":"2024-11-02T12:24:33Z","content_type":"text/html","content_length":"59516","record_id":"<urn:uuid:767bface-a7b9-42b7-838f-b10952d75cfb>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00468.warc.gz"} |
Holidays with grandmam - math word problem (8286)
Holidays with grandmam
We packed three white, red, and orange T-shirts and five pairs of pants - blue, green, black, pink, and yellow. How many days can we spend with the old mother if we put on a different combination of
clothes every day?
Correct answer:
Did you find an error or inaccuracy? Feel free to
write us
. Thank you!
You need to know the following knowledge to solve this word math problem:
Related math problems and questions: | {"url":"https://www.hackmath.net/en/math-problem/8286","timestamp":"2024-11-02T12:13:34Z","content_type":"text/html","content_length":"47675","record_id":"<urn:uuid:9431184e-3c64-4492-a56d-5b2830835e3c>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00395.warc.gz"} |
Functions: Contracts, Examples & Definitions
(Also available in Pyret)
Students learn to connect function descriptions across three representations: Contracts (a mapping between Domain and Range), Examples (a list of discrete inputs and outputs), and Definitions
Students will be able to:
• identify patterns where a function would be useful
Lesson Goals
• define their own function
• match examples, contracts, and definitions for the same function
• I can explain why a function is useful.
Student-Facing Lesson Goals • I can write my own function.
• I can connect contracts, examples, and definitions for a function.
a statement of the name, domain, and range of a function
shows the use of a function on specific inputs and the computation the function should perform on those inputs
a relation from a set of inputs to a set of possible outputs, where each input is related to exactly one output
function definition
code that names a function, lists its variables, and states the expression to compute when the function is used
a name or symbol that stands for some value or expression, often a value or expression that changes
🔗Three Representations of a Function 55 minutes
Students will practice describing functions using all 3 representations (in programming syntax), and test their code against the examples in the editor.
• Open the bc Starter File. Look at the Contract, some Examples, and the Function Definition for gt.
• What do you Notice? What do you wonder?
1 Every function has a contract.
2 We can write examples illustrating how a function should work to help us identify the pattern.
3 Function definitions replace whatever changes in the examples with a variable describing what changes.
If we use the correct syntax, we can include all three of these function representations in our WeScheme files. Let’s take a look!
• Click "Run". * Change (EXAMPLE (gt 10) (triangle 10 "solid" "green")) to (EXAMPLE (gt 10) (triangle 15 "solid" "green"))
• Click "Run". What happens?
□ The editor lets us know that the function doesn’t match the examples so that we can fix our mistake!
Examples not only help us to identify the pattern to define a function, they also let us double check that the functions we define do what we intend for them to do!
Think about these three representations of functions by completing:
For more practice, complete these Desmos card sort activities:
There are many more materials for students to work with in the Additional Practice section at the end of the lesson!
• What strategies did you use to match the examples with the contracts?
• What strategies did you use to match the examples with the function definitions?
🔗Defining bc and Other Functions
Using gt as an example, students will write the contract, examples, and definition for several other functions.
• On the top half of the page, you will see the contract, examples, and function definition for gt.
• Circle what is changing and label it with the word size.
• Using gt as a model, complete the contract, examples and function definition for bc.
When students have completed the above steps, direct them to type the Contract, Examples and Definition into the Definitions Area. They will then click “Run”, and make sure all of the examples pass!
Check-in with students to gauge their confidence level. (Thumbs up? Thumbs to the side? Thumbs down?) How confident do students feel in writing the contract, examples and function definition on their
own if they were given a word problem about another shape function?
As students work, walk around the room and make sure that they are circling what changes in the examples and labeling it with a variable name that reflects what it represents.
• How were each of the representations helpful?
• Why is it important to write examples in our code?
These materials were developed partly through support of the National Science Foundation, (awards 1042210, 1535276, 1648684, and 1738598). Bootstrap by the Bootstrap Community is licensed under a
Creative Commons 4.0 Unported License. This license does not grant permission to run training or professional development. Offering training or professional development with materials substantially
derived from Bootstrap must be approved in writing by a Bootstrap Director. Permissions beyond the scope of this license, such as to run training, may be available by contacting | {"url":"https://bootstrapworld.org/materials/fall2022/en-us/lessons/functions-examples-definitions-wescheme/index.shtml?pathway=algebra-wescheme","timestamp":"2024-11-07T05:47:12Z","content_type":"text/html","content_length":"29055","record_id":"<urn:uuid:2e284462-c108-4f0d-9810-479d3a2affbb>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00448.warc.gz"} |
There are many interesting problems that can be studied with probability theory. Here I will discuss a few of my favorites.
Maxima in Data
Suppose we have a sequence of values of random variables \(\left(X_{k} \right )_{k=1}^{n}\)that are independent and identically distributed with density function \(f_{X}(x)\). We wish to find the
expected number of local maxima in the data, that is, values of \(X_{k}\) such that \(X_{k-1} < X_{k} > X_{k+1}\). Let \(y=X_k\). Then, the probability \(p\) that a certain value y is a maximum is
that of having \(X_{k-1} < y > X_{k+1}\). As all the \(X\)s are independent and identically distributed, we can calculate this probability by finding \[p= \int_{-\infty}^{\infty}f_{X}(x)F_{X}^2(x)dx
\] Where \(F_{X}(x)=\int_{-\infty}^{x}f_{X}(y)dy\) is the cumulative distribution function of \(f_{X}\). The form above is like \(\sum_y P(X_k=y)P(X_{k-1} < y)P(X_{k+1} < y)\), except we use the
continuous formulation. However, clearly \(F_{X}'(x)=f_X(x)\), and so the formula becomes \[p= \int_{-\infty}^{\infty}F_{X}'(x)F_{X}^2(x)dx\] But, by elementary calculus, we can change this to \[p= \
int_{0}^{1}F^2 dF=\frac{1}{3}\] (the change in bounds arises from the fact that \(F_{X}(-\infty)=0\) and \(F_{X}(+\infty)=1\)). Thus, we expect one-third of the values in the sequence to be local
maxima. Likewise, we expect one-third to be local minima, and the remaining third to be neither. By the same method, we can look for other patterns. For instance, the fraction of data points that are
higher than all four of their closest neighbors is \(\frac {1}{5}\). The fraction of data points that are bigger than their closest neighbors and smaller than their next-closest neighbors is \(\frac
{1}{30}\). In fact, all the calculations can be made by evaluating integrals of the form \( \int_{0}^{1} x^m (1-x)^n dx\). We can also use results like this to test for non-independent or
non-identically distributed data. It may even be possible to use it in fraud or bias detection. Based on next-to-nothing-at-all, I would expect human generated data to fail some of these tests. We
can also find that the distribution of the number of maxima in \(n\) data points, regardless of the probability distribution \(f_{X}\), is approximately normally distributed with mean \(\frac{n}{3}\)
and variance \(\frac{2 n}{45}\). Thus, if we found fewer than 2960 or more than 3040 maxima in a list of 9000 data points, we could be 95% confident that the list was not of independent and
identically distributed values. We can also run the same test for minima, but for non-maxima-non-minima, the variance is instead \(\frac{2 n}{15}\). The values for the variances were found
empirically. I don't really know how one would go about finding them analytically. We can also find the distribution of the values of the maxima, which is easily found to be \[g(x)=3 f_{X}(x)F_{X}^2
\] Other distributions are similarly found.
Joint Lives
Suppose we stat with \(N\) couples (\(2N\) people), and at a later time, \(M\) of the original \(2N\) people remain. We want to find the expected number of intact couples remaining. Let \(C(M)\) and
\(W(M)\) be the expected remaining number of couples and widows respectively when M total people are left. We then note that, as any remaining person is equally likely to be eliminated next, we have:
\[ C(M-1)=C(M)-2 \frac{C(M)}{M} \\ W(M-1)=W(M)-\frac{W(M)}{M}+2 \frac{C(M)}{M}\] We can solve this recurrence relation, subject to the constraints \(W(M)+2 C(M)=M\) and \(C(2N)=N, W(2N)=0\), and find
that \[ C(M)=\frac{M(M-1)}{2(2N-1)} \\ W(M)=\frac{M(2N-M)}{2N-1} \] If we express M as a fraction of the total starting population: \(M=2xN\), and express \(C\) and \(W\) as fractions of the total
population, we find, for \(N\) big: \[ C(x)=x^2 \\ W(x)=x(1-x) \] Also, for the general case of starting out with \(kN\) \(k\)-tuples, the expected number of intact \(k\)-tuples when \(M\)
individuals remain is given by: \[K(M)=N \frac{\binom{M}{k}}{\binom{kN}{k}}\] For the case of triples, we have the number of triples, doubles and singles when M individuals remain is given by: \[K_3
(M)= \frac{M(M-1)(M-2)}{3(3N-1)(3N-2)} \\ K_2 (M)= \frac{M(M-1)(3N-M)}{(3N-1)(3N-2)} \\ K_1 (M)= \frac{M(3N-M)(3N-M-1)}{(3N-1)(3N-2)} \] Generally, with the same sense as discussed above, the
fraction of the population in a m-tuple, beginning with only k-tuples, when fraction \(x\) of the population remains, is given by: \[K_m (x)=\binom{k-1}{m-1} x^m (1-x)^{k-m}\] In fact, the general
form for the expected number can be given as \[ K_m (M)= N \frac{\binom{M}{m}\binom{kN-M}{k-m}}{\binom{kN}{k}} \]
Random Finite Discrete Distribution
Suppose we have a discrete random variable that can take on the values \(1,2,3,...,n\) with probabilities \(p_1,p_2,p_3,...p_n\) respectively, subject to the constraint \(\sum_{k=1}^n p_k=1\). Let \
(p\) be an arbitrary value among the \(p\)s. We will take any combination of values for the \(p\)s as equally likely. By looking at the cases of n=2 and n=3, we find that the probability density
function of \(p\) is given by \[ f_P(p)=(n-1)(1-p)^{n-2} \] And the cumulative distribution function is given by \[ F_P(p)=1-(1-p)^{n-1} \] The average value of \(p\) is then \[\int_{0}^{1} p(n-1)
(1-p)^{n-2}dp=\frac{1}{n}\] And the variance is \[\int_{0}^{1} p^2 (n-1)(1-p)^{n-2}dp-\frac{1}{n^2}=\frac{n-1}{n^2 (n+1)}\] We thus find that the chance that \(p\) is above the average value is \[P\
left ( p > \frac{1}{n} \right )=\left ( 1-\frac{1}{n} \right )^{n-1}\] In the limit as n becomes large, this value tends to \(\frac{1}{e}\). A confidence interval containing fraction x of the total
probability, for large n, is given by: \[ \frac{1}{n} \ln \left(\frac{e}{e x +1-x} \right) \leq p \leq \frac{1}{n} \ln \left(\frac{e}{1-x} \right) \] For instance, a \(50 \%\) confidence interval is
given by \(\frac{1}{n}\ln \left(\frac{2 e}{1+e}\right) \leq p \leq \frac{1}{n}\ln(2 e)\). We can also extend this to continuous distributions with finite support if we only consider the net
probability of landing in equally-sized bins. While the calculation may break down if the number of possible values is actually infinite, it can be used to get some information about distributions
with an arbitrarily large number of possible values.
Maximum of Exponential Random Variables
Suppose we have \(N\) independent and identically distributed exponential random variables \(X_1,X_2,...X_N\) with means \(\mu\). That is, \(f_X (x_k)=\frac{1}{\mu} e^{-\frac{x}{\mu}}\) when \(x \geq
0\) and zero otherwise. Let us interpret the random values as lifetimes for \(N\) units. The exponential distribution has the interesting property of
, which means that \(P(x > a+b| x > b)=P(x > a)\). We can show this by using the definition: \[ P(x>a+b|x>b)=\frac{P(x>a+b \cap x>b)}{P(x>b)}=\frac{P(x>a+b)}{P(x>b)} \\ P(x>a+b|x>b)=\frac{\int_{a+b}^
{\infty}e^{-\frac{x}{\mu}}dx}{\int_{b}^{\infty}e^{-\frac{x}{\mu}}dx}=\frac{e^{-\frac{a+b}{\mu}}}{e^{-\frac{b}{\mu}}}=e^{-\frac{a}{\mu}}=P(x>a) \] In other words, given that a unit lasted \(b\)
minutes, the chance that it will last another \(a\) minutes is the same as that it would last \(a\) minutes. We now calculate the probability distribution of the minimum of the \(N\) random
variables. The probability that the minimum of \(X_1,X_2,...X_N\) is no less than \(x\) is the same as the probability that \(X_1 \geq x \cap X_2 \geq x \cap...X_N \geq x \). As all the \(X\)s are
independent, this can be simplified to a product, and as all the \(X\)s are identically-distributed, we can simplify this further: \[ P(\min(X_1,X_2,...)\geq x)=\left ( P(X_1 \geq x) \right )^N= \
left ( \int_{x}^{\infty}\frac{1}{\mu} e^{-\frac{x}{\mu}}\right )^N \\ P(\min(X_1,X_2,...)\geq x)= e^{-\frac{xN}{\mu}} \\ P(\min(X_1,X_2,...)\leq x)= 1-e^{-\frac{xN}{\mu}} \\ f_{\min(X)}(x)=\frac{N}{\
mu}e^{-\frac{xN}{\mu}} \] Thus, the average of the minimum of \(X_1,X_2,...X_N\) is \(\frac{\mu}{N}\). We now combine these two facts, the mean minimum vale and the memorylessness. We start with all
units operational, and we have to wait an average of \(\frac{\mu}{N}\) until the first one fails. However, given that the first one fails, the expected additional wait time until the next one fails
is just \(\frac{\mu}{N-1}\), that is, the expected minimum of \(N-1\) units. Thus, the expected time that the \(m\)th unit fails is given by \[\mu\sum_{k=0}^{m-1}\frac{1}{N-k}\] Thus, the expected
maximum time, when the \(N\)th unit fails is \[\mu\sum_{k=1}^{N}\frac{1}{k}\] More generally, we can look at the distributions of the kth
order statistic
of \(X_1,X_2,...X_N\). The kth order statistic, denoted \(X_{(k)}\), is defined as the kth smallest value, so that \(X_{(1)}\) is the smallest (minimum) value, and \(X_{(N)}\) is the largest
(maximum) value. The pdf is easily found to be: \[ f_{X_{(k)}}(x)=k {N \choose k} F_X^{k-1}(x)\left[1-F_X(x)\right]^{N-k}f_X(x) \] Where \(F_X(x)\) is the cdf of X, and \(f_X(x)\) is the pdf of X.
So, in this case, \[ f_{X_{(k)}}(x)=\frac{k}{\mu} {N \choose k} e^{-(N-k+1)x/\mu}\left[1-e^{-x/\mu}\right]^{k-1} \] Thus, the
moment generating function
is given by: \[ g(t)=\frac{k}{\mu} {N \choose k} \int _0 ^\infty e^{-(N-k+1-\mu t)x/\mu}\left[1-e^{-x/\mu}\right]^{k-1} dx \] By a simple transformation, we find that: \[ g(t)=k {N \choose k} \int _0
^1 u^{N-k-\mu t}(1-u)^{k-1} du \] This puts the integral in a
well-known form
, which has the value \[ g(t)=\frac{N!}{\Gamma(N+1-\mu t)}\frac{\Gamma(N-k+1-\mu t)}{(N-k)!} \] By a simple calculation, the
are then given by the surprisingly simple form: \[ \kappa_n=\mu^n(n-1)!\sum_{j=N-k+1}^{N} \frac{1}{j^n} \] Several interesting results follow from this:
• For the Nth order statistic (the maximum), we already know that the mean value goes as \(\sum_{j=1}^{N} \frac{1}{j}\). But now we see that the other cumulants go as \((n-1)!\sum_{j=1}^{N} \frac
{1}{j^n}\). Thus, the variance converges, in the limit, to \(\mu^2 \frac{\pi^2}{6}\). The skewness converges, in the limit, to \(\frac{12 \sqrt{6}}{\pi^3}\zeta(3)\), and the excess kurtosis
converges to \(\frac{12}{5}\). In fact, if we shift to take into account the unbounded mean, the distribution of the maximum converges to a Gumbel distribution. This is a special case of a
fascinating result known as the extreme value theorem.
• For any given, fixed, finite \(k\geq 0\), \(X_{(N-k)}\) converges, as N goes to infinity, to a non-degenerate distribution with finite, positive variance, if we shift it to account for the
unbounded mean.
• For k of the form \(k=\alpha N\) (or the nearest integer thereto), for some fixed alpha between 0 and 1, for \(\alpha \neq 1\), in the limit at N goes to infinity, the distribution of \(X_{(\
alpha N)}\) become degenerate distributions with all the probability density located at \(\mu\ln\left(\frac{1}{1-\alpha}\right)\). These are, of course, the locations of the \(100\alpha \%\)
quantiles, and so \(X_{(\alpha N)}\) is a consistent estimator for the \(100\alpha \%\) quantile.
As a more general result, let us find the cdf of \(X_{(\alpha N)}\) for an arbitrarily distributed X, in the limit as N goes to infinity. The cdf of \(X_{(\alpha N)}\) is given by: \[ F_{X_{(\alpha
N)}}(y)=\alpha N {N \choose \alpha N} \int _{-\infty} ^{y} F_X^{\alpha N-1}(x)\left[1-F_X(x)\right]^{N-\alpha N}f_X(x) dx \] As \(f_X(x)=\frac{d}{dx}F_X(x)\), we then have, by a simple substitution:
\[ F_{X_{(\alpha N)}}(y)=\alpha N {N \choose \alpha N} \int _{0} ^{F_X(y)} u^{\alpha N-1}\left[1-u\right]^{N-\alpha N} du \] This is the cdf of a
Beta distributed
random variable, with mean \(\mu=\frac{\alpha N}{N+1}\) and variance \(\sigma^2=\frac{\alpha N (N-\alpha N+1)}{(N+1)^2(N+2)}\). Thus, as N goes to infinity, this will converge in distribution to a
degenerate distribution with all the density at \(y=F_X^{-1}(\alpha)\), that is, at the \(100\alpha \%\) quantile of the distribution.
Choosing a Secretary
Suppose we need to hire a secretary. We have \(N\) applicants arrive and we interview them sequentially: once we interview and dismiss an applicant, we cannot hire her. The applicants all have
differing skill levels, and we want to pick as qualified an applicant as we can. We want to find the optimal strategy for choosing whom to hire. We easily see that the optimal strategy is something
like the following. We consider and reject the first \(K\) applicants. We then choose the first applicant who is better than all the preceding ones. Thus, our problem reduces to finding the optimal
value for \(K\). We will do so in a way that maximizes the probability that the most qualified secretary is selected. We thus have the probability: \[ P(\mathrm{best\, is\, chosen})=\sum_{n=1}^{N}P(\
mathrm{n^{th}\, is\, chosen} \cap \mathrm{n^{th}\, is\, best}) \\ P(\mathrm{best\, is\, chosen})=\sum_{n=1}^{N}P(\mathrm{n^{th}\, is\, chosen}| \mathrm{n^{th}\, is\, best})P(\mathrm{n^{th}\, is\,
best}) \] We then note that each applicant in line is the best applicant with equal probability. That is, \(P(\mathrm{n^{th}\, is\, best})=\frac{1}{N}\). Also, we can find the conditional
probabilities. If \(M \leq K\), then \(P(\mathrm{M^{th}\, is\, chosen}| \mathrm{M^{th}\, is\, best})=0\). If the \((K+1)\)th applicant is best, she will certainly be chosen, that is \(P(\mathrm{(K+1)
^{th}\, is\, chosen}| \mathrm{(K+1)^{th}\, is\, best})=1\). Also, we find that \(P(\mathrm{(K+m)^{th}\, is\, chosen}| \mathrm{(K+m)^{th}\, is\, best})=\frac{K}{K+m}\), as that is the chance that the
second-best applicant among the first \(K+m\) applicants is in the first \(K\) applicants. We thus have \[ P(\mathrm{best\, is\, chosen})=\frac{K}{N}\sum_{n=K+1}^{N}\frac{1}{n} \] Let us assume we
are dealing with a relatively large number of applicants. In that case, we can approximate \(\sum_{n=A+1}^{B}\frac{1}{n} \approx \ln \left(\frac{B}{A} \right )\). Thus \[ P(\mathrm{best\, is\,
chosen})=\frac{K}{N}\ln \left(\frac{N}{K}\right )=-\frac{K}{N}\ln \left(\frac{K}{N}\right ) \] If we then let \(x=\frac{K}{N}\), we just need to maximize \(-x\ln(x)\), which happens at \(x=e^{-1}\).
From this, we find that \(P(\mathrm{best\, is\, chosen})=e^{-1}\). Thus, the best strategy is to interview and reject the first \(36.8 \%\) of the applicants, and then choose the next applicant who
is better than all the preceding ones. This will get us the best applicant with a probability of \(36.8 \%\). A related problem involves finding a strategy that minimizes the expected rank of the
selected candidate (the best candidate has rank 1, the second best rank 2, etc.).
Chow, Moriguti, Robbins and Samuels
have found that the optimal strategy involves the following (in the limit of large \(N\)): skip the first \(c_0 N\) applicants, then, for all applicants before the number \(c_1 N\), we stop looking
if the applicant is the best so far. If we have not yet selected an applicant, we choose the best or second best so far before the number \(c_2 N\). If we have not yet selected an applicant, we
choose the best or second best or third best so far before the number \(c_3 N\). And so on. By choosing the \(c_n\) optimally, we can get an expected rank of \(3.8695\). This is quite surprising: we
can expect an applicant in the top 4, even among billions of applicants! The optimal values for the \(c_n\) are \[ c_0=0.2584... \\ c_1=0.4476... \\ c_2=0.5639... \] The
general formula
for \(c_n\) is \[ c_n=\prod_{k=n+2}^{\infty}\left ( \frac{k-1}{k+1} \right )^{1/k}=\frac{1}{3.86951924...}\prod_{k=2}^{n+1}\left ( \frac{k+1}{k-1} \right )^{1/k} \] | {"url":"https://www.hyperphronesis.com/2014/07/","timestamp":"2024-11-14T01:09:28Z","content_type":"text/html","content_length":"61804","record_id":"<urn:uuid:cd1f3d0e-ec47-42ec-a49b-10f309ed22a6>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00154.warc.gz"} |
Problems on Trains Online Test, Train Problems: Concepts, Example
Problems on Trains Online Test in hindi and english language. take free problems on trains mock test, quiz, & MCQ from below. Check out Train Problems: Concepts, Examples and Practice Questions on
this page. Where can I get Aptitude Problems on Trains questions and answers with explanation? CAknowledge provides you fully solved Aptitude Problems on Trains questions and answers with Proper
Problems on Trains Online Test
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Problems on Trains Online Test Series 3
Problems on Trains concept
Due to the small size of the cars and other objects, we take them as point objects. Questions on trains are solved using the concept of time, speed and distance i.e. we use the formulas of time,
speed and distance to solve questions on trains. Since the size of the trains is comparable to the distances that they may travel, then we will have to take the size or the length of the trains into
account too. The same formulae that we saw already are applicable here too. Here we shall see some examples of this concept and then learn some tricks from other examples.
Problems on Trains Formula
Train Problems form an interesting portion of the time-distance problems. The Train Problems are a bit different than the regular problems on the motion of the objects. This is due to the finite size
of the trains. As a result of the length of the trains, many interesting train problems originate. Keep same units for all values mentioned in the problem i.e. as per the units of the given answers
convert kilometer per hour (km/hr) to meters per second (m/s) and vice versa. In a similar way, convert meter (m) into centimeter (cm) and vice versa. See the examples given below:
The formula to convert Km/hr into m/s:
• 1km is equal to 1000 meters
• 1 hours is equal to 3600 seconds
• 1Km/hr is equal to meter/sec or m/s
So, to convert a value in Km/hr to m/s, we need to multiply it with See the example given below:
60 km/hr* = 16.7 m/s
The formula to convert m/s to Km/hr
• 1 meter is equal to 1/1000 km
• 1 sec is equal to 1/3600 hours
• 1 m/s is equal to
So, to covert a value in m/s to Km/hr, we will multiply it with 18/5. See the example given below:
20 m/s* =72 km/hr
Important facts about moving trains:
• The distance traveled by a train to cross a pole or person is equal to the length of the train.
• The distance traveled by train when it crosses a platform is equal to the sum of the length of the train and length of the platform.
• When two trains are travelling in opposite directions at speeds V1 m/s and V2 m/s then their relative speed is the sum of their individual speeds (V1+V2) m/s.
• Two trains are travelling in the same direction at V1 m/s and V2 m/s where V1 > V2 then their relative speed will be equal to the difference between their individual speeds (V1-V2) m/s.
• When two trains of length X meters and Y meters are moving in opposite direction at V1 m/s and V2 m/s then the time taken by the trains to cross each other is =
• When two trains of length X meters and Y meters are moving in same direction at V1 and V2 where V1> V2 then the time taken by the faster train to cross the slower train =
• When two trains X and Y start moving towards each other at the same time from points A and B and after crossing each other the train X reaches point B in a seconds and train Y reaches points A in
b seconds, then
Train X speed: Train Y speed = b^1/2: a^1/2
Here we will learn certain tricks and see the various forms of the train problems.
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How to assemble the cross in a cube RubikHow to assemble the cross in a cube Rubik 🚩 How to assemble a Rubik's cube for dummies 🚩 More.
Let each side of your letter. F — front side, bottom side, top side, P — right side, B — back, L — left side.
Denote rotation of the cube faces : clockwise a quarter turn (90 deg.) — F, N, B, P, Z, L, counter-clockwise a quarter turn (90 deg.) — F1, H1, B1, L1, P1, L1, rotating half-turn in either direction
(180 deg.) — F2, N2, B2, P2, M2, L2.
Choose the color of the top face of the cube. It needs to remain at the top throughout the build procedures.
Collecting the cross of the color that was chosen as the basis. Follow the procedure spins P1,C, P, B1 — if the piece is in the first layer and deployed correctly, only is not in place or follow a
different algorithm spins A1, B1, F1, — if a piece is in the first layer, but is deployed incorrectly and should be moved to another location.
Running the algorithm spins B2, F1, B2 if a piece is in the second layer or the mirror image of b, P, B1 if the item is in the second layer, but the adjacent faces of the cube.
Make rotation N2, Z2 if the piece is in the third layer of the lower plane or one of the variants of rotations N1, F1, P, f, or P, V, F1,W1 if the option is in the third layer and deployed correctly.
Rotate the top face of the cube until you match the two colors of the side faces and side centers, then take one of the algorithms of rotation. This is P, B, P1, B1,P if you need to swap two adjacent
element crossa or A2, L2, N2, P2, L2 if you need to swap two opposite element of the cross.
Trying to understand, first note our notation to avoid confusion.
Useful advice
These algorithms of rotation is not necessary to remember, it is important to understand how they operate. Because you will be able to choose other options build right cross that you will be more | {"url":"https://eng.kakprosto.ru/how-16791-how-to-assemble-the-cross-in-a-cube-rubik","timestamp":"2024-11-11T00:27:41Z","content_type":"text/html","content_length":"29492","record_id":"<urn:uuid:122d389c-76d1-41fd-a2ae-61e10b0ab029>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00050.warc.gz"} |
FREE 11 Sample Order Of Operations Worksheet Templates In PDF | Order of Operation Worksheets
FREE 11 Sample Order Of Operations Worksheet Templates In PDF
FREE 11 Sample Order Of Operations Worksheet Templates In PDF
FREE 11 Sample Order Of Operations Worksheet Templates In PDF – You may have heard of an Order Of Operations Worksheet, yet what exactly is it? In addition, worksheets are a wonderful means for
students to practice new abilities and testimonial old ones.
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The main purpose of an order of operations worksheet is to help pupils learn the right means to resolve math equations. If a student doesn’t yet recognize the principle of order of operations, they
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One of the most vital points you can learn in math is the order of operations. The order of operations ensures that the mathematics troubles you address correspond. This is vital for tests as well as
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An order of operations worksheet is a great way to instruct pupils the proper method to fix mathematics formulas. Before pupils begin utilizing this worksheet, they might need to examine ideas
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An order of operations worksheet can assist trainees establish their skills in addition and subtraction. Natural born player’s worksheets are a best means to help pupils discover regarding the order
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Order Of Operations Integers Worksheet
Order Of Operations PEDMAS With Integers 1 Worksheet
Order Of Operations Integers Worksheet
Order Of Operations Integers Worksheet supply a wonderful resource for young learners. These worksheets can be easily personalized for particular requirements. They can be located in three levels of
problem. The initial degree is straightforward, requiring students to practice making use of the DMAS technique on expressions including four or even more integers or 3 operators. The second degree
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The Order Of Operations Integers Worksheet can be downloaded absolutely free and also can be published out. They can then be reviewed making use of addition, division, multiplication, as well as
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Quasi-Static solution with Timesteppers
Hello together,
the new timesteppers are great. But wouldn’t it be nice if the u.dt term evaluates to zero in case of classical use of bilinear form and solver.
Consider e.g. following problem:
from ngsolve import *
from ngsolve.webgui import Draw
from netgen.occ import *
shape = Rectangle(1,1).Face()
mesh = Mesh(OCCGeometry(shape, dim=2).GenerateMesh(maxh=0.05))
fes = H1(mesh, order=4, dirichlet="left|right")
u,v = fes.TnT()
eq = ( grad(u) * grad(v) + u.dt * v) * dx
gfu = GridFunction(fes)
def init_gfu(gfu):
t_left = 10
t_right = 20
t_ref = 15
gfu.Interpolate(mesh.BoundaryCF({"left": t_left, "right": t_right}, default=t_ref),definedon=mesh.Boundaries("left|right"))
works as expected in case of timestepper usage:
ts = timestepping.ImplicitEuler(equation=eq, dt=1e-1)
scene = Draw(gfu)
def callback(t, gfu):
ts.Integrate(gfu, start_time=0, end_time=2, callback=callback)
But if we plug eq directly into a bfa and solve the same problem, the u.dt evaluates to u and we get a different result instead of the quasi-static solution:
a = BilinearForm(fes, symmetric=True)
a += eq.Compile()
solvers.Newton(a=a, u=gfu, inverse="sparsecholesky", printing=False)
scene = Draw (gfu)
import matplotlib.pyplot as plt
import numpy as np
X = np.linspace(0, 1, num=100)
Y = np.ones_like(X) * 0
plt.plot(X, gfu(mesh(X, Y)))
Hi Nils,
thank you for the hint. We will throw an exception if there is a u.dt term in a Newton solver (which assumes to solve a stationary problem).
You have to replace the u.dt by a Zero-CF of the same shape.
Hi Joachim,
probably you are right, and an exception is the desired behavior.
I will replace the .dt on my own. Seems to be more or less straight forward, with the timestepping implementations in hand.
back again to timesteppers. A couple of equations do not work with “equation.Replace(…)” and raise an NgException: Transform not overloaed NgException. There are at least three (maybe many more) that
I would like to use:
NgException: Transform not overloaed, desc = symmetric,
NgException: Transform not overloaed, desc = class ngfem::IfPosCoefficientFunction,
NgException: Transform not overloaed, desc = scalar-vector multiply
Is it possible to implement them?
Best regards, Nils
Should now be there:
Thanks for the modifications.
Now I get the next exception:
Transform not overloaed, desc = matrix-vector multiply
Sorry for bothering | {"url":"https://forum.ngsolve.org/t/quasi-static-solution-with-timesteppers/2967","timestamp":"2024-11-07T23:38:09Z","content_type":"text/html","content_length":"24159","record_id":"<urn:uuid:dc2bf0da-2525-4552-a82e-92fee9f6330c>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00507.warc.gz"} |
Sorting Array of Strings C | HackerRank Solutions
Sorting Array of Strings C
Problem Statement :
To sort a given array of strings into lexicographically increasing order or into an order in which the string with the lowest length appears first, a sorting function with a flag indicating the type of comparison strategy can be written. The disadvantage with doing so is having to rewrite the function for every new comparison strategy.
A better implementation would be to write a sorting function that accepts a pointer to the function that compares each pair of strings. Doing this will mean only passing a pointer to the sorting function with every new comparison strategy.
Given an array of strings, you need to implement a string_sort function which sorts the strings according to a comparison function, i.e, you need to implement the function :
void string_sort(const char **arr,const int cnt, int (*cmp_func)(const char* a, const char* b)){
The arguments passed to this function are:
1. an array of strings : arr
2. length of string array: count
3. pointer to the string comparison function: You also need to implement the following four string comparison functions: camp_fun
Input Format
You just need to complete the function string\_sort and implement the four string comparison functions.
1 <= No. of Strings <= 50
1<= Total Length of all the strings <= 2500
You have to write your own sorting function and you cannot use the inbuilt qsort function
The strings consists of lower-case English Alphabets only.
Output Format
The locked code-stub will check the logic of your code. The output consists of the strings sorted according to the four comparsion functions in the order mentioned in the problem statement.
Solution :
Solution in C :
int lexicographic_sort(const char* a, const char* b) {
return strcmp(a, b);
int lexicographic_sort_reverse(const char* a, const char* b) {
return strcmp(b, a);
#define CHARS 26
int distinct_chars(const char *a)
int dist = 0;
int chars[CHARS] = {0};
while (*a != '\0') {
int chr = (*a++) - 'a';
if (chr < CHARS)
for (int i = 0; i < CHARS; i++)
if (chars[i])
return dist;
int sort_by_number_of_distinct_characters(const char* a, const char* b) {
int res = distinct_chars(a) - distinct_chars(b);
return (res) ? res : lexicographic_sort(a, b);
int sort_by_length(const char* a, const char* b) {
int res = strlen(a) - strlen(b);
return (res) ? res : lexicographic_sort(a, b);
/* simple bubble sort :) */
void string_sort(char** arr, const int len,int (*cmp_func)(const char* a, const char* b)) {
int sorted = 0;
int top = len - 1;
while (!sorted) {
sorted = 1;
for (int i = 0; i < top; i++) {
if (cmp_func(arr[i], arr[i + 1]) > 0) {
char *tmp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = tmp;
sorted = 0;
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Python code to multiply octonions and sedenions
Python code for octonion and sedenion multiplication
The previous post discussed octonions. This post will implement octonion multiplication in Python, and then sedenion multiplication.
Cayley-Dickson construction
There’s a way to bootstrap quaternion multiplication into octonion multiplication, so we’ll reuse the quaternion multiplication code from an earlier post. It’s called the Cayley-Dickson construction.
For more on this construction , see John Baez’s treatise on octonions.
If you represent an octonion as a pair of quaternions, then multiplication can be defined by
(a, b) (c, d) = (ac − d*b, da + bc*)
where a star superscript on a variable means (quaternion) conjugate.
Note that this isn’t the only way to define multiplication of octonions. There are many (480?) isomorphic ways to define octonion multiplication.
(You can take the Cayley-Dickson construction one step further, creating sedenions as pairs of octonions. We’ll also provide code for sedenion multiplication below.)
Code for octonion multiplication
import numpy as np
# quaternion multiplication
def qmult(x, y):
return np.array([
x[0]*y[0] - x[1]*y[1] - x[2]*y[2] - x[3]*y[3],
x[0]*y[1] + x[1]*y[0] + x[2]*y[3] - x[3]*y[2],
x[0]*y[2] - x[1]*y[3] + x[2]*y[0] + x[3]*y[1],
x[0]*y[3] + x[1]*y[2] - x[2]*y[1] + x[3]*y[0]
# quaternion conjugate
def qstar(x):
return x*np.array([1, -1, -1, -1])
# octonion multiplication
def omult(x, y):
# Split octonions into pairs of quaternions
a, b = x[:4], x[4:]
c, d = y[:4], y[4:]
z = np.zeros(8)
z[:4] = qmult(a, c) - qmult(qstar(d), b)
z[4:] = qmult(d, a) + qmult(b, qstar(c))
return z
Update: See this post for refactored code. Handles quaternions, octonions, sedenions, etc. all with one function.
Unit tests
Here are some unit tests to verify that our multiplication has the expected properties. We begin by generating three octonions with norm 1.
from numpy.linalg import norm
def random_unit_octonion():
x = np.random.normal(size=8)
return x / norm(x)
x = random_unit_octonion()
y = random_unit_octonion()
z = random_unit_octonion()
We said in the previous post that octonions satisfy the “alternative” condition, a weak form of associativity. Here we verify this property as a unit test.
eps = 1e-15
# alternative identities
a = omult(omult(x, x), y)
b = omult(x, omult(x, y))
assert( norm(a - b) < eps )
a = omult(omult(x, y), y)
b = omult(x, omult(y, y))
assert( norm(a - b) < eps )
We also said in the previous post that the octonions satisfy the “Moufang” conditions.
# Moufang identities
a = omult(z, omult(x, omult(z, y)))
b = omult(omult(omult(z, x), z), y)
assert( norm(a - b) < eps )
a = omult(x, omult(z, omult(y, z)))
b = omult(omult(omult(x, z), y), z)
assert( norm(a - b) < eps )
a = omult(omult(z,x), omult(y, z))
b = omult(omult(z, omult(x, y)), z)
assert( norm(a - b) < eps )
a = omult(omult(z,x), omult(y, z))
b = omult(z, omult(omult(x, y), z))
assert( norm(a - b) < eps )
And finally, we verify that the product of unit length octonions has unit length.
# norm condition
n = norm(omult(omult(x, y), z))
assert( abs(n - 1) < eps )
The next post uses the octionion multiplication code above to look at the distribution of the associator (xy)z – x(yz) to see how far multiplication is from being associative.
Sedenion multiplication
The only normed division algebras over the reals are the real numbers, complex numbers, quaternions, and octonions. These are real vector spaces of dimension 1, 2, 4, and 8.
You can proceed analogously and define a real algebra of dimension 16 called the sedenions. When we go from complex numbers to quaternions we lose commutativity. When we go from quaternions to
octonions we lose full associativity, but retain a weak version of associativity. Even weak associativity is lost when we move from octonions to sedenions. Non-zero octonions form an alternative
loop with respect to multiplication, but sedenions do not.
Sedenions have multiplicative inverses, but there are also some zero divisors, i.e. non-zero vectors whose product is zero. So senenions do not form a division algebra. If you continue the
Cayley-Dickson construction past the sedenions, you keep getting zero divisors, so no more division algebras.
Here’s Python code for sedenion multiplication, building on the code above.
def ostar(x):
mask = -np.ones(8)
mask[0] = 1
return x*mask
# sedenion multiplication
def smult(x, y):
# Split sedenions into pairs of octonions
a, b = x[:8], x[8:]
c, d = y[:8], y[8:]
z = np.zeros(16)
z[:8] = omult(a, c) - omult(ostar(d), b)
z[8:] = omult(d, a) + omult(b, ostar(c))
return z
As noted above, see this post for more concise code that also generalizes further.
2 thoughts on “Python code for octonion and sedenion multiplication”
1. A sedenion multiplier – the world was waiting for this!
One thing you can check is that every nonzero sedenion x has a multiplicative inverse: a sedenion y such that xy = yx = 1. Just let y = x*/(xx*), which makes sense because xx* is a positive real
“But wait! I thought the sedenions weren’t a division algebra!”
They’re not. The definition of division algebra requires that ab = 0 implies either a = 0 or b = 0. In the nonassociative case, this is not equivalent to the existence of multiplicative inverses.
There are nonassociative algebras with multiplicative inverses that aren’t division algebras (like the sedenions) but also division algebras that don’t have multiplicative inverses!
Mercifully, the octonions is a division algebra that does have multiplicative inverses, given by the same formula as for the sedenions.
2. Not quite sure whether the whole world was waiting for
a sedenion multiplier but I was :) | {"url":"https://www.johndcook.com/blog/2018/07/09/octonioin-multiplication/","timestamp":"2024-11-09T07:38:58Z","content_type":"text/html","content_length":"57601","record_id":"<urn:uuid:dda53c48-ca0a-4e39-959c-ac102882d05d>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00177.warc.gz"} |
Paving the cowpath in research within pure mathematics – A medium level model based on text driven variations.
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graph colorings. Mathematicians learn to systematically look for subsequent problems around a given problem. We argue that this toy-model captures a nontrivial part of professional mathematical
research within the pure fields and conjecture that it even grasps high level developments in mathematics. By doing this, we implicitly encourage a very simplistic view on criteria, so to speak a
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may also just plant grass and see where people choose to walk. Those pathways are also self-enforcing, since we are less hesitant to walk on those rather than criss-cross the landscape. | {"url":"https://research.uni-luebeck.de/de/publications/paving-the-cowpath-in-research-within-pure-mathematics-a-medium-l","timestamp":"2024-11-02T09:00:08Z","content_type":"text/html","content_length":"46789","record_id":"<urn:uuid:cd7ea6f3-96e9-4401-abae-f752cb83b264>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00356.warc.gz"} |
Which of the following statements isn't true regarding functions? a. The horizontal test may be used to determine whether a function is one-to-one b. A sequence is a function who's domain is the set of natural numbers. c.
Which of the following statements isn't true regarding functions?
a. The horizontal test may be used to determine whether a function is one-to-one
b. A sequence is a function who's domain is the set of natural numbers.
c. A function is a relation in which each value of the input variable is paired with at least one value of the output value.
d. The vertical line test may be used to determine whether a function is one-to-one.
Find an answer to your question 👍 “Which of the following statements isn't true regarding functions? a. The horizontal test may be used to determine whether a function is ...” in 📗 Mathematics if the
answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers | {"url":"https://cpep.org/mathematics/662140-which-of-the-following-statements-isnt-true-regarding-functions-a-the-.html","timestamp":"2024-11-13T19:34:36Z","content_type":"text/html","content_length":"23823","record_id":"<urn:uuid:f04170d7-3660-401d-822b-1bccf95f3976>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00456.warc.gz"} |
Water vapor pressure over complex particles .1. Sulfuric acid solution effect
• The equilibrium vapor pressures of water are calculated for two different geometric configurations: a liquid cap formed on a single substrate sphere and a liquid pendular ring formed about the
contact point of a pair of adhering, identical spheres. The substrate is a structureless, macroscopic (i.e., radius R > 50 nm), relatively hydrophobic sphere. For each configuration, pure water
and sulfuric acid solution are used separately as the interface liquid. In addition to the available surface tension measurements of sulfuric acid solution against air, our calculations utilize
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Then, by using Young's equation as a constraint in our calculation, we can determine the contact angle of the surface liquid residing on substrate spheres for both configurations. We apply
Kelvin's equation in combination with both water activity of sulfuric acid solution and combining relations (semiquantitative relations describing molecular forces) to perform the calculations in
the macroscopic picture. The calculations show, for example, that the equilibrium water vapor pressure over a pendular ring containing relatively dilute sulfuric acid solution (e.g., 0.5–10%) is
always less than the equilibrium vapor pressure over the same configuration with only pure water when both sphere radii are 100 nm and contact angle is around 20°. The results also show that if
all conditions are the same, except geometric configuration, the pendular ring of condensation has a lower equilibrium vapor pressure than the cap of condensation does. Even more significantly,
the graph of equilibrium vapor pressure vs volume of condensed water for the pendular ring configuration indicates unconstrained condensational growth at subsaturation relative humidity. In
contrast, in the cap configuration, condensational growth is usually limited for any subsaturation relative humidity. © 1997 American Association for Aerosol Research. | {"url":"https://vivo.library.tamu.edu/vivo/display/n186284SE","timestamp":"2024-11-05T06:44:58Z","content_type":"text/html","content_length":"22792","record_id":"<urn:uuid:6b703510-65bc-4049-9a9c-425164628f08>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00244.warc.gz"} |
Top Machine Learning Courses - Learn Machine Learning Online
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Searches related to machine learning
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industries such as healthcare, finance, technology, and automotive, earning a machine learning certificate can significantly enhance your career prospects and opportunities for advancement.‎ | {"url":"https://www.coursera.org/courses?query=machine%20learning&skills=Python%20Programming","timestamp":"2024-11-07T06:44:55Z","content_type":"text/html","content_length":"872434","record_id":"<urn:uuid:704652a5-301e-4f03-8243-4bbf6c547fbd>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00097.warc.gz"} |
How to Calculate the Beta of a Portfolio to a Factor | Man Institute
To understand risk, you have to understand the common metrics used to measure it.
September 2021
1. Introduction
Autumn 2020 was largely impacted by two major risk events: the US elections and the Covid-19 vaccine trial results. Both scenarios acted as a catalyst for a violent equity factor reversal.
Accurately measuring risk exposures to a specific factor and understanding the limitations of the risk metrics used is one of the major challenges of risk management. In this article, we present
different ways to measure risk exposure to the general markets and to a specific index with a focus on:
• Defining beta exposure;
• Discussing quality control toolkits;
• Illustrating the beta exposure concept with simple practical examples;
• Reviewing some important considerations when using beta.
A traditional risk measure employed in the asset-management industry is the market value exposure. While useful, it does not accurately explain the portfolio’s behaviour as the market moves.
2. Beta Exposure
A traditional risk measure employed in the asset-management industry is the market value exposure, which represents the notional exposure in percentage allocation of the fund. This measure is not
subjective. While useful, it does not accurately explain the portfolio’s behaviour as the market moves. Understanding and accurately quantifying this behaviour is captured by a risk metric that is
called the beta (‘β’).
The beta metric for a portfolio with respect to a pre-defined index, called X, captures the sensitivity of the fund to X. Basically, the fund’s beta to X tries to capture how much money the fund
makes as X goes up (or down) by a specified amount. For complex and large portfolios, this calculation is far from simple, because to be accurate, one needs to capture precisely how all positions
move with respect to each other as X moves by a pre-defined shock.
X can represent anything one needs to calculate the exposure to, such as the S&P 500 Index, the GBPUSD currency cross or any other index. Likewise, X can relate to more complex indices, such as a
Growth versus Value index, or its sector-neutral version. Whatever the index, accurately computing a portfolio’s beta to it is a powerful risk management tool.
3. Types of Beta
There are two types of beta measures:
• Ex-post beta focuses on the actual, realised and observed return of a fund and compares it, after its realisation, to X;
• Ex-ante beta tries to capture the expectation of a portfolio change upon a move in X – the ex-ante measure is based only on the current positions and not historical holdings.
Generally, ex-ante measures will input current positions in the calculation and simulate hypothetical profit and loss returns assuming constant positions. The ex-ante measure estimates forward risk
by using the current position and assumes that past market conditions such as returns, correlations and volatilities are a good representation of future market returns.
Both ex-ante and ex-post are important measures and tend to differ. For a specific date, their main difference lies in that the ex-ante will attempt to describe the current portfolio behaviour, while
the ex-post attempts to capture the historical portfolio behaviour. For example, an ex-post measure will capture the historical change in positions between that date and the start of the lookback
period, whilst an ex-ante does not.
More details on the technical computation of ex-post and ex-ante beta exposures are provided in the appendix.
4. Quality Control Toolkit
Beta is usually computed using simple linear regression statistical tools. As statistical tools, the results have limitations and are valid in a specific domain of application. There are at least
four risk management tools that can be used to control the quality of the output:
• P-value: gives a sense of whether the regression is meaningful;
• R-squared: gives a sense of how much of the risk is explained;
• Analysis of errors: gives a sense of the domain validation^1;
• Robustness: when dealing with non-normal data, a robust way to compute the beta.
There are further details in the appendix about the quality control toolkit and its technical aspects.
Ex-post beta is useful for investors as it can determine how the fund has historically behaved versus specific factors.
5. Illustrations
5.1. An Example of Ex-Post Beta
As discussed above, ex-post beta is useful for investors as it can determine how the fund has historically behaved versus specific factors such as the market, Momentum^2, Value, interest rates, gold
or oil. Below, we show the ex-post beta of a long-only fund that perfectly replicates the returns of the FTSE 250 Index to GBPUSD. In this case, we define the ex-post beta as the solution of the
simple linear regression using a least square method approach. For our purpose, we used a 6-month lookback coupled with daily returns as parameters.
FTSE250_returns[t] = β * GBPUSD_returns[t] + ε
Figure 1 shows the beta (illustrated by the grey line) of the regression of historical FTSE 250 returns on GBPUSD returns using a least square method. We overrode the beta to zero (‘adjusted’,
illustrated by the yellow line in Figure 1) whenever it was not statistically significant.^3 This was done to emphasise the danger of naively using the beta when it is not statistically significant
based on its P-value and R-squared.
Problems loading this infographic? - Please click here
Source: Bloomberg, Man Group; as of 19 October 2020.
The ex-post beta of the FTSE 250 Index depends on the past levels of the FTSE 250, which in turn is affected by participants’ beliefs – beliefs impact actions, which impacts supply and demand, which
impacts observed market moves, which therefore impacts beta.
In addition, investors’ belief is a complicated function which is difficult to reduce to a simpler form. However, it is reasonable to assume it depends on:
• What other market participants do;
• Market fundamentals and their interpretation;
• Price action which can reinforce pre-existing behaviour;
• Newsflow.
None of these inputs are static, with investors updating their investment theses through time. As they change their beliefs, and therefore their actions, beta will also change over time.
Ultimately, supply and demand for both the British pound and the FTSE 250 will decide how they evolve in relation to each other. We still try to simplify this complex reality into three regimes,
1. GBPUSD is non-significant (i.e. p-value is less than 5%): In this case, there is no point trying to explain the exposure of FTSE 250 with respect to GBPUSD since there is no significant
relationship between the two in that regime. Any price relationship between the two of them is non-significant, and chances are that beta is in fact zero. This is the case for example from December
2017 to December 2018 (illustrated by Zone D in Figure 1). If an investor is exposed to this kind of beta, it is certainly worth asking if this should be hedged or not;
2. Beta > 0 and is significant: In this case, the FTSE 250 and GBPUSD, on average, tend to move in the same direction. We can intuitively think of two different market environments which could
support positive beta:
• Investment cycles where foreign investors are attracted by mid-cap UK shares could support this kind of regime. Indeed, in such an environment, the foreign demand for GBP to buy UK stocks can be
high – all else equal. This seems to be the case before October 2012 (Zone A in figure 1);
• A large exogenous market shock could also support this kind of regime, as is demonstrated by the large spikes in 2016 (Zone B) and 2020 (Zone E). This was especially evident during Brexit- and
Covid-related market events, with daily correlation between the FTSE 250 and GBP starting to increase i.e. the FTSE 250 increased as GBP increased, and vice versa. In the case of Covid-19, and to
some extend Brexit, the US dollar acted as a safe haven for investors to reduce their exposure to UK risk. As more and more of those types of days kick in and are included in the lookback period,
the beta consistently increases and eventually becomes significant. In this regime, the British pound and the FTSE 250 tend to move in a pair. Another way to describe this regime is to recognise
that upon a large systemic shock, correlations go to one.
□ It is worth diving into the Brexit phase to understand the dynamic in detail. Between 23 June 2016 and 27 June 2016, when the results of the Brexit referendum were announced, the FTSE 250
sold off by 5% and GBPUSD by 10%. Those three days materially increased the ex-post beta from a nonsignificant level to a significant level and explains the Brexit spike observed in Zone B.
Subsequently, between 27 June 2016 and 30 December 2016, the relationship reversed and the FTSE 250 rallied by 20% while cable continued to sell off by -7%. This 6-month period eventually led
to a material reduction in beta (Point Z). Furthermore, once the Brexit vote was out of the lookback period, the adjusted beta dropped to zero since it was not significant at this point (Zone
3. Beta < 0 and is significant: In this case, the FTSE 250 and GBPUSD tend to move in opposite directions. One potential explanation is that the FTSE 250 is made up of some large international
companies which tend to improve their revenues when the British pound depreciates. This is a simple argument which justifies that if investors focus on this type of investment thesis, then we should
not be surprised to get negative ex-post beta (Zone C).
5.2. An Example of Ex-Ante Beta
To compute the ex-ante beta, we use current positions for a specific date and keep it unchanged in the calculation (see the appendix for further details on the computation). In this case, we focus on
a portfolio which tracks the FTSE 250 Index (that we will call π) and calculate its ex-ante beta with respect to GBPUSD in Figure 2.
Understanding how the beta behaves when changing parameters is a critical aspect of risk management.
6. Comparison of Ex-Post and Ex-Ante Beta
The results in Figure 2 show the ex-ante and ex-post beta exposures of the FTSE 250 to GBPUSD. The yellow line represents the ex-ante beta while the blue line is the ex-post beta of the same
portfolio. It appears that the two betas are relatively similar except during shock periods (Brexit in Zone A and Covid in Zone B) for which the expost beta is larger than the ex-ante beta. This
suggests that periods of large stress for which GBPUSD dropped, the FTSE 250 fell more than portfolio π.^5
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Source: Bloomberg, Man Group; as of 31 December 2020.
7. Important Considerations
Understanding how the beta behaves when changing parameters is a critical aspect of risk management. Besides ex-ante or ex-post decisions, the beta is also sensitive to different parameters.
Practitioners should be aware of potential limitations related to the usage of the betas. Listed below are important questions we think makes sense to focus on when assessing beta:
1. Should the beta be computed ex-ante or ex-post?;
2. Is the beta significant and can the null hypothesis be rejected?;
3. Should indices be decomposed or not?;
4. How to treat options? Does it result in a risk approximation?;
5. Which lookback period should be taken and why?;
6. Should a weighting be applied in the most recent days?;
7. Does the lookback period represent the best expectation of future market regime?;
8. How does the beta change when the index is updated for a given theme (pure versus non-pure)?;
9. How does the beta change when using daily returns versus 3-day or 5-day returns?;
10. Do overlapping returns be used? How does it impact the results?;
11. Do robust regressions be used? How does it impact the results?;
12. How much of the P&L is explained by the beta (R-squared)?;
13. For which market shock in standard deviation terms is the beta an acceptable representation of the portfolio behaviour? At which point does the beta not accurate anymore?;
14. What information does backtesting give? Does the beta explain the actual P&L and, if not, why not?;
15. How to deal with autocorrelation and how to deal with lookback be dealt with?
Whichever beta one decides to use, the most important aspect of risk management is to have a good understanding of the assumptions underlying the calculation of beta and the regimes for which it
8. Conclusion
There are many subtleties to consider when applying a beta analysis to a portfolio. One important distinction that needs to be made is between ex-post and ex-ante. The ex-post beta of a fund to an
index is particularly useful for an investor because it tries to capture how the fund has performed historically versus the index. The ex-ante beta is particularly useful for portfolio managers
because it tries to capture how the current position would behave versus the index. Whichever beta one decides to use, the most important aspect of risk management is to have a good understanding of
the assumptions underlying the calculation of beta and the regimes for which it works.
9. Appendix
9.1. Beta Computation
9.1.1. Ex-Post Beta
From a computational perspective, one could define the ex-post beta as the solution of the simple linear regression below using a least square method approach.
historical_portfolio_returns[t] = β * X_returns[t] + ε
9.1.2. Ex-Ante Beta
From a computational perspective, a simple way to compute an ex-ante beta is to compute the risk of every asset in the portfolio to X. Once you have the covariance between X and every asset in the
portfolio, you can then define the ex-ante beta to X as:
β = ∑^n [k=1] w[k] * Cov(X,s[k]) * h
i) s[k] represents returns of stock k
ii) w[k] represents the current weight kept constant of the stock k
iii) h is the pre-defined given shock to compute the β to the factor X
A mathematically equivalent way to rewrite the above equation consists in computing the beta as a simple linear regression, using hypothetical portfolio returns and assuming constant weights. This is
the method which was used in our illustration for portfolio π. In this approach, one would solve for the equation below using a least square method approach:
hypothetical_simulated_returns[t] = β * X_returns[t] + ε
i) X_returns represents the daily returns of factor X
ii) hypothetical_simulated_returns represents the returns of a hypothetical portoflio with constant holding
9.2. Quality Control Toolkit
9.2.1. P-Value
In statistics, the p-value represents the probability of obtaining results at least as extreme as the observed results assuming that the null hypothesis^6 is correct. A smaller p-value means that
there is stronger evidence in favour of the alternative hypothesis. Practitioners usually set the p-value threshold at 5% below which one could fairly reject the null hypothesis. We then say the test
is significant.
While it is simple to know if a beta is significant at a specified quantile (‘μ’), by checking if if the p-value is less than μ, it is relatively difficult to know which quantile μ to use.
Practitioners usually use μ = 5%. In addition, regime shifts can occur as we demonstrated in Figure 1, even though the beta exposure approach does not capture when regimes can change from
non-significant to significant. In our opinion, caution is usually a good approach and taking time to ask whether the null hypothesis should be rejected is a good place to start. If in doubt, we
recommend testing another similar hypothesis to be sure; for example, by changing slightly the index or the parameters or using weekly returns whilst correcting for autocorrelation.
Understanding if the beta is significant is a critical aspect of risk management. Unfortunately, our view is that this question is often viewed by practitioners as a mathematical consideration. From
a statistical and practical standpoint, however, this technical detail matters. Indeed, if the beta is non-significant, then one can’t reject the null hypothesis, which means that beta is likely to
be zero. It is then easy to understand that hedging a non-significant beta exposure potentially adds risk to a portfolio.
9.2.2. R-Squared
R-squared is a statistical measure that represents the proportion of the variance for a dependent variable that is explained by an independent variable or variables in a regression. In simpler terms,
it can be viewed as the percentage of risk explained by the factors used in the model input. A higher R-squared corresponds to more explanatory power of the factor used in the regression. When
computing a single linear regression, the R-squared happens to be equal to the correlation squared.
9.2.3. Analysis of Errors
The ordinary least square (‘OLS’) method is a statistical method for estimating the unknown parameters in a linear regression model. The OLS minimises the sum of the squares of the error ε i.e. the
differences between the observed dependent variable (values of the variable being observed) in the given dataset and those predicted by the linear function.
It is customary to assume:
• Homoscedasticity: E[ε^2 [i] | X ] = σ^2, which means that the error term has the same variance σ^2; in each observation. When this requirement is violated, it is called heteroscedasticity. In
this case, robust estimation techniques are recommended (see below);
• No autocorrelation: the errors are uncorrelated between observations: E[ε[i] * ε[j] | X ]=0 for i ≠ j.
9.2.4. Robustness
Robust regression is a form of regression analysis designed to overcome some limitations of traditional linear regressions such as the OLS method. For instance, least square estimates are highly
sensitive to outlier data points, such as multi-standard deviation moves. This is not normally a problem if the outlier is an extreme observation drawn from the tail of a normal distribution, but if
the outlier results from nonnormal measurement error or some other violation of standard ordinary least squares assumptions, then it compromises the validity of the regression results if a non-robust
regression technique is used.
Robust regression techniques include M-estimators (’M’ for ‘maximum likelihood-type’) popularised by Huber.^7
1. When model developers build a model, they generally define a domain application. The domain application specifies how the model is supposed to be used by the model user. In particular, domain
application describes for which input the output of the model can be trusted. Domain validation describes the set of inputs for which the model behaves adequately.
2. By Momentum we mean a signal defined by the performance of a long / short basket. The portfolio is long the stocks which performance is positive over a pre-defined time horizon, typically 12
months, and short the stocks which performance is negative over the same time horizon. This basket also rebalances over a different time horizon, typically every month. The performance of this basket
is the Momentum signal.
3. When the p-value was over 5%.
4. Whilst the goal of this article is not to dig into the relationship between FX and equities during large FX shocks, it is nevertheless worth highlighting that on 15 January 2015, as the Swiss
National Bank removed the 1.20 price peg from the EURCHF pair, a similar effect in the correlation between FX and equities was observed. On that day, correlation spiked as the EURCHF sold off by
~-2000 bps or ~-16.88% from 1.2010 to 0.9982 whilst SMI Index sold off by -8.67% from 9,198 to 8,400.
5. Ex-Ante Beta calculation assumes constant weights for portfolio π over the period.
6. The null hypothesis is the hypothesis that the beta is zero.
7. https://www.statsmodels.org/stable/rlm.html. | {"url":"https://www.man.com/maninstitute/calculate-beta-of-portfolio-to-factor","timestamp":"2024-11-06T18:23:04Z","content_type":"text/html","content_length":"93668","record_id":"<urn:uuid:8a286d8d-bdc2-44fd-b1cf-cb5c41bdf2a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00513.warc.gz"} |
ball mill variable speed kc engineer
This approach is based on a hybrid numerical model of a 24degreeoffreedom gearbox, simulating one gear train and two drive shafts. The impact forces of the mill drum are modelled by a discrete
element method (DEM). The ballfilling rate (Fr), the mill speed (Nr), and the ball size (Db) are considered to study this phenomenon. | {"url":"https://www.petite-venise-chartres.fr/ball/mill/variable/speed/kc/engineer-7983.html","timestamp":"2024-11-13T14:15:31Z","content_type":"text/html","content_length":"48536","record_id":"<urn:uuid:bd535031-068f-4d95-8539-44790f13c773>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00730.warc.gz"} |
Bard Does a Lesson Plan - If You Are Still Teaching, You Should Use This. - Dr. Doug Green
Bard Does a Lesson Plan – If You Are Still Teaching, You Should Use This.
I asked Google’s Bard to “Prepare a lesson plan to teach the quadratic equations at the high school level.” Below is what I got back. My conclusion is that if I were still teaching, I would always
ask for Bard’s plan before I made my own. I might not use Bard’s plan, but I would look to it for ideas. I would certainly share the Additional Resources links with my students. If you are still
teaching, why aren’t you doing this? This is probably more detailed than your real lesson plans and will impress your principal.
Lesson Plan: Quadratic Equations
Grade Level: High School
Subject: Algebra II
Time: 50 minutes
Learning Objectives
Students will be able to identify and define quadratic equations.
Students will be able to graph quadratic equations.
Students will be able to solve quadratic equations using factoring, completing the square, and the quadratic formula.
Students will be able to apply quadratic equations to real-world problems.
Whiteboard or projector
Markers or pens
Graph paper
Calculators (optional)
Worksheets (optional)
Introduction (5 minutes)
Begin by asking students what they know about quadratic equations.
Elicit responses from students and provide a brief overview of the topic.
Explain that quadratic equations are polynomial equations of the second degree, meaning they have a highest power of 2.
Identifying Quadratic Equations (10 minutes)
Display a variety of equations, some quadratic and some not.
Ask students to identify which equations are quadratic and explain their reasoning.
Provide examples of quadratic equations in various forms, including standard form, factored form, and vertex form.
Graphing Quadratic Equations (10 minutes)
Review the concept of graphing equations in general.
Introduce the concept of the axis of symmetry for quadratic equations.
Demonstrate how to graph quadratic equations using vertex form.
Allow students to practice graphing quadratic equations on their own or in pairs.
Solving Quadratic Equations (20 minutes)
Introduce three methods for solving quadratic equations: factoring, completing the square, and the quadratic formula.
Explain each method step by step, using examples to illustrate the process.
Provide students with practice problems to solve using each method.
Real-World Applications (5 minutes)
Discuss real-world situations where quadratic equations can be used to model problems.
Provide examples of such problems, such as calculating the height of an object launched into the air or determining the profit or loss of a business venture.
Encourage students to think of their own examples of real-world applications of quadratic equations.
Closure (5 minutes)
Summarize the key points of the lesson.
Review the different methods for solving quadratic equations.
Encourage students to ask any remaining questions.
Observe students’ participation in class discussions and activities.
Collect and review worksheets or homework assignments.
Administer a quiz or test to assess students’ understanding of the material.
For students who need more support, provide additional practice problems with worked-out solutions.
For students who are ready for more challenging material, introduce the concept of complex numbers and solving quadratic equations with complex roots.
Additional Resources
Quadratic Equations
Solving Quadratic Equations
Real-World Applications of Quadratic Equations
DrDougGreen.com If you like the summary, buy the book | {"url":"https://drdouggreen.com/2023/bard-does-a-lesson-plan-if-your-are-still-teaching-you-should-use-this/","timestamp":"2024-11-13T14:44:57Z","content_type":"application/xhtml+xml","content_length":"41983","record_id":"<urn:uuid:b8daf66d-0603-4e3c-9ed8-9bf3c3b7099b>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00324.warc.gz"} |